ONE   1  4
NUMBER SYSTEMS
1. Find the maximum value of n such that 157! Is perfectly divisible by 10n.
1) 37   2) 38   3) 16   4) -1.15
Ans. (2) [157/5] = 31. [31/5] = 6.[6/5] = 1.31+6+1 = 38. Option (2) is correct.
2. Find the maximum value of n such that 157! Is perfectly divisible by 12n.
1) 77   2) 76   3) 75   4) 78
Ans. (3) No. of 2’s in 157! = [157/2] + [157/4] + [157/8] …. + [157/128] = 78 + 39 + 19 + 9 + 4 + 2 + 1 = 152. Hence, the number of 22s would be [152/2] = 76. Number of 3’s in 157! = 52 + 17 + 5 + 1 = 75. The answer would be given by the lower of these values. Hence, 75 (Option c) is correct.
3. Find the maximum value of n such that 157! Is perfectly divisible by 18n.
1) 37   2) 38   3) 39   4) 40
Ans. (1) From the above solution:
Number of 2’s in 157! = 152
Number of 32s in 157! = [75/2] = 37.
Hence, option (1) is correct.
4. Find the maximum value of n such that 50! Is perfectly divisible by 2520n.
1) 6   2) 8   3) 7   4) None of these
Ans. (2) 2520 = 7 x 32 x 23 x 5.
The value of n would be given by the value of the number of 7s in 50!
This value is equal to [50/7] + [50/49] = 7 + 1 = 8
Option (2) is correct
5. Find the maximum value of n such that 50! is perfectly divisible by 12600n.
1) 7   2) 6   3) 8   4) None of these
Ans. (2) 12600 = 7 x 32 x 23 x 52
The value of ‘n’ would depend on which of number of 7s and number of 52 is lower in 50!.
Number of 7’s in 50! = 8. Note here that if we check for 7’s we do not need to check for 32s as there would be at least two 3’s before a 7 comes in every factorial’s value. Similarly, there would always be at least three 2’s before a 7 comes in any factorial’s value. Thus, the number of 32s and the number of 23s can never be lower than the number of 7s in any factorial’s value.
Number of 5s in 50! = 10 + 2 = 12. Hence, the number of 52s in 50! = [12/2] = 6.
6 will be the answer as the number of 52s is lower than the number of 7’s.
Option (2) is correct.

AVERAGES     1   2   3
1. The average age of 24 students and the principal is 15 years. When the principal’s age is excluded, the average age decreases by 1 year. What is the age of the principal?
1) 38   2) 40   3) 39   4) 37
Ans. (3) P =
2. The average weight of 3 men A, B and C is 84 kg. Another man D joins the group and the average now becomes 80 Kg. If another man E, whose weight is 3 kg more than that of D, replaces A then the average weight of B, C, D and E becomes 78 kg. The weight of A is
1) 70 kg    2) 72 kg   3) 79 kg    4) 78 kg
Ans. (3) D’s weight =  E’s weight = 68 + 3 = 71.
Now, we know that A + B + C + D =  and B + C + D + E =  Hence, A’s weight is 8 kg more than E’s weight. A = 71 + 8 = 79.
3. The mean temperature of Monday to Wednesday was  and of Tuesday to Thursday was . If the temperature on Thursday was 4/5 that of Monday, the temperature on Thursday was

1)    2)    3)    4)
Ans. (2) Monday + Tuesday + Wednesday =  Tuesday + Wednesday + Thursday =  Thus, Monday – Thursday = 9 and Thursday =  Monday/5  Thursday = 36 and Monday = 45.
4. Three years ago, the average age of A, B and C was 27 years and that of B and C 5 years ago was 20 years. A’s present age is
1) 30 years   2) 35 years   3) 40 years   4) 48 years
Ans. (3) Today’s total age of A, B and C =  Today’s total age for B and C =  C’s age = 90 - 50 = 40.
5. Ajit Tendulkar has a certain average for 9 innings. In the tenth inning, he scores 100 runs thereby increasing his average by 8 runs. His new average is
1) 20   2) 24   3) 28   4) 32
Ans. (3)  (average after 9 innings). Hence, new average = 20 + 8 = 28.
ALLIGATIONS
1. If 5 kg of salt costing  5/kg and 3 kg of salt costing  4/kg are mixed, find the average cost of the mixture per kilogram.
1)  4.5   2) 4.625   3)  4.75   4)  4.125
Ans. (2) Solving the following allegation figure:
4                                  ?                      5

3          :           5
2. Two types of oils having the rates of  4/kg and  5/kg respectively are mixed in order to produce a mixture having the rate of  4.60/kg. What should be the amount of the second type of oil if the amount of the first type of oil in the mixture is 40 kg?
1) 75 kg   2) 50 kg   3) 60 kg   4) 40 kg
Ans. (3) Mixing  4/kg and  5/kg to get  4.6 per kg we get that the ratio of mixing is 2:3. If the first oil is 40 kg, the second would be 60 kg.
3. How many kilograms of sugar worth  3.60 per kg should be mixed with 8 kg of sugar worth  4.20 per kg, such that by selling the mixture at  4.40 per kg, there may be a gain of 10%?
1) 6 kg   2) 3 kg    3) 2 kg   4) 4 kg
Ans. (4) Since by selling at  4.40 we want a profit of 10%, it means that the average cost required is  4 per kg. Mixing sugar worth  3.6/kg and  4.2/kg to get  4/kg means a mixture ratio of 1:2. Thus, to 8 kg of the second variety we need to add 4 kg of the first variety to get the required cost price.
4. A mixture of 125 gallons of wine and water contains 20% water. How much water must be added to the mixture in order to increase the percentage of water to 25% of the new mixture?
1) 10 gals   2) 8.5 gals   3) 8 gals   4) 8.33 gals
Ans. (4) In 125 gallons we have 25 gallons water and 100 gallons wine. To increase the percentage of water to 25%, we need to reduce the percentage of wine to 75%. This means that 100 gallons of wine = 75% of the new mixture. Thus the total mixture = 133.33 gallons. Thus, we need to mx 133.33 – 125 = 8.33 gallons of water in order to make the water equivalent to 25% of the mixture.
5. Ravi lends  3600 on simple interest to Harsh for a period of 5 years. He lends a part of the amount at 4% interest and the rest at 6% and receives  960 as the amount of interest. How much money did he lend on 4% interest rate?
1)  2800   2)  2100   3)  2400   4)  1200
Ans. (4) Since, Ravi earns  960 in 5 years, it means that he earns an interest of 960/5 =  192 per year. On an investment of 3600, an annual interest of 192 represents an average interest rate of 5.33%.
Then using the allegation figure below:

Three
4                                              5.33                 6

We get the ratio of investments as 1:2. Hence, he lent  3600/3 = 1200 at 4% per annum.
PERCENTAGE      1   2
1. Which of the following is the largest number?
1) 20% of 200   2) 7% of 500   3) 1300% of 3   4) 700% of 9
Ans. (4) It can be clearly seen that 700% of 9 = 63 is the highest number.
2. If 25% of a number is 75, then 45% of that number is:
1) 145   2) 125   3) 150   4) 135
Ans. (4) 0.25 N = 75  = 300. Thus, 0.45  300 = 135.
3. What is 20% of 50% of 75% of 70?
1) 5.25   2) 6.75   3) 7.25   4) 5.5
Ans. (1) 20% of 50% of 75% of 70% = 20/100  50/100  75/100  70 = 0.2  0.5  0.75  70 = 5.25.
A quicker way to think here would be: 20% of 70 = 14  50% of 14 = 7  75% of 7 = 5.25
4. If we express 41(3/17)% as a fraction, then it is equal to
1)    2)    3)    4)
Ans. (2) 41(3/17)% = 700/17%. As a fraction, the value = 700/(17  100) = 7/17.
5. Mr. Abhimanyu Banerjee is worried about the balance of his monthly budget. The price of pertrol has increased by 40%. By what percent should he reduce the consumption of petrol so that he is able to balance his budget?
1) 33.33   2) 28.56   3) 25   4) 14.28
Ans. (2) The following PCG will give the answer:
40%
100                              140                              100
Price effect               Consumption effect
Hence, the percentage reduction required is 28.56% (40/140)
PROFIT & LOSS   1    4
1. By selling a watch for  495, a shopkeeper incurs a loss of 10%. Find the cost price of the watch for the shopkeeper.
1)  545   2)  550   3)  555   4)  565
Ans. (2) 0.9  Price = 495  Price 550.
2. By selling a cap for  34.40, a man gains 7.5%. What will be the CP of the cap?
1)  32.80   2)  32   3)  32.40   4)  28.80
Ans. (2) The SP = 107.5% of the CP. Thus, CP = 34.4/1.075 =  32.
3. A cellular phone when sold for  4600 fetches a profit of 15%. Find the cost price of the cellular phone.
1)  4300   2)  4150   3)  4000   4)  4500
Ans. (3) 1.15  Price = 4600  Price = 4000.
4. A machine costs  375. If it is sold at a loss of 20%, what will be its cost price as a percentage of its selling price?
1) 80%   2) 120%   3) 110%   4) 125%
Ans. (4) A lost of 20% means a cost price of 100 corresponding to a selling price of 80. CP as a percentage of the SP would then be 125%
5. A shopkeeper sold goods for  2400 and made a profit of 25% in the process. Find his profit per cent if he had sold his goods for  2040.
1) 6.25%   2) 7%   3) 6.20%   4) 6.5%
Ans. (1) 2400 = 1.25  cost price  Cost price = 1920 Profit at 2040 =  120
Percentage profit = (120/2040) 100 = 6.25%
Four
INTEREST    4    5
1.  1200 is lent out at 5% per annum simple interest for 3 years. Find the amount after 3 years.
1)  1380   2)  1290   3)  1470   4)  1200
Ans. (1) The annual interest would be  60. After 3 years the total value would be 1200 + 60  3 = 1380
2. Interest obtained on a sum of  5000 for 3 years is  1500. Find the rate percent.
1) 8%   2) 9%   3) 10%   4) 11%
Ans. (3) The interest earned per year would be 1500/3 = 500. This represents a 10% rate of interest.
3.  2100 is lent at compound interest of 5% per annum for 2 years. Find the amount after two years.
1)  2300   2)  2315.25   3)  2310   4)  2320
Ans. (2) 2100 + 5% of 2100 = 2100 + 105 = 2205 (after 1 year). Next year it would become: 2205 + 5% of 2205 = 2205 + 110.25 = 2315.25
4.  1694 is repaid after two years at compound interest. Which of the following is the value of the principal and the rate?
1)  1200,20%   2)  1300,15%   3)  1400,10%   4)  1500,12%
Ans. (3)           10%                10%
1400                1540                1694.
5. Find the difference between the simple and the compound interest at 5% per annum for 2 years on a principal of  2000.
1) 5   2) 105   3) 4.5   4) 5.5
Ans. (1) Simple Interest for 2 years = 100 + 100 = 200. Compound interest for 2 years: Year 1 = 5% of 2000 = 100.
Year 2:5% of 2100 = 105  Total compound interest =  205.
Difference between the Simple and Compound interest = 205 – 200 =  5
RATIO, PROPORTION AND VARIATION
1. Divide  1870 into three parts in such a way that half of the first part, one-third of the second part and one-sixth of the third part are equal.
1) 241,343,245   2) 400,800,670   3) 470,640,1160   4) None of these
Ans. (4) Solve this question using options. ½ of the first part should equal 1/3rd of the second part and  of the thid part. This means that the first part should be divisible by 2, the second one by 3, and the third one by 6. Looking at the options, none of the first 3 options has its third number divisible by 6. Thus, option (4) is correct.
2. Divide  500 among A, B, C and D so that A and B together get thrice as much as C and D together, B gets four times of what C gets and C gets 1.5 times as much as D. Now the value of what B gets is
1) 300   2) 75   3) 125   4) 150
Ans. (a) (A + B) = 3 (C + D)  A + B   = 375 and C + D = 125. Also, since C gets 1.5 times D we have
C = 75 and D = 50, and B = 4 C = 300.
3. If , then each fraction is equal to
1)    2) ½   3) ¼   4) 0
Ans. (2) The given condition has a, b and c symmetrically placed. Thus, if we use a = b = c = 2 (say) we get each fraction as ½.
4. If 6  + 6  = 13 , what is the ratio of ?
1) 1:4   2) 3:2   3) 4:5   4) 1:2 (Hint: Use options to solve fast)
Ans. (2) Solve using options, Since  it is clear that a ratio of  as 3:2 fits the equation.
5. If  then the value of  is
1) ½   2)    3)    4)
Ans. (4) 1:2 = 3 : 6 So,    From the given options, only ab/cd gives us this value.
Five
TIME AND WORK    1   5
1. Raju can do 25% of a piece of work in 5 days. How many days will he take to complete the work ten times?
1) 150 days   2) 250 days   3) 200 days   4) 180 days
Ans. (3) He will complete the work in 20 days. Hence, he will complete ten times the work in 200 days.
2. 6 men can do a piece of work in 12 days. How many men are needed to do the work in 18 days.
1) 3 men   2) 6 men   3) 4 men   4) 2 men
Ans. (3) 6 men for 12 days means 72 mandays. This would be equal to 4 men for 18 days.
3. A can do a piece of work in 20 days and B can do it in 15 days. How long will they take if both work together?
1)    2)    3)    4)
Ans. (2) A’s one day work will be 5%, while B will do 6.66% of the work in one day. Hence, their total work will be 11.66% in a day.
In 8 days they will complete  11.66  8 = 93.33% This will leave 6.66% of the work. This will correspond to 4/7 of the ninth day since in 6.66/11.66 both the numerator and the denominator are divisible by 1.66.
4. In question 3 if C, who can finish the same work in 25 days, joins them, then how long will they take to complete the work?
1)    2) 12 days   3) 4)
Ans. (1)           A’s work = 5% per day
B’s work = 6.66% per day
C’s work = 4% per day.
Total no. of days = 100/15.66 = 300/47 = 6(18/47)
5. Nishu and Archana can do a piece of work in 10 days and Nishu alone can do it in 12 days. In how many days can Archana do it alone?
1) 60 days   2) 30 days   3) 50 days   4) 45 days
Ans. (1) N + A = 10%   N = 8.33%   Hence A = 1.66%  60 days.
TIME, SPEED AND DISTANCE  2  3
1. The Sinhagad Express left Pune at noon sharp. Two hours later, the Deccan Queen started from Pune in the same direction. The Deccan Queen started from Pune in the same direction. The Deccan Queen overtook the Sinhagad Express at 8 p.m. Find the average speed of the two trains over this journey if the sum of their average speeds is 70 km/h.
1) 34.28 km/h   2) 35 km/h   3) 50 km/h   4) 12 km/h
Ans. (1) The ratio of time for the travel is 4:3 (Sinhagad to Deccan Queen). Hence, the ratio of speeds would be 3:4. Since, the sum of their average speeds is 70 kmph, their respective speeds would be 30 and 40 kmph respectively. Use allegation to get the answer as 34.28 kmph.
2. Walking at ¾ of his normal speed, Abhishek is 16 minutes late in reaching his office. The usual time taken by him to cover the distance between his home and his office is
1) 48 minutes   2) 60 minutes   3) 42 minutes   4) 62 minutes
Ans. (1) When speed goes down to three fourth (i.e. 75%) time will go up to 4/3rd (or 133.33%) of the original time. Since, the extra time required is 16 minutes, it should be equated to 1/3rd of the normal time. Hence, the usual time required will be 48 minutes.
3. Ram and Bharat travel the same distance at the rate of 6 km per hour and 10 km per hour respectively. If Ram takes 30 minutes longer than Bharat, the distance travelled by each is
1) 6 km   2)10 km   3) 7.5 km   4) 20 km
Ans.(3) Since, the ratio of speeds is 3:5, the ratio of times would be 5:3. The difference in the times would be 2 (if looked at in the 5:3 ratio context). Further, since Ram takes 30 minutes longer, 2 corresponds to 30. Hence, using unitary method, 5 will correspond to 75 and 3 will correspond to 45 minutes. Hence at 10 kmph, Bharat would travel 7.5km.
Six
4. Two trains for Mumbai leave Delhi at 6:00 a.m. and 6:45 am and travel at 100 kmph and 136 kmph respectively. How many kilometers from Delhi will the two trains be together?
1) 262.4 km   2) 260 km   3) 283.33 km   4) 275 km
Ans. (3) The train that leaves at 6 am would be 75 km ahead of the other train when it starts. Also, the relative speed being 36 kmph, the distance from Mumbai would be : (75/36) 136 = 283.33 km
5. Two trains, Calcutta Mail and Bombay Mail, start at the same time from stations Kolkata and Mumbai respectively towards each other. After passing each other, they take 12 hours and 3 hours to reach Mumbai and Kolkata respectively. If the Calcutta Mail is moving at the speed of 48 km/h, the speed of the Bombay Mail is
1) 24 km/h   2) 22 km/h   3) 21 km/h   4) 96 km/h
Ans. (4) If you assume that the initial stretch of track is covered by the two trains in time t each, the following figure will give you a clearer picture.
Calcutta Mail               t                       12
3                      t                       Bombay Mail

Kolkata                               Meeting Point               Mumbai
From the above figure, we can deduce that, t/3 = 12/t.
Hence, t2 = 36, gives us t = 6.
Hence, the distance between Kolkata to the starting point is covered by the Calcutta Mail in 6 hours, while the same distance is covered by the Bombay Mail in 3 hours.
Hence, the ratio of their speeds would be 1:2. Hence, the Bombay Mail would travel at 96 kmph.
GEOMETRY AND MENSURATION
1. A vertical stick 20 m long casts a shadow 10 m long on the ground. At the same time, a tower casts the shadow 50 m long on the ground. Find the height of the tower.
1)100 m   2) 120 m   3) 25 m   4) 200 m
Ans.(1) When the length of stick = 20 m, then length of shadow = 10 m i.e. in this case length = 2  shadow with the same angle of inclination of the sun, the length of tower that casts a shadow of 50 m  2  50 m = 100 m
i.e. height of tower = 100 m
2. In the figure,  is similar to
A

B                                                          D
C
E
If we have AB = 4 cm, ED = 3 cm, CE = 4.2 and CD = 4.8 cm, find the value of CA and CB
1) 6 cm, 6.4 cm   2) 4.8 cm, 6.4 cm    3) 5.4 cm, 6.4 cm   4) 5.6 cm, 6.4 cm
Ans. (4)
Then       Then
3. The area of similar triangles, ABC and DEF are 144 cm2 and 81 cm2 respectively. If the longest side of larger  be 36 cm, the the longest side of larger  be 36cm, then the longest side of smaller  is
1) 20 cm   2) 26 cm   3) 27 cm   4) 30 cm
Ans. (3) For similar triangles (Ratio of sides)2 = Ratio of areas
Then as per question =     {Let the longest side of DEF =
4. Two isosceles  have equal angles and their areas are in the ratio 16:25. Find the ratio of their corresponding heights.
Seven
1) 4/5   2) 5/4   3) 3/2   4) 5/7
Ans. (1) (Ratio of corresponding sides)2 = Ratio of area of similar triangle
Ratio of corresponding sides in this question =
5. The areas of two similar  are respectively 9 cm2 and 16 cm2. Find the ratio of their corresponding sides.
1) 3:4   2)  4:3   3) 2:3   4) 4:5
Ans. (1) Ratio of corresponding sides =
COORDINATE GEOMETRY
1. Find the distance between the points (3, 4) and (8, -6).
1)    2)    3)    4)
Ans. (2)
2. Find the distance between the points (5, 2) and (0, 0).
1)    2)   3)  4)
Ans. (3)
3. Find the value of p if the distance between the points (8, p) and (4, 3) is 5.
1) 6   2) 0   3) Both (1) and (2)   4) None of these
Ans. (3)
4. Find the value of c if the distance between the point (c, 4) and the origin is 5 units.
1) 3   2) -3   3) Both (1) and (2)    4) None of these
Ans. (3)
5. Find the mid-point of the line segment made by joining the points (3, 2) and (6, 4).
1)    2)    3)    4)
Ans. (1)
PERMUTATIONS AND COMBINATIONS   1
1. How many numbers of 3-digits can be formed with the digits 9, 8, 7, 6, 5 (repetition of digits not allowed)?
1) 125   2) 120   3) 60   4) 150
Ans. (3) The number of numbers formed would be given by 5  4  3 (given that the first digit can be filled in 5 ways, the second in 4 ways and the third in 3 ways – MNP rule).
2. How many numbers between 2000 and 3000 can be formed with the digits 0, 1, 2, 3, 4, 5, 6, 7 (repetition of digits not allowed?
1) 42   2) 210   3) 336   4) 440
Ans. (2) The first digit can only be 2 (1 way), the second digit can be filled in 7 ways, the third in 6 ways and the fourth in 5 ways. A total of 1 7  6  5 = 210 ways.
3. In how many ways can a person send invitation cards to 6 of his friends if he has four servants to distribute the cards?
1) 64   2) 46   3) 24   4) 120
Ans. (2) Each invitation card can be sent in 4 ways. Thus, .
4. In how many ways can 5 prizes be distributed to 8 students if each student can get any number of prizes?
1) 40   2) 58   3) 85   4) 120
Ans. (3) In this case since nothing is mentioned about whether the prizes are identical or distinct we can take the prizes to be distinct (the most logical thought given the situation). Thus, each prize can be given in 8 ways – thus a total of 85 ways.
5. In how many ways can 7 Indians, 5 Pakistanis and 6 Dutch be seated in a row so that all persons of the same nationality sit together?
1) 3!   2) 7!5!6!   3) 3!7!5!6!   4) 182
Eight
Ans. (3) We need to assume that the 7 Indians are 1 person, so also for the 6 Dutch and the 5 Pakistanis. These 3 groups of people can be arranged amongst themselves in 3! Ways. Also, within themselves the 7 Indians the 6 Dutch and the 5 Pakistanis can be arranged in 7!, 6! And 5! Was respectively. Thus, the answer is .
PROBABILITY
1. In throwing a fair dice, what is the probability of getting the number ‘3’?
1)    2)    3)    4)
Ans. (2) Out of a total of 6 occurences, 3 is one possibility = 1/6.
2. What is the chance of throwing a number greater than 4 with an ordinary dice whose faces are numbered from 1 to 6?
1)    2)    3)    4)
Ans. (1) 5 or 6 out of a sample space of 1, 2, 3, 4, 5 or 6 = 2/6 = 1/3
3. Find the chance of throwing at least one ace in a simple throw with two dice.
1)    2)    3)    4)
Ans. (4) Event definition is:
(1 and 1) or (1 and 2) or (1 and 3) or (1 and 4) or (1 and 5) or (1 and 6) or (2 and 1) or (3 and 1) or (4 and 1) or (5 and 1) or (6 and 1). Total 11 out of 36 possibilities = 11/36
4. Find the chance of drawing 2 blue balls in succession from a bag containing 5 red and 7 blue balls, if the balls are not being replaced.
1)    2)    3)    4)
Ans. (3) Event definition: First is blue and second is blue = 7/12  6/11 = 7/22.
5. From a pack of 52 cards, two are drawn at random. Find the chance that one is a knave and the other a queen.
1)    2)    3)    4)
Ans. (1) Knave and queen or Queen and Knave  4/52  4/51 + 4/52  4/51 = 8/663
SET THEORY
1. The percentage of those who like coffee or tea but not smoking among those who like at least one of these is
1) more than 30   2) less than 30   3) less than 25   4) None of these
Ans. (1) 30/94  more than 30%. Option (1) is correct.
2. The percentage of those who like at least one of these is
1) 100   2) 90   3) Nil   4) 94
Ans. (4) 94%. Option (4) is correct.
3. The two items having the ratio 1:2 are
1) Tea only and tea and smoking only.   2) Coffee and smoking only and tea only.
3) Coffee and tea but not smoking and smoking but not coffee and tea.   4) None of these
Ans. (3) Option (3) is correct as the ratio turns out to be 10:20 in that case.
4. The number of persons who like coffee and smoking only and the number who like tea only bear a ratio
1) 1:2   2) 1:1   3) 5:1   4) 2:1
Ans. (2) 12:12 = 1:1  Option (2) is correct.
5. Percentage of those who like tea and smoking but not coffee is
1) 14   2) 14.9   3) less than 14   4) more than 15
Ans. (1) 14%. Option (1) is correct.