Question

$\text{Find the value for p, q for :}\frac{4}{p}+\frac{15}{q}=72,\frac{15}{p}-\frac{1}{q}=41$

• A. 3 , 4
• B. 1/3, 1/4
• C. 5, 7
• D. 1/5, 1/7

Answer 1

The correct option is B

1/3, 1/4

$\frac{4}{p}+\frac{15}{q}=72...................\left(i\right)\frac{15}{p}-\frac{1}{q}=41.................\left(ii\right)Let\frac{1}{p}=xand\frac{1}{q}=y$

4x+15y=72 .............(iii)

15x-y=41 ................(iv)

After (iii) x 15 - (iv) x 4, we get

60x+225y - (60x-4y) =1080 - 164

$\Rightarrow$ 229y=916

$\Rightarrow$ $y=\frac{916}{229}=4$

Substituting y=4 in equation (iv) by 4 we get

15x-4=41

$\Rightarrow 15x=45\Rightarrow x=\frac{45}{15}=3$

So, x=3 and y=4

$But\frac{1}{p}=3and\frac{1}{q}=4[\because \frac{1}{p}=x,\frac{1}{q}=y]\therefore p=\frac{1}{3},q=\frac{1}{4}$