What number must be added to each term of the ratio 7 ratio 12 to make the ratio 4 ratio 5?

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What must be added to each term of the ratio 2: 3, so that it may become equal to 4: 5 ?

Question

A

3

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B

2

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C

9

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D

4

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Solution

The correct option is B2Let the number to be added be x, then (2 + x):(3 + x) = 4 : 5⇒ 2+x3+x=45⇒5(2+x)=4(3+x)⇒10+5x=12+4x⇒5x−4x=12−10⇒x=2

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4

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ML Aggarwal Solutions Class 10 Mathematics Solutions for Ratio and Proportion Exercise 7.2 in Chapter 7 - Ratio and Proportion

Question 10 Ratio and Proportion Exercise 7.2

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

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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Hint:To find the term needed to add to \[9:16\] to make \[2:3\], we assume that the terms given are \[x\] in both the denominator and the numerator with the ratio \[9:16\] added to which is equal to \[2:3\] . After placing the values in the ratio we find the value of \[x\] by cross multiplying the ratios.

Complete step by step solution:
Let us assume that the number needed to be added is taken as \[x\].
Now to find the term\ratio needed i.e. \[2:3\] to add the unknown term of \[x\] to the numerator and denominator of the previous ratio of \[9:16\].
The value of the ratio on the LHS of the equation down below is equal to the ratio on the RHS when \[x\] is added in both the numerator and denominator of the LHS given.
\[\Rightarrow \dfrac{9+x}{16+x}=\dfrac{2}{3}\]
Cross multiplying the value of the ratio of the LHS and RHS, we get the value of the unknown variable of \[x\] as:
\[\Rightarrow 3\left( 9+x \right)=2\left( 16+x \right)\]
\[\Rightarrow 27+3x=32+2x\]
\[\Rightarrow x=5\]
Therefore, the value needed to be added with the ratio of \[9:16\] to get \[2:3\] is \[5\].

Note: Student may go wrong if they try to add a single number instead of adding two number in both the denominator and numerator as given below:
\[\Rightarrow \dfrac{9+x}{16+x}=\dfrac{2}{3}\] correct form
\[\Rightarrow \dfrac{9}{16}+x=\dfrac{2}{3}\] Incorrect form

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