## 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.