ICSE Class 10 Mathematics 2022 Semester 1 Question Paper with Solutions

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Last Updated on November 13, 2026 by sanjjeett

ICSE Class 10 Mathematics PYQ

Practice the complete ICSE Class 10 Mathematics 2022 Semester 1 Question Paper with Solutions. This 40-mark objective paper includes algebra, commercial mathematics, matrices, arithmetic progression, similarity, recurring deposits and linear inequations with expandable explanations for focused board-exam revision.

Year
2022 Semester 1
Maximum Marks
40
Time Allowed
One hour and a half
📄
Paper Instructions
ICSE Class 10 Mathematics 2022 Semester 1

All questions are compulsory.

The marks intended for questions are given in brackets [ ].

Select the correct option for each of the following questions.

✓
Solved Objective Paper

ICSE Mathematics 2022 Semester 1 Questions with Solutions

40 marks • All questions are compulsory.
40
Marks
1
Question 1
ICSE Class 10 Mathematics 2022 Semester 1
1

If \((\boldsymbol{x}+\mathbf{2})\) is a factor of the polynomial \(\boldsymbol{x}^{\mathbf{3}}-\boldsymbol{k} \boldsymbol{x}^{\mathbf{2}}-\mathbf{5} \boldsymbol{x}\) + 6 then the value of \(\boldsymbol{k}\) is [1]

A1
B2
C3
D-2
Answer & Solution

Ans. Option (b) is correct.

Explanation:

\[ \begin{aligned} x+2 & =0 \\ x & =-2 \\ p(x)=x^{3}-k x^{2}-5 x+6 & =0 \end{aligned} \]

Remainder \(=p(-2)=0\)

\[ \begin{aligned} (-2)^{3}-k(-2)^{2}-5(-2)+6 & =0 \\ -8-4 k+10+6 & =0 \\ -4 k & =-8 \\ k & =2 \end{aligned} \]
2
Question 2
ICSE Class 10 Mathematics 2022 Semester 1
2

The solution set of the inequation \(\boldsymbol{x}-\mathbf{3} \geq-\mathbf{5}, \boldsymbol{x} \in \boldsymbol{R}\) is [1]

A\(\{x: x>-2, x \in R\}\)
B\(\{x: x \leq-2, \mathrm{x} \in R\}\)
C\(\{x: x \geq-2, x \in R\}\)
D\{-2, -1, 0, 1, 2\}
Answer & Solution

Ans. Option (c) is correct.

Explanation: \(x-3 \geq-5\)

\[ \begin{aligned} & x \geq-5+3 \\ & x \geq-2 \end{aligned} \]
3
Question 3
ICSE Class 10 Mathematics 2022 Semester 1
3

The product \(\boldsymbol{A} \boldsymbol{B}\) of two matrices \(\boldsymbol{A}\) and \(\boldsymbol{B}\) is possible if [1]

A\(A\) and \(B\) have the same number of rows.
Bthe number of columns of \(A\) is equal to the number of rows of \(B\).
Cthe number of rows of \(A\) is equal to the number of columns of \(B\).
D\(A\) and \(B\) have the same number of columns.
Answer & Solution

Ans. Option (b) is correct.

Explanation: The product of \(A B\) of two matrices \(A\) and \(B\) is possible if the number of column in \(A\) is equal to the number of rows in \(B\).

4
Question 4
ICSE Class 10 Mathematics 2022 Semester 1
4

If 70, 75, 80, 85 are the first four terms of an Arithmetic Progression, then the \(\mathbf{1 0}^{\text {th }}\) term is [1]

A35
B25
C115
D105
Answer & Solution

Ans. Option (c) is correct.

Explanation: \(a=70, d=75-70=5\)

\[ \begin{aligned} a_{n} & =a+(n-1) d \\ a_{10} & =70+(10-1) 5 \\ & =70+45 \end{aligned} \]
\[ a_{10}=115 \]
5
Question 5
ICSE Class 10 Mathematics 2022 Semester 1
5

The selling price of a shirt excluding GST is ₹ 800. If the rate of GST is 12% then the total price of the shirt is [1]

A₹704
B₹96
C₹896
D₹848
Answer & Solution

Ans. Option (c) is correct.

Explanation: \(\mathrm{SP}=\text{₹} 800, \mathrm{GST}=12 \%\)

\[ \begin{aligned} \text { GST } & =\text{₹} \frac{12}{100} \times 800=\text{₹} 96 \\ \text { Total Cost } & =800+96=\text{₹} 896 \end{aligned} \]
6
Question 6
ICSE Class 10 Mathematics 2022 Semester 1
6

Which of the following quadratic equations has 2 and 3 as its roots? [1]

A\(x^{2}-5 x+6=0\)
B\(x^{2}+5 x+6=0\)
C\(x^{2}-5 x-6=0\)
D\(x^{2}+5 x-6=0\)
Answer & Solution

Ans. Option (a) is correct.

Explanation: \(\because \quad x^{2}-(\) sum of roots \() x-(\) product of roots \()=0\)

\[ \begin{array}{lr} \therefore & x^{2}-(2+3) x-(2 \times 3)=0 \\ \Rightarrow & x^{2}-5 x+6=0 \end{array} \]
7
Question 7
ICSE Class 10 Mathematics 2022 Semester 1
7

If \(\boldsymbol{x}\), 5.4, 5, 9 are in proportion then \(\boldsymbol{x}\) is [1]

A3
B9.72
C25
D25/3
Answer & Solution

Ans. Option (a) is correct.

Explanation: Numbers are in proportion (given)

\[ \begin{aligned} \Rightarrow & & \frac{x}{5.4} & =\frac{5}{9} \\ \Rightarrow & & 9 \times x & =5.4 \times 5 \\ \Rightarrow & & x & =\frac{5.4 \times 5}{9}=0.6 \times 5 \\ \Rightarrow & & x & =3 \end{aligned} \]
8
Question 8
ICSE Class 10 Mathematics 2022 Semester 1
8

Mohit opened a Recurring deposit account in a bank for 2 years. He deposits ₹1000 every month and receives ₹25500 on maturity. The interest he earned in 2 years is [1]

A₹13500
B₹3000
C₹24000
D₹1500
Answer & Solution

Ans. Option (d) is correct.

Explanation:

\[ \begin{aligned} \text { Maturity Amount } & =\text{₹} 25500 \\ \text { Total deposit } & =1000 \times 24=\text{₹} 24000 \\ \therefore \quad \text { Interest } & =\text{₹} 25,500-\text{₹} 24,000 \\ & =\text{₹} 1,500 \end{aligned} \]
9
Question 9
ICSE Class 10 Mathematics 2022 Semester 1
9

In the given figure \(\boldsymbol{A} \boldsymbol{B}=\mathbf{2 4 ~ c m}, \boldsymbol{A} \boldsymbol{C}=\mathbf{1 8 ~ c m}\), \(\boldsymbol{D E}=\mathbf{1 2 ~ c m}, \boldsymbol{D F}=\mathbf{9 ~ c m}\) and \(\angle \boldsymbol{B} A C=\angle E D F\). Then \(\triangle A B C \sim \triangle D E F\) by the condition [1]

ICSE Class 10 Mathematics 2022 Semester 1 Question 9 figure figure
ICSE Class 10 Mathematics 2022 Semester 1 Question 9 figure figure
A\(A A A\)
BSAS
C\(S S S\)
D\(A A S\)
Answer & Solution

Ans. Option (b) is correct.

Explanation: In \(\triangle D E F\) \& \(\triangle A B C\),

ICSE Class 10 Mathematics 2022 Semester 1 Question 9 solution figure
ICSE Class 10 Mathematics 2022 Semester 1 Question 9 solution figure
\[ \begin{aligned} \frac{A C}{D F} & =\frac{A B}{D E}=\frac{2}{1} \\ \angle B A C & =\angle E D F \end{aligned} \]

Hence, \(\triangle A B C \sim \triangle D E F\) (by SAS criterion)

10
Question 10
ICSE Class 10 Mathematics 2022 Semester 1
10

If \(\boldsymbol{A}=\left[\begin{array}{cc}5 & 10 \\ 3 & -4\end{array}\right]\) and \(\boldsymbol{I}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\) then \(\boldsymbol{A} \boldsymbol{I}\) is equal to [1]

A\(\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\)
B\(\left[\begin{array}{cc}5 & 10 \\ -3 & 4\end{array}\right]\)
C\(\left[\begin{array}{cc}5 & 10 \\ 3 & -4\end{array}\right]\)
D\(\left[\begin{array}{cc}15 & 15 \\ -1 & -1\end{array}\right]\)
Answer & Solution

Ans. Option (c) is correct.

Explanation: \(A \times I=A\)

11
Question 11
ICSE Class 10 Mathematics 2022 Semester 1
11

The polynomial \(\boldsymbol{x}^{\mathbf{3}}-\mathbf{2} \boldsymbol{x}^{\mathbf{2}}+\boldsymbol{a} \boldsymbol{x}+\mathbf{1 2}\) when divided by \((x+1)\) leaves, remainder 20, then ‘ \(a\) ‘ is equal to [1]

A-31
B9
C11
D– 11
Answer & Solution

Ans. Option (d) is correct.

Explanation:

\[ \begin{aligned} x+1 & =0 \\ x & =-1 \end{aligned} \]

Remainder \(=p(-1)\) Let

\[ p(x)=x^{3}-2 x^{2}+a x+12 \]

Now, \(p(-1)=20\)

\[ \begin{array}{lr} \Rightarrow & (-1)^{3}-2(-1)^{2}+a(-1)+12=20 \\ \Rightarrow & -1-2-a+12=20 \\ \Rightarrow & -a=20-9 \\ \Rightarrow & a=-11 \end{array} \]
12
Question 12
ICSE Class 10 Mathematics 2022 Semester 1
12

In an Arithmetic Progression (A.P.) if first term is 5, common difference is – 3 and the \(\boldsymbol{n}^{\text {th }}\) term is – 7, then \(\boldsymbol{n}\) is equal to [1]

A5
B17
C– 13
D7
Answer & Solution

Ans. Option (a) is correct.

Explanation: Given, \(a=5, d=-3\) and \(a_{n}=-7\)

\[ \begin{array}{rlrl} \text { Since, } & & a_{n} & =a+(n-1) d \\ \therefore & & -7 & =5+(n-1)(-3) \\ \Rightarrow & -7-5 & =-3 n+3 \\ \Rightarrow & -12-3 & =-3 n \\ \Rightarrow & -15 & =-3 n \\ \Rightarrow & n & =5 \end{array} \]
13
Question 13
ICSE Class 10 Mathematics 2022 Semester 1
13

In the given figure \(\boldsymbol{P} \boldsymbol{Q}\) is parallel to \(\boldsymbol{T} \boldsymbol{R}\), then by using condition of similarity [1]

ICSE Class 10 Mathematics 2022 Semester 1 Question 13 figure figure
A\(\frac{P Q}{R T}=\frac{O P}{O T}=\frac{O Q}{O R}\)
B\(\frac{P Q}{R T}=\frac{O P}{O R}=\frac{O Q}{O T}\)
C\(\frac{P Q}{R T}=\frac{O R}{O P}=\frac{O Q}{O T}\)
D\(\frac{P Q}{R T}=\frac{O P}{O R}=\frac{O T}{O Q}\)
Answer & Solution

Ans. Option (b) is correct.

Explanation: \(P Q \| T R, P R\) is transversal,

ICSE Class 10 Mathematics 2022 Semester 1 Question 13 solution figure
\[ \Rightarrow \quad \angle P=\angle R \quad \text { (alternate angles) } \]

Also, \(Q T\) is transversal,

\[ \begin{array}{lcc} \Rightarrow & \angle Q=\angle T & \text { (alternate angles) } \\ \therefore & \triangle R T O \sim \triangle P Q O & \text { (by AA criterion) } \end{array} \]

∴

\[ \frac{P Q}{R T}=\frac{P O}{R O}=\frac{Q O}{T O} \]

(by C.P. C.T.)

14
Question 14
ICSE Class 10 Mathematics 2022 Semester 1
14

If \(\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}\), and \(\boldsymbol{d}\) are proportional then \(\frac{\boldsymbol{a}+\boldsymbol{b}}{\boldsymbol{a}-\boldsymbol{b}}\) is equal to [1]

A\(\frac{c}{d}\)
B\(\frac{c-d}{c+d}\)
C\(\frac{d}{c}\)
D\(\frac{c+d}{c-d}\)
Answer & Solution

Ans. Option (d) is correct.

Explanation: Given,

\[ \frac{a}{b}=\frac{c}{d} \]

By compodendo-dividendo

\[ \frac{a+b}{a-b}=\frac{c+d}{c-d} \]
15
Question 15
ICSE Class 10 Mathematics 2022 Semester 1
15

The first four terms of an Arithmetic Progression (A. P.), whose first term is 4 and common difference is -6, are [1]

A4, – 10, – 16, – 22
B4, 10, 16, 22
C4, – 2, – 8, – 14
D4, 2, 8, 14
Answer & Solution

Ans. Option (c) is correct.

Explanation: \(a=4 ; d=-6\) First four terms of an A.P. are:

\[ \begin{array}{ll} & a, a+d, a+2 d, a+3 d \\ \Rightarrow & 4,4+(-6), 4+2(-6), 4+3(-6) \\ \Rightarrow & 4,-2,-8,-14 \end{array} \]
16
Question 16
ICSE Class 10 Mathematics 2022 Semester 1
16

One of the roots of the quadratic equation \(\boldsymbol{x}^{\mathbf{2}}\) – \(8 x+5=0\) is 7.3166. The root of the equation correct to 4 significant figures is [1]

A7.3166
B7.317
C7.316
D7.32
Answer & Solution

Ans. Option (b) is correct.

Explanation: Correct to 4 significant figures is 7.317.

17
Question 17
ICSE Class 10 Mathematics 2022 Semester 1
17

\((x+2)\) and \((x+3)\) are two factors of the polynomial \(\boldsymbol{x}^{\mathbf{3}}+\mathbf{6} \boldsymbol{x}^{\mathbf{2}}+\mathbf{1 1} \boldsymbol{x}+\mathbf{6}\). If this polynomial is completely factorised the result is [2]

A\((x-2)(x+3)(x+1)\)
B\((x+2)(x-3)(x-1)\)
C\((x+2)(x+3)(x-1)\)
D\((x+2)(x+3)(x+1)\)
Answer & Solution

Ans. Option (d) is correct.

Explanation: Let

\[ p(x)=x^{3}+6 x^{2}+11 x+6 \]

If

\[ x+1=0 \]

then

\[ x=-1 \]

Now,

\[ \begin{aligned} p(-1) & =(-1)^{3}+6(-1)^{2}+11(-1)+6 \\ & =-1+6-11+6=0 \end{aligned} \]

Hence, \((x+1)\) is a third factor.

18
Question 18
ICSE Class 10 Mathematics 2022 Semester 1
18

The sum of the first 20 terms of the Arithmetic Progression 2, 4, 6, 8, ____ is [2]

A400
B840
C420
D800
Answer & Solution

Ans. Option (c) is correct.

Explanation: \(S_{n}=\frac{n}{2}[2 a+(n-1) d]\) Here, \(a=2, d=4-2=2\) and \(n=20\) ∴

\[ \begin{aligned} S_{20} & =\frac{20}{2}[2 \times 2+(20-1) 2] \\ & =10[4+38] \\ & =10 \times 42=420 \end{aligned} \]
19
Question 19
ICSE Class 10 Mathematics 2022 Semester 1
19

The solution set on the number line of the linear inequation [2]

A
ICSE Class 10 Mathematics 2022 Semester 1 Question 19 solution figure option A figure
B
ICSE Class 10 Mathematics 2022 Semester 1 Question 19 solution figure option B figure
C
ICSE Class 10 Mathematics 2022 Semester 1 Question 19 solution figure option C figure
D
ICSE Class 10 Mathematics 2022 Semester 1 Question 19 solution figure option D figure
Answer & Solution

Ans. Option (b) is correct.

Explanation:

\[ \begin{aligned} 2 y-6 & <y+2 \\ 2 y-y & <2+6 \\ y & <8 \end{aligned} \]

and,

\[ \begin{aligned} y+2 & \leq 2 y \\ 2 & \leq 2 y-y \\ 2 & \leq y \end{aligned} \]

So,

\[ 2 \leq y<8 \]
20
Question 20
ICSE Class 10 Mathematics 2022 Semester 1
20

If \(\boldsymbol{x}, \boldsymbol{y}, \boldsymbol{z}\) are in continued proportion then \(\left(\boldsymbol{y}^{\mathbf{2}}+\boldsymbol{z}^{\mathbf{2}}\right)\) : \(\left(x^{2}+y^{2}\right)\) is equal to [2]

A\(z: x\)
B\(x: z\)
C\(z x\)
D\((y+z):(x+y)\)
Answer & Solution

Ans. Option (a) is correct.

Explanation: \(x, y, z\) are in proportion

\[ y^{2}=x z \]

∴

\[ \frac{y^{2}+z^{2}}{x^{2}+y^{2}}=\frac{x z+z^{2}}{x^{2}+x z}=\frac{z(x+z)}{x(x+z)}=\frac{z}{x} \]

i.e.,

\[ z: x \]
21
Question 21
ICSE Class 10 Mathematics 2022 Semester 1
21

The marked price of an article is ₹ 5000. The shopkeeper gives a discount of 10%. If the rate of GST is 12%, then the amount paid by the customer including GST is [2]

A₹5040
B₹6100
C₹6272
D₹6160
Answer & Solution

Ans. Option (a) is correct.

Explanation:

\[ \begin{aligned} \mathrm{SP} & =\mathrm{MP}-\text { Discount } \\ & =5000-\frac{10}{100} \times 5000 \\ & =5000-500=4500 \\ \text { GST } & =12 \% \end{aligned} \]
\[ \text { GST amount }=\frac{12}{100} \times 4500=540 \]

Total amount paid \(=4500+540=\text{₹} 5040\)

22
Question 22
ICSE Class 10 Mathematics 2022 Semester 1
22

If \(\boldsymbol{A}=\left[\begin{array}{ll}3 & 5 \\ 1 & 4\end{array}\right], \boldsymbol{B}=\left[\begin{array}{ll}2 & 4 \\ 0 & 3\end{array}\right]\) and \(\boldsymbol{C}=\left[\begin{array}{cc}1 & -1 \\ 2 & 1\end{array}\right]\), then \(\mathbf{5} \boldsymbol{A}-\boldsymbol{B} \boldsymbol{C}\) is equal to [2]

A\(\left[\begin{array}{cc}-5 & -23 \\ 1 & 17\end{array}\right]\)
B\(\left[\begin{array}{cc}5 & 23 \\ 1 & 17\end{array}\right]\)
C\(\left[\begin{array}{ll}-2 & 8 \\ -3 & 3\end{array}\right]\)
D\(\left[\begin{array}{cc}5 & 23 \\ -1 & 17\end{array}\right]\)
Answer & Solution

Ans. Option (d) is correct.

Explanation:

\[ \begin{aligned} 5 A-B C & =5\left[\begin{array}{ll} 3 & 5 \\ 1 & 4 \end{array}\right]-\left[\begin{array}{ll} 2 & 4 \\ 0 & 3 \end{array}\right]\left[\begin{array}{cc} 1 & -1 \\ 2 & 1 \end{array}\right] \\ & =\left[\begin{array}{cc} 15 & 25 \\ 5 & 20 \end{array}\right]-\left[\begin{array}{cc} 10 & 2 \\ 6 & 3 \end{array}\right] \\ & =\left[\begin{array}{cc} 5 & 23 \\ -1 & 17 \end{array}\right] \end{aligned} \]
23
Question 23
ICSE Class 10 Mathematics 2022 Semester 1

In the given figure \(A B C D\) is a trapezium in which \(D C\) is parallel to \(A B\). \(A B=\mathbf{1 6 ~ c m}\) and \(D C=\mathbf{8 ~ c m}, O D=\mathbf{5 ~ c m}\), \(O B=(\boldsymbol{y}+\mathbf{3}) \mathbf{~ c m}, \boldsymbol{O} \boldsymbol{A}=\mathbf{1 1 ~ c m}\) and \(O C=(\boldsymbol{x}-\mathbf{1}) \mathbf{c m}\). Using the given information answer the following questions.

ICSE Class 10 Mathematics 2022 Semester 1 Question 23 figure
i

From the given figure name the pair of similar triangles [1]

A\(\triangle O A B, \triangle O B C\)
B\(\triangle C O D, \triangle A O B\)
C\(\triangle A D B, \triangle A C B\)
D\(\triangle C O D, \triangle C O B\)
Answer & Solution
ICSE Class 10 Mathematics 2022 Semester 1 Question 23(i) solution figure

Option (b) is correct.

Explanation: \(D C \| A B\) (given)

\[ \begin{array}{rll} & \angle C D O & =\angle A B O \\ \therefore D C O & =\angle B A O & \text { (alternate angle) } \\ \therefore & \triangle C O D \sim \triangle A O B & \text { (alternate angle) } A A \text { criterion) } \end{array} \]
ii

The corresponding proportional sides with respect to the pair of similar triangles obtained in (i) [1]

A\(\frac{C D}{A B}=\frac{O C}{O A}=\frac{O D}{O B}\)
B\(\frac{A D}{B C}=\frac{O C}{O A}=\frac{O D}{O B}\)
C\(\frac{A D}{B C}=\frac{B D}{A C}=\frac{A B}{D C}\)
D\(\frac{O D}{O B}=\frac{C D}{C B}=\frac{O C}{O A}\)
Answer & Solution

Option (a) is correct.

Explanation:

\[ \triangle C O D \sim \triangle A O B \]

∴

\[ \frac{C D}{A B}=\frac{O D}{O B}=\frac{C O}{A O} \text { (by C.P. C.T.) } \]
iii

The ratio of the sides of the pair of similar triangles is [1]

A1 : 3
B1 : 2
C2 : 3
D3 : 1
Answer & Solution

Option (b) is correct.

Explanation:

\[ \frac{C D}{A B}=\frac{O D}{O B}=\frac{C O}{A O}=\frac{8}{16}=\frac{1}{2} \]
iv

Using the ratio of sides of the pair of similar triangles the values of \(\boldsymbol{x}\) and \(\boldsymbol{y}\) are respectively [1]

A\(x=4.6, y=7\)
B\(x=7, y=7\)
C\(x=6.5, y=7\)
D\(x=6.5, y=2\)
Answer & Solution

Option (c) is correct.

Explanation: Here, \(\frac{x-1}{11}=\frac{5}{y+3}=\frac{1}{2}\)

\[ \begin{aligned} & \therefore \frac{x-1}{11}=\frac{1}{2} \text { and } \frac{5}{y+3}=\frac{1}{2} \\ & \Rightarrow 2 x-2=11 \text { and } y+3=10 \\ & \Rightarrow \quad x=\frac{13}{2}=6.5 \text { and } y=7 \end{aligned} \]
24
Question 24
ICSE Class 10 Mathematics 2022 Semester 1

Two cars X and Y use 1 litre of diesel to travel \(\boldsymbol{x} \mathbf{~ k m}\) and \((\boldsymbol{x}+\mathbf{3}) \mathbf{k m}\) respectively. If both the cars covered a distance of 72 km, then:

i

The number of litres of diesel used by car X is [1]

A\(\frac{72}{x-3}\) litres
B\(\frac{72}{x+3}\) litres
C\(\frac{72}{x}\) litres
D\(\frac{12}{x}\) litres
Answer & Solution

Option (c) is correct.

Explanation:

\[ \begin{aligned} x \mathrm{~km} & =1 \text { litre } \\ 72 \mathrm{~km} & =\frac{72}{x} \text { litres } \end{aligned} \]
ii

The number of litres of diesel used by car Y is [1]

A\(\frac{72}{x-3}\) litres
B\(\frac{72}{x+3}\) litres
C\(\frac{72}{x}\) litres
D\(\frac{12}{x+3}\) litres
Answer & Solution

Option (b) is correct.

Explanation:

\[ \begin{aligned} (x+3) \mathrm{km} & =1 \text { litre } \\ 72 \mathrm{~km} & =\frac{72}{x+3} \text { litres } \end{aligned} \]
iii

If car X used 4 litres of diesel more than car Y in the journey, then [1]

A\(\frac{72}{x-3}-\frac{12}{x}=4\)
B\(\frac{72}{x+3}-\frac{72}{x}=4\)
C\(\frac{72}{x}-\frac{72}{x+3}=4\)
D\(\frac{72}{x-3}-\frac{72}{x+3}=4\)
Answer & Solution

Option (c) is correct.

Explanation: Since, given car X used 4 litres of diesel more than car Y. Therefore,

\[ \frac{72}{x}-\frac{72}{x+3}=4 \]
iv

The amount of diesel used by the car X is [1]

A6 litres
B12 litres
C18 litres
D24 litres
Answer & Solution

Option (b) is correct.

Explanation: We have, \(\frac{72}{x}-\frac{72}{x+3}=4\)

\[ \begin{array}{lrl} \Rightarrow & 72(x+3)-72 x & =4 x^{2}+12 x \\ \Rightarrow & 72 x+216-72 x & =4 x^{2}+12 x \\ \Rightarrow & 4 x^{2}+12 x-216 & =0 \\ \Rightarrow & x^{2}+3 x-54 & =0 \\ \Rightarrow & x^{2}+9 x-6 x-54 & =0 \\ \Rightarrow & x(x+9)-6(x+9) & =0 \\ \Rightarrow & (x-6)(x+9) & =0 \\ \Rightarrow & x & =6 \\ & & \text { (distance cannot be -ve) } \end{array} \]

∴ Amount of diesel used by car X is \(\frac{72}{6}=12\) litres

25
Question 25
ICSE Class 10 Mathematics 2022 Semester 1

Joseph has a recurring deposit account in a bank for two years at the rate of 8%per annum simple interest.

i

If at the time of maturity Joseph receives ₹ 2000 as interest then the monthly installment is [1]

A₹ 1200
B₹ 600
C₹ 1000
D₹ 1600
Answer & Solution

Option (c) is correct.

Explanation: Given, \(R=8 \% ; I=\text{₹} 2000, n=24\)

\[ \begin{aligned} I & =P \times \frac{n(n+1)}{2} \times \frac{R}{1200} \\ 2000 & =P \times \frac{24 \times 25}{2} \times \frac{8}{1200} \\ 2000 & =P \times 12 \times \frac{8}{48} \\ P & =\frac{2000}{2}=\text{₹} 1,000 \end{aligned} \]
ii

The total amount deposited in the bank [1]

A₹ 25000
B₹ 24000
C₹ 26000
D₹ 23000
Answer & Solution

Option (b) is correct.

\[ \begin{aligned} \text { Explanation: Total amount } & =\text{₹} 1000 \times 24 \\ & =\text{₹} 24,000 \end{aligned} \]
iii

The amount Joseph receives on maturity is [1]

A₹ 27000
B₹ 25000
C₹ 26000
D₹ 28000
Answer & Solution

Option (c) is correct.

Explanation:

\[ \begin{aligned} \text { Maturity amount } & =\text { Total amount }+ \text { Interest } \\ & =\text{₹} 24000+\text{₹} 2000 \\ & =\text{₹} 26,000 \end{aligned} \]
iv

If the monthly installment is ₹ 100 and the rate of interest is 8%, in how many months Joseph will receive ₹ 52 as interest? [1]

A18
B30
C12
D6
Answer & Solution

Option (c) is correct.

Explanation: \(R=8 \% ; I=\text{₹} 52, P=\text{₹} 100\)

\[ \begin{aligned} I & =P \times \frac{n(n+1)}{2} \times \frac{R}{1200} \\ 52 & =100 \times \frac{n(n+1)}{2} \times \frac{8}{1200} \\ n(n+1) & =3 \times 52 \end{aligned} \]
\[ \begin{array}{r} n^{2}+n-156=0 \\ n^{2}+13 n-12 n-156=0 \\ n(n+13)-12(n+13)=0 \end{array} \]
\[ \begin{aligned} & \therefore(n+13)(n-12)=0 \\ & \therefore n=12 \text { months } \\ & \\ & \square \square \square \end{aligned} \]

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