What must be added to each of the number 7/11 19 so that the resulting numbers may be in continued proportion?

Question 10 Ratio and Proportion Exercise 7.2

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Answer:

Consider x be added to each number

16 + x , 26 + x and 40 + x are in continued proportion

It can be written as

(16 + x)/ (26 + x) = (26 + x)/ (40 + x)

By cross multiplication

(16 + x) (40 + x) = (26 + x) (26 + x)

On further calculation

\begin{array}{l} 640+16 x+40 x+x^{2}=676+26 x+26 x+x^{2} \\ 640+56 x+x^{2}=676+52 x+x^{2} \\ 56 x+x^{2}-52 x-x^{2}=676-640 \end{array}

So we get

4x = 36

x = 36/4 = 9

Hence, 9 is the number to be added to each of the numbers.

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Answer & Solution

Answer: Option 4

Solution:

(7 + X) (19 + X) = (11 + X)2

What number must be added to each of the numbers 16, 26 and 40 so that the resulting numbers may be in continued proportion?

Let x be added to each number then16 + x, 26 + x and 40 + xare in continued proportion.⇒ `(16 + x)/(26 + x) = (26 + x)/(40 + x)`Cross Multiplying(16 + x) (40 + x) = (26 + x) (26 + x)

⇒ 640 + 16x + 40x + x2 = 676 + 26x + 26x + x2

∴ 9 is to be added.

Concept: Concept of Proportion

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What must be added to each of the numbers 7, 11 and 19, so that the resulting numbers may be in continued proportion?

Explanation:

Let X be the required number, then (7 + X) : (11 + X) :: (11 +X) : (19 + X) (7 + X) (19 + X) = (11 + X)2 X2 + 26X + 133 = X2 + 22X + 121

4X = - 12 or X = - 3

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