Selina Concise Class 6 Math Chapter 11 Ratio, Proportion And Arithmetic Problems Exercise 11B Solution
EXERCISE 11B
(1) Salary of the person = Rs 12000
Expenditure = Rs 8500
Savings = Rs (12000 – 8500) = Rs 3500
(i) Rs (12000 : 8500)
= Rs (12000 ÷ 100) : (8500 ÷ 100)
= Rs 120 : 85
= Rs (120 ÷ 5) : (85 ÷ 5)
= Rs 24 : 17
(ii) Rs 8500 : 3500
= Rs (8500 ÷ 100) : (3500 ÷ 100)
= Rs 85 : 35
= Rs (85 ÷ 5) : (35 ÷ 5)
= Rs 17 : 7
(iii) Rs 3500 : 12000
= Rs (3500 ÷ 100) : (12000 ÷ 100)
= Rs 35 : 120
= Rs (35 ÷ 5) : (120 ÷ 5)
= Rs 7 : 24
(2) Strength of the class = 65
Number of girls = 30
Number of boys = (65 – 30) = 35
(i) 30 : 35
= (30 ÷ 5) : (35 ÷ 5)
= 6 : 7
(ii) 35 : 65
= (35 ÷ 5) : (65 ÷ 5)
= 7 : 13
(ii) 65 : 30
= (65 ÷ 5) : (30 ÷ 5)
= 13 : 6
(3) Before increasing = Rs 1500
After increasing = Rs 2250
Increase = Rs (2250 – 1500) = Rs 750
(i) Rs 750 : 1500
= Rs (750 ÷ 10) : (1500 ÷ 10)
= Rs 75 : 150
= Rs (75 ÷ 75) : (150 ÷ 75)
= Rs 1 : 2
(ii) Rs 1500 : 2250
= Rs (1500 ÷ 10) : (2250 ÷ 10)
= Rs 150 : 225
= Rs (150 ÷ 25) : (225 ÷ 25)
= Rs 6 : 9
= Rs 2 : 3
(iii) Rs 2250 : 750
= Rs (2250 ÷ 10) : (750 : 10)
= Rs 225 : 75
= Rs (225 ÷ 75) : (75 ÷ 75)
= Rs 3 : 1
(4) Reduce each of the following ratios to their lowest terms:
(i) 1 hour 20 min = (60 + 20) = 80 min
2 hours = (2 × 60) = 120 min
Then, 80 : 120
= (80 ÷ 40) : (120 ÷ 40)
= 2 : 3
(ii) 4 weeks = (4 × 7) = 28 days
Then, 28 : 49
= (28 ÷ 7) : (49 ÷ 7)
= 4 : 7
(iii) 3 years 4 months = (3 × 12) + 4 = 40 months
5 years 5 months = (5 × 12) + 5 = 65 months
Then, 40 : 65
= (40 ÷ 5) : (65 ÷ 5)
= 8 : 13
(iv) 2 m 40 cm = (2 × 100) + 40 = 240 cm
1 m 44 cm = 100 + 44 = 144 cm
Then, 240 : 144
= (240 ÷ 12) : (144 : 12)
= 20 : 12
= 10 : 6
= 5 : 3
(v) 5 kg 500 gm = (5 × 1000) + 500 = 5500 gm
2 kg 750 gm = (2 × 1000) + 750 = 2750 gm
Then, 5500 : 2750
= 550 : 275
= (550 ÷ 25) : (275 ÷ 25)
= 22 : 11
= 2 : 1
(5) Let the number be 9x and 2x.
∴ 2x = 320
⇒ x = 160
Therefore the larger number are (9 × 160) = 1440.
(7) Total time of the class = 3.30 p.m – 10 a.m = 5 hours 30 mint
Time of the class period = Total time of school – interval = 5 hrs 30 min – 30 min = 5 hours = (5 × 60) = 300 mins
Ratio between time interval and time of class period
= 30 : 300
= 1 : 10
(9) Total members of the club = 360
Number of members who played carom = 40
Number of members who played table tennis = 96
Number of members who played badminton = 144
Number of members who played volley-ball
= 360 – (40 + 96 + 144) = 360 – 280 = 80
(i) 40 : 96
= (40 ÷ 8) : (96 ÷ 8)
= 5 : 12
(ii) 144 : 96
= (144 ÷ 12) : (96 ÷ 12)
= 12 : 8
= (12 ÷ 4) : (8 ÷ 4)
= 3 : 2
(iii) 96 : 80
= (96 ÷ 8) : (80 ÷ 8)
= 12 : 10
= 6 : 5
(iv) 80 : 280
= 8 : 28
= (8 ÷ 4) : (28 ÷ 4)
= 2 : 7
(11) Let the distance of the school from A’s house be 2x km.
And the B’s house = 1x km.
∴ 2x = 4
⇒ x = 2
Therefore,
Distance (in km) from A’s house to school | 2x =4 | x = 2×9 =18 | 2x = 8 | x = 2 ×8 = 16 | 2x = 6 |
Distance (in km) from B’s house to school | x =2 | x =9 | x = 4 | x = 8 | x = 3 |
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