RS Aggarwal Class 8 Math Eighth Chapter Linear Equations Exercise 8C Solution
EXERCISE 8C
OBJECTIVE QUESTIONS
Tick (√) the correct answer in each of the following:
(1) If 2x – 3 = x + 2, then x =?
⇒ 2x – 3 = x + 2
⇒ 2x – x = 2 + 3
⇒ x = 5
Ans: (c)
⇒ 10x – 3x = – 28 – 7
⇒ 7x = – 35
⇒ x = – 5
Ans: (b)
(5) If 5t – 3 = 3t – 5, then t =?
⇒ 5t – 3 = 3t – 5
⇒ 5t – 3t = – 5 + 3
⇒ 2t = – 2
⇒ t = – 1
Ans: (b)
⇒ 12x + 8 = x – 3
⇒ 12x – x = – 3 – 8
⇒ 11x = – 11
⇒ x = – 1
Ans: (b)
(12) If 3(t – 3) = 5(2t + 1), then t =?
⇒ 3(t – 3) = 5(2t + 1)
⇒ 3t – 9 = 10t + 5
⇒ 10t + 5 = 3t – 9
⇒ 10t – 3t = – 9 – 5
⇒ 7t = – 14
⇒ t = – 2
Ans: (a)
(13) Four – fifths of a number is greater than three-fourths of the number by 4. The number is
⇒ Let the number be x.
(14) The ages of A and B are in the ratio 5 : 7. Four years from now the ratio of their ages will be 3 : 4. The present age of B is
⇒ Let the age of the A and B is 5x years and 7x years respectively.
⇒ x = 4
Therefore the age of the B = (7 × 4) = 28 years
Ans: (b)
(15) The base of an isosceles triangle is 6 cm and perimeter is 16 cm. length of each of the equal sides is
⇒ Let the equal side of the triangle be x cm.
The perimeter of the triangle = (x + x+ 6) = (2x + 6) cm
∴ 2x + 6 = 16
⇒ 2x = 16 – 6
⇒ 2x = 10
⇒ x = 5
Therefore, the length of each side = 5 cm.
Ans: (b)
(16) Sum of three consecutive integers is 51. The middle one is
⇒ Let the three consecutive integers be x, (x+ 1) and (x + 2).
∴ x + (x + 1) + (x + 2) = 51
⇒ x + x + 1 + x + 2 = 51
⇒ 3x = 51 – 3
⇒ 3x = 48
⇒ x = 16
Therefore, the middle integer = (16 + 1) = 17
Ans: (d)
(17) The sum of two numbers is 95. If one exceeds the other by 15, then the smaller of the two is
⇒ Let the number of be x and x + 15.
∴ x + x+ 15 = 95
⇒ 2x = 95 – 15
⇒ 2x = 80
⇒ x = 40
Therefore, the smaller number is 40.
Ans: (a)
(18) Number of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the number of girls. The total class strength is
⇒ Let the number of boys be x and the girls be (x – 8).
Therefore, the number of boys = 28 and the number of girls = (28 – 8) = 20.
Total strength of the class = (28 + 20) = 48.
Ans: (c)
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