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

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

It will take train Y 60/50 + 1/4 = 29/20 hours to leave station C (which is 60 km from B)
In this time train X will travel 70 x 29/20 = 101.5 km
So the total distance between A and B is now reduced to 180  (101.5 + 60) = 18.5 km
Relative speed of trains (travelling in opposite direction) = 70 + 50 = 120 kmph
So they will meet after 18.5/120 hours.
Distance travelled by X in this time = 70 * 18.5/120 = 10.8 km
So they will meet at a distance 101.5 + 10.8 = 112 km (approx)

Q96. (CAT 2001)
Fresh grapes contain 90% water by weight while dry grapes contain 20% water by weight. What is the weight of dry grapes available from 20 kg of fresh grapes?
(1) 2 kg
(2) 2.4 kg
(3) 2.5 kg
(4) None of these

Fruit content in Fresh grapes = Fruit content in dried grapes
20 * 10/100 = x * 80/100
x = 20/8 = 2.5

Q97. (CAT 2003 Leaked)
In a 4000 metre race around a circular stadium having a circumference of 1000 metres, the fastest runner and the slowest runner reach the same point at the end of the 5th minute, for the first time after the start of the race. All the runners have the same starting point and each runner maintains a uniform speed throughout the race. If the fastest runner runs at twice the speed of the slowest runner, what is the time taken by the fastest runner to finish the race?
(1) 20 min
(2) 15 min
(3) 10 min
(4) 5 min

Q98. (CAT 2003 Leaked)
At the end of year 1998, Shepard bought nine dozen goats. Henceforth, every year he added p% of the goats at the beginning of the year and sold q% of the goats at the end of the year where p > 0 and q > 0. If Shepard had nine dozen goats at the end of year 2002, after making the sales for that year, which of the following is true?
(1) p = q
(2) p < q
(3) p > q
(4) p = q/2

Q99. (CAT 2001)
The owner of an art shop conducts his business in the following manner: Every once in a while he raises his prices by X %, then a while later he reduces all the new prices by X %. After one such updown cycle, the price of a painting decreased by Rs. 441. After a second updown cycle the painting was sold for Rs. 1,944.81. What was the original price of the painting?
(1) Rs 2,756.25
(2) Rs 2,256.25
(3) Rs 2,500
(4) Rs 2,000

Q100. (CAT 2001)
Three runners A, B and C run a race, with runner A finishing 12 metres ahead of runner B and 18 metres ahead of runner C, while runner B finishes 8 metres ahead of runner C. Each runner travels the entire distance at a constant speed. What was the length of the race?
(1) 36 meters
(2) 48 meters
(3) 60 meters
(4) 72 meters

Let speed of the slowest runner be x and fastest runner be 2x
Let d be the distance covered by the slowest runner in 5 min
Then the distance covered by the fastest runner in 5 min is (d + 1000) metre
d/x = (d + 1000)/2x
=> d = 1000 metre
fastest runner covers (d + 1000) = 2000 metre in 5 minute
therefore he needs 10 minute to cover 4000 metre

If we start forming equations here, we are trapped!
If you increase and decrease by the same percentage (say x) then the final result will be always lesser than than the initial value (net change is x^2/100%). So to get the same result the increase percentage should be greater than the decrease percentage. Answer is p > q
Note  Number of years doesn't matter here. So it would be easy to just take one year and observe the trend. For example, let say he had 100 goats at the beginning and increased by 25%, to make the final count as 125. Now to bring it back to 100, we have to decrease by 20%. 25% > 20%

Let d = distance of race.
Between A's finish and B's finish, B runs 12 meters while C runs 10 meters.
B's rate to C's rate = 6/5.
When B finishes, he is 8 meters ahead of C.
6/5 = d/(d8)
d = 48 meters

First cycle :
P(1  x/100)(1 + x/100) = P  441
so (x^2/100) = 441/P  (1)Second cycle :
(P  441) (1  x^2/100) = 1944.81  (2)From (1) & (2)
(P  441) (1  441/P) = 1944.81
(P  441)(P  441) = 1944.81P
P^2  2826.81P + 194481 = 0
P = 2756.25

@master_shifu ...IS their any other way to solve this question?

Work with ratios
Ratio of speed of N & S = 4 : 1
=> Ratio of time taken by N & S = 1 : 4
Total time taken = 1 hour
Time taken by N to cover the distance AB = 1/5 hour = 12 minutes
Time taken by S to cover the distance AB = 4/5 hour = 48 minutes
When the train was late by 20 minutes, total time available = 40 minutes
N doubled the speed, so time taken to cover distance AB = 12/2 = 6 minutes
So, S should cover the distance in 40  6 = 34 minutes
ratio = 6 : 34 = 1 : 6 (approx)

@master_shifu
Can we not calculate the average speed by adding the total distance, i.e 9000 kms and then adding the total time taken , i.e. 25 hours + 24 hours + 1.55(halt in India) and then dividing those figures. Total distance/total time?

@master_shifu In 19 th question CAT 2002 in arithmetic, the answer comes is 43.5