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

say there were n workers in total. So it took n days to finish the job
If 1 unit of work is completed by each worker then
day 1 = n units of work is done
day 2 = n  1 units of work is done
day 3 = n  2 units of work is done
..
day n  2 = 2 units of work is done
day n  1 = 1 unit of work is done
day n  work completedSo total = 1 + 2 + 3 + ... + n = n(n + 1)/2 units of work
If all worked together, then work would be completed in 2n/3 days.
=> unit of work = ( 1 + 1 + ... n times) 2n/3 = 2n^2/32n^2/3 = n(n + 1)/2
4n = 3n + 3
n = 3

Q81. (CAT 1993)
A ship leave on a long voyage. When it is 18 miles from the shore, a seaplane, whose speed is ten times that of the ship, is sent to deliver mail. How far from the shore does the seaplane catch up with the ship?
(a) 24 miles
(b) 25 miles
(c) 22 miles
(d) 20 miles

Speed of ship = S
Speed of seaplane = 10S
Let say we consider the travel starts 18 miles from the shore (both ship and plane are travelling now)
When they meet Ship travels x (say) miles and plane travels 18 + x miles.
Time is same for both
So, x/S = (18 + x)/10S
10x = 18 + x
x = 2
So they will meet 18 + 2 = 20 miles from the shore.

Q82. (CAT 1991)
Three machines, A, B and C can be used to produce a product. Machine A will take 60 hours to produce a million units. Machine B is twice as fast as Machine A. Machine C will take the same amount of time to produce a million units as A and B running together. How much time will be required to produce a million units if all the three machines are used simultaneously?
(a) 12 hours
(b) 10 hours
(c) 8 hours
(d) 6 hour

Let say A does n unit of work per day
So B does 2n units of work per day
C does 2n + n = 3n units of work per day
So combined A, B and C can do 6n units of work per day
As A (n units of work per day) takes 60 hours to produce a million units, we can say it will take 60/6 = 10 hours to produce a million units by all three machines working together (6n units of work per day)

Q83. (CAT 1991)
A player rolls a die and receives the same number of rupees as the number of dots on the face that turns up. What should the player pay for each roll if he wants to make a profit of one rupee per throw of the die in the long run?
(a) Rs. 2.50
(b) Rs. 2
(c) Rs.3.50
(d) Rs. 4

For an unbiased dice, we can get 1, 2, 3, 4, 5 and 6
so for long run, the average earning could be (1 + 2 + 3 + 4 + 5 + 6)/6 = 21/6 = 3.5
To make a profit of 1 rupee per throw, he should pay 3.5  1 = 2.50 for each throw.

Q84. (CAT 1991)
Every day Neera’s husband meets her at the city railway station at 6.00 p.m. and drives her to their residence. One day she left early from the office and reached the railway station at 5.00 p.m. She started walking towards her home, met her husband coming from their residence on the way and they reached home 10 minutes earlier than the usual time. For how long did she walk?
(a) 1 hour
(b) 50 minutes
(c) 1/2 hour
(d) 55 minutes

They arrived home 10 minutes earlier. So they saved 5 minutes in each direction.
Means they saved 5 minutes in the Station  Home journey
As the husband met her 5 minutes early, Neeraja would have walked for 1 hour  5 minutes = 55 minutes

Q85. (CAT 1991)
A sum of money compounded annually becomes Rs.625 in two years and Rs.675 in three years. The rate of interest per annum is
(a) 7%
(b) 8%
(c) 6%
(d) 5%

625 * 1 * r/100 = 675  625 = 50
r = 8%

Q86. (CAT 1990)
In a stockpile of products produced by three machines M1, M2 and M3, 40% and 30% were manufactured by M1 and M2 respectively. 3% of the products of M1 are defective, 1% of products of M2 defective, while 95% of the products of M3 are not defective. What is the percentage of defective in the stockpile?
(a) 3%
(b) 5%
(c) 2.5%
(d) 4%

Let say we have 100 products
40 are from M1 and 30 are from M2 => 30 are from M3
1.2 product from M1 are defective
0.3 product from M2 are defective
1.5 product from M3 are defective
So total 1.2 + 0.3 + 1.5 = 3 out of 100 are defective
3% is the answer.

Q87. (CAT 1990)
A car after traveling 18 km from a point A developed some problem in the engine and speed became 4/5 of its original speed As a result, the car reached point B 45 minutes late. If the engine had developed the same problem after traveling 30 km from A, then it would have reached B only 36 minutes late. The original speed of the car (in km per hour) and the distance between the points A and B (in km.) is
(a) 25, 130
(b) 30,150
(c) 20, 90
(d) None of these

Let's consider the points between 18 km and 30 km from A (the magic happens here!)
Speed of Car = S
In the first case, due to the problem, car covered this distance in 4S/5 speed. (1/5 decrease in speed)
This should amount to 1/4 increase in time.
We know this difference in time is 45  36 = 9 minutes
so t/4 = 9 => t = 36 (t being the time required to cover the 30  18 = 12 km stretch)
12 km in 36 minutes => speed = 20.Let d be the distance between A and B
In case 1, first 18 km was covered in 20 kmph
and the rest, (d  18) km was covered in 4/5th of speed = 16 kmph.
Had the total distance is covered in 20 kmph, we would have reached 45 minutes early.
18/20 + (d  18)/16  d/20 = 45 minutes = 3/4 hours
d = 78 km.Answer is none of these.

Q88. (CAT 1991)
A man starting at a point walks one km east, then two km north, then one km east, then one km north, then one km east and then one km north to arrive at the destination. What is the shortest distance from the starting point to the destination?
a) 2√2 km
b) 7 km
c) 3√2 km
d) 5 km

Shortest distance is √(3^2 + 4^2) = 5 km

Q89. (CAT 1990)
A, B and C individually can finish a work in 6, 8 and 15 hours respectively. They started the work together and after completing the work got Rs.94.60 in all. When they divide the money among themselves, A, B and C will respectively get (in Rs.)
(a) 44, 33, 17.60
(b) 43, 27.20, 24.40
(c) 45, 30, 19.60
(d) 42, 28, 24.60

Work done by A, B and C would be in the ratio 1/6 : 1/8 : 1/15 = 20 : 15 : 8
We can see C will get little more than half of what B got. Only Option A satisfies (no need to solve!)If there were confusing options, then
A will get 94.60 x 20/43 = 44
B will get 94.60 x 15/43 = 33
C will get 94.60 x 8/43 = 17.60

Q90. (CAT 1990)
Two trains are traveling in opposite direction at uniform speed 60 and 50 km per hour respectively. They take 5 seconds to cross each other. If the two trains had traveled in the same direction, then a passenger sitting in the faster moving train would have overtaken the other train in 18 seconds. What are the lengths of trains (in metres)?
(a) 112.78
(b) 97.78, 55
(c) 102.78, 50
(d) 102.78, 55