Selina Concise Class 7 Math Chapter 6 Ratio And Proportion Exercise 6A Solution
EXERCISE 6A
(1) Express each of the given ratios in its simplest form:
(i) 22 : 66
= (22 ÷ 22) : (66 ÷ 22)
= 1 : 3
(ii) 1.5 : 2.5
= (15/10) : (25/10)
= 15 : 25
= (15 ÷ 5) : (25 ÷ 5)
= 3 : 5
(iv) 40 kg : 1 quintal
= 40 kg : 100 kg
= (40 ÷ 20) kg : (100 ÷ 20) kg
= 2 kg : 5 kg
(v) 10 paise : Rs 1
= 10 paise : 100 paise
= (10 ÷ 10) paise : (100 ÷ 10) paise
= 1 paise : 10 paise
(vi) 200 m : 5 km
= 200 m : 5000 m
= (200 ÷ 200) m : (5000 ÷ 200) m
= 1 m : 25 m
(vii) 3 hours : 1 day
= 3 hours : 24 hours
= (3 ÷ 3) : (24 ÷ 3)
= 1 : 8
(2) Let the length of two parts are 5x and 3x cm.
∴ 5x + 3x = 64
⇒ 8x = 64
⇒ x = 8
Therefore, Length of two part is (5 × 8) = 40 cm and (3 × 8) = 24 cm.
(3) Let x get = Rs 4p and y get = Rs 5p
∴ 4p + 5p = 720
⇒ 9p = 720
⇒ p = 80
Therefore x get = Rs (4 × 80) = Rs 320 and y get = Rs (5 × 80) = Rs 400.
(4) Let the measure of triangles be 3x, 2x and 7x.
We know that all angles of a triangle is 180o.
∴ 3x + 2x + 7x = 180
⇒ 12x = 180
⇒ x = 15
Therefore, measure of required angles are (3 × 15) = 45o; (2 × 15)o = 30o and (7 × 15) = 105o.
(5) Length = 100 m and breadth = 80 m
(i) Ratio between length to breadth,
= 100 : 80
= (100 ÷ 20) : (80 ÷ 20)
= 5 ÷ 4
(ii) Its perimeter = 2 (100 + 80) = 2 × 180 = 360 m
Ratio between breadth to perimeter,
= 80 : 360
= (80 ÷ 40) : (360 ÷ 40)
= 2 : 9
(7) Let the two numbers be 3x and 4x.
Given first quantities (3x) = Rs 810
∴ 3x = 810
⇒ x = 270
Therefore second quantities is = Rs (4 × 270) = Rs 1080
(8) Let the two numbers be 5x and 7x.
Their difference is = (7x – 5x) = 10
∴ 7x – 5x = 10
⇒ 2x = 10
⇒ x = 5
Therefore the numbers are (5 × 5) = 25 and (7 × 5) = 35.
(9) Let the two numbers be 10x and 11x.
Their sum is = 10x + 11x = 168
∴ 10x + 11x = 168
⇒ 21x = 168
⇒ x = 8
The required numbers are (10 × 8) = 80 and (11 × 8) = 88.
(10) Let the divided two parts be 2.5x and 1.3x cm.
The smaller is = 1.3x = 35.1
∴ 1.3x = 35.1
⇒ x = (35.1 ÷ 1.3) = 27
The bigger is = (2.5 × 27) = 67.5 cm
Therefore the length of the line is (67.5 + 35.1) = 102.6 cm
(11) Sum of the ratio is = (7 + 8) = 15
.Therefore girls of the whole class are = 8/15.
(13) Total weight of alloy = 10 g
A metals contain = 7.5 g and B contains = (10 – 7.5) = 2.5 g
(i) Ration between A and B,
= 7.5 : 2.5
= (7.5 ÷ 2.5) : (2.5 ÷ 2.5)
= 3 : 1
(ii) Ratio between B and the total alloy
= 2.5 : 10
= (2.5 ÷ 2.5) : (10 ÷ 2.5)
= 1 : 4
(14) Age of A is = (6 × 12) + 8 = 72 + 8 = 80 months
Age of B is = (7 × 12) + 4 = 84 + 4 = 88 months
Ratio between = 80 : 88 = 10 : 11
Sum of their ratio = (10 + 11) = 21
(15) Ratio between spend amounts of three persons,
= 25000 : 15000 : 40000
= 25 : 15 : 40
= 5 : 3 : 8
Sum of their ratio = (5 + 3 + 8) = 16
The total profit = Rs 14400
(17) Let the lengths of two poles be 2x and 3x.
The height of smaller pole is = 2x = 7.5 m.
∴ 2x = 7.5
⇒ x = (7.5 ÷ 2) = 3.75
Therefore the length of the bigger pole is = (3 × 3.75) = 11.25 m.
(18) Let the two numbers be 4x and 7x
Their L.C.M = 4x × 7x = 28x
⇒ 28x = 168
⇒ x = (168 ÷ 28)
⇒ x = 6
Therefore, the required numbers are (4 × 6) = 24 and (7 × 6) = 42.
Hence, B get = Rs 200 and A gets = Rs (200/2) = Rs 100.
(i) Ratio between the shares of A and B
= 100 : 200
= 1 : 2
(ii) Share of A = Rs 100 and Share of B = Rs 200.
(20) Let the two numbers be 5x and 9x.
Their H.C.F is = 5x, 9x = x
∴ x = 16
The required numbers are (5 × 16) = 80 and (9 × 16) = 144.
(21) Let the numbers of 10 rupee notes be 2x and 20 rupee notes 3y.
Amount of 10 rupees notes = 10 × 2x = 20x
Amount of 20 rupees notes = 20 × 3x = 60x
∴ 20x + 60x = 1600
⇒ 80x = 1600
⇒ x = 20
Therefore, numbers are 10 rupees notes = (2 × 20) = 40
And numbers are 20 rupees notes = (3 × 20) = 60
Therefore the total number of notes in all = (40 + 60) = 100.
(22) Let the prices of scooter be 4x and prices of refrigerator be 1x.
∴ 4x – 1x = 45000
⇒ 3x = 45000
⇒ x = 15000
Therefore the price of the refrigerator is Rs 15000.
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