Aptitude Questions – 2 (for CAT and Bank Exams)

Aptitude Questions – 2 (for CAT and Bank Exams)

1. If a-3 = b3/2 and b = 4 then ‘a’ can take which of the following values?
(a) ½ (b) -1/8 (c) ¼ (d) 1/8 (e) 2
Ans: (a)
a-3 = b3/2 = 43/2 = 23. Thus a-3 = 23 -> a-1 = 2 -> a = ½.
Hence a = ½.

2. An empty basket, with a capacity for 50 oranges costs Rs 7.25. For the first 15 oranges bought each orange costs Rs 0.75 and every extra orange bought costs Rs 0.50 each. If a man has Rs 31.00 with him, what is the maximum number of oranges along with the requisite number of baskets for putting them in, he can buy?
(a) 31 (b) 62 (c) 40 (d) 50 (e) 47.
Ans: (c)
Let ‘n’ be the number of oranges that he bought along with one basket.
(option (b) alone requires more than one basket and it does not fit into the information given)
Now we have an equation —– 7.25 + (15 x 0.75) + (n – 15) x 0.50 = 31.00 (all figures in rupees)
7.25 + 11.25 + (n-15)x0.50 = 31.00 -> (n-15)x0.50 = 31.00 – 18.50 -> 12.50.
Thus (n-15) = 25 -> n = 25 + 15 = 40.

3. In a certain party, the ratio of boys to girls was 5 : 3 initially. After some time 10 boys left the party and the ration now became 1 : 1. How many people were originally there at the party?
(a) 48 (b) 32 (c) 64 (d) 40 (e) 56
Ans: (d)
No working required and the question can be answered from the choices.

4. If x3 = -27 and if y3 = (x2 + 7)(x – 1) then which of the following value ‘y’ can take?
(a) 2 x 3√4 (b) -2 (c) -4 (d) 2 (e) none of these
Ans: (c)
X3 = -27. Hence ‘x’ = -3. Y3 = (x2 + 7)(x – 1) -> (9 + 7)(-3-1) -> 16 x (-4) = -64. (after substituting
The value of ‘x’)
Thus ‘y3 = -64 and ‘y’ = -4.

5. a/4 + 4/a + a/4 + 4/a + a/4 + 4/a —————–
In the above sequence the odd numbered term is a/4 and the even numbered term is 4/a.
What is the sum of the first 40 terms of the sequence?
(a) 80 + 5a2 (b) (80 + 5a)/a (c) (80 + 5a2)/a (d) (80 + a2)/a (e) (5 + a)/80
Ans: (c)
Sum of 20 odd terms —— a/4 x 20 = 20a/4 -> 5a………… (i)
Sum of 20 even terms——4/a x 20 = 80/a…………………… (ii)
The sum of 40 terms in the sequence is — (i) + (ii) ——— 5a + 80/a -> (5a2 + 80)/a

6. A and B together worked for 4 hours and completed ½ of the job. A worked thrice as fast as B did. B left and A was joined by C and they finished the remaining job in 1 hour. How long C would have taken to complete the whole job by himself?
(a) 3 hours (b) 7/2 hours (c) 32/13 hours (d) 4 hours (e) 16/5 hours
Ans: (c)
Let ‘x’ be the time taken by A to complete the job while B takes 3x time to complete.
In 4 hours A & B together would have done —- 4/x + 4/3x = 16/3x job.
In four hours they have done 16/3x job that is equal to ½. -> 3x = 32 and x = 32/3
Let C take ‘c’ hours to complete the job.
In one hour A and C together finished the remaining job.
ie 1/x + 1/c = ½ -> 1/(32/3) + 1/c = ½ -> 3/32 + 1/c = ½
1/c = ½ – 3/32 = 13/32
Hence C will take 32/13 hours to complete the job.

7. In a party there are 12 boys and 15 girls. In how many different dancing pairs can be made out of the group?
(a) 60 (b) 180 (c) 150 (d) 300 (e) 250
Ans: (b)
The dancing pairs are —– 12C1 x 15C1 -> 12 x 15 = 180

8. A family consists of father, mother, son and daughter. The father is five years elder to his wife. The daughter is 23 years younger to her father and the son is 16 years younger to his mother. What is the difference in age between the son and the daughter?
(a) 17 years (b) 7 years (c) 5 years (d) 2 years (e) None of these
Ans: (d)
Let F, M, S and D be the ages of the father, mother, son and daughter.
F = M + 5. D = M + 5 – 23 ( F – 23 ) = M – 18.
S = M – 16
D = M – 18. Hence the difference between B and D is 2.

9. A shopkeeper marks the price of an item 60% over its cost price. However, during a sale promotion period he offered a discount of 40% on the marked price. If a customer pays Rs 24.00 for the item what is the profit/loss in the transaction for the trader?
(a) Rs 4 profit (b) Rs 2 profit (c) Rs 16 loss (d) Re 1 loss (e) Rs 4 loss
Ans: (d)
Let the cost price be ‘x’. Then the marked price is —- 1.6x (60% up)
On 1.6x the discount allowed is 0.64x (40% discount on the marked price)
Thus the realized value of the item is —- 0.96x and this equals to Rs 24 (the amount customer paid)
Hence the cost price of the item is x = 24 x 100/0.96 = Rs 25.
CP = 25. SP = 24. Hence the loss is Re 1 in the transaction.

10. A tank of volume 432 cu.ft.has one inlet and two outlet pipes. The inlet pipe fill the tank at the rate of 4 cu.in per minute, while the two outlet pipes drain the tank at the rate of 14 cu. In and 6 cu. In per minute respectively. If all the three pipes are opened when the tank is full, in how much time the tank will become empty?
(a) 777.6 hrs (b) 3 days (c) 15 days (d) 77.76 hrs (e) None of these.
Ans: (a)
This is a tricky question. Please note the volume of the tank is given in cubic feet while the inlet pipes filling and draining is given in cubic inches per minute. Hence the first thing to do is to convert the volume into cubic inches.
Volume of the tank — 432 cu.ft -> 432 x 12 x 12 x 12 = 746496 cubic inches ( 12 inches make one foot)
In one minute when the three pipes are opened the effective reduction in the volume is
—– 4 – 14 – 6 = -16 (When the tank is full in each minute the volume gets reduced by 16 cubic inches.
Hence, to drain 746496 cubic inches it would take — 746496/16 = 46656 minutes or
46656/60 = 777.6 hours.

Aptitude Questions (for CAT and Bank Exams)

Aptitude Questions ( for CAT and Bank exams)

1. A regular working day is of 8 hours and a regular week is 5 working days. A worker is paid Rs 24 per regular hour and Rs 32 per hour over time. If the worker earns Rs 4320.00 in four weeks then what is the total number of hours he worked?
(a) 180 (b) 175 (c) 160 (d) 195 (e) 200
Ans: (b)
During the four weeks the worker would have worked normal hours of — 5 x 8 x 4 = 160 hours.
For this normal working of 160 hours his wages will be —– 24 x 160 = Rs 3840.00
He actually earned Rs 4320 during the four weeks of working.
Thus the difference amount — Rs 4320 – Rs 3840 = Rs 480 represents his overtime.
His overtime wages for one hour is Rs 32 and hence he had worked overtime
For —— 480 / 32 = 15 hours.
Thus totally in 4 weeks he worked ——- 160 + 15 = 175 hours.

2. Five apples and four oranges cost the same as three apples and seven oranges. What is the ratio of the following —– cost of one orange / cost of one apple ?
(a) 2 : 3 (b) 3 : 4 (c) 3 : 1 (d) 4 : 3 (e) 2 : 5
Ans: (a)
5 A + 4 O = 3 A + 7 O -> 2 A = 3 O -> O/A = 2/3 Ratio O : A = 2 : 3

3. The average of 2, 7, 6, and X is 5.The average of 18, 1, 6, X, and Y is 10. What is the value of ‘Y’?
(a) 20 (b) 30 (c) 10 (d) 5 (e) 15
Ans: (a)
The total of —- 2 + 7 + 6 + x = 5 * 4 = 20. Hence the value of x = 20 – 15 = 5
The total of —- 18 + 1 + 6 + x + y = 10 * 5 = 50. Now substituting the value of x in this we get
— 18 + 1 + 6 + 5 + y = 50.
Hence y = 50 – 30 = 20.

4. In a certain boy’s camp, 20% of boys are from Maharashtra State and 30% of those are from Mumbai city. If there are 49 boys in the camp who are from Maharashtra State but not from Mumbai city, then what is the total number of boys in the camp?
(a) 70 (b) 245 (c) 163 (d) 350 (e) 817
Ans: (d)
Boys from Maharashtra State in the camp are 20%.
Of this, Boys from Mumbai city is 30% of 20% -> 6%
Excluding Mumbai city boys from Maharashtra State is —- 20% – 6% = 14% this is equal to 49 boys.
Hence the total number of boys in the camp is —- (49 x 100) / 14 = 350 boys.

5. The sum of the quotient and the reminder obtained when a number is divided by 4 is 8 and the sum of their squares is 34. Which of the following is the number?
(a) 17 (b) 26 (c) 21 (d) 23 (e) 29
Ans: (d)
23 when divided by 4 will give a quotient 5 and reminder 3. Sum of these aggregates to 8
The square of 5 is 25 and the square of 3 is 9. Aggregate of these two amounts to 34.

6. If the average of ‘m’ numbers is ‘a’ and when ‘x’ is added to the ‘m’ numbers the average of (m+1) numbers is ‘b’. then ‘x’ is equal to which of the following?
(a) Ma+b (b) m(a+b) (c) 2ma – mb (d) b-a+ma (e) (m+1)b – ma.
Ans: (e)
(ma + x) / m + 1 = b -> ma + x = mb + b -> x = mb + b – ma -> m(b – a) + b -> (m + 1)b – ma

7. If the ratio of A to B and ratio of X to Y are both equal to 1/3, then the ratio of (A + X) to (B + Y) is equal to:
(a) 5/6 (b) 1/12 (c) 1/3 (d) ½ (e) 1/6
Ans: (c)
A/B = X/Y = I/3 —– (A+X) / (B+Y) = ? Please observe the following:
2/4 = 3/6 = ½ —– (2+3) / (4+6) = 5/10 = ½.

8. P, Q and R, are all multiples of 5. Q is greater than P and less than R and R is greater than P. What is the value of (P-Q)(P-R) / (Q-R)
(a) -10 (b) -5 (c) 1 (d) 10 (e) cannot be determined
Ans: (a)
Let the values of P, Q and R be respectively 5, 10 and 15.
Then we have —- (5 – 10)(5-15) / (10-15) —— solving we get the answer — (-10)

9. Twice some value of ‘a’ is greater than another value ‘b’. But, twice ‘b’ is greater than twice ‘a’. If ‘c’ is less than one half of “b’ hen which of the following is correct?
(a) c < a < b (b) c > b > a (c) a < c < b (d) a < 2 < c < b/2 (e) c > a > b
Ans: (a)
Let us apportion value 4 for ‘a’. then, 2a = 8
Let the value of ‘b’ be 6. Then 2a is greater than ‘b’. But, 2b = 12 is greater than 2a.
Half of ‘b’ is ‘c’ and this equals to 3.
Hence we find, c (3) < a (4) < b (6) 10. A company manufactures two products A and B. Product A requires 3 units of material 1 and 4 units of material 2 for producing one unit of A. Product B requires 5 units of material 1 and 7 units of material 2 for producing one unit of B. If 26 units of A and B were produced, how many units of material 2 were used in the process? (a) 213 (b) 293 (c) 384 (d) 286 (e) None of these. Ans: (d) A simple question. Totally 4 + 7 = 11 units of material 2 is required in the process of producing one unit of A and B. Hence, for producing 26 units of each A and B the total quantity of material 2 required is ----- 26 x 11 = 286 units.

Time & Work (Selective Placement Questions)

Time & Work (Selective Placement Questions)

1. 5 men working 6 hours a day can make 10 toys in 6 days. Then in how many days 12 men working 8 hours per day can make 16 toys?
Ans: 3 days.
(Points to note in these types of questions:
More work, more days. More hours, less number of days, More men, less number of days and conversely the same applies.)
The easy way of answering the above question is to segment the given information keeping the one that is to be arrived at the end.
Men Toys Hours Days
5 10 6 6
12 16 8 x (Now relate each segment with the last keeping the above in mind)

X = 6 x 5/12 x 16/10 x 6/8 —– Reducing we get X = 3 days.

2. 12 men and 16 boys can do a piece of work in 5 days. 13 men and 24 boys can do the same work in 4 days. How long will 7 men and 10 boys take to do the same work?
Ans: 8 1/3 days.
12 M + 16 B = 5 days In 1 day —— 1/12 M + 1/16 B = 1/5 —– (i)

13 M + 24 B = 4 days In 1 day —— 1/13 M + 1/24 B = 1/4 —– (ii) Equating both (i) and (ii)
We get
60 M + 80 B = 52 M + 96 B -> 8 M = 16B -> 1 M = 2B

7M + 10B -> 7M + 5M = 12M. From Equation (i) we find – 12M + 8 M = 20M takes 5 days.

Hence 12 M will take —– 5 x 20/12 = 8 1/3 days.

3. If 5 men and 3 boys can reap 23 acres in 4 days and 3 men and 2 boys can reap 7 acres in 2 days then how many boys should help 7 men reap 45 acres in 6 days?
Ans: 2 boys.
Let ‘m’ and ‘b’ be unknown work done by a man and a boy in one day. Thus we now have
4 x (5m + 3b) = 23 ——- (i) and
2 x (3m + 2b) = 7 ——–(ii)
Equating the two we get —- 20m + 12b = 23 —— (i) and
6m + 4b = 7 ——(ii)
Multiplying (ii) by 3 and subtracting from (i) we get — 2m = 2 and one ‘m’ = 1
Substituting this value of ‘m’ in equation (ii) we get
2 x (3*1 + 2b)= 7 —– 6 + 4b = 7 -> 4b = 1 and b = 1/4
Thus we now have — 1man in one day can reap 1 acre and 1boy in one day can reap 1/4 acre.
Hence, 7 men in 6 days will reap —- 7x1x6 = 42 acres.
Total acres to be reaped is 45 acres and the boys in 6 days will have to reap 3 acres.
So the number of boys required to reap in one day —– 3/ (1/4 *6) -> 2
Hence, 2 boys will have to assist the 7 men in reaping 45 acres of land in 6 days.

4. A and B are working on your car. A alone can complete the work in six hours while B takes eight hours to complete the same work. A and B together start the work and after 2 hours A leaves to attend some other work. In how much more time B will be able to complete the job?
Ans: 3 1/3 hours.
A in one hour can do —– 1/6 job. In 2 hours can do —- 2/6 -> 1/3 job.
B in one hour can do —– 1/8 job. In 2 hours can do —- 2/8 -> 1/4 job.
Thus in 2 hours A & B together would have completed —– 1/3 + 1/4 = 7/12 work.
The remaining work is — 1 – 7/12 = 5/12. This will have to be completed by B alone.
B will take —– (5/12) / (1/8) —- 3 1/3 hours.

5. A and B together can finish a job in T days. A alone can complete the job in (T+5) days while B takes (T+45) days to finish the job. What is the value of T?
Ans: 15 days
A in one day will do –—- 1/T+5 work.
B in one day will do —— 1/T+45 work.
Thus we have an equation ——————- 1/T+5 + 1/T+45 = 1/T
(T+45) + (T + 5) / (T+5)(T+45) = 1/T proceeding further we get
T(T+45) + T(T+5) = (T+5)(T+45) -> T2 + 45T + T2 + 5T = T2 + 45T + 5T + 225
Reducing the equation we get —— T2 = 225 and T = 15.

6. Two workers A & B manufactured a batch of identical parts. A worked for 2 hours and B worked for 3 hours and they did half the job. They worked together for another 3 hours and 1/20 of the job is left out. How much time B would have taken if he had worked alone to complete the job?
Ans: 15 hours
Let ‘x’ and ‘y’ be the time taken by A and B separately to complete the job alone.
We now have an equation —- 2/x + 3/y = 1/2 (job) ——– (i)
After A and B working together for 3 hours the remaining job is —1/20
Thus 19/20 job was already done by A and B individually and jointly.
A and B in 3 hours have completed —- 1 – 1/2 – 1/20 = 9/20 job.
We now have a second equation —- 3/x + 3/y = 9/20 —— (ii)
Solving (i) and (ii) we get the value of ‘y’ as 15 hours.

7. A can do a piece of work in 80 days. He works on it for 10 days and then leave. B then finishes the remaining work in 42 days. If A and B together had worked then in how many days they would have completed the work?
Ans: 30 days.
A in one day can do 1/80 job. So in 10 days he had finished 10/80 -> 1/8 job.
The remaining job was 7/8 and this B completed in 42 days.
So, B in one day can do —- (7/8) / 42 -> 7/ 336 -> 1/48
Thus A and B in one day can do —– 1/80 + 1/48 = (3 + 5)/ 240 -> 8/240 -> 1/30.
Hence A and B together will complete the job in 30 days.

8. 15 men working 8 hours per day take 21 days to complete a work. In how many days 21 women working at 6 hours per day would take to complete the same job if 3 women do as much work as 2 men?
Ans: 30 days.
3 women do as much work as 2 men. So, the 21 women will be equivalent to 14 men.
We can now segment the information as under:
15 M 8 Hours 21 days.
14 M* 6 Hours x days (* relates to 21 women)

X = 21 x 15/14 x 8/6 = 30 days.

Pipes & Cistern (Some Selective Questions)

Pipes & Cistern – (Some Selective Questions)

Short methods for answering some of the questions:
Let there be three pipes A, B and C. Let us assume that:
Pipe A takes ‘x’ minutes to fill
Pipe B takes ‘y’ minutes to fill/drain
Pipe C takes ‘z’ minutes to fill/drain.

If A and B are opened, the time taken to fill is given by —- xy/(x + y)

If A, B and C are opened the time taken to fill is given by — xyz/ (xz + yz + xy)

If A fills and B drains the time taken to fill is given by ——- xy/ (y-x)

If A and B fills and C drains the time taken to fill is given by — xyz/ (xz + yz – xy)

1. Two pipes A and B can fill a tank in 36 hours and 45 hours respectively. If both the pipes are opened simultaneously, then, in how much time the tank will be full?
Ans: 20 hours. ( Use the formula — xy/x+y)

2. A pipe can fill the tank in 15 hours. But due to a leak at the bottom it takes 20 hours to fill.If the tank is full then in how much time the leak will empty the tank?
Ans: 60 hours. (Use the formula — xy/y-x. Treat the leak as another pipe that drains the tank)

3. Pipe A and pipe B can fill a tank in 20 hours and 30 hours respectively while pipe C can drain a full tank in 40 hours. If all the three pipes are opened in how much time the tank will become full?
Ans: 17.14 hours (approx.) (use the formula — xyz/ (xz+yz-xy)

4. Pipe A fills a tank of capacity 700 liters at the rate of 40 liters per minute. Pipe B fills the same tank at the rate of 30 liters per minute. Pipe C at the bottom of the tank drains at the rate of 20 liters per minute. If pipe A is kept open for a minute and closed, followed by pipe B also kept open for a minute and closed and finally pipe C kept open for a minute and closed. If these successive operations continue in how much time the tank will be full?
Ans: 42 minutes.
Pipe A in one minute fills —– 40 liters
Pipe B in one minute fills —– 30 liters
Pipe C in one minute drains – 20 liters.
Thus in 3 minutes the tank gets filled ——– 40 + 30 – 20 = 50 liters. (This is one cycle of operation)
The tank capacity is 700 liters and to fill completely will require
700/50 = 14 cycles. Since each cycle of operation is 3 minutes the
total time required to fill the tank is —– 14 x 3 = 42 minutes.

5. Nine large pipes will drain a pond in eight hours and 6 small pipes will drain the same pond in sixteen hours. How long will it take 3 large pipes and 5 small pipes to drain the pond?
Ans: 10 hours 40 minutes or 10 2/3 hours.
9 large pipes take 8 hours to drain the tank.
In one hour 9 large pipes will drain 1/8 tank and
One large pipe in one hour will drain —-1/8 x 1/9 = 1/72 tank.
3 large pipes in one hour will drain — 3 x 1/72 -> 3/72 -> 1/24 tank. ——— (i)
6 small pipes take 16 hours to drain the tank.
In one hour 6 small pipes will drain 1/16 tank and
One small pipe in one hour will drain —- 1/16 x 1/6 = 1/96 tank.
5 small pipes in one hour will drain — 5 x 1/96 -> 5/96 tank. ——————–(ii)
Thus in one hour
3 large pipes and 5 small pipes will drain (i) + (ii) —— 1/24 + 5/96 -> 27/288 -> 3/32 tank.
Hence to drain the entire tank the time taken will be — 32/3 -> 10 2/3 hours.

6. Two pipes A and B can fill a cistern in 12 minutes and 15 minutes respectively, but a third pipe C can empty the full cistern in 6 minutes. A and B are kept open for 5 minutes after which C also is opened. In how much more time the cistern will be empty?
Ans: 45 minutes.
Pipe A in one minute fills — 1/12 . In 5 minutes it will fill —- 5/12 cistern.—— (i)
Pipe B in one minute fills — 1/15 . In 5 minutes it will fill —5/15 -> 1/3 cistern — (ii)
Thus in 5 minutes A and B would have filled —– (i) + (ii) — 5/12 + 1/3 -> (5+4)/12 = ¾ cistern
Hence after 5 minutes the cistern is full up to ¾ level.
Now pipe C also is opened.
In one minute all the three pipes would fill/drain —
1/12 + 1/15 – 1/6 -> (5 + 4 – 10)/60 = -1/60. (minus)
So in each minute the cistern gets drained up 1/60 level.
The cistern was ¾ full when C was opened.
So the time to drain this quantity ——– (3/4) / (1/60) = 45 minutes.

7. 12 buckets of water with each bucket capacity 13.5 liters will fill a tank. How many buckets of capacity 9 liters will be required to fill the tank?
Ans: 18
The capacity of the tank is —– 13.5 x 12 = 162 liters.
The number of buckets of capacity 9 liters required to fill this quantity — 162/9 = 18 buckets.

8. A leak in the bottom of a tank can empty the full tank in 6 hours. Let this leak be called B.
An inlet pipe A can fill the tank at the rate of 4 liters per minute. When the tank is full the inlet pipe is opened and due to the leak it now takes 8 hours to drain. What is the capacity of the tank in liters?
Ans: 5760 liters.
Let the capacity be ‘x’ . Then A takes —– x/4×60 -> x/240 minutes to fill. —– (i)
B takes ——- x/ 6 x60 -> x/360 minutes to drain.
When these two are in operation, the tank gets filled —- x/240 – x/360 = x/720 liters.
Now it is given the time to drain a full tank now is – 8 hours.
Thus we have — x/720 = 8 hours — or ‘x’ = 5760 liters.

Tech Mahindra – Recent Aptitude and Reasoning Questions

Tech Mahindra – Recent Aptitude and Reasoning Questions

1. Consider a group of working men. If 1/10th of the workers are absent then work of each person would be increased by:
(a) 20%
(b) 22.22%
(c) 10%
(d) 11.11%
Ans: (d)
Assume there are 10 workers and they do 100 jobs. Each worker does 10 jobs.
When 1/10 of work force leave, the remaining is 9 workers and they will have to now do 100/9 = 11.11 job. The increase in work load for each person now is – 11.11 – 10 = 1.11
Hence the percentage increase in work load is — 11.11%.

2. Consider the following statements:
(I) All members of team A are good tall guys with an average height of 6 feet.
(II) Average height of members of team B is less than that of team A.
Which could be true among the two options given below?
(a) Every member of team B is shorter than each member of team A.
(b) Few members of team B could be over 6 feet.
Ans: (b)
The average height of team B is less than that of team A which is 6 feet. This does not mean that members of team B are all shorter than members of team A.
It is possible that some members of team B can be more than 6 feet in height though the average of the team is less than 6 feet.

3. John is supposed to walk from his house to park every morning. One morning, he is in real hurry and wants to save at least 1/3rd of the time. By how much percentage he should increase his speed?
(a) 100%
(b) 33%
(c) 66%
(d) 50%
Ans: (d)
Let us assume that John walks 6 km to the park covering in one hour or 60 minutes.
Hence his speed is 6 km/hr
He now wants to cover the distance in 2/3 hour or 40 minutes to save 1/3 of the time.
His new speed will be — 6 / (2/3) Distance/ time gives the speed.
Thus his ne speed will be —- 6 * 3/2 = 9 km/hr.
Thus the increase in speed over the normal speed of 6 km/hr is 3 km/hr or 50%

4. Pointing to the woman in the picture, Govind said, “Her mother has only one grand -child whose mother is my wife”. How is the woman in the picture related to Govind?
(a) Sister
(b) Wife
(c) cousin
(d) data inadequate
Ans: (b)
5. Anand is the uncle of Bimala who is the daughter of Cauvery and Cauvery is the daughter in law of Palani. How is Anand related to Palani?
(a) Son
(b) son in law
(c) Brother
(d) Nephew
Ans: (a)

6. Pointing to a man, Lakshmi says, “This man’s son’s sister is my mother-in-law”. How is the Lakshmi’s husband related to the man?
(a) son
(b) son-in-law
(c) grandson
(d)nephew
Ans: (c)

7. A cylinder of diameter 14 and height 7 is converted into a cone of radius 6. Now, what could be the percentage height of the new shape?
(a) 308%
(b) 32%
(c) 2150%
(d) 215%
Ans: (a)
Volume of the cylinder should equal volume of the cone.
Volume of the cylinder is: πr2h -> π * 72 * 7 -> 343π ————- (i)
Volume of the cone is : 1/3 (πr2h) -> 1/3 (π * 62 * h) -> 12πh —— (ii)
Since (i) and (ii) are equal we get the value of h as —– h = 343/12
Percentage increase in height ——- [(343/12) – 7] / 7 * 100 = 308% (approx.)

8. Find the largest number which gives out reminders 2 and 5 when used to divide 80 and 122.
(a) 19
(b) 17
(c) 13
(d) infinite number of solutions
Ans: (c)
Find the HCF of (80 -2) and (122 – 5) -> 78 and 117. HCF is 13 and that’s the answer.

9. Find the ratio of the volumes of a sphere and cone of same radius. Also the height of the cone is twice its radius.
(a) 3 : 1
(b) 4 : 3
(c) 4 : 1
(d) 2 : 1
Ans: (d)
The volume of a sphere is 4/3 πr3.
Volume of a cone is 1/3 (πr2h). It is given in the question that h is equal to 2r.
Thus we get the volume of the cone as –—- 1/3 ( πr2*2r) -> 2/3πr3
Hence the ratio of volume of sphere and cone ——- 4/3 πr3 / 2/3 πr3 -> 2 : 1
10. Find the missing number in the sequence 504 , _ , 990, 1320, 1716
(a) 720
(b) 1000
(c) 721
(d) 790
Ans: (a)
The series is (n3-n). The first is 83-8, the second is 93-9, the third is 103-10 and so on.

11. The perimeter of a square and a rectangle is the same. If the rectangle is 12 cm by 10 cm, then by what percentage is the area of the square more than that of the rectangle?
(a) 1
(b) 3
(c) 5/6
(d) 1/2
Ans: (c)
Since the sides of the rectangle are 12 cm and 10 cm, its perimeter is – 2*12 + 2*10 = 44 cm
The perimeter of the Square is the same as the rectangle and is — 44 cm
Hence each side of the square is – 44/4 = 11 cm.
The area of the square is — 11*11 = 121 sq.cm.
The area of the rectangle is — 12 * 10 = 120 sq.com
Thus the Area of the square is 1 cm more than the area of the rectangle.
The percentage increase in the area —— 1/120 * 100 = 5/6%

12. Capacity of a cylindrical vessel is 25,872 litres. If the height of the cylinder is 200% more than the radius of its base, what is the area of the base in square cms?
(a) 336
(b) 1232
(c) 616
(d) cannot be determined
Ans: (c)
The volume of the cylinder is — πr2h. Now from the information h = r +(200% of r) = 3r
Thus the volume of the cylinder is — 3πr3 and this is equal to — 25872 litres.
Hence – r3 = 25872/3π -> 25872 / 3*22/7 -> 2744 and r = 14.
Hence the area of the base is —- πr2 -> 22/7 * 14 * 14 = 616 sq.cm

13. If the radius of a circle is increased by 12%, then the area of the circle
(a) decreases by 25.44%
(b) increases by 25.44%
(c) no change in area
(d) decreases by 12%
Ans: (b) (No working is required)

14. Aravamudhan, Balakrishnan and Chinnan work in a developing software company in starting level positions. However their salaries are different. Aravamudhan’s salary to Balakrishnan’s salary and Balakrishnan’s salary to Chinnan’s salary are in the ratio 4:3. If the total salary of all the three employees is Rs.29230, what is the salary of Chinnan?
(a) Rs.12640
(b) Rs.9480
(c) Rs.7110
(d) Rs.8660
Ans: (c)
Ratio of Aravamudhan : Balakrishnan :: 4 : 3
Ratio of Balakrishnan : Chinnan :: 4 : 3
Now assuming Aravamudhan salary as Rs 4, then Balakrishnan’s salary will be Rs 3.
Now Balakrishnan/Chinnan = 4/3 -> 3/Chinnan = 4/3 -> Chinnan salary will be 9/4.
Thus the total salary of all the three is ———- 4 + 3 + 9/4 -> 37/4
We know the total salary is Rs 29230 which is equal to 37/4 (as arrived above)
Chinnan’s salary is 9/4 which is equal to —- 29230 * (9/4)/(37/4) -> Rs 7,110/-

15. Three cricket players — Ramesh, Suresh and Ganesh play for three different cricket Teams – Haryana, Delhi and Rajasthan respectively. Ramesh was born in Chennai but brought up in Karnal, Haryana. Suresh was born in Delhi and brought up in Delhi. Ganesh’s fore-fathers were from Madurai but settled in Rajasthan. All the three participated in a cricket tournament. Ramesh’s runs to Suresh’s runs and Suresh’s runs to Ganesh’s runs are in the ratio 5:7. If the total runs scored by all the three players in the tournament is 327, find the total runs scored by Suresh in the tournament?
(a) 95
(b) 85
(c) 65
(d) 105
Ans: (d)
This question is similar to the earlier question and we shall proceed from the last ratio:
Suresh : Ganesh :: 5 : 7
Ramesh : Suresh :: 5 : 7
Assuming Ganesh scored 7 runs, then Suresh scored 5 runs.
Now, Ramesh/Suresh :: 5/7 -> 7 Ramesh = 5 x 5 = 25. Hence, Ramesh = 25/7
We know the total runs scored as 327.
Thus we have —– (R) 25/7 + (S) 5 + (G) 7 = 327 -> 109/7 = 327.
Runs scored by Suresh is ——- 327/(109/7) * 5 -> 105.