Substitution Method PYQs

Class 10 Previous Year Questions (2014 – 2026)

Topic Overview

Substitution is a fundamental algebraic method. It is particularly useful for solving word problems where one variable can be easily expressed in terms of another, such as in fraction or age-related problems.

Q1 2024

00:00

Sum of two numbers is 105 and their difference is 45. Find the numbers.

Let numbers be $x$ and $y$.

$x + y = 105$

$x - y = 45$

Adding: $2x = 150 \Rightarrow x = 75$.

Subtracting: $2y = 60 \Rightarrow y = 30$.

Ans: 75 and 30

Q2 2023

00:00

A fraction becomes $1/3$ when 1 is subtracted from numerator. It becomes $1/4$ when 8 is added to denominator. Find the fraction.

Let fraction be $x/y$.

$(x-1)/y = 1/3 \Rightarrow 3x - 3 = y$

$x/(y+8) = 1/4 \Rightarrow 4x = y + 8$

Substitute $y$: $4x = (3x-3) + 8 \Rightarrow 4x = 3x + 5 \Rightarrow x = 5$.

$y = 3(5) - 3 = 12$. Fraction is $5/12$.

Ans: 5/12

Q3 2020

00:00

Solve by substitution method: $7x - 15y = 2$ and $x + 2y = 3$

From Eq 2: $x = 3 - 2y$.

Substitute in Eq 1: $7(3 - 2y) - 15y = 2$.

$21 - 14y - 15y = 2 \Rightarrow 29y = 19 \Rightarrow y = 19/29$.

$x = 3 - 2(19/29) = (87 - 38)/29 = 49/29$.

Ans: $x = 49/29, y = 19/29$

Q4 2019

00:00

Find two numbers whose sum is 75 and difference is 15.

$x + y = 75, x - y = 15$.

$2x = 90 \Rightarrow x = 45$.

$2y = 60 \Rightarrow y = 30$.

Ans: 45 and 30

Q5 2018

00:00

Solve: $\sqrt{x} + \sqrt{y} = 4$ and $1/\sqrt{x} + 1/\sqrt{y} = 4/3$ ($x, y > 0$)

Let $a = \sqrt{x}, b = \sqrt{y}$.

$a + b = 4$

$1/a + 1/b = 4/3 \Rightarrow (a+b)/ab = 4/3 \Rightarrow 4/ab = 4/3 \Rightarrow ab = 3$.

Numbers whose sum is 4 and product is 3 are 1 and 3.

If $a=1, b=3 \Rightarrow x=1, y=9$. If $a=3, b=1 \Rightarrow x=9, y=1$.

Ans: (1, 9) or (9, 1)

Q6 2017

00:00

Solve: $x + y = 5$ and $2x - 3y = 5$

From Eq 1: $x = 5 - y$.

Substitute in Eq 2: $2(5 - y) - 3y = 5$.

$10 - 2y - 3y = 5 \Rightarrow 5y = 5 \Rightarrow y = 1$.

$x = 5 - 1 = 4$.

Ans: $x = 4, y = 1$

Q7 2015

00:00

Solve: $\sqrt{3}x - \sqrt{8}y = 0$ and $\sqrt{5}x + \sqrt{2}y = 0$

This is a homogeneous system. Since $D = \sqrt{6} - (-\sqrt{40}) \neq 0$, the only solution is $(0, 0)$.

Ans: $x = 0, y = 0$

Q8 2025

00:00

Sum of digits of a two-digit number is 12. Reversing digits gives a number 36 more than original. Find the original number.

Let tens digit $= x$, units digit $= y$. Number $= 10x + y$.

$x + y = 12$

$(10y + x) - (10x + y) = 36 \Rightarrow 9y - 9x = 36 \Rightarrow y - x = 4$.

Adding: $2y = 16 \Rightarrow y = 8, x = 4$. Number is 48.

Ans: 48

Q9 2022

00:00

A library has a fixed charge for first 3 days, extra per day after. Saritha paid ₹27 for 7 days and Susy paid ₹21 for 5 days. Find the charges.

Let fixed charge $= x$, extra per day $= y$.

Saritha: $x + 4y = 27$ (7 days - 3 fixed).

Susy: $x + 2y = 21$ (5 days - 3 fixed).

Subtract: $2y = 6 \Rightarrow y = 3$. $x = 21 - 2(3) = 15$.

Ans: Fixed = ₹15, Extra = ₹3/day

Q10 2022

00:00

Solve: $2/x + 3/y = 13$ and $5/x - 4/y = -2$ ($x \neq 0, y \neq 0$)

Let $1/x = a, 1/y = b$.

$2a + 3b = 13$

$5a - 4b = -2$

Multiply Eq 1 by 4, Eq 2 by 3: $8a + 12b = 52, 15a - 12b = -6$.

Adding: $23a = 46 \Rightarrow a = 2 \Rightarrow x = 1/2$.

$b = (13-4)/3 = 3 \Rightarrow y = 1/3$.

Ans: $x = 1/2, y = 1/3$

Q11 2020

00:00

Sum of a two-digit number and its reverse is 110. If 10 is subtracted, new number is 4 more than 5 times the sum of digits. Find the number.

Let number $= 10x + y$. Reverse $= 10y + x$.

$(10x+y) + (10y+x) = 110 \Rightarrow 11x + 11y = 110 \Rightarrow x + y = 10$.

$(10x+y) - 10 = 5(x+y) + 4 \Rightarrow 10x+y-10 = 5(10) + 4 = 54$.

$10x + y = 64$. Solving with $x+y=10$: $9x = 54 \Rightarrow x = 6, y = 4$.

Ans: 64

Q12 2018

00:00

The larger of two supplementary angles exceeds the smaller by 18°. Find them.

$x + y = 180$

$x - y = 18$

$2x = 198 \Rightarrow x = 99^{\circ}$.

$y = 180 - 99 = 81^{\circ}$.

Ans: 99° and 81°

Q13 2017

00:00

Sum of a two-digit number and its reverse is 99. Digits differ by 3. Find the number.

$(10x+y) + (10y+x) = 99 \Rightarrow x+y = 9$.

$x-y = 3$ or $y-x = 3$.

Case 1: $2x=12 \Rightarrow x=6, y=3$. Number 63.

Case 2: $2y=12 \Rightarrow y=6, x=3$. Number 36.

Ans: 63 and 36

Q14 2021

00:00

Sum of ages of father and son is 45 years. Five years ago, product of their ages was 4 times the father's age then. Find their present ages.

Let father $= x$, son $= y$. $x+y=45$.

Five years ago: father $= x-5$, son $= y-5$.

$(x-5)(y-5) = 4(x-5) \Rightarrow y-5 = 4 \Rightarrow y = 9$.

Father $x = 45 - 9 = 36$.

Ans: Father = 36, Son = 9

Q15 2022

00:00

PASSAGE: Book rental shop: fixed charge for first 2 days, then ₹y per day. Latika paid ₹22 for 6 days; Anand paid ₹16 for 4 days.
(i) Write the pair of equations.
(ii) Fixed charge for first 2 days.
(iii) (a) Daily charge after 2 days + Ramesh's cost for 7 days. OR (b) Solve graphically.

(i) $x+4y=22, x+2y=16$.

(ii) Subtracting: $2y=6 \Rightarrow y=3, x=10$.

(iii) (a) Ramesh for 7 days (2 fixed + 5 extra) $= 10 + 5(3) = ₹25$.

Q16 2021

00:00

PASSAGE: Sneha and Trisha together have 45 marbles. Each loses 5, product of new counts = 128. Their counts satisfy x+y=45 and x-y=k (k>0).
(i) Write the equations.
(ii) Find marbles each has.
(iii) (a) Find k and verify. OR (b) If each gains 5 instead, what are the new equations?

(i) $x+y=45, (x-5)(y-5)=128$.

(ii) $y=45-x \Rightarrow (x-5)(40-x)=128 \Rightarrow 40x-x^2-200+5x=128 \Rightarrow x^2-45x+328=0$.

Solving: $x=24, y=21, k=3$.

00:00

Sum of two numbers is 105 and their difference is 45. Find the numbers.

Let numbers be $x$ and $y$.

$x + y = 105$

$x - y = 45$

Adding: $2x = 150 \Rightarrow x = 75$.

Subtracting: $2y = 60 \Rightarrow y = 30$.

Ans: 75 and 30

Q2 2023

00:00

A fraction becomes $1/3$ when 1 is subtracted from numerator. It becomes $1/4$ when 8 is added to denominator. Find the fraction.

Let fraction be $x/y$.

$(x-1)/y = 1/3 \Rightarrow 3x - 3 = y$

$x/(y+8) = 1/4 \Rightarrow 4x = y + 8$

Substitute $y$: $4x = (3x-3) + 8 \Rightarrow 4x = 3x + 5 \Rightarrow x = 5$.

$y = 3(5) - 3 = 12$. Fraction is $5/12$.

Ans: 5/12

Q3 2020

00:00

Solve by substitution method: $7x - 15y = 2$ and $x + 2y = 3$

From Eq 2: $x = 3 - 2y$.

Substitute in Eq 1: $7(3 - 2y) - 15y = 2$.

$21 - 14y - 15y = 2 \Rightarrow 29y = 19 \Rightarrow y = 19/29$.

$x = 3 - 2(19/29) = (87 - 38)/29 = 49/29$.

Ans: $x = 49/29, y = 19/29$

Q4 2019

00:00

Find two numbers whose sum is 75 and difference is 15.

$x + y = 75, x - y = 15$.

$2x = 90 \Rightarrow x = 45$.

$2y = 60 \Rightarrow y = 30$.

Ans: 45 and 30

Q5 2018

00:00

Solve: $\sqrt{x} + \sqrt{y} = 4$ and $1/\sqrt{x} + 1/\sqrt{y} = 4/3$ ($x, y > 0$)

Let $a = \sqrt{x}, b = \sqrt{y}$.

$a + b = 4$

$1/a + 1/b = 4/3 \Rightarrow (a+b)/ab = 4/3 \Rightarrow 4/ab = 4/3 \Rightarrow ab = 3$.

Numbers whose sum is 4 and product is 3 are 1 and 3.

If $a=1, b=3 \Rightarrow x=1, y=9$. If $a=3, b=1 \Rightarrow x=9, y=1$.

Ans: (1, 9) or (9, 1)

Q6 2017

00:00

Solve: $x + y = 5$ and $2x - 3y = 5$

From Eq 1: $x = 5 - y$.

Substitute in Eq 2: $2(5 - y) - 3y = 5$.

$10 - 2y - 3y = 5 \Rightarrow 5y = 5 \Rightarrow y = 1$.

$x = 5 - 1 = 4$.

Ans: $x = 4, y = 1$

Q7 2015

00:00

Solve: $\sqrt{3}x - \sqrt{8}y = 0$ and $\sqrt{5}x + \sqrt{2}y = 0$

This is a homogeneous system. Since $D = \sqrt{6} - (-\sqrt{40}) \neq 0$, the only solution is $(0, 0)$.

Ans: $x = 0, y = 0$

Q8 2025

00:00

Sum of digits of a two-digit number is 12. Reversing digits gives a number 36 more than original. Find the original number.

Let tens digit $= x$, units digit $= y$. Number $= 10x + y$.

$x + y = 12$

$(10y + x) - (10x + y) = 36 \Rightarrow 9y - 9x = 36 \Rightarrow y - x = 4$.

Adding: $2y = 16 \Rightarrow y = 8, x = 4$. Number is 48.

Ans: 48

Q9 2022

00:00

A library has a fixed charge for first 3 days, extra per day after. Saritha paid ₹27 for 7 days and Susy paid ₹21 for 5 days. Find the charges.

Let fixed charge $= x$, extra per day $= y$.

Saritha: $x + 4y = 27$ (7 days - 3 fixed).

Susy: $x + 2y = 21$ (5 days - 3 fixed).

Subtract: $2y = 6 \Rightarrow y = 3$. $x = 21 - 2(3) = 15$.

Ans: Fixed = ₹15, Extra = ₹3/day

Q10 2022

00:00

Solve: $2/x + 3/y = 13$ and $5/x - 4/y = -2$ ($x \neq 0, y \neq 0$)

Let $1/x = a, 1/y = b$.

$2a + 3b = 13$

$5a - 4b = -2$

Multiply Eq 1 by 4, Eq 2 by 3: $8a + 12b = 52, 15a - 12b = -6$.

Adding: $23a = 46 \Rightarrow a = 2 \Rightarrow x = 1/2$.

$b = (13-4)/3 = 3 \Rightarrow y = 1/3$.

Ans: $x = 1/2, y = 1/3$

Q11 2020

00:00

Sum of a two-digit number and its reverse is 110. If 10 is subtracted, new number is 4 more than 5 times the sum of digits. Find the number.

Let number $= 10x + y$. Reverse $= 10y + x$.

$(10x+y) + (10y+x) = 110 \Rightarrow 11x + 11y = 110 \Rightarrow x + y = 10$.

$(10x+y) - 10 = 5(x+y) + 4 \Rightarrow 10x+y-10 = 5(10) + 4 = 54$.

$10x + y = 64$. Solving with $x+y=10$: $9x = 54 \Rightarrow x = 6, y = 4$.

Ans: 64

Q12 2018

00:00

The larger of two supplementary angles exceeds the smaller by 18°. Find them.

$x + y = 180$

$x - y = 18$

$2x = 198 \Rightarrow x = 99^{\circ}$.

$y = 180 - 99 = 81^{\circ}$.

Ans: 99° and 81°

Q13 2017

00:00

Sum of a two-digit number and its reverse is 99. Digits differ by 3. Find the number.

$(10x+y) + (10y+x) = 99 \Rightarrow x+y = 9$.

$x-y = 3$ or $y-x = 3$.

Case 1: $2x=12 \Rightarrow x=6, y=3$. Number 63.

Case 2: $2y=12 \Rightarrow y=6, x=3$. Number 36.

Ans: 63 and 36

Q14 2021

00:00

Sum of ages of father and son is 45 years. Five years ago, product of their ages was 4 times the father's age then. Find their present ages.

Let father $= x$, son $= y$. $x+y=45$.

Five years ago: father $= x-5$, son $= y-5$.

$(x-5)(y-5) = 4(x-5) \Rightarrow y-5 = 4 \Rightarrow y = 9$.

Father $x = 45 - 9 = 36$.

Ans: Father = 36, Son = 9

Q15 2022

00:00

(i) $x+4y=22, x+2y=16$.

(ii) Subtracting: $2y=6 \Rightarrow y=3, x=10$.

(iii) (a) Ramesh for 7 days (2 fixed + 5 extra) $= 10 + 5(3) = ₹25$.

Q16 2021

00:00

(i) $x+y=45, (x-5)(y-5)=128$.

(ii) $y=45-x \Rightarrow (x-5)(40-x)=128 \Rightarrow 40x-x^2-200+5x=128 \Rightarrow x^2-45x+328=0$.

Solving: $x=24, y=21, k=3$.