Class 8 Maths


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


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 pigment1471220
Parts of base8????

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`


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`


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`


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`


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`


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


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

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