Question Bank - 100 Arithmetic Questions From Previous CAT Papers (Solved)
master_shifu last edited by zabeer
Let speed of S = x. Then E = 2x and N = 4x.
Let the distance between A & B be d.
On a normal day :
N goes from A to B in (d/4x) hrs.
S goes from B to A in (d/x) hrs.
S cannot start before N reaches there, as there is only a single track.
So S can start immediately after N reaches there, or after some time.
Hence the time of travel of ‘S + N’ is either equal to 1hr or lesser (depending on when S starts from B )
So (d/4x) + (d/x) ≤ 1.
(5d/4x) ≤ 1
(d/x) ≤ 4/5
On the late day:
N is 20 minutes late.
So, the entire distance needs to be covered in 40 minutes, or 2/3 hrs.
N goes from A to B in d/8x time (As speed of N is doubled)
S goes from B to A in d/y time (where y is the new speed of S)
Hence, the total time is (d/8x) + (d/y) = 2/3
(d/y) = (2/3) – (d/8x)
(d/y) ≥ (2/3) – (4/40) (As d/x ≤ 4/5)
(d/y) ≥ (2/3) – (1/10)
(d/y) ≥ 17/30
Now, we have (d/x) ≤ 4/5 and (d/y) ≥ 17/30
Since (d/x) ≤ 5/4, we have (x/d) ≥ 5/4.
So, we have (x/d) ≥5/4 and (d/y) ≥17/30.
Hence (x/d) * (d/y) ≥ (5/4) * (17/30)
Hence (x/y) ≥ 17/24.
We need to find y/8x
(y/x) ≤ 24/17
(y/8x) ≤ 3/17
In the given options, only 1/6 is less than 3/17.
Hence, the answer is 1/6.
master_shifu last edited by master_shifu
Q15. (CAT 2002)
A string of length 40 metres is divided into three parts of different lengths. The first part is three times the second part, and the last part is 23 metres smaller than the first part. Find the length of the largest part.
Solution: Let the 2nd piece length be x.
1st piece length is 3x.
3rd piece length is 3x – 23.
Sum of all the pieces is 40.
3x + x + 3x – 23 = 40.
7x = 63.
x = 9.
Lengths of the pieces are 27, 9, 4.
Q16. (CAT 2002)
On a 20km tunnel connecting two cities A and B there are three gutters. The distance between gutter 1 and 2 is half the distance between gutter 2 and 3. The distance from city A to its nearest gutter, gutter 1 is equal to the distance of city B from gutter 3. On a particular day the hospital in city A receives information that an accident has happened at the third gutter. The victim can be saved only if an operation is started within 40 minutes. An ambulance started from city A at 30 km/hr and crossed the first gutter after 5 minutes. If the driver had doubled the speed after that, what is the maximum amount of time the doctor would get to attend the patient at the hospital? Assume 1 minute is elapsed for taking the patient into and out of the ambulance.
(1) 4 minutes
(2) 2.5 minutes
(3) 1.5 minutes
(4) Patient died before reaching the hospital
master_shifu last edited by zabeer
Solution: Get the diagram correct first, where G1, G2 and G3 are the positions of the 3 gutters.
From the diagram, it is clear that:
AG1 = BG3 = x (given in question)
2 * G1G2 = G2G3 = 2y (given in question).
At 30kmph, AG1 takes 5 minutes.
So, AG1 = 30 * 5/60 = 2.5kms.
So, 2.5 + y + 2y + 2.5 = 20 (As total distance is 20km)
Hence y = 5.
So, G1G2 = 5 and G2G3 = 10.
Now, let us calculate the time taken:
AG1 = 5 mins (given in the question)
From G1, velocity is 60kmph, and to reach G3, he has to travel 15 kms.
It takes 15 minutes for that.
Now, from G3 to A, it is a 17.5 km journey at 60kmph. It takes 17.5 minutes.
So, total time taken is 5 + 15 + 17.5 = 37.5
Also, 1 minute for taking patient in and out of ambulance.
So, total time taken is 38.5
He should be saved in 40 minutes.
So, the doctor gets 40 – 38.5 = 1.5 minutes.
Q17. (CAT 2002)
It takes 6 technicians a total of 10 hours to build a new server from Direct Computer, with each working at the same rate. If six technicians start to build the server at 11 am and one technician per hour is added beginning at 5 pm, at what time will the server be complete?
(1) 6:40 pm
(2) 7:00 pm
(3) 7:20 pm
(4) 8:00 pm
Solution: 6 technicians take 10 hours to build the server. So, total man-hours required are 60.
So, let us take here that the work that needs to be completed is “60”.
In 10hrs, 6 technicians complete 10*6 = “60” amount of work, which is the whole work.
So, in 10hrs, 1 technician completes “10” amount of work.
Hence, each technician completes in 1hr, “1” amount of work.
Now, from 11am – 5pm that is for 6hrs there are 6 technicians.
1 technician 1 hr = “1” amount of work.
Hence 6 technicians in 6hrs complete “36” amount of work.
Remaining is 60 – 36 = “24” amount of work.
Next hour, there are 7 employees.
Hence the work done is “7”, and total work till now is 36 + 7 = 43.
Next hour, there are 8 employees.
Hence the work done is “7”, and total work till now is 43 + 8 = 51.
Next hour, there are 9 employees.
Hence the work done is “9”, and total work till now is 51 + 9 = 60.
Hence, the work is completed 3hrs after 5pm.
Q18. (CAT 2002)
Three small pumps and one large pump are filling a tank. Each of the three small pumps works at 2/3rd the rate of the large pump. If all 4 pumps work at the same time, then they should fill the tank in what fraction of time that it would have taken the large pump alone?
Solution : Let the large pump fill at the rate of x litres/hour.
So, each of the small tank fills at the rate of 2x/3 litres/hour.
The problem can be solved in many ways, I will do it in the least time consuming way.
We have :
L = x litres/hr (Large tank)
S1 = S2 = S3 = 2x/3 litres/hr ( 3 small tank).
Let x = 3.
Then S1 = S2 = S3 = 2.
Let the capacity of the tank be 9l ( 3 + 2 + 2 + 2)
Large tank alone will take 3hrs (as its rate is 3 litres/hr)
Large + 3 small tank will take 1hr (As rate will be 3 +2 + 2 + 2 = 9 litres/hr )
Hence, ratio is 1:3
Q19. (CAT 2002)
There is a tunnel connecting city A & B. There is a cat which is standing at 3/8 thelength of the tunnel from A. It listens a whistle of the train and starts running towards the entrance where, the train and the cat meet. In another case, the cat started running towards the exit and the train again met the cat at the exit. What is the ratio of their speeds?
(4) None of these
Solution: Let ‘c’ and ‘t’ be the velocity of train and cat.
Get the diagram right (if diagram is required)
1st case :
Train and cat meet at point A in the diagram.
Let train be at a distance of “y” from the entrance of the tunnel.
Cat is at a distance of 3x/8 from the entrance of the tunnel (Cat is between A&B. i.e. 3x/8 from A and 5x/8 from B )
So, by the time train runs a distance of y, cat runs a distance of 3x/8 (As they reach entrance at the same time)
Hence, equating the time taken, we have, y/t = 3x/8c
Train and cat meet at point B in the diagram.
Train travels a distance of y + x (Till the end of the tunnel)
Cat travels a distance of 5x/8 (Till the end of the tunnel)
Both reach at the same time, hence we can equate the time. We get, y + x/t = 5x/8c.
Subtracting the 2 equations, we have, (2nd case – 1st case) :
[(y +x)/t] – [y/t] = (5x/8c) – (3x/8c)
x/t = 2x/8c
c/t = 1/4.
t/c = 4/1.
Q20. (CAT 2001)
A student took five papers in an examination, where the full marks were the same for each paper. His marks in these papers were in the proportion of 6 : 7 : 8 : 9 : 10. In all papers together, the candidate obtained 60% of the total marks. Then the number of papers in which he got more than 50% marks is:
Solution: Let the maximum mark in each subject be 100. Hence, total marks (maximum) is 500, as there are 5 subjects.
Candidate obtained 60% of total marks, hence 60% of 500 = 300.
His individual marks were 6x, 7x, 8x, 9x and 10x.
So, total marks will be 6x + 7x + 8x + 9x + 10x = 40x
40x = 300.
x = 300/40 = 7.5
His marks are 6x, 7x, 8x, 9x and 10x.
So his marks were 45, 52.5 and all others are 50+ (as 7*7.5 itself is 50+)
So, 4 subjects he has 50+ marks out of 100, which means more than 50%.
Q21. (CAT 2001)
A can complete a piece of work in 4 days. B takes double the time taken by A, C takes double that of B, and D takes double that of C to complete the same task. They are paired in groups of two each. One pair takes two-thirds the time needed by the second pair to complete the work. Which is the first pair?
a. A, B
b. A, C
c. B, C
d. A, D
Solution: In a day, A would do 1/4 th of work ; B would d 1/8th of work, C would do 1/16th of work, D would do 1/32th of work.
One pair would take > > 2/3 rd of time take by second pair.
Out of all combinations ; we get A+D together in a day can do 9/32 th of work.
B + C can do 3/16th of work.
3/16 = 2/3 * 9/32
B + C = 2/3 (A + D )
Hence the first pair is B,C and the second pair is A,D.
Note: A, D or B, C should be the answer because for other combinations this ratio will be less than 1/2.now we can safely assume it to be A,D combination and start checking with it - saves time!
Q22 (CAT 1997)
A certain race is made up of three stretches: A, B and C, each 2 km long, and to be covered by a certain mode of transport. The following table gives these modes of transport for the stretches, and the minimum and maximum possible speeds (in km/hr) over these stretches. The speed over a particular stretch is assumed to be constant. The previous record for the race is 10 min.
Mr Tortoise completes the race at an average speed of 20 km/hr. His average speed for the first two stretches is four times that for the last stretch. Find the speed over stretch C.
a. 15 km/hr
b. 12 km/hr
c. 10 km/hr
d. This is not possible
Mr Hare completes the first stretch at the minimum speed and takes the same time for stretch B. He takes 50% more time than the previous record to complete the race. What is Mr Hare's speed for the stretch C?
a. 10.9 km/hr
b. 13.3 km/hr
c. 17.1 km/hr
d. None of these
Anshuman travels at minimum speed by car over A and completes stretch B at the fastest speed. At what speed should he cover stretch C in order to break the previous record?
a. Maximum speed for C
b. Minimum speed for C
c. This is not possible
d. None of these
Part 1 : Solution: Total distance = 2 + 2 + 2 = 6KM
Average speed for the whole journey = 20 KM/hr
Time taken = 6/20 = 0.3 hr ----> (1)
Speed for C = S
Then avg speed for A + B = 4S
Time taken for A + B = (2+2)/4S = 1/S
Time Taken for C = 2/S
Total Time = 1/S + 2/S = 0.3 ( from (1) )
=> S = 10 Km/hr
Part 2 : Solution : Hare is taking 2/40 hours = 1/20 hours = 3 minutes each for first 2 sets.
Total time taken = 10 + 10/2 = 15 min.
Time taken for third set = 9 min.
2 km in 9 min for C - speed = 13.3kmph
Part 3 : Solution : Previous record is 10 min.
Stretch A is covered at minimum speed - 2 km at 40 kmph = 3 minutes
Stretch B is covered at fastest speed - 2 km at 50 kmph = 2.4 minutes
Minimum time required to cover C = 2 km at 20 kmph = 6 minutes
So we cannot break the previous record!
Q23 (CAT 1997)
The Weirdo Holiday Resort follows a particular system of holidays for its employees. People are given holidays on the days where the first letter of the day of the week is the same as the first letter of their names. All employees work at the same rate.
Raja starts working on February 25, 1996, and finishes the job on March 2, 1996. How much time would T and J take to finish the same job if both start on the same day as Raja?
a. 4 days
b. 5 days
c. Either (a) or (b)
d. Cannot be determined
Starting on February 25, 1996, if Raja had finished his job on April 2, 1996, when would T and S together likely to have completed the job, had they started on the same day as Raja?
a. March 15, 1996
b. March 14, 1996
c. March 22, 1996
d. Data insufficient
Part 1 : Solution: 1996 is a leap year.
Raja takes 7 days to complete the job (So our work package is 7 person days)
Now we'll apply Zeller's rule (Rule is explained as part of Q7) to determine that Feb 25th,1996 is a Sunday.
If T and J starts on Sunday, by Wednesday they can complete the work. (Sunday - 2 Person days, Monday - 2 Person days, Tuesday - 1 Person day (T is on leave) and Wednesday - 2 Person days)
So 4 days is sufficient.
Part 2 : Solution: Raja doesn't have his name's first letter common to any of the day, so he won't rest.
He started his work on 25th Feb and finished on April 2 => He took 38 days to complete the job ( 1996 is a leap year) and he finished 1/38th of the work each day.
Now S will take break on Saturday & Sunday and T will take a break on Tuesday & Thursday.
So every week each of them work for 5 days and together they work for 10 days.
so 10/38 of the work will be completed in one week. In three weeks they complete 30/38 of the work.
Feb 25th, 1996 is a Sunday. (Refer Zeller's rule in Q7)
8 person days is required to finish the job.
Monday - Both work together - 2 person days
Tuesday - Only S - 1 person day
Wednesday - Both work together - 2 person days
Thursday - Only S - 1 person day
Friday - 2 person days
So by Friday they will complete the work.
Total 21 + 6 = 27 days is required. March 22, 1996 would be the answer.
Q24. (CAT 2001)
If 09/12/2001 happens to be Sunday, then 09/12/1971 would have been a