Question Bank  100 Arithmetic Questions From Previous CAT Papers (Solved)

Solution: There is more than 1 approach to the question.
First method:
Wind is from A to B. Hence on the forward journey, velocity is v – 50, and on return journey, velocity is v + 50, where v is the velocity of the plane.
Plane cruises from B to A, stays in A for 1hr, then starts the return journey.
B to A = time = 3000 / (v – 50)
1 hr in A.
A to B = time = 3000 / (v + 50)
Total time = 3000 / (v – 50) + 1 + 3000 / (v + 50).
We can see that the onward flight is at 8 AM from B and return flight reaches back B at 8PM. So, total time is 12hrs. So 3000 / (v – 50) + 1 + 3000 / (v + 50) = 12.
Substitute for ‘v’ from the options of the 2nd sub question, and we get v = 550 as the answer.
So, we now know v = 550, B to A time is 3000/v – 50 = 3000/500 = 6hrs. So B to A flight starts from B at 8 AM, and reaches A at 2PM (B’s local time). However A’s local time is 3PM then. So, time difference is 1hr.Second method:
Let the time difference be x hrs. So, time for onward journey is 7 – x (8 AM to 3PM)
Time for return journey is 4 + x (4pm to 8pm)
Velocity for onward is v – 50, and for return is v + 50.
(v50) (7x) = 3000 (Distance = time * velocity)
(v +50) (4+x) = 3000
1st equation = 7v + 50x – vx = 3350
2nd equation = 4v + 50x + vx = 2800
Subtracting these 2 equations, we get, 3v – 2vx = 550
v (3 – 2x) = 550. x can only be 1 among the given options, which means v = 550.

Q2. (CAT 2006)
Arun, Barun and Kiranmala start from the same place and travel in the same direction at speeds of 30, 40 and 60 km per hour respectively. Barun starts two hours after Arun. If Barun and Kiranmala overtake Arun at the same instant, how many hours after Arun did Kiranmala start?
(1) 3
(2) 3.5
(3) 4
(4) 4.5
(5) 5

Solution: B & K overtake A at the same instant. So all the 3 have travelled the same distance so far.
Distance = Time * Speed.
Let time taken by A be x.
Since B starts 2hrs later, time taken by B to reach the same point is x – 2.
Distance travelled by A and B are same, and speed of A & B are 30 and 40 respectively.
30x = 40(x2), solving, we get x = 8.
So A takes 8hrs and B takes 6 hrs, and they travel 30*8 = 240kms.
Time taken by K to travel 240kms is 240/60 = 4hrs. (Speed of K is 60)
So K starts 84 = 4hrs later than A.

Q3. (CAT 2007)
Ten years ago, the ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born during the same year. After another three years, one more member died, again at 60, and a child was born during the same year. The current average age of this eight member joint family is nearest to:
(1) 23 years
(2) 22 years
(3) 21 years
(4) 25 years
(5) 24 years

Solution: 231 + (8 * 3)  60 + 0 = 195
195 + (8 * 3)  60 + 0 = 159
To get the current age, add 8 * 4 = 32 ( 10 years  3 years  3 years = 4 years to reach the current time!)
159 + 8 * 4 = 191
Avg age =191/8 ~ 24

Q4. (CAT 2007)
Mr. David manufactures and sells a single product at a fixed price in a niche market. The selling price of each unit is Rs. 30. On the other hand, the cost, in rupees, of producing x units is 240 + bx + cx2, where b and c are some constants. Mr. David noticed that doubling the daily production from 20 to 40 units increases the daily production cost by 66.66%. However, an increase in daily production from 40 to 60 units results in an increase of only 50% in the daily production cost. Assume that demand is unlimited and that Mr. David can sell as much as he can produce. His objective is to maximize the profit.How many units should Mr. David produce daily?
(1) 130
(2) 100
(3) 70
(4) 150
(5) Cannot be determinedWhat is the maximum daily profit, in rupees, that Mr. David can realize from his business?
(1) 620
(2) 920
(3) 840
(4) 760
(5) Cannot be determined

Solution: Cost price function f(x) = cx^2 + bx + 240.
When x = 20, f(20) = 400c + 20b + 240
f(40) = 1600c + 40b + 240
66.67% is 2/3.
f(40) – f(20) = 2/3 * f(20)
Substituting f(40) and f(20), we get, 1200c+ 20b = (2/3) * (400c + 20b + 240)
Solving, we get 20b + 2800c = 480Now, f(60) = 3600c + 60b + 240.
f(60) – f(40) = 1/2 * f(40)
Substituting, and solving, we get, 360 + 60b + 2400c = 240 + 60b + 3600c
c = 1/10.Substituting we get, b = 10.
So, the cost function f(x) = 0.1x^2 + 10x + 240.
Selling price = 30 * x ( Each unit SP is 30)
So, profit = SP – CP = 30x – (0.1x^2 + 10x + 240) = x^2/10 + 20x – 240.For maximum profit, differentiating this should give 0.
dp/dt = 0, dp/dt = x/5 + 20
20 – x/5 = 0. x= 100.Also double differentiating d2p/dt2, we get a negative number, so profit is maximum. So, profit is maximum when 100 units are produced daily. The maximum profit can be calculated with the equation x^2/10 + 20x – 240, with x = 100, and it is 760.

Q5. (CAT 2008)
Rahim plans to drive from city A to station C, at the speed of 70 km per hour, to catch a train arriving there from B. He must reach C at least 15 minutes before the arrival of the train. The train leaves B, located 500 km south of A, at 8:00 am and travels at a speed of 50 km per hour. It is known that C is located between west and northwest of B, with BC at 60° to AB. Also, C is located between south and southwest of A with AC at 30° to AB. The latest time by which Rahim must leave A and still catch the train is closest to
(1) 6:15 am
(2) 6:30 am
(3) 6:45 am
(4) 7:00 am
(5) 7:15 am

Solution: Basically, we have ABC forming a triangle with right angle at C. (∠ABC = 600 and ∠BAC = 300, So, ∠ACB = 900)
We have AB = 500km. tan 60 = AC/CB. tan 60 = √3.
So, √3 = AC/CB => AC = √3 * CB.
Also in a right triangle, AC^2 + CB^2 = BA^2. (Pythagoras theorem)
So, CB^2 + (√3 * CB)^2 = 500^2
4(CB)^2 = 500^2
2CB = 500. So CB = 250, AC = 250√3.Now, the train travels along CB, at a speed of 50kmph. CB distance is 250kms. So, time for train to reach C is 250/50 = 5hrs. Train starts at 8AM, and hence reaches C by 1PM. So, Rahim needs to reach C minimum by 12:45PM. He has to travel a distance of 250√3 at a speed of 70kmph. This takes a time of 250√3/70 approximately is 6.2hrs. 6.2 hrs is. 6hrs 12 minutes. He cannot start at 6:45, as he will reach only by 12:57. He needs to reach by 12:45. So, he can start by 6:30, so that he reaches before 12:45.

Q6. (CAT 2005)
Ram and Shyam run a race between points A and B, 5 km apart. Ram starts at 9 a.m. from A at a speed of 5 km/hr, reaches B, and returns to A at the same speed. Shyam starts at 9:45 a.m. from A at a speed of 10 km/hr, reaches B and comes back to A at the same speed.At what time do Ram and Shyam first meet each other?
(1) 10:00 a.m.
(2) 10:10 a.m.
(3) 10:20 a.m.
(4) 10:30 a.m.At what time does Shyam overtake Ram?
(1) 10:20 a.m.
(2) 10:30 a.m.
(3) 10:40 a.m.
(4) 10:50 a.m.

Solution: We have 2 points A & B 5kms apart. R runs at 5kmph and S runs at 10kmph. So R reaches B in 1hr and S reaches B in 30 minutes. So, R starts from A by 9am, reaches B by 10am. At 10am, R starts the return journey to A, and reach there by 11am. S starts from A at 9:45am and reaches B at 10:15 am. So, here, we see that R meets S for the 1st time during R’s return journey and S’s onward journey, at some time after 10. We need to find that first.
At 10am, R reaches B and starts to A from B. At 10am, S has travelled for 15 minutes, and hence he has travelled 10 * 1/4 = 2.5 kms (Speed = 10, time = 15 minutes = 1/4 hr). So, at 10 am, the distance between R & S is 2.5 km (R is at B, and S is 2.5 kms away from A). We know that speed of S is double that of R, and hence in a “fixed” time S can travel twice the distance that R travels. So, from 10am, S travels twice the distance that R has travelled, when they both meet each other.
The distance separating R & S is 2.5km (at 10am), and S has travelled twice the distance of R. So S travelled 2.5 * 2/3 when he meets R for the 1st time. Subsequently, R has travelled 2.5 * 1/3 kms. So R travelled 2.5/3 kms at a speed of 5kmph. So, it has travelled in a time of 2.5/ (3 * 5) hours, which is 1/6hrs or 10 minutes. So, they meet for the 1st time at 10:10.
Now, we need to know at what time S overtakes R. We know R starts back from B at 10 and reaches A at 11. S starts back from B at 10:15 and reaches A at 10:45. So, in the return journey, S has overtaken R. R starts at 10 am, let it travel for a time of t hrs. S starts at 10:15 am, so it travels for a time of t hrs – 15 minutes or (t  1/4) hrs when it meets R.
Distance will be same.
So, 5 * t = 10 * (t – 1/4)
On solving, we get t = 1/2 , that is 30 minutes. So, S overtakes R at 10:30am.

Q7. (CAT 2004)
Two boats, traveling at 5 and 10 kms per hour, head directly towards each other. They begin at a distance of 20 kms from each other. How far apart are they (in kms) one minute before they collide?
a. 1/12
b. 1/6
c. 1/4
d. 1/3

Solution: There are more than 1 way to solve the problem, one is by calculating the distance each travel before colliding, then by calculating the time, and proceeding. However, we can calculate the answer just based on the speed information of the two boats.
Let them collide at time ‘t’ at a place ‘x’.
Boat 1 travels at 10kmph, and in 1 minute it travels 1/6km.
Boat 2 travels at 5kmph, and in 1 minute it travels 1/12 km.
During the time of collision, both the boats are together. 1 minute before colliding, so the distance between them is 1/6 + 1/12 = 1/4

Q8. (CAT 2004)
Karan and Arjun run a 100metre race, where Karan beats Arjun by 10 metres. To do a favour to Arjun, Karan starts 10 metres behind the starting line in a second 100 metre race. They both run at their earlier speeds. Which of the following is true in connection with the second race?
a. Karan and Arjun reach the finishing line simultaneously.
b. Arjun beats Karan by 1 metre.
c. Arjun beats Karan by 11 metres.
d. Karan beats Arjun by 1 metre.

Solution: Karan beats Arjun by 10m in the 1st race.
So when K runs 100, A runs 90.
When K runs 1m, A runs 90/100 = 9/10 m.
In the 2nd race, K runs 110m.
So, A runs 110 * 9/10 = 99m.
So K beats A by 1m.

Q9. (CAT 2004)
A milkman mixes 20 litres of water with 80 litres of milk. After selling onefourth of this mixture, he adds water to replenish the quantity that he has sold. What is the current proportion of water to milk?
a. 2 : 3
b. 1 : 2
c. 1 : 3
d. 3 : 4

Solution: Initially, there is 20L of water and 80L of milk. So totally 100L of mixture with milk and water in the ratio 4:1.
He sells 25L of this mixture. So he sells 5L of water and 20L of milk (As the ratio is 4:1)
He has 75L of the mixture left with him in which 15L is water and 60L is milk.
Now, he replenishes the quantity with water, that means he adds 25L of water.
So, now there is totally 15 + 25 = 40L of water and 60L of milk, i.e. water : milk ratio is 2:3.

Q10. (CAT 2004)
If a man cycles at 10 km/hr, then he arrives at a certain place at 1 p.m. If he cycles at 15 km/hr, he will arrive at the same place at 11 a.m. At what speed must he cycle to get there at noon?
a. 11 km/hr
b. 12 km/hr
c. 13 km/hr
d. 14 km/hr

Solution: At 10kmph, he reaches at 1pm, and at 15kmph, he reaches at 11am.
The difference in time is 2hrs. Let the distance be dkm
So, d/10 – d/15 = 2 (Difference in time)
d/30 = 2.
d = 60kms.
So, the total distance is 60kms, time for 1st journey is 6hrs.
He reaches at 1pm when at 10kmph and takes 6hrs, so he starts at 7am.
Now, he needs to reach by noon. i.e. Travel time = 5hrs, and distance = 60kms.
So, the speed by which he should travel is 60/5 = 12kmph.

Q11. (CAT 2004)
A sprinter starts running on a circular path of radius r metres. Her average speed (in metres/minute) is πr during the first 30 seconds, πr/2 during next one minute, πr/4 during next 2 minutes, πr/8 during next 4 minutes, and so on. What is the ratio of the time taken for the nth round to that for the previous round?
a. 4
b. 8
c. 16
d. 32