CAT Arithmetic Marathon Questions 2026 with Video Solutions

We are given that the bus travels at an average speed of $$60$$ km/h when it is moving continuously without any stoppages but including stoppages, its average speed reduces to $$45$$ km/h.

Let's assume the bus travels for $$1$$ hour.

Now, considering the stoppages, the bus will travel $$45$$ km in 1 hour.

Without stoppages, the bus would have covered $$45$$ km in time $$ = \dfrac {45}{60}=0.75$$ hours.

Now, the travel time with stoppages is nothing but the sum of the bus's actual travel time and the stoppage time.

$$\therefore 1=\text{stoppage time}+0.75$$

$$\text{stoppage time}=0.25$$ hours or $$15$$ min.

This means bus stop for $$15$$ min after running for $$45$$ min.

Thus, the stoppage time is $$\dfrac{1}{3}$$​ of the running time.

Therefore, for every $$1$$ hour ($$60$$ minutes) of actual running:

Stoppage time per hour of running time$$=\dfrac{1}{3}*60=20$$ minutes.

Hence, Option B is correct.

Let us assume that the person's usual speed is $$v$$ km/hr and the usual time he takes to cover the distance is $$t$$ hours.

Therefore, the distance travelled by the person will be $$=vt$$ km

Now, we are given that if he were to travel at $$6$$ km/h slower than his usual speed, he would take $$4$$ hours longer to complete the same distance.

Therefore, the distance travelled by the person can be written as $$=(v-6)(t+4)$$ km

Also, we are given that if he were to travel at $$4$$ km/h faster than his usual speed, he would take $$1$$ hour less than the usual time.

Therefore, the distance travelled by the person can be written as $$=(v+4)(t-1)$$ km

We know that the distance travelled by the person in each case is the same.

$$\therefore vt=(v-6)(t+4)=(v+4)(t-1)$$

Now firstly, we have

$$vt=(v-6)(t+4)$$

or $$vt=vt+4v-6t-24$$

or $$4v=6t+24$$

or $$2v=3t+12$$....(1)

Secondly, we have,

$$ vt=(v+4)(t-1)$$

$$vt=vt-v+4t-4$$

$$v=4t-4$$..........(2)

From 1 and 2, we get

$$3t+12=2(4t-4)$$

$$3t+12=8t-8$$

$$5t=20$$

$$t=4$$ hours

Also, $$v=4t-4=4(4)-4=12$$ km/hr

Hence, the usual speed of the person is $$12$$ km/hr

Hence, Option A is correct.

Let the speed of the faster car be $$v$$ kmph and the distance between point A and B be $$x$$ km. The time taken in hours is

$$1 + \dfrac{2}{60}+\dfrac{30}{3600} = \dfrac{25}{24}$$ hours

$$\dfrac{25}{24} = \dfrac{x}{v} \implies v = \dfrac{24}{25}x $$

Distance travelled by the slower car during this time is

$$\dfrac{25}{24} \times 24 = 25 $$km

Now the distance between the two cars is $$(x-25)$$km. Since they are facing opposite directions, using relative speed, we can write

$$\text{time to meet} = \dfrac{x-25}{24+v}$$

$$\dfrac{20}{60} + \dfrac{50}{3600} = \dfrac{25}{72} = \dfrac{x-25}{24+v}$$

$$\dfrac{25}{72}(24+\dfrac{24}{25}x) =x-25$$

$$\dfrac{25}{3}(1+\dfrac{1}{25}x) = x-25$$

$$25+x = 3(x-25)$$

$$x = 50$$ km

Let $$N$$ be the total visible steps when the escalator is off.
Let $$p$$ be the person’s speed (steps/sec).
Let $$e$$ be the escalator’s speed (steps/sec).
Let $$t$$ be the time taken.

When the escalator moves upward and the person walks upward, both move in the same direction: $$N=(p+e)t $$

Given time (=20) sec, $$N=(p+e)\cdot 20 \quad ...(1) $$

When the person stands still, only the escalator moves: $$N=et $$

Given time (=60) sec, $$N=e\cdot 60 \quad ...(2) $$

From (2), $$e=\dfrac{N}{60} $$

Substitute in (1): $$N=\left(p+\dfrac{N}{60}\right)20 $$

or, $$ \dfrac{N}{20}=p+\dfrac{N}{60} $$

or, $$ p=\dfrac{N}{20}-\dfrac{N}{60} =\dfrac{3N-N}{60} =\dfrac{N}{30} $$

Now the escalator moves downward while the person walks upward.
So they move in opposite directions: $$N=(p-e)t $$

or, $$N=\left(\dfrac{N}{30}-\dfrac{N}{60}\right)t $$

or, $$ N=\dfrac{N}{60}t ; t=60 $$

So the required time is 60 seconds .

Amal and Bimal are running in the same direction with their speeds in the ratio 4:7. So, they will meet at |4 - 7|, i.e., 3 distinct points.
Chetan is running in the opposite direction of Amal but twice his speed, which means that the ratio of their speeds is 2:1. This implies that they will meet at (2 + 1) = 3 distinct points.

Now, as Chetan runs at twice the speed of Amal and the ratio of speeds of Amal and Bimal is 4:7, the ratio of speeds of Chetan and Bimal is 8:7. This implies that they will meet at (8 + 7) = 15 distinct points.

As all the meeting points are equidistant, we can divide the track into 15 and 3 equal areas:

X, Y and Z are the 3 points on the track that will be common in all the pairwise meeting points.
Since they are counted thrice (in 3 pairs), we must subtract them twice, giving a total of (3 + 3 + 15 - 2*3) = 15 distinct meeting points.

Since the ratios are preserved, the distance between the current meeting point to B would also increase by 20% if the track length increases by 20%. Thus, the distance between the meeting point to be should be $$\dfrac{840}{1.2}=700$$ meters. This point is in the ratio 1:7 away from A and B respectively. Thus, we get that, $$7k=700$$, and hence $$(1+7)k=800$$, is the total length AB.

Let the speeds of the two runners be $$x$$ and $$7x$$ meters/min respectively.  If they run in opposite directions, the time taken for them to meet will be

$$\Rightarrow \dfrac{800}{x+7x} = \dfrac{100}{x}$$ minutes. This is provided as $$\dfrac{100}{x}=5$$  giving us  $$x=20$$ meters/min.

If the speed of Shyam was twice of what it is now, the two speeds would've been $$x$$ and $$14x$$, giving us the meeting time,

$$\dfrac{800}{15x} = \dfrac{160}{3x} = \dfrac{160}{20\times 3} = \dfrac{8}{3}$$ minutes, or $$\dfrac{8}{3}\times 60 = 160$$ seconds.

Let us assume Manoj and Rushi cover 3x and 5x metres per hop.
The length of the race is equal to 60 hops of Rushi, i.e., $$60\times\ 5$$x = 300x.

Now, assume Manoj hopped N times before Rushi started, and the race ended in a dead heat.
This implies that in the time Rushi covered the full 300x distance, Manoj covered only 300x - N*3x.

As they both took an equal amount of time to cover these distances, the ratio of distance covered will be equal to the ratio of their speeds.

$$\dfrac{300x}{300x\ -\ 3Nx}=\dfrac{5}{3}$$
$$900x\ =\ 1500x-15Nx$$
$$15Nx\ =\ 600x$$
N = 40 hops.
So, the correct answer is 40.

Speed of the boat in still water (B) = 10 km/hr

Speed of the stream/river current (S) = 2 km/hr

$$\Rightarrow$$ Downstream speed of the boat = B + S = 10 + 2 = 12 km/hr

The boat covers the downstream distance in 2 hours. Let the distance be D.

$$\Rightarrow$$ D = 12 $$\times$$ 2 = 24 km

Now the boat has to travel double the distance upstream.

$$\Rightarrow$$ New distance = 2D = 2 $$\times$$ 24 = 48 km

$$\Rightarrow$$ Upstream speed of the boat = B - S = 10 - 2 = 8 km/hr

$$\Rightarrow$$ Time taken = 48 / 8 = 6 hours

Hence, option C is the correct choice.

Let the length of the tunnel be $$d$$ meters.

The time it takes for train A's last carriage to pass the tunnel after the engine has already passed is $$=\dfrac{\text{length of train A - length of tunnel}}{\text{speed of train A}}$$ seconds

The speed of train A in metres per second is $$\dfrac{5}{18}\times 108 = 30$$ metres per second.

Thus, we have, $$x=\dfrac{200-d}{30}$$ seconds

Similarly, the speed of train B in metres per second is $$\dfrac{5}{18}\times 90= 25$$ metres per second.

Thus, we further have, $$y=\dfrac{250-d}{25}$$ seconds

We are provided that $$x+y=5\dfrac{2}{3}$$ or $$x+y= \dfrac{17}{3}$$

Therefore, we get,

$$\dfrac{200-d}{30}+\dfrac{250-d}{25} = \dfrac{17}{3}$$

$$\Rightarrow 5000-25d+7500-30d = 4250$$

$$\Rightarrow 8250 = 55d$$  or  $$d=150$$ metres

10 men can complete the work in 12 days. Therefore, the total work is equal to 120 man days.
The same work can be completed by 20 women in 8 days. Therefore, the total work is equal to 160 women days.

120m = 160w
=> m = 4w/3.

If a man and woman work together, then 1.75m units of work will be completed in a day.
Total work = 120m*1.75
Time taken = 120*1.75/1.75 = 120 days. Therefore, option B is the right answer.

Work = 20m*15 = 300m
Also, work = 35w*10 = 350w.
300 man days = 350 woman days.
6m = 7w
Therefore, the work completed by a man in 6 days is equal to the work completed by a woman in 7 days.

13 persons can complete the work in 25 days. Let the number of women be ‘x’ and the number of men be 13-x.

25*((13-x)m + xw) = 300m
(13-x)m + 6xm/7 = 12m
7*(13-x)m + 6xm = 84m
91m - 7xm + 6xm = 84m
91 - x = 84
x = 7

Therefore, option B is the right answer.

Let the total work be 60 units (LCM of 10, 15 and 20)
Efficiency of A = 60/10 = 6 units/day
Efficiency of B = 60/15 = 4 units/day
Efficiency of C = 60/20 = 3 units/day
They all can do 13 units of work together.
Then, Work done in 3 days = 13*3 = 39 units
Remaining work = 60-39 = 21 units.
21 units of work will be finished by B and C together in 21/7 = 3 days
Hence, Total work will be completed in 3+3 = 6 days.

Let P, Q and R alone fill the tank in p, q and r hours respectively.
Thus, $$\frac{8}{p} + \frac{8}{q} + \frac{8}{r} = 1$$
Also,
$$\frac{12}{p} + \frac{12}{q} + \frac{6}{r} = 1$$
$$\frac{12}{8} - \frac{12}{r} + \frac{6}{r} = 1$$
0.5 = $$\frac{6}{r}$$
r = 12 hours
Hence, option D is the correct choice.

Let the cost price of one unit of the article be $$100x$$.
He marks up the second article by 60% over cost price and offers successive discounts of 20% and 12.5%.
So, Selling price = $$100x\times1.6\times0.8\times0.875=112x\ $$
He also charges a 10% service fee on the selling price.
So, final selling price $$=112x\times1.1=123.2x\ \ $$

We are given that customer pays Rs. 1386 for this article.

So, $$123.2x=1386\ \ $$
So, $$x=11.25$$
Cost price of each unit$$=100x=1125$$
Cost price of all three units$$=3*1125=₹3375$$

Total profit on all 3 units combined is to be at least 20%.
Let the total selling price be $$y$$.
So, $$y\ge1.2\times3375\ \ $$
$$y\ge4050$$

The selling price of the damaged unit$$=0.75\times1125=843.75\ $$
The selling price of the second unit$$=1386\ $$
So, the minimum selling price of the third unit$$=4050-\left(1386+843.75\right)=1820.25\ $$
Option D.

Let the cost price of each car be $$x$$.

Profit of $$10$$% on the cost price would imply a selling price of $$1.1x$$ and a loss of $$10%$$ would imply a selling price of $$0.9x$$.

From the first transaction, $$15(1.1x) + 5(0.9x) - 20x = 100 \implies 21x - 20x = 100 \implies x = 100.$$

Thus, the cost price of each car is $$100$$.

Since the wholesaler sells the $$20$$ cars for $$2332$$ at a $$10\%$$ profit, his cost price is $$\dfrac{2332}{1.1} = 2120.$$

Hence, the shopkeeper sold the $$20$$ cars to the wholesaler for $$2120$$.

Since $$100$$ is the cost price, the selling prices of the cars would be $$1.1\times 100 = 110$$ and  $$0.9\times 100 = 90$$

Let $$a$$ be the number of cars sold at a $$10\%$$ profit by the shopkeeper in the second transaction. Then,

$$110a + 90(20-a) = 2120 \implies 20a = 320 \implies a = 16.$$

Therefore, the required ratio is $$15:16$$.

The profit ratio A, B and C is
5000 x 12 : (10000 x 3) + (8000 x 9) : (7000 x 6) + (10000 x 6)
60 : (30 + 72) : (42 + 60)
60 : 102 : 102
10 : 17 : 17
Total profit at the end of year = ₹13,200
$$\therefore$$ Profit earned by C = $$\frac{17}{10+17+17}\times$$13200 = $$\frac{17}{44}\times$$13200 = ₹5100
Hence, the correct answer is Option C

Shopkeeper pays for 1000 g but actually receives 1200 g.

Let the original cost price be 1000 Rs per 1000 gm.

Thus, the cost incurred by the shopkeeper = Rs 1000.

Now 10% of the goods are wasted.

Remaining goods = 1200 ( 1 - 10%) = 1080 gm.

Marked price = 1000 ( 1 + 25%) = Rs 1250

Discount = 1250 $$\times$$ 20% = Rs 250

Stated selling price = 1250 - 250 = Rs 1000

He claims to sell at Rs 1000 per 1000 gm but gives only 800 g per transaction.

$$\Rightarrow$$ Revenue per 800 gm = Rs 1000

$$\Rightarrow$$ Total revenue from 1080 gm = 1080 $$\times$$ (1000/800) = Rs 1350

Overall profit = Total revenue - Total investment = 1350 - 1000 = Rs 350

% profit = 350 / 1000  $$\times$$ 100 = 35%

Hence, option B is the correct choice.

Let 1000 mL (1 litre) of milk originally cost Rs. 1200.

Now, the milkman pays for 1000 mL but actually gets 20% extra.

$$\Rightarrow$$ Milkman receives 1200 mL for Rs. 1200

$$\Rightarrow$$ Cost incurred by the milkman = Rs 1200 / 1200 mL = Rs 1/mL

Now, he replaces the milk with water, so the final mixture has a milk-to-water ratio of 3:1.

He also cheats while selling by giving 20% less. Thus, for every 1 litre, he is actually selling only 800 mL.

In this 800 mL mixture, the actual milk is 75% of the total (3:1 ratio).

Thus, for the price of 1000 mL of mixture, he is actually selling (800 $$\times$$ 0.75) = 600 mL

He marks up the cost price of pure milk by 25% and gives a discount of 10%

$$\Rightarrow$$ Selling price = 1200 $$\times$$ 1.25 $$\times$$ 0.9 = Rs 1350

Thus, he sells 600 mL of pure milk at Rs 1350

$$\Rightarrow$$ Actual selling price = Rs 1350 / 600 mL = Rs 2.25/mL

$$\Rightarrow$$ Profit = Rs 2.25 $$-$$ Rs 1 = Rs 1.25 /mL

Thus, percentage profit = 1.25 $$\times$$ 100 = 125%

Hence, option C is the correct choice.

Let the profit earned on a domestic rose be 100p. Therefore the profit earned on International rose will be 200p.

Let "x" be the number of imported roses the shopkeeper mistakenly delivers.Therefore the number of domestic roses delivered will be (100-x)

Therefore the profit he earns will be equal to 100p(100-x)+200px.

Now we know that this is 12.5% less than what he would have earned had he delivered the right number of roses in which case since the number of roses delivered is exchanged profit will be 100px+ 200(100-x).

Therefore 87.5%(100px+200(100-x))= 100p(100-x)+200px

Therefore 7(100px + 200p(100-x))=8(200px+100p(100-x))\

Therefore 800px+80000p=-700px+140000p

Therefore 1500 px= 60000p

Therefore x=40.

Since that is the number of roses the shopkeeper accidentally delivered, the number of imported roses he should have delivered will be equal to 60.Therefore the number of domestic roses he should have delivered will be equal to 40.

The cost price of a smartphone is ₹ $$9,000$$

Now the shopkeeper marks the cost price by $$40$$% and also offers a discount of $$10$$ %

So, the selling price of the smartphone is $$=1.4*0.9*9000=$$Rs $$11,340$$

Now, let us assume that the yearly instalment is Rs $$i$$

$$11,340(1.1)^3=i(1.1)^2+i(1.1)+i$$

$$11,340*1.331=1.21i+1.1i+i$$

$$11,340*1.331=3.31i$$

$$i=$$ Rs $$4,559.98$$

Hence, the total amount paid by the customer is $$=3*4559.98=$$ Rs $$13679.94$$

The profit percent made by the shopkeeper is$$=\dfrac{(13679.94-9000)*100}{9000}=51.99$$%

Hence, Option C is correct.

The joint variation relationship: $$Y = k \cdot F \cdot \sqrt{R}$$

First, find the constant of proportionality ($$k$$) using the initial conditions ($$Y = 2400$$, $$F = 50$$, $$R = 64$$):

$$2400 = k \cdot 50 \cdot \sqrt{64}$$
$$2400 = k \cdot 50 \cdot 8$$

$$2400 = 400k \Rightarrow k = 6$$

Now, use the formula $$Y = 6 \cdot F \cdot \sqrt{R}$$ to find the new fertilizer requirement ($$Y = 3240$$, $$R = 81$$):

$$3240 = 6 \cdot F \cdot \sqrt{81}$$

$$3240 = 6 \cdot F \cdot 9$$

$$3240 = 54F \Rightarrow F = \dfrac{3240}{54} = 60 \text{ kg}$$

Let the weights of the three pieces be $$x$$, $$2x$$, and $$3x$$.
The original total weight of the gemstone was $$x + 2x + 3x = 6x$$.

The value formula is $$V = k \cdot (\text{Weight})^2$$.

Original Value: $$V_{\text{original}} = k \cdot (6x)^2 = 36kx^2$$

We are given that the original value is Rs. 72,000:

$$36kx^2 = 72000 \Rightarrow kx^2 = 2000$$

Combined value of the three broken pieces:
$$V_{\text{piece1}} = k(x)^2 = kx^2$$
$$V_{\text{piece2}} = k(2x)^2 = 4kx^2$$
$$V_{\text{piece3}} = k(3x)^2 = 9kx^2$$
$$\text{Total New Value} = kx^2 + 4kx^2 + 9kx^2 = 14kx^2$$

$$\text{Loss} = 36kx^2 - 14kx^2 = 22kx^2$$

Since we know $$kx^2 = 2000$$:

$$\text{Loss} = 22 \times 2000 = 44,000$$

Let a be the speed with no passengers. Let t be the reduction in speed with each person.
Hence, a-t=60 and a-10t=54.
Hence, 9t=6.
So the bus decelerates by 6/9 kmph per passenger. Hence, with 90 passengers the bus will come to a stand-still.
Max number of passengers = 90-1=89.

The surface area of 2m x 2m x 2m cube is $$6a^2$$ or $$6a^2=6\times\left(2\right)^2=24\ m^2$$

As the profit is proportional to the surface area, assume the constant of proportionality to be $$k$$

So, $$24k=10000$$

or $$k=\dfrac{10000}{24}=\dfrac{1250}{3}$$

Now, the combined area of the five blocks of wood is calculated as,

The surface area of cuboid is $$2\left(lb+bh+hl\right)$$

i.e. $$2\left(lb+bh+hl\right)=2\left(4+2+2\right)=16\ m^2$$

The surface area of the four cubes is $$4\times\left(6\times1^2\right)=24$$

So, the combined area now is 16+24 = 40

Hence, the profit now will be $$40k=40\times\dfrac{1250}{3}=16667$$

Hence, the answer is Rs. 16667

Let us assume the quantity of Local coffee mixed is X kg.
Total mixture sold = 80 kgs + X kgs.
Cost Price of imported coffee = 80*220 = 17600 Rs.
Cost Price of local coffee = 50*X = 50X.
Total Cost Price = Rs. 17600 + 50X.
Total Selling Price = 220*(80 + X) = Rs. 17600 + 220X.
Profit Earned = (17600 + 220X) - (17600 + 50X) = Rs. 170X.
Profit Percentage = $$\frac{170X}{1760+50X}\times\ 100$$
==> $$\frac{170X}{17600+50X}\times\ 100\ =\ 20$$
==> 850X = 17600 + 50X
==> 800X = 17600
==> X = 22.
The quantity of local coffee used in the mixture is 22 Kgs.

The man first pours 200 mL of pure alcohol into his glass, then refills the bottle with 200 mL of water. The bottle now contains 800 mL of alcohol and 200 mL of water, giving an alcohol-to-water ratio of 4:1.

When serving the eldest son, the man pours enough of this mixture so that the son receives exactly 200 mL of alcohol. Since the mixture is in a 4:1 ratio, the eldest son receives 200 mL of alcohol and 50 mL of water.

The same process continues for each subsequent son. Although the amount of water in each drink changes, every son receives exactly 200 mL of pure alcohol.

Before serving the youngest son, the man has already distributed drinks to himself and the three older sons. Since each of those four glasses contains 200 mL of alcohol, a total of 4 × 200 = 800 mL of alcohol has been removed from the bottle. Therefore, only 200 mL of alcohol remains.

Because the bottle is refilled with water after each serving, the bottle still contains 1 litre of solution when the youngest son is served. To receive the remaining 200 mL of alcohol, he must be given the entire contents of the bottle: 1,000 mL of solution containing 200 mL of alcohol and 800 mL of water.

Therefore, the youngest son receives 800 mL of water.

On the first day, drink sold= 100 ml. And on every subsequent day, sale increases by 50 ml.

So, sales of drink follows an A.P with first term= 100 ml and ends on the day when sales= 10L or 10000 ml.

With common difference= 50 ml,

the nth day when the entire drink will be sold can be found using a+ (n-1)d= 10000

=> 100+ 50(n-1)= 10000

=> 10+ 5n-5= 1000

=> 5n= 995

.'. n= 995/5= 199 days.

Total drinks sold= Sum of A.P. with n= 199, first term= 100 ml and last term= 10000 ml= $$\ \frac{\ n}{2}\left[a+l\right]=\ \frac{\ 199}{2}\left[100+10000\right]=199\left[50+5000\right]=199\times5050=1004950\ ml\ \ $$ or 1004.95 L.

Therefore, % of Taaza sold= $$\ \frac{\ 10}{1004.95}\times100\%=\ 0.995\%\ $$