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CAT 2020 Slot 3 — QA Question 7

IndicesEasy

If a, b, c are non-zero and 14a = 36b = 84c, then 6b(1c-1a) is equal to

Answer & solution

Correct answer: 3

Answer: 3

Solution

Medium

Set the common value to kk and write each base as a power of kk. The key relation 84=6×1484=6\times14 links the exponents, and 36=6236=6^2 pins down 66 as a power of kk.

1

Introduce kk. Let k=14a=36b=84ck=14^{a}=36^{b}=84^{c}. Taking each to its reciprocal power:

14=k1/a36=k1/b ⇒ 6=k1/(2b)(36=62)84=k1/c\begin{aligned} &14=k^{1/a}\\ &36=k^{1/b}\ \Rightarrow\ 6=k^{1/(2b)} \quad\text{(}36=6^2)\\ &84=k^{1/c} \end{aligned}
2

Use 84=6×1484=6\times14. Substitute the powers of kk from step 1.

k1/c=k1/(2b)×k1/a⇒ k1/c=k 1/(2b)+1/a(add exponents)⇒ 1c=12b+1a(equate exponents)\begin{aligned} &k^{1/c}=k^{1/(2b)}\times k^{1/a}\\ &\Rightarrow\ k^{1/c}=k^{\,1/(2b)+1/a} \quad\text{(add exponents)}\\ &\Rightarrow\ \frac{1}{c}=\frac{1}{2b}+\frac{1}{a} \quad\text{(equate exponents)} \end{aligned}
3

Isolate the required expression.

1c−1a=12b⇒ 2b(1c−1a)=1(multiply by 2b)⇒ 6b(1c−1a)=3(multiply by 3)\begin{aligned} &\frac{1}{c}-\frac{1}{a}=\frac{1}{2b}\\ &\Rightarrow\ 2b\left(\frac{1}{c}-\frac{1}{a}\right)=1 \quad\text{(multiply by }2b)\\ &\Rightarrow\ 6b\left(\frac{1}{c}-\frac{1}{a}\right)=3 \quad\text{(multiply by }3) \end{aligned}
6b(1c−1a)=36b\left(\frac{1}{c}-\frac{1}{a}\right)=3

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CAT 2020 Slot 3 QA Q7: If a, b, c are non-zero and 14 a = 36 b = 84 c , then 6 b 1 c - 1 a is equal to — Solution | TheCATExam