Word Problems on Proportion

We will learn how to solve the word problems on proportion. We know if the phone numbers are the ratio of the first two is equal to the ratio of the last two then the phone numbers are said to be in proportional and the four numbers are said to be in proportion.

1. Which number is to be added to each of 2, 4, 6 and 10 to make the sums proportional?  

Solution:

Let the required number k be added to each.

Then, according to the question

2 + k, 4 + k, 6 + k and 10 + k will be proportional.

Therefore,

\(\frac{2 + k}{4 + k}\) = \(\frac{6 + k}{10 + k}\)

⟹ (2 + k)(10 + k) = (4 + k)(6 +k)

⟹ 20 + 2k + 10k + k\(^{2}\) = 24 + 4k + 6k + k\(^{2}\)

⟹ 20 + 12k + k\(^{2}\) = 24 + 10k + k\(^{2}\)

⟹ 20 + 12k = 24 + 10k

⟹ 12k - 10k = 24 - 20

⟹ 2k = 4

⟹ k = \(\frac{4}{2}\)

⟹ k = 2

Therefore, the required number is 2.

2. What number should be added to 6, 15, 20 and 43 to make the numbers proportional?

Solution:

Let the required number be k.

Then, according to the problem

6 + k, 15 + k, 20 + k and 43 + k are proportional numbers.

Therefore, \(\frac{6 + k}{15 + k}\) = \(\frac{20 + k}{43 + k}\)

⟹ (6 + k)(43 + k) = (15 + k)(20 + k)

⟹ 258 + 6k + 43k + k\(^{2}\) = 300 + 15k + 20k + k\(^{2}\)

⟹ 258 + 49k = 300+ 35k

⟹ 49k – 35k = 300 - 258

⟹ 14k = 42

⟹ k = \(\frac{42}{14}\)

⟹ k = 3

Therefore, the required number is 3.

3. Find the third proportional of 2m\(^{2}\) and 3mn.

Solution:

Let the third proportional be k.

Then, according to the problem

2m\(^{2}\), 3mn and k are in continued proportion.

Therefore,

\(\frac{2m^{2}}{3mn}\) = \(\frac{3mn}{k}\)

⟹ 2m\(^{2}\)k = 9m\(^{2}\)n\(^{2}\)

⟹ 2k = 9n\(^{2}\)

⟹ k = \(\frac{9n^{2}}{2}\)

Therefore, the third proportional is \(\frac{9n^{2}}{2}\).

4. John, David and Patrick have $ 12, $ 15 and $ 19 respectively with them. Their father asks them to give him equal amount so that the money held by them now are in continued proportion. Find the amount taken from each of them.

Solution:

Let the amount taken from each of them is $ p.

Then, according to the problem

12 – p, 15 – p and 19 – p are in continued proportion.

Therefore,

\(\frac{12 - p}{15 - p}\) = \(\frac{15 - p}{19 - p}\)

⟹ (12 – p)(19 – p) = (15 – p)\(^{2}\)

⟹ 228 – 12p – 19p + p\(^{2}\) = 225 – 30p + p\(^{2}\)

⟹ 228 – 31p = 225 – 30p

⟹ 228 – 225 = 31 p – 30p

⟹ 3 = p

⟹ p = 3

Therefore, the required amount is $ 3.

5. Find the fourth proportional of 6, 9 and 12.

Solution:

Let the fourth proportional be k.

Then, according to the problem

6, 9, 12 and k are in proportional

Therefore,

\(\frac{6}{9}\) = \(\frac{12}{k}\)

⟹ 6k = 9 × 12

⟹ 6k = 108

⟹ k = \(\frac{108}{6}\)

⟹ k = 18

Therefore, the fourth proportional is 18.

6. Find two numbers whose mean proportional is 16 and the third proportional is 128.

Solution:

Let the required number be a and b.

Then, according to the question,

\(\sqrt{ab}\) = 16, [Since, 16 is the mean proportional of a, b]

and \(\frac{b^{2}}{a}\) = 128, [Since, the third proportional of a, b is 128]

Now, \(\sqrt{ab}\) = 16

⟹ ab = 16\(^{2}\)

⟹ ab = 256

Again, \(\frac{b{2}}{a}\) = 128

⟹ b\(^{2}\) = 128a

⟹ a = \(\frac{b^{2}}{128}\)

Substituting a = \(\frac{b^{2}}{128}\) in ab = 256

⟹\(\frac{b^{2}}{128}\) × b = 256

⟹\(\frac{b^{3}}{128}\) = 256

⟹ b\(^{3}\) = 128 × 256

⟹  b\(^{3}\) = 2\(^{7}\)  × 2\(^{8}\)

⟹  b\(^{3}\) = 2\(^{7 + 8}\)

⟹  b\(^{3}\) = 2\(^{15}\)

⟹ b = 2\(^{5}\)

⟹ b = 32

So, from equation a = \(\frac{b^{2}}{128}\) we get

a = \(\frac{32^{2}}{128}\)

⟹ a = \(\frac{1024}{128}\)

⟹ a = 8

Therefore, the required numbers are 8 and 32.

10th Grade Math

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