# 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.

**Solution:**

Let the parts of red pigments used be x and parts of base used be y.

As the number of parts of red pigment increases, the amount of base also increases in the same ratio. So they are in direct proportion.

Two numbers x and y are said in direct proportion if, the relationship between two quantities is such that if we increase one, the other will also increase, and if we decrease one the other quantity will also decrease.

x/y = k, x = ky

where k is a constant.

Thus, x₁_{ }/ y₁ = x₂ / y₂

x₁ = 1, x₂ = 4

y₁ = 8, y₂ = ?

x₁/y₁ = x₂/y₂

1/8 = 4/y₂

y₂ = 8 × 4 = 32 parts

32 parts of base is needed for 4 parts of red pigment.

Here,

x₁ = 1, x₂ = 7

y₁ = 8, y₂ = ?

x₁/y₁ = x₂/y₂

1/8 = 7/y₂

y₂ = 8 × 7 = 56 parts

56 parts of base is needed for 7 parts of red pigment.

Here

x₁ = 1, x₂ = 12

y₁ = 8, y₂ = ?

x₁/y₁ = x₂/y₂

1/8 = 12/y₂

y₂ = 8 × 12 = 96

96 parts of base is needed for 12 parts of red pigment.

Here

x₁ = 1, x₂ = 20

y₁ = 8, y₂ = ?

x₁/y₁ = x₂_{ }/ y₂

1/8 = 20/y₂

y₂ = 8 × 20 = 160

160 parts of base are needed for 20 parts of red pigment.

Thus, the complete table is shown below.

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

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

**☛ Check: **Class 8 Maths NCERT Solutions Chapter 13

**Video Solution:**

## 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

NCERT Solutions Class 8 Maths Chapter 13 Exercise 13.1 Question 2

**Summary:**

A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, 32 parts of base is needed for 4 parts of red pigment, 56 parts of base is needed for 7 parts of red pigment, 96 parts of base is needed for 12 parts of red pigment and 160 parts of base are needed for 20 parts of red pigment.

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