Let’s solve EXERCISE 3.3 from chapter Pair Of Linear Equations In Two Variables. This is from the 3rd chapter of the Mathematics book of class 10. Here, we would primarily use the Elimination and substitution method to solve the linear equations. Let’s checkout the step-by-step solution for the questions below. In the end, 10 questions for practice are more challenging than the NCERT book questions.
Class 10 EXERCISE 3.3 – Pair Of Linear Equations In Two Variables
1. Solve the following pair of linear equations by the elimination method and the substitution method :
(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
(iv) x/2 + 2y/3 = -1 and x – y/3 =3
(i) x + y = 5 and 2x – 3y = 4
Elimination Method:
Multiply the first equation by 3 and the second by 1:
3(x + y) = 3(5) → 3x + 3y = 15 (1)
2x – 3y = 4 (2)
Add equations (1) and (2):
(3x + 2x) + (3y – 3y) = 15 + 4
5x = 19
x = 19/5
Substitute x = 19/5 in x + y = 5:
19/5 + y = 5
y = 5 – 19/5 = 6/5
So, x = 19/5, y = 6/5.
Substitution Method:
From x + y = 5, we get x = 5 – y.
Substitute this in 2x – 3y = 4:
2(5 – y) – 3y = 4
10 – 2y – 3y = 4
-5y = -6
y = 6/5
Substitute y = 6/5 in x = 5 – y:
x = 5 – 6/5 = 19/5
So, x = 19/5, y = 6/5.
(ii) 3x + 4y = 10 and 2x – 2y = 2
Elimination Method:
Multiply the second equation by 2:
2(2x – 2y) = 2(2) → 4x – 4y = 4 (1)
3x + 4y = 10 (2)
Add equations (1) and (2):
(4x + 3x) + (-4y + 4y) = 4 + 10
7x = 14
x = 2
Substitute x = 2 in 3x + 4y = 10:
3(2) + 4y = 10
4y = 4
y = 1
So, x = 2, y = 1.
Substitution Method:
From 2x – 2y = 2, we get x = y + 1.
Substitute this in 3x + 4y = 10:
3(y + 1) + 4y = 10
3y + 3 + 4y = 10
7y = 7
y = 1
Substitute y = 1 in x = y + 1:
x = 1 + 1 = 2
So, x = 2, y = 1.
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
Elimination Method:
Rewrite the equations:
3x – 5y = 4 (1)
9x – 2y = 7 (2)
Multiply equation (1) by 3:
3(3x – 5y) = 3(4) → 9x – 15y = 12 (3)
Subtract equation (3) from equation (2):
(9x – 9x) – (-15y + 2y) = 7 – 12
-13y = -5
y = 5/13
Substitute y = 5/13 in 3x – 5y = 4:
3x – 5(5/13) = 4
3x = 4 + 25/13
x = (4*13 + 25) / 39
x = 77/39
So, x = 77/39, y = 5/13.
Substitution Method:
From 9x = 2y + 7, we get x = (2y + 7)/9.
Substitute this in 3x – 5y – 4 = 0:
3((2y + 7)/9) – 5y = 4
(2y + 7)/3 – 5y = 4
2y + 7 – 15y = 12
-13y = 5
y = 5/-13 = -5/13
Substitute y = -5/13 in x = (2y + 7)/9:
x = (2(-5/13) + 7)/9
x = (7 – 10/13)/9
x = 81/117 = 27/39
So, x = 27/39, y = -5/13.
(iv) x/2 + 2y/3 = -1 and x – y/3 = 3
Elimination Method:
Rewrite the equations:
x/2 + 2y/3 = -1 (1)
x – y/3 = 3 (2)
Multiply equation (1) by 6 and equation (2) by 2:
6(x/2 + 2y/3) = 6(-1) → 3x + 4y = -6 (3)
2(x – y/3) = 2(3) → 2x – 2y/3 = 6 (4)
Subtract equation (4) from equation (3):
(3x – 2x) + (4y + 2y/3) = -6 – 6
x + 14y/3 = -12
14y/3 = -12 – x
y = -3(12 + x)/14
Substitute y in x – y/3 = 3:
x – (-3(12 + x)/14)/3 = 3
Solve for x and y.
Substitution Method:
From x/2 + 2y/3 = -1, we get x = -2 – 4y/3.
Substitute this in x – y/3 = 3:
(-2 – 4y/3) – y/3 = 3
-2 – 5y/3 = 3
-5y/3 = 5
y = -3
Substitute y = -3 in x = -2 – 4y/3:
x = -2 – 4(-3)/3
x = -2 + 4
x = 2
So, x = 2, y = -3.
2. Form the pair of linear equations in the following problems, and find their solutions
(if they exist) by the elimination method :
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
(iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?
Let the numerator be x and the denominator be y. The fraction is x/y.
According to the problem, (x + 1)/(y – 1) = 1 (1)
and x/(y + 1) = 1/2 (2)
From equation (1): x + 1 = y – 1 → x – y = -2 (3)
From equation (2): 2x = y + 1 → 2x – y = 1 (4)
Multiply equation (3) by 2:
2(x – y) = 2(-2) → 2x – 2y = -4 (5)
Now subtract equation (5) from equation (4):
(2x – 2x) – (-2y – y) = 1 – (-4)
-3y = 5
y = -5/3
Substitute y = -5/3 in x – y = -2:
x – (-5/3) = -2
x = -2 – 5/3
x = -11/3
So, the fraction is -11/3 divided by -5/3, which simplifies to 11/5.
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
Let Nuri’s current age be N and Sonu’s current age be S.
According to the problem, N – 5 = 3(S – 5) (1)
and N + 10 = 2(S + 10) (2)
From equation (1): N – 5 = 3S – 15 → N – 3S = -10 (3)
From equation (2): N + 10 = 2S + 20 → N – 2S = 10 (4)
Subtract equation (4) from equation (3):
(N – N) – (3S – 2S) = -10 – 10
-S = -20
S = 20
Substitute S = 20 in N – 2S = 10:
N – 2(20) = 10
N = 50
So, Nuri is 50 years old, and Sonu is 20 years old.
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Let the tens digit be x and the units digit be y. The number is 10x + y.
According to the problem, x + y = 9 (1)
and 9(10x + y) = 2(10y + x) (2)
From equation (2): 90x + 9y = 20y + 2x → 90x – 2x = 20y – 9y → 88x = 11y (3)
Now, substitute y = 9 – x in equation (3):
88x = 11(9 – x)
88x = 99 – 11x
99x = 99
x = 1
Substitute x = 1 in x + y = 9:
1 + y = 9
y = 8
So, the number is 10(1) + 8 = 18.
(iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.
Let the number of ₹ 50 notes be x and the number of ₹ 100 notes be y.
According to the problem, x + y = 25 (1)
and 50x + 100y = 2000 (2)
From equation (1): y = 25 – x (3)
Now, substitute y in equation (2):
50x + 100(25 – x) = 2000
50x + 2500 – 100x = 2000
-50x = -500
x = 10
Substitute x = 10 in y = 25 – x:
y = 25 – 10
y = 15
So, Meena received 10 notes of ₹ 50 and 15 notes of ₹ 100.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Let the fixed charge be ₹ x and the charge for each extra day be ₹ y.
According to the problem, for Saritha (7 days): x + 4y = 27 (1)
and for Susy (5 days): x + 2y = 21 (2)
Subtract equation (2) from equation (1):
(x + 4y) – (x + 2y) = 27 – 21
2y = 6
y = 3
Substitute y = 3 in x + 2y = 21:
x + 2(3) = 21
x = 21 – 6
x = 15
So, the fixed charge is ₹ 15 and the charge for each extra day is ₹ 3.
Worksheet for EXERCISE 3.3 NCERT Solutions Pair Of Linear Equations In Two Variables for Class 10
1. The age of a father is three times the sum of the ages of his two children. After 5 years, his age will be twice the sum of the ages of the children. Find the present ages of the father and each child.
Hint
Let the father’s age be F and the sum of the children’s ages be C.
F = 3C (1)
F + 5 = 2(C + 5) (2)
Solve these equations to find F and C.
2. The sum of a two-digit number and the number obtained by reversing its digits is 110. If the tens digit exceeds the units digit by 3, find the original number.
Hint
Let the tens digit be x and the units digit be y. The number is 10x + y.
10x + y + 10y + x = 110 (1)
x – y = 3 (2)
Solve these equations to find x and y.
3. Two water taps together can fill a tank in 6 hours. If one tap takes 5 hours longer than the other to fill the tank alone, find the time each tap takes to fill the tank alone.
Hint
Let the time taken by the faster tap be x hours. The slower tap takes x + 5 hours.
According to the problem: 1/x + 1/(x + 5) = 1/6
Solve for x to find the times taken by each tap.
4. A boat’s speed in still water is 15 km/hr, and the speed of the stream is 3 km/hr. Find the time taken by the boat to go 68 km downstream and return.
Hint
Let the time taken to go downstream be x hours and upstream be y hours.
According to the problem: 15 + 3 = 68/x (downstream) and 15 – 3 = 68/y (upstream)
Solve for x and y to find the total time.
5. A rectangle’s length is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is reduced by 75 cm². Find the original dimensions of the rectangle.
Hint
Let the breadth be x cm. The length is 2x cm.
According to the problem: (2x – 5)(x + 5) = 2x*x – 75
Solve for x to find the original dimensions.
6. The denominator of a fraction is 4 more than twice the numerator. If 3 is added to both the numerator and the denominator, the fraction becomes 3/5. Find the original fraction.
Hint
Let the numerator be x. The denominator is 2x + 4.
According to the problem: (x + 3)/(2x + 7) = 3/5
Solve for x to find the original fraction.
7. In a class of 40 students, the number of girls is two-thirds the number of boys. If 5 girls leave the class, how many boys must leave to maintain the same ratio of girls to boys?
Hint
Let the number of boys be B and girls be G. G = 2/3 B.
According to the problem: B + G = 40 and G – 5 = 2/3(B – x)
Solve for x to find how many boys must leave.
8. The present ages of A and B are in the ratio 4:3. After 6 years, the ratio of their ages will be 5:4. Find their present ages.
Hint
Let A’s age be 4x and B’s age be 3x.
According to the problem: 4x + 6 : 3x + 6 = 5 : 4
Solve for x to find their present ages.
9. A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less to cover the same distance. Find the original speed of the train.
Hint
Let the original speed be x km/hr.
According to the problem: 360/x – 360/(x + 5) = 1
Solve for x to find the original speed.
10. Two angles are complementary. The larger angle is 30 degrees more than twice the smaller angle. Find the measures of the angles.
Hint
Let the smaller angle be x degrees. The larger angle is then 2x + 30 degrees.
According to the problem: x + 2x + 30 = 90
Solve for x to find the measures of the angles.