Page 154 - ICSE Math 7

⇒ 2{(3x + 3) + (x – 2)} = 90
90
⇒ 2{4x + 1} = 90 ⇒ 4x + 1 = 2
⇒ 4x + 1 = 45

⇒ 4x = 44 ⇒ x = 11
Hence, width = 11 cm; length = 3x – 4 = 29 cm

Example 9: Sohail gave one-third of his income to his wife, one-fifth to his son and donated
` 5,000. If he is left with ` 6,200, find Sohail’s income.
Solution: Let Sohail’s income be ` x.
x x
So, his wife gets ` and son gets ` .
3 5
According to the question,
x x
+ + 5,000 + 6,200 = x
3 5
x
x
⇒ 11,200 = x – – (Using the rule of transposition)
3 5
x x
⇒ x – – = 11,200 (Interchanging the sides)
3 5
15x – 5x – 3x
⇒ = 11,200
15

7x
⇒ = 11,200
15
⇒ 7x = 11,200 × 15

1,68,000
⇒ x = = 24,000
7
\ Sohail’s income is ` 24,000.

Example 10: How many kilograms of sugar at ` 70 per kilogram should be mixed with 30 kilograms
of sugar at ` 90 per kilogram to get a mixture of ` 80 per kilogram?

Solution: Let the required amount of sugar be x kilograms.
Cost of x kilograms of sugar at ` 70 per kilogram = ` 70x
Cost of 30 kilograms of sugar at ` 90 per kilogram = ` (30 × 90) = ` 2,700

\ Total cost of (x + 30) kilograms of sugar = ` 70x + ` 2,700 = ` (70x + 2,700)
Cost of (x + 30) kilograms of sugar at the rate of ` 80 per kilogram = ` (x + 30)(80)
= ` (80x + 2,400)
According to the question,

80x + 2,400 = 70x + 2,700
⇒ 80x – 70x = 2,700 – 2,400 (Using the rule of transposition)
⇒ 10x = 300

300
⇒ x = = 30
10
Thus, the required amount of sugar is 30 kg.

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