Question

Mark the correct alternative in the following question:

What least number is to be subtracted from each term of the ratio 15 : 19 to make the ratio 3 : 4?

(a) 3 (b) 5 (c) 6 (d) 9

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{least}\mathrm{number}\mathrm{that}\mathrm{is}\mathrm{to}\mathrm{be}\mathrm{subtracted}\mathrm{from}\mathrm{each}\mathrm{term}\mathrm{of}\mathrm{the}\mathrm{ratio}15:19\mathrm{be}x.\mathrm{Then}, \\ \left(15-x\right):\left(19-x\right)=3:4 \\ \Rightarrow \frac{\left(15-x\right)}{\left(19-x\right)}=\frac{3}{4} \\ \Rightarrow 4\left(15-x\right)=3\left(19-x\right) \\ \Rightarrow 60-4x=57-3x \\ \Rightarrow 3x-4x=57-60 \\ \Rightarrow -x=-3 \\ \therefore x=3 \end{gathered}$$

So, 3 is the least number to be subtracted from each term of the ratio 15 : 19 to make the ratio 3 : 4.

Hence, the correct alternative is option (a).