Solved CAT Questions (Arithmetic) - Set 2
Q1. (CAT 2004)
A sprinter starts running on a circular path of radius r metres. Her average speed (in metres/minute) is πr during the first 30 seconds, πr/2 during next one minute, πr/4 during next 2 minutes, πr/8 during next 4 minutes, and so on. What is the ratio of the time taken for the nth round to that for the previous round?
Solution: There is more than 1 way to approach the solution; however, I will detail the easiest way to go about it here.
We want to find the ratio of time taken for nth round : time taken for (n-1)th round
It will be same as finding the ratio of time taken for 2nd round : Time taken for 1st round.
1 round = Circumference of the circle = 2πr
1st round :
Speed = πr for 30 seconds. So, total distance travelled = πr/2.
Speed = πr/2 for 1 minute. So, total distance travelled = πr/2.
Speed = πr/4 for 2 minutes. So, total distance travelled = πr/2.
Speed = πr/8 for 4 minutes. So, total distance travelled = πr/2.
So, for a distance of 2πr, time taken is 7.5 minutes.
Speed = πr/16 for 8 minutes. So, total distance travelled = πr/2.
Speed = πr/32 for 16 minutes. So, total distance travelled = πr/2.
Speed = πr/64 for 32 minutes. So, total distance travelled = πr/2.
Speed = πr/128 for 64 minutes. So, total distance travelled = πr/2.
So, for a distance of 2πr, time taken is 120 minutes.
Ratio is 120:7.5 = 16:1.
P.S: Essentially the time taken is a GP, as we see that speed is halved, then quartered and so on. So, we can also calculate by the following method:
1st round: Sum of 1st 4 terms of GP.
2nd round: Sum of 1st 8 terms – Sum of 1st 4 terms.
Then calculate the ratio.
Q2. (CAT 2004)
In an examination, there are 100 questions divided into three groups A, B and C such that each group contains at least one question. Each question in group A carries 1 mark, each question in group B carries 2 marks and each question in group C carries 3 marks. It is known that the questions in group A together carry at least 60% of the total marks.
If group B contains 23 questions, then how many questions are there in group C?
d. Cannot be determined
If group C contains 8 questions and group B carries at least 20% of the total marks, which of the following best describes the number of questions in group B?
a. 11 or 12
b. 12 or 13
c. 13 or 14
d. 14 or 15
Solution: Let there be x questions in A, y in B and z in C. So, x + y + z = 100 (As there are 100 questions). Also, questions in A have 1 mark, B have 2 marks and C have 3 marks. So, total marks is x + 2y + 3z
1st sub question :
y = 23. So, x + z = 77. z = 77 – x.
Total marks = x + 2y + 3z = x + 46 + 3z
Substituting for z = 77 – x, we have,
Total marks = z + 3z + 46 = x + 3* (77 – x) + 46 = 277 – 2x.
We know that A should have atleast 60% of the total marks. i.e. x should be atleast 60% of total marks.
So, x ≥ 60/100 * (277 – 2x)
10x ≥ 277 * 6 – 12x
x ≥ 277 * 6/22
x ≥ 75.
So, x should be minimum 76, and y = 23. So z can be only 1 (100 – 76 – 23).
2nd sub question :
z = 8. So, x + y = 92. y = 92 – x.
Total marks = x + 2y + 3z = x + 2y + 24.
Substituting for y = 92 – x, we have,
Total marks = x + 2y + 24 = x + 2* (92 – x) + 24 = 208 –x.
We know that A should have atleast 60% of the total marks. i.e. x should be atleast 60% of total marks.
So, x ≥ 60/100 * (208 –x)
10x ≥ 208 * 6 – 6x
x ≥ 208 * 6/16
x ≥ 78.
So x can be 78, 79 etc etc. z is 8.
So y can be a maximum of 14.
This option is there only in option 3, where y can be 13 or 14. (If you substitute y for 13 or 14, we see that total marks in y is 26 or 28, and it will be atleast 20% of the total marks)
Q3. (CAT 2003)
In a coastal village, every year floods destroy exactly half of the huts. After the flood water recedes, twice the number of huts destroyed are rebuilt. The floods occurred consecutively in the last three years namely 2001, 2002 and 2003. If floods are again expected in 2004, the number of huts expected to be destroyed is:
a. Less than the number of huts existing at the beginning of 2001.
b. Less than the total number of huts destroyed by floods in 2001 and 2003.
c. Less than the total number of huts destroyed by floods in 2002 and 2003.
d. More than the total number of huts built in 2001 and 2002.
Let there be 2x buildings in 2001. Let us form a table
Huts destroyed in 2004 = 3.375x:
1st option : Total huts in 2001 = 2x
2nd option: Total huts destroyed in 2001 and 2003 = 3.25x
Total huts destroyed in 2002 and 2003 : 3.75x. 3.375 < 3.75. Hence this is true.
Huts built in 2001 and 2002 = 5x.
Q4. (CAT 2002)
Only a single rail track exists between station A and B on a railway line. One hour after the north bound superfast train N leaves station A for station B, a south passenger train S reaches station A from station B. The speed of the superfast train is twice that of a normal express train E, while the speed of a passenger train S is half that of E. On a particular day N leaves for station B from station A, 20 minutes behind the normal schedule. In order to maintain the schedule both N and S increased their speed. If the superfast train doubles its speed, what should be the ratio (approximately) of the speed of passenger train to that of the superfast train so that passenger train S reaches exactly at the scheduled time at the station A on that day.
(1) 1 : 3
(2) 1 : 4
(3) 1 : 5
(4) 1 : 6
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.
Q5. (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.
Q6. (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
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 17.5kms.
It takes 17.5 minutes for that.
Now, from G3 to A, it is a 20km journey at 60kmph. It takes 20 minutes.
So, total time taken is 5 + 17.5 + 20 = 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.
Q7. (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.
Q8. (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
Q9. (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.
Q10. (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%.