Exercise 11.1

1. Following are the car parking charges near a railway station upto:

4 hours → 60

8 hours → 100

12 hours → 140

24 hours → 180

Check if the parking charges are in direct proportion to the parking time.

Instructions

Calculating charges per hour: C1 = 604 = Rs.

Similarly, C2 = 1008 = Rs. ; C3 = 14012 = Rs. ; C4 = 18024 = Rs.

Thus, the charges per hour the same as C1

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.

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

Instructions

Let the ratio of parts of red pigment and parts of the base be ab.

Case 1: Here, a1 = 1, b1 = 8. So, a1b1 = (assume to be 'k')

Case 2: When a2 = 4, b2 = ? ⇒ b2 = a2k = 418 = × =

Case 3: When a3 = 7, b3 = ? ⇒ b3 = a3k = 718 = × =

Case 4: When a4 = 12, b4 = ? ⇒ b4 = a4k = 1218 = × =

Case 5: When a5 = 20, b5 = ? ⇒ b5 = a5k = 2018 = × =

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?

Instructions

Let the parts of red pigment mix with 1800 mL base be x. Since it is in direct proportion, 175 =

So, 75x = 1800 i.e. x = 180075 =

Hence, with the base 1800 mL, 24 parts of the 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?

Instructions

Let the number of bottles filled in five hours be x. Here, the ratio of hours and bottles is in direct proportion.

So, 6840 = which gives us: = 5 ×

Thus, x = 5×8406 =

Hence, the 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?

Instructions

Let the enlarged length of bacteria be x.

Actual length of bacteria = = cm = cm

Here, the length and enlarged length of bacteria are in direct proportion. So, 550000 = x20000 ⇒ x = 5 × 20000050000 = cm.

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

Instructions

Let the length of the model ship be x. Here, the length of the mast and the actual length of the ship are in direct proportion.

We have: 129 = which = 9 ×

So, x = 9×2812 = cm

Hence, the 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?

Instructions

(i) Let sugar crystals be x. Using proportions concept: we can say = 5x

Further on: x = 9×106×52 = × 106 =

Hence, the number of sugar crystals is 2.25 × 107.

(ii) Let sugar crystals be x. Here, the weight of sugar and the number of crystals are in direct proportion.

So, 29×106 =

x = 9×106×1.22 = × 106

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?

Instructions

Let the distance covered in the map be x. Here, the actual distance and distance covered in the map are in direct proportion.

So, 181 = which gives: = 1 ×

Thus, x = 7218 = cm

Hence, the distance covered on 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 5m long.

Instructions

Here, the height of the pole and the length of the shadow are in direct proportion.

We know: 1 m = cm

So, 5 m 60 cm = × 100 + = cm and 5 m = × 100 = cm

3 m 20 cm = × 100 + = cm and 10 m 50 cm = × 100 + = cm

(i) Let the length of the shadow of another pole be x. Thus, = 1050x which gives us: = × 320

x = 1050×320560 = cm = m

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

(ii) Let the height of the pole be x. Thus, 560320 =

This gives us: = 560 ×

So, x = 560×500320 = cm = m cm

Hence, the 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?

Instructions

Let the distance covered in 5 hours be x km. We know: 1 hour = minutes

Therefore, 5 hours = 5 × 60 = minutes. Here, the distance covered and time are in direct proportion.

So, 1425 = . This gives us: = 300(14)

x = 300×1425 =

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

Chapter 11: Direct and Inverse Proportions

Glossary

Exercise 11.1

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Chapter 11: Direct and Inverse Proportions

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Glossary

Exercise 11.1