CAT Arithmetic Questions

Master CAT Arithmetic Questions with practice questions and detailed solutions.

Question 1.

Shruti travels a distance of 224 km in four parts for a total travel time of 3 hours. Her speeds in these four parts follow an arithmetic progression, and the corresponding time taken to cover these four parts follow another arithmetic progression. If she travels at a speed of 960 meters per minute for 30 minutes to cover the first part, then the distance, in meters, she travels in the fourth part is:

Text Explanation:

According to question, Shruti travels a distance of 224 km in four parts for a total travel time of 3 hours.
Given, in the first part time required is 30 minutes.
Also, the time taken to cover these four parts follow arithmetic progression.
So, let us say the times be 30 minutes, (30+d) minutes, (30+2d) minutes and (30+3d) minutes
So, $30+\left(30+d\right)+\left(30+2d\right)+\left(30+3d\right)=180$
or, $6d=180-120$
or, $6d=60$
or, $d=\dfrac{60}{6}=10$ minutes.
So, the time required in these four parts are 30 minutes, 40 minutes, 50 minutes and 60 minutes respectively.
Now, in the first part, speed is 960 metres per minute.
Speed in the four parts is also in arithmetic progression.
Let's say the speeds be $(960+x)$,$(960+2x)$,$(960+3x)$ metres per minute
So, total distance covered = $960\times\ 30+\left(960+x\right)\times\ 40+\left(960+2x\right)\times\ 50+\left(960+3x\right)\times\ 60$ metres
So, $960\times\ 30+\left(960+x\right)\times\ 40+\left(960+2x\right)\times\ 50+\left(960+3x\right)\times\ 60=224000$
or, $960\left(30+40+50+60\right)+40x+\left(2x\right)\left(50\right)+\left(3x\right)\left(60\right)=224000$
or, $172800+320x=224000$
or, $320x=224000-172800=51200$
or, $x=\dfrac{51200}{320}=160$
So, the speed in the fourth part is $960+3x=960+3\times\ 160=1440$ metres per minute
Time in the fourth part is 60 minutes
So, distance covered in the fourth part = $1440\times\ 60=86400$ metres
So, option D is the correct answer.

Question 2.

The ratio of the number of students in the morning shift and afternoon shift of a school was $13 : 9$. After 21 students moved from the morning shift to the afternoon shift, this ratio became $19 : 14$. Next, some new students joined the morning and afternoon shifts in the ratio $3 : 8$ and then the ratio of the number of students in the morning shift and the afternoon shift became $5 : 4$. The number of new students who joined is:

Video Explanation

Text Explanation:

Let M=13k, A=9k.
After 21 students were moved $\frac{13k-21}{9k+21}=\frac{19}{14}\Rightarrow 14(13k-21)=19(9k+21)$
So $182k-294=171k+399\Rightarrow 11k=693\Rightarrow k=63$
So, the number of students in the morning and afternoon shifts are $819-21=798, 567+21=588$ respectively.
Let's assume that $3t$ and $8t$ students joined the respective sessions.
$\dfrac{798+3t}{588+8t}=\dfrac{5}{4}\Rightarrow 4(798+3t)=5(588+8t)$
So $3192+12t=2940+40t\Rightarrow 28t=252\Rightarrow t=9$
Number of new students $=11t=99$

Question 3.

Rita and Sneha can row a boat at 5 km/h and 6 km/h in still water, respectively. In a river flowing with a constant velocity, Sneha takes 48 minutes more to row 14 km upstream than to row the same distance downstream. If Rita starts from a certain location in the river, and returns downstream to the same location, taking a total of 100 minutes, then the total distance, in km, Rita will cover is

Video Explanation

Text Explanation:

Let the river speed be v km/h. For Sneha still-water speed = 6 km/h:
Upstream speed = (6-v), downstream speed = (6+v).
Given $\frac{14}{(6-v)}-\frac{14}{(6+v)}=48$minutes =0.8 hours.
So, $\frac{14(6+v-6+v)}{36-v^2}=0.8 \Rightarrow \frac{28v}{36-v^2}=0.8$
$28v=0.8(36-v^2)=28.8-0.8v^2 \Rightarrow 0.8v^2+28v-28.8=0$
Multiply by 5: $4v^2+140v-144=0$
$v^2+35v-36=0$
We get v = 1
Thus, the river speed is 1 km/h. For Rita, the still water speed is 5 km/h:
Upstream speed = 5-1=4 km/h, Downstream Speed = (5+1=6) km/h.
If she rows d km upstream and returns d km downstream, the total time
$\frac{d}{4}+\frac{d}{6}=d\Big(\frac{1}{4}+\frac{1}{6}\Big)=d\cdot\frac{5}{12}\ \text{hours}$
This equals 100 minutes $=100/60=\dfrac{5}{3}$ hours.
So, $d\cdot\frac{5}{12}=\frac{5}{3}\ \Rightarrow\ d=4\ \text{km}$
Total distance covered 2d=8 km.

Question 4.

The ratio of expenditures of Lakshmi and Meenakshi is 2 : 3, and the ratio of income of Lakshmi to expenditure of Meenakshi is 6 : 7. If excess of income over expenditure is saved by Lakshmi and Meenakshi, and the ratio of their savings is 4 : 9, then the ratio of their incomes is

Video Explanation

Text Explanation:

Let Lakshmi's income = A and expenditure = B.
Let Meenakshi's income = C and expenditure = D.
From B:D = 2:3, take B = 2k and D = 3k.
From A:D = 6:7, $A = \tfrac{6}{7} D = \tfrac{6}{7}\cdot 3k = \tfrac{18}{7}k$.
Lakshmi's saving = $A - B = \tfrac{18}{7}k - 2k = \tfrac{4}{7}k$.
Let Meenakshi's saving $= C - D = C - 3k$. Given the savings ratio $\frac{\frac{4k}{7}}{C-3k} = \tfrac{4}{9}$.
Solve: $\dfrac{4}{7}k\cdot\dfrac{9}{4} = C - 3k \Rightarrow \dfrac{9}{7}k = C - 3k \Rightarrow C = \dfrac{30}{7}k$
So incomes ratio $A:C = \dfrac{18}{7}k : \dfrac{30}{7}k = 18:30 = 3:5.$

Question 5.

Rahul starts on his journey at 5 pm at a constant speed so that he reaches his destination at 11 pm the same day. However, on his way, he stops for 20 minutes, and after that, increases his speed by 3 km per hour to reach on time. If he had stopped for 10 minutes more, he would have had to increase his speed by 5 km per hour to reach on time. His initial speed, in km per hour, was

Video Explanation

Text Explanation:

Let the initial speed of Rahul be $x$ kilometres per hour. Since he travels the distance usually in $6$ hours, (5 pm to 11 pm), the total distance should be $6\times x = 6x$ kilometres.

Let the distance after which Rahul stops for some duration in both scenarios be $y$ kilometres.

In the first scenario, he stops for $20$ minutes or $\dfrac{1}{3}$ hours, therefore, the travel time would be $6-\dfrac{1}{3} = \dfrac{17}{3}$ hours. We have,

$\dfrac{k}{x} + \dfrac{6x-k}{x+3} = \dfrac{17}{3}$

$\Rightarrow \dfrac{6x^2+3k}{x^2+3x} = \dfrac{17}{3}$

$\Rightarrow 18x^2 + 9k = 17x^2 + 51x$

$\Rightarrow x^2 = 51x - 9k$ .....(1)

Similarly, in the second scenario, he stops for $20+10=30$ minutes or $\dfrac{1}{2}$ hours, therefore, the travel time would be $6-\dfrac{1}{2} = \dfrac{11}{2}$ hours. We have,

$\dfrac{k}{x} + \dfrac{6x-k}{x+5} = \dfrac{11}{2}$

$\Rightarrow \dfrac{6x^2 + 5k}{x^2 + 5x} = \dfrac{11}{2}$

$\Rightarrow 12x^2 + 10k = 11x^2 + 55x$

$\Rightarrow x^2 = 55x - 10k$ .....(2)

From equations (1) and (2), we have;

$55x - 10k = 51x - 9k$ or

$4x = k$.

Substituting the value of $k$ in equation (1), we have

$x^2 = 51x - 36x$ or, since $x$ is positive, $x= 15$. Therefore, option B is the correct answer.

Question 6.

Ankita walks from A to C through B, and runs back through the same route at a speed that is 40% more than her walking speed. She takes exactly 3 hours 30 minutes to walk from B to C as well as to run from B to A. The total time, in minutes, she would take to walk from A to B and run from B to C, is

Video Explanation

Text Explanation:

Let the walking speed of Ankita be $5x$, implying that her running speed (which is $40\%$ more than her walking speed), is $1.4\times 5x = 7x$.
Thus, the ratio of her walking and running speeds is $5:7$. Therefore, the ratio of the time taken by Ankita to cover a fixed distance walking and running would by $7:5$.
She takes 3 hours 30 minutes, or $3.5$ hours to walk from B to C. Since Ankita is running from B to C in the second scenario, her time will reduce inversely to the ratio of her speed. The time taken by her to run from B to C would be $\dfrac{3.5}{7}\times 5 = 2.5$ hours.
She takes 3 hours 30 minutes, or $3.5$ hours to run from A to B. Since Ankita is walking from A to B in the second scenario, her time will increase inversely to the ratio of her speed. The time taken by her to walk from A to B would be $\dfrac{3.5}{5}\times 7 = 4.9$ hours.
Therefore, the total time it takes for Ankita in the second scenario is $4.9+2.5 = 7.4$ hours. This gives $7.4*60 = 444$ minutes.

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CAT Arithmetic Questions

Question 1.

Question 2.

Question 3.

Question 4.

The ratio of expenditures of Lakshmi and Meenakshi is 2 : 3, and the ratio of income of Lakshmi to expenditure of Meenakshi is 6 : 7. If excess of income over expenditure is saved by Lakshmi and Meenakshi, and the ratio of their savings is 4 : 9, then the ratio of their incomes is

Question 5.

Question 6.

Ankita walks from A to C through B, and runs back through the same route at a speed that is 40% more than her walking speed. She takes exactly 3 hours 30 minutes to walk from B to C as well as to run from B to A. The total time, in minutes, she would take to walk from A to B and run from B to C, is

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