WORD PROBLEMS RELATED TO
LINEAR EQUATIONS
CLASS7th, 8th, 9th AND 10th
Q.1) Divide Rs. 350 into two parts such that half of the greater exceeds the smaller part by 100.
SOLUTION:
Given:
Let the smaller part be x
Let the greater part be 350 – x
According to the problem,
350 / 2 – x / 2 = x + 100
350 – x = 2 ( x + 100 )
350 – x = 2x + 200
350 – 200 = 2x + x
150 = 3x
150 / 3 = x
50 = x
Thus,
Smaller part = Rs. 50
Greater part = Rs. 300
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Q.2) Seventy tickets of a theatre show were sold for Rs. 8000. If the tickets were sold in the denomination of Rs. 125 and Rs. 75, how many tickets of each denomination were sold ?
SOLUTION:
Given:
Total ticket = 70
Amount = Rs. 8000
Let the no. of tickets of Rs. 75 be = x
Let the no. of tickets of Rs. 125 be = ( 70 – x)
Amount from selling tickets of Rs. 75 denomination = 75x
Amount from selling tickets of Rs. 125 denomination = 125 ( 70 – x)
75x + 125 ( 70 – x ) = 8000
75x + 8750 – 125x = 8000
75x – 125x = 8000 – 8750
-50x = -750
x = -750 / -50
x = 15
Thus,
No. of tickets of Rs. 75 = x = 15
No. of tickets of Rs. 125 = ( 70 – x )
= ( 70 – 15 )
= 55
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Q.3) In an examination, a student requires 40% marks to pass. If a student gets 274 marks and fails by 16 marks, find the total marks.
SOLUTION:
Given:
Total marks = x
Fails by 16 = ( x – 16 )
According to the problem,
Gets mark = 274
Less by 16 = + 16
Passing marks = 290
Passing percentage ( % ) = 40%
40% of x = 290
40 / 100 * x = 290
x = 290 – 100 / 40
x = 7x
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Q.4) Mr. Kapoor withdrew Rs. 25,000 from an ATM. If he received 150 notes in denominations of Rs. 500 and Rs. 100, find the number of notes of each denomination.
SOLUTION:
Given:
Total amount withdrew = Rs. 25,000
Total notes received = 150 notes
Let no. of denominations of Rs. 500 = x
Let no. of denominations of Rs. 25,000 = ( 150 – x )
Amount of notes of denominations of Rs. 500 = Rs. ( 500x )
Amount of notes of denominations of Rs. 100 = 100 ( 150 – x )
= Rs. ( 15000 – 100x )
According to the problem
( 500x ) + ( 15000 – 100x ) = 25000
500x + 15000 – 100x = 25000
400x = 25000 – 15000
400x = 10000
x = 10000 / 400
x = 25
Thus,
No. of notes of denomination of Rs. 500 = x = 25
No. of notes of denomination of Rs. 100 = ( 150 – x )
= ( 150 – 25 )
= 125
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Q.5) Ramesh’s father’s present age if three times Ramesh’s age. After 12 years, his father’s age will be twice his age. Find their present ages.
SOLUTION:
Given:
Let Ramesh’s present age = x years
Ramesh’s father’s age = 3x years
After 12 years,
Ramesh’s age = ( x + 12 ) years
Ramesh’s father’s age = ( 3x +12 ) years
According to the problem,
( 3x +12 ) = 2 ( x + 12 )
3x – 2x = 24 – 12
3x + 12 = 2x + 24
x = 12
Thus,
Ramesh’s age = x = 12 years
Ramesh’s father’s age = 3x = 3 * 12 = 36 years
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Q.6) If 7 / 3 of a number is 13 more than 1 / 6 of the number, find the number.
SOLUTION:
Given:
Let the number be x
According to the problem,
7x / 3 = 1x / 6 + 13
7x / 3 – 1x / 6 = 13
14x / 6 – 1x / 6 = 13
13x / 6 = 13
x = 13 * 6 / 13
x = 6
Thus,
The number is 6.
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