CAT 2020 Slot 3 — QA questions with answers

Let the volume of the initial mixture of A and B be $V$. Since they are mixed in the proportion 1:3 by volume, the initial volume of A is $0.25V$ and B is $0.75V$.

The volume of the mixture is doubled by adding solution A. So we add $V$ volume of solution A.
Total volume becomes $2V$.
Total volume of solution A in the mixture = $0.25V + V = 1.25V$.
Total volume of solution B in the mixture = $0.75V$.

Solution A has 60% alcohol.
Let solution B have $x\%$ alcohol.

Amount of alcohol in solution A = $1.25V \times 60\% = 0.75V$
Amount of alcohol in solution B = $0.75V \times x\% = 0.75V \times \frac{x}{100}$

The resulting mixture has 72% alcohol.
Total amount of alcohol = $2V \times 72\% = 1.44V$

Equating the total amount of alcohol:
$0.75V + 0.75V \times \frac{x}{100} = 1.44V$

Divide by $V$:
$0.75 + 0.75 \times \frac{x}{100} = 1.44$
$0.75 \times \frac{x}{100} = 1.44 - 0.75 = 0.69$
$\frac{3}{4} \times \frac{x}{100} = 0.69$
$\frac{3x}{400} = 0.69$
$3x = 0.69 \times 400 = 276$
$x = \frac{276}{3} = 92$

So, the percentage of alcohol in solution B is 92%.

Hence, Option C is correct.

A batsman played $n + 2$ innings and his average score was 29 runs.
Total score in $n + 2$ innings = $29(n + 2)$.

He scored 38 and 15 runs in the last two innings.
Total score in the first $n$ innings = $29(n + 2) - 38 - 15 = 29n + 58 - 53 = 29n + 5$.

We are given that in the first $n$ innings, his average score was 30 runs.
Total score in the first $n$ innings = $30n$.

Equating the two expressions for the total score in the first $n$ innings:
$30n = 29n + 5 \implies n = 5$.

So, the batsman played $5$ innings initially, and his total score in these 5 innings was $30 \times 5 = 150$.

We are given that he scored less than 38 runs in each of the first $n$ innings.
This means his maximum possible score in any of these 5 innings is 37.
Let the lowest score in these 5 innings be $x$.

To find the smallest possible value of $x$, we must maximize the scores in the other 4 innings.
The maximum possible score for the other 4 innings is 37 each.
Sum of the scores in the 5 innings: $x + 37 + 37 + 37 + 37 = 150$
$x + 148 = 150 \implies x = 2$.

The smallest possible value of $x$ is 2.

Hence, Option C is correct.

Total work to be done = 6 km of road.
Total time allowed = 200 days.
Initial workforce = 140 persons.

After 60 days, work completed = 1.5 km.
Work remaining = $6 - 1.5 = 4.5$ km.
Time remaining = $200 - 60 = 140$ days.

Let the work done by 1 person in 1 day be $E$.
Work done by 140 persons in 60 days:
$140 \times 60 \times E = 1.5$ km.
$E = \frac{1.5}{140 \times 60} = \frac{1.5}{8400}$.

Let the total number of persons required to finish the remaining work on time be $N$.
Work done by $N$ persons in the remaining 140 days must be equal to 4.5 km.
$N \times 140 \times E = 4.5$
Substitute the value of $E$:
$N \times 140 \times \frac{1.5}{140 \times 60} = 4.5$
$N \times \frac{1.5}{60} = 4.5$
$N = \frac{4.5 \times 60}{1.5} = 3 \times 60 = 180$.

Total 180 persons are needed. Since 140 persons are already employed, the number of additional people required is:
$180 - 140 = 40$.

We are given $x_1 = -1$ and $x_m = x_{m+1} + (m+1)$ for every positive integer $m$.
We can rewrite the recurrence relation to find the next term:
$x_{m+1} = x_m - (m+1)$.

Let's find the first few terms:
For $m=1$: $x_2 = x_1 - (1+1) = x_1 - 2 = -1 - 2 = -3$.
For $m=2$: $x_3 = x_2 - (2+1) = x_2 - 3 = -3 - 3 = -6$.
For $m=3$: $x_4 = x_3 - (3+1) = x_3 - 4 = -6 - 4 = -10$.

We can see a pattern:
$x_1 = -1$
$x_2 = -1 - 2$
$x_3 = -1 - 2 - 3$
$x_4 = -1 - 2 - 3 - 4$

By induction, the $n$-th term is:
$x_n = -(1 + 2 + 3 + \dots + n)$

The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$.
So, $x_n = -\frac{n(n+1)}{2}$.

We need to find $x_{100}$:
$x_{100} = -\frac{100(100+1)}{2} = -\frac{100 \times 101}{2} = -50 \times 101 = -5050$.

Hence, Option A is correct.

Given:
$\log_a 30 = A$ --- (1)
$\log_a (5/3) = -B$ --- (2)
$\log_2 a = 1/3$

From $\log_2 a = 1/3$, we can deduce $\log_a 2 = \frac{1}{\log_2 a} = 3$.

Using the properties of logarithms, we can expand equations (1) and (2):
(1) $\log_a (2 \times 3 \times 5) = \log_a 2 + \log_a 3 + \log_a 5 = A$
Substitute $\log_a 2 = 3$:
$3 + \log_a 3 + \log_a 5 = A \implies \log_a 3 + \log_a 5 = A - 3$ --- (3)

(2) $\log_a (5/3) = \log_a 5 - \log_a 3 = -B$ --- (4)

We have a system of two linear equations (3) and (4) with variables $\log_a 3$ and $\log_a 5$.
We need to find the value of $\log_3 a$. Let's first find $\log_a 3$.

Subtract equation (4) from equation (3):
$(\log_a 3 + \log_a 5) - (\log_a 5 - \log_a 3) = (A - 3) - (-B)$
$2 \log_a 3 = A + B - 3$
$\log_a 3 = \frac{A + B - 3}{2}$

We need $\log_3 a$, which is the reciprocal of $\log_a 3$:
$\log_3 a = \frac{1}{\log_a 3} = \frac{2}{A + B - 3}$.

Hence, Option A is correct.

Let Tom's age be $x$.
Dick is thrice as old as Tom, so Dick's age is $3x$.
Harry is twice as old as Dick, so Harry's age is $2 \times 3x = 6x$.

The average age of all three is:
$\text{Average} = \frac{\text{Tom} + \text{Dick} + \text{Harry}}{3} = \frac{x + 3x + 6x}{3} = \frac{10x}{3}$.

We are given that Dick's age is 1 year less than the average age:
$3x = \frac{10x}{3} - 1$

Solve for $x$:
$3x + 1 = \frac{10x}{3}$
Multiply by 3:
$9x + 3 = 10x \implies x = 3$.

Tom's age is 3 years.
Dick's age is $3 \times 3 = 9$ years.
Harry's age is $6x = 6 \times 3 = 18$ years.

Hence, Harry's age is 18 years.

Let the total distance to the office be $d$ km.
Usual speed = 40 km/hr. Time taken at usual speed = $\frac{d}{40}$ hours.
If she drives at 35 km/hr, time taken = $\frac{d}{35}$ hours.
She is late by 6 minutes ($\frac{6}{60} = \frac{1}{10}$ hours).
$\frac{d}{35} - \frac{d}{40} = \frac{1}{10}$
$\frac{8d - 7d}{280} = \frac{1}{10} \implies \frac{d}{280} = \frac{1}{10} \implies d = 28$ km.

Usual total time to reach the office = $\frac{d}{40} = \frac{28}{40} = \frac{7}{10}$ hours = $42$ minutes.
She covers two-thirds of her distance ($\frac{2}{3} \times 28 = \frac{56}{3}$ km) in one-third of her usual time ($\frac{1}{3} \times 42 = 14$ minutes).
Then she stops for 8 minutes.

Total time spent so far = $14 + 8 = 22$ minutes.
Remaining time to reach exactly on time = $42 - 22 = 20$ minutes = $\frac{20}{60} = \frac{1}{3}$ hours.

Remaining distance = $28 - \frac{56}{3} = \frac{84 - 56}{3} = \frac{28}{3}$ km.

Required speed for the remaining distance = $\frac{\text{Remaining distance}}{\text{Remaining time}} = \frac{28/3}{1/3} = 28$ km/hr.

Hence, Option C is the correct answer.

The set $\{100, 101, 102, \dots, 999\}$ contains all the 3-digit numbers.
Total number of integers in the set = $999 - 100 + 1 = 900$.

We want to find the number of integers that have at least one digit repeated.
This is equal to (Total number of 3-digit numbers) - (Number of 3-digit numbers with all distinct digits).

Let's calculate the number of 3-digit numbers with all distinct digits.
The first digit (hundreds place) can be any digit from 1 to 9 (9 choices, as it cannot be 0).
The second digit (tens place) can be any digit from 0 to 9 except the digit chosen for the hundreds place (9 choices).
The third digit (units place) can be any digit from 0 to 9 except the two digits already chosen (8 choices).

Total 3-digit numbers with distinct digits = $9 \times 9 \times 8 = 648$.

Number of integers with at least one digit repeated = $900 - 648 = 252$.

Let the total maximum marks in the examination be $T$.

Marks obtained by Bishnu = $52\%$ of $T = 0.52T$.
Marks obtained by Asha = $64\%$ of $T = 0.64T$.
Let the marks obtained by Ramesh be $R$.

We are given two conditions about Ramesh's marks:
1. Bishnu's marks is 23 less than Ramesh's marks: $0.52T = R - 23 \implies R = 0.52T + 23$.
2. Asha's marks is 34 more than Ramesh's marks: $0.64T = R + 34 \implies R = 0.64T - 34$.

Equating the two expressions for $R$:
$0.52T + 23 = 0.64T - 34$
$0.64T - 0.52T = 23 + 34$
$0.12T = 57$
$T = \frac{57}{0.12} = \frac{5700}{12} = 475$.

The total marks is 475.

We need to find the marks obtained by Geeta, who scored 84%.
Marks of Geeta = $84\%$ of $T = 0.84 \times 475 = 399$.

Hence, Option D is the correct answer.

Let $14^a = 36^b = 84^c = k$.

Taking log on all sides, or converting to exponential form:
$14 = k^{1/a}$
$36 = k^{1/b}$
$84 = k^{1/c}$

We know that $84 = 14 \times 6$.
Since $36 = 6^2$, we have $6 = 36^{1/2} = (k^{1/b})^{1/2} = k^{1/(2b)}$.

So, $84 = 14 \times 6$ can be written in terms of $k$ as:
$k^{1/c} = k^{1/a} \times k^{1/(2b)}$
$k^{1/c} = k^{1/a + 1/(2b)}$

Equating the powers of $k$ (since $a, b, c$ are non-zero, $k \neq 1$):
$\frac{1}{c} = \frac{1}{a} + \frac{1}{2b}$
$\frac{1}{c} - \frac{1}{a} = \frac{1}{2b}$

We need to find the value of $6b \left(\frac{1}{c} - \frac{1}{a}\right)$.
Substitute $\left(\frac{1}{c} - \frac{1}{a}\right) = \frac{1}{2b}$:
$6b \times \frac{1}{2b} = \frac{6}{2} = 3$.

The value is 3.

Let the principal amount invested be $P$.
Annual rate of interest, $R = 10\%$.
Since it is compounded half-yearly, the rate per half-year is $r = \frac{10}{2} = 5\%$.
Time period = $1.5$ years.
Number of half-yearly cycles, $n = 1.5 \times 2 = 3$.

The formula for amount with compound interest is:
$A = P \left(1 + \frac{r}{100}\right)^n$

Given the amount (principal + interest) is Rs 18522:
$18522 = P \left(1 + \frac{5}{100}\right)^3$
$18522 = P \left(1 + \frac{1}{20}\right)^3 = P \left(\frac{21}{20}\right)^3$
$18522 = P \times \frac{9261}{8000}$

$P = \frac{18522 \times 8000}{9261}$

Notice that $9261 \times 2 = 18522$.
$P = 2 \times 8000 = 16000$.

The amount invested is Rs 16000.

Let the cost price (CP) of 1 kg of sugar be Rs 100.
Total CP for 35 kg of sugar = $35 \times 100 = 3500$.

He sets a marked price to make a 20% profit.
Marked Price (MP) per kg = $100 \times 1.2 = 120$.

He sells 5 kg at this marked price:
Revenue from first 5 kg = $5 \times 120 = 600$.

He sells 15 kg at a 10% discount on the marked price.
Selling Price per kg = $120 \times 0.9 = 108$.
Revenue from next 15 kg = $15 \times 108 = 1620$.

3 kg of sugar is wasted. So it generates Rs 0 revenue.
Total sugar accounted for = $5 + 15 + 3 = 23$ kg.
Remaining sugar = $35 - 23 = 12$ kg.

He wants to make an overall profit of 15% on his total investment.
Overall target Revenue = Total CP $\times 1.15 = 3500 \times 1.15 = 4025$.

Revenue already generated = $600 + 1620 = 2220$.
Required revenue from the remaining 12 kg = $4025 - 2220 = 1805$.

New Selling Price per kg for the remaining 12 kg = $\frac{1805}{12} \approx 150.41$.
He raised the original marked price (Rs 120) by $p\%$.
$p = \frac{150.41 - 120}{120} \times 100 = \frac{30.41}{120} \times 100 = \frac{3041}{120} \approx 25.34\%$.

This is nearest to 25%.

Hence, Option C is correct.

Let the given opposite vertices of the parallelogram be $A(2, 1)$ and $C(-3, -4)$.
In a parallelogram, the diagonals bisect each other. Therefore, the midpoint of the diagonal $AC$ is the same as the midpoint of the diagonal $BD$ (where $B$ and $D$ are the other two vertices).

Midpoint of $AC = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$
$= \left(\frac{2 + (-3)}{2}, \frac{1 + (-4)}{2}\right) = \left(-\frac{1}{2}, -\frac{3}{2}\right)$.

We are given that the other two vertices $B$ and $D$ lie on the line $x + 9y + c = 0$.
Since both $B$ and $D$ lie on this line, the entire diagonal $BD$ lies on this line.
Therefore, the midpoint of $BD$ must also lie on this line.
Substitute the coordinates of the midpoint $\left(-\frac{1}{2}, -\frac{3}{2}\right)$ into the line equation:

$x + 9y + c = 0$
$-\frac{1}{2} + 9\left(-\frac{3}{2}\right) + c = 0$
$-\frac{1}{2} - \frac{27}{2} + c = 0$
$-\frac{28}{2} + c = 0$
$-14 + c = 0 \implies c = 14$.

Hence, Option D is the correct answer.

Distance between stations A and B is 90 km.

Train 1 leaves A at 9:00 am at a speed of 40 km/hr.
Train 2 leaves B at 10:30 am at a speed of 20 km/hr.

By 10:30 am, Train 1 has been travelling for 1.5 hours (from 9:00 am to 10:30 am).
Distance travelled by Train 1 in this 1.5 hours = $40 \times 1.5 = 60$ km.

At 10:30 am, the remaining distance between the two trains is $90 - 60 = 30$ km.
Now, both trains are moving towards each other.
Relative speed = Speed of Train 1 + Speed of Train 2 = $40 + 20 = 60$ km/hr.

Time taken to meet after 10:30 am = $\frac{\text{Remaining Distance}}{\text{Relative Speed}} = \frac{30}{60} = 0.5$ hours = 30 minutes.

So, they will meet 30 minutes after 10:30 am, which is at 11:00 am.

Hence, Option C is the correct answer.

We are given $N = x + y$, where $x$ and $y$ are positive integers.
The conditions are $2 < x < 10$ and $14 < y < 23$.
This means $x$ can take integer values from 3 to 9 (inclusive): $x \in \{3, 4, 5, 6, 7, 8, 9\}$.
And $y$ can take integer values from 15 to 22 (inclusive): $y \in \{15, 16, 17, 18, 19, 20, 21, 22\}$.

We are also given $N > 25$. Let's find the possible distinct values of $N$.
$N_{min}$ subject to $N > 25$ is 26.
Let's check if 26 is possible: $N = x + y$. To get 26, we could have $x = 4, y = 22$. (Possible).
Let's find the maximum possible value of $N$:
$N_{max} = x_{max} + y_{max} = 9 + 22 = 31$.

Let's check if all integer values from 26 to 31 are possible:
If $N = 26$: $x=4, y=22$ (valid)
If $N = 27$: $x=5, y=22$ (valid)
If $N = 28$: $x=6, y=22$ (valid)
If $N = 29$: $x=7, y=22$ (valid)
If $N = 30$: $x=8, y=22$ (valid)
If $N = 31$: $x=9, y=22$ (valid)

The possible distinct values for $N$ are 26, 27, 28, 29, 30, and 31.
There are 6 distinct values.

We have a system of two linear equations:
$kx + y = 3$
$4x + ky = 4$

A system of linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ has a unique solution if and only if the lines are intersecting, meaning their slopes are not equal.
The condition for a unique solution is:
$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

Applying this condition to our equations:
$\frac{k}{4} \neq \frac{1}{k}$

Cross-multiplying gives:
$k^2 \neq 4$
$k \neq \pm 2$

This can be written as $|k| \neq 2$.

Hence, Option A is correct.

We need to find the number of integers from 1 to 120 that are not divisible by 2, 5, or 7.
Let $U$ be the set of integers from 1 to 120. $n(U) = 120$.
Let $A$ be the set of integers divisible by 2. $n(A) = \lfloor 120 / 2 \rfloor = 60$.
Let $B$ be the set of integers divisible by 5. $n(B) = \lfloor 120 / 5 \rfloor = 24$.
Let $C$ be the set of integers divisible by 7. $n(C) = \lfloor 120 / 7 \rfloor = 17$.

Now find the intersections:
$n(A \cap B)$ = multiples of 10 = $\lfloor 120 / 10 \rfloor = 12$.
$n(B \cap C)$ = multiples of 35 = $\lfloor 120 / 35 \rfloor = 3$.
$n(C \cap A)$ = multiples of 14 = $\lfloor 120 / 14 \rfloor = 8$.
$n(A \cap B \cap C)$ = multiples of 70 = $\lfloor 120 / 70 \rfloor = 1$.

Using the Principle of Inclusion-Exclusion, the number of integers divisible by at least one of 2, 5, or 7 is:
$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C)$
$= 60 + 24 + 17 - 12 - 3 - 8 + 1$
$= 101 - 23 + 1 = 79$.

The number of integers divisible by none of them is:
$n(A' \cap B' \cap C') = n(U) - n(A \cup B \cup C)$
$= 120 - 79 = 41$.

Hence, Option B is correct.

The length of the circular track is $D = 3$ km.
Speeds are: $S_A = 15$ km/hr, $S_S = 10$ km/hr, $S_R = 8$ km/hr.
They are all running in the same direction.

Time taken by Anil to complete one round $t_A = \frac{3}{15} = \frac{1}{5}$ hours.
Time taken by Sunil to complete one round $t_S = \frac{3}{10}$ hours.
Time taken by Ravi to complete one round $t_R = \frac{3}{8}$ hours.

We need to find the time when Anil and Sunil meet again for the first time at the starting point.
This time $T$ is the LCM of their individual times to complete one round.
$T = LCM(t_A, t_S) = LCM\left(\frac{1}{5}, \frac{3}{10}\right) = \frac{LCM(1, 3)}{GCD(5, 10)} = \frac{3}{5}$ hours.

So Anil and Sunil meet for the first time at the starting point after $\frac{3}{5}$ hours.
In this time, the distance run by Ravi is:
$\text{Distance} = \text{Speed} \times \text{Time} = 8 \times \frac{3}{5} = \frac{24}{5} = 4.8$ km.

Hence, Option A is correct.

Numerator: $2 \times 4 \times 8 \times 16 = 2^{1+2+3+4} = 2^{10} = 1024$.

The logarithms: $\log_2 4 = 2$,   $\log_4 8 = \frac{3}{2}$,   $\log_8 16 = \frac{4}{3}$.

Denominator: $2^2 \cdot \left(\frac{3}{2}\right)^3 \cdot \left(\frac{4}{3}\right)^4 = 4 \cdot \frac{27}{8} \cdot \frac{256}{81} = \frac{128}{3}$.

So the expression is $1024 \div \frac{128}{3} = \frac{1024 \times 3}{128} = \mathbf{24}$.

The vertices of the triangle are $A(0,0)$, $B(4,0)$ and $C(3,9)$.
We need to find the area of the circumcircle of this triangle. The circumcircle passes through $A, B$, and $C$.

Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.
Since it passes through $(0,0)$, substituting $x=0, y=0$ gives $c = 0$.

Since it passes through $(4,0)$, substitute $x=4, y=0$:
$4^2 + 0^2 + 2g(4) + 0 + 0 = 0$
$16 + 8g = 0 \implies g = -2$.

Since it passes through $(3,9)$, substitute $x=3, y=9$, $c=0$, and $g=-2$:
$3^2 + 9^2 + 2(-2)(3) + 2f(9) + 0 = 0$
$9 + 81 - 12 + 18f = 0$
$78 + 18f = 0$
$18f = -78 \implies f = -\frac{78}{18} = -\frac{13}{3}$.

The radius $r$ of the circle is given by $r^2 = g^2 + f^2 - c$:
$r^2 = (-2)^2 + \left(-\frac{13}{3}\right)^2 - 0$
$r^2 = 4 + \frac{169}{9} = \frac{36 + 169}{9} = \frac{205}{9}$.

The area of the circle is $\pi r^2$:
$\text{Area} = \pi \left(\frac{205}{9}\right) = \frac{205\pi}{9}$.

Hence, Option D is correct.