# Proportion

## Exercise 13.1

Question 1: Following are the car parking charges near a railway station upto
4 hours Rs 60
8 hours Rs 100
12 hours Rs 140
24 hours Rs 180
Check if the parking charges are in direct proportion to the parking time.

Answer: The charges are not increasing in direct proportion to the parking time because

(x_1)/(y_1)≠(x_2)/(y_2)

Or, (4)/(60)≠(8)/(100)

Question 2: A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.

 Parts of red pigment 1 4 7 12 20 Parts of base 8 ? ? ? ?

Answer: k=(x_1)/(y_1)=1/8

Or, y_1=x_1xx8

Or, y_2=4xx8=32

Or, y_3=7xx8=56

Or, y_4=12xx8=96

Or, y_5=20xx8=160

Qeustion 3: In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?

Answer: x_1=1 and y_1 = 75
x_2 = ? and y_2 = 1800

(x_1)/(y_1)=(x_2)/(y_2)

Or, (1)/(75)=(x_2)/(1800)

Or, x_2=(1800)/(75)=24

Question 4: A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?

Answer: x_1=840 and y_1=6
x_2=? And y_2=5

(x_1)/(y_1)=(x_2)/(y_2)

Or, (840)/(6)=(x_2)/(5)

Or, x_2=(840xx5)/(6)=700

Alternate Method:

Since, in 6 hours the machine fills 840 bottles

Hence, in 1 hour the machine fills (840)/(6) bottles

Hence, in 5 hour the machine fills (840)/(6)xx5=700 bottles

Question 5: A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?

Answer: Original size of bacteria =(5)/(50000) cm

Size after 20,000 times enlargement =(5)/(50000)xx20000=2 cm

Question 6: In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship?

Answer: x_1= 9 cm and y_1 = ?
x_2= 12 and y_2 = 28

(x_1)/(y_1)=(x_2)/(y_2)

Or, (9)/(y_1)=(12)/(28)

Or, y_1=(9xx28)/(12)=21 cm

Question 7: Suppose 2 kg of sugar contains 9 xx 10^6 crystals. How many sugar crystals are there in (i) 5 kg of sugar? (ii) 1.2 kg of sugar?

Answer: Since 2 kg sugar contains 9xx10^6 crystals

Hence, 1 kg sugar contains (9xx10^6)/(2)=4.5xx10^6 crystals

Hence, 5 kg sugar contains 4.5xx10^6xx5=22.5xx10^6 crystals

Hence, 1.2 kg sugar contains 4.5xx10^6xx1.2=5.4xx10^6 crystals

Question 8: Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?

Answer: x_1=1 and y_1 = 18
x_2 = ? and y_2 = 72

(x_1)/(y_1)=(x_2)/(y_2)

Or, (1)/(18)=(x_2)/(72)

Or, x_2=(72)/(18)=4 cm

Question 9: A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time

(i) the length of the shadow cast by another pole 10 m 50 cm high

(ii) the height of a pole which casts a shadow 5m long.

Answer: x_1 = 5.6, x_2 = 10.5 and x_3 = ?
y_1 = 3.2, y_2 = ? and y_3 = 5

(x_1)/(y_1)=(x_2)/(y_2)=(x_3)/(y_3)

Or, (5.6)/(3.2)=(10.5)/(y_2)

Or, y_2=(10.5xx3.2)/(5.6)=6 cm

Or, (5.6)/(3.2)=(x_3)/(5)

Or, x_3=(5.6xx5)/(3.2)=8.75 m

Question 10: A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?

Answer: x_1= 14 km and y_1 = 25 minutes
x_2 = ? and y_2 = 300 minutes

(x_1)/(y_1)=(x_2)/(y_2)

Or, (14)/(25)=(x_2)/(300)

Or, x_2=(14xx300)/(25)=168 km