Selina Concise Class 6 Math Chapter 11 Ratio, Proportion And Arithmetic Problems Exercise 11C Solution
EXERCISE 11C
(1) Let the Hari’s and Gopi’s amount 5x and 3x respectively.
Then, 5x + 3x = 120
⇒ 8x = 120
⇒ x = 15
Therefore, Hari got = Rs (5 × 15) = Rs 75 and Gopi got = Rs (3 × 15) = Rs 45.
Now, let the number be 5x and 3x.
Then, 5x + 3x = 72
⇒ 8x = 72
⇒ x = 9
Hence, the numbers are (5 × 9) = 45 and (3 × 9) = 27.
(3) Let the numbers be 2x, 3x and 4x.
Then, 2x + 3x + 4x = 81
⇒ 9x = 81
⇒ x = 9
Therefore, the numbers are (2 × 9) = 18, (3 × 9) = 27 and (4 × 9) = 36.
(4) Ratio of A, B, C
= 6 : 4 : 3
Now, let the numbers be 6x, 4x and 3x.
Then, 6x + 4x + 3x = 10400
⇒ 13x = 10400
⇒ x = (10400 ÷ 13) = 800
Therefore A got =Rs (6 × 800) = Rs 4800
B got = Rs (4 × 800) = Rs 3200
C got = Rs (3 × 800) = Rs 2400
(5) Sum of ratio = (6 + 9 + 10) = 25
(7) Sum of the required ratio = (3 + 2 + 4) = 9
(9) Let C get be Rs x.
Then, B gets = 2x and A gets = 4x.
Then, 4x + 2x+ x = 31500
⇒ 7x = 31500
⇒ x = (31500 ÷ 7) = 4500
Therefore, A get = Rs (4 × 4500) = Rs 18000
B gets = Rs (2 × 4500) = Rs 9000
C gets = Rs 4500
(10) Let Geeta gets be Rs x, Mohits gets = 2.5x and Ashok gets = (4 ×2.5x)= 10x.
Then, 10x + 2.5x + x = 81000
⇒ 13.5 x = 81000
⇒ x = (81000 ÷ 13.5) = 6000
Therefore, Ashok gets = Rs (6000 × 10) = Rs 60000
Mohit gets = Rs (6000 × 2.5) = 15000
Geeta gets = Rs 6000