RS Aggarwal Class 8 Math Eighth Chapter Linear Equations Exercise 8C Solution

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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