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General suggestions for solving such problems:

• Read the statement carefully and determine what quantity is to be found.
• Represent the unknown quantity by a variable.

• Determine which expressions are equal and write an equation.

• Solve the resulting equation.

In the following examples, some more word problems are solved using the concept of linear
equations.
Example 6: In a vegetable shop, the cost of 1 kg of potatoes is `5 less than the cost of 1 kg of brinjals.
Naveen bought 5 kg potatoes and 3 kg of brinjals from the shop. If he paid `127 to the shopkeeper,
what is the cost of each item?
Solution: Let the cost of 1 kg of brinjals be `x.

∴ The cost of 1 kg of potatoes = `(x – 5).
So, cost of 5 kg potatoes = `5(x – 5) and cost of 3 kg brinjals = `3x

According to the question,
Cost of 5 kg potatoes + cost of 3 kg brinjals = `127

∴ 5(x – 5) + 3x = 127 ⇒ 5x – 25 + 3x = 127
⇒ 8x – 25 = 127 ⇒ 8x = 127 + 25

152
⇒ 8x = 152 ⇒ x = = 19
8
So, the cost 1 kg of brinjals = `19

And, the cost 1 kg of potatoes = x – 5 = 19 – 5 = `14
Example 7: Mr Shyam has a rectangular garden, whose length is 4 m more than the thrice the
breadth. If the perimeter of the garden is 1128 m, find the dimensions of the rectangular garden.
Also, find the cost of adding fertiliser in the garden at the rate of `1000 per 100 m .
2
Solution: Let the breadth of the garden be x m.

So, the length of the garden will be (3x + 4) m.
Given: The perimeter of the garden = 1128 m

∴ 2(length + breadth) = 1128 ⇒ 2[(3x + 4) + x] = 1128
⇒ 2(3x + 4 + x) = 1128 ⇒ 2(4x + 4) = 1128

1128
⇒ 4 x + 4 = ⇒ 4x + 4 = 564
2
560
⇒ 4 x = 564 – 4 = 560 ⇒ x = = 140
4
So, the breadth of the garden = 140 m

And the length of the garden = (3 × 140 + 4) m = 424 m
2
Now, area of the garden = length × breadth = (140 × 424) m = 59360 m 2
2
Thus, the cost of fertilising the garden at the rate of `1000 per 100 m = 59360 × 1000 = `5,93,600.
100

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