aptitude

2.

The price of 2 sarees and 4 shirts is Rs. 1600. With the same money one can buy 1

saree and 6 shirts. If one wants to buy 12 shirts, how much shall he have to pay ?

A fires 5 shots to B's 3 but A kills only once in 3 shots while B kills once in 2 shots.

When B has missed 27 times, A has killed:

Let the total number of shots be x. Then, Shots fired by

A =

5 x

8

Shots fired by

B =

3 x

8

Killing shots by

A =

1

of

5

x = 5

x 3 8

2

4

Shots missed by

B =

1

of

3

x = 3

x 2 8

1

6

3

x = 27 or x =

27 x

16

=

144. 1

6 3

Birds killed by

A =

5

x =

5 x

144

=

30. 2

4

2

4

3. If 5 = 2.236, then the

value of

5 -

1

0 + 125 is

equal to: 2 5

5 -

1

0 +

125 =

(5)2 - 20 + 25

x 55

2 5 25

=

5 - 20 +

50

25

=

3

5 x

5

2

5 5

=

35

5

10

7 x

2.236

2

= 7 x

1.118

=

7.826

4.

62

5 x

1

4 x

11

is equal

to: 11

2

5

19

6

Given

Expression =

2

5 x

1

4 x

1

1 =

5. 1

1 5

1

4

5

.

A two-digit number is such that the product of the digits is 8. When 18 is added to the

number, then the digits are reversed. The number is:

Let the ten's and unit digit be x and 8

respectively. x

The

n,

10x +

8

+ 18 =

10 x

8 +

x x x

10x2 + 8 + 18x = 80 + x2

9x2 + 18x - 72 = 0

x2 + 2x - 8 = 0

(x + 4)(x - 2) = 0

x = 2.

6. A batsman scored 110 runs which included 3 boundaries and 8 sixes. What

percent of his total score did he make by running between the wickets?

Number of runs made by running = 110 - (3 x 4 + 8 x 6) = 50.

Required

percentage =

50 x

100 % =

45

5

% 11

0

1

1

7. Rajeev buys good worth Rs. 6650. He gets a rebate of 6% on it. After getting the

rebate, he pays sales tax @ 10%. Find the amount he will have to pay for the

goods.

Rebate = 6% of Rs.

6650 = Rs.

6 x

6650

= Rs.

399. 10

0

Sales tax = 10% of Rs.

(6650 - 399) = Rs.

10 x

6251

= Rs.

625.10 10

0

Final amount = Rs. (6251 + 625.10) = Rs. 6876.10

8. In a flight of 600 km, an aircraft was slowed down due to bad weather. Its

average speed for the trip was reduced by 200 km/hr and the time of flight

increased by 30 minutes. The duration of the flight is:

Let the duration of the flight be x hours.

The

n,

60

0 -

600 =

200 x

x + (1/2)

60

0 -

120

0 =

200 x

2x + 1

x(2x + 1) = 3

2x2 + x - 3 = 0

(2x + 3)(x - 1) = 0

x = 1 hr. [neglecting the -ve value of x]

9. A goods train runs at the speed of 72 kmph and crosses a 250 m long platform in

26 seconds. What is the length of the goods train?

Speed

=

72

x

5

m/sec

= 20

m/sec. 1

8

Time = 26 sec.

Let the length of the train be x metres.

The

n,

x + 250

=

20 26

x + 250 = 520

x = 270.

10. How many kilogram of sugar costing Rs. 9 per kg must be mixed with 27 kg of

sugar costing Rs. 7 per kg so that there may be a gain of 10% by selling the

mixture at Rs. 9.24 per kg?

S.P. of 1 kg of mixture = Rs. 9.24, Gain 10%.

C.P. of 1 kg of

mixture = Rs.

10

0 x

9.24

= Rs.

8.40 11

0

By the rule of allilation, we have:

C.P. of 1 kg sugar of 1st kind Cost of 1 kg

sugar of 2nd kind

Rs. 9 Mean Price

Rs. 8.40

Rs. 7

1.40 0.60

Ratio of quantities of 1st and 2nd kind = 14 : 6 = 7 : 3.

Let x kg of sugar of 1st be mixed with 27 kg of 2nd kind.

Then, 7 : 3 = x : 27

x =

7 x

27

= 63

kg. 3

11. A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854

in 4 years. The sum is:

S.I. for 1 year = Rs. (854 - 815) = Rs. 39.

S.I. for 3 years = Rs.(39 x 3) = Rs. 117.

Principal = Rs. (815 - 117) = Rs. 698.

12. A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest

at the same rate of interest and received Rs. 2200 in all from both of them as

interest. The rate of interest per annum is:

Let the rate be R% p.a.

The

n,

5000 x R

x 2

+

3000 x R

x 4

=

2200. 100 100

100R + 120R = 2200

R

=

220

0

=

10. 220

Rate = 10%.

13. The percentage increase in the area of a rectangle, if each of its sides is

increased by 20% is:

Let original length = x metres and original breadth = y metres.

Original area = (xy) m2.

New

length =

12

0 x m

=

6

x m.

10

0 5

New

12 y =

6 y

breadth = 0 m m.

10

0 5

New

Area =

6 x x

6

y m2

=

3

6 xy

m2. 5 5 2

5

Increase

% =

1

1 xy x

1 x

100 %

=

44%. 2

5

xy

14. What is the least number of squares tiles required to pave the floor of a room

15 m 17 cm long and 9 m 2 cm broad?

Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.

Area of each tile = (41 x 41) cm2.

Required number of

tiles =

1517 x

902

=

814. 41 x 41

15. A large cube is formed from the material obtained by melting three smaller

cubes of 3, 4 and 5 cm side. What is the ratio of the total surface areas of the

smaller cubes and the large cube?

Volume of the large cube = (33 + 43 + 53) = 216 cm3.

Let the edge of the large cube be a.

So, a3 = 216 a = cm.

Required

ratio =

6 x (32 + 42

+ 52)

=

5

0 = 25 :

18. 6 x 62

3

6

16. At what angle the hands of a clock are inclined at 15 minutes past 5?

Angle traced by hour 2 hrs

36 x 2

= 15 1

hand in 1 = 0 1 7

4 12 4 2

Angle traced by min. hand

in 15 min. =

36

0 x

15

=

90. 60

Required

angle =

15

7

1

- 90 =

67

1

2 2

17. In how many ways a committee, consisting of 5 men and 6 women can be

formed from 8 men and 10 women?

Required number of

ways = (8C5 x 10C6)

= (8C3 x 10C4)

=

8 x 7

x 6 x

10 x 9 x 8

x 7

3 x 2

x 1

4 x 3 x 2 x

1

= 11760.

18. In how many ways can a group of 5 men and 2 women be made out of a total of

7 men and 3 women?

Required number of ways =

(7C5 x 3C2) = (7C2 x 3C1) =

7

x

6 x

3

=

63. 2

x

1

19. The true discount on a bill due 9 months hence at 16% per annum is Rs. 189. The

amount of the bill is:

Let P.W. be Rs. x.

Then, S.I. on Rs. x at 16% for 9 months = Rs. 189.

x x 16 x

9

x

1 = 186 or x = 1575.

1

2

10

0

P.W. = Rs. 1575.

Sum due = P.W. + T.D. = Rs. (1575 + 189) = Rs. 1764.

20. The banker's gain on a sum due 3 years hence at 12% per annum is Rs. 270.

The is:

T.D.

=

B.G. x

100

=

Rs.

270 x

100

= Rs.

750. R x T 12 x 3

B.D. = Rs.(750 + 270) = Rs. 1020.

1. A fires 5 shots to B's 3 but A kills only once in 3 shots while B kills once in 2

shots. When B has missed 27 times, A has killed:

Let the total number of shots be x. Then, Shots fired by

A =

5 x

8

Shots fired by

B =

3 x

8

Killing shots by

A =

1

of

5

x = 5

x 3 8

2

4

Shots missed by

B =

1

of

3

x = 3

x 2 8

1

6

3

x = 27 or x =

27 x

16

=

144. 1

6 3

Birds killed by

A =

5

x =

5 x

144

=

30.

2. A starts business with Rs. 3500 and after 5 months, B joins with A as his partner.

After a year, the profit is divided in the ratio 2 : 3. What is B's contribution in the

capital?

Let B's capital be Rs. x.

Then,

3500 x 12 =

2

7x 3

14x = 126000

x = 9000.

3. A and B started a partnership business investing some amount in the ratio of 3 :

5. C joined then after six months with an amount equal to that of B. In what

proportion should the profit at the end of one year be distributed among A, B and

C?

Let the initial investments of A and B be 3x and 5x.

A : B : C = (3x x 12) : (5x x 12) : (5x x 6) = 36 : 60 : 30 = 6 : 10 : 5.

4. 36 men can complete a piece of work in 18 days. In how many days will 27 men

complete the same work?

Let the required number of days be x.

Less men, More days (Indirect Proportion)

27 : 36 :: 18 : x 27 x x = 36 x 18

x = 36 x 18

27

x = 24.

Direction (for Q.No. 5):

Each of the questions given below consists of a question followed by three

statements. You have to study the question and the statements and decide which of

the statement(s) is/are necessary to answer the question.

5. How many workers are required for completing the construction work in 10 days?

I. 20% of the work can be completed by 8 workers in 8 days.

II. 20 workers can complete the work in 16 days.

III. One-eighth of the work can be completed by 8 workers in 5 days.

A. I only

B. II and III only

C. III only

D. I and III only

E. Any one of the three

I. 20 work can be completed by (8 x 8) workers

in 1 day. 100

Whole work can be completed by (8 x 8 x 5) workers in 1 day.

= 8 x 8 x 5

workers in 10 days = 32 workers in 10 days. 10

II. (20 x 16) workers can finish it in 1 day.

(20 x 16)

workers can finish it in 10 days. 10

32 workers can finish it in 10 days.

III. 1

work can be completed by (8 x 5) workers in 1 day. 8

Whole work can be completed by (8 x 5 x 8) workers in 1 day.

= 8 x 5 x 8

workers in 10 days = 32 workers in 10 days. 10

Any one of the three gives the answer.

Correct answer is (E).

6. Two trains 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr

respectively in opposite directions on parallel tracks. The time (in seconds) which

they take to cross each other, is:

Relative speed = (60 + 40) km/hr =

100 x 5

m/sec =

250

m/sec. 18 9

Distance covered in crossing each other = (140 + 160) m = 300 m.

Required time =

300 x 9

sec =

54 sec = 10.8 sec.

250 5

7. A boat running upstream takes 8 hours 48 minutes to cover a certain distance,

while it takes 4 hours to cover the same distance running downstream. What is

the ratio between the speed of the boat and speed of the water current

respectively?

Let the man's rate upstream be x kmph and that downstream be y kmph.

Then, distance covered upstream in 8 hrs 48 min = Distance covered downstream

in 4 hrs.

x x 8 4

= (y x 4)

5

44 x =4y

5

y = 11

x. 5

Required ratio =

y + x

:

y - x

2 2

=

16x x 1

:

6x x 1

5 2 5 2

= 8

: 3

5 5

= 8 : 3.

8. A container contains 40 litres of milk. From this container 4 litres of milk was

taken out and replaced by water. This process was repeated further two times.

How much milk is now contained by the container?

Amount of milk left after 3 operations =

40

1 - 4

3 litres 40

=

40 x 9

x 9

x 9

= 29.16 litres. 10 10 10

9. What will be the ratio of simple interest earned by certain amount at the same

rate of interest for 6 years and that for 9 years?

Let the principal be P and rate of interest be R%.

Required ratio =

P x R x 6

100 =

6PR

=

6

= 2 : 3.

P x R x 9

100

9PR 9

10. If the simple interest on a sum of money for 2 years at 5% per annum is Rs. 50,

what is the compound interest on the same at the same rate and for the same

time?

Sum = Rs.

50 x 100

= Rs. 500. 2 x 5

Amount

=

Rs.

500

x

1

+

5

2

100

= Rs.

500 x 21

x 21

20 20

= Rs. 551.25

C.I. = Rs. (551.25 - 500) = Rs. 51.25

11. The slant height of a right circular cone is 10 m and its height is 8 m. Find the

area of its curved surface.

l = 10 m,

h = 8 m.

So, r = l2 - h2 = (10)2 - 82 = 6 m.

Curved surface area = rl = ( x 6 x 10) m2 = 60 m2.

12. In a 100 m race, A beats B by 10 m and C by 13 m. In a race of 180 m, B will beat

C by:

A : B = 100 : 90.

A : C = 100 : 87.

B =

B x A

= 90

x 100

= 30

. C A C 100 87 29

When B runs 30 m, C runs 29 m.

When B runs 180 m, C runs

29 x 180

m = 174 m.

30

B beats C by (180 - 174) m = 6 m.

13. What will be the day of the week 15th August, 2010?

15th August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)

Odd days in 1600 years = 0

Odd days in 400 years = 0

9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 odd days 4 odd

days.

Jan. Feb. March April May June July Aug.

(31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days

227 days = (32 weeks + 3 days) 3 odd days.

Total number of odd days = (0 + 0 + 4 + 3) = 7 0 odd days.

Given day is Sunday.

14. If 6th March, 2005 is Monday, what was the day of the week on 6th March, 2004?

The year 2004 is a leap year. So, it has 2 odd days.

But, Feb 2004 not included because we are calculating from March 2004 to

March 2005. So it has 1 odd day only.

The day on 6th March, 2005 will be 1 day beyond the day on 6th March, 2004.

Given that, 6th March, 2005 is Monday.

6th March, 2004 is Sunday (1 day before to 6th March, 2005).

15. At what time between 9 and 10 o'clock will the hands of a watch be together?

To be together between 9 and 10 o'clock, the minute hand has to gain 45 min.

spaces.

55 min. spaces gained in 60 min.

45 min. spaces are gained in

60 x 45

min or 49

1 min.

55 11

The hands are together at 49 1

min. past 9. 11

16. A watch which gains uniformly is 2 minutes low at noon on Monday and is 4

min. 48 sec fast at 2 p.m. on the following Monday. When was it correct?

Time from 12 p.m. on Monday to 2 p.m. on the following Monday = 7 days 2

hours = 170 hours.

The watch gains

2 + 4 4

min. or

34 min. in 170 hrs.

5 5

Now, 34

min. are gained in 170 hrs. 5

2 min. are gained in

170 x 5

x 2 hrs

= 50 hrs. 34

Watch is correct 2 days 2 hrs. after 12 p.m. on Monday i.e., it will be correct at 2 p.m. on Wednesday.

17. A man buys Rs. 20 shares paying 9% dividend. The man wants to have an

interest of 12% on his money. The market value of each share is:

Dividend on Rs. 20 = Rs.

9 x 20

= Rs. 9

. 100 5

Rs. 12 is an income on Rs. 100.

Rs. 9

is an income on Rs.

100 x 9

= Rs. 15. 5 12 5

18. A 12% stock yielding 10% is quoted at:

To earn Rs. 10, money invested = Rs. 100.

To earn Rs. 12, money invested = Rs.

100 x 12

= Rs. 120. 10

Market value of Rs. 100 stock = Rs. 120.

19. The on a certain sum due 2 years hence is

11 of the true discount.

10

The rate percent is:

Let T.D. be Re. 1.

Then, B.D. = Rs. 11

= Rs. 1.10. 10

Sum = Rs.

1.10 x 1

= Rs.

110

= Rs. 11. 1.10 - 1 10

S.I. on Rs. 11 for 2 years is Rs. 1.10

Rate =

100 x 1.10

% = 5%.

11 x 2

Find out the wrong number in the series.

20. 2880, 480, 92, 24, 8, 4, 4

Go on dividing by 6, 5, 4, 3, 2, 1 respectively to obtain the next number.

Clearly, 92 is wrong.

1. Find the lowest common multiple of 24, 36 and 40.

2 | 24 - 36 - 40

--------------------

2 | 12 - 18 - 20

--------------------

2 | 6 - 9 - 10

-------------------

3 | 3 - 9 - 5

-------------------

| 1 - 3 - 5

L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360.

2. Sachin is younger than Rahul by 7 years. If their ages are in the respective ratio

of 7 : 9, how old is Sachin?

Let Rahul's age be x years.

Then, Sachin's age = (x - 7) years.

x - 7 =

7

x 9

9x - 63 = 7x

2x = 63

x = 31.5

Hence, Sachin's age =(x - 7) = 24.5 years.

3. The value of [(10)150 (10)146]

(10)150 (10)146 = 10150

10146

= 10150 - 146

= 104

= 10000.

4. 1 +

1 = ?

1 + a(n - m) 1 + a(m - n)

1

+

1

=

1

+

1

1 + an

am

1 + am

an

1 + a(n - m) 1 + a(m - n)

= am

+ an

(am + an) (am + an)

= (am + an) (am + an)

= 1.

5

.

A student multiplied a

number by

3 instead

of

5 .

5 3.

What is the percentage error in the calculation?

Let the number be x.

Then, error = 5

x - 3

x = 16

x. 3 5 15

Error% =

16x x 3

x 100 % = 64%. 15 5x

6. A shopkeeper sells some articles at the profit of 25% on the original price. What is

the exact amount of profit? To find the answer, which of the following information

given in Statements I and II is/are necessary?

I. Sale price of the article

II. Number of articles sold

A. Only I is necessary

B. Only II is necessary

C. Either I or II is necessary

D. Both I and II are necessary

E. None of these

Gain = 25% of C.P.

In order to find gain, we must know the sale price of each article and the number

of articles sold.

Correct answer is (D).

7. The ratio of the number of boys and girls in a college is 7 : 8. If the percentage

increase in the number of boys and girls be 20% and 10% respectively, what will

be the new ratio?

Originally, let the number of boys and girls in the college be 7x and 8x respectively.

Their increased number is (120% of 7x) and (110% of 8x).

120 x 7x

and

110 x 8x

100 100

42x and

44x 5 5

The required ratio =

42x : 44x

= 21 : 22. 5 5

8. If 40% of a number is equal to two-third of another number, what is the ratio of

first number to the second number?

Let 40% of A = 2

B 3

Then, 40A

= 2B

100 3

2A =

2B

5 3

A =

2 x 5

= 5

B 3 2 3

A : B = 5 : 3.

9. Three partners shared the profit in a business in the ratio 5 : 7 : 8. They had

partnered for 14 months, 8 months and 7 months respectively. What was the ratio

of their investments?

Let their investments be Rs. x for 14 months, Rs. y for 8 months and Rs. z for 7 months respectively.

Then, 14x : 8y : 7z = 5 : 7 : 8.

Now, 14x

= 5

98x = 40y y = 49

x 8y 7 20

And, 14x

= 5

112x = 35z z = 112

x = 16

x. 7z 8 35 5

x : y : z = x : 49

x : 16

x = 20 : 49 : 64. 20 5





10. How much did Rohit get as profit at the year-end in the business done by Nitin,

Rohit and Kunal?

I. Kunal invested Rs. 8000 for nine months, his profit was times that of

Rohit's and his investment was four times that of Nitin.

II. Nitin and Rohit invested for one year in the proportion 1 : 2 respectively.

III. The three together got Rs. 1000 as profit at the year end.

A. Only I and II

B. Only I and III

C. Question cannot be answered even with the information in all the three

statements.

D. All I, II and III

E. None of these

I and II give:

K = Rs. (8000 x 9) for 1 month = Rs. 72000 for 1 month.

N = Rs.

1 x 8000 x 12

for 1 month = Rs. 24000 for 1 month. 4

R = Rs. 48000 for 1 month.

K : N : R = 72000 : 24000 : 48000 = 3 : 1 : 2.

III gives, total profit = Rs. 1000.

Rohit's share = Rs.

1000 x 2

= Rs. 333 1

6 3


11. It takes eight hours for a 600 km journey, if 120 km is done by train and the rest

by car. It takes 20 minutes more, if 200 km is done by train and the rest by car.

The ratio of the speed of the train to that of the cars is:

Let the speed of the train be x km/hr and that of the car be y km/hr.

Then, 120

+ 480

= 8 1

+ 4

= 1

....(i) x y x y 15

And, 200

+ 400

= 25

1

+ 2

= 1

....(ii) x y 3 x y 24

Solving (i) and (ii), we get: x = 60 and y = 80.

Ratio of speeds = 60 : 80 = 3 : 4.

Each of the questions given below consists of a statement and / or a question and

two statements numbered I and II given below it. You have to decide whether the

data provided in the statement(s) is / are sufficient to answer the given question.

Read the both statements and

G

ive answer (A) if the data in Statement I alone are sufficient to answer the

question, while the data in Statement II alone are not sufficient to answer

the question.

G

ive answer (B) if the data in Statement II alone are sufficient to answer the

question, while the data in Statement I alone are not sufficient to answer the

question.

G

ive answer (C) if the data either in Statement I or in Statement II alone are

sufficient to answer the question.

G

ive answer (D) if the data even in both Statements I and II together are not


G

ive answer(E) if the data in both Statements I and II together are necessary

to answer the question.

12. The towns A, B and C are on a straight line. Town C is between A and B. The

distance from A to B is 100 km. How far is A from C?

I. The distance from A to B is 25% more than the distance from C to B.

II. The distance from A to C is of the distance C to B.

A. I alone sufficient while II alone not sufficient to answer

B. II alone sufficient while I alone not sufficient to answer

C. Either I or II alone sufficient to answer

D. Both I and II are not sufficient to answer

E. Both I and II are necessary to answer

.__________.______________________________________.

A x C (100 - x) B

Let AC = x km. Then, CB = (100 -x) km.

I. AB = 125% of CB

100 = 125

x (100 - x) 100

100 - x = 100 x 100

= 80 125

x = 20 km.

AC = 20 km.

Thus, I alone gives the answer.

II. AC = 1

CB 4

x = 1

(100 - x) 4

x = 100

x = 20.

AC = 20 km.

Thus, II alone gives the answer.

Correct answer is (C).

13. Two trains running in opposite directions cross a man standing on the platform

in 27 seconds and 17 seconds respectively and they cross each other in 23

seconds. The ratio of their speeds is:

Let the speeds of the two trains be x m/sec and y m/sec respectively.

Then, length of the first train = 27x metres,

and length of the second train = 17y metres.

27x + 17y = 23

x+ y

27x + 17y = 23x + 23y

4x = 6y

x =

3 .

y 2

14. The diagonal of a rectangle is 41 cm and its area is 20 sq. cm. The perimeter of

the rectangle must be:

l2 + b2 = 41.

Also, lb = 20.

(l + b)2 = (l2 + b2) + 2lb = 41 + 40 = 81

(l + b) = 9.

Perimeter = 2(l + b) = 18 cm.





G



the question.

G



question.

G



G



G



15. Is a given rectangular block, a cube?

I. At least 2 faces of the rectangular block are squares.

II. The volume of the block is 64.






I gives, any two of l, b, h are equal.

II gives, lbh = 64.

From I and II, the values of l, b, h may be (1 ,1 , 64), (2 ,2 ,16), (4, 4, 4).

Thus, the block may be a cube or cuboid.


16. How many times in a day, are the hands of a clock in straight line but opposite in

direction?

The hands of a clock point in opposite directions (in the same straight line) 11

times in every 12 hours. (Because between 5 and 7 they point in opposite

directions at 6 o'clcok only).

So, in a day, the hands point in the opposite directions 22 times.

17. In a class, there are 15 boys and 10 girls. Three students are selected at random.

The probability that 1 girl and 2 boys are selected, is:

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.

Then, n(S) = Number ways of selecting 3 students out of 25

= 25C3 `

=

(25 x 24 x 23)

(3 x 2 x 1)

= 2300.

n(E) = (10C1 x 15C2)

=

10 x (15 x 14)

(2 x 1)

= 1050.

P(E) = n(E)

= 1050

= 21

. n(S) 2300 46

18. One card is drawn at random from a pack of 52 cards. What is the probability

that the card drawn is a face card?

Clearly, there are 52 cards, out of which there are 12 face cards.

P (getting a face card) = 12

= 3

. 52 13

Insert the missing number.

19. 11, 13, 17, 19, 23, 29, 31, 37, 41, (....)

Numbers are all primes. The next prime is 43.


20. 15, 16, 34, 105, 424, 2124, 12576

2nd term = (1st term) x 1 + 1 = 15 x 1 + 1 = 16.

3rd term = (2nd term) x 2 + 2 = 16 x 2 + 2 = 34.

4th term = (3th term) x 3 + 3 = 34 x 3 + 3 = 105.

5th term = (4th term) x 4 + 4 = 105 x 4 + 4 = 424

6th term = (5th term) x 5 + 5 = 424 x 5 + 5 = 2125

6th term should 2125 instead of 2124.

1. The G.C.D. of 1.08, 0.36 and 0.9 is:

Given numbers are 1.08, 0.36 and 0.90. H.C.F. of 108, 36 and 90 is 18,

H.C.F. of given numbers = 0.18.

2. Three numbers which are co-prime to each other are such that the product of the

first two is 551 and that of the last two is 1073. The sum of the three numbers is:

Since the numbers are co-prime, they contain only 1 as the common factor.

Also, the given two products have the middle number in common.

So, middle number = H.C.F. of 551 and 1073 = 29;

First number =

551

= 19; Third number =

1073

= 37. 29 29

Required sum = (19 + 29 + 37) = 85.

3. Which of the following has the most number of divisors?

99 = 1 x 3 x 3 x 11

101 = 1 x 101

176 = 1 x 2 x 2 x 2 x 2 x 11

182 = 1 x 2 x 7 x 13

So, divisors of 99 are 1, 3, 9, 11, 33, .99

Divisors of 101 are 1 and 101

Divisors of 176 are 1, 2, 4, 8, 16, 22, 44, 88 and 176

Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.

Hence, 176 has the most number of divisors.

4. There are two examinations rooms A and B. If 10 students are sent from A to B,

then the number of students in each room is the same. If 20 candidates are sent

from B to A, then the number of students in A is double the number of students in

B. The number of students in room A is:

Let the number of students in rooms A and B be x and y respectively.

Then, x - 10 = y + 10 x - y = 20 .... (i)

and x + 20 = 2(y - 20) x - 2y = -60 .... (ii)

Solving (i) and (ii) we get: x = 100 , y = 80.

The required answer A = 100.

5. If one-third of one-fourth of a number is 15, then three-tenth of that number is:


Then, 1

of 1

of x = 15 x = 15 x 12 = 180. 3 4

So, required number =

3 x 180

= 54. 10

6. The percentage profit earned by selling an article for Rs. 1920 is equal to the

percentage loss incurred by selling the same article for Rs. 1280. At what price

should the article be sold to make 25% profit?

Let C.P. be Rs. x.

Then, 1920 - x

x 100 = x - 1280

x 100 x x

1920 - x = x - 1280

2x = 3200

x = 1600

Required S.P. = 125% of Rs. 1600 = Rs.

125 x 1600

= Rs 2000. 100

7. X and Y can do a piece of work in 20 days and 12 days respectively. X started the

work alone and then after 4 days Y joined him till the completion of the work.

How long did the work last?

Work done by X in 4 days =

1 x 4

= 1

. 20 5

Remaining work =

1 - 1

= 4

. 5 5

(X + Y)'s 1 day's work =

1 +

1

= 8

= 2

. 20 12 60 15

Now, 2

work is done by X and Y in 1 day. 15

So, 4

work will be done by X and Y in

15 x 4

= 6 days. 5 2 5

Hence, total time taken = (6 + 4) days = 10 days.

8. A tap can fill a tank in 6 hours. After half the tank is filled, three more similar

taps are opened. What is the total time taken to fill the tank completely?

Time taken by one tap to fill half the tank = 3 hrs.

Part filled by the four taps in 1 hour =

4 x 1

= 2

. 6 3

Remaining part =

1 - 1

= 1

. 2 2

2 : 1

:: 1 : x 3 2

x =

1 x 1 x

3

= 3

hours i.e., 45 mins. 2 2 4

So, total time taken = 3 hrs. 45 mins.

9. A car travelling with of its actual speed covers 42 km in 1 hr 40 min 48 sec. Find

the actual speed of the car.

Time taken = 1 hr 40 min 48 sec = 1 hr 40 4

min = 1 51

hrs = 126

hrs. 5 75 75

Let the actual speed be x km/hr.

Then, 5

x x 126

= 42 7 75

x =

42 x 7 x 75

= 35 km/hr. 5 x 126

10. A jogger running at 9 kmph alongside a railway track in 240 metres ahead of the

engine of a 120 metres long train running at 45 kmph in the same direction. In

how much time will the train pass the jogger?

Speed of train relative to jogger = (45 - 9) km/hr = 36 km/hr.

=

36 x 5

m/sec 18

= 10 m/sec.

Distance to be covered = (240 + 120) m = 360 m.

Time taken =

360

sec = 36 sec.

10

11. A certain amount earns simple interest of Rs. 1750 after 7 years. Had the

interest been 2% more, how much more interest would it have earned?

We need to know the S.I., principal and time to find the rate.

Since the principal is not given, so data is inadequate.

12. The difference between simple interest and compound on Rs. 1200 for one year at

10% per annum reckoned half-yearly is:

S.I. = Rs

1200 x 10 x 1

= Rs. 120. 100

C.I. = Rs.

1200 x

1 + 5

2 - 1200

= Rs. 123. 100

Difference = Rs. (123 - 120) = Rs. 3.

13. If log10 2 = 0.3010, then log2 10 is equal to:

log2 10 = 1

= 1

= 10000

= 1000

. log10 2 0.3010 3010 301

14. 66 cubic centimetres of silver is drawn into a wire 1 mm in diameter. The length

of the wire in metres will be:

Let the length of the wire be h.

Radius = 1

mm = 1

cm. Then, 2 20

22 x 1

x 1

x h = 66. 7 20 20

h =

66 x 20 x 20 x 7

= 8400 cm = 84 m. 22

15. In a 100 m race, A can give B 10 m and C 28 m. In the same race B can give C:

A : B = 100 : 90.

A : C = 100 : 72.

B : C = B

x A

= 90

x 100

= 90

. A C 100 72 72

When B runs 90 m, C runs 72 m.

When B runs 100 m, C runs

72 x 100

m = 80 m.

90

B can give C 20 m.

16. The calendar for the year 2007 will be the same for the year:

Count the number of odd days from the year 2007 onwards to get the sum equal

to 0 odd day.

Year : 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017

Odd day : 1 2 1 1 1 2 1 1 1 2 1

Sum = 14 odd days 0 odd days.

Calendar for the year 2018 will be the same as for the year 2007.

17. January 1, 2008 is Tuesday. What day of the week lies on Jan 1, 2009?

The year 2008 is a leap year. So, it has 2 odd days.

1st day of the year 2008 is Tuesday (Given)

So, 1st day of the year 2009 is 2 days beyond Tuesday.

Hence, it will be Thursday.

18. How many times in a day, the hands of a clock are straight?

In 12 hours, the hands coincide or are in opposite direction 22 times.

In 24 hours, the hands coincide or are in opposite direction 44 times a day.

Find the odd man out.

19. 1, 4, 9, 16, 20, 36, 49

The pattern is 12, 22, 32, 42, 52, 62, 72. But, instead of 52, it is 20 which to be

turned out.

Find out the wrong number in the given sequence of numbers.

20. 56, 72, 90, 110, 132, 150

The numbers are 7 x 8, 8 x 9, 9 x 10, 10 x 11, 11 x 12, 12 x 13.

So, 150 is wrong.

1. Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the

same remainder in each case. Then sum of the digits in N is:

N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)

= H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

2. The value of

489.1375 x 0.0483 x 1.956 is closet to:

0.0873 x 92.581 x 99.749

489.1375 x 0.0483 x 1.956

489 x 0.05 x 2

0.0873 x 92.581 x 99.749 0.09 x 93 x 100

= 489

9 x 93 x 10

= 163

x 1

279 10

= 0.58

10

= 0.058 0.06.

3. The fraction 101

27 in decimal for is:

100000

101 27

= 101 + 27

= 101 + .00027 = 101.00027 100000 100000

4. The price of 10 chairs is equal to that of 4 tables. The price of 15 chairs and 2

tables together is Rs. 4000. The total price of 12 chairs and 3 tables is:

Let the cost of a chair and that of a table be Rs. x and Rs. y respectively.

Then, 10x = 4y or y = 5

x. 2

15x + 2y = 4000

15x + 2 x 5 x = 4000

2

20x = 4000

x = 200.

So, y =

5 x 200

= 500. 2

Hence, the cost of 12 chairs and 3 tables = 12x + 3y

= Rs. (2400 + 1500)

= Rs. 3900.

5. The average of 20 numbers is zero. Of them, at the most, how many may be

greater than zero?

Average of 20 numbers = 0.

Sum of 20 numbers (0 x 20) = 0.

It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a).





6. What is Arun's present age?

I. Five years ago, Arun's age was double that of his son's age at that time.

II. Present ages of Arun and his son are in the ratio of 11 : 6 respectively.

III. Five years hence, the respective ratio of Arun's age and his son's age will

become 12 : 7.

A. Only I and II

B. Only II and III

C. Only I and III

D. Any two of the three

E. None of these

II. Let the present ages of Arun and his son be 11x and 6x years respectively.

I. 5 years ago, Arun's age = 2 x His son's age.

III. 5 years hence, Arun's Age

= 12

Son's age 7

Clearly, any two of the above will give Arun's present age.


7. A, B, C rent a pasture. A puts 10 oxen for 7 months, B puts 12 oxen for 5 months

and C puts 15 oxen for 3 months for grazing. If the rent of the pasture is Rs. 175,

how much must C pay as his share of rent?

A : B : C = (10 x 7) : (12 x 5) : (15 x 3) = 70 : 60 : 45 = 14 : 12 : 9.

C's rent = Rs.

175 x 9

= Rs. 45. 35

8. An industrial loom weaves 0.128 metres of cloth every second. Approximately,

how many seconds will it take for the loom to weave 25 metres of cloth?

Le the required time be x seconds.

More metres, More time (Direct Proportion)

0.128 : 25 :: 1 : x 0.128x = 25 x 1

x = 25

= 25 x 1000

0.128 128

x = 195.31.

Required time = 195 sec (approximately).

9. A train 125 m long passes a man, running at 5 kmph in the same direction in

which the train is going, in 10 seconds. The speed of the train is:

Speed of the train relative to man =

125

m/sec 10

=

25

m/sec. 2

=

25 x 18

km/hr 2 5

= 45 km/hr.

Let the speed of the train be x kmph. Then, relative speed = (x - 5) kmph.

x - 5 = 45 x = 50 kmph.

10. A train passes a station platform in 36 seconds and a man standing on the

platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of

the platform?

Speed =

54 x 5

m/sec = 15 m/sec. 18

Length of the train = (15 x 20)m = 300 m.

Let the length of the platform be x metres.

Then, x + 300

= 15 36

x + 300 = 540

x = 240 m.

11. Two trains are moving in opposite directions @ 60 km/hr and 90 km/hr. Their

lengths are 1.10 km and 0.9 km respectively. The time taken by the slower train

to cross the faster train in seconds is:

Relative speed = (60+ 90) km/hr

=

150 x 5

m/sec 18

=

125

m/sec. 3

Distance covered = (1.10 + 0.9) km = 2 km = 2000 m.

Required time =

2000 x 3

sec = 48 sec. 125

12. A train 110 metres long is running with a speed of 60 kmph. In what time will it

pass a man who is running at 6 kmph in the direction opposite to that in which

the train is going?

Speed of train relative to man = (60 + 6) km/hr = 66 km/hr.

=

66 x 5

m/sec 18

=

55

m/sec. 3

Time taken to pass the man =

110 x 3

sec = 6 sec. 55


Each of these questions is followed by three statements. You have to study the

question and all the three statements given to decide whether any information

provided in the statement(s) is redundant and can be dispensed with while

answering the given question.

13. What is the length of a running train P crossing another running train Q?

I. These two trains take 18 seconds to cross each other.

II. These trains are running in opposite directions.

III. The length of the train Q is 180 metres.

A. I only

B. II only

C. III only

D. All I, II and III are required

E. Even with I, II and III, the answer cannot be obtained.

Let the length of the train P be x metres.

II. These trains are running in opposite directions.

III. Length of the train Q is 180 m.

I. Time taken by P to cross Q = (180 + x)

18 = (180 + x)

Relative speed Relative speed

Thus, even with I, II and III, the answer cannot be obtained.



In each of the following questions, a question is asked and is followed by three

statements. While answering the question, you may or may not require the data

provided in all the statements. You have to read the question and the three

statements and then decide whether the question can be answered with any one or

two of the statements or all the three statements are required to answer the

question. The answer number bearing the statements, which can be dispensed with,

if any, while answering the question is your answer.

14. What is the compound interest earned at the end of 3 years?

I. Simple interest earned on that amount at the same rate and for the same

period is Rs. 4500.

II. The rate of interest is 10 p.c.p.a.

III. Compound interest for 3 years is more than the simple interest for that

period by Rs. 465.

A. I and II only

B. II and III only

C. I and III only

D. Either II or III only

E. Any two of the three

I. gives, S.I for 3 years = Rs. 4500.

II. gives, Rate = 10% p.a.

III. gives, (C.I.) - (S.I.) = Rs. 465.

Clearly, using I and III we get C.I. = Rs. (465 + 4500).

Thus, II is redundant.

Also, from I and II, we get sum =

100 x 4500

= 15000. 10 x 3

Now C.I. on Rs. 15000 at 10% p.a. for 3 years may be obtained.

Thus, III is redundant.

Either II or III is redundant.

15. If log10 2 = 0.3010, the value of log10 80 is:

log10 = log10 (8 x 10)

= log10 8 + log10 10

= log10 (23 ) + 1

= 3 log10 2 + 1

= (3 x 0.3010) + 1

= 1.9030.






G



the question.

G



question.

G



G



G



16. The area of a rectangle is equal to the area of right-angles triangle. What is the

length of the rectangle?

I. The base of the triangle is 40 cm.

II. The height of the triangle is 50 cm.






Given: Area of rectangle = Area of a right-angles triangle.

l x b = 1 x B x H

2

I gives, B = 40 cm.

II gives, H = 50 cm.

Thus, to find l, we need b also, which is not given.

Given data is not sufficient to give the answer.


17. In a game of 100 points, A can give B 20 points and C 28 points. Then, B can give

C:

A : B = 100 : 80.

A : C = 100 : 72.

B =

B x A

=

80 x 100

= 10

= 100

= 100 : 90. C A C 100 72 9 90

B can give C 10 points.

18. In a 300 m race A beats B by 22.5 m or 6 seconds. B's time over the course is:

B runs 45

m in 6 sec. 2

B covers 300 m in

6 x 2

x 300 sec

= 80 sec. 45

19. The present worth of a sum due sometime hence is Rs. 576 and the banker's gain

is Rs. 16. The true discount is:

T.D. = P.W. x B.G. = 576 x 16 = 96.



20. 64, 71, 80, 91, 104, 119, 135, 155

Go on adding 7, 9, 11, 13, 15, 19 respectively to obtain the next number.

So, 135 is wrong.

1. The greatest possible length which can be used to measure exactly the lengths 7

m, 3 m 85 cm, 12 m 95 cm is:

Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm.

2. (0.1667)(0.8333)(0.3333) is approximately equal to:

(0.2222)(0.6667)(0.1250)

Given expression

= (0.3333)

x (0.1667)(0.8333)

(0.2222) (0.6667)(0.1250)

=

3333

x

1 x 5

6 6

2222 2

x 125

3 1000

=

3 x 1

x 5

x 3

x 8

2 6 6 2

=

5

2

= 2.50

3. If

144 =

14.4 , then the value of x is:

0.144 x

144 =

14.4

0.144 x

144 x 1000 =

14.4

44 x

x = 14.4

= 0.0144 1000

4. The average weight of 8 person's increases by 2.5 kg when a new person comes in

place of one of them weighing 65 kg. What might be the weight of the new person?

Total weight increased = (8 x 2.5) kg = 20 kg.

Weight of new person = (65 + 20) kg = 85 kg.





5. What is the two-digit number?

I. Sum of the digits is 7.

II. Difference between the number and the number obtained by interchanging

the digits is 9.

III. Digit in the ten's place is bigger than the digit in the unit's place by 1.

A. I and II only

B. II and III only

C. I and III only


E. None of these

Let the tens and units digit be x and y respectively.

I. x + y = 7.

II. (10x + y) - (10y + x) = 9 x - y = 1.

III. x - y = 1.

Thus, I and II as well as I and III give the answer.


6. A father said to his son, "I was as old as you are at the present at the time of your

birth". If the father's age is 38 years now, the son's age five years back was:

Let the son's present age be x years. Then, (38 - x) = x

2x = 38.

x = 19.

Son's age 5 years back (19 - 5) = 14 years.





7. What is Ravi's present age?

I. The present age of Ravi is half of that of his father.

II. After 5 years, the ratio of Ravi's age to that of his father's age will be 6 : 11.

III. Ravi is 5 years younger than his brother.

A. I and II only

B. II and III only

C. I and III only


E. Even with all the three statements answer cannot be determined.

I. Let Ravi's present age be x years. Then, his father's present age = 2x years.

II. After 5 years, Ravi's age

= 6

Father's age 11

III. Ravi is younger than his brother.

From I and II, we get x + 5

= 6

. This gives x, the answer. 2x + 5 11

Thus, I and II together give the answer. Clearly, III is redundant.

Correct answer is (A).

8. Two students appeared at an examination. One of them secured 9 marks more

than the other and his marks was 56% of the sum of their marks. The marks

obtained by them are:

Let their marks be (x + 9) and x.

Then, x + 9 = 56

(x + 9 + x) 100

25(x + 9) = 14(2x + 9)

3x = 99

x = 33

So, their marks are 42 and 33.

9. A man buys a cycle for Rs. 1400 and sells it at a loss of 15%. What is the selling

price of the cycle?

S.P. = 85% of Rs. 1400 = Rs.

85 x 1400

= Rs. 1190 100

10. 4 mat-weavers can weave 4 mats in 4 days. At the same rate, how many mats

would be woven by 8 mat-weavers in 8 days?

Let the required number of bottles be x.

More weavers, More mats (Direct Proportion)

More days, More mats (Direct Proportion) Wavers 4 : 8

:: 4 : x Days 4 : 8

4 x 4 x x = 8 x 8 x 4

x = (8 x 8 x 4)

(4 x 4)

x = 16.

11. A can contains a mixture of two liquids A and B is the ratio 7 : 5. When 9 litres of

mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7

: 9. How many litres of liquid A was contained by the can initially?

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left =

7x - 7

x 9

litres =

7x - 21

litres. 12 4

Quantity of B in mixture left =

5x - 5

x 9

litres =

5x - 15

litres. 12 4

7x - 21

4 =

7

5x - 15

+ 9 4

9

= 28x - 21

= 7

20x + 21 9

252x - 189 = 140x + 147

112x = 336

x = 3.

So, the can contained 21 litres of A.

12. A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a. in 5

years. What is the sum?

Principal

= Rs.

100 x 4016.25

9 x 5

= Rs.

401625

45

= Rs. 8925.

13. A person borrows Rs. 5000 for 2 years at 4% p.a. simple interest. He immediately

lends it to another person at 6 p.a for 2 years. Find his gain in the transaction

per year.

Gain in 2 years = Rs.

5000 x 25

x 2

-

5000 x 4 x 2

4 100 100

= Rs. (625 - 400)

= Rs. 225.

Gain in 1 year = Rs.

225

= Rs. 112.50 2

14. What is the total surface area of a right circular cone of height 14 cm and base

radius 7 cm?

h = 14 cm, r = 7 cm.

So, l = (7)2 + (14)2 = 245 = 75 cm. Total surface area = rl + r2

=

22 x 7 x 75 +

22 x 7 x 7

cm2 7 7

= [154(5 + 1)] cm2

= (154 x 3.236) cm2

= 498.35 cm2.

15. How much does a watch lose per day, if its hands coincide ever 64 minutes?

55 min. spaces are covered in 60 min.

60 min. spaces are covered in

60 x 60

min. = 65

5 min.

55 11

Loss in 64 min. =

65 5

- 64

= 16

min. 11 11

Loss in 24 hrs =

16 x 1

x 24 x 60 min.

= 32 8

min. 11 64 11

16. Rs. 20 is the true discount on Rs. 260 due after a certain time. What will be the

true discount on the same sum due after half of the former time, the rate of

interest being the same?

S.I. on Rs. (260 - 20) for a given time = Rs. 20.

S.I. on Rs. 240 for half the time = Rs. 10.

T.D. on Rs. 250 = Rs. 10.

T.D. on Rs. 260 = Rs.

10 x 260

= Rs. 10.40 250

17. The interest on Rs. 750 for 2 years is the same as the true discount on Rs. 960

due 2 years hence. If the rate of interest is the same in both cases, it is:

S.I. on Rs. 750 = T.D. on Rs. 960.

This means P.W. of Rs. 960 due 2 years hence is Rs. 750.

T.D. = Rs. (960 - 750) = Rs. 210.

Thus, S.I. on R.s 750 for 2 years is Rs. 210.

Rate =

100 x 210

% = 14%

750 x 2

18. The present worth of a certain bill due sometime hence is Rs. 800 and the true

discount is Rs. 36. The is:

B.G. = (T.D.)2

= Rs.

36 x 36

= Rs. 1.62 P.W. 800

B.D. = (T.D. + B.G.) = Rs. (36 + 1.62) = Rs. 37.62



19. 2, 5, 10, 17, 26, 37, 50, 64

(1*1)+1 , (2*2)+1 , (3*3)+1 , (4*4)+1 , (5*5)+1 , (6*6)+1 , (7*7)+1 , (8*8)+1

But, 64 is out of pattern.



20. 6, 13, 18, 25, 30, 37, 40

The differences between two successive terms from the beginning are 7, 5, 7, 5, 7,

5.

So, 40 is wrong.

1. The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and

28 is:

Required number = (L.C.M. of 12,16, 18, 21, 28) + 7

= 1008 + 7

= 1015

2. What should come in place of both x in the equation

x =

162 .

128 x

Let x

= 162

128 x

Then x2 = 128 x 162

= 64 x 2 x 18 x 9

= 82 x 62 x 32

= 8 x 6 x 3

= 144.

x = 144 = 12.

3. A group of students decided to collect as many paise from each member of group

as is the number of members. If the total collection amounts to Rs. 59.29, the

number of the member is the group is:

Money collected = (59.29 x 100) paise = 5929 paise.

Number of members = 5929 = 77.

4. In Arun's opinion, his weight is greater than 65 kg but less than 72 kg. His

brother doest not agree with Arun and he thinks that Arun's weight is greater

than 60 kg but less than 70 kg. His mother's view is that his weight cannot be

greater than 68 kg. If all are them are correct in their estimation, what is the

average of different probable weights of Arun?

Let Arun's weight by X kg.

According to Arun, 65 < X < 72

According to Arun's brother, 60 < X < 70.

According to Arun's mohter, X

2

8x + 16 = 5x + 40

3x = 24

x = 8.

Hence, required ratio = (4x + 16)

= 48

= 2. (x + 16) 24






G



the question.

G



question.

G



G



G



7. Two towns are connected by railway. Can you find the distance between them?

I. The speed of the mail train is 12 km/hr more than that of an express train.

II. A mail train takes 40 minutes less than an express train to cover the distance.






Let the distance between the two stations be x km.

I. Then, speed of the mail train = (y + 12) km/hr.

II. x - x

= 40

. y (y + 12) 60

Thus, even I and II together do not give x.


8. A train 360 m long is running at a speed of 45 km/hr. In what time will it pass a

bridge 140 m long?

Formula for converting from km/hr to m/s: X km/hr =

X x 5

m/s. 18

Therefore, Speed =

45 x 5

m/sec =

25 m/sec.

18 2

Total distance to be covered = (360 + 140) m = 500 m.

Formula for finding Time =

Distance

Speed

Required time =

500 x 2

sec = 40 sec.

25

9. A boat covers a certain distance downstream in 1 hour, while it comes back in 1

hours. If the speed of the stream be 3 kmph, what is the speed of the boat in still

water?

Let the speed of the boat in still water be x kmph. Then,

Speed downstream = (x + 3) kmph,

Speed upstream = (x - 3) kmph.

(x + 3) x 1 = (x - 3) x 3

2

2x + 6 = 3x - 9

x = 15 kmph. Direction (for Q.No. 10):





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10. What percentage of simple interest per annum did Anand pay to Deepak?

I. Anand borrowed Rs. 8000 from Deepak for four years.

II. Anand returned Rs. 8800 to Deepak at the end of two years and settled the

loan.






Let the rate be R% p.a.

I gives, P = Rs. 8000 and T = 4 years.

II gives, S.I. = Rs. (8800 - 8000) = Rs. 800.

R =

100 x S.I.

=

100 x 800

% = 2

1 % p.a.

P x T 8000 x 4 2

Thus, I and II both are needed to get the answer.


11. What is the difference between the compound interests on Rs. 5000 for 1 years

at 4% per annum compounded yearly and half-yearly?

C.I. when interest

compounded yearly

= Rs.

5000 x

1 + 4

x

1 + x 4

100 100

= Rs.

5000 x 26

x 51

25 50

= Rs. 5304.

C.I. when interest is

compounded half-yearly

= Rs.

5000 x

1 + 2

3

100

= Rs.

5000 x 51

x 51

x 51

50 50 50

= Rs. 5306.04

Difference = Rs. (5306.04 - 5304) = Rs. 2.04

12. Albert invested an amount of Rs. 8000 in a fixed deposit scheme for 2 years at

compound interest rate 5 p.c.p.a. How much amount will Albert get on maturity

of the fixed deposit?

Amount

= Rs.

8000 x

1 + 5

2

100

= Rs.

8000 x 21

x 21

20 20

= Rs. 8820.

13. If log 2 = 0.30103, the number of digits in 264 is:

log (264) = 64 x log 2

= (64 x 0.30103)

= 19.26592

Its characteristic is 19.

Hence, then number of digits in 264 is 20.

14. The value of log2 16 is:

Let log2 16 = n.

Then, 2n = 16 = 24 n = 4.

log2 16 = 4.





15. What is the capacity of the cylindrical tank?

I. The area of the base is 61,600 sq. cm.

II. The height of the tank is 1.5 times the radius.

III. The circumference of base is 880 cm.

A. Only I and II

B. Only II and III

C. Only I and III


E. Only II and either I or III

Capacity = r2h.

I gives, r2 = 61600. This gives r.

II gives, h = 1.5 r.

Thus, I and II give the answer.

Again, III gives 2 r = 880. This gives r.

So, II and III also give the answer.


16. A runs 1 times as fast as B. If A gives B a start of 80 m, how far must the

winning post be so that A and B might reach it at the same time?

Ratio of the speeds of A and B = 5

: 1 = 5 : 3. 3

Thus, in race of 5 m, A gains 2 m over B.

2 m are gained by A in a race of 5 m.

80 m will be gained by A in race of

5 x 80

m = 200 m.

2

Winning post is 200 m away from the starting point.

17. A watch which gains 5 seconds in 3 minutes was set right at 7 a.m. In the

afternoon of the same day, when the watch indicated quarter past 4 o'clock, the

true time is:

Time from 7 a.m. to 4.15 p.m. = 9 hrs 15 min. = 37

hrs. 4

3 min. 5 sec. of this clock = 3 min. of the correct clock.

37 hrs of this clock =

1 hrs of the correct clock.

720 20

37 hrs of this clock =

1 x 720

x 37

hrs of the correct clock. 4 20 37 4

= 9 hrs of the correct clock.

The correct time is 9 hrs after 7 a.m. i.e., 4 p.m.

18. A man wants to sell his scooter. There are two offers, one at Rs. 12,000 cash and

the other a credit of Rs. 12,880 to be paid after 8 months, money being at 18%

per annum. Which is the better offer?

P.W. of Rs. 12,880 due 8 months hence = Rs.

12880 x 100

100 +

18 x 8

12

= Rs.

12880 x 100

112

= Rs. 11500.



19. 8, 27, 64, 100, 125, 216, 343

The pattern is 23, 33, 43, 53, 63, 73. But, 100 is not a perfect cube.


Insert the missing number.

20. 165, 195, 255, 285, 345, (....)

Each number is 15 multiplied by a prime number i.e., 15 x 11, 15 x 13, 15 x 17, 15 x 19, 15 x 23.

So, the next number is 15 x 29 = 435.

1. 5 x 1.6 - 2 x 1.4 = ?

1.3

Given Expression = 8 - 2.8

= 5.2

= 52

= 4. 1.3 1.3 13

2. If a - b = 3 and a2 + b2 = 29, find the value of ab.

2ab = (a2 + b2) - (a - b)2

= 29 - 9 = 20

ab = 10.

3. A man has some hens and cows. If the number of heads be 48 and the number of

feet equals 140, then the number of hens will be:

Let the number of hens be x and the number of cows be y.

Then, x + y = 48 .... (i)

and 2x + 4y = 140 x + 2y = 70 .... (ii)

Solving (i) and (ii) we get: x = 26, y = 22.

The required answer = 26.

4. A number consists of two digits. If the digits interchange places and the new

number is added to the original number, then the resulting number will be

divisible by:

Let the ten's digit be x and unit's digit be y.

Then, number = 10x + y.

Number obtained by interchanging the digits = 10y + x.

(10x + y) + (10y + x) = 11(x + y), which is divisible by 11.

5. A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 :

3. If C gets Rs. 1000 more than D, what is B's share?

Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.

Then, 4x - 3x = 1000

x = 1000.

B's share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000.

6. Two number are in the ratio 3 : 5. If 9 is subtracted from each, the new numbers

are in the ratio 12 : 23. The smaller number is:

Let the numbers be 3x and 5x.

Then, 3x - 9

= 12

5x - 9 23

23(3x - 9) = 12(5x - 9)

9x = 99

x = 11.

The smaller number = (3 x 11) = 33.

7. 4 men and 6 women can complete a work in 8 days, while 3 men and 7 women can

complete it in 10 days. In how many days will 10 women complete it?

Let 1 man's 1 day's work = x and 1 woman's 1 day's work = y.

Then, 4x + 6y = 1

and 3x + 7y = 1

. 8 10

Solving the two equations, we get: x = 11

, y = 1

400 400

1 woman's 1 day's work = 1

. 400

10 women's 1 day's work =

1 x 10

= 1

. 400 40

Hence, 10 women will complete the work in 40 days.

8. A tank is filled by three pipes with uniform flow. The first two pipes operating

simultaneously fill the tank in the same time during which the tank is filled by

the third pipe alone. The second pipe fills the tank 5 hours faster than the first

pipe and 4 hours slower than the third pipe. The time required by the first pipe is:

Suppose, first pipe alone takes x hours to fill the tank .

Then, second and third pipes will take (x -5) and (x - 9) hours respectively to fill the tank.

1 +

1 =

1

x (x - 5) (x - 9)

x - 5 + x =

1

x(x - 5) (x - 9)

(2x - 5)(x - 9) = x(x - 5)

x2 - 18x + 45 = 0

(x - 15)(x - 3) = 0

x = 15. [neglecting x = 3]

9. Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2

hours, C is closed and A and B can fill the remaining part in 7 hours. The number

of hours taken by C alone to fill the tank is:

Part filled in 2 hours = 2

= 1

6 3

Remaining part =

1 - 1

= 2 .

3 3

(A + B)'s 7 hour's work = 2

3

(A + B)'s 1 hour's work = 2

21

C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }

=

1 - 2

= 1

6 21 14

C alone can fill the tank in 14 hours.

10. A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr.

The average speed for the first 320 km of the tour is:

Total time taken =

160 +

160

hrs. =

9 hrs.

64 80 2

Average speed =

320 x 2

km/hr = 71.11 km/hr.

9






11. At what time will the train reach city X from city Y?

I. The train crosses another train of equal length of 200 metres and running in

opposite directions in 15 seconds.

II. The train leaves city Y and 7.15 a.m. for city X situated at a distance of 558

km.

III. The 200 metres long train crosses a signal pole in 10 seconds.

A. I only

B. II only

C. III only

D. II and III only

E. All I, II and III are required.

From the statement I, we get length of the train is 200 metres (Redundant info

while comparing with Statement III). The rest of the info given in this statement

cannot be used for calculating the speed of the train, because the two trains

might run at different speed.

III gives, speed = 200

m/sec = 20 m/sec =

20 x 18

km/hr = 72 km/hr. 10 5

II gives, time taken =

558

hrs =

31 hrs = 7

3 hrs = 7 hrs 45 min.

72 4 4

So, the train will reach city X at 3 p.m.

Hence II and III only gives the answer.

12. The difference between the length and breadth of a rectangle is 23 m. If its

perimeter is 206 m, then its area is:

We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103.

Solving the two equations, we get: l = 63 and b = 40.

Area = (l x b) = (63 x 40) m2 = 2520 m2.






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13. The area of playground is 1600 m2. What is the perimeter?

I. It is a perfect square playground.

II. It costs Rs. 3200 to put a fence around the playground at the rate of Rs. 20

per metre.






Area = 1600 m2.

I. Side = 1600 m = 40 m. So, perimeter = (40 x 4) m = 160 m.

I alone gives the answer.

II. Perimeter = Total cost

= 3200

m = 160 m. Cost per metre 20

II alone gives the answer.






14. What is the area of the hall?

I. Material cost of flooring per square metre is Rs. 2.50

II. Labour cost of flooring the hall is Rs. 3500

III. Total cost of flooring the hall is Rs. 14,500.

A. I and II only

B. II and III only

C. All I, II and III


E. None of these

I. Material cost = Rs. 2.50 per m2

II. Labour cost = Rs. 3500.

III. Total cost = Rs. 14,500.

Let the area be A sq. metres.

Material cost = Rs. (14500 - 3500) = Rs. 11,000.

5A = 11000 A =

11000 x 2

= 4400 m2. 2 5

Thus, all I, II and III are needed to get the answer.


15. In a race of 200 m, A can beat B by 31 m and C by 18 m. In a race of 350 m, C

will beat B by:

A : B = 200 : 169.

A : C = 200 : 182.

C =

C x A

=

182 x 200

= 182 : 169. B A B 200 169

When C covers 182 m, B covers 169 m.

When C covers 350 m, B covers

169 x 350

m = 325 m.

182

Therefore, C beats B by (350 - 325) m = 25 m.

16. If the true discount on s sum due 2 years hence at 14% per annum be Rs. 168,

the sum due is:

P.W. = 100 x T.D.

= 100 x 168

= 600. R x T 14 x 2

Sum = (P.W. + T.D.) = Rs. (600 + 168) = Rs. 768.

17. An observer 1.6 m tall is 203 away from a tower. The angle of elevation from his

eye to the top of the tower is 30. The heights of the tower is:

Let AB be the observer and CD be the tower.

Draw BE CD.

Then, CE = AB = 1.6 m,

BE = AC = 203 m.

DE = tan 30 =

1

BE 3

DE = 203

m = 20 m. 3

CD = CE + DE = (1.6 + 20) m = 21.6 m.


18. 22, 33, 66, 99, 121, 279, 594

Each of the number except 279 is a multiple of 11.


19. 3, 7, 15, 39, 63, 127, 255, 511

Go on multiplying 2 and adding 1 to get the next number.

So, 39 is wrong.

20. 10, 26, 74, 218, 654, 1946, 5834

2nd term = (1st term) x 3 - 4 = 10 x 3 - 4 = 26.

3rd term = (2nd term) x 3 - 4 = 26 x 3 - 4 = 74.

4th term = (3th term) x 3 - 4 = 74 x 3 - 4 = 218.

5th term = (4th term) x 3 - 4 = 218 x 3 - 4 = 650.

5th term must be 650 instead of 654.

1. One-third of Rahul's savings in National Savings Certificate is equal to one-half of

his savings in Public Provident Fund. If he has Rs. 1,50,000 as total savings, how

much has he saved in Public Provident Fund ?

Let savings in N.S.C and P.P.F. be Rs. x and Rs. (150000 - x) respectively. Then, 1

x = 1

(150000 - x) 3 2

x +

x = 75000

3 2

5x = 75000

6

x = 75000 x 6

= 90000 5

Savings in Public Provident Fund = Rs. (150000 - 90000) = Rs. 60000

2. 1.5625 = ?

1|1.5625( 1.25 |1

|-------

22| 56

| 44

|-------

245| 1225

| 1225

|-------

| X

|-------

1.5625 = 1.25.

3. The square root of 64009 is:

2|64009( 253 |4

|----------

45|240

|225

|----------

503| 1509

| 1509

|----------

| X

|----------

64009 = 253.






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4. What is the average age of children in the class?

I. The age of the teacher is as many years as the number of children.

II. Average age is increased by 1 year if the teacher's age is also included.






Let there be x children.

I gives, age of teacher = x years.

II gives, average age of (x + 1) persons = (x + 1) years.

Teacher's age = (x + 1) (x + 1) - x2 = (x2 + 1 + 2x) - x2 = (1 + 2x)

Thus, teacher's age cannot be obtained.

Correct answer is (D)

5. The sum of the digits of a two-digit number is 15 and the difference between the

digits is 3. What is the two-digit number?

Let the ten's digit be x and unit's digit be y.

Then, x + y = 15 and x - y = 3 or y - x = 3.

Solving x + y = 15 and x - y = 3, we get: x = 9, y = 6.

Solving x + y = 15 and y - x = 3, we get: x = 6, y = 9.

So, the number is either 96 or 69.

Hence, the number cannot be determined.

6. Find a positive number which when increased by 17 is equal to 60 times the

reciprocal of the number.


Then, x + 17 = 60

x

x2 + 17x - 60 = 0

(x + 20)(x - 3) = 0

x = 3.






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7. What is the two-digit number whose first digit is a and the second digit is b?. The number is greater than 9.

I. The number is multiple of 51.

II. The sum of the digits a and b is 6.






From statement I:

A two digit number, greater than 9 and multiple of 51 should be 51 itself.

Because, 2 x 51 = 102 (3 digit number). Therefore, I alone sufficient to answer.

From statement II:

A two digit number, greater than 9 and sum of the digit is 6.

It can be 15, 24, 33, 42, 51. So we cannot determine the required answer from the

statement II alone.

Thus, I alone give the answer while II alone not sufficient to answer.






8. What will be the ratio between ages of Sam and Albert after 5 years?

I. Sam's present age is more than Albert's present age by 4 years.

II. Albert's present age is 20 years.

III. The ratio of Albert's present age to Sam's present age is 5 : 6.

A. I or II or III only

B. II only

C. III only

D. I or III only

E. II or III only

Clearly, any two of the given statements will give the answer and in each case,

the third is redundant.

Correct answer is (A).

9. In a certain school, 20% of students are below 8 years of age. The number of

students above 8 years of age is of the number of students of 8 years age which

is 48. What is the total number of students in the school?

Let the number of students be x. Then,

Number of students above 8 years of age = (100 - 20)% of x = 80% of x.

80% of x = 48 + 2

of 48 3

80 x = 80

100

x = 100.

10. When a plot is sold for Rs. 18,700, the owner loses 15%. At what price must that

plot be sold in order to gain 15%?

85 : 18700 = 115 : x

x =

18700 x 115

= 25300. 85

Hence, S.P. = Rs. 25,300.

11. A man completes of a job in 10 days. At this rate, how many more days will it

takes him to finish the job?

Work done = 5

8

Balance work =

1 - 5

= 3

8 8

Let the required number of days be x.

Then, 5

: 3

= :: 10 : x 5

x x = 3

x 10 8 8 8 8

x =

3 x 10 x

8

8 5

x = 6.

12. A is thrice as good as workman as B and therefore is able to finish a job in 60

days less than B. Working together, they can do it in:

Ratio of times taken by A and B = 1 : 3.

The time difference is (3 - 1) 2 days while B take 3 days and A takes 1 day.

If difference of time is 2 days, B takes 3 days.

If difference of time is 60 days, B takes

3 x 60

= 90 days. 2

So, A takes 30 days to do the work.

A's 1 day's work = 1

30

B's 1 day's work = 1

90

(A + B)'s 1 day's work =

1 +

1

= 4

= 2

30 90 90 45

A and B together can do the work in 45

= 22 1

days. 2 2

13. One pipe can fill a tank three times as fast as another pipe. If together the two

pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to

fill the tank in:

Let the slower pipe alone fill the tank in x minutes.

Then, faster pipe will fill it in x

minutes. 3

1 +

3 =

1

x x 36

4 =

1

x 36

x = 144 min.

14. Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is

45 kmph. For how many minutes does the bus stop per hour?

Due to stoppages, it covers 9 km less.

Time taken to cover 9 km

9 x 60

min = 10 min.

54

15. The ratio between the speeds of two trains is 7 : 8. If the second train runs 400

kms in 4 hours, then the speed of the first train is:

A. 70 km/hr

B. 75 km/hr

C. 84 km/hr

D. 87.5 km/hr

Let the speed of two trains be 7x and 8x km/hr. Then, 8x =

400

= 100

4

x =

100

= 12.5 8

Speed of first train = (7 x 12.5) km/hr = 87.5 km/hr.






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16. A boat takes a total time of three hours to travel downstream from P and Q and

upstream back from Q to P. What is the speed of the boat in still water?

I. The speed of the river current is 1 km per hour.

II. The distance between P and Q is 4 km.






I. Speed of the current = 1 km/hr.

II. PQ = 4 km.

Let the speed of the boat in still water be x km/hr. Then, 4

+ 4

= 3. This gives x. (x + 1) (x - 1)

Co

aptitude

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