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 workedout 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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