Proportion Problems

We will learn how to solve proportion problems. We know, the first term (1st) and the fourth term (4th) of a proportion are called extreme terms or extremes, and the second term (2nd) and the third term (3rd) are called middle terms or means.

Therefore, in a proportion, product of extremes  = product of middle terms.

Solved examples:

1. Check whether the two ratios form a proportion or not:

(i) 6 : 8 and 12 : 16;                           (ii) 24 : 28 and 36 : 48

Solution:

(i) 6 : 8 and 12 : 16

6 : 8 = 6/8 = 3/4

12 : 16 = 12/16 = 3/4

Thus, the ratios 6 : 8 and 12 : 16 are equal.

Therefore, they form a proportion.

(ii) 24 : 28 and 36 : 48

24 : 28 = 24/28 = 6/7

36 : 48 = 36/48 = 3/4

Thus, the ratios 24 : 28 and 36 : 48 are unequal.

Therefore, they do not form a proportion.

2. Fill in the box in the following so that the four numbers are in proportion.

5, 6, 20, ____

Solution:

5 : 6 = 5/6

20 : ____ = 20/____

Since the ratios form a proportion.

Therefore, 5/6 = 20/____

To get 20 in the numerator, we have to multiply 5 by 4. So, we also multiply the denominator of 5/6, i.e. 6 by 4

Thus, 5/6 = 20/6 × 4 = 20/24

Hence, the required numbers is 24

3. The first, third and fourth terms of a proportion are 12, 8 and 14 respectively.  Find the second term.

Solution:

Let the second term be x.

Therefore, 12, x, 8 and 14 are in proportion i.e., 12 : x = 8 : 14

⇒ x × 8 = 12 × 14, [Since, the product of the means = the product of the extremes]

⇒ x = (12 × 14)/8

⇒ x = 21

Therefore, the second term to the proportion is 21.

More worked-out proportion problems:

4. In a sports meet, groups of boys and girls are to be formed. Each group consists of 4 boys and 6 girls. How many boys are required, if 102 girls are available for such groupings?

Solution:

Ratio between boys and girls in a group = 4 : 6 = 4/6 = 2/3 = 2 : 3

Let the number of boys required = x

Ratio between boys and girls = x : 102

So, we have, 2 : 3 = x : 102

Now, product of extremes = 2 × 102 = 204

Product of means = 3 × x

We know that in a proportion product of extremes = product of means

i.e., 204 = 3 × x

If we multiply 3 by 68, we get 204 i.e., 3 × 68 = 204

Thus, x = 68

Hence, 68 boys are required.

5. If a : b = 4 : 5 and b : c = 6 : 7; find a : c.

Solution:

a : b = 4 : 5

⇒ a/b = 4/5

b : c = 6 : 7

⇒ b/c  = 6/7

Therefore, a/b × b/c = 4/5 × 6/7

⇒ a/c = 24/35

Therefore, a : c = 24 : 35

6. If a :  b = 4 : 5 and b : c = 6 : 7; find a : b : c.

Solution:

We know that of both the terms of a ratio are multiplied by the same number; the ratio remains the same.

So, multiply each ratio by such a number that the value of b (the common term in both the ratios) acquires the same value.

Therefore, a :  b = 4 : 5 = 24 : 30, [Multiplying both the terms by 6]

And, b : c = 6 : 7 = 30 : 35, [Multiplying both the terms by 5]

Clearly,; a : b : c = 24 : 30 : 35

Therefore, a : b : c = 24 : 30 : 35

From, the above solved proportion problems we get the clear concept how to find whether the two ratios form a proportion or not and word problems.

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