WORD PROBLEMS RELATED TO
LINEAR EQUATIONS
CLASS 7TH, 8TH, 9TH & 10TH
Q.1) Seven more than half of a number is 42. Find the number.
SOLUTION :
Let the number be = x
Half of the number = 1x/2
According to the problem,
1x/2 + 7 = 42
1x/2 = 42 -7
1x/2 = 35
x = 35*2
x = 70
Thus, the number is 70
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Q.2) The sum of three consecutive multiples of 8 is 96. Find the multiples.
SOLUTION :
Let the three consecutive multiples of 8 be = 8x , (8x + 8), (8x +16)
Their sum = 96
According to the problem,
8x + ( 8x + 8 ) + ( 8x + 16 ) = 96
8x + 8x + 8 + 8x + 16 = 96
24x + 24 = 96
24x = 96 -24
24x = 72
x = 72 / 24
x = 3
Thus, the 1st multiple of 8 = 8x = 8*x = 8*3 = 24
the 2nd multiple of 8 = 8x + 8 = 8 * 3 + 8 = 24 + 8 = 32
the 3rd multiple of 8 = 8x + 16 = 8 * 3 + 16 = 24 + 16 = 40
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Q.3) Two angles of a right-angled triangle are in the ratio 1 : 5. Find each angle of the triangle.
SOLUTION :
The 1st angle of the right- angled triangle = 90°
Ratio of two angles = 1 :5
Let the two angles be = 1x and 5x
Sum of the angles of a triangle = 180
According to the problem,
90 + 1x + 5x =180
6x = 180 – 90
6x = 90
x = 90 /6
x = 15
1x = 15 5x = 5 * 15 = 75
Thus, the angles of the triangle are 90°, 15° , 75°
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Q.4) Rs. 1260 is divided into two parts. The first part is four more three times the second part. Find the two parts.
SOLUTION :
Let the second part be =x
the first part = ( 3x + 4 )
Total amount = 1260
According to the problem,
x + ( 3x + 4 ) = 1260
x + 3x + 4 = 1260
4x + 4 = 1260
4x = 1260 – 4
4x = 1256
x = 1256 / 4
x = 314
Thus, second part = x= Rs. 314 first part = 3x +4 = 3* 314 +4 = Rs. 946
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Q.5) The present ages of Sagar and Rahul are in the ratio 2: 3. Five years from now, the sum of their ages will be 55. Find their present ages.
SOLUTION :
Let the present ages of Sagar be 2x years and Rahul be 3x years.
Five years from now,
Sagar’s age = ( 2x + 5 ) years
Rahul’s age = ( 3x + 5 ) years
According to the problem,
( 2x + 5) + ( 3x + 5) = 55
2x + 5 + 3x + 5 = 55
5x + 10 = 55
5x = 55 -10
5x = 45
x = 45/5 x = 9
Thus, the present age of Sagar = 2x = 2 * 9 = 18 years
Rahul = 3x = 3 * 9 + 27 years
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Q.6) Each base angle of an isosceles triangle is four times the vertex angle. Find all the angles of the triangle.
SOLUTION:
Let the vertex angle be = x° Each base angle = 4x°
According to the problem,
sum of the angles of the triangle = 180
x + 4x + 4x =180
9x = 180
x = 180 /9 x = 20
Thus, vertex angle = 20° Each base angle = 4x = 4 * 20 = 80°
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Q.7) The breadth of a rectangle is five less than one-third of the length of the rectangle. Find its dimensions if the perimeter is 142.
SOLUTION:
Let the length of rectangle = x cm. Breadth = 1x/3-5 Perimeter = 142cm
According to the problem,
Perimeter = 2 ( L + B )
142 = 2[ ( x) +( 1x/3-5)]
142/2 = x + 1x/3 – 5
71 + 5 = 3x/3 + 1x/3
76 * 3 = 4x
76 * 3/4 = X
19 * 3 = X X = 57 cm.
Thus, length = x = 57 cm Breadth = 1x/3 – 5 = 1 * 57/3 -5 = 19 -5 = 14 cm.
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