Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹105 and for a journey of 15 km, the charge paid is ₹155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes 9/11 , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6 . Find the fraction. (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solution:
(i) Let the first (larger) number = x
And the second number = y
The difference between two numbers is 26.
x - y = 26 ....(1)
One number is three times the other
x = 3y ....(2)
Substituting x = 3y in equation (1), we get
3y - y = 26
2y = 26
y = 13
Substituting y = 13 in equation (2)
x = 3 × 13
x = 39
The two numbers are 39 and 13.
(ii) Let the larger angle = x
And the smaller angle = y
Since the angles are supplementary,
x + y = 180 ....(1)
Larger angle exceeds the smaller by 18
x = y + 18 ....(2)
Substituting x = y + 18 in equation (1), we get y + 18 + y = 180
2y = 180 - 18
y = 162/2
y = 81
Substituting y = 81 in equation (2), we get
x = 81 + 18
x = 99
The angles are 99 degrees and 81 degrees.
(iii) Let the cost of 1 bat = ₹ x
And the cost of 1 ball = ₹ y
7x + 6y = 3800 ....(1)
3x + 5y = 1750 ....(2)
By solving equation (1),
7x + 6y = 3800
6y = 3800 - 7x
y = (3800 - 7x) / 6 ....(3)
Substituting y = (3800 - 7x) / 6 in equation (2), we get
3x + 5 [(3800 - 7x) / 6] = 1750
(18x + 19000 - 35x) / 6 = 1750
-17x + 19000 = 1750 × 6
17x = 19000 - 10500
x = 8500/17
x = 500
Substituting x = 500 in equation (3), we get
y = (3800 - 7 × 500) / 6
y = 300/6
y = 50
Cost of 1 bat is ₹ 500
Cost of 1 ball is ₹ 50
(iv) Let the fixed charge = ₹ x
And charge per km = ₹ y
Charge for a distance of 10 km
x + 10y = 105 ....(1)
Charge for a distance of 15 km
x + 15y = 155 ....(2)
By solving equation (1),
x +10y = 105
x = 105 - 10y ....(3)
Substituting x = 105 - 10y in equation (2), we get
105 - 10y + 15y = 155
5y = 155 - 105
y = 50/5
y = 10
Substituting y = 10 in equation (3)
x = 105 - 10 × 10
x = 105 - 100
x = 5
Now, charge for a distance of 25 km = x + 25y
= 5 + 25 × 10
= 5 + 250
= 255
Fixed charge = ₹ 5
Charge per km = ₹ 10
Charge for 25 km = ₹ 255
(v) Let the numerator = x
And denominator = y
Then the fraction is x / y
When 2 is added to both numerator and denominator,
(x + 2) / (y + 2) = 9/11
11(x + 2) = 9( y + 2)
11x + 22 = 9y + 18
11x - 9y + 22 - 18 = 0
11x - 9y + 4 = 0 ...(1)
When 3 is added to both numerator and denominator
(x + 3) / (y + 3) = 5/6
6(x + 3) = 5( y + 3)
6x + 18 = 5y + 15
6x - 5y + 18 - 15 = 0
6x - 5y + 3 = 0 ....(2)
5y = 6x + 3
y = (6x + 3) / 5 ....(3)
Substituting y = (6x + 3) / 5 in equation (1)
11x - 9 [(6x + 3) / 5] + 4 = 0
[55x - 9(6x + 3) + 20] / 5 = 0
55x - 54x - 27 + 20 = 0
x - 7 = 0
x = 7
Substituting x = 7 in equation (3) we get,
y = (6 × 7 + 3) / 5
y = (42 + 3) / 5
y = 45/5
y = 9
Thus, the fraction is 7/9.
(vi) Let the present age of Jacob = x years
And present age of his son = y years
5 years from now,
Jacob’s age = (x + 5) years
Son’s age = (y + 5) years
(x + 5) = 3(y + 5)
x + 5 = 3y + 15
x - 3y + 5 - 15 = 0
x - 3y - 10 = 0 ...(1)
5 years ago, Jacob’s age = (x - 5) years
Son’s age = (y - 5) years
(x - 5) = 7(y - 5)
x - 5 = 7y - 35
x - 7y - 5 + 35 = 0
x - 7y + 30 = 0 ...(2)
7y = x + 30
y = (x + 30) / 7 ....(3)
Substituting y = (x + 30) / 7 in equation (1)
x - 3[(x + 30) / 7] - 10 = 0
[7x - 3(x + 30) - 70] / 7 = 0
7x - 3x - 90 - 70 = 0
4x - 160 = 0
x = 160/4
x = 40
Substituting x = 40 in equation (3)
y = (40 + 30) / 7
y = 70 / 7
y = 10
Thus, the present age of Jacob is 40 years and his son is 10 years.
☛ Check: NCERT Solutions for Class 10 Maths Chapter 3
Video Solution:
Form the pair of linear equations for the following problems and find their solution by substitution method.(i) The difference between two numbers is 26 and one number is three times the other. Find them.(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750. Find the cost of each bat and each ball.(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹105 and for a journey of 15 km, the charge paid is ₹155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?(v) A fraction becomes 9/11 , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6 . Find the fraction. (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
NCERT Solutions for Class 10 Maths - Chapter 3 Exercise 3.3 Question 3
Summary:
The pair of linear equations for the following problems have been formed and their solution is found by substitution method. (i) The difference between two numbers is 26 and one number is three times the other. The two numbers are 39 and 13, (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. The angles are 99 degrees and 81 degrees, (iii) The coach of a cricket team buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750. Cost of 1 bat is ₹ 500 and cost of 1 ball is ₹ 50, (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹105 and for a journey of 15 km, the charge paid is ₹155. Fixed charge = ₹ 5, charge per km = ₹ 10, charge for 25 km = ₹ 255, (v) A fraction becomes 9/11 , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6 . The fraction is 7/9, (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. The present age of Jacob is 40 years and his son is 10 years.
☛ Related Questions:
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