NCERT Solutions Class 8th Maths Chapter 11 Direct and Indirect proportions Exercise 11.1

NCERT Solutions Class 8th Maths Chapter 11 Direct and Indirect proportions

TextbookNCERT
Class 8th
Subject Mathematics
Chapter11th
Chapter NameDirect and Indirect proportions
CategoryClass 8th Maths Solutions 
Medium English
SourceLast Doubt

NCERT Solutions Class 8th Maths Chapter 11 Direct and Indirect proportions

Chapter – 11

Direct and Indirect proportions

Exercise 11.1

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

Ch 13 13.1

Check if the parking charges are in direct proportion to the parking time.
Solution:
Charges per hour:
C1 = 60/4 = Rs. 15
C2 = 100/8 = Rs. 12.50
C3 =  140/12 = Rs. 11.67
C4 = 180/24  = Rs.7.50
Here, the charges per hour are not same, i.e., C1 ≠ C2 ≠ C3 ≠ C4
Therefore, the parking charges are not in direct proportion to the parking time.

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.

Ch 13 13.1

Solution:

Ch 13 13.1
Let the ratio of parts of red pigment and parts of base be a/b .
Case 1: Here, a1 = 1, b1 = 8
a1/b1 = 1/8 = k (say)
Case 2: When  a2 = 4 , b2 = ?
b2 = a2/k = 4/(1/8) = 4×8 = 32
Case 3: When  a3 = 7 , b3 = ?
b3 = a3/k = 7/(1/8) = 7×8 = 56
Case 4: When a4 = 12 , b4 =?
b4 = a4/k = 12/(1/8) = 12×8 = 96
Case 5: When  a5 = 20 , b5 = ?
b5 = a5/k = 20/(1/8) = 20×8 = 160
Combine results for all the cases, we have

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?

Solution:
Let the parts of red pigment mix with 1800 mL base be  x.

Ch 13 13.1

Since it is in direct proportion.

Ch 13.1

Hence with base 1800 mL, 24 parts red pigment should be mixed.

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

Ch 13 13.1

Solution:
Let the number of bottles filled in five hours be x.
Here ratio of hours and bottles are in direct proportion.

Ch 13.1

6x = 5×840
x = 5×840/6 = 700

Hence machine will fill 700 bottles in five hours.

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

Solution:
Let enlarged length of bacteria be  x .
Actual length of bacteria = 5/50000 = 1/10000  cm = 10-4  cm

Ch 13 13.1

Here length and enlarged length of bacteria are in direct proportion.

ch 13 13.1

Hence the enlarged length of bacteria is 2 cm.

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 if the ship is 28 m, how long is the model ship?

Ch 13 13.1

Solution:
Let the length of model ship be  x .

Ch 13 13.1

Here length of mast and actual length of ship are in direct proportion.

Ch 13 13.1

Hence length of the model ship is 21 cm.

7. Suppose 2 kg of sugar contains 9×106 crystals. How many sugar crystals are there in

(i) 5 kg of sugar? (ii) 1.2 kg of sugar?

Solution:
(i) Let sugar crystals be  x.

Ch 13 13.1

Here, weight of sugar and number of crystals are in direct proportion.

Ch 13 13.1

Hence the number of sugar crystals is

Ch 13 13.1

(ii)  Let sugar crystals be  x.

Ch 13 13.1

Here weight of sugar and number of crystals are in direct proportion.

Ch 13 13.1

Hence the number of sugar crystals is  5.4×106

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?

Solution:
Let distance covered in the map be  x.

Ch 13 13.1

Here actual distance and distance covered in the map are in direct proportion.

Ch 13 13.1

Hence distance covered in the map is 4 cm.

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 5 m long.

Solution: Here height of the pole and length of the shadow are in direct proportion.
And 1 m = 100 cm
5 m 60 cm = 5×100+60 = 560 cm
3 m 20 cm = 3×100+20 = 320 cm
10 m 50 cm = 10×100+50 = 1050 cm
5 m = 5×100 = 500 cm

(i) Let the length of the shadow of another pole be  x.

Ch 13 13.1

Ch 13 13.1

Hence length of the shadow of another pole is 6 m.

(ii) Let the height of the pole be  x.

Ch 13 13.1

Ch 13 13.1

Hence height of the pole is 8 m 75 cm.

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

Solution: Let distance covered in 5 hours be x km.
1 hour = 60 minutes
Therefore, 5 hours = 5×60 = 300 minutes

Ch 13 13.1

Here distance covered and time in direct proportion.

Ch 13 13.1

Therefore, a truck can travel 168 km in 5 hours.

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