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Free Online Aptitude Test: Placement

General Instructions:

1. Each Aptitude test contains 15 questions.
2. Total allotted time is 25 minutes for each test.
3. No negative marking.
4. Try to finish it on time.
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Aptitude Test 5

1 / 15

1. The difference between two parallel sides of a trapezium is 4 cm. perpendicular distance between them is 19 cm. If the area of the trapezium is 475 find the lengths of the parallel sides.

A. 27 cm and 23 cm

B. 26 cm and 24 cm

C. 29 cm and 21 cm

D. 25 cm and 25 cm

Solution:

Let the two parallel sides of the trapezium be a em and b em.

Then, a – b = 4

And, (1/2) x (a + b) x 19 = 475 ó (a + b) =((475 x 2)/19) ó a + b = 50

Solving (i) and (ii), we get: a = 27, b = 23.

So, the two parallel sides are 27 cm and 23 cm

2 / 15

2. 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:

A. 4

B. 5

C. 6

D. 8

Solution:

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

3 / 15

3. The banker’s gain on a bill due 1 year hence at 12% per annum is Rs. 6. The true discount is:

A. Rs. 72

B. Rs. 36

C. Rs. 54

D. Rs. 50

Solution:

T.D. =	B.G. x 100	= Rs.		6 x 100		= Rs. 50.
	R x T			12 x 1

4 / 15

4. A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:

A. 45 km/hr

B. 50 km/hr

C. 54 km/hr

D. 55 km/hr

Solution:

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 km/hr. Then, relative speed = (x – 5) km/hr.

x – 5 = 45 x = 50 km/hr.

5 / 15

5. A batsman in his 18th innings makes a score of 150 runs and thereby increasing his Average by 6. Find his average after 18th innings.

A. 45

B. 42

C. 39

D. 40

Solution:

Let the average for 17 innings is x runs

Total runs in 17 innings = 17x

Total runs in 18 innings = 17x + 150

Average of 18 innings = 17x + 150/18

: . 17x + 150/18 = x + 6 — > x = 42

Thus, average after 18 innings = 42

6 / 15

6. Which of the following statements is not correct?

A. log10 10 = 1

B. log (2 + 3) = log (2 x 3)

C. log10 1 = 0

D. log (1 + 2 + 3) = log 1 + log 2 + log 3

Solution:

(a) Since log_a a = 1, so log₁₀ 10 = 1.

(b) log (2 + 3) = log 5 and log (2 x 3) = log 6 = log 2 + log 3

log (2 + 3) log (2 x 3)

(d) log (1 + 2 + 3) = log 6 = log (1 x 2 x 3) = log 1 + log 2 + log 3.

So, (b) is incorrect.

7 / 15

7. Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?

A. Rs. 6400

B. Rs. 6500

C. Rs. 7200

D. Rs. 7500

Solution:

Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 – x).

Then,		x x 14 x 2		+		(13900 – x) x 11 x 2		= 3508
		100				100

28x – 22x = 350800 – (13900 x 22)

6x = 45000

x = 7500.

So, sum invested in Scheme B = Rs. (13900 – 7500) = Rs. 6400.

8 / 15

8. The price of commodity X increases by 40 paise every year, while the price of commodity Y increases by 15 paise every year. If in 2001, the price of commodity X was Rs. 4.20 and that of Y was Rs. 6.30, in which year commodity X will cost 40 paise more than the commodity Y ?

A. 2010

B. 2011

C. 2012

D. 2013

Solution:

Suppose commodity X will cost 40 paise more than Y after z years.

Then, (4.20 + 0.40z) – (6.30 + 0.15z) = 0.40

0.25z = 0.40 + 2.10

z =	2.50	=	250	= 10.
	0.25		25

X will cost 40 paise more than Y 10 years after 2001 i.e., 2011.

9 / 15

9. If 2x+3y+z=55,x-y=4 and y – x + z=12,then what are the values of x , y and z?

A. x=12,y=7,z=8

B. x=8,y=11,z=7

C. x=11,y=7,z=8

D. x=7,y=11,z=8

Solution:

The given equations are:

2x+3y+z=55 …(i); x + z – y=4 …(ii); y -x + z =12 …(iii)

Subtracting (ii) from (i), we get: x+4y=51 …(iv)

Subtracting (iii) from (i), we get: 3x+2y=43 …(v)

Multiplying (v) by 2 and subtracting (iv) from it, we get: 5x=35 or x=7.

Putting x=7 in (iv), we get: 4y=44 or y=11.

Putting x=7,y=11 in (i), we get: z=8

10 / 15

10. P is able to do a piece of work in 15 days and Q can do the same work in 20 days. If they can work together for 4 days, what is the fraction of work left?

A. 7/15

B. 8/15

C. 11/15

D. 2/11

Solution:

Amount of work P can do in 1 day = 1/15

Amount of work Q can do in 1 day = 1/20

Amount of work P and Q can do in 1 day = 1/15 + 1/20 = 7/60

Amount of work P and Q can together do in 4 days = 4 × (7/60) = 7/15

Fraction of work left = 1 – 7/15= 8/15

11 / 15

11. Find the sum of all 2 digit numbers divisible by 3

A. 1665

B. 1666

C. 1668

D. 1669

Solution:

All 2 digit numbers divisible by 3 are :

12, 51, 18, 21, …, 99.

This is an A.P. with a = 12 and d = 3.

Let it contain n terms. Then,

12 + (n – 1) x 3 = 99 or n = 30.

\Required sum = 30/2 x (12+99) = 1665.

12 / 15

12. The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:

A. 1677

B. 1683

C. 2523

D. 3363

Solution:

L.C.M. of 5, 6, 7, 8 = 840.

Required number is of the form 840k + 3

Least value of k for which (840k + 3) is divisible by 9 is k = 2.

Required number = (840 x 2 + 3) = 1683.

13 / 15

13. A train 240 m long passes a pole in 24 seconds. How long will it take to pass a platform 650 m long?

A. 65 sec

B. 89 sec

C. 100 sec

D. 150 sec

Solution:

Speed =		240	m/sec = 10 m/sec.
		24

Required time =

240 + 650

sec = 89 sec.

14 / 15

14. After getting 2 successive discounts, a shirt with a list price of Rs 150 is available at Rs 105. If the second discount is 12.55, find the first discount.

A. 24%

B. 25%

C. 20%

D. 22%

Solution:

Let the first discount be x%

Then,87.5% of (100-x)% of 150= 105

87.5/100*(100-x)/100*450=150=>105=>100-x=(105*100*100)/(150*87.5)=80

x=(100-80)=20

first discount = 20%

15 / 15

15. In a caravan, in addition to 50 hens there are 45 goats and 8 camels with some keepers. If the total number of feet be 224 more than the number of heads, find the number of keepers

A. 17

B. 16

C. 14

D. 15

Solution:

Let the number of keepers be x then,

Total number of heads =(50 + 45 + 8 + x)= (103 + x).

Total number of feet = (45 + 8) x 4 + (50 + x) x 2 =(312 +2x).

(312 + 2x)-(103 + x) =224ó x =15.

Hence, number of keepers =15.

Your score is

The average score is 38%

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Aptitude Test 4

1 / 15

1. A 270 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?

A. 230 m

B. 240 m

C. 260 m

D. 320 m

E. None of these

Solution:

Relative speed = (120 + 80) km/hr

=		200 x	5	m/sec
			18

=		500	m/sec.
		9

Let the length of the other train be x metres.

Then,	x + 270	=	500
	9		9

x + 270 = 500

x = 230.

2 / 15

2. f 3^(x – y) = 27 and 3^(x + y) = 243, then x is equal to

A. 0

B. 2

C. 4

D. 6

Solution:

3^x^– y = 27 = 3³ x – y = 3 ….(i)

3^x^+ y = 243 = 3⁵ x + y = 5 ….(ii)

On solving (i) and (ii), we get x = 4.

3 / 15

3. In an examination , 80% of the students passed in English , 85% in Mathematics and 75% in both English and Mathematics. If 40 students failed in both the subjects , find the total number of students

A. 400

B. 500

C. 450

D. 420

Solution:

Let the total number of students be x .

Let A and B represent the sets of students who passed in English and Mathematics respectively

Then, number of students passed in one or both the subjects

= n (AÈB) =n (A)+n(B)- n(AÇB)=80% of x + 85% of x –75% of x

=[(80/100)x+(85/100)x-(75/100)x]=(90/100)x=(9/10)x

Students who failed in both the subjects = [x-(9x/10)]=x/10.

So, x/10=40 of x=400.

Hence, total number of students = 400

4 / 15

4. A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is:

A. 10%

B. 12%

C. 20%

D. 28%

Solution:

Let original length = x and original breadth = y.

Decrease in area

= xy –		80	x	x	90	y
		100			100

=		xy –	18	xy
			25

=	7	xy.
	25

Decrease % =

xy x

x 100

= 28%.

5 / 15

5. Two trains 100 meter’s and 120 meter’s long are running in the same direction with speeds of 72 km/hr ,In how much time will the first train cross the second?

A. 40 sec

B. 44 sec

C. 42 sec

D. 41 sec

Solution:

Relative speed of the trains = (72 – 54) km/hr = 18 km/hr

= (18 * 5/18) m/sec = 5 m/sec.

Time taken by the trains to cross each other

= Time taken to cover (100 + 120) m at 5 m /sec = (220/5) sec = 44 sec.

6 / 15

6. In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

A. 1/3

B. 3/4

C. 7/19

D. 8/21

Solution:

Your Answer: Option (Not Answered)

Correct Answer: Option A

Explanation:

Total number of balls = (8 + 7 + 6) = 21.

Let E	= event that the ball drawn is neither red nor green
	= event that the ball drawn is blue.

n(E) = 7.

P(E) =	n(E)	=	7	=	1	.
	n(S)		21		3

7 / 15

7. 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?

A. 76 kg

B. 76.5 kg

C. 85 kg

D. None of these

Solution:

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

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

8 / 15

8. Of the three numbers, second is twice the first and also thrice the third. If the average of the three numbers is 44, the largest number is

A. 24

B. 36

C. 72

D. 108

Solution:

Let the 3rd number be x, Then, second number = 3x

: . First number = 3x/2

: . x+3x + 3x/2 = (44*3)

— > 11x/2 = 132 — > x = 24

Largest number — > 3x

— > 3*24 = 72

9 / 15

9. If Rs. 10 be allowed as true discount on a bill of Rs. 110 due at the end of a certain time, then the discount allowed on the same sum due at the end of double the time is:

A. 20

B. 21.81

C. 22

D. 18.33

Solution:

S.I. on Rs. (110 – 10) for a certain time = Rs. 10.

S.I. on Rs. 100 for double the time = Rs. 20.

T.D. on Rs. 120 = Rs. (120 – 100) = Rs. 20.

T.D. on Rs. 110 = Rs.		20	x 110		= Rs. 18.33
		120

10 / 15

10. Find the odd man out.
	41, 43, 47, 53, 61, 71, 73, 81

A. 61

B. 71

C. 73

D. 81

Solution:

Each of the numbers except 81 is a prime number

11 / 15

11. A man is standing on a railway bridge which is 180 m long. He finds that a train crosses the bridge in 20 seconds but himself in 8 seconds. Find the length of the train and its speed?

A. 52 km

B. 53 km

C. 54 km

D. 55 km

Solution:

Let the length of the train be x meters,

Then, the train covers x meters in 8 seconds and (x + 180) meters in 20 sec

x/8=(x+180)/20 ó 20x = 8 (x + 180) <=> x = 120.

Length of the train = 120 m.

Speed of the train = (120/8) m / sec = m / sec = (15 *18/5) kmph = 54 km

12 / 15

12. A wheel that has 6 cogs is meshed with a larger wheel of 14 cogs. When the smaller wheel has made 21 revolutions, then the number of revolutions mad by the larger wheel is:

A. 4

B. 9

C. 12

D. 49

Solution:

Let the required number of revolutions made by larger wheel be x.

Then, More cogs, Less revolutions (Indirect Proportion)

14 : 6 :: 21 : x 14 x x = 6 x 21

x =	6 x 21
	14

x = 9.

13 / 15

13. If a man walks at the rate of 5 kmph, he misses a train by 7 minutes. However, if he walks at the rate of 6 kmph, he reaches the station 5 minutes before the arrival of the train. Find the distance covered by him to reach the station.

A. 8 km

B. 7 km

C. 6 km

D. 9 km

Solution:

Let the required distance be x km

Difference in the time taken at two speeds=1 min =1/2 hr

Hence x/5-x/6=1/5<=>6x-5x=6

x=6

Hence, the required distance is 6 km

14 / 15

14. If a quarter kg of potato costs 60 paise, how many paise will 200 gm cost?

A. 48 paise

B. 54 paise

C. 56 paise

D. 72 paise

Solution:

Let the required weight be x kg.

Less weight, Less cost (Direct Proportion)

250 : 200 :: 60 : x 250 x x = (200 x 60)

x=	(200 x 60)
	250

x = 48.

15 / 15

15. A room is half as long again as it is broad. The cost of carpeting the at Rs. 5 per sq. m is Rs. 270 and the cost of papering the four walls at Rs. 10 per m² is Rs. 1720. If a door and 2 windows occupy 8 sq. m, find the dimensions of the room

A. 8 m

B. 9 m

C. 6 m

D. 5 m

Solution:

Let breadth = x metres, length = 3x metres, height = H metres.

Area of the floor=(Total cost of carpeting)/(Rate/m²)=(270/5)m²=54m².

x* (3x/2) = 54 <=> x^2 = (54*2/3) = 36 <=> x = 6.

So, breadth = 6 m and length =(3/2)*6 = 9 m.

Now, papered area = (1720/10)m² = 172 m².

Area of 1 door and 2 windows = 8 m².

Total area of 4 walls = (172 + 8) m² = 180 m²

2*(9+ 6)* H = 180 <=> H = 180/30 = 6 m.

Your score is

The average score is 45%

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Aptitude Test 3

1 / 15

1. If A`s salary is 20% less then B`s salary , by how much percent is B`s salary more than A`s ?

A. 25%

B. 26%

C. 30%

D. 27%

Solution:

Required percentage = [(20*100)/(100-20)]%=25%.

2 / 15

2. A sum of Rs. 12,500 amounts to Rs. 15,500 in 4 years at the rate of simple interest. What is the rate of interest?

A. 3%

B. 4%

C. 5%

D. 6%

Solution:

S.I. = Rs. (15500 – 12500) = Rs. 3000.

Rate =		100 x 3000	%	= 6%
		12500 x 4

3 / 15

3. A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

A. 2.91 m

B. 3 m

C. 5.82 m

D. None of these

Solution:

Area of the park = (60 x 40) m² = 2400 m².

Area of the lawn = 2109 m².

Area of the crossroads = (2400 – 2109) m² = 291 m².

Let the width of the road be x metres. Then,

60x + 40x – x² = 291

x² – 100x + 291 = 0

(x – 97)(x – 3) = 0

x = 3.

4 / 15

4. The sum of two numbers is 16 and the sum of their squares is 113. Find the numbers

A. 5 and 6

B. 6 and 7

C. 7 and 8

D. 8 and 9

Solution:

Let the numbers be x and (15 – x).

Then, x² + (15 – x)² = 113 => x² + 225 + x² – 30x = 113

=> 2x² – 30x + 112 = 0 => x² – 15x + 56 = 0

=> (x – 7) (x – 8) = 0 => x = 7 or x = 8.

So, the numbers are 7 and 8.

5 / 15

5. Two unbiased coin are tossed .what is the probability of getting at most one head?

A. 3/5

B. 3/4

C. 3/6

D. 3/7

Solution:

S={HH,HT,TH,TT}

Let EE=event of getting one head

E={tt,ht,th}

P(e)=n(e)/n(s)=3/4

6 / 15

6. Present ages of Sameer and Anand are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Anand’s present age in years?

A. 24

B. 27

C. 40

D. Cannot be determined

Solution:

Let the present ages of Sameer and Anand be 5x years and 4x years respectively.

Then,	5x + 3	=	11
	4x + 3		9

9(5x + 3) = 11(4x + 3)

45x + 27 = 44x + 33

45x – 44x = 33 – 27

x = 6.

Anand’s present age = 4x = 24 years.

7 / 15

7. 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?

A. 5 : 7 : 8

B. 20 : 49 : 64

C. 38 : 28 : 21

D. None of these

Solution:

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 :

= 20 : 49 : 64.

8 / 15

8. A train starts full of passengers at the first station it drops 1/3 of the passengers and takes 280 more at the second station it drops one half the new total and takes twelve more .on arriving at the third station it is found to have 248 passengers. Find the no of passengers in the beginning?

A. 282

B. 285

C. 280

D. 288

Solution:

Let no of passengers in the beginning be x

After first station no passengers=(x-x/3)+280=2x/3 +280

After second station no passengers =1/2(2x/3+280)+12

½(2x/3+280)+12=248

2x/3+280=2*236

2x/3=192

x=288

9 / 15

9. Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats?

A. 2 : 3 : 4

B. 6 : 7 : 8

C. 6 : 8 : 9

D. None of these

Solution:

Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.

Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).

		140	x 5x		,		150	x 7x		and		175	x 8x
		100					100					100

7x,	21x	and 14x.
	2

The required ratio = 7x :	21x	: 14x
	2

14x : 21x : 28x

2 : 3 : 4.

10 / 15

10. A Cone and a sphere have equal radii and equal volumes. Find the ratio of the sphere of the diameter of the sphere to the height of the cone

A. 1:3

B. 2:4

C. 1:2

D. 2:8

Solution:

Let radius of each be R and height of the cone be H.

Then, (4/3) ∏ R³= (1/3) ∏ R²H (or) R/H = ¼ (or) 2R/H = 2/4 =1/2

\Required ratio = 1:2.

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11. A can do a certain job in 12 days. B is 60% more efficient than A. How many days does B alone take to do the same job?

A. 6 1/2 days

B. 7 1/2 days

C. 7 days

D. 8 1/2 days

Solution:

Ratio of times taken by A and B = 160 : 100 = 8 : 5.

Suppose B alone takes x days to do the job.

Then, 8 : 5 :: 12 : x = 8x = 5 x 12 =x = 7 1/2 days

12 / 15

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

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

A. 25

B. 30

C. 37

D. 40

Solution:

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

So, 40 is wrong.

13 / 15

13. Mr. Jones gave 40% of the money he had to his wife. He also gave 20% of the remaining amount to his 3 sons. Half of the amount now left was spent on miscellaneous items and the remaining amount of Rs.12000 was deposited in the bank. How much money did Mr. jones have initially?

A. 150000

B. 110000

C. 100000

D. 99990

Solution:

Let the initial amount with Mr.Jones be Rs.x

Then,(1/2)[100-(3*20)]% of x=12000

ó (1/2)*(40/100)*(60/100)*x=12000

óx=((12000*25)/3)=100000

14 / 15

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

A. 20

B. 22

C. 24

D. 48

Solution:

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.

15 / 15

15. 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?

A. 69

B. 78

C. 96

D. Cannot be determined

Solution:

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.

Your score is

The average score is 48%

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Aptitude Test 2

1 / 15

1. The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

A. 74

B. 94

C. 184

D. 364

Solution:

L.C.M. of 6, 9, 15 and 18 is 90.

Let required number be 90k + 4, which is multiple of 7.

Least value of k for which (90k + 4) is divisible by 7 is k = 4.

Required number = (90 x 4) + 4 = 364.

2 / 15

2. Rajeev’s age after 15 years will be 5 times his age 5 years back. What is the Present age of Rajeev?

A. 11 years

B. 12 years

C. 10 years

D. 15 years

Solution:

Let Rajeev’s present age be x years. Then,

Rajeev’s age after 15 years = (x + 15) years.

Rajeev’s age 5 years back = (x – 5) years.

:. X + 15 = 5 (x – 5) óx + 15 = 5x – 25 ó 4x = 40 ó x = 10.

hence, Rajeev’s present age = 10 years.

3 / 15

3. A cyclist covers a distance of 750 m in 2 min 30 sec. What is the speed in km/hr. of the cyclist?

A. 12km/hr

B. 16km/hr

C. 17km/hr

D. 18km/hr

Solution:

Speed = {750/150} m/sec =5 m/sec = { 5 * 18 } km/hr =18km/hr

4 / 15

4. The average age of a class of 39 students is 15 years. If the age of the teacher be included, then the average increases by3 months. Find the age of the teacher.

A. 35 years

B. 28 years

C. 25 years

D. 30 years

Solution :

Total age of 39 persons = (39 x 15) years = 585 years.

Average age of 40 persons= 15 yrs 3 months = 61/4 years.

Total age of 40 persons = (_(61/4 )x 40) years= 610 years.

:. Age of the teacher = (610 – 585) years=25 years.

5 / 15

5. Find the unit’s digit in (264)¹⁰² + (264)¹⁰³

A. 0

B. 1

C. 2

D. 3

Solution:

Required unit’s digit = unit’s digit in (4)¹⁰² + (4)^103.

now, 4² gives unit digit 6.

(4)¹⁰²gives unjt digit 6.

(4)103 gives unit digit of the product (6 x 4) i.e., 4.

hence, unit’s digit in (264)m + (264)103 = unit’s digit in (6 + 4) = 0

6 / 15

6. 8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of water is 16 : 65. How much wine did the cask hold originally?

A. 18 liters

B. 24 liters

C. 32 liters

D. 42 liters

Solution:

Let the quantity of the wine in the cask originally be x liters.

Then, quantity of wine left in cask after 4 operations =		x		1 –	8		4	liters.
					x

		x(1 – (8/x))⁴		=	16
		x			81

		1 –	8		4	=		2		4
			x					3

		x – 8		=	2
		x			3

3x – 24 = 2x

x = 24.

7 / 15

7. If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:

A. 55/601

B. 601/55

C. 11/120

D. 120/11

Solution:

Let the numbers be a and b.

Then, a + b = 55 and ab = 5 x 120 = 600.

The required sum =	1	+	1	=	a + b	=	55	=	11
	a		b		ab		600		120

8 / 15

8. In an election between two candidates, 75% of the voters cast their votes, out of which 2% of the votes were declared invalid. A candidate got 9261 votes which were 75% of the total valid votes. Find the total number of votes enrolled in that election.

A. 16800

B. 15800

C. 16600

D. 15800

Solution:

Let the number of votes enrolled be x. Then,

Number of votes cast =75% of x. Valid votes = 98% of (75% of x).

75% of (98% of (75%of x)) =9261.

[(75/100)*(98/100)*(75/100)*x] =9261.

X = [(9261*100*100*100)/(75*98*75)] =16800

9 / 15

9. A train travelling at a speed of 75 mph enters a tunnel 3 miles long. The train is mile long. How long does it take for the train to pass through the tunnel from the moment the front enters to the moment the rear emerges?

A. 2.5 min

B. 3 min

C. 3.2 min

D. 3.5 min

Solution:

Total distance covered

=		7	+	1		miles
		2		4

=	15	miles.
	4

Time taken

=		15		hrs
		4 x 75

=	1	hrs
	20

=		1	x 60		min.
		20

= 3 min.

10 / 15

10. A crate of mangoes contains one bruised mango for every thirty mango in the crate. If three out of every four bruised mango are considerably unsaleable and there are 12 unsaleable mangoes in the crate then how many mango are there in the crate?

A. 450

B. 480

C. 500

D. 490

Solution:

Let the total no of mangoes in the crate be x

Then the no of bruised mango = 1/30 x

Let the no of unsalable mangoes =3/4 (1/30 x)

1/40 x =12

x=480

11 / 15

11. The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team?

A. 23 years

B. 24 years

C. 25 years

D. None of these

Solution:

Let the average age of the whole team by x years.

11x – (26 + 29) = 9(x -1)

11x – 9x = 46

2x = 46

x = 23.

So, average age of the team is 23 years.

12 / 15

12. 66 cubic centimeters of silver is drawn into a wire 1 mm in diameter. The length of the wire in metres will be:

A. 84

B. 90

C. 168

D. 336

Solution:

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

13 / 15

13. The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

A. 1

B. 2

C. 3

D. 4

Solution:

Let the numbers 13a and 13b.

Then, 13a x 13b = 2028

ab = 12.

Now, the co-primes with product 12 are (1, 12) and (3, 4).

[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]

So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).

Clearly, there are 2 such pairs.

14 / 15

14. Excluding stoppages, the speed of the bus is 54kmph and including stoppages, it is 45kmph.for how many min does the bus stop per hr.

A. 10 min

B. 12 min

C. 11 min

D. 15 min

Solution:

Due to stoppages, it covers 9km less.

time taken to cover 9 km is [9/54 *60] min = 10min

15 / 15

15. A train 150 m long is running with a speed of 68 kmph. In what time will it pass a man who is running at 8 kmph in the same direction in which the train is going?

A. 7 sec

B. 8 sec

C. 9 sec

D. 10 sec

Solution:

Speed of the train relative to man = (68 – 8) kmph

= (60* 5/18) m/sec = (50/3)m/sec

Time taken by the train to cross the man I

= Time taken by It to cover 150 m at 50/3 m / sec = 150 *3/ 50 sec = 9sec

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Aptitude Test 1

1 / 15

1. If 12 man can do a piece of work in 36 days. In how many days 18 men can do the same work?

A. 22 days

B. 26 days

C. 36 days

D. 24 days

Solution:

12 men can do a work in 36 days.

18 men can do the work in (12/18 ) x 36 = 24 days.

2 / 15

2. A batsman makes a score of 87 runs in the 17th inning and thus increases his average by 3. Find his average after 17th innings.

A. 30

B. 39

C. 48

D. 87

Solution:

Let the average after 17th inning = x.

Then, average after 16th inning = (x – 3).

==> 16 (x – 3) + 87 = 17x or x = (87 – 48) = 39.

3 / 15

3. The ratio between the speeds of two trains is 7: 8. If the second train runs 440 kms in 4 hours, then the speed of the first train is:

A. 96.25 km/hr

B. 98.25 km/hr

C. 90.75 km/hr

D. 96.50 km/hr

Solution:

Let the speed of two trains be 7x and 8x km/hr.
Then, 8x = (440/4) = 110
x = (110/8) = 13.75
Hence, speed of first train = (7 x 13.75) km/hr = 96.25 km/hr

4 / 15

At present, the ratio between the ages of Arun and Deepak is 4 : 3. After 6 years, Arun’s age will be 26 years. What is the age of Deepak at present ?

A. 12 years

B. 15 years

C. 19 and half

D. 21 years

Solution:

Let the present ages of Arun and Deepak be 4x years and 3x years respectively. Then,

4x + 6 = 26 4x = 20

x = 5.

Deepak’s age = 3x = 15 years.

5 / 15

5. A pupil’s marks were wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by half (1/2). The number of pupils in the class is:

A. 10

B. 20

C. 40

D. 73

Solution:

Let there be x pupils in the class.

Total increase in marks =		x x	1		=	x
			2			2

	x	= (83 – 63)	x	= 20 x= 40.
	2		2

6 / 15

6. A man sitting in a train which is traveling at 50 kmph observes that a goods train, traveling in opposite direction, takes 9 seconds to pass him. If the goods train is 280 m long, find its speed?

A. 62 kmph

B. 72 kmph

C. 64 kmph

D. 56 kmph

Solution:

Relative speed = 280/9 m / sec = ((280/9)*(18/5)) kmph = 112 kmph.

Speed of goods train = (112 – 50) kmph = 62 kmph.

7 / 15

7. 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

Solution:

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

8 / 15

8. 3 pumps, working 8 hours a day, can empty a tank in 2 days. How many hours a day must 4 pumps work to empty the tank in 1 day?

A. 9

B. 10

C. 11

D. 12

Solution:

Let the required number of working hours per day be x.

More pumps, Less working hours per day (Indirect Proportion)

Less days, More working hours per day (Indirect Proportion)

Pumps	4	:	3		:: 8 : x
Days	1	:	2		:: 8 : x

4 x 1 x x = 3 x 2 x 8

x =	(3 x 2 x 8)
	(4)

x = 12.

9 / 15

9. A wheel makes 1000 revolutions in covering a distance of 88 km. Find the radius of the wheel.

A. 12m

B. 88m

C. 14m

D. 16m

Solution:

Distance covered in one revolution =((88 x 1000)/1000)= 88m

2∏r = 88 ==> 2 x (22/7) x r = 88 ==> r = 88 x (7/44) = 14 m

10 / 15

10. A mixture contains alcohol and water in the ratio 4 : 3. If 5 liters of water is added to the mixture, the ratio becomes 4: 5. Find the quantity of alcohol in the given mixture.

A. 2 liters

B. 4 liters

C. 2.5 liters

D. 3.5 liters

Solution:

Let the quantity of alcohol and water be 4x litres and 3x litres respectively

==> 4x/(3x+5) =4/5 ==> 20x = 4(3x+5) ==> 8x=20==> x = 2.5 liters.

11 / 15

11. The effective annual rate of interest corresponding to a nominal rate of 6% per annum payable half-yearly is:

A. 6.06%

B. 6.07%

C. 6.08%

D. 6.09%

Solution:

Amount of Rs. 100 for 1 year when compounded half-yearly		= Rs.		100 x		1 +	3		2		= Rs. 106.09
							100

Effective rate = (106.09 – 100)% = 6.09%

12 / 15

12. What is the cost painting the two adjacent walls of a hall at Rs. 5 per m² which has no windows or doors?

I.	The area of the hall is 24 sq. m.
II.	The breadth, length and height of the hall are in the ratio of 4 : 6 : 5 respectively.
III.	Area of one wall is 30 sq. m.

A. I only

B. II only

C. III only

D. Either I or III

E. All I, II and III are required

Solution:

From II, let l = 4x, b = 6x and h = 5x.

Then, area of the hall = (24x²) m².

From I. Area of the hall = 24 m².

From II and I, we get 24x² = 24 x = 1.

l = 4 m, b = 6 and h = 5 m.

Thus, area of two adjacent walls = [(l x h) + (b x h)] m² can be found out and so the cost of painting two adjacent walls may be found out.

Thus, III is redundant.

13 / 15

13. The square root of 64009 is:

A. 253

B. 347

C. 363

D. 803

Solution:

2|64009( 253

|———-

45|240

|225

|———-

503| 1509

| 1509

|———-

| X

|———-

64009 = 253.

14 / 15

14. Insert the missing number.

8, 24, 12, 36, 18, 54, (….)

A. 27

B. 108

C. 68

D. 72

Solution:

Numbers are alternatively multiplied by 3 and divided by 2.

So, the next number = 54 ÷ 2 = 27.

15 / 15

A man took loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was:

A. Rs. 2000

B. Rs. 10000

C. Rs. 15000

D. Rs. 20000

Solution:

Principal = Rs.		100 x 5400		= Rs. 15000.
		12 x 3

Free Online Aptitude Test: Placement

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