Solved CAT Questions (Arithmetic) - Set 3
Q1. (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!
Q2 - Q4 (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.
Q2. 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
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
Q3. 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
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
Q4. 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
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!
Q5 - Q6 (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.
Q5. 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
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.
Q6. 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
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.
Q7. (CAT 2001)
If 09/12/2001 happens to be Sunday, then 09/12/1971 would have been a
Solution: Apply Zeller's rule directly and we'll get the answer as Thursday.
Wait, if you don't know Zeller's rule then it is a very handy trick for Date related problems. Rule is explained below [credits - MathForum]
[x] means the greatest integer that is smaller than or equal to x. You can find this number by just dropping everything after the decimal point. For example, [3.79] is 3.
Here's the formula:
f = k + [(13 * m-1)/5] + D + [D/4] + [C/4] - 2 * C.
Eg: January 29, 2064 is a ?
-- k is the day of the month. Let's use January 29, 2064 as an example. For this date, k = 29.
-- m is the month number. Months have to be counted specially for Zeller's Rule: March is 1, April is 2, and so on to February, which is 12. (This makes the formula simpler, because on leap years February 29 is counted as the last day of the year.) Because of this rule, January and February are always counted as the 11th and 12th months of the previous year. In our example, m = 11.
-- D is the last two digits of the year. Because in our example we are using January (see previous bullet) D = 63 even though we are using a date from 2064.
-- C stands for century: it's the first two digits of the year. In our case, C = 20.
Now let's substitute our example numbers into the formula.
f = k + [(13 * m-1)/5] + D + [D/4] + [C/4] - 2 * C
= 29 + [(13 * 11-1)/5] + 63 + [63/4] + [20/4] - 2 * 20
= 29 + [28.4] + 63 + [15.75] +  - 40
= 29 + 28 + 63 + 15 + 5 - 40
Once we have found f, we divide it by 7 and take the remainder. Note that if the result for f is negative, care must be taken in calculating the proper remainder. Suppose f = -17. When we divide by 7, we have to follow the same rules as for the greatest integer function; namely we find the greatest multiple of 7 less than -17, so the remainder will be positive (or zero). -21 is the greatest multiple of 7 less than -17, so the remainder is 4 since -21 + 4 = -17. Alternatively, we can say that -7 goes into -17 twice, making -14 and leaving a remainder of -3, then add 7 since the remainder is negative, so -3 + 7 is again a remainder of 4.
A remainder of 0 corresponds to Sunday, 1 means Monday, etc. For our example, 100/7 = 14, remainder 2, so January 29, 2064 will be a Tuesday.
Q8. (CAT 2001)
At his usual rowing rate, Rahul can travel 12 miles downstream in a certain river in six hours less than it takes him to travel the same distance upstream. But if he could double his usual rowing rate for this 24 mile round trip, the downstream 12 miles would then take only one hour less than the upstream 12 miles. What is the speed of the current in miles per hour?
Solution: Let speed in stagnant water is x and speed of current=y.
12/(x - y) - 12/(x + y) = 6..
1/(x - y) - 1/(x + y) = 1/2
2y/(x^2-y^2) = 1/2 --- (1)
Similarly from second condition 2y/(4x^2 - y^2) = 1/12 -- (2)
(1) / (2) = (4x^2 - y^2)/(x^2-y^2) = 6
2x^2 = 5y^2
Substituting this value in any of the condition we'll get y=8/3.
so option D
Q9. (CAT 2000)
A truck travelling at 70 kilometres per hour uses30% more diesel to travel a certain distance than it does when it travels at the speed of 50 kilometres per hour. If the truck can travel 19.5 kilometres on a litre of diesel at 50 kilometres per hour, how far can the truck travel on 10 litres of diesel at a speed of 70 kilometres per hour?
Solution: Diesel consumption increases by 30% for 70kmph speed. If diesel consumption is fixed (say 1 litre) then it'll go 19.5/1.3 = 15km when it's speed is 70kmph.
With 10 litres ,it will go 10 x 15 = 150 km.
Q10. (CAT 2000)
A, B, C are three numbers. Let
@ (A, B ) = average of A and B,
/ (A, B ) = product of A and B, and
X (A, B ) = the result of dividing A by B
Average of A, B and C is given by
(1) @(/(@(/(B, A), 2), C), 3)
(2) X(@(/(@(B, A), 3), C), 2)
(3) /(@(X(@(B, A), 2), C), 3)
(4) /(X(@(/(@(B, A), 2), C), 3), 2)
Solution: Solve via options.
Simplify options further to see which one yield (A + B + C)/3
Only Option 4 does. (explained below)
/(X(@ (/(@(B, A), 2), C), 3), 2) = /(X(@ (/(A+B)/2), 2), C), 3), 2)
= /(X(@ (A+B), C), 3), 2) = /(X((A+B, C)/2, 3), 2)
= (A + B + C)/3