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

• 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

• 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.

• 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

• Look at the options. Length of the slower train is unique for every options. So we will go for it first
Length of trains = x (Faster) and y (Slower)

While they travel in same direction,
Distance covered is y
Relative speed = 60 - 50 = 10 kmph = 50/18 m/s
Time = 18 s
y = (50/18) x 18 = 50 m
Only option C satisfies!

If there were confusing options, let see how we could proceed.
While they travel in opposite directions,
Distance covered = x + y
Relative speed = 50 + 60 = 110 kmph = 550/18 m/s
time taken = 5 seconds
(x + y) = (550/18) x 5
x = (550/18) x 5 - 50
x = 50 ( 55/18 - 1) = 50 x (37/18) = 102.77 m.

• Q91. (CAT 1990)
If equal numbers of people are born on each day, find the approximate percentage of the people whose birthday will fall on 29thFebruary.(if we are to consider people born in 20thcentury and assuming no deaths).
(a) 0.374
(b) 0.5732
(c) 0.0684
(d) None of these

• Total number of days = 365 x 100 + 25 (25 extra days from leap years) = 36525
We are asked to find (25/36525) x 100 = 100/1461 = 0.0684

• Q92. (CAT 2001)
Shyama and Vyom walk up an escalator (moving stairway). The escalator moves at a constant speed. Shyama takes three steps for every two of Vyom's steps. Shyama gets to the top of the escalator after having taken 25 steps. While Vyom (because his slower pace lets the escalator do a little more of the work) takes only 20 steps to reach the top. If the escalator were turned off, how many steps would they have to take to walk up?
(1) 40
(2) 50
(3) 60
(4) 80

• Let number of steps in the escalator = N
For a complete journey to top, Syama takes 25 steps and escalator takes N - 25 steps
Vyom takes 20 steps and escalator takes N - 20 steps
Ratio of speed of Syama and Vyom = 3 : 2
25/(N - 25) : 20/(N - 20) = 3 : 2
25 (N - 20)/20 (N - 25) = 3/2
N = 50 steps

Alternate approach from Chandra sir (takshzila)

Since the ratio of speeds of Shyam and Vyom is 3 : 2 and the distance that each travels is 25 : 20, the ratio of the time in which they reach the top is (25/3) : (20/2) i.e. 5 : 6.
Thus, if the escalator has a total of n steps, the escalator covers the balance (n – 25) and (n – 20) steps in time intervals in the ratio 5 : 6.
Since speed of escalator is same in both the cases, the ratio of distances covered is same as ratio of time i.e. (n – 25) : (n – 20) is in the ratio 5 : 6.
The difference of 5 parts and 6 parts i.e. 1 part corresponds to an actual difference between (n – 25) and (n – 20) i.e. 5. Thus (n – 25), corresponding to 5 parts, is 25 (or n – 20, corresponding to 6 parts is 30), giving n = 50.

• Q93. (CAT 2001)
Three math classes; X, Y, and Z, take an algebra test.
The average score in class X is 83.
The average score in class Y is 76.
The average score in class Z is 85.
The average score of all students in classes X and Y together is 79.
The average score of all students in classes Y and Z together is 81.
What is the average for all three classes?
(1) 81
(2) 81.5
(3) 82
(4) 84.5

• Say a, b and c are the number of students in X, Y and Z respectively.
We know 83a + 76b = 79 (a + b)
=> 4a = 3b
Also, 76b + 85c = 81(b + c)
=> 4c = 5b
=> 4a : 4b : 4c = 3b : 4b : 5b
=> a : b : c = 3 : 4 : 5
(83a + 76b + 85c) / (a + b + c)
= (83 x 3 + 76 x 4 + 85 x 5)/(3 + 4 + 5) = 978/12 = 81.5

• Q94. (CAT 2001)
There's a lot of work in preparing a birthday dinner. Even after the turkey is in oven, there are still the potatoes and gravy, yams, salad, and cranberries, not to mention setting the table. Three friends, Asit, Arnold, and Afzal, work together to get all of these chores done. The time it takes them to do the work together is six hours less than Asit would have taken working alone, one hour less than Arnold would have taken, and half the time Afzal would have taken working alone. How long did it take them to do these chores working together?
(1) 20 minutes
(2) 30 minutes
(3) 40 minutes
(4) 50 minutes

• Let t is the total hours to do the chores if all friends work together.
Also x, y and z are the time taken by Asit, Arnold and Afzal respectively.
t = x - 6 = y - 1 = z/2
t = 1/x + 1/y + 1/z = xyz / (xy + yz + zx) = x - 6
we know y = x - 5 and z = 2(x - 6)
convert all in terms of x and solve for x.
we get x = 20/3 and t = (20/3) - 6 = 2/3 hours = 40 minutes.

Above solution is very calculation intensive and the smart way would be to use options and solve it. As options are all in terms of the total time (t)
x = t + 6
y = t + 1
z = 2t
1/t = 1/(t + 6) + 1/(t - 1) + 1/2t
Substitute options and t = 2/3 hours (40 minutes) satisfies.

• Q95. (CAT 2001)
A train X departs from station A at 11.00 a.m. for station B, which is 180 km away. Another train Y departs from station B at 11.00 a.m. for station A. Train X travels at an average speed of 70 km/hr and does not stop anywhere until it arrives at station B. Train Y travels at an average speed of 50 km/hr, but has to stop for 15 minutes at station C, which is 60 km away from station B enroute to station A. Ignoring the lengths of the trains, what is the distance, to the nearest km, from station A to point where the trains cross other?
(1) 112
(2) 118
(3) 120
(4) None of these

127

11

5

16

22

19

6

6