Linear Equations Made Simple : With Word Problems For Class 5th,6th,7th,8th,9th,10th[1]

LINEAR EQUATIONS
WORD PROBLEMS FOR
CLASS 5TH T0 10TH





Q.1) Ina stationery shop, the cost of a pen is Rs.5 less then the cost of a notebook. Jasmeet bought 5 pens and 3 notebooks from the shop. If she paid Rs.127 to the shopkeeper, what is the cost of each item ?

SOLUTION:

Given:
Find the cost of each item = ?
Let the cost of a notebook be Rs. x
Let the cost of a pen = Rs. ( x – 5 )
No. of notebooks = 3
Hence, cost of 3 notebooks = 3x

No. of pens = 5 Hence, cost of 5 pens = 5 ( x – 5 )
= 5x – 25
Total amount paid by Jasmeet = Rs. 127

According to the question,

3x + ( 5x – 25 ) = 127
3x + 5x – 25 = 127
8x = 127 + 25
8x = 152
x = 152 / 8
x = 19

ANS: Cost of a notebook = x = Rs.19
Cost of a pen = x – 5 = 19 – 5
= Rs. 14
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Q.2) Mr. Rana bought a rectangular plot whose length is 4m more than thrice the breadth. If the perimeter of the plot is 1128m, find the dimensions of the rectangular plot. Also find the cost of cementing the plot at Rs.1000 per 100 m2 .

SOLUTION:

Given:
Find the dimensions = ?
Cost of cementing = ?
Let the breadth of the plot = x m
Let the length of the plot = ( 3x + 4 ) m
Perimeter = 1128 m

STEP – 1) Perimeter = 1128 m
2 ( L + B ) = 1128
2 [ ( 3x + 4 ) + x ] = 1128
Hence, 3x + 4 + x = 1128 / 2
4x + 4 = 564
4x = 564 – 4
4x = 560
x = 560 / 4
x = 140

Hence, Breadth = x = 140 m
Length = 3x + 4 = 420 + 4
= 3 * 140 + 4
= 420 + 4
= 424 m

STEP – 2) Cost of cementing per 100 m2 = Rs.1000
Hence, Area of plot = L * B = 140 * 424
= 59360 m2

Cost of cementing per 100 m2 —- Rs.1000
Cost of cementing – 59360 m2 —- ?

= 59360 * 1000 / 100
= Rs. 5,93,600
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Q.3) The adjacent angles of a parallelogram are in the ratio 7 : 5. Find the measure of each of these angles.

SOLUTON:

Given:
Find the measures of adjacent angles = ?
Ratio of the adjacent = 7 : 5
Let the adjacent angles be 7x , 5x

Adjacent angle of a parallelogram supplementary
7x + 5x = 180 12x = 180
x = 15 x = 180 / 12

ANS: The requires adjacent angles are:
7x = 7 * 15 = 105°
5x = 5 * 15 = 75°
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Q.4) Shreya collected a sum of Rs.150 in a piggy bank. She had put into it only Rs.2 , Rs. 5 and Rs.10 coins. If Rs.10 and Rs.5 coins are in the ratio 3 : 7 and there are a total of 30 coins, find the number of coins of each type.

SOLUTON:

Given:
Find the coins of each type
Total amount = Rs.150
Total coins = 30
Types of coins = Rs.2, Rs.5, Rs.10
Ratio of Rs.10 coins and Rs.5 coins = 3 : 7

Hence, Let the no. of Rs.10 coins = 3x
Rs.5 coins = 7x

The total no. of coins of Rs.10 and Rs.5 = 3x + 7x
= 10x

No. of coins of Rs.2 = Total coins – Total no. of coins of Rs.10 and Rs.5
= ( 30 – 10x )

Amount from Rs.10 coins = 10 * 3x = 30x
Amount from Rs.5 coins = 5 * 7x = 35x
Amount from Rs.2 coins = 2 * ( 30 – 10x )
= 60 – 20x

According to the question

30x + 35x + 60 – 20x = 150
45x + 60 = 150
45x = 150 – 60
45x = 90
x = 90 / 2
x = 2

Hence, No. of Rs.10 coins = 3x = 3 * 2 = 6
No. of Rs.5 coins = 7x = 7 * 2 = 14
No. of Rs.2 coins = 30 – 10x
= 30 – 10 * 2
= 30 – 20
= 10
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