**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.

Solution:

We have the ratio of time period and parking 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

Hence, the parking charges are not in directly proportion.

**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 | …… | …….. | ……… | ……. |

**Solution:**

Let the required field be a, b, c and d respectively.

1/8 = 4/a

So, a = 32

1/8 = 7/b

So, b = 56

1/8 = 12/c

So, c = 96

1/8 = 20/d

So, d = 160

Parts of red pigment | 1 | 4 | 7 | 12 | 20 |

Parts of base | 8 | 32 | 56 | 96 | 160 |

**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.

1/x = 75 ml/ 1800 ml

So, x = 1800/75

= 24

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

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

**Solution:**

Let the required number of bottles filled in five hours be x.

No. of bottles | Time in hours |

840 | 6 |

X | 5 |

Now,

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 en**l**arged length?**

**Solution:**

Let enlarged length of bacteria be x cm.

Enlargement | Length |

50,000 | 5 cm |

20,000 | X cm |

Now,

50,000/20,000 = 5/x

X = (5 x 20,000)/ 50,000

= 2 cm

Now,

Original length of bacteria = 5/50,000

= 1/10,000

= 10^{-4}

**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?**

**Solution:**

Let the length of the model ship be *X* m.

Height of mast | Length of ship |

9 cm | 12 m |

X cm | 28 |

Here, length of mast and actual length of ship are in directly proportion.

Now,

9/*X* = 12/28

12 *X* = 9 x 12

*X* = 21

Hence, required length of the model ship is 21 cm.

**7. Suppose 2 kg of sugar contains 9×10 ^{6} crystals. How many sugar crystals are there in**

(i)5 kg of sugar?

(ii) 1.2 kg of sugar?

**Solution:**

**(i)** Let the required number of sugar crystals be x.

No. of sugar crystals | Weight of sugar |

9 x 10^{6} | 2 kg |

x | 5 kg |

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

Now,

(9 x 10^{6}) /*x* = 2/5

2*x* = 5 x 9 x 10^{6}

x = 2.25 x 10^{7}

**(ii)** Let sugar crystals be y.

No. of sugar crystals | Weight of sugar |

9 x 10^{6} | 2 kg |

y | 1.2 kg |

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

(9 x 10^{6})/y = 1/1.2

2y = 1.2 x 9 x 10^{6}

Y = 5.4 x 9 x 10^{6}

**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 required distance covered in the map be x.

Scale in cm | Distance in km |

1 | 18 |

x | 72 |

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

1/x = 18/72

18x = 72

x = 4 cm

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:**

(i)

Let the required length of shadow be x.

Height of Pole | Length of Shadow |

5 m 60 cm = 5.60 m | 3 m 20 cm = 3.20 m |

10 m 50 cm = 10.50 m | X m |

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

Now,

5.60/10.50 = 3.20/x

x = (3.20 x 10.50)/5.60

= [(320 x 1050)/560] x [1/100]

= 6 m

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

Height of Pole | Length of shadow |

5.60 | 3.20 |

y | 5 |

5.60/y = 3.20/5

Y = (5.60 x 5)/3.20

= 35/4

= 8 m 75 cm

Hence, required 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 the required distance covered in 5 hours be x km.

Distance | Time |

14 km | 25 min |

X km | 5 hours = 300 min |

1 hour = 60 minutes

5 hours = 5×60 = 300 minutes

Here distance covered and time in directly proportion.

14/x = 25/300

X = (14 x 300)/25

x = 168 km

Hence, the required distance travelled by truck in 5 hours = 168 km

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