What least number must be subtracted from each of 13, 16, 17, and 21 to make them in proportion?

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Question 7 Ratio and Proportion Exercise 7.2

Answer:

Consider x be subtracted from each term

23 – x, 30 – x, 57 – x and 78 – x are proportional

It can be written as

23 – x: 30 – x :: 57 – x: 78 – x

(23 – x)/ (30 – x) = (57 – x)/ (78 – x)

By cross multiplication

(23 – x) (78 – x) = (30 – x) (57 – x)

By further calculation

$\begin{array}{l} 1794-23 x-78 x+x^{2}=1710-30 x-57 x+x^{2} \\ x^{2}-101 x+1794-x^{2}+87 x-1710=0 \end{array}$

So we get

-14x + 84 = 0

14x = 84

x = 84/14 = 6

Therefore, 6 is the number to be subtracted from each of the numbers.

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Hint: To answer this type of problem suppose the least number is x then subtract x from each of the given numbers. At last apply the given conditions that the remainders are in continued proportion then calculate the value of by solving the algebraic equation.

Complete step-by-step answer:

The given numbers are 7, 17, and 47We have to calculate the least number when subtracted from each of the numbers such that the remainders are in continued proportion.Let x be the number that needs to be subtracted from these numbers to make it proportional.The new numbers will be \[7 - x,17 - x,{\text{ }}47 - x\] Since these numbers should be proportional,So their ratios will be in proportion\[\dfrac{{17 - x}}{{7 - x}} = \dfrac{{47 - x}}{{17 - x}}\] By simplification we get,\[\Rightarrow {\left( {17 - x} \right)\left( {17 - x} \right) = \left( {47 - x} \right)\left( {7 - x} \right)} \\ \Rightarrow {{x^2} - 34x + 289 = 329 + {x^2} - 54x} \\ \Rightarrow {20x = 40 \Rightarrow x = 2} \] Now put the value of x in \[7 - x,17 - x,{\text{ }}47 - x\] we get the numbers asThe new numbers will be 5,15,45Therefore ′2 ′ is the least number to be subtracted so that the numbers will be proportional.

Note: If there are two values of x then we will choose that value of x which satisfy the condition of proportionality as the question says and discard the other one.