# Free Direct and Inverse Proportions 03 Practice Test - 8th Grade

### Question 1

Given that x and y are in inverse proportion, and y1, y2 are values of y corresponding to the values x1, x2 of x respectively. Which of the following is correct?

x1y2=x2y1

x1y1=x2y2

x1x2=y1y2

x1y1 = x2y2

#### SOLUTION

Solution :B

If x and y are in inverse proportion, and y1, y2are values of y corresponding to the values x1, x2 of x respectively, then

x1y1=x2y2

### Question 2

If a box of sweets is divided among 24 children, they will get 5 sweets each. How many would each get, if the number of children is reduced by 4?

5

6

7

9

#### SOLUTION

Solution :B

As the number of children increases, the number of sweets for each decreases. It is an inverse proportion.

Let the required number of sweets be x.

AB = constant

A1×B1 = A2×B2

5×24 = x×20

5x = 2024x = 5×2420 = 6

So, each child gets 6 sweets.

### Question 3

In the equation y = x + 3, y is directly proportional to x.

True

False

#### SOLUTION

Solution :B

Condition for direct proportionality is xy= constant and condition for inverse proportionality is xy= constant

So, the given equation y = x + 3; y is neither directly nor inversely proportional to x.

### Question 4

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?

4 cm

6 cm

9 cm

8.2 cm

#### SOLUTION

Solution :A

Let us assume that her distance covered in cm in the map be x.

The scale and kilometer it represent are directly proportional

AB= Constant

A1B1 = A2B218 x = 72

x = 7218=4 cm

### Question 5

The Claude crayon company can make 2400 crayons in 4 minutes. If each crayon box has 12 crayons, how many crayon boxes can they make in 15 minutes?

750

800

900

1000

#### SOLUTION

Solution :A

Let n be the number of boxes.

As there are 12 crayons in each box, 2400 crayons fits into

240012=200 boxes

Now the number of boxes produced will be directly proportional to time taken.

AB= Constant

A1B1 = A2B2

No. of crayons boxestime=2004=n15So, n = 750

### Question 6

Working at the same constant rate, 7 identical machines can produce a total of 490 toys per minute.At this rate, how many toys could 10 such machines produce in 5 minutes?

2,700

3,500

4,200

4,500

#### SOLUTION

Solution :B

Let the required number of toys produced by 7 machines in 5 minutes be

n.As the time increases, the number of toys increases. Hence they are in direct proportion.

n = 490 x 5= 2450

So, 7 machines can make 2,450 toys in 5 minutes.

As the number of machines increases, number of toys increases. They are in direct proportion.

Let k is the number of toys produced by 10 machines in 5 minutes.

Then, k = 2450×107

⟹ k = 3500

So, the number of toys produced = 3,500

### Question 7

A batch of bottles were packed in 25 boxes with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, 15 boxes are required.

True

False

#### SOLUTION

Solution :A

The number of bottles and the number of boxes are in inverse proportion.

AB = constant

A1×B1 = A2×B2

Suppose the required number of boxes is xThen, 25x = 2012

Number of boxes, x = (25×12)20=15

### Question 8

In the equation x(y+4)=1, y is

Directly proportional to x

#### SOLUTION

Solution :B

Expanding the given equation, we get

xy+4x=1

Here, xy=1−4x,which is not a constant. Similarly,

xy=(14y−x4),which is not a constant.

So, y is not proportional to x.

### Question 9

Two quantities x and y are said to vary in inverse proportion, if there exists a relation of the type xy = k between them, k being a constant.

True

False

#### SOLUTION

Solution :A

Given, two quantities x and y are said to vary in inverse proportion.

Then, x∝1y or y∝1x.

In any case, we get xy=constant

Hence the given statemnet is true.

### Question 10

A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find the cost of painting a pole that casts a shadow 5 m long. Painting is done at the rate of ₹ 5 per cm.

₹4375

₹ 5425

₹3450

₹ 3680

#### SOLUTION

Solution :A

The ratio of the length of the pole to the length of the shadow remains constant.

The length of pole and the length of the shadow are directly proportional

AB= Constant

A1B1 = A2B2Length of polelength of shadow=560320=n500 (where n is length the second pole)

n = 875 cm

∴ the height of the pole = 875 cm

Cost of painting the pole = 875 x 5 = ₹ 4375