Concise Mathematics Selina Solutions Class 10 ICSE Chapter 27 Mixed Practice

Our comprehensive ICSE Class 10 Maths Mixed Practice Solutions provide clear, step-by-step guidance for this crucial revision chapter from the Class – 10 Concise Mathematics Selina textbook. This chapter is designed to test your combined understanding of several key Commercial Mathematics topics you have learned. It brings together problems from GST, Banking (especially recurring deposits), and Shares & Dividends into a single set of exercises. By working through these problems, you will strengthen your ability to identify which formula to apply and how to structure your answers correctly, which is excellent practice for the board examinations.

Are you trying to verify your answers or feeling stuck on a complex question from the Mixed Practice sets? You’ve come to the right place. This page contains detailed, expert-verified solutions for all 81 questions from both SET A and SET B. Each solution is crafted to follow the exact method and presentation style that the ICSE board expects, helping you learn how to score full marks. Here, you will find reliable and easy-to-follow walkthroughs for every single problem in the chapter.

SET A

Question 1

A dealer in Calcutta supplied the following goods/services to another dealer in Banaras. Find the total amount of bill :

To determine the total bill, we need to adjust the prices for any discounts and then add GST.

First, calculate the discounted prices using:

Discounted price = (100 – Discount)% × MRP

Since the transaction is inter-state, Integrated GST (IGST) is applicable.

Calculate IGST as follows:

IGST = 18% of 5657.50 = \dfrac{18}{100} \times 5657.50 = 1018.35

Therefore, the total amount payable is:

Total amount = ₹ 5657.50 + ₹ 1018.35 = ₹ 6675.85

Hence, amount of bill = ₹ 6675.85

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Question 2

Some goods/services are supplied for ₹ 20,000 from Mathura (U.P.) to Ratlam (M.P.) and then from Ratlam to Indore (M.P.). If at each stage, the rate of tax under G.S.T. system is 12% and the profit made by the dealer in Ratlam is ₹ 3,750; find the cost of the article (in Indore) under GST.

The selling price in Mathura is ₹ 20,000.

For an interstate transaction, the GST applied is IGST, calculated as 12% of ₹ 20,000. This amounts to:
\text{GST} = \dfrac{12}{100} \times 20000 = ₹ 2400.
This GST collected by the Mathura dealer becomes the input GST for the dealer in Ratlam.

Now, the selling price in Ratlam, after including the profit of ₹ 3,750, is:
₹ 20000 + ₹ 3750 = ₹ 23750.
In Ratlam, the transaction is intra-state, so GST is divided into CGST and SGST, each at 6% of ₹ 23,750.

Calculate CGST:
\text{CGST} = \dfrac{6}{100} \times 23750 = ₹ 1425.
SGST is equal to CGST, so:
\text{SGST} = ₹ 1425.
Thus, the total GST for this stage is:
\text{GST} = ₹ 1425 + ₹ 1425 = ₹ 2850.
Finally, the cost of the article in Indore is the selling price in Ratlam plus the GST:
₹ 23750 + ₹ 2850 = ₹ 26600.
Hence, cost of article in Indore = ₹ 26600.

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Question 3

Mr. Kumar has a recurring deposit account in a bank for 4 years at 10% p.a. rate of interest. If he gets ₹ 21,560 as interest at the time of maturity, find :

(i) the monthly instalment paid by Mr. Kumar.

(ii) the amount of maturity of this recurring deposit account.

(i) Assume the monthly deposit is ₹ P.

Here, the total duration is 4 years, which translates to 4 \times 12 = 48 months.

The interest for a recurring deposit is calculated using the formula:

Interest (I) = P \times (\dfrac{n(n + 1)}{2 \times 12}) \times \dfrac{r}{100}

Plugging in the known values:

\Rightarrow 21560 = P \times \dfrac{48 \times (48 + 1)}{2 \times 12} \times \dfrac{10}{100} \Rightarrow 21560 = P \times \dfrac{48 \times 49}{24} \times \dfrac{1}{10} \Rightarrow P = \dfrac{21560 \times 24 \times 10}{48 \times 49} \Rightarrow P = \dfrac{107800}{49} = ₹ 2,200.

Thus, the monthly installment is ₹ 2,200.

(ii) The maturity value is given by the formula:

Maturity value = P \times n + Interest

Substituting the values:

= ₹ 2200 \times 48 + ₹ 21560

= ₹ 105600 + ₹ 21560

= ₹ 1,27,160.

Therefore, the maturity amount is ₹ 1,27,160.

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Question 4

The maturity value of a cumulative deposit account is ₹ 1,20,400. If each monthly installment for this account is ₹ 1,600 and the rate of interest is 10% per year, find the time for which the account was held.

Assume the duration of the account in months is n.

We know:

Maturity Value (M.V.) = ₹ 1,20,400

The formula for the maturity value is:

\text{M.V.} = P \times n + P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}

Plugging in the given values:

\Rightarrow 120400 = 1600 \times n + 1600 \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{10}{100}

This simplifies to:

\Rightarrow 120400 = 1600\Big(n + \dfrac{n(n + 1)}{24} \times \dfrac{10}{100}\Big)

Dividing both sides by 1600:

\Rightarrow \dfrac{120400}{1600} = n + \dfrac{n(n + 1)}{240}

Simplifying further:

\Rightarrow \dfrac{301}{4} = \dfrac{240n + n^2 + n}{240}

Multiply through by 240 to clear the fraction:

\Rightarrow \dfrac{301 \times 240}{4} = 240n + n^2 + n

Simplifying gives:

\Rightarrow 301 \times 60 = n^2 + 241n

Rearranging terms:

\Rightarrow n^2 + 241n - 18060 = 0

Using factorization:

\Rightarrow n^2 + 301n - 60n - 18060 = 0

Grouping terms:

\Rightarrow n(n + 301) - 60(n + 301) = 0

Factoring out common terms:

\Rightarrow (n - 60)(n + 301) = 0

The solutions are:

\Rightarrow n = 60 \text{ or } n = -301.

Since time cannot be negative, we select n = 60 months.

∴ The account was held for \dfrac{60}{12} = 5 years.

Hence, time = 5 years.

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Question 5

Rajat invested ₹ 24,000 in 7% hundred rupee shares at 20% discount. After one year, he sold these shares at ₹ 75 each and invested the proceeds (including dividend of first year) in 18% twenty five rupee shares at 64% premium. Find :

(i) his gain or loss after one year.

(ii) his annual income from the second investment.

(iii) the percentage of increase in return on his original investment.

Rajat begins with an investment of ₹ 24,000 in shares with a nominal value (N.V.) of ₹ 100 and a dividend rate of 7%. The shares are purchased at a market value (M.V.) which is calculated by applying a 20% discount on the nominal value:

\text{M.V.} = \text{N.V.} - \text{discount} = ₹ 100 - \frac{20}{100} \times 100 = ₹ 80.

(i) Calculating Gain or Loss After One Year:

The number of shares Rajat buys is given by:

\text{Number of shares} = \frac{\text{Investment}}{\text{M.V. of each share}} = \frac{₹ 24000}{₹ 80} = 300.

The dividend per share is 7% of ₹ 100, which is ₹ 7. Hence, Rajat’s total dividend for the first year is:

\text{Dividend} = ₹ 7 \times 300 = ₹ 2100.

Thus, Rajat gains ₹ 2100 in the first year.

(ii) Calculating Annual Income from the Second Investment:

Rajat sells each share at ₹ 75. Therefore, the total proceeds from selling the shares, including the dividend, are:

300 \times ₹ 75 + ₹ 2100 = ₹ 22500 + ₹ 2100 = ₹ 24600.

The new shares have a nominal value of ₹ 25 and are bought at a market value with a 64% premium:

\text{M.V. of each share} = ₹ 25 + \frac{64}{100} \times 25 = ₹ 25 + ₹ 16 = ₹ 41.

With a dividend rate of 18%, the number of shares Rajat can buy is:

\text{Number of shares} = \frac{\text{Investment}}{\text{M.V. of each share}} = \frac{₹ 24600}{₹ 41} = 600.

The dividend per share is 18% of ₹ 25, which is:

\frac{18 \times 25}{100} = ₹ 4.50.

Thus, the annual income from the second investment is:

600 \times ₹ 4.50 = ₹ 2700.

Hence, the annual income from the second investment is ₹ 2700.

(iii) Calculating the Percentage Increase in Return:

The increase in return from the original investment is:

₹ 2700 - ₹ 2100 = ₹ 600.

To find the percentage increase in return, we use:

\text{Percentage increase} = \frac{600}{24000} \times 100 = \frac{100}{40} = 2.5 \%.

Therefore, the percentage increase in return is 2.5%.

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Question 6

A man sold some ₹ 20 shares, paying 8% dividend, at 10% discount and invested the proceeds in ₹ 10 shares, paying 12% dividend, at 50% premium. If the change in his annual income is ₹ 600, find the number of shares sold by the man.

The nominal value (N.V.) of each share is ₹ 20.

To find the market value (M.V.) after a 10% discount, calculate:

M.V. = N.V. – Discount = ₹ 20 – 10% of ₹ 20

= ₹ 20 – \dfrac{10}{100} \times 20

= ₹ 20 – ₹ 2 = ₹ 18.

∴ The dividend per share is 8% of ₹ 20, which is:

\dfrac{8}{100} \times 20 = ₹ 1.60

Assume the number of shares sold is x.

The total dividend from these shares is 1.6x.

The selling price of each share is ₹ 18, so the total proceeds from selling x shares is:

₹ 18x

Now, for the second investment:

The nominal value of each new share is ₹ 10.

With a 50% premium, the market value becomes:

M.V. = N.V. + Premium = ₹ 10 + 50% of ₹ 10

= ₹ 10 + ₹ 5 = ₹ 15

The number of new shares purchased is given by:

\dfrac{\text{Investment}}{\text{M.V. of share}} = \dfrac{18x}{15}

∴ The dividend per new share is 12% of ₹ 10, which is:

\dfrac{12}{100} \times 10 = ₹ 1.2

The total income from these new shares is:

1.2 \times \dfrac{18x}{15} = \dfrac{21.6x}{15} = 1.44x

According to the problem, the change in income is ₹ 600:

⇒ 1.6x - 1.44x = 600

⇒ 0.16x = 600

⇒ x = \dfrac{600}{0.16} = 3750

Therefore, the number of shares sold is 3750.

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Question 7

Find the value of x which satisfies the inequation :

-2 ≤ \dfrac{1}{2} - \dfrac{2x}{3} \le 1\dfrac{5}{6}, x ∈ W.

Also, graph the solution on a number line.

To tackle this inequality, we’ll start by focusing on the left-hand side (L.H.S.):

\begin{aligned}\Rightarrow -2 \le \dfrac{1}{2} - \dfrac{2x}{3} \\\Rightarrow \dfrac{2x}{3} \le \dfrac{1}{2} + 2 \\\Rightarrow \dfrac{2x}{3} \le \dfrac{5}{2} \\\Rightarrow x \le \dfrac{5}{2} \times \dfrac{3}{2} \\\Rightarrow x \le \dfrac{15}{4} \text{ .........(1) }\end{aligned}

Now, let’s solve the right-hand side (R.H.S.):

\begin{aligned}\Rightarrow \dfrac{1}{2} - \dfrac{2x}{3} \le 1\dfrac{5}{6} \\\Rightarrow \dfrac{1}{2} - \dfrac{2x}{3} \le \dfrac{11}{6} \\\Rightarrow \dfrac{2x}{3} \ge \dfrac{1}{2} - \dfrac{11}{6}\\\Rightarrow \dfrac{2x}{3} \ge \dfrac{3 - 11}{6} \\\Rightarrow x \ge -\dfrac{8}{6} \times \dfrac{3}{2} \\\Rightarrow x \ge -2 \text{ .......(2)}\end{aligned}

Combining results from (1) and (2), we have:

-2 \le x \le \dfrac{15}{4}

Since x \in W, which means x is a whole number, the possible values for x are:

x = \{0, 1, 2, 3\}

Hence, (x = {0, 1, 2, 3}.

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Question 8

Solve (using formula) the equation :

\dfrac{x}{x + 1} + \dfrac{x + 1}{x} = 2\dfrac{4}{15}.

We start with the equation:

\Rightarrow \dfrac{x}{x + 1} + \dfrac{x + 1}{x} = 2\dfrac{4}{15}

This simplifies to:

\Rightarrow \dfrac{x^2 + (x + 1)^2}{x(x + 1)} = \dfrac{34}{15}

Multiplying both sides by 15x(x + 1) gives:

\Rightarrow 15[x^2 + (x + 1)^2] = 34x(x + 1)

Expanding the terms, we have:

\Rightarrow 15x^2 + 15(x + 1)^2 = 34x^2 + 34x

Breaking it further:

\Rightarrow 15x^2 + 15(x^2 + 1 + 2x) = 34x^2 + 34x

Simplifying, we find:

\Rightarrow 15x^2 + 15x^2 + 15 + 30x = 34x^2 + 34x

Rearranging terms, we get:

\Rightarrow 34x^2 - 15x^2 - 15x^2 + 34x - 30x - 15 = 0

This simplifies to:

\Rightarrow 4x^2 + 4x - 15 = 0

We compare this quadratic equation to the standard form ax^2 + bx + c = 0 and identify:

a = 4, b = 4, and c = -15.

Using the quadratic formula x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}, we substitute the values:

x = \dfrac{-4 \pm \sqrt{4^2 - 4 \times 4 \times (-15)}}{2 \times 4}

This simplifies to:

= \dfrac{-4 \pm \sqrt{16 + 240}}{8}

Further simplifying:

= \dfrac{-4 \pm \sqrt{256}}{8}

Which gives us:

= \dfrac{-4 \pm 16}{8}

So, we have two solutions:

= \dfrac{-4 + 16}{8}, \dfrac{-4 - 16}{8}

Which evaluate to:

= \dfrac{12}{8}, -\dfrac{20}{8}

Simplifying the fractions, we find:

= \dfrac{3}{2}, -\dfrac{5}{2}

Hence, x = -\dfrac{5}{2}, \dfrac{3}{2}

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Question 9

By selling an article for ₹ 24, a man gains as much percent as its cost price. Find the cost price of the article.

Assume the cost price of the article is ₹ x.

The profit made on selling the article is ₹ (24 – x).

As per the problem, the profit percentage equals the cost price percentage, which is x%.

Using the formula for profit percentage:

\Rightarrow \text{Profit \%} = \dfrac{\text{Profit}}{\text{C.P.}} \times 100

Substituting the given values, we have:

\begin{aligned}\Rightarrow x = \dfrac{24 - x}{x} \times 100 \\\Rightarrow x^2 = 100(24 - x) \\\Rightarrow x^2 = 2400 - 100x \\\Rightarrow x^2 + 100x - 2400 = 0 \\\Rightarrow x^2 + 120x - 20x - 2400 = 0 \\\Rightarrow x(x + 120) - 20(x + 120) = 0 \\\Rightarrow (x - 20)(x + 120) = 0 \\\Rightarrow x - 20 = 0 \text{ or } x + 120 = 0 \\\Rightarrow x = 20 \text{ or } x = -120.\end{aligned}

Since the cost price cannot be negative, we discard x = -120.

∴ x = ₹ 20.

Hence, cost price = ₹ 20.

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Question 10

B takes 16 days less than A to do a certain piece of work. If both working together can complete the work in 15 days, in how many days will B alone complete the work ?

Assume A completes the work in x days. Then, B would take (x - 16) days.

The work done by A in one day is \dfrac{1}{x}, while B completes \dfrac{1}{x - 16} in one day.

Together, A and B’s combined one-day work is \dfrac{1}{x} +\dfrac{1}{x - 16}.

Since they finish the work together in 15 days, we have:

15\Big(\dfrac{1}{x} + \dfrac{1}{x - 16}\Big) = 1

Simplifying, we find:

\Rightarrow \dfrac{x - 16 + x}{x(x - 16)} = \dfrac{1}{15}

This leads to:

\Rightarrow \dfrac{2x - 16}{x^2 - 16x} = \dfrac{1}{15}

Cross-multiplying gives:

\Rightarrow 15(2x - 16) = x^2 - 16x

Rearranging terms, we have:

\Rightarrow 30x - 240 = x^2 - 16x

Bringing all terms to one side results in:

\Rightarrow x^2 - 16x - 30x + 240 = 0

Simplifying further:

\Rightarrow x^2 - 46x + 240 = 0

Factoring the quadratic gives:

\Rightarrow x^2 - 40x - 6x + 240 = 0

Grouping terms, we find:

\Rightarrow x(x - 40) - 6(x - 40) = 0

Thus:

\Rightarrow (x - 6)(x - 40) = 0

Solving for x, we get:

\Rightarrow x - 6 = 0 \text{ or } x - 40 = 0

So, x = 6 or x = 40.

Since x must be greater than 16, we have x = 40. Therefore, x - 16 = 40 - 16 = 24.

Hence, B alone can finish the work in 24 days.

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Question 11

Divide ₹ 1870 into three parts in such a way that half of the first part, one-third of the second part and one-sixth of the third part are all equal.

Assume that half of the first portion, one-third of the second portion, and one-sixth of the third portion are all equal to x.

⇒ \dfrac{1}{2} of the first portion = x

⇒ First portion = 2x

⇒ \dfrac{1}{3} of the second portion = x

⇒ Second portion = 3x

⇒ \dfrac{1}{6} of the third portion = x

⇒ Third portion = 6x

Given that the total sum is ₹ 1870,

⇒ 2x + 3x + 6x = ₹ 1870

⇒ 11x = ₹ 1870

⇒ x = \dfrac{1870}{11} = ₹ 170.

Now, calculate each part:

First portion = 2x = 2 \times 170 = ₹ 340

Second portion = 3x = 3 \times 170 = ₹ 510

Third portion = 6x = 6 \times 170 = ₹ 1020

Therefore, the first portion is ₹ 340, the second portion is ₹ 510, and the third portion is ₹ 1020.

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Question 12

If a + c = be and \dfrac{1}{b} + \dfrac{1}{d} = \dfrac{e}{c}, prove that : a, b, c and d are in proportion.

We start with the given equations:

a + c = be
\dfrac{1}{b} + \dfrac{1}{d} = \dfrac{e}{c}

First, consider the equation a + c = be. By dividing every term by b, we have:

\Rightarrow \dfrac{a}{b} + \dfrac{c}{b} = e \quad \text{...........(1)}

Next, let’s work with the second equation:

\dfrac{1}{b} + \dfrac{1}{d} = \dfrac{e}{c}

Multiplying each term by c gives us:

\Rightarrow \dfrac{c}{b} + \dfrac{c}{d} = e

Now, substitute the expression for e from equation (1) into the above equation:

\Rightarrow \dfrac{c}{b} + \dfrac{c}{d} = \dfrac{a}{b} + \dfrac{c}{b}

This simplifies to:

\Rightarrow \dfrac{c}{d} = \dfrac{a}{b}

Thus, since \dfrac{a}{b} = \dfrac{c}{d}, we conclude that a, b, c, and d are in proportion.

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Question 13

If \dfrac{4x + 3y}{4x - 3y} = \dfrac{7}{4}, use the properties to find the value of \dfrac{2x^2 - 11y^2}{2x^2 + 11y^2}.

We’re given the equation:

\dfrac{4x + 3y}{4x - 3y} = \dfrac{7}{4}

To simplify, apply the componendo and dividendo property:

\begin{aligned}\Rightarrow \dfrac{4x + 3y + (4x - 3y)}{4x + 3y - (4x - 3y)} = \dfrac{7 + 4}{7 - 4} \\\Rightarrow \dfrac{8x}{6y} = \dfrac{11}{3} \\\Rightarrow \dfrac{x}{y} = \dfrac{6 \times 11}{3 \times 8} \\\Rightarrow \dfrac{x}{y} = \dfrac{11}{4}\end{aligned}

Now, square both sides:

\begin{aligned}\Rightarrow \Big(\dfrac{x}{y}\Big)^2 = \dfrac{121}{16} \\\Rightarrow \dfrac{x^2}{y^2} = \dfrac{121}{16}\end{aligned}

Multiply both sides by \dfrac{2}{11}:

\begin{aligned}\Rightarrow \dfrac{2x^2}{11y^2} = \dfrac{121}{16} \times \dfrac{2}{11} \\\Rightarrow \dfrac{2x^2}{11y^2} = \dfrac{11}{8}\end{aligned}

Once again, use componendo and dividendo:

\begin{aligned}\Rightarrow \dfrac{2x^2 + 11y^2}{2x^2 - 11y^2} = \dfrac{11 + 8}{11 - 8} \\\Rightarrow \dfrac{2x^2 + 11y^2}{2x^2 - 11y^2} = \dfrac{19}{3} \\\Rightarrow \dfrac{2x^2 - 11y^2}{2x^2 + 11y^2} = \dfrac{3}{19}.\end{aligned}

Thus, the value of \dfrac{2x^2 - 11y^2}{2x^2 + 11y^2} is \dfrac{3}{19}.

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Question 14

Use the properties of proportionality to solve : \dfrac{\sqrt{12x + 1} + \sqrt{2x - 3}}{\sqrt{12x + 1} - \sqrt{2x - 3}} = \dfrac{3}{2}.

We start with the equation:

\dfrac{\sqrt{12x + 1} + \sqrt{2x - 3}}{\sqrt{12x + 1} - \sqrt{2x - 3}} = \dfrac{3}{2}.

By applying the componendo and dividendo rule, we can simplify the expression:

\Rightarrow \dfrac{\sqrt{12x + 1} + \sqrt{2x - 3} + (\sqrt{12x + 1} - \sqrt{2x - 3})}{\sqrt{12x + 1} + \sqrt{2x - 3} - (\sqrt{12x + 1} - \sqrt{2x - 3})} = \dfrac{3 + 2}{3 - 2}.

This simplifies to:

\Rightarrow \dfrac{2\sqrt{12x + 1}}{2\sqrt{2x - 3}} = 5.

Thus, we have:

\Rightarrow \dfrac{\sqrt{12x + 1}}{\sqrt{2x - 3}} = 5.

Next, squaring both sides gives us:

\Rightarrow \Big(\dfrac{\sqrt{12x + 1}}{\sqrt{2x - 3}}\Big)^2 = 5^2.

This results in:

\Rightarrow \dfrac{12x + 1}{2x - 3} = 25.

Multiplying both sides by (2x - 3), we get:

\Rightarrow 12x + 1 = 25(2x - 3).

Expanding the right side, we have:

\Rightarrow 12x + 1 = 50x - 75.

Rearranging terms, we find:

\Rightarrow 50x - 12x = 75 + 1.

Simplifying further:

\Rightarrow 38x = 76.

Finally, solving for x gives:

\Rightarrow x = \dfrac{76}{38} = 2.

Hence, x = 2.

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Question 15

What number should be added to 2x^3 – 3x^2 – 8x + 3 so that the resulting polynomial leaves the remainder 10 when divided by 2x + 1 ?

Let’s start by setting the divisor equal to zero:
2x + 1 = 0
Solving for x, we get:
x = -\dfrac{1}{2}
Assume we add a number a to the polynomial. The polynomial becomes:
2x^3 - 3x^2 - 8x + 3 + a
According to the remainder theorem, when this polynomial is divided by 2x + 1, the remainder is the polynomial evaluated at x = -\dfrac{1}{2}. Thus,
2\Big(-\dfrac{1}{2}\Big)^3 - 3\Big(-\dfrac{1}{2}\Big)^2 - 8\Big(-\dfrac{1}{2}\Big) + 3 + a = 10
Simplifying each term, we find:
2 \times -\dfrac{1}{8} - 3 \times \dfrac{1}{4} + 4 + 3 + a = 10
This simplifies to:
-\dfrac{1}{4} - \dfrac{3}{4} + a = 10 - 4 - 3
Combine the fractions:
\dfrac{-4}{4} + a = 3
Which simplifies to:
-1 + a = 3
Solving for a, we add 1 to both sides:
a = 3 + 1 = 4
Therefore, the number that must be added is 4.

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Question 16

If A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}, \text{ B =} \begin{bmatrix} x & 1 \\ 4 & -1 \end{bmatrix} and A^2 + B^2 = (A + B)^2, find the value of x.

State, whether A^2 + B^2 and (A + B)^2 are always equal or not.

We start with the equation given:

A^2 + B^2 = (A + B)^2

Substitute the matrices:

\begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}^2 + \begin{bmatrix} x & 1 \\ 4 & -1 \end{bmatrix}^2 = \Big(\begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} + \begin{bmatrix} x & 1 \\ 4 & -1 \end{bmatrix}\Big)^2

Calculate A^2 and B^2:

\begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} + \begin{bmatrix} x & 1 \\ 4 & -1 \end{bmatrix}\begin{bmatrix} x & 1 \\ 4 & -1 \end{bmatrix}

The sum of matrices A and B is:

\Big(\begin{bmatrix} 1 + x & 0 \\ 6 & -2 \end{bmatrix}\Big)^2

Now, let’s compute each element:

For A^2:

\begin{bmatrix} 1 \times 1 + (-1) \times 2 & 1 \times -1 + (-1) \times(-1) \\ 2 \times 1 + (-1) \times 2 & 2 \times -1 + (-1) \times (-1) \end{bmatrix}

For B^2:

\begin{bmatrix} x \times x + 1 \times 4 & x \times 1 + 1 \times(-1) \\ 4 \times x + (-1) \times 4 & 4 \times 1 + (-1) \times (-1) \end{bmatrix}

The expanded form of (A + B)^2 is:

\begin{bmatrix} (1 + x)^2 + 0 \times 6 & (1 + x) \times 0 + 0 \times -2 \\ 6(1 + x) + (-2) \times 6 & 6 \times 0 + (-2) \times(-2) \end{bmatrix}

Simplifying further, we have:

\begin{bmatrix} 1 - 2 & -1 + 1 \\ 2 - 2 & -2 + 1 \end{bmatrix} + \begin{bmatrix} x^2 + 4 & x - 1 \\ 4x - 4 & 4 + 1 \end{bmatrix} = \begin{bmatrix} (1 + x)^2 & 0 \\ 6 + 6x - 12 & 4 \end{bmatrix}

Now, simplify both sides:

\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} + \begin{bmatrix} x^2 + 4 & x - 1 \\ 4x - 4 & 5 \end{bmatrix} = \begin{bmatrix} (1 + x)^2 & 0 \\ 6x - 6 & 4 \end{bmatrix}

This leads to:

\begin{bmatrix} x^2 + 4 - 1 & x - 1 \\ 4x - 4 & 5 - 1 \end{bmatrix} = \begin{bmatrix} (1 + x)^2 & 0 \\ 6x - 6 & 4 \end{bmatrix}

Finally, we have:

\begin{bmatrix} x^2 + 3 & x - 1 \\ 4x - 4 & 4 \end{bmatrix} = \begin{bmatrix} (1 + x)^2 & 0 \\ 6x - 6 & 4 \end{bmatrix}

Solving for x:

x - 1 = 0

⇒ x = 1.

It is important to note that A^2 + B^2 and (A + B)^2 are not always the same, as demonstrated by the identity:

(A + B)^2 = A^2 + B^2 + 2AB.

Hence, x = 1.

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Question 17

Find the 10^th term of the sequence 10, 8, 6, ……..

Consider the sequence: 10, 8, 6, ……. This sequence forms an arithmetic progression (A.P.). Here, the first term, denoted as a, is 10. The common difference d can be calculated by subtracting the first term from the second term: 8 - 10 = -2.

To find the 10^{th} term of this sequence, we use the formula for the n^{th} term of an A.P.:

a_n = a + (n - 1) \times d

Substituting the known values into the formula, we have:

a_{10} = 10 + (10 - 1) \times -2

Simplifying further:

a_{10} = 10 + 9 \times -2 = 10 - 18 = -8

Thus, the 10^{th} term of the sequence is -8.

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Question 18

If the 5^th and 11^th terms of an A.P. are 16 and 34 respectively. Find the A.P.

Assume the first term of the arithmetic progression (A.P.) is a and the common difference is d.

Using the formula for the n-th term of an A.P., we have:

a_n = a + (n - 1)d

For the 5th term, we know:

a_5 = 16

This gives us:

a + (5 - 1)d = 16
a + 4d = 16 ( \text{……..(1)} )

For the 11th term, we have:

a_{11} = 34

This results in:

a + (11 - 1)d = 34
a + 10d = 34 ( \text{……..(2)} )

Now, subtract equation (1) from equation (2):

(a + 10d) - (a + 4d) = 34 - 16
a - a + 10d - 4d = 18
6d = 18
d = 3

Substitute d = 3 back into equation (1):

a + 4(3) = 16
a + 12 = 16
a = 4

Thus, the A.P. is given by the sequence:

a, (a + d), (a + 2d), \ldots

Substituting the values of a and d:

4, (4 + 3), (4 + 2 \times 3), \ldots

This simplifies to:

4, 7, 10, \ldots

Hence, A.P. = 4, 7, 10, \ldots

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Question 19

If p^th term of and A.P. is q and its q^th term is p, show that its r^th term is (p + q – r).

Consider an arithmetic progression (A.P.) with the first term as a and the common difference as d. The general formula for the n^{th} term of an A.P. is:

a_n = a + (n - 1)d

For the p^{th} term being q, we have:

a + (p - 1)d = q \quad \text{...(1)}

Similarly, for the q^{th} term being p, we get:

a + (q - 1)d = p \quad \text{...(2)}

By subtracting equation (2) from equation (1), we find:

a + (p - 1)d - [a + (q - 1)d] = q - p

Simplifying, this becomes:

a - a + (p - 1)d - (q - 1)d = q - p

This reduces to:

d[(p - 1) - (q - 1)] = q - p

Further simplification gives:

d(p - q) = q - p

Recognizing that d(p - q) = -(p - q), we conclude:

d = -1

Substitute d = -1 back into equation (1):

a + (p - 1)(-1) = q

This simplifies to:

a - p + 1 = q

Thus, solving for a, we find:

a = p + q - 1

Now, for the r^{th} term, a_r, we use:

a_r = a + (r - 1)d

Substituting the known values, we have:

a_r = p + q - 1 + (r - 1)(-1)

Simplifying further, we get:

p + q - 1 - r + 1

Finally, this results in:

p + q - r

Thus, it is shown that the r^{th} term is p + q - r.

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Question 20

If n^th term of an A.P. is (2n – 1), find its 7^th term.

We know that the general formula for the n-th term of the arithmetic progression (A.P.) is given by:

⇒ a_n = (2n – 1)

To find the 7th term, substitute n = 7 into the formula:

⇒ a_7 = (2 \times 7 – 1) = 13.

Hence, 7^\text{th} term = 13.

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Question 21

If the sum of first n terms of an A.P. is 3n^2 + 2n, find its r^th term.

The formula for the sum of the first n terms of an arithmetic progression (A.P.) is given by:

S~n = \dfrac{n}{2}[2a + (n - 1)d]

Here, we are given:

S~n = 3n^2 + 2n

By equating the two expressions, we have:

\begin{aligned}\Rightarrow 3n^2 + 2n = \dfrac{n}{2}[2a + (n - 1)d] \\\Rightarrow 3n^2 + 2n = \dfrac{n}{2}[a + a + (n - 1)d] \\\Rightarrow n(3n + 2) = \dfrac{n}{2}[a + a_n] \\\Rightarrow 3n + 2 = \dfrac{a + a_n}{2} \\\Rightarrow 2(3n + 2) = a + a_n \\\Rightarrow 6n + 4 = a + a_n \text{ .........(1)}\end{aligned}

Let’s find the first term, a. Since:

S~n = 3n^2 + 2n

For the first term, S~1 = 3(1)^2 + 2(1) = 3 + 2 = 5.

∴ a = 5.

Now, substitute this value into equation (1):

⇒ 6n + 4 = 5 + a~n

⇒ a~n = 6n – 1

Thus, for the r-th term, we have:

⇒ a~r = 6r – 1

Hence, r~th term = 6r – 1.

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Question 22

For an A.P. the sum of its terms is 60, common difference is 2 and last term is 18. Find the number of terms in the A.P.

We know that the common difference d is 2 and the last term l is 18. Let us denote the number of terms by n.

Using the formula for the last term in an arithmetic progression:

l = a + (n - 1)d

Substituting the given values, we have:

18 = a + 2(n - 1)
18 = a + 2n - 2
a + 2n = 20
a = 20 - 2n ( \text{…(1)} )

The sum of the terms S~n is given by:

S~n = \dfrac{n}{2}(a + l)

Substituting the known values:

60 = \dfrac{n}{2}(a + 18)
120 = n(a + 18)

Now, substituting the expression for a from equation (1):

120 = n(20 - 2n + 18)
120 = n(38 - 2n)
120 = 38n - 2n^2
2n^2 - 38n + 120 = 0
2(n^2 - 19n + 60) = 0
n^2 - 19n + 60 = 0

To factorize the quadratic equation:

n^2 - 15n - 4n + 60 = 0
n(n - 15) - 4(n - 15) = 0
(n - 4)(n - 15) = 0

Thus, n - 4 = 0 or n - 15 = 0, which gives:

n = 4 \text{ or } n = 15.

Hence, no. of terms = 4 or 15.

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Question 23

Find the geometric progression whose 5^th term is 48 and 8^th term is 384.

Consider the first term of the geometric progression as a and the common ratio as r.

Using the formula for the n^{th} term of a geometric progression:

a_n = ar^{n-1}

For the 5th term:

a_5 = ar^{5-1}
48 = ar^4
(1)

For the 8th term:

a_8 = ar^{8-1}
ar^7 = 384
(2)

Now, divide equation (2) by equation (1):

\begin{aligned}\Rightarrow \dfrac{ar^7}{ar^4} = \dfrac{384}{48} \\\Rightarrow r^3 = 8 \\\Rightarrow r^3 = (2)^3 \\\Rightarrow r = 2.\end{aligned}

Substitute r = 2 back into equation (1):

\begin{aligned}a(2)^4 = 48 \\16a = 48 \\a = \dfrac{48}{16} = 3.\end{aligned}

Thus, the geometric progression is:

G.P. = a, ar, ar^2, \ldots
= 3, 3 \times 2, 3 \times 2^2, \ldots
= 3, 6, 12, \ldots

Hence, G.P. = 3, 6, 12, \ldots

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Question 24

Sum of how many terms of the G.P. \dfrac{2}{9} - \dfrac{1}{3} + \dfrac{1}{2} - ......\text{ is } \dfrac{55}{72} ?

The common ratio r for the given geometric progression can be calculated as follows:

r = \dfrac{-\dfrac{1}{3}}{\dfrac{2}{9}} = -\dfrac{9}{2 \times 3} = -\dfrac{3}{2}.

Assume the number of terms is n. Using the sum formula for a geometric progression where |r| > 1, we have:

\text{Sum of G.P.} = \dfrac{a(r^n - 1)}{(r - 1)}.

Substituting the values into the formula, we get:

\Rightarrow \dfrac{\dfrac{2}{9}\Big\Big(-\dfrac{3}{2}\Big)^n – 1\Big}{\Big(-\dfrac{3}{2} - 1\Big)} = \dfrac{55}{72}.

Simplifying further:

\Rightarrow \dfrac{\dfrac{2}{9}\Big\Big(-\dfrac{3}{2}\Big)^n – 1\Big}{\Big(\dfrac{-3 - 2}{2}\Big)} = \dfrac{55}{72}.

\Rightarrow \dfrac{\dfrac{2}{9}\Big\Big(-\dfrac{3}{2}\Big)^n – 1\Big}{\Big(-\dfrac{5}{2}\Big)} = \dfrac{55}{72}.

\Rightarrow \dfrac{2}{9}\Big\Big(-\dfrac{3}{2}\Big)^n – 1\Big = \dfrac{55}{72} \times -\dfrac{5}{2}.

\Rightarrow \Big\Big(-\dfrac{3}{2}\Big)^n – 1\Big = \dfrac{55}{72} \times -\dfrac{5}{2} \times \dfrac{9}{2}.

\Rightarrow \Big\Big(-\dfrac{3}{2}\Big)^n – 1\Big = -\dfrac{275}{32}.

\Rightarrow \Big(-\dfrac{3}{2}\Big)^n = -\dfrac{275}{32} + 1. \Rightarrow \Big(-\dfrac{3}{2}\Big)^n = \dfrac{-275 + 32}{32}. \Rightarrow \Big(-\dfrac{3}{2}\Big)^n = -\dfrac{243}{32}. \Rightarrow \Big(-\dfrac{3}{2}\Big)^n = \Big(-\dfrac{3}{2}\Big)^5.

Thus, n = 5.

Hence, sum of 5 terms = \dfrac{55}{72}.

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Question 25

Find the sum of n terms of the sequence : 5 + 55 + 555 + …….

Consider the sequence: 5, 55, 555, and so on.

Notice that this can be expressed as:

⇒ 5(1 + 11 + 111 + \timesext{...})

This is equivalent to:

⇒ \dfrac{5}{9} \times 9(1 + 11 + 111 + \timesext{... up to n terms})

Which can be rewritten as:

⇒ \dfrac{5}{9}(9 + 99 + 999 + \timesext{... up to n terms})

Breaking this down further:

⇒ \dfrac{5}{9}[(10 - 1) + (10^2 - 1) + (10^3 - 1) + \timesext{... up to n terms}]

This simplifies to:

⇒ \dfrac{5}{9}[(10 + 10^2 + 10^3 + \text{ up to n terms}) - n]

We can use the formula for the sum of a geometric series:

⇒ \dfrac{5}{9}\Big[\Big(\dfrac{10(10^n - 1)}{10 - 1} - n\Big)\Big]

Simplifying further, we have:

⇒ \dfrac{5}{9} \times \dfrac{10(10^n - 1)}{9} - \dfrac{5}{9}n

Finally, the sum of the sequence is:

⇒ \dfrac{50(10^n - 1)}{81} - \dfrac{5}{9}n

Hence, sum = \dfrac{50(10^n - 1)}{81} - \dfrac{5}{9}n.

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Question 26

What point on x-axis is equidistant from the points (6, 7) and (4, -3) ?

Consider a point on the x-axis, which has a y-coordinate of 0. Let’s denote this point as (x, 0).

According to the distance formula, the distance between two points (x_1, y_1) and (x_2, y_2) is given by:

\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Since the point (x, 0) is equidistant from the points (6, 7) and (4, -3), we can set up the equation:

\therefore \sqrt{(x - 6)^2 + (0 - 7)^2} = \sqrt{(x - 4)^2 + [0 - (-3)]^2}

Simplifying both sides, we have:

\Rightarrow \sqrt{x^2 + 36 - 12x + (-7)^2} = \sqrt{x^2 + 16 - 8x + 3^2}

This simplifies further to:

\Rightarrow \sqrt{x^2 + 36 - 12x + 49} = \sqrt{x^2 + 16 - 8x + 9}

By squaring both sides, we eliminate the square roots:

\Rightarrow x^2 - 12x + 85 = x^2 - 8x + 25

Cancel out the x^2 terms and simplify:

\Rightarrow -12x + 85 = -8x + 25

Rearranging gives:

\Rightarrow 4x = 60

Dividing both sides by 4, we find:

\Rightarrow x = 15.

Thus, the point on the x-axis is (x, 0) = (15, 0).

Hence, (15, 0) is equidistant from (6, 7) and (4, -3).

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Question 27

Calculate the ratio in which the line segment A(6, 5) and B(4, -3) is divided by the line y = 2.

Consider that any point on the line y = 2 will have a y-coordinate of 2. Let’s denote this point as (x, 2).

Assume that this point divides the line segment joining A(6, 5) and B(4, -3) in the ratio k : 1.

According to the section formula, the coordinates of a point dividing a line segment in the ratio m_1 : m_2 are given by:

(x, y) = \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

Substituting the known values for the y-coordinate, we have:

\Rightarrow 2 = \dfrac{k \times (-3) + 1 \times 5}{k + 1}

Simplifying the equation:

\Rightarrow 2(k + 1) = -3k + 5

Expanding and rearranging terms:

\Rightarrow 2k + 2 = -3k + 5

Bringing like terms together:

\Rightarrow 2k + 3k = 5 - 2

This simplifies to:

\Rightarrow 5k = 3

Thus, solving for k, we find:

\Rightarrow k = \dfrac{3}{5}

Therefore, the ratio k : 1 is \dfrac{3}{5} : 1, which simplifies to 3 : 5.

Hence, the line segment joining A(6, 5) and B(4, -3) is divided by the line y = 2 in the ratio 3 : 5.

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Question 28

Find the equations of the diagonals of a rectangle whose sides are x + 1 = 0, x – 4 = 0, y + 1 = 0 and y – 2 = 0.

Consider the rectangle formed by the lines x + 1 = 0, x – 4 = 0, y + 1 = 0, and y – 2 = 0. These lines intersect to form the vertices I, J, K, and L of the rectangle.

To find the equation of the diagonal IK, we use the two-point formula for a line:

Equation: y - y_1 = \dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1)

For diagonal IK:

\Rightarrow y - 2 = \dfrac{-1 - 2}{4 - (-1)}[x - (-1)] \\Rightarrow y - 2 = -\dfrac{3}{5}[x + 1] \\Rightarrow 5(y - 2) = -3[x + 1] \\Rightarrow 5y - 10 = -3x - 3 \\Rightarrow 5y + 3x - 10 + 3 = 0 \\Rightarrow 5y + 3x - 7 = 0.

Next, we find the equation of diagonal LJ:

\Rightarrow y - (-1) = \dfrac{2 - (-1)}{4 - (-1)}[x - (-1)] \\Rightarrow y + 1 = \dfrac{3}{5}[x + 1] \\Rightarrow 5(y + 1) = 3[x + 1] \\Rightarrow 5y + 5 = 3x + 3 \\Rightarrow 5y - 3x + 5 - 3 = 0 \\Rightarrow 5y - 3x + 2 = 0.

Therefore, the equations of the diagonals are 5y + 3x – 7 = 0 and 5y – 3x + 2 = 0.

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Question 29

The line 4x + 5y + 20 = 0 meets x-axis at point A and y-axis at point B. Find :

(i) the coordinates of points A and B.

(ii) the co-ordinates of point P in AB such that AB : BP = 5 : 3.

(iii) the equation of the line through P and perpendicular to AB.

(i) To find the coordinates where the line intersects the axes, let’s start with point A on the x-axis. Here, the y-coordinate is 0. Substitute y = 0 into the equation 4x + 5y + 20 = 0:

⇒ 4x + 5(0) + 20 = 0

⇒ 4x = -20

⇒ x = -5.

Thus, point A is (-5, 0).

For point B on the y-axis, the x-coordinate is 0. Substitute x = 0 into the equation:

⇒ 4(0) + 5y + 20 = 0

⇒ 5y = -20

⇒ y = \dfrac{-20}{5}

⇒ y = -4.

Thus, point B is (0, -4).

Hence, A = (-5, 0) and B = (0, -4).

(ii) Given the ratio AB : BP = 5 : 3, let AB = 5a and BP = 3a. From the relationship AB = AP + PB, we have:

⇒ 5a = AP + 3a

⇒ AP = 5a – 3a = 2a.

Thus, AP : BP = 2a : 3a = 2 : 3.

Let the coordinates of point P be (x, y). Using the section formula:

(x, y) = \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)

For the x-coordinate:

⇒ x = \dfrac{2 \times 0 + 3 \times -5}{2 + 3} \
⇒ x = \dfrac{0 – 15}{5} \
⇒ x = \dfrac{-15}{5} \
⇒ x = -3.

For the y-coordinate:

⇒ y = \dfrac{2 \times -4 + 3 \times 0}{2 + 3} \
⇒ y = \dfrac{-8 + 0}{5} \
⇒ y = \dfrac{-8}{5} = -1\dfrac{3}{5}.

Thus, P = (x, y) = \Big(-3, -1\dfrac{3}{5}\Big).

(iii) To find the equation of the line through P and perpendicular to AB, first determine the slope of AB:

Slope = \dfrac{y_2 – y_1}{x_2 – x_1}

⇒ Slope of AB = \dfrac{-4 – 0}{0 – (-5)} = -\dfrac{4}{5}.

Let m be the slope of the line perpendicular to AB. Since the product of the slopes of perpendicular lines is -1:

∴ -\dfrac{4}{5} \times m = -1 \
⇒ m = \dfrac{5}{4}.

Using the point-slope form for the equation of the line:

Equation of line: y – y_1 = m(x – x_1)

Substitute the values:

⇒ y – \Big(-1\dfrac{3}{5}\Big) = \dfrac{5}{4}[x – (-3)] \
⇒ y + \Big(\dfrac{8}{5}\Big) = \dfrac{5}{4}[x + 3] \
⇒ \dfrac{5y + 8}{5} = \dfrac{5x + 15}{4} \
⇒ 4(5y + 8) = 5(5x + 15) \
⇒ 20y + 32 = 25x + 75 \
⇒ 25x – 20y + 75 – 32 = 0 \
⇒ 25x – 20y + 43 = 0

Hence, the equation of the line through P and perpendicular to AB is 25x – 20y + 43 = 0.

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Question 30

In the given figure, DE || BC and AE : EC = 5 : 4. Find :

(i) DE : BC

(ii) DO : DC

(iii) area of △DOE : area of △ DCE.

(i) Since DE is parallel to BC, we have:

In triangles ADE and ABC:

∠ADE = ∠ABC because they are corresponding angles.

∠DAE = ∠BAC as they are the same angle.

Thus, △ADE is similar to △ABC by the AA similarity criterion.

Given that AE : EC = 5 : 4, we can set AE = 5x and EC = 4x.

Therefore, AC = AE + EC = 5x + 4x = 9x.

Since corresponding sides of similar triangles are proportional:

\Rightarrow \dfrac{DE}{BC} = \dfrac{AE}{AC} \Rightarrow \dfrac{DE}{BC} = \dfrac{5x}{9x} = \dfrac{5}{9}

Hence, DE : BC = 5 : 9.

(ii) Considering triangles DOE and BOC:

∠DOE = ∠BOC as they are vertically opposite angles.

∠ODE = ∠OCB because they are alternate angles.

Thus, △DOE is similar to △BOC by the AA similarity criterion.

We know DE : BC = 5 : 9.

Therefore, for corresponding sides of similar triangles:

\Rightarrow \dfrac{DO}{OC} = \dfrac{DE}{BC} \Rightarrow \dfrac{DO}{OC} = \dfrac{5}{9}

This implies DO : OC = 5 : 9.

Let DO = 5y and OC = 9y.

From the figure, DC = DO + OC = 5y + 9y = 14y.

Thus, DO : DC = 5y : 14y = 5 : 14.

Hence, DO : DC = 5 : 14.

(iii) We understand that the ratio of the areas of two triangles with the same height is equal to the ratio of their corresponding bases.

\Rightarrow \dfrac{\text{Area of △DOE}}{\text{Area of △DCE}} = \dfrac{DO}{DC} = \dfrac{5}{14}

Hence, area of △DOE : area of △ DCE = 5 : 14.

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Question 31

If chords AB and CD of a circle intersect each other at a point P inside the circle, prove that : PA × PB = PC × PD.

Consider the triangles △PAC and △PDB.

Notice that ∠APC and ∠DPB are equal because they are vertically opposite angles.

Also, ∠CAP is equal to ∠BDP since they are angles subtended by the same arc in the circle.

∴ △PAC is similar to △PDB by the Angle-Angle (AA) similarity criterion.

For similar triangles, the corresponding sides are in proportion.

∴ \dfrac{PA}{PD} = \dfrac{PC}{PB}

⇒ PA × PB = PC × PD.

Hence, proved that PA × PB = PC × PD.

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Question 32

P and Q are centers of a circle of radii 9 cm and 2 cm respectively. PQ = 17 cm. R is the centre of a circle of radius x cm which touches the above circles externally. Given that ∠PRQ = 90°, write an equation in x and solve it.

Consider the triangle △PQR. We are given that ∠PRQ = 90°.

According to the Pythagorean theorem, the relationship between the sides is given by:

∴ PQ^2 = PR^2 + QR^2

Substituting the known values, we have:

⇒ 17^2 = (x + 9)^2 + (x + 2)^2

Calculating further, we get:

⇒ 289 = x^2 + 81 + 18x + x^2 + 4 + 4x

Simplifying, this becomes:

⇒ 289 = 2x^2 + 22x + 85

Rearranging the terms, we have:

⇒ 2x^2 + 22x + 85 - 289 = 0

⇒ 2x^2 + 22x - 204 = 0

Factoring out the 2, we get:

⇒ 2(x^2 + 11x - 102) = 0

Dividing by 2, the equation simplifies to:

⇒ x^2 + 11x - 102 = 0

To solve for x, we factor the quadratic:

⇒ x^2 + 17x - 6x - 102 = 0

Grouping the terms, we find:

⇒ x(x + 17) - 6(x + 17) = 0

This gives us:

⇒ (x - 6)(x + 17) = 0

Setting each factor to zero, we solve:

⇒ x - 6 = 0 or x + 17 = 0

⇒ x = 6 or x = -17

Since a radius cannot be negative, we conclude:

⇒ x = 6 cm.

Hence, equation is x^2 + 11x - 102 = 0 and x = 6 cm.

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Question 33

Use ruler and a pair of compasses only in this question :

(i) Draw a circle on AB = 6.4 cm as diameter.

(ii) Construct another circle of radius 2.5 cm to touch the circle in (i) above and the diameter AB produced.

(i) Follow these construction steps:

1. Begin by marking a line segment AB with a length of 6.4 cm.
2. Construct the perpendicular bisector of AB, which will intersect AB at the midpoint, labeled as point O.
3. Using O as the center, draw a circle with a radius equal to OA or OB.
4. Draw a line LM that is parallel to AB and positioned 2.5 cm away from it.
5. With O as the center, draw an arc with a radius of 5.7 cm (since 3.2 + 2.5 = 5.7 cm) that intersects line LM at point D.
6. Using D as the center, draw a circle with a radius of 2.5 cm. This circle should touch the previously drawn circle at point E and extend to touch the extended line of AB at point C.

Hence, above is the required circle.

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Question 34

The internal and external diameters of a hollow hemispherical vessel are 14 cm and 21 cm respectively. The cost of silver plating of 1 cm^2 of its surface is ₹ 32. Find the total cost of silver plating the vessel all over.

To solve this, we start by determining the radii of the hollow hemispherical vessel:

• The external radius R is given by \dfrac{21}{2} = 10.5 cm.
• The internal radius r is \dfrac{14}{2} = 7 cm.

The total surface area T of the hollow hemisphere is calculated using the formula:

T = \text{Curved surface area of inner hemisphere} + \text{Curved surface area of outer hemisphere} + \text{Area of annular disc}.

This simplifies to:

T = 2\pi r^2 + 2\pi R^2 + \pi R^2 - \pi r^2 = 3\pi R^2 + \pi r^2.

Substituting the values of R and r into the equation, we have:

T = 3 \times \dfrac{22}{7} \times (10.5)^2 - \dfrac{22}{7} \times 7^2.

Calculating further:

T = \dfrac{66}{7} \times 110.25 + 154 = 66 \times 15.75 + 154 = 1039.5 + 154 = 1193.5 \text{ cm}^2.

The cost of silver plating per square centimeter is ₹ 32. Therefore, the total cost of plating the entire surface is:

\text{Total cost} = 1193.5 \times ₹ 32 = ₹ 38192.

Thus, the total cost for silver plating the vessel is ₹ 38192.

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Question 35(i)

Prove that :

\dfrac{\text{cosec A + 1}}{\text{cosec A - 1}} = (sec A + tan A)^2

We need to demonstrate that:

\dfrac{\text{cosec A + 1}}{\text{cosec A - 1}} = (\text{sec A + tan A})^2

Starting with the left-hand side:

\Rightarrow \dfrac{\dfrac{1}{\text{sin A}} + 1}{\dfrac{1}{\text{sin A}} - 1}

This can be rewritten as:

\Rightarrow \dfrac{\dfrac{\text{1 + sin A}}{\text{sin A}}}{\dfrac{\text{1 - sin A}}{\text{sin A}}}

Simplifying further, we have:

\Rightarrow \dfrac{\text{1 + sin A}}{\text{1 - sin A}}

To eliminate the fraction, multiply both the numerator and the denominator by (1 + \text{sin A}):

\Rightarrow \dfrac{\text{1 + sin A}}{\text{1 - sin A}} \times \dfrac{\text{1 + sin A}}{\text{1 + sin A}}

This results in:

\Rightarrow \dfrac{(\text{1 + sin A})^2}{\text{1 - sin}^2 A}

Using the identity 1 - \text{sin}^2 A = \text{cos}^2 A, substitute to get:

\Rightarrow \dfrac{\text{(1 + sin A)}^2}{\text{cos}^2 A}

This can be expressed as:

\Rightarrow \Big(\dfrac{\text{1 + sin A}}{\text{cos A}}\Big)^2

Breaking it down further gives:

\Rightarrow \Big(\dfrac{1}{\text{cos A}} + \dfrac{\text{sin A}}{\text{cos A}}\Big)^2

Which simplifies to:

\Rightarrow (\text{sec A + tan A})^2.

Thus, the left-hand side equals the right-hand side, confirming that:

Hence, proved that \dfrac{\text{cosec A + 1}}{\text{cosec A - 1}} = (\text{sec A + tan A})^2.

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Question 35(ii)

Prove that :

cot A – tan A = \dfrac{\text{2 cos}^2 A - 1}{\text{sin A cos A}}

Let’s start with the right side of the equation:

\begin{aligned}\Rightarrow \dfrac{\text{2 cos}^2 A - 1}{\text{sin A cos A}} \\\Rightarrow \dfrac{\text{cos}^2 A - 1 + \text{cos}^2 A}{\text{sin A cos A}} \\\Rightarrow \dfrac{\text{cos}^2 A - \text{(1 - cos}^2 A)}{\text{sin A cos A}}\end{aligned}

Recall the identity:

1 - \text{cos}^2 A = \text{sin}^2 A

Substituting this, we have:

\begin{aligned}\Rightarrow \dfrac{\text{cos}^2 A - \text{sin}^2 A}{\text{sin A cos A}} \\\Rightarrow \dfrac{\text{cos}^2 A}{\text{sin A cos A}} - \dfrac{\text{sin}^2 A}{\text{sin A cos A}} \\\Rightarrow \dfrac{\text{cos A}}{\text{sin A}} - \dfrac{\text{sin A}}{\text{cos A}} \\\Rightarrow \text{cot A - tan A}.\end{aligned}

Therefore, the left-hand side equals the right-hand side.

Hence, proved that cot A – tan A = \dfrac{\text{2 cos}^2 A - 1}{\text{sin A cos A}}.

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Question 35(iii)

Prove that :

\sqrt{\dfrac{\text{1 + cos A}}{\text{1 - cos A}}} + \sqrt{\dfrac{\text{1 - cos A}}{\text{1 + cos A}}} = 2 cosec A

We aim to show that:

\sqrt{\dfrac{\text{1 + cos A}}{\text{1 - cos A}}} + \sqrt{\dfrac{\text{1 - cos A}}{\text{1 + cos A}}} = 2 \text{ cosec A}

Let’s work through the left-hand side:

\Rightarrow \sqrt{\dfrac{\text{1 + cos A}}{\text{1 - cos A}}} + \sqrt{\dfrac{\text{1 - cos A}}{\text{1 + cos A}}}

To simplify, multiply each fraction by its conjugate:

\Rightarrow \sqrt{\dfrac{\text{1 + cos A}}{\text{1 - cos A}} \times \dfrac{\text{1 + cos A}}{\text{1 + cos A}}} + \sqrt{\dfrac{\text{1 - cos A}}{\text{1 + cos A}} \times \dfrac{\text{1 - cos A}}{\text{1 - cos A}}}

This gives us:

\Rightarrow \sqrt{\dfrac{(\text{1 + cos A})^2}{\text{1 - cos}^2 A}} + \sqrt{\dfrac{(\text{1 - cos A})^2}{\text{1 - cos}^2 A}}

Recall the identity:

1 - \cos^2 A = \sin^2 A

Substitute this identity:

\Rightarrow \sqrt{\dfrac{(\text{1 + cos A})^2}{\text{sin}^2 A}} + \sqrt{\dfrac{(\text{1 - cos A})^2}{\text{sin}^2 A}}

Simplify the square roots:

\Rightarrow \dfrac{\text{1 + cos A}}{\text{sin A}} + \dfrac{\text{1 - cos A}}{\text{sin A}}

Combine the terms:

\Rightarrow \dfrac{1}{\text{sin A}} + \dfrac{\text{cos A}}{\text{sin A}} + \dfrac{1}{\text{sin A}} - \dfrac{\text{cos A}}{\text{sin A}}

Notice how the terms simplify:

\Rightarrow \text{cosec A + cot A + cosec A - cot A}

This reduces to:

\Rightarrow \text{2 cosec A}

Thus, the left-hand side equals the right-hand side.

Hence, proved that \sqrt{\dfrac{\text{1 + cos A}}{\text{1 - cos A}}} + \sqrt{\dfrac{\text{1 - cos A}}{\text{1 + cos A}}} = 2 \text{ cosec A}.

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Question 35(iv)

Prove that :

\dfrac{\text{cos A}}{\text{cosec A + 1}} + \dfrac{\text{cos A}}{\text{cosec A - 1}} = 2 tan A

We need to establish that:

\dfrac{\text{cos A}}{\text{cosec A + 1}} + \dfrac{\text{cos A}}{\text{cosec A - 1}} = 2 \tan A

Let’s focus on the left-hand side (LHS) of the equation:

\Rightarrow \dfrac{\text{cos A}}{\text{cosec A + 1}} + \dfrac{\text{cos A}}{\text{cosec A - 1}}

Combine the two fractions by finding a common denominator:

\Rightarrow \dfrac{\text{cos A(cosec A - 1) + cos A(cosec A + 1)}}{\text{(cosec A + 1)(cosec A - 1)}}

Simplify the numerator:

\Rightarrow \dfrac{\text{cos A.cosec A - cos A + cos A.cosec A + cos A}}{\text{cosec}^2 A - 1}

Notice that the denominator can be rewritten using the identity \text{cosec}^2 A - 1 = \text{cot}^2 A:

\Rightarrow \dfrac{\text{2 cos A cosec A}}{\text{cot}^2 A}

Since \text{cosec A} = \dfrac{1}{\text{sin A}}, substitute to get:

\Rightarrow \dfrac{\text{2 cos A} \times \dfrac{1}{\text{sin A}}}{\text{cot}^2 A}

This simplifies to:

\Rightarrow \dfrac{\text{2 cot A}}{\text{cot}^2 A}

Further simplification gives:

\Rightarrow \dfrac{2}{\text{cot A}}

Since \tan A = \dfrac{1}{\text{cot A}}, it follows that:

\Rightarrow \text{2 tan A}

Thus, we have shown that the LHS equals the right-hand side (RHS):

Hence, proved that \dfrac{\text{cos A}}{\text{cosec A + 1}} + \dfrac{\text{cos A}}{\text{cosec A - 1}} = 2 \tan A.

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Question 36

Solve for x, 0° ≤ x ≤ 90°

(i) 4 cos^2 2x – 3 = 0

(ii) 2 sin^2 x – sin x = 0

(i) We start with the equation:

⇒ 4 \cos^2 2x – 3 = 0

Rearranging gives:

⇒ 4 \cos^2 2x = 3

Dividing both sides by 4:

⇒ \cos^2 2x = \dfrac{3}{4}

Taking the square root, we have:

⇒ \cos 2x = \sqrt{\dfrac{3}{4}}

⇒ \cos 2x = \dfrac{\sqrt{3}}{2}

Recognizing this as a standard angle:

⇒ \cos 2x = \cos 30°

Therefore, equating the angles:

⇒ 2x = 30°

Solving for x:

⇒ x = \dfrac{30°}{2} = 15°.

Hence, x = 15°.

(ii) Consider the equation:

⇒ 2 \sin^2 x – \sin x = 0

Factoring out \sin x:

⇒ \sin x(2 \sin x – 1) = 0

This gives us two possible cases:

⇒ \sin x = 0 \text{ or } 2 \sin x – 1 = 0

Solving each case:

⇒ \sin x = 0 \text{ or } 2 \sin x = 1

Thus, we have:

⇒ \sin x = 0 \text{ or } \sin x = \dfrac{1}{2}

Recognizing these values:

⇒ \sin x = \sin 0° \text{ or } \sin x = \sin 30°

Therefore, the solutions are:

⇒ x = 0° \text{ or } x = 30°.

Hence, x = 0° or 30°.

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Question 37

A ladder rests against a wall at an angle α to the horizontal. Its foot is pulled away from the wall through a distance a, so that it slides a distance b down the wall making an angle β with the horizontal. Show that :

\dfrac{a}{b} = \dfrac{\text{cos α - cos β}}{\text{sin β - sin α}}

Consider the ladder represented by AB and CD in the diagram. Since both represent the same ladder, their lengths are identical.

Let the ladder’s length be denoted by h.

∴ AB = CD = h

Now, examine △AEB:

For angle α, we have:

\sin α = \dfrac{AE}{AB}

This implies:

AE = AB \sin α = h \sin α.

Similarly, for the cosine of angle α:

\cos α = \dfrac{BE}{AB}

Hence:

BE = AB \cos α = h \cos α.

Next, consider △DEC:

For angle β, we find:

\sin β = \dfrac{DE}{CD}

Thus:

DE = CD \sin β = h \sin β

And for the cosine of angle β:

\cos β = \dfrac{CE}{CD}

Therefore:

CE = CD \cos β = h \cos β

From the figure, we observe:

\Rightarrow \dfrac{a}{b} = \dfrac{BC}{AD} \\Rightarrow \dfrac{a}{b} = \dfrac{CE - BE}{AE - DE} \\Rightarrow \dfrac{a}{b} = \dfrac{\text{h cos β} - \text{h cos α}}{\text{h sin α} - \text{h sin β}} \\Rightarrow \dfrac{a}{b} = \dfrac{\text{cos β} - \text{cos α}}{\text{sin α} - \text{sin β}}

Thus, it is demonstrated that \dfrac{a}{b} = \dfrac{\text{cos β} - \text{cos α}}{\text{sin α} - \text{sin β}}.

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Question 38

For the following frequency distribution, draw an ogive and then use it to estimate the median.

For the same distribution, as given above, draw a histogram and then use it to estimate the mode.

To determine the median using an ogive, we first create a cumulative frequency table:

The total number of terms is 450, which is an even number.

Using the formula for the median:

Median = \dfrac{n}{2} th term = 225.

Steps to construct the ogive:

1. On the x-axis, use a scale of 2 cm for 100 units (C.I.).
2. On the y-axis, use a scale of 1 cm for 50 units (frequency).
3. Begin the ogive at (450, 0), indicating the lower limit of the first class.
4. Plot the following points: (550, 40), (650, 108), (750, 194), (850, 314), (950, 404), (1050, 444), and (1150, 450).
5. Connect these points with a smooth, free-hand curve.
6. Draw a horizontal line from the y-axis at the cumulative frequency of 225 to intersect the ogive at point J. From J, draw a vertical line down to the x-axis at point K.

From the graph, point K corresponds to 780.

Therefore, the median is 780.

Steps to estimate the mode using a histogram:

1. Construct a histogram for the given frequency distribution.
2. Identify the tallest rectangle, which represents the modal class.
3. Draw lines AC and BD diagonally from the top corners of the rectangles adjacent to the tallest one.
4. The intersection of these diagonals is point K. Draw a perpendicular line from K to the x-axis, meeting it at L.
5. The x-coordinate of point L represents the mode.

∴ Mode = 803.

Hence, the mode is 803.

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Question 39(i)

Find the probability of drawing an ace or a jack from a pack of 52 cards.

In a standard deck, there are 52 cards in total.

∴ The total number of possible outcomes is 52.

Within this deck, there are 4 aces and 4 jacks.

∴ The number of favourable outcomes, which are either an ace or a jack, is 8.

The probability of drawing either an ace or a jack is given by the formula:

P(\text{drawing an ace or jack}) = \dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{8}{52} = \dfrac{2}{13}.

Hence, probability of drawing an ace or a jack = \dfrac{2}{13}.

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Question 39(ii)

A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of red ball, find the number of blue balls in the bag.

Assume the number of blue balls is x.

The total number of balls becomes (x + 5), which means the number of possible outcomes is x + 5.

Since there are 5 red balls, the number of favourable outcomes for picking a red ball is 5. Thus, the probability of selecting a red ball is given by:

P(\text{drawing a red ball}) = \dfrac{5}{5 + x}.

With x blue balls, the number of favourable outcomes for a blue ball is x. So, the probability of drawing a blue ball is:

P(\text{drawing a blue ball}) = \dfrac{x}{5 + x}.

According to the problem, the probability of getting a blue ball is twice that of a red ball. Therefore, we have:

\begin{aligned} \dfrac{x}{5 + x} = 2 \times \dfrac{5}{5 + x} \\\Rightarrow \dfrac{x}{5 + x} = \dfrac{10}{5 + x} \\\Rightarrow x = 10. \end{aligned}

Hence, no. of blue balls = 10.

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Question 40

In a single throw of two dice, what is the probability of getting

(i) a total of 9

(ii) two aces (ones)

(iii) at least one ace

(iv) a doublet

(v) five on one die and six on the other

(vi) a multiple of 2 on one and a multiple of 3 on the other ?

When rolling two dice together, the total number of possible outcomes is 6 \times 6 = 36.

(i) To achieve a total of 9, the successful outcomes are: (3, 6), (4, 5), (5, 4), and (6, 3).

∴ Number of successful outcomes = 4.

Thus, the probability of getting a total of 9 is \dfrac{4}{36} = \dfrac{1}{9}.

Hence, probability of getting a total of 9 = \dfrac{1}{9}.

(ii) For rolling two aces (both ones), the only successful outcome is (1, 1).

∴ Number of successful outcomes = 1.

Hence, the probability of rolling two aces is \dfrac{1}{36}.

Hence, probability of getting two aces = \dfrac{1}{36}.

(iii) To get at least one ace, the successful outcomes are: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (3, 1), (4, 1), (5, 1), and (6, 1).

∴ Number of successful outcomes = 11.

Thus, the probability of getting at least one ace is \dfrac{11}{36}.

Hence, probability of getting at least one ace = \dfrac{11}{36}.

(iv) For rolling a doublet (both dice showing the same number), the successful outcomes are: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), and (6, 6).

∴ Number of successful outcomes = 6.

Therefore, the probability of getting a doublet is \dfrac{6}{36} = \dfrac{1}{6}.

Hence, probability of getting a doublet = \dfrac{1}{6}.

(v) To roll a 5 on one die and a 6 on the other, the successful outcomes are: (5, 6) and (6, 5).

∴ Number of successful outcomes = 2.

Thus, the probability of rolling a 5 and a 6 is \dfrac{2}{36} = \dfrac{1}{18}.

Hence, probability of getting 5 on one die and 6 on other = \dfrac{1}{18}.

(vi) For rolling a multiple of 2 on one die and a multiple of 3 on the other, the successful outcomes are: (2, 3), (2, 6), (3, 2), (3, 4), (3, 6), (4, 3), (4, 6), (6, 2), (6, 3), (6, 4), and (6, 6).

∴ Number of successful outcomes = 11.

Therefore, the probability of this event is \dfrac{11}{36}.

Hence, probability of getting a multiple of 2 on one die and a multiple of 3 on other = \dfrac{11}{36}.

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SET B

Question 41

A dealer sells goods/services, worth ₹ 30,000 to some other dealer in the same town at a discount of 25%. If the rate of GST is 12%, find the amount of bill.

Let’s determine the final bill amount after applying the discount and adding GST.

Given:

• Marked Price (M.R.P.) = ₹ 30,000
• Discount offered = 25%

First, calculate the discount amount:

Discount = \dfrac{25}{100} \times 30000

⇒ Discount = ₹ 7,500

Now, subtract the discount from the M.R.P. to find the price after discount:

Amount after discount = ₹ 30,000 – ₹ 7,500

⇒ Amount after discount = ₹ 22,500

Next, add the GST to this discounted amount. The GST rate is 12%:

GST = \dfrac{12}{100} \times 22500

⇒ GST = 12 × 225

⇒ GST = ₹ 2,700

Therefore, the total amount including GST is:

Total bill amount = ₹ 22,500 + ₹ 2,700

⇒ Total bill amount = ₹ 25,200

Hence, amount of bill = ₹ 25,200.

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Question 42

The monthly installment of a recurring deposit account is ₹ 2400. If the account is held for 1 year 6 months and its maturity value is ₹ 47,304, find the rate of interest.

We know that the account is held for 1 year and 6 months, which converts to 18 months in total.

The monthly deposit amount, denoted as P, is ₹ 2400.

The maturity value of the account is ₹ 47,304.

Let’s assume the rate of interest is r\%.

The formula for maturity value is:

\text{Maturity value} = P \times n + P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}

Substitute the given values into the formula:

47304 = 2400 \times 18 + 2400 \times \dfrac{18 \times 19}{2 \times 12} \times \dfrac{r}{100}

Simplifying further:

47304 = 43200 + 2400 \times \dfrac{3 \times 19}{2 \times 2} \times \dfrac{r}{100}

Subtract 43200 from 47304:

47304 - 43200 = 2400 \times \dfrac{57}{4} \times \dfrac{r}{100}

This simplifies to:

4104 = 6 \times 57r

To find r, divide 4104 by 6 \times 57:

r = \dfrac{4104}{6 \times 57}

Calculate r:

r = \dfrac{4104}{342} = 12\%

Therefore, the rate of interest is 12%.

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Question 43

The maturity value of a recurring deposit account is ₹ 42,400. If the account is held for 2 years and the rate of interest is 10% per annum, find the amount of each monthly installment.

Let’s denote the monthly installment as ₹ P.

We know:
– Duration (n) = 2 years = 24 months
– Interest rate (r) = 10%
– Maturity value = ₹ 42,400

According to the formula for the maturity value of a recurring deposit:

\text{Maturity value} = P \times n + P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}

Substituting the given values:

\begin{aligned}\Rightarrow 42400 = P \times 24 + P \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{10}{100} \\\Rightarrow 42400 = 24P + P \times 25 \times \dfrac{1}{10} \\\Rightarrow 42400 = 24P + 2.5P \\\Rightarrow 26.5P = 42400 \\\Rightarrow P = \dfrac{42400}{26.5} \\\Rightarrow P = 1600.\end{aligned}

Thus, the monthly installment is ₹ 1600.

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Question 44

A man invests equal amounts of money in two companies A and B. Company A pays a dividend of 15% and its ₹ 100 shares are available at 20% discount. The shares of company B has a nominal value of ₹ 25 and are available at 20% premium. If at the end of one year, the man gets equal dividends from both the companies, find the rate of dividend paid by company B.

For company A, consider the investment amount to be ₹ x.

The nominal value (N.V.) of each share is ₹ 100, and they are sold at a 20% discount. Therefore, the market value (M.V.) of each share is calculated as:

\text{M.V.} = \text{N.V.} - \text{Discount} = ₹ 100 - \dfrac{20}{100} \times 100 = ₹ 100 - ₹ 20 = ₹ 80.

The number of shares he can buy is:

\text{Number of shares} = \dfrac{\text{Invested money}}{\text{M.V.}} = \dfrac{x}{80}.

Given the dividend rate is 15%, the total dividend from company A is:

\text{Total dividend for company A} = \text{Number of shares} \times \text{Dividend} \times 100 = \dfrac{x}{80} \times \dfrac{15}{100} \times 100 = \dfrac{3x}{16}.

Now, for company B, again assume the investment amount to be ₹ x.

The nominal value of each share is ₹ 25, and they are sold at a 20% premium. Thus, the market value (M.V.) is:

\text{M.V.} = \text{N.V.} + \text{Premium} = ₹ 25 + \dfrac{20}{100} \times 25 = ₹ 25 + ₹ 5 = ₹ 30.

The number of shares he can buy is:

\text{Number of shares} = \dfrac{\text{Invested money}}{\text{M.V.}} = \dfrac{x}{30}.

Let the rate of dividend be d\%. The total dividend from company B is:

\text{Total dividend for company B} = \text{Number of shares} \times \text{Dividend} \times 100 = \dfrac{x}{30} \times \dfrac{d}{100} \times 25 = \dfrac{xd}{120}.

Since the dividends from both companies are equal:

\Rightarrow \dfrac{3x}{16} = \dfrac{xd}{120}

Solving for d:

\Rightarrow d = \dfrac{3x \times 120}{x \times 16} \Rightarrow d = \dfrac{360}{16} = 22.5\%.

Hence, dividend paid by company B = 22.5%.

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Question 45

A sum of ₹ 54000 is invested partly in shares paying 6% dividend at 40% premium and partly in 5% shares at 25% premium. If the nominal value of one share in each company is ₹ 100 and the total income of the man is ₹ 2,240, find the money invested in the second company.

To determine the investment in the first company, assume ₹ x is used.

The nominal value (N.V.) of a share is ₹ 100. Given a 40\% premium, the market value (M.V.) becomes:

\text{M.V.} = 100 + \frac{40}{100} \times 100 = 100 + 40 = ₹ 140.

Thus, the number of shares purchased is:

\frac{x}{140}.

With a dividend rate of 6\%, the total dividend from the first company is:

\frac{x}{140} \times \frac{6}{100} \times 100 = \frac{3x}{70}.

For the second company, let the investment be ₹ (54000 - x).

Here, the nominal value is also ₹ 100, but with a 25\% premium, the market value is:

\text{M.V.} = 100 + \frac{25}{100} \times 100 = 100 + 25 = ₹ 125.

The number of shares becomes:

\frac{54000 - x}{125}.

With a 5\% dividend rate, the total dividend from the second company is:

\frac{54000 - x}{125} \times \frac{5}{100} \times 100 = \frac{54000 - x}{25}.

The total income from both investments is ₹ 2240. Therefore, we have:

\frac{54000 - x}{25} + \frac{3x}{70} = 2240.

Solving this equation:

\frac{14(54000 - x) + 3x \times 5}{350} = 2240 \Rightarrow \frac{756000 - 14x + 15x}{350} = 2240 \Rightarrow 756000 + x = 2240 \times 350 \Rightarrow 756000 + x = 784000 \Rightarrow x = 784000 - 756000 = 28000.

Thus, the investment in the second company is:

₹ (54000 - x) = ₹ 54000 - ₹ 28000 = ₹ 26000.

Hence, money invested in second company = ₹ 26000.

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Question 46

Solve and graph the solution set of :

(i) 2x – 9 < 7 and 3x + 9 ≤ 25; x ∈ R

(ii) 3x – 2 > 19 or 3 – 2x ≥ 7; x ∈ R

(i) Let’s tackle the first inequality:

2x - 9 < 7

Adding 9 to both sides, we have:

2x < 16

Dividing by 2, it follows that:

x < \dfrac{16}{2}

x < 8 ………..(1)

Now consider the second inequality:

3x + 9 \leq 25

Subtracting 9 from both sides gives:

3x \leq 16

Dividing by 3, we get:

x \leq \dfrac{16}{3}

x \leq 5\dfrac{1}{3}……….(2)

Combining results from (1) and (2), we conclude:

x \leq 5\dfrac{1}{3}

Hence, solution = {x : x \leq 5\dfrac{1}{3}, x \in R}.

(ii) For the second part, let’s start with:

3x - 2 > 19

Adding 2 to both sides, we have:

3x > 21

Dividing by 3, it follows that:

x > \dfrac{21}{3}

x > 7 ……..(1)

Now, take the second inequality:

3 - 2x \geq 7

Subtracting 3 from both sides gives:

-2x \geq 4

Dividing by -2 (and reversing the inequality sign), we get:

x \leq -\dfrac{4}{2}

x \leq -2 ……..(2)

Combining results from (1) and (2), we conclude:

x > 7 \text{ or } x \leq -2

Hence, solution = {x : x > 7 or x \leq -2, x \in R}.

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Question 47

Use formula to solve the quadratic equation :

x^2 + x – (a + 1)(a + 2) = 0

The quadratic equation given is:

x^2 + x – (a + 1)(a + 2) = 0

To solve it using the quadratic formula, we first compare this with the standard form ax^2 + bx + c = 0. Here, we identify:

a = 1, b = 1, and c = -(a + 1)(a + 2)

The quadratic formula is:

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting the identified values, we have:

x = \dfrac{-1 \pm \sqrt{1^2 - 4 \times 1 \times -(a + 1)(a + 2)}}{2 \times 1}

This simplifies to:

x = \dfrac{-1 \pm \sqrt{1 + 4(a + 1)(a + 2)}}{2}

Expanding ((a + 1)(a + 2)) gives us:

x = \dfrac{-1 \pm \sqrt{1 + 4(a^2 + 2a + a + 2)}}{2}

Simplifying the expression further:

x = \dfrac{-1 \pm \sqrt{1 + 4(a^2 + 3a + 2)}}{2} x = \dfrac{-1 \pm \sqrt{1 + 4a^2 + 12a + 8}}{2} x = \dfrac{-1 \pm \sqrt{4a^2 + 12a + 9}}{2}

Recognizing 4a^2 + 12a + 9 as a perfect square:

x = \dfrac{-1 \pm \sqrt{(2a + 3)^2}}{2}

Taking the square root:

x = \dfrac{-1 \pm (2a + 3)}{2}

This results in two possible solutions:

x = \dfrac{-1 + (2a + 3)}{2}, \dfrac{-1 - (2a + 3)}{2}

Simplifying each:

x = \dfrac{2a + 3 - 1}{2}, \dfrac{-1 - 2a - 3}{2} x = \dfrac{2a + 2}{2}, \dfrac{-2a - 4}{2}

Further simplification gives:

x = \dfrac{2(a + 1)}{2}, \dfrac{-2(a + 2)}{2}

Finally, we have:

x = (a + 1), -(a + 2).

Hence, x = (a + 1) or -(a + 2).

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Question 48

By selling an article for ₹ 96, a man gains as much percent as its cost price. Find the cost price of the article.

Assume the cost price of the article is ₹ x.

The selling price is given as ₹ 96.

The profit made is calculated as the difference between the selling price and the cost price, which is ₹ (96 – x).

The problem states that the percentage gain is equal to the cost price. Thus, the profit percentage is x%.

Setting up the equation for profit percentage, we have:

\Rightarrow \dfrac{\text{Profit}}{\text{C.P.}} \times 100 = x

Substituting the known values, we get:

\Rightarrow \dfrac{96 - x}{x} \times 100 = x

Simplifying, we arrive at:

\Rightarrow 100(96 - x) = x^2

This leads to the equation:

\Rightarrow x^2 = 9600 - 100x

Rearranging terms gives:

\Rightarrow x^2 + 100x - 9600 = 0

Next, factorizing the quadratic equation:

\Rightarrow x^2 + 160x - 60x - 9600 = 0

Grouping terms, we get:

\Rightarrow x(x + 160) - 60(x + 160) = 0

Factoring completely, we find:

\Rightarrow (x - 60)(x + 160) = 0

This gives the solutions:

\Rightarrow x - 60 = 0 \text{ or } x + 160 = 0

Solving these, we find:

\Rightarrow x = 60 \text{ or } x = -160.

Since a cost price cannot be negative, we have:

∴ x = 60.

Hence, C.P. = ₹ 60.

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Question 49

A trader brought a number of articles for ₹ 900, five were damaged and he sold each of the rest at ₹ 2 more than what he paid for it. If on the whole he gains ₹ 80, find the number of articles brought.

Assume the trader initially purchased x articles.

The cost of each article is ₹ \dfrac{900}{x}.

Since 5 articles were damaged, the remaining articles are (x - 5).

The selling price for each remaining article is ₹ \dfrac{900}{x} + 2, which is ₹ 2 more than the cost price.

It’s given that the trader gains ₹ 80 in total.

Thus, the equation for total gain is:

(x - 5)\left(\dfrac{900}{x} + 2\right) - 900 = 80

Simplifying this:

x \times \dfrac{900}{x} + 2x - 5 \times \dfrac{900}{x} - 10 = 980

This simplifies further to:

\dfrac{900x + 2x^2 - 4500 - 10x}{x} = 980

Which leads to:

900x + 2x^2 - 4500 - 10x = 980x

Rearranging terms, we get:

2x^2 + 900x - 980x - 10x - 4500 = 0

This simplifies to:

2x^2 - 90x - 4500 = 0

Divide the entire equation by 2:

x^2 - 45x - 2250 = 0

Factoring the quadratic equation:

x^2 - 75x + 30x - 2250 = 0 x(x - 75) + 30(x - 75) = 0 (x + 30)(x - 75) = 0

The solutions are x = -30 or x = 75.

Since the number of articles cannot be negative, we conclude:

Hence, no. of articles = 75.

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Question 50

1077 boxes of oranges were loaded in three trucks. While unloading them 7, 12 and 8 boxes were found rotten in the trucks respectively. If the number of remaining boxes in the three trucks are in the ratio 4 : 6 : 5, find the number of boxes loaded originally in each truck.

Here, the remaining boxes in the three trucks maintain a ratio of 4 : 6 : 5. To find the original number of boxes, let us assume:

• First truck has 4x remaining boxes
• Second truck has 6x remaining boxes
• Third truck has 5x remaining boxes

Now, consider the total boxes including the rotten ones:

• First truck: 4x + 7
• Second truck: 6x + 12
• Third truck: 5x + 8

The sum of all boxes is given as 1077. Therefore, we set up the equation:

4x + 7 + 6x + 12 + 5x + 8 = 1077

Simplifying, we have:

15x + 27 = 1077

Subtract 27 from both sides:

15x = 1050

Dividing by 15, we find:

x = \dfrac{1050}{15} = 70

Thus, the original number of boxes in each truck were:

• First truck: (4x + 7 = 4(70) + 7 = 280 + 7 = 287)
• Second truck: (6x + 12 = 6(70) + 12 = 420 + 12 = 432)
• Third truck: (5x + 8 = 5(70) + 8 = 350 + 8 = 358)

Hence, the number of boxes originally in each truck were 287, 432, and 358.

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Question 51

If a ≠ b and a : b is the duplicate ratio of (a + c) and (b + c), show that a, c and b are in continued proportion.

We start with the information that the ratio a : b is the duplicate ratio of (a + c) and (b + c).

This tells us that:

\therefore \dfrac{a}{b} = \dfrac{a + c}{b + c} \times \dfrac{a + c}{b + c}

Let’s simplify this expression:

\Rightarrow \dfrac{a}{b} = \dfrac{(a + c)^2}{(b + c)^2}

Expanding the squares gives us:

\Rightarrow \dfrac{a}{b} = \dfrac{a^2 + 2ac + c^2}{b^2 + 2bc + c^2}

Cross-multiplying, we get:

\Rightarrow a(b^2 + 2bc + c^2) = b(a^2 + 2ac + c^2)

Simplifying both sides leads to:

\Rightarrow ab^2 + 2abc + ac^2 = ba^2 + 2abc + bc^2

By subtracting similar terms, we arrive at:

\Rightarrow ac^2 - bc^2 = ba^2 - ab^2

Rearranging gives:

\Rightarrow c^2(a - b) = ab(a - b)

Since a \neq b, we can divide through by (a - b):

\Rightarrow c^2 = ab.

This result, c^2 = ab, confirms that a, c, and b are in continued proportion, as required.

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Question 52

If 16\Big(\dfrac{a - x}{a + x}\Big)^3 = \Big(\dfrac{a + x}{a - x}\Big); show that : a = 3x.

We start with the equation:

16\Big(\dfrac{a - x}{a + x}\Big)^3 = \Big(\dfrac{a + x}{a - x}\Big)

This can be rewritten as:

16\Big(\dfrac{a - x}{a + x}\Big)^3 = \dfrac{1}{\dfrac{a - x}{a + x}}

Introduce t such that t = \Big(\dfrac{a - x}{a + x}\Big). Substituting t in the equation gives:

16t^3 = \dfrac{1}{t}

Multiply both sides by t to clear the fraction:

16t^4 = 1

This implies:

t^4 = \dfrac{1}{16}

Recognizing \dfrac{1}{16} as \Big(\dfrac{1}{2}\Big)^4, we find:

t = \dfrac{1}{2}

Substitute back for t:

\dfrac{a - x}{a + x} = \dfrac{1}{2}

Cross-multiply to solve for a:

2(a - x) = a + x

Simplifying, we have:

2a - 2x = a + x

Rearranging terms gives:

2a - a = x + 2x

Thus, we arrive at:

a = 3x

Hence, proved that a = 3x.

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Question 53

Solve for x, using the properties of proportionality :

\dfrac{1 + x + x^2}{1 - x + x^2} = \dfrac{62(1 + x)}{63(1 - x)}

To solve this equation using the componendo and dividendo method, notice:

\Rightarrow \dfrac{1 + x + x^2 + 1 - x + x^2}{1 + x + x^2 - (1 - x + x^2)} = \dfrac{62(1 + x) + 63(1 - x)}{62(1 + x) - 63(1 - x)}

Simplifying further, we have:

\Rightarrow \dfrac{2 + 2x^2}{1 - 1 + x + x + x^2 - x^2} = \dfrac{62 + 62x + 63 - 63x}{62 + 62x - 63 + 63x}

This reduces to:

\Rightarrow \dfrac{2(1 + x^2)}{2x} = \dfrac{125 - x}{125x - 1}

Dividing both sides by 2, we get:

\Rightarrow \dfrac{1 + x^2}{x} = \dfrac{125 - x}{125x - 1}

Cross-multiplying gives us:

\Rightarrow (125x - 1)(1 + x^2) = x(125 - x)

Expanding both sides, we find:

\Rightarrow 125x + 125x^3 - 1 - x^2 = 125x - x^2

After simplifying, we have:

\Rightarrow 125x - 125x - x^2 + x^2 + 125x^3 - 1 = 0

This simplifies to:

\Rightarrow 125x^3 - 1 = 0

Solving for x^3:

\Rightarrow 125x^3 = 1

Dividing both sides by 125:

\Rightarrow x^3 = \dfrac{1}{125}

Recognizing the cube, we have:

\Rightarrow x^3 = \Big(\dfrac{1}{5}\Big)^3

Thus, taking the cube root:

\Rightarrow x = \dfrac{1}{5}.

Hence, x = \dfrac{1}{5}.

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Question 54

Show that 2x + 7 is a factor of 2x^3 + 7x^2 – 4x – 14. Hence, solve the equation :

2x^3 + 7x^2 – 4x – 14 = 0.

Let’s start by finding the roots of the equation 2x + 7 = 0. Solving this, we have:

⇒ 2x = -7

⇒ x = –\dfrac{7}{2}

Next, let’s substitute x = -\dfrac{7}{2} into the polynomial 2x^3 + 7x^2 – 4x – 14 to verify if it results in zero:

\begin{aligned}2 \times \left(-\dfrac{7}{2}\right)^3 + 7 \times \left(-\dfrac{7}{2}\right)^2 - 4 \times \left(-\dfrac{7}{2}\right) - 14 \\= 2 \times -\dfrac{343}{8} + 7 \times \dfrac{49}{4} + 14 - 14 \\= -\dfrac{343}{4} + \dfrac{343}{4} + 14 - 14 \\= 0.\end{aligned}

Since the left-hand side equals the right-hand side, 2x + 7 is indeed a factor of the given polynomial.

Now, let’s divide 2x^3 + 7x^2 – 4x – 14 by 2x + 7:

\begin{array}{l} \phantom{2x + 7)}{x^2 - 2} \\2x + 7\overline{\smash{\big)}2x^3 + 7x^2 - 4x - 14} \\\phantom{2x + 7}\underline{\underset{-}{}2x^3 \underset{-}{+}7x^2} \\\phantom{{2x + 7}2x^3} \times \phantom{+25^2} - 4x - 14 \\\phantom{{2x + 7}x^3+2+25^2}\underline{\underset{+}{-}4x \underset{+}{-} 14} \\\phantom{{2x + 7}x^3+2-5x^23-3}\times\end{array}

Thus, we can express the polynomial as:

(2x^3 + 7x^2 – 4x – 14 = (2x + 7)(x^2 – 2))

Breaking down further, we get:

((2x + 7)(x + \sqrt{2})(x – \sqrt{2}))

Let’s solve each factor:

For 2x + 7 = 0:

⇒ 2x = -7

⇒ x = –\dfrac{7}{2}

For x + \sqrt{2} = 0:

⇒ x = –\sqrt{2} = -1.41

For x – \sqrt{2} = 0:

⇒ x = \sqrt{2} = 1.41

Hence, x = –\dfrac{7}{2}, 1.41, -1.41

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Question 55(a)

What number should be subtracted from 2x^3 – 5x^2 + 5x + 8 so that the resulting polynomial has a factor 2x – 3 ?

Assume the number we need to subtract is a. This means the new polynomial becomes 2x^3 - 5x^2 + 5x + 8 - a.

According to the factor theorem, a polynomial f(x) will have a factor (x - b) if f(b) = 0.

Here, the factor given is 2x - 3, which implies:

2x - 3 = 0

Solving for x, we get:

x = \dfrac{3}{2}

Substituting x = \dfrac{3}{2} into the polynomial 2x^3 - 5x^2 + 5x + 8 - a, we set the result equal to zero:

\Rightarrow 2 \times \left(\dfrac{3}{2}\right)^3 - 5 \times \left(\dfrac{3}{2}\right)^2 + 5 \times \dfrac{3}{2} + 8 - a = 0

This simplifies to:

\Rightarrow 2 \times \dfrac{27}{8} - 5 \times \dfrac{9}{4} + \dfrac{15}{2} + 8 - a = 0

Continuing with the simplification:

\Rightarrow \dfrac{27}{4} - \dfrac{45}{4} + \dfrac{15 + 16}{2} - a = 0

Further simplifying gives:

\Rightarrow -\dfrac{18}{4} + \dfrac{31}{2} - a = 0

Combine the fractions:

\Rightarrow \dfrac{-18 + 62}{4} - a = 0

Which simplifies to:

\Rightarrow \dfrac{44}{4} - a = 0

Thus:

\Rightarrow 11 - a = 0

Solving for a gives:

\Rightarrow a = 11

Thus, the number to be subtracted is 11.

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Question 55(b)

The expression 4x^3 – bx^2 + x – c leaves remainder 0 and 30 when divided by (x + 1) and (2x – 3) respectively. Calculate the values of b and c.

Utilizing the remainder theorem, we know that when a polynomial p(x) is divided by a linear factor (x - a), the remainder is p(a).

For the divisor x + 1, set x + 1 = 0, which gives x = -1. The problem states that dividing 4x^3 - bx^2 + x - c by (x + 1) yields a remainder of 0.

∴ 4(-1)^3 - b(-1)^2 + (-1) - c = 0

⇒ 4(-1) - b(1) - 1 - c = 0

⇒ -4 - b - 1 - c = 0

This simplifies to b + c = -5, leading to b = -5 - c ………(1)

Next, for the divisor 2x - 3, set 2x - 3 = 0, giving 2x = 3 or x = \dfrac{3}{2}. The polynomial 4x^3 - bx^2 + x - c leaves a remainder of 30 when divided by (2x - 3). Thus, substituting x = \dfrac{3}{2}:

\Rightarrow 4 \times \Big(\dfrac{3}{2}\Big)^3 - b \times \Big(\dfrac{3}{2}\Big)^2 + \Big(\dfrac{3}{2}\Big) - c = 30 \Rightarrow 4 \times \Big(\dfrac{27}{8}\Big) - b \times \Big(\dfrac{9}{4}\Big) + \Big(\dfrac{3}{2}\Big) - c = 30 \Rightarrow \dfrac{27}{2} - \dfrac{9b}{4} + \dfrac{3}{2} - c = 30 \Rightarrow \dfrac{54 - 9b + 6 - 4c}{4} = 30 \Rightarrow 60 - 9b - 4c = 120

Substitute b from equation (1):

⇒ 60 - 9(-5 - c) - 4c = 120

⇒ 60 + 45 + 9c - 4c = 120

⇒ 5c + 105 = 120

⇒ 5c = 120 - 105

⇒ 5c = 15

⇒ c = \dfrac{15}{5} = 3.

Substitute c back into equation (1):

⇒ b = -5 - c = -5 - 3 = -8.

Hence, b = -8 and c = 3.

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Question 56

If for two matrices M and N, N = \begin{bmatrix} 3 & 2 \\ 2 & -1 \end{bmatrix} and product M × N = [-1 4]; find matrix M.

Consider the matrix M with dimensions a × b.

Given that M_{a \times b} \times \begin{bmatrix} 3 & 2 \\ 2 & -1 \end{bmatrix}_{2 \times 2} = \begin{bmatrix} -1 & 4 \end{bmatrix}_{1 \times 2}.

For matrix multiplication to be feasible, the number of columns in the first matrix must match the number of rows in the second matrix.

∴ b = 2.

Additionally, the number of rows in the resulting matrix equals the number of rows in the first matrix.

∴ a = 1.

Thus, the order of matrix M is 1 × 2.

Assume M is \begin{bmatrix} x & y \end{bmatrix}.

This implies:

\Rightarrow \begin{bmatrix} x & y \end{bmatrix} \times \begin{bmatrix} 3 & 2 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 4 \end{bmatrix} \Rightarrow \begin{bmatrix} x \times 3 + y \times 2 & x \times 2 + y \times -1 \end{bmatrix} = \begin{bmatrix} -1 & 4 \end{bmatrix} \Rightarrow \begin{bmatrix} 3x + 2y & 2x - y \end{bmatrix} = \begin{bmatrix} -1 & 4 \end{bmatrix}

From the equality of matrices, we derive:

⇒ 3x + 2y = -1 and 2x – y = 4.

From 2x – y = 4, we find:

⇒ y = 2x – 4.

Substitute the expression for y into 3x + 2y = -1:

⇒ 3x + 2(2x – 4) = -1

⇒ 3x + 4x – 8 = -1

⇒ 7x = -1 + 8

⇒ 7x = 7

⇒ x = 1.

Now, substitute x back to find y:

⇒ y = 2x – 4 = 2(1) – 4 = 2 – 4 = -2.

∴ Matrix M is \begin{bmatrix} x & y \end{bmatrix} = \begin{bmatrix} 1 & -2 \end{bmatrix}.

Hence, M = \begin{bmatrix} 1 & -2 \end{bmatrix}.

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Question 57

If the sum of first 20 terms of an A.P. is same as the sum of its first 28 terms, find the sum of its 48 terms.

Consider the first term of the arithmetic progression (A.P.) to be a and the common difference to be d.

The formula for the sum of the first n terms of an A.P. is:

S_n = \dfrac{n}{2}[2a + (n - 1)d]

We know that:

S_{20} = S_{28}

This implies:
\begin{aligned}&\quad \dfrac{20}{2}[2a + (20 - 1)d] = \dfrac{28}{2}[2a + (28 - 1)d] \\&\Rightarrow 10[2a + 19d] = 14[2a + 27d] \\&\Rightarrow 20a + 190d = 28a + 378d \\&\Rightarrow 28a - 20a = 190d - 378d \\&\Rightarrow 8a = -188d \\&\Rightarrow a = -\dfrac{188}{8}d \quad \text{........(1)}\end{aligned}

Now, for the sum of 48 terms:

\begin{aligned}S_{48} &= \dfrac{48}{2}[2a + (48 - 1)d] \\&= 24 \times 2 \times -\dfrac{188}{8}d + 47d\begin{aligned} \\&= 24 \times \left\end{aligned}-\dfrac{376}{8}d + 47d\right\begin{aligned} \\&= 24 \times [-47d + 47d] \\&= 24 \times 0 \\&= 0.\end{aligned}\end{aligned}

Thus, the sum of the 48 terms, S_{48}, is 0.

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Question 58

If a, b, c are in A.P., show that : (b + c), (c + a) and (a + b) are also in A.P.

We start with the information that a, b, and c form an arithmetic progression (A.P.).

∴ b - a = c - b.

This implies b + b = a + c, or 2b = a + c.

Now, consider the sequence (b + c), (c + a), and (a + b). To show these are in A.P., we need to verify:

(a + b) - (c + a) = (c + a) - (b + c)

Simplifying the left side:

a + b - c - a = b - c.

And the right side:

c + a - b - c = a - b.

This gives us b - c = a - b.

Revisiting the equation b + b = a + c, we have 2b = a + c, which we already established.

Since 2b = a + c holds true, it confirms that (b + c), (c + a), and (a + b) are indeed in A.P.

Hence, proved that (b + c), (c + a), and (a + b) are also in A.P.

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Question 59

If a, b, c are in G.P; a, x, b are in A.P. and b, y, c are also in A.P.

Prove that : \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{2}{b}.

We’re given that a, x, b form an arithmetic progression (A.P.).

∴ b - x = x - a

⇒ 2x = a + b

⇒ x = \dfrac{a + b}{2} ………(1)

Similarly, for the sequence b, y, c in A.P., we have:

∴ c - y = y - b

⇒ 2y = b + c

⇒ y = \dfrac{b + c}{2} ………..(2)

Also, since a, b, c are in geometric progression (G.P.), we know:

⇒ \dfrac{b}{a} = \dfrac{c}{b}

⇒ b^2 = ac ……….(3)

Our goal is to show that:

\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{2}{b}

Substitute the expressions for x and y from equations (1) and (2) into the left-hand side:

\begin{aligned}\Rightarrow \dfrac{1}{\dfrac{a + b}{2}} + \dfrac{1}{\dfrac{b + c}{2}} \\\Rightarrow \dfrac{2}{a + b} + \dfrac{2}{b + c} \\\Rightarrow \dfrac{2(b + c) + 2(a + b)}{(a + b)(b + c)} \\\Rightarrow \dfrac{2b + 2c + 2a + 2b}{ab + ac + b^2 + bc} \\\Rightarrow \dfrac{2a + 2c + 4b}{ab + b^2 + b^2 + bc} \space [\text{From} (3)] \\\Rightarrow \dfrac{2(a + c + 2b)}{ab + 2b^2 + bc} \\\Rightarrow \dfrac{2(a + c + 2b)}{b(a + 2b + c)} \\\Rightarrow \dfrac{2}{b}.\end{aligned}

Therefore, it is proved that \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{2}{b}.

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Question 60

Evaluate : 9 + 99 + 999 + …….. upto n terms.

Consider the series:

∴ 9 + 99 + 999 + …….. up to n terms.

Notice that each term can be expressed as a power of 10 minus 1:

∴ (10 – 1) + (100 – 1) + (1000 – 1) + ……… up to n terms.

This can be split into two separate series:

∴ (10 + 100 + 1000 + ……. up to n terms) + (-1 + -1 + -1 + ……… up to n terms).

Observe that the first series, 10 + 10^2 + 10^3 + ……. up to n terms, forms a geometric progression (G.P.) where the first term (a) is 10 and the common ratio (r) is also 10.

Using the formula for the sum of a G.P.,

Sum of G.P. = \dfrac{a(r^n - 1)}{r - 1}.

Applying this formula, we have:

\begin{aligned}\Rightarrow \dfrac{10(10^n - 1)}{10 - 1} + (-n) \\\Rightarrow \dfrac{10(10^n - 1)}{9} - n.\end{aligned}

Thus, the sum of the series 9 + 99 + 999 + …….. up to n terms is \dfrac{10(10^n - 1)}{9} - n.

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Question 61

Find the point on the y-axis whose distances from the points (3, 2) and (-1, 1.5) are in the ratio 2 : 1.

Consider that the x-coordinate for any point on the y-axis is 0. Therefore, let the point be represented as (0, y).

Using the distance formula:

D = \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}

We know that the distances from (0, y) to the points (3, 2) and (-1, 1.5) must be in the ratio 2:1.

Thus, we can write:

\Rightarrow \dfrac{\sqrt{(2 - y)^2 + (3 - 0)^2}}{\sqrt{(1.5 - y)^2 + (-1 - 0)^2}} = \dfrac{2}{1} \Rightarrow \dfrac{\sqrt{4 + y^2 - 4y + 9}}{\sqrt{2.25 + y^2 - 3y + 1}} = \dfrac{2}{1} \Rightarrow \dfrac{\sqrt{y^2 - 4y + 13}}{\sqrt{y^2 - 3y + 3.25}} = \dfrac{2}{1}

Squaring both sides of the equation, we have:

\Rightarrow \dfrac{y^2 - 4y + 13}{y^2 - 3y + 3.25} = \dfrac{4}{1}

Simplifying further:

\Rightarrow y^2 - 4y + 13 = 4(y^2 - 3y + 3.25) \Rightarrow y^2 - 4y + 13 = 4y^2 - 12y + 13 \Rightarrow 4y^2 - y^2 - 12y + 4y + 13 - 13 = 0 \Rightarrow 3y^2 - 8y = 0

Factoring out y gives:

\Rightarrow y(3y - 8) = 0

This implies:

\Rightarrow y = 0 \text{ or } 3y - 8 = 0

Solving for y:

\Rightarrow y = 0 \text{ or } 3y = 8 \Rightarrow y = 0 \text{ or } y = \dfrac{8}{3} = 2\dfrac{2}{3}.

Given that the point must lie on the y-axis, y cannot be 0.

Thus, the point is (0, y) = \Big(0, 2\dfrac{2}{3}\Big).

Hence, required point = \Big(0, 2\dfrac{2}{3}\Big).

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Question 62

In what ratio does the point P(a, 2) divide the line segment joining the points A(5, -3) and B(-9, 4) ? Also, find the value of ‘a’.

Assume the point P divides the line segment in the ratio k : 1.

Using the section formula, the coordinates of the dividing point ((x, y)) are given by:

(x, y) = \left(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\right)

For the y-coordinate, substitute the values:

\begin{aligned}\Rightarrow 2 = \dfrac{k \times 4 + 1 \times (-3)}{k + 1} \\\Rightarrow 2(k + 1) = 4k - 3 \\\Rightarrow 2k + 2 = 4k - 3 \\\Rightarrow 4k - 2k = 2 + 3 \\\Rightarrow 2k = 5 \\\Rightarrow k = \dfrac{5}{2}\end{aligned}

Thus, the ratio k : 1 becomes \dfrac{5}{2} : 1, which simplifies to 5 : 2.

So, the ratio m_1 : m_2 is 5 : 2.

Now, for the x-coordinate, substitute the values:

\begin{aligned}\Rightarrow a = \dfrac{5 \times (-9) + 2 \times 5}{5 + 2} \\\Rightarrow a = \dfrac{-45 + 10}{7} \\\Rightarrow a = \dfrac{-35}{7} = -5.\end{aligned}

Hence, a = -5 and the ratio is 5 : 2.

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Question 63

A straight line makes on the coordinate axes positive intercepts whose sum is 5. If the line passes through the point P(-3, 4), find its equation.

We know that a straight line makes positive intercepts on the coordinate axes, and their sum is 5.

Assume the intercept on the x-axis is a and on the y-axis is b.

Thus, a + b = 5.

From this, we find a = 5 - b ……….(1)

The equation of the line in intercept form is given by: \dfrac{x}{a} + \dfrac{y}{b} = 1.

Since the line passes through the point P(-3, 4), substitute these values:

\begin{aligned}\Rightarrow \dfrac{-3}{a} + \dfrac{4}{b} = 1 \\\Rightarrow \dfrac{-3b + 4a}{ab} = 1 \\\Rightarrow \dfrac{4a - 3b}{ab} = 1 \\\Rightarrow 4a - 3b = ab\end{aligned}.

Now, replace a using equation (1):

\begin{aligned}\Rightarrow 4(5 - b) - 3b = (5 - b)b \\\Rightarrow 20 - 4b - 3b = 5b - b^2 \\\Rightarrow 20 - 7b = 5b - b^2 \\\Rightarrow b^2 - 7b - 5b + 20 = 0 \\\Rightarrow b^2 - 12b + 20 = 0 \\\Rightarrow b^2 - 10b - 2b + 20 = 0 \\\Rightarrow b(b - 10) - 2(b - 10) = 0 \\\Rightarrow (b - 2)(b - 10) = 0 \\\Rightarrow b = 2 \text{ or } b = 10\end{aligned}.

Plug these values of b back into equation (1):

For b = 2,

a = 5 - b = 5 - 2 = 3.

For b = 10,

a = 5 - b = 5 - 10 = -5.

Since both intercepts must be positive, discard a = -5.

∴ a = 3 and b = 2.

Substitute a and b back into the intercept form equation:

\begin{aligned}\Rightarrow \dfrac{x}{3} + \dfrac{y}{2} = 1 \\\Rightarrow \dfrac{2x + 3y}{6} = 1 \\\Rightarrow 2x + 3y = 6\end{aligned}.

Therefore, the equation of the line is 2x + 3y = 6.

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Question 64

The line 3x – 4y + 12 = 0 meets x-axis at point A and y-axis at point B. Find :

(i) the coordinates of A and B.

(ii) equation of perpendicular bisector of line segment AB.

(i) To find the coordinates of point A, where the line intersects the x-axis, we set the y-coordinate to 0. Thus, let A be at (a, 0).

Substituting into the equation 3x – 4y + 12 = 0, we have:

⇒ 3a – 4(0) + 12 = 0

⇒ 3a + 12 = 0

⇒ 3a = -12

⇒ a = -\dfrac{12}{3} = -4.

Thus, the coordinates for A are (-4, 0).

For point B, where the line intersects the y-axis, the x-coordinate is 0. So, let B be at (0, b).

Substitute into the equation:

⇒ 3(0) – 4b + 12 = 0

⇒ -4b + 12 = 0

⇒ 4b = 12

⇒ b = \dfrac{12}{4} = 3.

Thus, the coordinates for B are (0, 3).

Therefore, points A = (-4, 0) and B = (0, 3).

(ii) Now, let’s determine the equation of the perpendicular bisector of line segment AB.

First, calculate the midpoint of AB using the formula:

Mid-point = \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)

For AB, this becomes:

Mid-point = \Big(\dfrac{-4 + 0}{2}, \dfrac{0 + 3}{2}\Big) = \Big(\dfrac{-4}{2}, \dfrac{3}{2}\Big) = (-2, 1.5).

Next, find the slope of AB using:

Slope = \dfrac{y_2 - y_1}{x_2 - x_1}

For AB, the slope is:

Slope = \dfrac{3 - 0}{0 - (-4)} = \dfrac{3}{4}.

Since the product of the slopes of perpendicular lines is -1, the slope of the perpendicular bisector is:

⇒ \dfrac{3}{4} \times Slope of perpendicular line = -1

⇒ Slope of perpendicular line (m) = -\dfrac{4}{3}.

Using the point-slope form to find the equation of the perpendicular bisector:

⇒ y – y~1 = m(x – x~1)

⇒ y – 1.5 = -\dfrac{4}{3}[x – (-2)]

⇒ 3(y – 1.5) = -4[x + 2]

⇒ 3y – 4.5 = -4x – 8

⇒ 4x + 3y – 4.5 + 8 = 0

⇒ 4x + 3y + 3.5 = 0

⇒ 4x + 3y + \dfrac{35}{10} = 0

⇒ 4x + 3y + \dfrac{7}{2} = 0

⇒ \dfrac{8x + 6y + 7}{2} = 0

⇒ 8x + 6y + 7 = 0.

Thus, the equation of the perpendicular bisector of AB is 8x + 6y + 7 = 0.

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Question 65

In a triangle PQR, L and M are two points on the base QR, such that ∠LPQ = ∠QRP and ∠RPM = ∠RQP. Prove that :

(i) △PQL ~ △RPM

(ii) QL × RM = PL × PM

(iii) PQ^2 = QR × QL

(i)

Consider the given triangle PQR with points L and M on base QR. Notice that ∠QRP is equal to ∠MRP and ∠LPQ matches ∠MRP. Similarly, ∠RQP equals ∠LQP, and ∠LQP is the same as ∠RPM.

In triangles △PQL and △RPM, we have:

∠LPQ = ∠MRP (as shown above)

∠LQP = ∠RPM (as shown above)

∴ By the Angle-Angle (A.A.) criterion, △PQL is similar to △RPM.

Hence, proved that △PQL ~ △RPM.

(ii) Given that the triangles are similar, the corresponding sides are proportional.

∴ \dfrac{QL}{PM} = \dfrac{PL}{RM}

⇒ QL × RM = PL × PM

Hence, proved that QL × RM = PL × PM.

(iii) Consider triangles △LPQ and △PQR:

∠Q is common to both triangles.

∠QPL = ∠PRQ (as given)

∴ By the A.A. criterion, △LPQ is similar to △PQR.

The sides of similar triangles are proportional:

∴ \dfrac{PQ}{QR} = \dfrac{QL}{PQ}

⇒ PQ^2 = QL × QR

Hence, proved that PQ^2 = QL × QR.

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Question 66

In a rectangle ABCD, its diagonal AC = 15 cm and ∠ACD = α. If cot α = \dfrac{3}{2}, find the perimeter and the area of the rectangle.

We know that \cot \alpha = \dfrac{3}{2}.

Using the identity:

\cosec^2 \alpha = 1 + \cot^2 \alpha

we substitute \cot \alpha = \dfrac{3}{2}:

\cosec^2 \alpha = 1 + \left(\dfrac{3}{2}\right)^2 \cosec^2 \alpha = 1 + \dfrac{9}{4} \cosec^2 \alpha = \dfrac{4 + 9}{4} \cosec^2 \alpha = \dfrac{13}{4}

Thus, \cosec \alpha = \sqrt{\dfrac{13}{4}} = \dfrac{\sqrt{13}}{2}.

Now, using the definition of cosecant:

\cosec \alpha = \dfrac{\text{Hypotenuse}}{\text{Perpendicular}} \cosec \alpha = \dfrac{AC}{AD}

Substitute the known values:

\dfrac{\sqrt{13}}{2} = \dfrac{15}{AD}

Solving for AD:

AD = \dfrac{15 \times 2}{\sqrt{13}} = \dfrac{30}{\sqrt{13}}

Next, using the cotangent definition:

\cot \alpha = \dfrac{\text{Base}}{\text{Perpendicular}} \cot \alpha = \dfrac{CD}{AD}

Substitute the known values:

\dfrac{3}{2} = \dfrac{CD}{\dfrac{30}{\sqrt{13}}}

Solving for CD:

CD = \dfrac{3}{2} \times \dfrac{30}{\sqrt{13}} = \dfrac{45}{\sqrt{13}}

To find the perimeter:

\text{Perimeter} = 2(\text{length} + \text{breadth}) = 2(CD + AD) = 2\left(\dfrac{45}{\sqrt{13}} + \dfrac{30}{\sqrt{13}}\right) = 2\left(\dfrac{75}{\sqrt{13}}\right) = \dfrac{150}{\sqrt{13}}

Rationalising gives:

= \dfrac{150}{\sqrt{13}} \times \dfrac{\sqrt{13}}{\sqrt{13}} = \dfrac{150\sqrt{13}}{13} \text{ cm}

For the area:

\text{Area} = \text{length} \times \text{breadth} = CD \times AD = \dfrac{45}{\sqrt{13}} \times \dfrac{30}{\sqrt{13}} = \dfrac{1350}{13} = 103\dfrac{11}{13} \text{ cm}^2

Hence, perimeter = \dfrac{150\sqrt{13}}{13} cm and area = 103\dfrac{11}{13} cm^2.

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Question 67

In the given figure, AB is a diameter of the circle. Chords AC and AD produced meet the tangent to the circle at point B in points P and Q respectively. Prove that :

AB^2 = AC × AP

Connect point B to point C.

We have the property that the diameter of a circle creates a right angle at any point on the circle.

∴ ∠ACB = 90°

Additionally, a tangent to a circle forms a right angle with the radius at the point of contact.

∴ ∠ABP = 90°

Consider triangles △ACB and △ABP:

• Both have a right angle: ∠ACB = ∠ABP = 90°
• They share the angle at A: ∠A = ∠A

∴ △ACB is similar to △ABP by the Angle-Angle (A.A.) similarity criterion.

For similar triangles, the ratios of corresponding sides are equal.

Therefore, \dfrac{AB}{AP} = \dfrac{AC}{AB}.

⇒ AB^2 = AC × AP

Thus, it is proven that AB^2 = AC × AP.

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Question 68

Use ruler and compasses for this question.

(i) Construct an isosceles triangle ABC in which AB = AC = 7.5 cm and BC = 6 cm.

(ii) Draw AD, the perpendicular from vertex A to side BC.

(iii) Draw a circle with center A and radius 2.8 cm, cutting AD at E.

(iv) Construct another circle to circumscribe the triangle BCE.

Steps for the construction:

1. Begin by drawing line segment BC with a length of 6 cm.
2. Using a compass, set the radius to 7.5 cm. With B as the center, draw an arc. Repeat the same with C as the center, ensuring the arcs intersect at point A.
3. Connect points A to B and A to C to form triangle ABC.
4. Construct the perpendicular from A to line BC, labeling the intersection point on BC as D.
5. Draw a circle centered at A with a radius of 2.8 cm. This circle should intersect line AD at point E.
6. Connect E to B and E to C. Find the perpendicular bisector of line segment EB, and let it intersect AD at point O.
7. Finally, with O as the center and using the distance OE, OB, or OC as the radius, draw a circle. This circle should pass through points B and C and be tangent to the circle centered at A at point E.

Hence, above is the required circle.

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Question 69

In triangle ABC, ∠BAC = 90°, AB = 6 cm and BC = 10 cm. A circle is drawn inside the triangle which touches all the sides of the triangle (i.e. an incircle of △ABC is drawn). Find the area of the triangle excluding the circle.

Consider the right-angled triangle △ABC with ∠BAC = 90°, where AB = 6 cm and BC = 10 cm.

To find AC, apply the Pythagorean theorem:

⇒ BC² = AB² + AC²
⇒ 10² = 6² + AC²
⇒ 100 = 36 + AC²
⇒ AC² = 64
⇒ AC = \sqrt{64} = 8 cm.

The circle inside △ABC is an incircle, which means it touches all sides. The radius r of this circle is perpendicular to each side at the point of contact, so DF ⊥ BC, DG ⊥ AC, and DH ⊥ AB.

Thus, DF = DG = DH = r.

The area of a triangle is given by:

Area = \dfrac{1}{2} \times \text{base} \times \text{height}.

Here, the area of △ABC can be divided into three smaller triangles:

Area of △ABC = Area of △ADC + Area of △CDB + Area of △ADB.

⇒ \dfrac{1}{2} \times AC \times AB = \dfrac{1}{2} \times DG \times AC + \dfrac{1}{2} \times DF \times BC + \dfrac{1}{2} \times DH \times AB

⇒ \dfrac{1}{2} \times 8 \times 6 = \dfrac{1}{2} \times r \times 8 + \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 6

⇒ ( \dfrac{1}{2} \times 48 = \dfrac{1}{2}(8r + 10r + 6r) )

⇒ 48 = 8r + 10r + 6r

⇒ 24r = 48

⇒ r = \dfrac{48}{24} = 2 \text{ cm}.

To find the area of the triangle excluding the circle, subtract the area of the circle from the area of the triangle:

Area of triangle (excluding circle) = Area of triangle – Area of circle

= \dfrac{1}{2} \times AC \times AB – πr²

= ( \dfrac{1}{2} \times 8 \times 6 – \dfrac{22}{7} \times (2)^2 )

= 24 – \dfrac{88}{7}

= \dfrac{168 - 88}{7}

= \dfrac{80}{7}

= 11\dfrac{3}{7} \text{ cm}^2.

Hence, area of triangle excluding circle = 11\dfrac{3}{7} \text{ cm}^2.

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Question 70

A conical vessel of radius 6 cm and height 8 cm is completely filled with water. A sphere is lowered into the water and its size is such that when it touches the sides it is just immersed . What fraction of water overflows ?

Let’s consider the problem where we have a conical vessel filled with water. The radius of this cone, denoted as R, is 6 cm, and its height, denoted as H, is 8 cm.

Now, let’s introduce a sphere with radius r. When this sphere is lowered into the cone, it touches the sides and is just submerged. Therefore, we have PC = PD = r cm.

We know from geometry that the lengths of tangents drawn from an external point to a circle are equal. Thus, AC = AD = 6 cm. Since the radius at the point of tangency is perpendicular to the tangent, triangles \triangle OCA and \triangle OPD are right-angled.

First, consider \triangle OCA. Applying the Pythagorean theorem:

\begin{aligned}\Rightarrow OA^2 = OC^2 + AC^2 \\\Rightarrow OA = \sqrt{OC^2 + AC^2} \\\Rightarrow OA = \sqrt{8^2 + 6^2} \\\Rightarrow OA = \sqrt{64 + 36} \\\Rightarrow OA = \sqrt{100} = 10 \text{ cm}.\end{aligned}

Next, in \triangle OPD, again using the Pythagorean theorem:

\Rightarrow OP^2 = OD^2 + PD^2 \quad \text{...(1)}

From the figure, we know:
– OD = OA - AD = 10 - 6 = 4 cm.
– OP = OC - PC = 8 - r

Substituting OP, OD, and PD into equation (1), we have:

\begin{aligned}\Rightarrow (8 - r)^2 = 4^2 + r^2 \\\Rightarrow 64 + r^2 - 16r = 16 + r^2 \\\Rightarrow r^2 - r^2 + 64 - 16 - 16r = 0 \\\Rightarrow 16r = 48 \\\Rightarrow r = \dfrac{48}{16} = 3 \text{ cm}.\end{aligned}

The volume of water displaced by the sphere is equal to the volume of the sphere. Thus:

\text{Volume of water overflown} = \text{Volume of sphere} = \dfrac{4}{3} \pi r^3 = \dfrac{4}{3} \pi (3)^3 = \dfrac{4}{3} \pi \times 27 = 36 \pi \text{ cm}^3.

The initial volume of water in the cone is given by:

\text{Original volume of water} = \text{Volume of cone} = \dfrac{1}{3} \pi R^2 H = \dfrac{1}{3} \pi \times 6^2 \times 8 = 96 \pi \text{ cm}^3.

Thus, the fraction of water that overflows is:

\dfrac{\text{Volume of water overflown}}{\text{Original volume of water}} = \dfrac{36 \pi}{96 \pi} = \dfrac{3}{8}.

Hence, fraction of water overflown = \dfrac{3}{8}.

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Question 71(i)

Prove that :

\dfrac{\text{1 + cot A}}{\text{cos A}} + \dfrac{\text{1 + tan A}}{\text{sin A}} = 2(sec A + cosec A)

We need to demonstrate that the expression on the left-hand side equals the expression on the right-hand side.

Left-Hand Side (LHS):

Start with:

\dfrac{1 + \cot A}{\cos A} + \dfrac{1 + \tan A}{\sin A}

Combine into a single fraction:

\Rightarrow \dfrac{\sin A(1 + \cot A) + \cos A(1 + \tan A)}{\cos A \sin A}

Expand each term:

\Rightarrow \dfrac{\sin A + \sin A \cot A + \cos A + \cos A \tan A}{\cos A \sin A}

Substitute \cot A = \dfrac{\cos A}{\sin A} and \tan A = \dfrac{\sin A}{\cos A}:

\Rightarrow \dfrac{\sin A + \sin A \times \dfrac{\cos A}{\sin A} + \cos A + \cos A \times \dfrac{\sin A}{\cos A}}{\cos A \sin A}

Simplify the expression:

\Rightarrow \dfrac{\sin A + \cos A + \cos A + \sin A}{\cos A \sin A}

Combine like terms:

\Rightarrow \dfrac{2(\sin A + \cos A)}{\sin A \cos A}

Right-Hand Side (RHS):

Start with:

2(\sec A + \csc A)

Express in terms of sine and cosine:

\Rightarrow 2\left(\dfrac{1}{\cos A} + \dfrac{1}{\sin A}\right)

Combine into a single fraction:

\Rightarrow 2\left(\dfrac{\sin A + \cos A}{\sin A \cos A}\right)

Simplify:

\Rightarrow \dfrac{2(\sin A + \cos A)}{\sin A \cos A}

Since both LHS and RHS simplify to:

\dfrac{2(\sin A + \cos A)}{\sin A \cos A}

∴ The given expression is proven true: (\dfrac{1 + \cot A}{\cos A} + \dfrac{1 + \tan A}{\sin A} = 2(\sec A + \csc A)).

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Question 71(ii)

Prove that :

\sqrt{\dfrac{\text{1+ sin A}}{\text{1 - sin A}}} - \sqrt{\dfrac{\text{1 - sin A}}{\text{1 + sin A}}} = 2 tan A

We need to show that:

\sqrt{\dfrac{\text{1+ sin A}}{\text{1 - sin A}}} - \sqrt{\dfrac{\text{1 - sin A}}{\text{1 + sin A}}} equals 2 \tan A.

Let’s work on the left-hand side:

\Rightarrow \sqrt{\dfrac{\text{1 + sin A}}{\text{1 - sin A}} \times \dfrac{\text{1 + sin A}}{\text{1 + sin A}}} - \sqrt{\dfrac{\text{1 - sin A}}{\text{1 + sin A}} \times \dfrac{\text{1 - sin A}}{\text{1 - sin A}}} \Rightarrow \sqrt{\dfrac{(\text{1 + sin A})^2}{\text{1 - sin}^2 A}} - \sqrt{\dfrac{(\text{1 - sin A})^2}{\text{1 - sin}^2 A}}

Recall the identity:

1 - \sin^2 A = \cos^2 A

Thus, we have:

\Rightarrow \sqrt{\dfrac{(\text{1 + sin A})^2}{\text{cos}^2 A}} - \sqrt{\dfrac{(\text{1 - sin A})^2}{\text{cos}^2 A}}

This simplifies to:

\Rightarrow \dfrac{\text{1 + sin A}}{\text{cos A}} - \dfrac{\text{1 - sin A}}{\text{cos A}}

Now, combining the fractions:

\Rightarrow \dfrac{\text{1 + sin A - (1 - sin A)}}{\text{cos A}}

Simplify further:

\Rightarrow \dfrac{1 - 1 + \text{sin A + sin A}}{\text{cos A}} \Rightarrow \dfrac{\text{2 sin A}}{\text{cos A}}

\Rightarrow \text{2 tan A}.

∴ The left-hand side equals the right-hand side.

Hence, proved that \sqrt{\dfrac{\text{1+ sin A}}{\text{1 - sin A}}} - \sqrt{\dfrac{\text{1 - sin A}}{\text{1 + sin A}}} = 2 tan A.

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Question 72

Solve for x ∈ W, 0° ≤ x ≤ 90°.

(i) 3 tan^2 2x = 1

(ii) tan^2 x = 3(sec x – 1)

(i) We start with the equation:

∴ 3 \tan^2 2x = 1

Dividing both sides by 3, we have:

∴ \tan^2 2x = \dfrac{1}{3}

Taking the square root of both sides:

∴ \tan 2x = \sqrt{\dfrac{1}{3}}

Which simplifies to:

∴ \tan 2x = \dfrac{1}{\sqrt{3}}

Recognizing this as the tangent of 30°:

∴ \tan 2x = \tan 30°

Thus, we have:

∴ 2x = 30°

Solving for x gives:

∴ x = \dfrac{30°}{2} = 15°.

Hence, x = 15°.

(ii) For the second equation:

∴ \tan^2 x = 3(\sec x – 1)

Recall that \tan^2 x = \sec^2 x – 1, so:

∴ \sec^2 x – 1 = 3\sec x – 3

Rearranging terms, we get:

∴ \sec^2 x – 3\sec x – 1 + 3 = 0

Simplifying further:

∴ \sec^2 x – 3\sec x + 2 = 0

Factorizing the quadratic equation:

∴ \sec^2 x – 2\sec x – \sec x + 2 = 0

Group and factor:

∴ \sec x(\sec x – 2) – 1(\sec x – 2) = 0

This gives:

∴ (\sec x – 1)(\sec x – 2) = 0

Setting each factor to zero:

∴ \sec x – 1 = 0 \text{ or } \sec x – 2 = 0

Solving these, we find:

∴ \sec x = 1 \text{ or } \sec x = 2

Recognizing these as:

∴ \sec x = \sec 0° \text{ or } \sec x = \sec 60°

Thus, x can be:

∴ x = 0° \text{ or } x = 60°

Hence, x = 0° or 60°.

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Question 73

The angle of elevation of the top of a tower as observed from a point on the ground is ‘α’ and on moving a metre towards the tower, the angle of elevation is ‘β’.

Prove that the height of the tower is : \dfrac{\text{a tan α tan β}}{\text{tan β - tan α}}

Consider a tower CO with a height of h meters.

To find the height, analyze the triangles involved:

In \triangle AOC:

∴ \tan \alpha = \dfrac{\text{Perpendicular}}{\text{Base}}

∴ \tan \alpha = \dfrac{OC}{AO}

∴ \tan \alpha = \dfrac{h}{a + x}

∴ (a + x) \tan \alpha = h

∴ a \tan \alpha + x \tan \alpha = h

∴ x \tan \alpha = h - a \tan \alpha

∴ x = \dfrac{\text{h - a tan α}}{\text{tan α}} ……..(1)

In \triangle BOC:

∴ \tan \beta = \dfrac{\text{Perpendicular}}{\text{Base}}

∴ \tan \beta = \dfrac{OC}{BO}

∴ \tan \beta = \dfrac{h}{x}

∴ x = \dfrac{h}{\text{tan β}} …………(2)

Equating the expressions for x from equations (1) and (2):

\Rightarrow \dfrac{h}{\text{tan β}} = \dfrac{\text{h - a tan α}}{\text{tan α}} \Rightarrow \dfrac{h}{\text{tan β}} =\dfrac{h}{\text{tan α}} - \dfrac{\text{a tan α}}{\text{tan α}} \Rightarrow \dfrac{h}{\text{tan β}} - \dfrac{h}{\text{tan α}} = -a \Rightarrow \dfrac{\text{h tan α - h tan β}}{\text{tan α tan β}} = -a \Rightarrow \dfrac{\text{h (tan α - tan β)}}{\text{tan α tan β}} = -a \Rightarrow h = \dfrac{-\text{a tan α tan β}}{\text{(tan α - tan β)}} \Rightarrow h = \dfrac{\text{a tan α tan β}}{\text{(tan β - tan α)}}

Thus, we have shown that h = \dfrac{\text{a tan α tan β}}{\text{(tan β - tan α)}}. This confirms the height of the tower.

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Question 74

The mean of the following frequency distribution is 50, but the frequencies f~1 and f~2 in class 20-40 and 60-80 respectively are not known. Find these frequencies.

Given that the sum of frequencies is 120.

Let’s organize the data into a table to find the unknown frequencies f_1 and f_2:

\begin{array}{|c|c|c|c|}\hline\text{Class} & \text{Frequencies (f)} & \text{Class marks (x)} & fx \\\hline0-20 & 17 & 10 & 170 \\20-40 & f_1 & 30 & 30f_1 \\40-60 & 32 & 50 & 1600 \\60-80 & f_2 & 70 & 70f_2 \\80-100 & 19 & 90 & 1710 \\\hline\text{Total} & \Sigma f = f_1 + f_2 + 68 & & \Sigma fx = 3480 + 30f_1 + 70f_2 \\\hline\end{array}

We’re told the total frequency \Sigma f = 120. Thus:

f_1 + f_2 + 68 = 120

Simplifying, we find:

f_1 + f_2 = 120 - 68 = 52

This gives us:

f_1 = 52 - f_2 \quad \text{...(1)}

The formula for the mean is:

\text{Mean} = \dfrac{\Sigma fx}{\Sigma f}

Substituting the given values:

50 = \dfrac{3480 + 30f_1 + 70f_2}{120}

Multiplying through by 120, we have:

6000 = 3480 + 30f_1 + 70f_2

Using equation (1) to substitute for f_1:

6000 = 3480 + 30(52 - f_2) + 70f_2

Simplifying further:

6000 = 3480 + 1560 - 30f_2 + 70f_2 6000 = 5040 + 40f_2

Solving for f_2:

40f_2 = 6000 - 5040 40f_2 = 960 f_2 = \dfrac{960}{40} = 24

Now, substituting back to find f_1:

f_1 = 52 - f_2 = 52 - 24 = 28

Therefore, f_1 = 28 and f_2 = 24.

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Question 75

A card is drawn at random from a well-shuffled deck of 52 playing cards. Find the probability that it is :

(i) an ace

(ii) a jack of hearts

(iii) a three of clubs or a six of diamonds

(iv) a heart

(v) any suit except heart

(vi) a ten or a spade

(vii) neither a four nor a club

(viii) a picture card

(ix) a spade or a picture card.

The deck contains a total of 52 cards.

∴ Total possible outcomes = 52.

(i) There are 4 aces in the deck, one from each suit.

∴ Favourable outcomes = 4.

The probability of drawing an ace is given by:

P(\text{drawing an ace}) = \dfrac{\text{Favourable outcomes}}{\text{Total outcomes}} = \dfrac{4}{52} = \dfrac{1}{13}.

Hence, probability of drawing an ace = \dfrac{1}{13}.

(ii) Only one card is the jack of hearts.

∴ Favourable outcomes = 1.

The probability of drawing the jack of hearts is:

P(\text{drawing a jack of hearts}) = \dfrac{1}{52}.

Hence, probability of drawing a jack of hearts = \dfrac{1}{52}.

(iii) There is one three of clubs and one six of diamonds.

∴ Favourable outcomes = 2.

The probability of drawing either a three of clubs or a six of diamonds is:

P(\text{drawing a three of clubs or a six of diamonds}) = \dfrac{2}{52} = \dfrac{1}{26}.

Hence, probability of drawing a three of clubs or a six of diamonds = \dfrac{1}{26}.

(iv) The deck has 13 hearts.

∴ Favourable outcomes = 13.

The probability of drawing a heart is:

P(\text{drawing a heart}) = \dfrac{13}{52} = \dfrac{1}{4}.

Hence, probability of drawing a heart = \dfrac{1}{4}.

(v) There are 39 cards that are not hearts.

∴ Favourable outcomes = 39.

The probability of drawing a card from any suit except hearts is:

P(\text{drawing any suit except heart}) = \dfrac{39}{52} = \dfrac{3}{4}.

Hence, probability of drawing any suit except heart = \dfrac{3}{4}.

(vi) There are 13 spades and 3 other 10’s (one from each suit except spades).

∴ Favourable outcomes = 16.

The probability of drawing a ten or a spade is:

P(\text{drawing a ten or a spade}) = \dfrac{16}{52} = \dfrac{4}{13}.

Hence, probability of drawing a ten or a spade = \dfrac{4}{13}.

(vii) The deck has 13 clubs and 3 other fours (one from each suit except clubs).

Total cards that are either a club or a four = 16.

Remaining cards = 52 – 16 = 36.

∴ Favourable outcomes = 36.

The probability of drawing neither a club nor a four is:

P(\text{drawing neither a club nor 4}) = \dfrac{36}{52} = \dfrac{9}{13}.

Hence, probability of drawing neither a club nor 4 = \dfrac{9}{13}.

(viii) Picture cards include jacks, queens, and kings.

There are 4 jacks, 4 queens, and 4 kings in the deck.

∴ Favourable outcomes = 12.

The probability of drawing a picture card is:

P(\text{drawing a picture card}) = \dfrac{12}{52} = \dfrac{3}{13}.

Hence, probability of drawing a picture card = \dfrac{3}{13}.

(ix) There are 13 spade cards.

The deck has 3 jacks, 3 queens, and 3 kings that are not spades.

Total cards that are either spades or picture cards = 13 + 3 + 3 + 3 = 22.

∴ Favourable outcomes = 22.

The probability of drawing a spade or a picture card is:

P(\text{drawing a spade or a picture card}) = \dfrac{22}{52} = \dfrac{11}{26}.

Hence, probability of drawing a spade or a picture card = \dfrac{11}{26}.

Frequently Asked Questions

What topics are covered in the Class 10 Maths Mixed Practice chapter?

This chapter is a comprehensive revision exercise. It primarily covers concepts from Commercial Mathematics, including Goods and Services Tax (GST), Banking (Recurring Deposits), and Shares and Dividends. It tests your ability to apply formulas and solve problems from these interconnected topics.

How should I use these Mixed Practice solutions for board exam preparation?

First, attempt to solve the problems on your own to test your understanding. Then, use these solutions to verify your answers and, more importantly, to learn the correct step-by-step method expected by the ICSE board. This practice is crucial for improving speed and accuracy for your final exams.

Are questions from this Mixed Practice chapter important for the ICSE Class 10 board exams?

Yes, absolutely. The Mixed Practice chapter is designed to simulate the types of integrated problems that often appear in Section B of the board exam paper. Mastering these questions helps build confidence and ensures you are well-prepared to tackle complex questions that combine multiple concepts from commercial mathematics.

How many questions are there in this chapter?

The Mixed Practice chapter in the Concise Selina Mathematics textbook for Class 10 contains a total of 81 questions. These are divided into two sections, SET A and SET B, to provide extensive practice for students.

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• More Class 10 ICSE Solutions →