CAT 2025 Slot 2 Memory-Based Questions
Q1. If $N=23\times 37\times 57\times 79\times 10!$, how many factors of $N$ are perfect squares and also multiples of $420$?
First compute prime exponents in $10!$:
$,v_2(10!)=\left\lfloor\frac{10}{2}\right\rfloor+\left\lfloor\frac{10}{4}\right\rfloor+\left\lfloor\frac{10}{8}\right\rfloor=5+2+1=8.$
$,v_3(10!)=\left\lfloor\frac{10}{3}\right\rfloor+\left\lfloor\frac{10}{9}\right\rfloor=3+1=4.$
$,v_5(10!)=\left\lfloor\frac{10}{5}\right\rfloor=2,\qquad v_7(10!)=\left\lfloor\frac{10}{7}\right\rfloor=1.$
Note $57=3\times 19$, and $23,37,79$ are primes, so multiplying by $23,37,57,79$ adds $+1$ to exponents of $23,37,79,3,19$. Thus
$,N=2^8\cdot 3^{4+1}\cdot 5^2\cdot 7^1\cdot 19^1\cdot 23^1\cdot 37^1\cdot 79^1
=2^8\cdot 3^5\cdot 5^2\cdot 7^1\cdot 19^1\cdot 23^1\cdot 37^1\cdot 79^1.$
Prime factorization of $420$ is $420=2^2\cdot 3^1\cdot 5^1\cdot 7^1$.
Let a divisor $d$ of $N$ have exponent $e_p$ at prime $p$. Requirements for $d$:
$,d\mid N\Rightarrow 0\le e_p\le$ exponent in $N$.
$,d$ a multiple of $420\Rightarrow e_2\ge2,\ e_3\ge1,\ e_5\ge1,\ e_7\ge1$.
$,d$ a perfect square$\Rightarrow$ every $e_p$ is even.
Look at prime $7$: exponent of $7$ in $N$ is $1$. There is no even integer $e_7$ with $1\le e_7\le1$. Hence it is impossible to satisfy both “perfect square” and “multiple of $420$” simultaneously.
Therefore the number of such divisors is $ \boxed{0} $.
Q2. Adam’s savings $=$ Ben’s expenditure $=$ Mary’s savings $=50{,}000$. Incomes are in ratio $3:1:4$, so let incomes be $3k,\ k,\ 4k$.
Then
$,\text{Adam's expenditure}=3k-50{,}000,\quad \text{Ben's expenditure}=50{,}000,\quad \text{Mary's expenditure}=4k-50{,}000.$
Feasibility requires Ben's income $\ge$ Ben's expenditure, so
$,k\ge 50{,}000.\quad(1)$
Given: Mary’s expenditure is less than thrice Adam’s expenditure:
$,4k-50{,}000<3(3k-50{,}000).$
So $,4k-50{,}000<9k-150{,}000\Rightarrow 100{,}000<5k\Rightarrow k>20{,}000.\quad(2)$
Given: Twice Adam’s expenditure is less than two times Ben’s income:
$,2(3k-50{,}000)<2k.$
So $,6k-100{,}000<2k\Rightarrow 4k<100{,}000\Rightarrow k<25{,}000.\quad(3)$
From (2) and (3): $,20{,}000<k<25{,}000.$
This contradicts (1) $k\ge 50{,}000$. Hence no $k$ satisfies all conditions. The data are inconsistent, so the correct choice is $ \boxed{\text{D: Data Inconsistent}} $.
Q 3. A and B can complete a work in 12 days and 18 days respectively. They start together, but A leaves after 4 days. How many more days will B take to finish the remaining work?
A's rate is $1/12$ work/day and B's rate is $1/18$ work/day.
Together their rate is $1/12+1/18=5/36$ work/day.
Work done in 4 days $=4\cdot(5/36)=20/36=5/9$.
Remaining work $=1-5/9=4/9$.
B's rate is $1/18$, so days for B to finish $=(4/9)\div(1/18)=(4/9)\cdot 18=8$.
$\boxed{8\text{ days}}$
Q 4. If the quadratic $ax^2+bx+c=0$ has roots that differ by $4$, and $a+b+c=12$, what is the value of $b^2-4ac$?
For a quadratic $ax^2+bx+c=0$ the difference of roots satisfies $|r_1-r_2|=\dfrac{\sqrt{b^2-4ac}}{|a|}$.
Given $|r_1-r_2|=4$, we have $4=\dfrac{\sqrt{b^2-4ac}}{|a|}$, so
$b^2-4ac=16a^2$
Thus the discriminant is $b^2-4ac=16a^2$.
(Using $a+b+c=12$ alone does not fix a unique numeric value for $a$, so the discriminant is expressed in terms of $a$.)
Q 5. In a triangle $ABC$, the sides are in the ratio $3:4:5$. If the area of the triangle is $96$ sq units, what is its perimeter?
Let the sides be $3k,4k,5k$. This is a right triangle, area $=\tfrac{1}{2}\cdot 3k\cdot 4k=6k^2$.
Given $6k^2=96\Rightarrow k^2=16\Rightarrow k=4$.
Perimeter $=(3+4+5)k=12k=12\cdot 4=48$.
$\boxed{48\text{ units}}$
Read More: CAT 2025 Cutoff for IIM ABC
