Pair of Linear Equations Class 10th Important Questions with Answers Mathematics

Updated on June 1, 2025 | By Learnzy Academy

Q1. The sum of thrice the first and the second is 142 and four times the first exceeds the second by 138, then find the numbers.

Let the first number be x and the second number be y.

According to the question:

Three times the first number plus the second number is 142:
So, 3x + y = 142        -----------------------------(1)

Four times the first number exceeds the second number by 138:
So, 4x - y = 138       --------------------------------(2)

Now, add Equation (1) and Equation (2):

(3x + y) + (4x - y) = 142 + 138
3x + y + 4x - y = 280
7x = 280
x = 280 divided by 7
x = 40

Now substitute x = 40 into Equation (1) Then

3x + y = 142
3(40) + y = 142
120 + y = 142
y = 142 - 120
y = 22

Hence The first number is 40 & The second number is 22

Q2. The difference between two numbers is 14 and the difference between their squares is 448, then find the numbers.

Let the two numbers be x and y,  And we assume where x is greater than y.

So according to the question:

The difference between the two numbers is 14:
So, x - y = 14     --------------------------------------------(1)

The difference between their squares is 448:
So, x² - y² = 448     -----------------------------------------(2)

As we know that x² - y² = (x - y)(x + y).
So put this into Equation (2):

(x - y)(x + y) = 448

From Equation (1),    x - y = 14.
Now substitute:

14(x + y) = 448

Divide both sides by 14:
x + y = 32     ----------------------------------------(3)

Now add equation (1) & (3)

(x - y) + (x + y) = 14 + 32
2x = 46
x = 23

Now substitute x = 23 into Equation 1:
23 - y = 14
y = 9

Hence the first number is 23  &   the second number is 9

 

Q3. A and B each has some money. If A gives Rs. 30 to B then B will have twice the money left with A. But if B gives Rs. 10 to A then A will have thrice as much as is left with B. How much money does each have?

Let the amount of money A has be Rs. x and B has Rs. y.

First condition:  If A gives Rs. 30 to B:
A will have x - 30
B will have y + 30

According to the condition:
y + 30 = 2(x - 30)     ------------------------------------------------ (1)

Second condition:  If B gives Rs. 10 to A:
A will have x + 10
B will have y - 10

According to the condition:
x + 10 = 3(y - 10)   ---------------------------------------------------- (2)

Now solve the equations:

From Equation (1):
y + 30 = 2x - 60
y = 2x - 90    -----------------------------------------------(3)

Substitute y value  into Equation (2):
x + 10 = 3((2x - 90) - 10)
x + 10 = 3(2x - 100)
x + 10 = 6x - 300
310 = 5x
x = 62

Now to find y put x = 62 in equation (3):

y = 2(62) - 90 = 124 - 90 = 34

Hence A has Rs. 62 & B has Rs. 34

Q4. The students of a class are made to stand in rows. If 4 students are extra in a row, there would be 2 rows less. If 4 students are less in a row, there would be 4 rows more. Find the number of students in the class.

Let the number of students in each row be x and the number of rows be y.
So, the total number of students is x × y.

First condition:
If 4 students are added to each row, the number of rows becomes 2 less.

This gives the equation:
(x + 4)(y - 2) = x × y
 =>   xy - 2x + 4y - 8 = xy
=>    -2x + 4y = 8    ------------------------------------ (1)

Second condition:
If 4 students are removed from each row, the number of rows becomes 4 more.

This gives the equation:
(x - 4)(y + 4) = x × y
=>   xy + 4x - 4y - 16 = xy
=>   4x - 4y = 16     ---------------------------------- (2)

Now add both equations:

(−2x + 4x) + (4y − 4y) = 8 + 16
=>  2x = 24  =>  x = 12

Now put x = 12 into Equation 1:

−2(12) + 4y = 8
=>   −24 + 4y = 8
=>   4y = 32   =>    y = 8

Hence  Students in each row = 12 & Number of rows = 8

Total students = 12 × 8 = 96

So, there are 96 students in the class.

Q5. 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.

Let the cost of 1 pencil be Rs. x and the cost of 1 pen be Rs. y

From the Question:
5 pencils and 7 pens cost Rs. 50
=>  5x + 7y = 50   -------------------------------------------- (1)

7 pencils and 5 pens cost Rs. 46
=>  7x + 5y = 46    -------------------------------------------- (2)

Step 1:Eliminate one variable
Multiply both equations to make the coefficients of x the same:

Multiply the first equation by 7 then:
35x + 49y = 350 ------------------------------------- (3)

Multiply the second equation by 5:
=>  35x + 25y = 230  ----------------------------------(4)

Now equation(3) - eyuation(4)

=>   (35x + 49y) − (35x + 25y) = 350 − 230
=>    24y = 120
=>     y = 5

Step 2: Substitute y = 5 into equation (1)

 5x + 7y = 50
=>   5x + 7(5) = 50
=>   5x + 35 = 50
=>   5x = 15
=>    x = 3

Hence cost of one pencil = Rs. 3 & Cost of one pen = Rs. 5 

Q6. The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save Rs 2000 per month, find their monthly incomes.

The ratio of both person incomes is 9 : 7
So, let their incomes be 9x and 7x

The ratio of their expenditures is 4 : 3
So, let their expenditures be 4y and 3y

Both save Rs. 2000 per month

Use the formula:
Income = Expenditure + Savings

For the first person:
Income = 9x, Expenditure = 4y, Saving = 2000
=>  9x = 4y + 2000   --------------------------------------------------- (1)

For the second person:
Income = 7x, Expenditure = 3y, Saving = 2000
=>  7x = 3y + 2000   ------------------------------------------------- (2)

Solve the two equations:
From the first equation:
=> 4y = 9x − 2000   ---------------------------------------- (3)

From the second equation:
=> 3y = 7x − 2000  ------------------------------------------ (4)

Now multiply:

Third equation by 3:
12y = 27x − 6000   ------------------------------------- (5)

Fourth equation by 4:
12y = 28x − 8000  ------------------------------------(6)

Now from (5)th and (6)th equations
=> 27x − 6000 = 28x − 8000
=>  2000 = x

Now find the incomes:
First person's income = 9x = 9 × 2000 = Rs. 18,000

Second person's income = 7x = 7 × 2000 = Rs. 14,000

Hence  First person's income = Rs. 18,000 & Second person's income = Rs. 14,000

Q7. Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.

Let the number of Rs. 50 notes be x
Let the number of Rs. 100 notes be y

From the Question:
The total amount is Rs. 2000
So: 50x + 100y = 2000
Divide the whole equation by 50:
=>   x + 2y = 40   -------------------------------------------- (1)

The total number of notes is 25
So: x + y = 25

Now solve the two equations:
From x + y = 25, we get:
x = 25 - y    ------------------------------------------------------- (2)
Substitute this into the first equation:
(25 - y) + 2y = 40
=>  25 + y = 40
=>   y = 15

Now put y = 15 into equation (2)
=>  x = 25 - 15 = 10

Hence Rs. 50 notes = 10   &   Rs. 100 notes = 15 

Q8. A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day

Let fixed charge for the first 3 days = Rs. x
And extra charge for each additional day = Rs. y

From the Question:
Saritha kept the book for 7 days, so she paid:
Fixed charge for first 3 days = x
Extra charge for remaining 4 days = 4y
So:  x + 4y = 27     ----------------------------------------- (1)

She kept the book for 5 days, so she paid:
Fixed charge for first 3 days = x
Extra charge for remaining 2 days = 2y
So:  x + 2y = 21    --------------------------------------------(2)

Now solve the equations:
From Equation 1:
x + 4y = 27

From Equation 2:
x + 2y = 21

Now subtract Equation 2 from Equation 1:
(x + 4y) − (x + 2y) = 27 − 21
=>  2y = 6
=>   y = 3

Now substitute y = 3 into Equation 2:
x + 2(3) = 21
x + 6 = 21
x = 15
Hence Fixed charge = Rs. 15   &   Extra charge per day = Rs. 3

Q9. From a bus stand in Bangalore , if we buy 2 tickets to Malleswaram and 3 tickets to Yeshwanthpur, the total cost is Rs 46; but if we buy 3 tickets to Malleswaram and 5 tickets to Yeshwanthpur the total cost is Rs 74. Find the fares from the bus stand to Malleswaram, and to Yeshwanthpur.

Let fare from the bus stand to Malleswaram = Rs. x
And fare from the bus stand to Yeshwanthpur = Rs. y

From the Question:
2 tickets to Malleswaram and 3 to Yeshwanthpur cost Rs. 46
=>  2x + 3y = 46    ------------------------------------- (1)

3 tickets to Malleswaram and 5 to Yeshwanthpur cost Rs. 74
=>   3x + 5y = 74   -----------------------------------------(2)

Now solve the equations:
Multiply Equation 1 by 3:
=>  6x + 9y = 138   ----------------------------------(3)

Multiply Equation 2 by 2:
=> 6x + 10y = 148  -----------------------------------------------(4)

Now subtract Equation 3 from Equation 4:
(6x + 10y) − (6x + 9y) = 148 − 138
=>  y = 10
Now substitute y = 10 into Equation 1:
2x + 3(10) = 46
=> 2x + 30 = 46
=> 2x = 16
=> x = 8

Hence fare to Malleswaram = Rs. 8  &   fare to Yeshwanthpur = Rs. 10

Q10. One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital?

Let the first person has Rs. x
And the second person has Rs. y

The first person says, “Give me a hundred, friend! I shall then become twice as rich as you.”
This means: After the second person gives Rs. 100 to the first person, the first person will have x + 100, and the second person will have y - 100. According to the condition:
x + 100 = 2(y - 100)
x - 2y = -300  ------------------------------------ (1)

The second person says, “If you give me ten, I shall be six times as rich as you.”
This means: After the first person gives Rs. 10 to the second person, the first person will have x - 10, and the second person will have y + 10. According to the condition:
y + 10 = 6(x - 10)
- 6x + y = -70  ----------------------------------------(2)

Solve the two equations:
Multiply Equation 1 by 6 :

6x - 12y = -1800 ---------------------------------------- (3)

Now add Equation (2) and Equation (3):
-11y = -1870
=>  y = 170

Now substitute y = 170 into Equation 1:
x - 2(170) = -300
=> x - 340 = -300
=> x = 40

Hence the first person has Rs. 40  &  the second person has Rs. 194

Q11. The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Let the smaller angle be x degrees.
And the larger angle will be (x + 18) degrees (since it exceeds the smaller by 18 degrees).

Two angles are supplementary if their sum is 180°. So, we have the equation:

x + (x + 18) = 180
2x + 18 = 180
x + 9 = 90 
x = 81

The larger angle is x + 18:
Larger angle = 81 + 18 = 99

Hence the smaller angle is 81°  &  the larger angle is 99°.

Q12. The solution of the equations x + y = 14 and x – y =4 is

We are given two equations:

x + y = 14   --------------------------------- (1)
x - y = 4    ------------------------------------(2)

Step 1:Add the two equations to eliminate y:
(x + y) + (x - y) = 14 + 4
=> 2x = 18
=> x = 9

Step 2: Substitute x = 9 into the first equation:
9 + y = 14
=>  y = 5
Hence, the solution is:   x = 9 and y = 5

Q13. The sum of the digits of a two digit number is 9. If 27 is added to it, the digits of the numbers get reversed. The number is

Let the two-digit number have:

Tens digit = x
Units digit = y
So the number is: 10x + y

It is given that:
The sum of the digits is 9:
So   x + y = 9   ----------------------------------(1)

If 27 is added, the digits reverse:
10x + y + 27 = 10y + x 
9x - 9y = -27
x - y = -3    ---------------------------------------(2)

Now solve the two equations:

Equation 1:   x + y = 9
Equation 2:    x - y = -3

Add both equations:
(x + y) + (x - y) = 9 + (-3)
2x = 6 
=> x = 3
Substitute x = 3 into Equation 1:
3 + y = 9 
=>  y = 6

So, the number is: 10x + y = 10×3 + 6 = 36
Hence answer is 36

Q14. Find the value of x and y if x + y = 5xy and 3x + 2y = 13xy . Also, x ≠ 0 and y ≠ 0

It is given:
x + y = 5xy   ----------------------------- (1)
3x + 2y = 13xy    -----------------------------(2)
Also, x ≠ 0 and y ≠ 0

Step 1:   Divide both equations by xy
From equation 1:
x + y = 5xy
Divide both sides by xy:
(x/xy) + (y/xy) = 5
=>  1/y + 1/x = 5 ------------------------------------(3)

From equation 2:
3x + 2y = 13xy
Divide both sides by xy:
(3x/xy) + (2y/xy) = 13
=>  3/y + 2/x = 13 --------------------------------(4)

Step 2:Let
1/x = a and 1/y = b

Then Equation (3) becomes:
a + b = 5 ------------------------------------(5)
Equation (4) becomes:
2a + 3b = 13 -------------------------------------(6)

Step 3:Solve equations (5) and (6)

From Equation (3):   a = 5 - b
Substitute into Equation (6):
2(5 - b) + 3b = 13
10 - 2b + 3b = 13
10 + b = 13
b = 3 
now a = 5 - b = 5 - 3 = 2

Step 4: Back-substitute

a = 1/x = 2  =>  x = 1/2
b = 1/y = 3  =>  y = 1/3

Final Answer: x = 1/2 and y = 1/3

Q15. The sum of two numbers is 137 and their difference is 43. Find the numbers.

Let the two numbers be x and y.

So we have two equations:
x + y = 137  ---------------------------(1)
x - y = 43   ------------------------------(2)
Add both equations:
(x + y) + (x - y) = 137 + 43
2x = 180
x = 90

Now substitute x = 90 into the first equation:
90 + y = 137
y = 137 - 90
y = 47
Hence the two numbers are 90 and 47.

Q16. A two-digit number is 4 more than 6 times the sum of its digit. If 18 is subtracted from the number, the digits are reversed. Find the number.

Let the tens digit be x and the units digit be y.
Then the two-digit number is: 10x + y

First condition:
The number is 4 more than 6 times the sum of its digits.
So   10x + y = 6(x + y) + 4
=> 10x + y = 6x + 6y + 4
=> 10x - 6x + y - 6y = 4
=> 4x - 5y = 4   ---------------------------------(1)

Second condition:
If 18 is subtracted from the number, the digits are reversed.
The reversed number is: 10y + x
So   (10x + y) - 18 = 10y + x
=>  10x + y - 18 = 10y + x
=>  10x - x + y - 10y = 18
=>   9x - 9y = 18
=>    x - y = 2 ----------------------------------(2)

Solve the two equations:
From Equation 2:
x = y + 2
Substitute into Equation 1:
4(y + 2) - 5y = 4
4y + 8 - 5y = 4
-y + 8 = 4
y = 4

Then,   x = y + 2 = 4 + 2 = 6

So, the number is:
10x + y = 10×6 + 4 = 64

Q17. The denominator of a fraction is greater than its numerator by 11. If 8 is added to both its numerator and denominator, it reduces to 1/2. Find the fraction.

Let the numerator be x.
Then the denominator is x + 11 (since it is 11 more than the numerator).

So, the fraction is:  x / (x + 11)

According to the problem:
If 8 is added to both the numerator and the denominator, the fraction becomes 1/2.

That gives the equation:
(x + 8) / (x + 19) = 1/2

Now cross-multiply:
2(x + 8) = x + 19
2x + 16 = x + 19
x + 16 = 19
x = 3

So, the numerator is 3, and the denominator is:
x + 11 = 3 + 11 = 14

Final Answer: The fraction is 3/14.

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