Selina Concise Class 6 Math Chapter 12 PROPORTION (including Word Problems) Exercise 12B Solution
EXERCISE 12B
(1) If x, y and z are in continued proportion, then which of the following is true:
(iii) x : y = y : z
(2) Which of the following numbers are in continued proportion:
(iii) 6, 12 and 24 and (iv) 12, 18 and 27
(3) Find the mean proportional between
(4) Find the third proportional to:
(i) Let the third proportional be x.
∴ 36 : 18 = 18 : x
⇒ 36x = 18 × 18
⇒ x = 9
Hence, the required third proportional is 9.
(ii) Let the third proportional be x.
∴ 5.25 : 7 = 7 : x
⇒ 5.25x = 7 × 7
⇒ x = 49/5.25
⇒ x = 9(1/3)
Hence, the required third proportional is 9(1/3).
(iii) Let the third proportional be Rs x.
∴ 1.60 : 0.40 = 0.40 : x
⇒ 1.60x = 0.40 × 0.40
⇒ x = 0.16/1.60
⇒ x = 0.1
Hence, the third proportional is 0.1.
(5) 7 : 5 = x : 20.50
⇒ 5x = 20.50 × 7
⇒ x = 143.50/5
⇒ x = 28.70
(6) If (4x + 3y) : (3x + 5y) = 6 : 7
(i) 7x (4x + 3y) = 6x (3x + 5y)
⇒ 28x + 21y = 18x + 30y
⇒ 28x – 18x = 30y – 21y
⇒ 10x = 9y
⇒ x/y = 9/10
⇒ x : y = 9 : 10
(ii) (4x + 3y) : (3x + 5y) = 6 : 7
⇒ (4x + 3 × 10) : (3x + 5 × 10) = 6 : 7
⇒ (4x + 30) : (3x + 50) = 6 : 7
⇒ (4x+30)/(3x+50) =6/7
⇒ 28x + 210 = 18x + 300
⇒ 28x – 18x = 300 – 210
⇒ 10x = 90
⇒ x = 9
(iii) (4x + 3y) : (3x + 5y) = 6 : 7
⇒ (4 × 27 + 3y) : (3 × 27 + 5y) = 6 : 7
⇒ (108 + 3y)/(81 + 5y) = 6/7
⇒ 756 + 21y = 486 + 30y
⇒ 30y – 21y = 756 – 486
⇒ 9y = 270
⇒ y = 30