ICSE Class 10 Maths Chapter 2 Short Answer Solutions
ICSE Class 10 Maths Short Answer Questions 1: What this page covers
ICSE Class 10 Maths short answer questions test whether you can choose the correct formula, apply it in steps, and state the final answer clearly. This page gives step-by-step solutions for Chapter 2 competency-focused Short Answer Questions 1, covering GST, recurring deposits, shares, inequations, algebra, matrices, AP, GP, reflection, section formula and straight lines.
Students searching for ICSE Maths Solved Competency Focused Questions Solutions Class 10 Chapter 2 Short Answer Questions 1 should focus on method, not memorised answers. In each solution below, the formula, substitution and simplification are shown separately.
Formula and method reference
| Topic | Formula or rule | Use |
|---|---|---|
| GST | \text{Final price}=\text{taxable value}\left(1+\frac{\text{GST rate}}{100}\right) | Use when price includes GST. |
| Recurring deposit | I=\frac{P\times n(n+1)\times r}{2\times12\times100} | Use when equal monthly deposits are made. |
| Shares | \text{Dividend}=\frac{\text{rate}}{100}\times\text{nominal value} | Dividend is on nominal value, not market value. |
| AP | T_n=a+(n-1)d, S_n=\frac{n}{2}\{2a+(n-1)d\} | Use for constant difference. |
| GP | S_n=\frac{a(r^n-1)}{r-1}, r\ne1 | Use for constant ratio. |
| Section formula | P=\left(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}\right) | Use when P divides AB internally in m:n. |
Concept snapshot
A competency question is like a two-step lock. The first key is the formula. The second key is the condition in the wording: including GST, return percent, x\in W, invariant point, or equally inclined. If you miss the condition, the answer may look correct but the method will be incomplete.
Original worked examples
Worked example 1: GST inclusive price
Step 1: A product marked at \text{₹ }1000 is sold for \text{₹ }944 including 18\% GST after discount. Let the taxable value after discount be x.
x+\frac{18}{100}x=944
\frac{118}{100}x=944
x=800
Step 2: Discount =1000-800=200.
\text{Discount percent}=\frac{200}{1000}\times100=20\%
Final answer: The discount is 20\%.
Worked example 2: Recurring deposit interest
Step 1: Let P=500, n=12, and r=8.
I=\frac{500\times12\times13\times8}{2\times12\times100}
I=260
Final answer: The interest is \text{₹ }260.
Worked example 3: AP of common multiples
Step 1: Common multiples of 3 and 7 are multiples of 21. Taking the first positive term, the AP is 21,42,63,\ldots.
T_8=21+(8-1)21=168
S_5=\frac{5}{2}\{2(21)+4(21)\}=315
Final answer: T_8=168 and S_5=315.
Step-by-step solutions for Short Answer Questions 1
Question 50: GST discount on pressure cooker
Step 1: Marked price =\text{₹ }1800, GST =12\%, final price =\text{₹ }1792. Let the discounted price before GST be x.
x+\frac{12}{100}x=1792
\frac{112}{100}x=1792
x=1600
Step 2: Discount =1800-1600=200.
\text{Discount percent}=\frac{200}{1800}\times100=\frac{100}{9}\%=11\frac{1}{9}\%
Final answer: The discount is 11\frac{1}{9}\%, approximately 11.11\%.
Question 51: Recurring deposit rate
Step 1: Let the monthly deposit be \text{₹ }x. Time =24 months and interest =\text{₹ }x.
x=\frac{x\times24\times25\times r}{2\times12\times100}
x=\frac{25xr}{100}
Step 2: Since x\ne0, divide by x.
1=\frac{r}{4}
r=4
Final answer: The rate is 4\% per annum.
Question 52: Market value of shares
Step 1: Dividend on one share of nominal value \text{₹ }50 at 14\% is
\frac{14}{100}\times50=7
Step 2: Let market value be M. Return is 10\%.
\frac{7}{M}\times100=10
M=70
Final answer: Market value of each share is \text{₹ }70.
Question 53: Compound inequation
Step 1: Solve \frac{1}{2}(2x-1)\le2x+\frac{1}{2}\le5\frac{1}{2}+x.
x-\frac{1}{2}\le2x+\frac{1}{2}\Rightarrow x\ge-1
2x+\frac{1}{2}\le\frac{11}{2}+x\Rightarrow x\le5
-1\le x\le5
Final answer: If x\in R, minimum =-1, maximum =5. If x\in W, minimum =0, maximum =5.
Question 54: Equation with x\ne0
Step 1: Put t=\frac{1}{x}.
\frac{5}{x}+4\sqrt{3}=\frac{2\sqrt{3}}{x^2}\Rightarrow5t+4\sqrt{3}=2\sqrt{3}t^2
2\sqrt{3}t^2-5t-4\sqrt{3}=0
2\sqrt{3}t^2-8t+3t-4\sqrt{3}=0
(2t+\sqrt{3})(\sqrt{3}t-4)=0
t=-\frac{\sqrt{3}}{2}\quad\text{or}\quad t=\frac{4}{\sqrt{3}}
Final answer: x=-\dfrac{2}{\sqrt{3}} or x=\dfrac{\sqrt{3}}{4}.
Question 55: GST percentage equal to marked price number
Step 1: Marked price =\text{₹ }x, GST rate =x\%, final price =\text{₹ }24.
x+\frac{x}{100}\times x=24
x^2+100x-2400=0
(x-20)(x+120)=0
Step 2: Reject x=-120 because price cannot be negative.
Final answer: x=20. Marked price is \text{₹ }20, and GST rate is 20\%.
Question 56: Mean proportion and third proportion
Step 1: Let the numbers be a and b. Since 6 is the mean proportional,
\frac{a}{6}=\frac{6}{b}\Rightarrow ab=36\Rightarrow a=\frac{36}{b}
Step 2: Since 48 is the third proportional,
\frac{a}{b}=\frac{b}{48}\Rightarrow b^2=48a
b^2=48\times\frac{36}{b}\Rightarrow b^3=1728\Rightarrow b=12
a=\frac{36}{12}=3
Final answer: The two numbers are 3 and 12.
Question 57: Factor theorem check
Step 1: Let f(x)=2x^3+3x^2-3x-2.
f(-2)=2(-2)^3+3(-2)^2-3(-2)-2=0
f(1)=2(1)^3+3(1)^2-3(1)-2=0
f(2)=2(2)^3+3(2)^2-3(2)-2=20\ne0
Step 2: So x+2 and x-1 are factors, but x-2 is not a factor.
(x+2)(x-1)=x^2+x-2
(x^2+x-2)(2x+1)=2x^3+3x^2-3x-2
Final answer: Pamela’s factorisation is incorrect. Correct factorisation is (x+2)(x-1)(2x+1).
Question 58: Matrix value
Step 1: Given M=\frac{1}{2}A-2B+5I.
\frac{1}{2}A=\begin{bmatrix}-3&0\\2&1\end{bmatrix},\quad2B=\begin{bmatrix}2&0\\2&6\end{bmatrix},\quad5I=\begin{bmatrix}5&0\\0&5\end{bmatrix}
M=\begin{bmatrix}-3&0\\2&1\end{bmatrix}-\begin{bmatrix}2&0\\2&6\end{bmatrix}+\begin{bmatrix}5&0\\0&5\end{bmatrix}
M=\begin{bmatrix}0&0\\0&0\end{bmatrix}
Final answer: M=\begin{bmatrix}0&0\\0&0\end{bmatrix}.
Question 59: AP of numbers divisible by 3 and 7
Step 1: The common multiples of 3 and 7 are multiples of 21. Taking the first positive common multiple, the AP is 21,42,63,\ldots.
a=21,\quad d=21
T_n=21+(n-1)21=21n
Answer to (a): T_n=21n, assuming the first term is 21.
Answer to (b): The two-digit terms are 21,42,63,84; hence there are 4.
S_{10}=\frac{10}{2}\{2(21)+9(21)\}=5(231)=1155
Answer to (c): S_{10}=1155.
Edge case: If 0 is explicitly included as the first whole-number term, the AP becomes 0,21,42,\ldots, so T_n=21(n-1).
Question 60: Sequence (\sqrt{3})^n
Step 1: First five terms are
\sqrt{3},\;3,\;3\sqrt{3},\;9,\;9\sqrt{3}
\frac{3}{\sqrt{3}}=\sqrt{3}
Answer to (a): It is a GP with common ratio \sqrt{3}.
S_{10}=\frac{\sqrt{3}\{(\sqrt{3})^{10}-1\}}{\sqrt{3}-1}=\frac{\sqrt{3}(243-1)}{\sqrt{3}-1}
S_{10}=\frac{242\sqrt{3}(\sqrt{3}+1)}{3-1}=121(3+\sqrt{3})
Final answer: p=121.
Question 61: Reflection of triangle coordinates
Step 1: From the question, take A(4,5), B(0,3), and C(3,0).
Step 2: Reflection in the origin maps (x,y) to (-x,-y).
A(4,5)\mapsto A'(-4,-5)
B(0,3)\mapsto B'(0,-3)
C(3,0)\mapsto C'(-3,0)
Answer to (a): A'(-4,-5), B'(0,-3), C'(-3,0).
Answer to (b): C(3,0) is invariant on reflection in the x-axis, and B(0,3) is invariant on reflection in the y-axis.
Question 62: Ratio in which a line divides a segment
Step 1: For A(-3,9) and B(4,2), slope of AB is
\frac{2-9}{4-(-3)}=-1
y-9=-1(x+3)\Rightarrow x+y=6
Step 2: Intersect with y=2+3x.
x+(2+3x)=6\Rightarrow x=1
y=2+3(1)=5
Step 3: Let P(1,5) divide AB in m:n.
1=\frac{4m-3n}{m+n}\Rightarrow m+n=4m-3n\Rightarrow 3m=4n
Final answer: The ratio is 4:3.
Question 63: Slope of diagonal and equation of another diagonal
Step 1: Since A(-3,-1) and P(-2,-2) lie on AC,
\text{slope of }AC=\frac{-2-(-1)}{-2-(-3)}=-1
Step 2: Diagonals of a square are perpendicular.
(\text{slope of }AC)(\text{slope of }BD)=-1
(-1)(\text{slope of }BD)=-1\Rightarrow \text{slope of }BD=1
Answer to (a): Slope of BD=1.
Step 3: Equation of AC through P(-2,-2) with slope -1:
y+2=-(x+2)
x+y+4=0
Answer to (b): Equation of AC is x+y+4=0.
Quick answer index
| Question | Answer |
|---|---|
| 50 | 11\frac{1}{9}\% |
| 51 | 4\% per annum |
| 52 | \text{₹ }70 |
| 53 | For R: -1 to 5; for W: 0 to 5 |
| 54 | -\frac{2}{\sqrt{3}}, \frac{\sqrt{3}}{4} |
| 55 | x=20 |
| 56 | 3, 12 |
| 57 | (x+2)(x-1)(2x+1) |
| 58 | Zero matrix |
| 59 | T_n=21n, S_{10}=1155 |
| 60 | GP; p=121 |
| 61 | A'(-4,-5), B'(0,-3), C'(-3,0) |
| 62 | 4:3 |
| 63 | Slope 1; x+y+4=0 |
Examiner’s mindset
In ICSE Class 10 Maths short answers, the final number alone is not enough. A good answer shows the formula, the substituted values, the simplification and the final statement. Conditions such as x\ne0, x\in W, including GST, and invariant point should be written in the solution because they affect the answer.
Common mistakes students make
- GST base error: GST is calculated after discount, so use x+\frac{12}{100}x=1792 in Question 50.
- RD formula error: Do not omit the monthly conversion factor 12 in the denominator.
- Shares error: Dividend is on nominal value; return percent is on market value.
- Whole-number AP edge case: If 0 is included, T_n changes from 21n to 21(n-1).
- Coordinate sign error: Check the quadrant and axis condition before finalising reflected points or slopes.
Related ICSE Class 10 Maths study pages
- ICSE Class 10 solutions and study material
- ICSE Class 10 Maths solutions index
- ratio and proportion solutions for ICSE Class 10 Maths
- quadratic equations solutions for ICSE Class 10 Maths
Sources and syllabus alignment
This page follows the standard ICSE Class 10 Mathematics treatment of commercial mathematics, algebra, matrices, sequences and coordinate geometry. For syllabus confirmation, use the CISCE official website. For overlapping school mathematics concepts, the NCERT official website is also a suitable reference.
Frequently Asked Questions
Are these ICSE Class 10 Maths short answer solutions enough for competency practice?
They are useful for practising method-based short answers, but students should also solve the full textbook exercise and CISCE-style papers because competency questions can combine two topics in one problem.
How should I write a three-mark ICSE Maths solution?
Write the formula or rule first, substitute the values, show simplification, and then state the final answer with the correct unit or ratio.
Why is zero treated carefully in the AP question on numbers divisible by 3 and 7?
Many school AP solutions take the first positive common multiple as 21. If 0 is explicitly included, the AP begins 0,21,42,\ldots, so the nth term and sum change.
What is the quickest way to check factor theorem answers?
For a factor such as x-a, substitute x=a in the polynomial. If the value is zero, x-a is a factor.
How do I avoid coordinate geometry mistakes in short answers?
Find the slope, point or ratio first, then check its sign against the quadrant, axis or diagram condition. Most errors happen because one coordinate sign is copied incorrectly.