Solved CAT Questions (Arithmetic) - Set 11
Q1. (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 5 km ( = √(3^2 + 4^2) )
Q2. (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
Q3. (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)?
(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.
Q4. (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).
(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
Q5. (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?
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.
Q6. (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?
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
Q7. (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.
Q8. (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?
(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)
Q9. (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 up-down cycle, the price of a painting decreased by Rs. 441. After a second up-down 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
Q10. (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