WORD PROBLEMS RELATED TO
LINEAR EQUATIONS
CLASS 7th, 8th, 9th AND 10th
Q1 ]. Sahil rides his bicycle at the speed of 15km /h and reaches school late by 2 minutes. On increasing the speed by 5 km / h , he reaches 8 minutes early . Find the distance between the school and his house .
Solution :
Let d be the distance between Sahil’s house and school .
And t be the time it takes to reach school on time .
Given :
1] Sahil rides at speed of 15km /hr and is 2 min late . 2 min = 2/60 = 1/30 hrs .
2] On increasing the speed by 5km / hr i.e ( 15 + 5 = 20 km /hr ) , he reaches 8 min early
8 km = 8/60 = 2 / 15 hrs .
Step 1] When Sahil is riding at 15km / hr .
Time taken = ( t + 1/ 30 ) hrs
Step 2 ] When Sahil is riding at 20 km / hr .
Time Taken = ( t – 2/15) hrs
Distance = Speed x Time => D = 15 ( t + 1/30 ) —— [1]
Distance = Speed x Time => D = 20 ( t – 2/15 ) ——[2]
3] Since the distance remains the same
=15(t+ 1/30) = 20( t- 2/15)
=15t + 15 / 30 = 20t – 40 / 15
= 15 t + 1/2 = 20t – 8/3
= 1/2 + 8/3 = 20t – 15t
= (3+ 16 )/ 6 = 5t
= 19 /6 = 5t
= 19 / 6 x 5 = 19 / 30 hrs = t
4] Substitute t = 19 /30 in equation [1]
= D = 15( t + 1 / 30)
= 15 ( 19/30 + 1/30)
= 15 x 20 / 30
= 20 / 2
D = 10 km
Ans : Thus the distance between the school and his house is 10 km .
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Q 2]. By selling a table for Rs. 4500 , a shopkeeper gains one – third of its cost price . Find the cost price .
Solution :
Let Cost Price = C.P. = Rs . x
Therefore , Gain = 1/3x and Selling price = S.P. =4500
P = S.P. – C.P.
= 1/3 x + x/ 1 = 4500
= (1x + 3x) / 3 = 4500
= 4x = 4500 x 3
= 4x = 13500
= x = 13500 / 4 = Rs . 3375
Ans : Thus , the cost price is Rs.3375 .
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Q.3] Ankita’s grandfather’s age is four times her age. Her grandfather is also 42 years older than her. Find the present age of her grandfather.
Solution:
Let Ankita’s present age = x years
Let grandfather’s present age = 4x years
Her grandfather’s present age is 42 years older than her.
According to the problem,
4x – x = 42
3x = 42
x = 42 / 3
x = 14
Ans: Ankita’s present age = x = 14 years
Grandfather’s present age = 4 * x
= 4 * 14
= 56 years
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Q.4) The denominator of a rational number if three less than the numerator. If the numerator id doubled and the denominator is increased by 15, the new number obtained is 4/ 5. Find the original rational number.
Solution:
Given:
Let the numerator be = x
Original rational no. = x / x – 3
Let the denominator be = ( x – 3 )
The number is doubled = 2x
The denominator is increased by 15 = x – 3 + 15 = ( x + 12 )
New rational no. = 2x / x + 12
According the problem,
2x / x + 12 = 4 / 5
2x * 5 = 4 ( x + 12 )
10x = 4x + 48
10x – 4x = 48
6x = 48
x = 48 / 6
x = 8
Ans: Original rational no. x / x – 3 = 8 / 8 – 3 = 8 / 5
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Q.5) The ratio of the ages of Ram and Rahim two years ago was 2 : 3. Four years from now, the ratio of their ages will be 3 : 4. Find their present ages.
Solution:
Given:
Let x be the age of Ram 2 years age = 2x years = x
Ratio of Ram to Rahim’s age 2 years ago = 2 : 3
Hence, Rahim’s age 2 years ago = 3x / 2 years
Ago —- 2 years — present — after 4 years
x x + 2 x + 2 + 4
3x / 2 ( 3x / 2 + 2 ) 3x / 2 + 2 + 4
Present age of Ram = ( x+ 2 ) years
Present age of Rahim = ( 3x / 2 + 2 ) years
Four years from now,
Ram’s age = x + 2 + 4 = ( x + 6 ) years
Rahim’s age = 3x / 2 + 2 + 4 = ( 3x / 2 + 6 ) years
Ratio of ages after 4 years = 3 : 4
According to the problem,
x + 6 / ( 3x / 2 + 6 ) = 3 / 4
4 ( x + 6 ) = 3 ( 3x / 2 + 6 )
4x + 24 = 9x + 18
24 – 18 = 9x / 2 – 4x
6 = 9x / 2 – 8x / 2
6 * 2 = x
12 = x
Present age of Ram = x + 12 = 12 + 2 = 14 years
Present age of Rahim = 3x / 2 + 2 = 3 * 12 / 2 + 2 = 18 + 2 = 20 years
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