# ML Aggarwal Direct and Inverse Variation Exe-9.1 Class 8 ICSE Ch-9 Maths Solutions

ML Aggarwal Direct and Inverse Variation Exe-9.1 Class 8 ICSE Ch-9 Maths Solutions. We Provide Step by Step Answer of Exe-9.1 Questions for Direct and Inverse Variation as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.

**ML Aggarwal Direct and Inverse Variation Exe-9.1 Class 8 ICSE Maths Solutions**

Board | ICSE |

Publications | Avichal Publishig Company (APC) |

Subject | Maths |

Class | 8th |

Chapter-9 | Direct and Inverse Variation |

Writer | ML Aggarwal |

Book Name | Understanding |

Topics | Solution of Exe-9.1 Questions |

Edition | 2023-2024 |

**Direct and Inverse Variation Exe-9.1**

ML Aggarwal Class 8 ICSE Maths Solutions

Page-152

**Question 1. Observe the following tables and find if x and y are directly proportional:**

(i)

x | 5 | 8 | 12 | 15 | 18 | 20 |

y | 15 | 24 | 36 | 60 | 72 | 100 |

(ii)

x | 3 | 5 | 7 | 9 | 10 |

y | 9 | 15 | 21 | 27 | 30 |

**Answer:**

**(i) x/y = 5/15 = 1/3**

x/y = 8/24 = 1/3

x/y = 12/36 = 1/3

But, x/y = 15/60 = 1/4

x/y = 18/72 = 1/4

x/y = 20/100 = 1/5 which is not equal to 1/3

So, x/y is not constant.

Therefore, x and y are not directly proportional.

**(ii) x/y = 3/9 = 1/3**

x/y = 5/15 = 1/3

x/y = 7/21 = 1/3

x/y = 9/27 = 1/3

x/y = 10/30 = 1/3

so, x/y is constant.

Therefore, x and y are directly proportional.

**Direct and Inverse Variation Exe-9.1**

ML Aggarwal Class 8 ICSE Maths Solutions

Page-153

**Question 2. If x and y are in direct variation, complete the following tables:**

(i)

x | 3 | 5 | … | … | 10 |

y | 45 | … | 90 | 120 | ….. |

(ii)

x | 4 | 8 | … | 20 | 28 |

y | 7 | … | 21 | …… | ….. |

**Answer:**

**(i) x/y = 3/45 = 1/15**

x/y is constant with 1/15.

Hence, x and y are directly proportional.

Now,

x_{1}/y_{1} = 1/15 => 5/y_{1} = 1/15

y_{1} = 5 × 15 = 75

x_{2}/y_{2} = x_{2}/90 = 1/15

x_{2} = 90/15 = 6

x_{3}/y_{3} = x_{3}/120 = 1/15

x_{3} = 120/15 = 8

x_{4}/y_{4} = 10/y_{4} = 1/15

y_{4} = 10 × 15 = 150

x | 3 | 5 | 6 | 8 | 10 |

y | 45 | 75 | 90 | 120 | 150 |

**(ii)** **x/y = 4/7**

x/y is constant with 4/7.

Hence, x and y are directly proportional.

x_{1}/y_{1} = 8/y_{1} = 4/7

y_{1} = (8 × 7)/4 = 14

x_{2}/y_{2} = x_{2}/21 = 4/7

x_{2} = (21 × 4)/7 = 12

x_{3}/y_{3} = 21/y_{3} = 4/7

y_{3} = (21 × 7)/4 = 35

x_{4}/y_{4} = 28/y_{4} = 4/7

y_{4} = (28 × 7)/4 = 49

x | 4 | 8 | 12 | 20 | 28 |

y | 7 | 14 | 21 | 35 | 49 |

**Question 3. If 8 metres cloth costs ₹250, find the cost of 5·8 metres of the same cloth.**

**Answer:**

Let the cost of 5.8 m cloth = Rs. x

length in m=8, 5.8

cost of cloth (Rs.) =250 , x

hence it is the case of direct variation

Therefore 8:250 = 5.8:x

8/250 = 5.8/x

x = (5.8 × 250)/8

= ₹181.25

**Question 4. If a labourer earns ₹672 per week, how much will he earn in 18 days?**

**Answer:**

Let the labourer earns Rs. x in 18 days

days= 7, 18

Money earned (in Rs.)=672,x

Hence in the case of direct variation

So, 7: 672 = 18: x

7/672 = 18/x

x = (18 × 672)/7

= ₹1728

**Question 5. If 175 dollars cost ₹7350, how many dollars can be purchased in ₹24024?**

**Answer:**

Let x dollars be purchased in Rs.24024

Cost (in Rs.)=7350,24024

Dollars=175,x

Hence the question is of direct variation.

So, 7350: 175 = 24024: x

7350/175 = 24024/x

x = (24024 × 175)/7350

= 572 Dollars

**Question 6. If a car travels 67·5 km in 4·5 litres of petrol, how many kilometres will it travel in 26-4 litres of petrol?**

**Answer:**

Let car travels x km in 26.4 lts. of petrol.

Distance (in km) =x , 67.5

Petrol (in 1)=26.4,4.5

Hence the question is of direct variation .

So, x: 26.4 = 67.5: 4.5

x/26.4 = 67.5/4.5

x = (67.5 × 26.4)/4.5

= 396 km

**Question 7. If the thickness of a pile of 12 cardboard sheets is 45 mm, then how many sheets of the same cardboard would be 90 cm thick?**

**Answer:**

Thickness of 12 cardboard = 45 mm

Let ‘x’ be the number of cardboard whose thickness is 90 cm = 900 m

So, 12: x = 45: 900

12/x = 45/900

x = (12 × 900)/45

= 240 mm

Thickness of 90 cardboard is 240 mm.

**Question 8. In a model of a ship, the mast (flagstaff) is 6 cm high, while the mast of the actual ship is 9 m high. If the length of the ship is 33 m, how long is the model of the ship?**

**Answer:**

Height of a model of ship=6 cm.

But height of actual ship=9 m.

If length of ship=33 m

Let length of model =x

Therefore 6:x=9m :33

6/x = 9/33

x = (6 × 33)/9

∴ Length of model is 22 cm.

**Question 9. The mass of an aluminum rod varies directly with its length. If a 16 cm long rod has a mass of 192 g, find the length of the rod whose mass is 105 g.**

**Answer:**

length of rod=16 cm and mass=192 kg.

if mass is 105 g,then

then let length of rod=x cm.

Therefore 16:x=192:105

16/x = 192/105

x = (16 × 105)/192

= 35/4

= 8.75 cm

∴ Length of rod = 8.75cm.

**Question 10. Anita has to drive from village A to village B. She measures a distance of 3.5 cm between these villages on the map. What is the actual distance between the villages if the map scale is 1 cm = 20 km?**

**Answer:**

distance from village A to b on the map=3.5 cm.

scale of map 1 cm.=20 km.

therefore 1:3.5=20:x

1/3.5 = 20/x

x = (20 × 3.5)/1

x = 70 km

Distance between village A and village B is 70 km.

**Question 11. A 23 m 75 cm high water tank casts a shadow 20 m long. Find at the same time:**

(i) the length of the shadow cast by a tree 9 m 50 cm high.

(ii) the height of the tree if the length of the shadow is 12 m.

**Answer:**

Height of a water tank = 23 m 75 cm

= 23 ¾ m = 95/4 m.

Its shadow = 20 m

**(i)** **If height of a tree = 9 m 50 cm**

= 9 ½ m

= 19/2 m

Now, let its shadow be ‘x’ m

So, 95/4 : 19/2 :: 20: x

(95/4) × (2/19) = 20/x

5/2 = 20/x

x = (2 × 20)/5

= 8 m

∴ Its shadow = 8 m

**(ii)** **Let the height of a tree be ‘x’ m**

and its shadow = 12 m

95/4 : x :: 20: 12

95/4x = 20/12

4x = (95 × 12)/20

x = 57/4 m

x =14 m 25 cm

Height of the tree is14 m 25 cm.

**Question 12. If 5 men or 7 women can earn ₹525 per day, how much would 10 men and 13 women will earn per day**

**Answer:**

Since in a day 5 men can earn Rs 525

Therefore in a day, 1 man can earn =Rs 525/5 =Rs. 105

Therefore in a day , 10 men will earn Rs 105 x 10 = Rs 1050

as in a day 7 women can earn Rs 525

Therefore in a day 1 woman can earn Rs 525

Therefore in a day , 1 women can earn Rs 525/7=Rs.75

In a day 13 women will earn Rs 75 x 13 = rs 975

Hence, Total earning of 7 men and 13 women in a day=Rs(1050+975)

=Rs 2025