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

• 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 determined

What 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 = 480

Now, 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 100-metre 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 one-fourth 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

• Solution: There is more than 1 way to approach the solution; however, I will detail the easiest way to go about it here.
We want to find the ratio of time taken for nth round : time taken for (n-1)th round
It will be same as finding the ratio of time taken for 2nd round : Time taken for 1st round.
1 round = Circumference of the circle = 2πr

1st round :
Speed = πr for 30 seconds. So, total distance travelled = πr/2.
Speed = πr/2 for 1 minute. So, total distance travelled = πr/2.
Speed = πr/4 for 2 minutes. So, total distance travelled = πr/2.
Speed = πr/8 for 4 minutes. So, total distance travelled = πr/2.
So, for a distance of 2πr, time taken is 7.5 minutes.

2nd round:
Speed = πr/16 for 8 minutes. So, total distance travelled = πr/2.
Speed = πr/32 for 16 minutes. So, total distance travelled = πr/2.
Speed = πr/64 for 32 minutes. So, total distance travelled = πr/2.
Speed = πr/128 for 64 minutes. So, total distance travelled = πr/2.
So, for a distance of 2πr, time taken is 120 minutes.

Ratio is 120:7.5 = 16:1.

P.S: Essentially the time taken is a GP, as we see that speed is halved, then quartered and so on. So, we can also calculate by the following method:

1st round: Sum of 1st 4 terms of GP.
2nd round: Sum of 1st 8 terms – Sum of 1st 4 terms.
Then calculate the ratio.

• Q12. (CAT 2004)
In an examination, there are 100 questions divided into three groups A, B and C such that each group contains at least one question. Each question in group A carries 1 mark, each question in group B carries 2 marks and each question in group C carries 3 marks. It is known that the questions in group A together carry at least 60% of the total marks.

If group B contains 23 questions, then how many questions are there in group C?
a. 1
b. 2
c. 3
d. Cannot be determined

If group C contains 8 questions and group B carries at least 20% of the total marks, which of the following best describes the number of questions in group B?
a. 11 or 12
b. 12 or 13
c. 13 or 14
d. 14 or 15

• Solution: Let there be x questions in A, y in B and z in C. So, x + y + z = 100 (As there are 100 questions). Also, questions in A have 1 mark, B have 2 marks and C have 3 marks. So, total marks is x + 2y + 3z

1st sub question :
y = 23. So, x + z = 77. z = 77 – x.
Total marks = x + 2y + 3z = x + 46 + 3z
Substituting for z = 77 – x, we have,
Total marks = z + 3z + 46 = x + 3* (77 – x) + 46 = 277 – 2x.
We know that A should have atleast 60% of the total marks. i.e. x should be atleast 60% of total marks.
So, x ≥ 60/100 * (277 – 2x)
10x ≥ 277 * 6 – 12x
x ≥ 277 * 6/22
x ≥ 75.
So, x should be minimum 76, and y = 23. So z can be only 1 (100 – 76 – 23).

2nd sub question :
z = 8. So, x + y = 92. y = 92 – x.
Total marks = x + 2y + 3z = x + 2y + 24.
Substituting for y = 92 – x, we have,
Total marks = x + 2y + 24 = x + 2* (92 – x) + 24 = 208 –x.
We know that A should have atleast 60% of the total marks. i.e. x should be atleast 60% of total marks.
So, x ≥ 60/100 * (208 –x)
10x ≥ 208 * 6 – 6x
x ≥ 208 * 6/16
x ≥ 78.
So x can be 78, 79 etc etc. z is 8.
So y can be a maximum of 14.
This option is there only in option 3, where y can be 13 or 14. (If you substitute y for 13 or 14, we see that total marks in y is 26 or 28, and it will be atleast 20% of the total marks)

• Q13. (CAT 2003)
In a coastal village, every year floods destroy exactly half of the huts. After the flood water recedes, twice the number of huts destroyed are rebuilt. The floods occurred consecutively in the last three years namely 2001, 2002 and 2003. If floods are again expected in 2004, the number of huts expected to be destroyed is:
a. Less than the number of huts existing at the beginning of 2001.
b. Less than the total number of huts destroyed by floods in 2001 and 2003.
c. Less than the total number of huts destroyed by floods in 2002 and 2003.
d. More than the total number of huts built in 2001 and 2002.

21

6

21

21

16

7

200

21