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CAT 2022 Slot 3 — QA Question 4

IndicesEasy

If (75)3x-y = 8752401 and (4ab)6x-y = (2ab)y-6x, for all non-zero real values of a and b, then the value of x + y is

Answer & solution

Answer: 14

Solution

Easy

Each equation forces a linear relation between xx and yy. The first comes from writing 8752401\dfrac{875}{2401} as a power of 75\dfrac75. The second holds "for all a,ba,b", which is only possible if its exponent is 00. Two equations, two unknowns.

1

First equation. Note 75=(75)1/2\sqrt{\tfrac75}=\left(\tfrac75\right)^{1/2}, and factor 8752401\dfrac{875}{2401}:

(75)3x−y2=8752401=125×774=5373=(75)−3⇒ 3x−y2=−3 ⇒ 3x−y=−6...(1)\begin{aligned} &\left(\frac75\right)^{\frac{3x-y}{2}}=\frac{875}{2401}=\frac{125\times7}{7^4}=\frac{5^3}{7^3}=\left(\frac75\right)^{-3}\\ &\Rightarrow\ \frac{3x-y}{2}=-3\ \Rightarrow\ 3x-y=-6 \quad\text{...(1)} \end{aligned}
2

Second equation. Move the right side over: since (2ab)y−6x=(b2a)6x−y\left(\tfrac{2a}{b}\right)^{y-6x}=\left(\tfrac{b}{2a}\right)^{6x-y},

(4ab)6x−y⋅(2ab)6x−y=1 ⇒ (8a2b2)6x−y=1\left(\frac{4a}{b}\right)^{6x-y}\cdot\left(\frac{2a}{b}\right)^{6x-y}=1\ \Rightarrow\ \left(\frac{8a^2}{b^2}\right)^{6x-y}=1
3

Kill the exponent. The base 8a2b2\tfrac{8a^2}{b^2} varies with a,ba,b, so the only way this holds for all non-zero a,ba,b is:

6x−y=0 ⇒ y=6x...(2)6x-y=0\ \Rightarrow\ y=6x \quad\text{...(2)}
4

Solve (1) and (2). Substitute y=6xy=6x into 3x−y=−63x-y=-6:

3x−6x=−6 ⇒ x=2,y=12 ⇒ x+y=143x-6x=-6\ \Rightarrow\ x=2,\quad y=12\ \Rightarrow\ x+y=14
x+y=14x+y=\mathbf{14}
CAT 2022 Slot 3 QA Q4: If 7 5 3 x - y = 875 2401 and 4 a b 6 x - y = 2 a b y - 6 x , for all non-zero real values of a and b, then th — Solution | TheCATExam