Question

Given 1176 = 2p , 3q, 7r, find
(i) The numerical values of p, q and r.
(ii)The value of 2p , 3q, 7 – r as a fraction.
WARNING: IT IS NOT A MCQ.​

Answer 1

Answer:

The value of p is 588,q is 392 and r is 168 .

Answer 2

Required answer :-

Question :

(i) The numerical values of p, q and r.

(ii)The value of 2p , 3q, 7 – r as a fraction.

Solution :

Given,

$1176 \: = \: 2 {}^{p}.3 {}^{q}.7 {}^{r}$

Formula used :

$1st \: Law \: of \: indices : a {}^{m} \times a {}^{n} \: = a {}^{m + n} \\ first \: law \: of \: indices \: also \: knows \: as \: product \: law$

$(i)\: \: \: 1176 \: = \: 2 {}^{p}.3 {}^{q}.7 {}^{r} \\→2 {}^{3} \times 3 {}^{1} \times 7 {}^{2} = \: 2 {}^{p}.3 {}^{ q}.7 {}^{r}\\→p \: = 3 \: ,q \: = 1,r \: = 2$

$(ii) \: 2 {}^{p}.3 {}^{q}.7 {}^{ - r} \: = \: 2 {}^{3}.3 {}^{1}.7 {}^{ - 2} \\ \\ as \: we \: have \: to \: find \: the \: value \: as \: a \: fraction \: so \: it \: will \: be \: \\ = \frac{8 \times 3}{7 {}^{2} } = \: \frac{24}{49}$