Chapter 11 – Ratio – Proportion
Practice set 28
1. In each example below, find the ratio of the first number to the second.
(1) 24, 56
Solution:
\(\large \frac {24}{56}\) = \(\large \frac {3\,×\,8}{7\,×\,8}\)
∴ \(\large \frac {24}{56}\) = \(\large \frac {3}{7}\)
(2) 63, 49
Solution:
\(\large \frac {63}{49}\) = \(\large \frac {9\,×\,7}{7\,×\,7}\)
∴ \(\large \frac {63}{49}\) = \(\large \frac {9}{7}\)
(3) 52, 65
Solution:
\(\large \frac {52}{65}\) = \(\large \frac {4\,×\,13}{5\,×\,13}\)
∴ \(\large \frac {52}{65}\) = \(\large \frac {4}{5}\)
(4) 84, 60
Solution:
\(\large \frac {84}{60}\) = \(\large \frac {7\,×\,12}{5\,×\,12}\)
∴ \(\large \frac {24}{56}\) = \(\large \frac {7}{5}\)
(5) 35, 65
Solution:
\(\large \frac {35}{65}\) = \(\large \frac {7\,×\,5}{13\,×\,5}\)
∴ \(\large \frac {35}{65}\) = \(\large \frac {7}{13}\)
(6) 121, 99
Solution:
\(\large \frac {121}{99}\) = \(\large \frac {11\,×\,11}{9\,×\,11}\)
∴ \(\large \frac {121}{99}\) = \(\large \frac {11}{9}\)
2. Find the ratio of the first quantity to the second.
(1) 25 beads, 40 beads
Solution:
\(\large \frac {25\, beads}{40\, beads}\) = \(\large \frac {5\,×\,5}{8\,×\,5}\)
∴ \(\large \frac {25 \,beads}{40\, beads}\) = \(\large \frac {5}{8}\)
(2) 40 rupees, 120 rupees
Solution:
\(\large \frac {40\, rupees}{120 \,rupees}\) = \(\large \frac {1\,×\,40}{3\,×\,40}\)
∴ \(\large \frac {40 \,rupees}{120 \,rupees}\) = \(\large \frac {1}{3}\)
(3) 15 minutes, 1 hour
Solution:
1 hour = 60 minutes
\(\large \frac {15 \,minutes}{60 \,minutes}\) = \(\large \frac {1\,×\,15}{4\,×\,15}\)
∴ \(\large \frac {15\, minutes}{1\, hour}\) = \(\large \frac {1}{4}\)
(4) 30 litres, 24 litres
Solution:
\(\large \frac {30\, litres}{24\, litres}\) = \(\large \frac {5\,×\,6}{4\,×\,6}\)
∴ \(\large \frac {30\, litres}{24\, litres}\) = \(\large \frac {5}{4}\)
(5) 99 kg, 44000 grams
Solution:
1000 grams = 1 kg
∴ 44000 grams = 44 kg
\(\large \frac {99\, kg}{44 \,kg}\) = \(\large \frac {9\,×\,11}{4\,×\,11}\)
∴ \(\large \frac {99 \,kg}{44000 \,grams}\) = \(\large \frac {9}{4}\)
(6) 1 litre, 250 ml
Solution:
1 litre = 1000 ml
\(\large \frac {1000\, ml}{250\, ml}\)\(\large \frac {4\,×\,250}{1\,×\,250}\)
∴ \(\large \frac {1\, litre}{250 \,ml}\) = \(\large \frac {4}{1}\)
(7) 60 paise, 1 rupee
Solution:
1 rupee = 100 paise
\(\large \frac {60\, paise}{100 \,paise}\) = \(\large \frac {3\,×\,20}{5\,×\,20}\)
∴ \(\large \frac {60 \,paise}{1\, rupee}\) = \(\large \frac {3}{5}\)
(8) 750 grams, \(\large \frac {1}{2}\) kg
Solution:
1 kg = 1000 grams
∴ \(\large \frac {1}{2}\) kg = 500 grams
\(\large \frac {750\, grams}{500\, grams}\) = \(\large \frac {3\,×\,250}{2\,×\,250}\)
∴ \(\large \frac {750 \,grams}{\frac {1}{2} \,kg}\) = \(\large \frac {3}{2}\)
(9) 125 cm, 1 metre
Solution:
1 metre = 100 cm
\(\large \frac {125 \,cm}{100 \,cm}\)\(\large \frac {5\,×\,25}{4\,×\,25}\)
∴ \(\large \frac {125\, cm}{1\, metre}\) = \(\large \frac {5}{4}\)
3. Reema has 24 notebooks and 18 books. Find the ratio of notebooks to books.
Solution:
Number of notebooks = 24
Number of books = 18
Ratio of notebooks to books
= \(\large \frac {24}{18}\)
= \(\large \frac {4\,×\,6}{3\,×\,6}\)
= \(\large \frac {4}{3}\)
Ans: The ratio of notebooks to books of Reema is \(\large \frac {4}{3}\).
4. 30 cricket players and 20 kho-kho players are training on a field. What is the ratio of cericket players to the total number of players?
Solution:
Number of cricket players = 30
Number of kho-kho players = 20
Total number of players
= Cricket players + Kho-kho players
= 30 + 20
= 50
Ratio of cricket players to the total number of players
= \(\large \frac {30}{50}\)
= \(\large \frac {3\,×\,10}{5\,×\,10}\)
= \(\large \frac {3}{5}\)
Ans: The ratio of cricket players to the total number of players is \(\large \frac {3}{5}\).
5. Snehal has a red ribbon that is 80 cm long and a blue ribbon, 2.20 m long. What is the ratio of the length of the red ribbon to that of the blue ribbon?
Solution:
1 metre = 100 cm
Length of the red ribbon = 80 cm
Length of the blue ribbon
= 2.20 m
= 2.20 × 100 cm
= 220 cm
∴ Ratio of length of the red ribbon to that of the blue ribbon
= \(\large \frac {80}{220}\)
= \(\large \frac {4\,×\,20}{11\,×\,20}\)
= \(\large \frac {4}{11}\)
Ans: The ratio of the length of the red ribbon to that of the blue ribbon is \(\large \frac {4}{11}\).
6. Shubham’s age today is 12 years and his father’s is 42 years. Shm’s mother is younger than his father by 6 years. Find the following ratios.
(1) Ratio of Shubham’s age today to his mother’s age today.
(2) Ratio of Shubham’s mother’s age today to his father’s age today
(3) The ratio of Shubham’s age to his mother’s age when Shubham was 10 years old.
Solution:
Shubham’s age today = 12 years
Shubham’s father’s age = 42 years
Shubham’s mother age
= Shubham’s father’s age – 6 years
= 42 years – 6 years
= 36 years
(1) Ratio of Shubham’s age today to his mother’s age today
= \(\large \frac {12}{36}\)
= \(\large \frac {1\,×\,12}{3\,×\,12}\)
= \(\large \frac {1}{3}\)
Ans: The ratio of Shubham’s age today to his mother’s age today is \(\large \frac {1}{3}\).
(2) Ratio of Shubham’s mother age today to his father’s age today
= \(\large \frac {36}{42}\)
= \(\large \frac {6\,×\,6}{7\,×\,7}\)
= \(\large \frac {6}{7}\)
Ans: The ratio of Shubham’s mother’s age today to his father’s age today is \(\large \frac {6}{7}\).
(3) Shubham’s age today is 12 years and his mothers age is 36 years.
So, when Shubham’s age was 10 years, his mother’s age was 36 – 2 years = 34 years
Ratio of Shubham’s age to his mother’s age when Shubham was 10 years old
= \(\large \frac {10}{34}\)
= \(\large \frac {5\,×\,2}{17\,×\,2}\)
= \(\large \frac {5}{17}\)
Ans: The ratio of Shubham’s age to his mother’s age when Shubham was 10 years old is \(\large \frac {5}{17}\).
Practice set 29
(1) If 20 metres of cloth cost ₹ 3600, find the cost of 16 m of cloth.
Solution:
Cost of 20 metres of cloth = Rs 3600
∴ Cost of 1 metre of cloth
= \(\large \frac {Cost of 20 metres of cloth}{20}\)
= \(\large \frac {3600}{20}\)
= Rs 180
∴ Cost of 16 metres of cloth
= Cost of 1 metre of a cloth × 16
= 180 × 16
= Rs 2880
Ans: The cost of 16 metres of cloth is Rs 2880.
(2) Find the cost of 8 kg of rice, if the cost of 10 kg is ₹ 325.
Solution:
Cost of 10 kg rice = Rs 325
∴ Cost of 1 kg rice
= \(\large \frac {Cost of 10 kg rice}{10}\)
= \(\large \frac {325}{10}\)
= Rs 32.5
Cost of 8 kg rice
= Cost of 1 kg rice × 8
= 32.5 × 8
= Rs 260
Ans: The cost of 8 kg rice is Rs 260.
(3) If 14 chairs cost ₹ 5992, how much will have to be paid for 12 chairs?
Solution:
Cost of 14 chairs = Rs 5992
∴ Cost of 1 chair
= \(\large \frac {Cost of 14 chair}{14}\)
= \(\large \frac {5992}{14}\)
= Rs 428
∴ Cost of 12 chairs
= Cost of 1 chair × 12
= 428 × 12
= Rs 5136
Ans: The amount to be paid for 12 chairs is Rs 5136.
(4) The weight of 30 boxes is 6 kg. What is the weight of 1080 such boxes?
Solution:
Weight of 30 boxes = 6 kg
∴ Weight of 1 box
= \(\large \frac {Weight of 30 boxes}{30}\)
= \(\large \frac {6}{30}\)
= 0.2 kg
∴ Weight of 1080 boxes
= Weight of 1 box × 1080
= 0.2 × 1080
= 216 kg
Ans: The weight of 1080 boxes is 216 kg.
(5) A car travelling at a uniform speed covers a distance of 165 km in 3 hours. At that same speed,
(a) How long will it take to cover a distance of 330 km?
(b) How far will it travel in 8 hours?
Solution:
Distance covered in 3 hours = 165 km
Distance covered in 1 hour
= \(\large \frac {Distance covered in 3 hours}{3}\)
= \(\large \frac {165}{3}\)
= 55 km
(a) Time required to covered a distance of 330 km
= \(\large \frac {Total distance}{Distance covered in 1 hour}\)
= \(\large \frac {330}{55}\)
= 6 hours
Ans: The time required to cover a distance of 330 km is 6 hours.
(b) Distance traveled in 8 hours
= Distance covered in 1 hour × 8
= 55 × 8
= 440 km
Ans: The distance traveled in 8 hours is 440 km.
(6) A tractor uses up 12 litres of diesel while ploughing 3 acres of land. How much diesel will be needed to plough 19 acres of land?
Solution:
Diesel required to plough 3 acres of land = 12 litres
∴ Diesel required to plough 1 acre of land
= \(\large \frac {Diesel required to plough 3 acres of land}{3}\)
= \(\large \frac {12}{3}\)
= 4 liters
∴ Diesel required to plough 19 acres of land
= Diesel required to plough 1 acre of land × 19
= 4 × 19
= 76 litres
Ans: Diesel needed to plough 19 acres of land is 76 litres.
(7) At a sugar factory, 5376 kg of sugar can be obtained from 48 tonnes of sugarcane. If Savitatai has grown 50 tonnes of sugarcane, how much sugar will it yield?
Solution:
Sugar obtained from 48 tonnes of sugarcane = 5376 kg
∴ Sugar obtained from 48 tonnes of sugarcane
= \(\large \frac {Sugar obtained from 48 tonnes of sugarcane}{48}\)
= \(\large \frac {5376}{48}\)
= 112 kg
∴ Sugar obtained from 50 tonnes of sugarcane
= Sugar obtained from 1 tonne of sugarcane × 50
= 112 × 50
= 5600 kg
Ans: 50 tonnes of sugarcane will yield 5600 kg of sugar.
(8) In an orchard, there are 128 mango trees in 8 rows. If all the rows have an equal number of trees, how many trees would there be in 13 rows?
Solution:
Number of mango trees in 8 rows = 128
Number of mango trees in 1 row
= \(\large \frac {Number of mango trees in 8 rows}{8}\)
= \(\large \frac {128}{8}\)
= 16 trees
∴ Number of mango trees in 13 rows
= Number of mango trees in 1 row × 13
= 16 × 13
= 208 trees
Ans: The number of mango trees in 13 rows are 208.
(9) A pond in a field holds 120000 litres of water. It costs 18000 rupees to make such a pond. How many ponds will be required to store 480000 litres of water, and what would be the expense?
Solution:
Capacity of 1 pond = 1,20,000 litres
Total quantity of water = 4,80,000 litres
∴ Number of ponds required
= \(\large \frac {Total quantity of water}{Capacity of 1 pond}\)
= \(\large \frac {480000}{120000}\)
= 4
Amount required to make 1 pond = Rs 18,000
∴ Amount required to make 4 ponds
= Amount required to make 1 pond × 4
= 18,000 × 4
= Rs 72,000
Ans: The number of ponds required to store 4,80,000 litres of water is 4, and the expense incurred in making the ponds is Rs 72,000.