CAT 2019 Slot 1 — QA questions with answers

Let the length of the racecourse be $D$ meters.
Let the speeds of the three horses be $S_1, S_2,$ and $S_3$.

When the first horse finishes the race, it has covered distance $D$.
In the same time, the second horse has covered $D - 11$ meters, and the third horse has covered $D - 90$ meters.
The ratio of the speeds of the second and third horse is the same as the ratio of the distances they cover in the same time:
$\frac{S_2}{S_3} = \frac{D - 11}{D - 90}$ --- (1)

When the second horse finishes the race, it has covered distance $D$.
In the same time, the third horse has covered $D - 80$ meters.
The ratio of their speeds is:
$\frac{S_2}{S_3} = \frac{D}{D - 80}$ --- (2)

Equating (1) and (2):
$\frac{D - 11}{D - 90} = \frac{D}{D - 80}$

Cross-multiplying:
$(D - 11)(D - 80) = D(D - 90)$
$D^2 - 80D - 11D + 880 = D^2 - 90D$
$D^2 - 91D + 880 = D^2 - 90D$
$-91D + 880 = -90D$
$880 = 91D - 90D$
$D = 880$

The length of the racecourse is 880 meters.

The function is defined as:
$f(n) = n(n + 1)$ if $n$ is even
$f(n) = n + 3$ if $n$ is odd

We are given $8f(m + 1) - f(m) = 2$ where $m$ is a positive integer.
We must check two cases based on whether $m$ is even or odd.

Case 1: $m$ is even
If $m$ is even, then $f(m) = m(m + 1)$.
Since $m$ is even, $m + 1$ is odd. So $f(m + 1) = (m + 1) + 3 = m + 4$.
Substitute into the given equation:
$8(m + 4) - m(m + 1) = 2$
$8m + 32 - m^2 - m = 2$
$-m^2 + 7m + 32 = 2$
$m^2 - 7m - 30 = 0$
$(m - 10)(m + 3) = 0$
Since $m$ is a positive integer, $m = 10$.
We check if $m=10$ is even. Yes, it is. So this is a valid solution.

Case 2: $m$ is odd
If $m$ is odd, then $f(m) = m + 3$.
Since $m$ is odd, $m + 1$ is even. So $f(m + 1) = (m + 1)((m + 1) + 1) = (m + 1)(m + 2)$.
Substitute into the given equation:
$8(m + 1)(m + 2) - (m + 3) = 2$
$8(m^2 + 3m + 2) - m - 3 = 2$
$8m^2 + 24m + 16 - m - 3 - 2 = 0$
$8m^2 + 23m + 11 = 0$
The discriminant $\Delta = 23^2 - 4(8)(11) = 529 - 352 = 177$.
Since $\sqrt{177}$ is not an integer, $m$ will not be an integer. Thus, no valid solution in this case.

The only solution is $m = 10$.

Let the three speeds be $S_1 = 10$, $S_2 = 20$, and $S_3 = 30$ kmph.

Amal's Journey:
He took each mode for $1/3$rd of his total journey time.
Let his total time be $3t$ hours.
He travels at 10 kmph for $t$ hours, at 20 kmph for $t$ hours, and at 30 kmph for $t$ hours.
Total distance covered by Amal $= 10t + 20t + 30t = 60t$ km.
Average speed of Amal $= \frac{60t}{3t} = 20$ kmph.
Amal's travel time $= \frac{D}{20}$.

Bimal's Journey:
He took each mode for $1/3$rd of the total distance.
Let the total distance be $D$. He travels $D/3$ distance at each speed.
Total time taken by Bimal $= \frac{D/3}{10} + \frac{D/3}{20} + \frac{D/3}{30}$
$= \frac{D}{3} \left(\frac{1}{10} + \frac{1}{20} + \frac{1}{30}\right)$
$= \frac{D}{3} \left(\frac{6 + 3 + 2}{60}\right) = \frac{D}{3} \left(\frac{11}{60}\right) = \frac{11D}{180}$.

We need to find the percentage by which Bimal's travel time exceeds Amal's travel time.
Amal's travel time = $\frac{D}{20} = \frac{9D}{180}$.
Bimal's travel time = $\frac{11D}{180}$.

Excess time = $\frac{11D}{180} - \frac{9D}{180} = \frac{2D}{180}$.

Percentage excess = $\frac{\text{Excess time}}{\text{Amal's time}} \times 100$
$= \frac{2D/180}{9D/180} \times 100 = \frac{2}{9} \times 100$
$= 22.22\%$.

The nearest integer is 22.

Hence, Option A is the correct answer.

Let $S_n$ be the sum of the first $n$ terms of the series.
$S_n = a_1 + a_2 + a_3 + \dots + a_n = 3(2^{n+1} - 2)$.

We need to find $a_{11}$, the 11th term of the series.
The $n$-th term of a sequence can be found using $a_n = S_n - S_{n-1}$.
So, $a_{11} = S_{11} - S_{10}$.

Substitute $n = 11$ into the sum formula:
$S_{11} = 3(2^{11+1} - 2) = 3(2^{12} - 2) = 3(4096 - 2) = 3(4094) = 12282$.

Substitute $n = 10$ into the sum formula:
$S_{10} = 3(2^{10+1} - 2) = 3(2^{11} - 2) = 3(2048 - 2) = 3(2046) = 6138$.

Now find $a_{11}$:
$a_{11} = 12282 - 6138 = 6144$.

Alternatively, algebraically:
$a_{11} = 3(2^{12} - 2) - 3(2^{11} - 2)$
$a_{11} = 3(2^{12} - 2 - 2^{11} + 2) = 3(2^{12} - 2^{11}) = 3 \cdot 2^{11}(2 - 1) = 3 \cdot 2^{11}$.
$2^{10} = 1024 \implies 2^{11} = 2048$.
$3 \times 2048 = 6144$.

Let the total number of students in the class be $x$.
Number of girls = 60% of $x = 0.6x$.
Number of boys = 40% of $x = 0.4x$.

We are given that there are 30 more girls than boys:
$0.6x - 0.4x = 30$
$0.2x = 30 \implies x = 150$.

Total number of students = 150.
Number of girls = $0.6 \times 150 = 90$.
Number of boys = $0.4 \times 150 = 60$.

Total number of students who pass the examination = 68% of 150.
$0.68 \times 150 = 102$.

Out of these 102 students who passed, 30 are boys.
So, the number of girls who passed = $102 - 30 = 72$.

Total number of girls is 90.
Number of girls who do not pass = $90 - 72 = 18$.

The percentage of girls who do not pass is:
$\frac{\text{Girls who did not pass}}{\text{Total girls}} \times 100$
$= \frac{18}{90} \times 100 = \frac{1}{5} \times 100 = 20\%$.

Let the work done by 1 man in 1 day be $M$, and the work done by 1 machine in 1 day be $E$.
We are given that 3 men and 8 machines can finish the job in half the time taken by 3 machines and 8 men.
Since time and efficiency (work done per day) are inversely proportional, this means 3 men and 8 machines are twice as efficient as 3 machines and 8 men.

$3M + 8E = 2(3E + 8M)$
$3M + 8E = 6E + 16M$
$8E - 6E = 16M - 3M$
$2E = 13M$

This means the work done by 2 machines in a day is exactly equal to the work done by 13 men in a day.

We are given that 2 machines can finish the job in 13 days.
Since 2 machines are equivalent to 13 men in terms of work rate, replacing the 2 machines with 13 men would still finish the job in exactly the same time (13 days).

Therefore, 13 men can finish the job in 13 days.

A step is either $(x, y) \to (x+1, y)$ [Right step, R] or $(x, y) \to (x, y+1)$ [Up step, U].
This is a standard grid path problem.

The number of paths from $(x_1, y_1)$ to $(x_2, y_2)$ is given by $\binom{\Delta x + \Delta y}{\Delta x}$ where $\Delta x = x_2 - x_1$ and $\Delta y = y_2 - y_1$.

We need to go from $(1, 1)$ to $(4, 6)$ and then from $(4, 6)$ to $(8, 10)$.

Path from (1, 1) to (4, 6):
$\Delta x_1 = 4 - 1 = 3$
$\Delta y_1 = 6 - 1 = 5$
Total steps = $3 + 5 = 8$.
Number of paths = $\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.

Path from (4, 6) to (8, 10):
$\Delta x_2 = 8 - 4 = 4$
$\Delta y_2 = 10 - 6 = 4$
Total steps = $4 + 4 = 8$.
Number of paths = $\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$.

Since these are sequential and independent choices, we multiply the number of paths:
Total paths = $56 \times 70 = 3920$.