Page 151 - ICSE Math 6

(g) 7y = –28 (h) –4x = –48 (i) 4(x + 1) + 2x = 3
(j) 5(1 – x) = 3x – 10 (k) 10(2y – 2) = 3y – 3 (l) 4(z – 2) + 3(2 – z) = 8(1 – z)

(m) 6(z + 3) + 4(2 – z) = 5z (n) –5(x – 1) – 2(x + 5) = 3(x + 5)
4. Solve the following and verify the solution.

(a) 3(z + 2) = 6 (b) 2(x + 5) = 10 (c) 9(z – 3) = 8z
5 5
(d) z + 3 = 2(5 + z) (e) 4(y – 4) = 3(y + 6) (f) ( + 5)x =
12 6
–2 1 4 1 1 y –10 7
(g) x +2 = (h) (7 – 2)x = (i) =
3 3 3 5 4 5 25

Word Problems

To solve word problems, form an equation according to the given conditions in an unknown variable.
Solve the equation formed to find the value of the variable.

Example 5: A number when decreased by 12 gives 30. Find the number.
Solution: Let the required number be x.
As the required number decreased by 12 gives 30.
\ x – 12 = 30
⇒ x = 30 + 12 = 42

Thus, the required number is 42.
Example 6: Twice of a number after increasing by 5 equals 200. Find the number.

Solution: Let the required number be x.
As twice of number after increasing by 5 equals 200.

\ 2(x + 5) = 200
2(x + 5) 200
⇒ = (Dividing both the sides by 2)
2 2
⇒ x + 5 = 100
⇒ x = 100 – 5 = 95
Thus, the required number is 95.

Example 7: The sum of three consecutive numbers is 63. Find the numbers.
Solution: Let the smallest of the three numbers be x. Then, the other two consecutive numbers are
x + 1 and x + 2.
As sum of the three consecutive numbers is 63.
\ x + (x + 1) + (x + 2) = 63
⇒ 3x + 3 = 63
⇒ 3x = 63 – 3 = 60
3x 60
⇒ = (Dividing both the sides by 3)
3 3
⇒ x = 20

As x = 20, therefore x + 1 and x + 2 are 21 and 22 respectively.
Thus, the required numbers are 20, 21 and 22.

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