ICSE Class 10 Maths Mixed Practice Selina Solutions
ICSE Class 10 Maths Mixed Practice solutions at a glance
ICSE Class 10 Maths Mixed Practice in Selina Concise Mathematics Chapter 27 brings together questions from Commercial Mathematics, algebra and ratio-based problem solving, so the main task is to identify the topic first and then apply the correct formula. This page gives worked solutions for Concise Mathematics Selina Solutions Class 10 ICSE Chapter 27 Mixed Practice, with each numerical answer shown step by step in MathJax format.
The mixed-practice format is different from a single-topic exercise. One question may use GST, the next may use recurring deposit interest, shares and dividends, a compound inequation, a quadratic equation or time and work. In timed practice, first classify the question, write the formula, substitute values carefully and then state the answer with the correct unit.
For connected revision, you may also practise ICSE Class 10 Maths MCQ solutions, ICSE Class 10 Maths short-answer competency questions and ICSE Class 10 Maths graph-based long-answer practice after finishing this chapter.
Formula and method reference for Mixed Practice
Use this table before the solutions. In Mixed Practice, marks are often lost not because the formula is hard, but because the student uses the right formula for the wrong topic.
| Question type | Formula or method | What to check |
|---|---|---|
| Discount and GST | Taxable value after discount: \text{value}=\text{marked price}\times\frac{100-\text{discount rate}}{100}. GST: \text{GST}=\frac{\text{rate}}{100}\times\text{taxable value}. | Inter-state supply uses IGST. Intra-state supply is split as CGST and SGST, but the total rate remains the same. |
| Recurring deposit | I=P\times\frac{n(n+1)}{2\times12}\times\frac{r}{100}, where P is the monthly instalment, n is the number of months and r is the annual rate. | Convert years to months before substituting n. |
| Maturity value | M=P n+I. | Do not forget to add the total deposits P n to the interest. |
| Shares and dividends | \text{Number of shares}=\frac{\text{investment}}{\text{market value per share}}. Dividend is calculated on nominal value, not market value. | Discount and premium affect market value only. Dividend rate is always on nominal value. |
| Compound inequation | Solve both sides together or separately, then intersect the ranges. | When multiplying or dividing by a negative number, reverse the inequality sign. |
| Quadratic equation | x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} for ax^2+bx+c=0. | Reject values that make a denominator zero. |
| Time and work | If a person completes work in x days, one-day work is \frac{1}{x}. | Add rates of work, not the number of days. |
Concept snapshot: choosing the right method
Think of Mixed Practice as a set of labelled drawers. GST questions ask for taxable value and tax; recurring deposit questions ask for monthly instalments, time or maturity; shares questions ask for market value, nominal value and dividend; algebra questions ask for solving a range or equation. Before writing the first line, name the drawer. That one decision usually tells you which formula to use.
Concise Mathematics Selina Solutions Class 10 ICSE Chapter 27 Mixed Practice
The solutions below are written in school-style steps. In your notebook, keep the formula line visible, substitute values in the next line and state the final answer with units such as rupees, days or the solution set.
Question 1: Find the total GST bill for goods supplied from Calcutta to Banaras
A dealer in Calcutta supplied the following goods or services to another dealer in Banaras. Find the total amount of the bill.
| Cost in Calcutta | Discount | GST rate |
|---|---|---|
| \text{₹ }1050 | Not given | 18\% |
| \text{₹ }1750 | Not given | 18\% |
| \text{₹ }1100 | 25\% | 18\% |
| \text{₹ }1425 | 30\% | 18\% |
| \text{₹ }1725 | 40\% | 18\% |
Step 1: Since Calcutta and Banaras are in different states, the supply is inter-state. Therefore IGST is applied at 18\%.
Step 2: Find the taxable value of each item after discount.
\text{Taxable values}=1050,\ 1750,\ 1100\times\frac{75}{100},\ 1425\times\frac{70}{100},\ 1725\times\frac{60}{100}
=1050,\ 1750,\ 825,\ 997.50,\ 1035
Step 3: Add the taxable values.
\text{Total taxable value}=1050+1750+825+997.50+1035=5657.50
Step 4: Calculate IGST at 18\%.
\text{IGST}=\frac{18}{100}\times5657.50=1018.35
Step 5: Add IGST to the taxable value.
\text{Bill amount}=5657.50+1018.35=6675.85
Final answer: The total amount of the bill is \text{₹ }6675.85.
Question 2: Cost of an article under GST from Mathura to Ratlam to Indore
Some goods or services are supplied for \text{₹ }20000 from Mathura in Uttar Pradesh to Ratlam in Madhya Pradesh and then from Ratlam to Indore in Madhya Pradesh. At each stage, the GST rate is 12\%. If the profit made by the dealer in Ratlam is \text{₹ }3750, find the cost of the article in Indore under GST.
Step 1: The supply from Mathura to Ratlam is inter-state, so IGST is charged on \text{₹ }20000.
\text{IGST}=\frac{12}{100}\times20000=2400
Step 2: The dealer in Ratlam sells after adding a profit of \text{₹ }3750.
\text{Selling price before GST}=20000+3750=23750
Step 3: Ratlam to Indore is intra-state within Madhya Pradesh. The total GST rate is still 12\%, split as CGST 6\% and SGST 6\%.
\text{CGST}=\frac{6}{100}\times23750=1425
\text{SGST}=\frac{6}{100}\times23750=1425
\text{Total GST}=1425+1425=2850
Step 4: Add GST to the Ratlam dealer’s selling price.
\text{Cost in Indore}=23750+2850=26600
Final answer: The cost of the article in Indore is \text{₹ }26600.
Question 3: Recurring deposit instalment and maturity amount
Mr. Kumar has a recurring deposit account in a bank for 4 years at 10\% per annum. If he gets \text{₹ }21560 as interest at maturity, find the monthly instalment and the maturity amount.
Step 1: Convert the time into months.
n=4\times12=48
Step 2: Let the monthly instalment be \text{₹ }P. Use the recurring deposit interest formula.
I=P\times\frac{n(n+1)}{2\times12}\times\frac{r}{100}
Step 3: Substitute I=21560, n=48 and r=10.
21560=P\times\frac{48(49)}{24}\times\frac{10}{100}
21560=P\times98\times\frac{1}{10}=9.8P
P=\frac{21560}{9.8}=2200
Step 4: Find the maturity amount.
\text{Maturity value}=P n+I=2200\times48+21560
=105600+21560=127160
Final answer: The monthly instalment is \text{₹ }2200, and the maturity amount is \text{₹ }127160.
Question 4: Time for which a cumulative deposit account was held
The maturity value of a cumulative deposit account is \text{₹ }120400. Each monthly instalment is \text{₹ }1600 and the rate of interest is 10\% per year. Find the time for which the account was held.
Step 1: Let the time be n months.
Step 2: Use the maturity value formula.
M=P n+P\times\frac{n(n+1)}{2\times12}\times\frac{r}{100}
Step 3: Substitute M=120400, P=1600 and r=10.
120400=1600n+1600\times\frac{n(n+1)}{24}\times\frac{10}{100}
Step 4: Divide by 1600.
\frac{120400}{1600}=n+\frac{n(n+1)}{240}
\frac{301}{4}=n+\frac{n(n+1)}{240}
Step 5: Multiply by 240 and simplify.
18060=240n+n(n+1)
n^2+241n-18060=0
Step 6: Factorise the quadratic.
n^2+301n-60n-18060=0
n(n+301)-60(n+301)=0
(n-60)(n+301)=0
Step 7: Since time cannot be negative, take n=60.
\text{Time}=\frac{60}{12}=5\text{ years}
Final answer: The account was held for 5 years.
Question 5: Shares, dividend, gain and second investment
Rajat invested \text{₹ }24000 in 7\% hundred-rupee shares at 20\% discount. After one year, he sold these shares at \text{₹ }75 each and invested the proceeds, including the first year’s dividend, in 18\% twenty-five-rupee shares at 64\% premium. Find his gain or loss after one year, his annual income from the second investment and the percentage increase in return on his original investment.
Step 1: Find the market value of each first share.
\text{Market value}=100-\frac{20}{100}\times100=80
Step 2: Find the number of first shares.
\text{Number of shares}=\frac{24000}{80}=300
Step 3: Dividend is calculated on nominal value.
\text{Dividend per share}=\frac{7}{100}\times100=7
\text{First year's dividend}=300\times7=2100
Step 4: Find the money received after selling the shares and adding dividend.
\text{Sale proceeds}=300\times75=22500
\text{Total amount available}=22500+2100=24600
Step 5: Compare this with the original investment.
\text{Gain}=24600-24000=600
Step 6: Find the market value of each second share.
\text{Market value}=25+\frac{64}{100}\times25=25+16=41
Step 7: Find the number of second shares.
\text{Number of shares}=\frac{24600}{41}=600
Step 8: Find the annual income from the second investment.
\text{Dividend per share}=\frac{18}{100}\times25=4.50
\text{Annual income}=600\times4.50=2700
Step 9: Find the increase in annual return on the original investment.
\text{Increase in annual income}=2700-2100=600
\text{Increase in return on original investment}=\frac{600}{24000}\times100=2.5\%
Final answer: Rajat gains \text{₹ }600 after one year; the annual income from the second investment is \text{₹ }2700; the increase in return on the original investment is 2.5\%.
Question 6: Number of shares sold when annual income changes
A man sold some \text{₹ }20 shares paying 8\% dividend at 10\% discount and invested the proceeds in \text{₹ }10 shares paying 12\% dividend at 50\% premium. If the change in his annual income is \text{₹ }600, find the number of shares sold.
Step 1: Let the number of old shares sold be x.
Step 2: Find the old market value and old annual income per share.
\text{Old market value}=20-\frac{10}{100}\times20=18
\text{Old dividend per share}=\frac{8}{100}\times20=1.6
\text{Old income}=1.6x
Step 3: The proceeds from selling x old shares are 18x.
Step 4: Find the new market value and new dividend per share.
\text{New market value}=10+\frac{50}{100}\times10=15
\text{Number of new shares}=\frac{18x}{15}
\text{New dividend per share}=\frac{12}{100}\times10=1.2
Step 5: Find the new income.
\text{New income}=1.2\times\frac{18x}{15}=1.44x
Step 6: The income decreases by \text{₹ }600, because 1.6x is greater than 1.44x.
1.6x-1.44x=600
0.16x=600
x=\frac{600}{0.16}=3750
Final answer: The number of shares sold is 3750.
Question 7: Solve a compound inequation in whole numbers
Find the value of x which satisfies -2\le\frac{1}{2}-\frac{2x}{3}\le1\frac{5}{6}, where x\in W. Also represent the solution on a number line.
Step 1: Convert the mixed number.
1\frac{5}{6}=\frac{11}{6}
Step 2: Subtract \frac{1}{2} from all three parts.
-2-\frac{1}{2}\le-\frac{2x}{3}\le\frac{11}{6}-\frac{1}{2}
-\frac{5}{2}\le-\frac{2x}{3}\le\frac{4}{3}
Step 3: Multiply all parts by -\frac{3}{2}. Since this is negative, reverse the inequality signs.
\frac{15}{4}\ge x\ge -2
-2\le x\le\frac{15}{4}
Step 4: Since x\in W, take only whole numbers in this interval.
x\in\{0,1,2,3\}
Final answer: x=\{0,1,2,3\}. On the number line, mark filled points at 0, 1, 2 and 3.
Question 8: Solve the equation using the quadratic formula
Solve \frac{x}{x+1}+\frac{x+1}{x}=2\frac{4}{15} using the formula.
Step 1: Note the restrictions: x\ne0 and x\ne-1.
Step 2: Convert the mixed number.
2\frac{4}{15}=\frac{34}{15}
Step 3: Combine the fractions.
\frac{x}{x+1}+\frac{x+1}{x}=\frac{x^2+(x+1)^2}{x(x+1)}
\frac{x^2+(x+1)^2}{x(x+1)}=\frac{34}{15}
Step 4: Cross-multiply and simplify.
15\{x^2+(x+1)^2\}=34x(x+1)
15\{x^2+x^2+2x+1\}=34x^2+34x
30x^2+30x+15=34x^2+34x
4x^2+4x-15=0
Step 5: Apply the quadratic formula with a=4, b=4 and c=-15.
x=\frac{-4\pm\sqrt{4^2-4(4)(-15)}}{2(4)}
x=\frac{-4\pm\sqrt{16+240}}{8}=\frac{-4\pm16}{8}
x=\frac{12}{8}\ \text{or}\ x=\frac{-20}{8}
x=\frac{3}{2}\ \text{or}\ x=-\frac{5}{2}
Final answer: x=\frac{3}{2} or x=-\frac{5}{2}. Both values satisfy the denominator restrictions.
Question 9: Cost price when gain percent equals cost price
By selling an article for \text{₹ }24, a man gains as much percent as its cost price. Find the cost price of the article.
Step 1: Let the cost price be \text{₹ }x.
Step 2: Profit is selling price minus cost price.
\text{Profit}=24-x
Step 3: The profit percent is equal to x.
x=\frac{24-x}{x}\times100
Step 4: Form and solve the quadratic equation.
x^2=100(24-x)
x^2+100x-2400=0
x^2+120x-20x-2400=0
x(x+120)-20(x+120)=0
(x-20)(x+120)=0
Step 5: Cost price cannot be negative, so reject x=-120.
Final answer: The cost price of the article is \text{₹ }20.
Question 10: Time and work problem
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, find how many days B alone will take.
Step 1: Let B alone complete the work in x days. Then A alone takes x+16 days.
Step 2: Write their one-day work rates.
\text{B's one-day work}=\frac{1}{x},\qquad \text{A's one-day work}=\frac{1}{x+16}
Step 3: Together they finish the work in 15 days.
\frac{1}{x}+\frac{1}{x+16}=\frac{1}{15}
Step 4: Solve the equation.
\frac{x+16+x}{x(x+16)}=\frac{1}{15}
15(2x+16)=x(x+16)
30x+240=x^2+16x
x^2-14x-240=0
x^2-24x+10x-240=0
x(x-24)+10(x-24)=0
(x-24)(x+10)=0
Step 5: Time cannot be negative, so x=24.
Final answer: B alone can complete the work in 24 days.
Question 11: Divide an amount in a given ratio
Divide \text{₹ }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.
Step 1: Let the common value be x.
\frac{1}{2}\text{ of first part}=x,\quad \frac{1}{3}\text{ of second part}=x,\quad \frac{1}{6}\text{ of third part}=x
Step 2: Express each part in terms of x.
\text{First part}=2x,\quad \text{second part}=3x,\quad \text{third part}=6x
Step 3: Add the three parts.
2x+3x+6x=1870
11x=1870
x=170
Step 4: Find each part.
2x=340,\quad 3x=510,\quad 6x=1020
Final answer: The three parts are \text{₹ }340, \text{₹ }510 and \text{₹ }1020.
Question 12: Prove that four quantities are in proportion
If a+c=be and \frac{1}{b}+\frac{1}{d}=\frac{e}{c}, prove that a, b, c and d are in proportion. Assume the denominators used are non-zero.
Step 1: From a+c=be, divide both sides by b.
e=\frac{a+c}{b}=\frac{a}{b}+\frac{c}{b} \qquad \text{...(1)}
Step 2: From \frac{1}{b}+\frac{1}{d}=\frac{e}{c}, multiply each term by c.
\frac{c}{b}+\frac{c}{d}=e \qquad \text{...(2)}
Step 3: Equate the two expressions for e.
\frac{a}{b}+\frac{c}{b}=\frac{c}{b}+\frac{c}{d}
Step 4: Subtract \frac{c}{b} from both sides.
\frac{a}{b}=\frac{c}{d}
Step 5: Therefore a:b=c:d.
Hence proved: a, b, c and d are in proportion.
Examiner’s mindset for Mixed Practice answers
In ICSE Class 10 Maths, mixed exercises test method selection as much as calculation. A clear answer usually earns credit through these visible parts: the correct formula, proper substitution, correct simplification and a final answer with unit or set notation.
- In GST questions, write whether the supply is inter-state or intra-state before calculating tax.
- In shares and dividends, mention that dividend is calculated on nominal value. This prevents a common wrong-method error.
- In recurring deposit questions, show n in months. Do not substitute years directly in the RD formula.
- In inequations, show the sign reversal when multiplying by a negative number.
- In quadratic equations with fractions, state excluded values such as x\ne0 before accepting roots.
Common mistakes students make in Mixed Practice
| Mistake | Correction |
|---|---|
| Calculating share dividend on market value. | Use nominal value for dividend. Market value is used only to find the number of shares bought or sold. |
| Using 4 instead of 48 for a 4-year recurring deposit. | Convert years into months first: n=4\times12=48. |
| Forgetting to reverse the sign in -\frac{2x}{3} inequations. | When multiplying or dividing by a negative number, reverse \le to \ge and \ge to \le. |
| Adding days in time and work problems. | Add work rates such as \frac{1}{x} and \frac{1}{x+16}, not the number of days. |
| Writing only the numerical value without unit. | End with \text{₹}, days, years, percent or set notation as required. |
Quick answer index
| Question | Answer |
|---|---|
| 1 | \text{₹ }6675.85 |
| 2 | \text{₹ }26600 |
| 3(i) | Monthly instalment =\text{₹ }2200 |
| 3(ii) | Maturity amount =\text{₹ }127160 |
| 4 | 5 years |
| 5(i) | Gain =\text{₹ }600 |
| 5(ii) | Annual income =\text{₹ }2700 |
| 5(iii) | Increase in return on original investment =2.5\% |
| 6 | 3750 shares |
| 7 | x=\{0,1,2,3\} |
| 8 | x=\frac{3}{2} or x=-\frac{5}{2} |
| 9 | Cost price =\text{₹ }20 |
| 10 | B alone takes 24 days |
| 11 | \text{₹ }340,\ \text{₹ }510,\ \text{₹ }1020 |
| 12 | \frac{a}{b}=\frac{c}{d}, so a:b=c:d |
Sources and syllabus alignment
This page is aligned with the standard ICSE Class 10 Maths treatment of Commercial Mathematics, algebra and ratio-type applications. For official syllabus framing, refer to the CISCE official website. For overlapping algebraic methods such as quadratic equations and inequations, the NCERT mathematics resources are useful supporting references, while the exercise scope follows Selina Concise Mathematics Class 10 Chapter 27 Mixed Practice.
Frequently Asked Questions
What is covered in ICSE Class 10 Maths Selina Chapter 27 Mixed Practice?
Selina Chapter 27 Mixed Practice brings together questions from different parts of ICSE Class 10 Maths, especially GST, recurring deposits, shares and dividends, inequations, quadratic equations, ratio and time and work. The purpose is to test whether you can choose the correct method without being told the chapter name.
How should I identify the formula in a Mixed Practice question?
Read the key words first. Words such as GST, IGST, CGST and SGST point to tax calculation; monthly instalment and maturity point to recurring deposit; nominal value, market value and dividend point to shares; and phrases such as together complete the work point to time and work.
Why is dividend calculated on nominal value and not market value?
In ICSE Class 10 Maths shares problems, the dividend rate is stated on the nominal value of the share. Market value decides how many shares are bought or sold, but the income per share is \frac{\text{dividend rate}}{100}\times\text{nominal value}.
What is the most common error in recurring deposit questions?
The most common error is using the number of years directly as n. In the recurring deposit formula I=P\times\frac{n(n+1)}{2\times12}\times\frac{r}{100}, n means the number of months, so 4 years must be written as 48 months.
Do I need to draw the number line for every inequation answer?
Draw the number line when the question asks for it. For x\in W, list only whole-number solutions and show them as filled points if the end values are included by the inequality.