cat 2024 Question 4

Question

If a, b and c are positive real numbers such that a > 10 \geq b \geq c and \cfrac{\log_8 (a + b)}{\log_2c} + \cfrac{\log_{27} (a - b)}{\log_3c} = \cfrac{2}{3}, then the greatest possible integer value of a is

Accepted Answer

The first term of the expression can be rewritten as $\frac{\frac{1}{3}\log_2\left(a+b\right)}{\log_2c}$

Using the property $\frac{m}{n}\log_ab=\log_ab^{\frac{m}{n}}$ this can be rewritten as

$\frac{\log_2\left(a+b\right)^{\frac{1}{3}}}{\log_2c}$

And finally using the property $\frac{\log_ba}{\log_bc}=\log_ca$ , we can rewrite the expression as

$\log_c\left(a+b\right)^{\frac{1}{3}}$

Doing identical operations in the second term, we get the entire left-hand side to be:

$\log_c\left(a+b\right)^{\frac{1}{3}}+\log_c\left(a-b\right)^{\frac{1}{3}}$

Using property $\log_ca+\log_cb=\log_c\left(ab\right)$ we get

$\log_c\left[\left(a+b\right)^{\frac{1}{3}}\left(a-b\right)^{\frac{1}{3}}\right]$

$\log_c\left[\left(a+b\right)\left(a-b\right)\right]^{\frac{1}{3}}$

$\log_c\left[\left(a^2-b^2\right)\right]^{\frac{1}{3}}$

This expression is given to be equal to 2/3

Using the definition of log: $\log_cN=a\$ which is $c^a=N$

we get: $c^{\frac{2}{3}}=\left(a^2-b^2\right)^{\frac{1}{3}}$

Cubing both sides:

$c^2=a^2-b^2$

Finally giving $a^2=b^2+c^2$

We have upper limits on b and c as 10, and we want to maximize the value of a squared.

This can be thought of as a right-angled triangle, and the value of a will be maximum when both b and c are equal to 10, giving $a^2=200$ , but this would not give an integer value of a

We need to adjust $a^2$ to the biggest square less than 200, which is 196

Giving the value of $a$ as 14.

Therefore, 14 is the correct answer.

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