# Quantitative Aptitude Test - 3 - PDF Flipbook

Quantitative Aptitude Test - 3

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GATE
EEE

Quantitative
Aptitude

Test-03Solutions

QUANTITATIVE APTITUDE
1. Jayesh is as much younger to Anil as he is older to Prashant if

the sum of the ages of anil and Prashant is 48 years, what is the
age of Jayesh?
a) 20 years
b) 24 years
c) 30 years
d) Cannot be determined
Solution:
Anil’s age = x years Prashant’s age = (48 – x) Years
The age of Jayesh be P years
P – (48 – x) = x – p
⇒ 2P = 48
⇒ P = 24 Years
2. What is the value of x when 81 × �2165� +2 + �53�2 +4 = 144?
a) 1
b) – 1
c) – 2
d) Cannot be determined
Solution:

81 × �2165�x + 2 ÷ �35�2x + 4 = 144

1

=�1265�x + 2 144 = (12)2 = �192�2
81 (9)2
�53�2x + 4

�54�x + 2 = �192�2
�53�2x + 4

(4)x + 2 × (5)2x + 4 = �192�2
(5)2x + 4 (3)x + 2

(4)2x + 4 = �192�2 = �34�2
(3)2x + 4

�43�2x + 4 = �43�2

2x + 4 = 2

2x = – 2

x=–1

3. A train covers the first 16 km at a speed of 20 km per hour

another 20 km at 40 km per hour and the last 10 km at 15 km

per hour. Find the average speed for the entire journey.

a) 24 km

b) 26 km

c) 21 km

d) 23(23/59) km

Solution:

Average speed = total distance covered/ total time

Total distance = (16 + 40 + 10) = 46 km

Time taken = (16/20) + (20/40) + (10/15) = 59/30

Average speed = 46 × 30/59 = 23(23/59) km/hr.

2

4. If Aug 15th, 2012 falls on Thursday then June 11th, 2013 falls
on which day?
a) Wednesday
b) Saturday
c) Monday
d) Tuesday
Solution:
First, we count the number of odd days for the left-over days in
the given period.
Here, given period is 15.8.2012 to 11.6.2013
Aug Sept Oct Nov Dec Jan Feb Mar Apr May Jun
16 30 31 30 31 31 28 31 30 31 11(left days)
2 + 2 + 3 + 2 + 3 + 3 + 0 + 3 + 2 + 3 + 4 (odd days) = 6 odd
days
So, given day Thursday + 6 = Wednesday is the required result.

5. 5 Skilled workers can build a wall in 20 days, 8 semi-skilled
workers can build a wall in 25 days; 10 unskilled workers can
build a wall in 30 days. If a team has 2 skilled 6 semi-skilled
and 5 unskilled workers, how long will it take to build the wall?
a) 20 days
b) 18 days
c) 16 days
d) 15 days

3

Solution:

5 skilled workers can build a wall in 20 days 1 skilled workers 5

× 20 = 100 days

1-day work of skilled worker = 1
100

8 semi-skilled workers can build wall = 25

1 semi-skilled worker = 8 × 25 = 200 days

1-day work of semi-skilled worker = 1
200

10 unskilled workers can build a wall = 30 days

1 unskilled worker = 10 × 30 = 300 days

1 unskilled worker can build a wall = 1
300

∴ 2 skilled + 6 semi-skilled + 5 unskilled

= 2 �1100� + 6 �2100� + 5 �3100�

1 + 3 + 1 = 6+9+5 = 1
50 100 60 300 15

∴ 2 skilled + 6 semi-skilled + 5 unskilled

Can build a wall = 15 days

6. A fort had provision of food for 150 men for 45 days. After 10

days, 25 men left the fort. Find out the number of days for

which the remaining food will last.

a) 44

b) 42

c) 40

d) 38

4

Solution:
Given that fort had provision of food for 150 men for 45 days
Hence, after 10 days, the remaining food is sufficient for 150
men for 35 days
Remaining men after 10 days = 150 – 25 = 125
Assume that after 10 days, the remaining food is sufficient for
125 men for x days
More men, Less days (Indirect Proportion)
⇒ Men 150: 125}:: x: 35
⇒ 150 × 35 = 125x
⇒ 6 × 35 = 5x
⇒ x = 6 × 7 = 42
7. At what time between 4 and 5 o'clock will the hands of a clock
be at right angle?
a) 40/11 min. past 4
b) 45/11 min. past 4
c) 50/11 min. past 4
d) 60/11 min. past 4
Solution:
At 4 o'clock, the minute hand will be 20 min. spaces behind the
hour hand
Now, when the two hands are at right angles, they are 15min.
spaces apart.
So, they are at right angles in following two cases.

5

Case I. When minute hand is 15 min. spaces behind the hour
hand:
In this case min. hand will have to gain (20 - 15) = 5-minute
spaces.
55 min. spaces are gained by it in 60 min.
5 min spaces will be gained by it in 60×5/55 min=60/11min.
They are at right angles at 60/11min. past 4.
Case II. When the minute hand is 15 min. spaces ahead of the
hour hand:
To be in this position, the minute hand will have to gain (20 +
15) = 35 minute spaces.
55 min. spaces are gained in 60 min.
35 min spaces are gained in (60 × 35)/55 min = 40/11
They are at right angles at 40/11 min. past 4.
8. If the product of two numbers is 84942 and their H.C.F. is 33,
find their L.C.M.
a) 2474
b) 2574
c) 2674
d) 2774
Solution:
HCF × LCM = 84942, because we know
Product of two numbers = Product of HCF and LCM
LCM = 84942/33 = 2574

6

9. Given digits 2, 2, 3, 3, 3, 4, 4, 4, 4 how many distinct 4digit
numbers greater than 3000 can be form
a) 50
b) 51
c) 52
d) 54
Solution:
More than 3000, 4-digit number means, the First digit may be 3
or 4, the second third and fourth digits are three in each (i.e.)
23 33
= 2 × 3 × 3 × 3 = 54

10. The circumferences of two circles are 264 meters and 352
meters. Find the difference between the areas of the larger and
the smaller circles.
a) 4192 sq m
b) 4304 sq m
c) 4312 sq m
d) 4360 sq m
Solution:
Let the radii of the smaller and the larger circles be s m and l m
respectively.
2πs = 264 and 2πl = 352
s = 264/2π and l = 352/2π

7

Difference between the areas = πl2 – πs2
= π {1762/π2 – 1322/π2}
= 1762/π – 1322/π
= (176 – 132) (176 + 132)/π
= (44) (308) / (22/7) = (2) (308) (7) = 4312 sq m
11. What is the remainder when 91 + 92 + 93 + ...? + 98 is divided
by 6?
a) 3
b) 2
c) 0
d) 5
Solution:
6 is an even multiple of 3. When any even multiple of 3 is
divided by 6, it will leave a remainder of 0. Or in other words it
is perfectly divisible by 6.
On the contrary, when any odd multiple of 3 is divided by 6, it
will leave a remainder of 3. For e.g. when 9 an odd multiple of 3
is divided by 6, you will get a remainder of 3.
9 is an odd multiple of 3. And all powers of 9 are odd multiples
of 3.
Therefore, when each of the 8 powers of 9 listed above is
divided by 6, each of them will leave a remainder of 3.
The total value of the remainder = 3 + 3 + .... + 3 (8 remainders)
= 24.

8

24 is divisible by 6. Hence, it will leave no remainder.
Hence, the final remainder when the expression 91 + 92 + 93 +
.... + 98 is divided by 6 will be equal to '0'.
12. A and B started business in partnership investing Rs. 20,000
and Rs. 15,000 respectively. After six months, C joined them
with Rs. 20,000. What will be B's share in the total profit of Rs.
25,000 earned at the end of 2 years from the starting of the
a) Rs. 7500
b) Rs. 9000
c) Rs. 9500
d) Rs. 10,000
Solution:
A: B: C = (20000 × 24): (15000 × 24): (20000 × 18) = 4:3:3
B's share = 25000 × 3/10 = Rs. 7500.
13. The price of raw materials has gone up by 15%, labor cost has
also increased from 25% of the cost of raw material to 30% of
the cost of raw material. By how much percentage should there
be reduction in the usage of raw materials so as to keep the cost
same?
a) 28%
b) 17%
c) 27%
d) 24%

9

Solution:
Let the initial cost of raw material be 100. So, initial labour cost
was 25 and net cost was 125.
Now,
15% increment in raw materials cost and labour cost has gone
up to 30% from 25 %.
Raw material cost = 115. And
Labour cost = (115 × 30) = 34.5
So, new net cost,

= 115 + 34.5 = 149.5
% reduction = 24.5/149.5 = 17 % (approx.)
14. There friends, R, S and T shared toffees from a bowl. R took
1/3rd of the toffees but returned four to the bowl. S took 1/4th of
what was left but returned three toffees to the bowl. T took half
of the remainder but returned two back into the bowl. If the
bowl had 17 toffees left, how many toffees were originally there
in the bowl?
a) 38
b) 31
c) 48
d) 41
Solution:

The total No. of toffees in a bowl = x

10

= 2x + 12 − �2x1−224�
3

→ for R ⇒ = 8x+48−2x+24 �3x − 4�
12

= 6x+72
12

The remaining toffees in a bowl

= x − �3x − 4�

= x − x + 4 = 3x – x + 12
3 3

= 2x + 12
3

→ for S ⇒

= 2x+12 − 2 ⇒2x1+212 − 3 = 2x+12−36
12
3

4

= 2x−24
12

The remaining toffees in a bowl

2x+12 − 3
12

= x − �3x − 4�

= x − x + 4 = 3x – x + 12
3 3

= 2x + 12
3

= 8x+48−2x+24
12

= 6x+72
12

→ for T ⇒

= 6x+72 − 2 ⇒ 6x+72 − 2
12 24

2

11

= 6x+72−48 = 6x+24
24 24

The remaining toffees in a bowl

6x+72 − �6x2+424�
12

= 12x+144−6x−24
24

= 6x+120
24

∴ 6x+120 = 17
24

6x + 120 = 17 × 24 = 408

6x = 408 – 120 = 288

∴x= 288 = 48
6

Short cut method

Among all given alternatives, divisible by 3 is 48 only

For R ⇒ 48 = 16 ⇒ 48 – 16 = 32 + 4 = 36
3

For S ⇒ 36 = 9 ⇒ 36 – 9 = 27 + 3 = 30
4

For T ⇒ 30 = 15 ⇒ 30 – 15 = 15 + 2 = 17
2

15. A student has to opt for 2 subjects out of 5 subjects for a

course. Namely commerce, economics, statistics, mathematics 1

and Mathematics 2, Mathematics 2 can be offered only if

mathematics 1 has also opted. The number of different

combinations of two subjects which can be opted is

a) 5

b) 6

12

c) 7

d) 18

Solution:

Number of ways of opting a subject other than Mathematics II =

4C2.

= 4 × 3 × 2! / 2! × 2 = 6.

Number of ways of selection of Mathematics II = 1

Therefore, Total Number of ways = 6 + 1 = 7.

16. Given that (y) : |y|/y , and q is any non-zero real number, the
value of |f (q) − r(−q)|is

a) 0

b) – 1

c) 1

d) 2

Solution:

Given that f(y) = 141/4

Q = any non-zero real number

= + q (or) – q

If q is the

f(q) = |q| = 1
q

f(–q) = |−q| = –1
−q

∴ |f(q) − f(−q)| = |1 − (−1)| = 2

13

If q is – ve

f(–q) = |−q| = –1
−q

f(–q) = |q| = 1
q

|f(q) − f(−q)| = |1 − (−1)| = |−1 − 1| = −2

17. What is the probability of getting a multiple of 2 or 5 when an

unbiased cubic die is thrown?

a) 1/3

b) 1/2

c) 2/3

d) 1/6

Solution:

Total numbers in die = 6

P (multiple of 2) = 1/2

P (multiple of 5) = 1/6

P (multiple of 2 or 5) = 2/3

18. The sale price of an article including the sales tax is Rs. 616.

The rate of sales tax is 10%. If the shopkeeper has made a profit

of 12%, then the cost price of the article is:

a) Rs. 500

b) Rs. 515

c) Rs. 550

d) Rs. 600

14

Solution:
110% of S.P. = 616
S.P. = (616 × 100)/110 = Rs. 560
C.P = (110 × 560)/112 = Rs. 500

19. Let p and q be the roots of the quadratic equation x2 – (α – 2)x
– α – 1 = 0. What is the minimum possible value of p2 + q2?
a) 0
b) 3
c) 4
d) 5
Solution:
We have the equation x2 – (α – 2)x – α – 1 = 0. its roots are p
and q.
So, we have sum of roots = p+q = a–2 and the product of roots,
pq = –a –1
Now p2 + q2 = (p + q)2 – 2pq = (a – 2)2 + 2(a + 1) = a2 + 4 – 4a +
2a + 2 = (a – 1)2 + 5
Since (α – 1)2 is a perfect square, so its minimum value will be
0, when α = 1.
In that case, the minimum value of p2 + q2 will be 5.

15

20. Two boys A and B run at 4 1⁄5 and 8 km an hour respectively.

A having 150 m start and the course being 1 km, B wins by a

distance of

a) 325 m

b) 60 m

c) 120 m

d) 275 m

Solution:

A has a start of 150 m. So, A has to run 1000 – 150 = 850 metre

while B has to run 1000 metre.

Speed of A = 4 1 km/hr = 21 km/hr
5 5

= 21 × 5 = 21
5 18 18

= 7 m/s
6

Speed of B = 8 km/hr = 8 × 5 = 21 = 20 m/s
18 18 9

Time taken by B to travel 1000 metre = distance
speed of B

= 1000 = 450 sec
�290�

Distance travelled by A by this time = time × speed of A = 450

× 7 = 525 metre
6

Hence, A win by 850 – 525 = 325 metre

16

21. Ratio of the earnings of A and B is 4:7. If the earnings of A
increases by 50% and those of B decreased by 25%, the new
ratio of their earnings becomes 8:7. What are A's earnings?
a) Rs. 21,000
b) Rs. 26,000
c) Rs. 28,000
Solution:
Let the original earnings of A and B be Rs. 4x and Rs. 7x.
New earnings of A = 150% 0f Rs. 4x = (150/100 × 4x) = Rs. 6x
New earnings of B = 75% of Rs. 7x = (75/100 × 7x) = Rs. 21x/4
6x: 21x/4 = 8:7
This does not give x. So, the given data is inadequate.

22. The sum of n terms of the series 4 + 44 + 444 + .... is
a) (4/81) [10n+1 – 9n – 1]
b) (4/81) [10n-1 – 9n – 1]
c) (4/81) [10n+1 – 9n – 10]
d) (4/81) [10n – 9n – 10]
Solution:
The sum of n terms of the series 4 + 44 + 444 + … is in the
given alternatives if n = 1, which will give 4 and if n = 2 which
will give sum of first two numbers (i.e.) 4 + 44 = 48 is our

17

Ex: if n = 1

(a) 4 [10n+1 − 9n − 1]
81

4 [101+1 − 9(1) − 1] = 4 [100 − 10] = 4 × 90
81 81 81

It is not

(b) 4 [10n−1 − 9n − 1]
81

4 [101−1 − 9(1) − 1] = 4 [1 − 10] = 4 × (−9)
81 81 81

It is not

(c) 4 [10n+1 − 9n − 10]
81

4 [101+1 − 9(1) − 10] = 4 [100 − 10] = 4 × 81 = 4
81 81 81

if n = 2

4 [10n+1 − 9n − 10]
81

4 [102+1 − 9(2) − 10] = 4 [1000 − 28] = 4 × 972
81 81 81

It is Correct

23. 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?

a) Rs. 51.25

b) Rs. 52

c) Rs. 54.25

d) Rs. 60

18

Solution:

Sum = Rs. �502××1500� = Rs. 500.
Amount = Rs. �500 + �1 + 1500�2�

= Rs. �500 × 21 × 21 �
20 20

= Rs. 551.25

∴ C.I. = Rs. (551.25 – 500) = Rs. 51.25

24. A transporter receives the same number of orders each day.

Currently, he has some pending orders (backlog) to be shipped.

If he uses 7 trucks, then at the end of the 4th day he can clear all

the orders. Alternatively, if he uses only 3 trucks, then all the

orders are cleared at the end of the 10th day. What is the

minimum number of trucks required so that there will be no

pending order at the end of the 5th day?

a) 4

b) 5

c) 6

d) 7

Solution:

7 trucks can clear all the orders at the end of 4th day

3 trucks can clear all the orders at the end of 10th day

Transporter receives every day same number order currently; he

has some pending orders (backlog).

19

What is the minimum number of trucks required so that there

will be no pending order at the end of the 5th day?

1 day = 7 × 4 = 28 No. of trucks

1 day = 3 × 10 = 30 No. of trucks

LCM of 28 and 30 = 420

Assume everyday received orders = 105

First case:

1 day = 420 orders+backlog (b)
28

420 = 105
4

Second case:

10 days = 1050

1 day = 1050 orders+backlog (b)
30

420+b = 1050+b
28 30

12,600 + 30b = 29400 + 28b

2b = 16,800

b = 8400

1 day = 8820 = 315
28

For 5 days case:

1 day = 105 × 5 + 8400 = 8925

No. of trucks required on 5th day = 8925 = 5.66 ≅ 6
315 × 5

∴ Minimum number of trucks required without pending orders

at the end of 5th day = 6

20

25. Find the value of 2 1 3 + 3 1 4 + 4 1 5 + 5 1 6 + ⋯ + 9 1
× × × × × 10

a) 2/5

b) 2/7

c) 3/7

d) 1/3

Solution:

Given expression = �12 − 13� + �31 − 14� + �14 − 15� + �15 − 61� +

⋯ + �19 − 110� .

= �21 − 110�

= 4 = 2
10 5

26. A container originally containsl0 litres of pure spirit. From this

container 1 litre of spirit is replaced with 1 litre of water,

Subsequently, l litre of the mixture is again replaced with l litre

of water and this process is repeated one more time. How much

spirit is now left in the container?

a) 6.29 lit

b) 9 lit

c) 5 lit

d) 7.29 lit

21

Solution:

Short cut:

10 × 9 × 9 × 9 = 7.29 lit
10 10 10

27. 1 + 1
1+P(n−m) 1+P(m−n)

a) 2

b) 1/1+P

c) 1

d) 1/P

Solution:

1 + 1
1+P(n−m) 1+P(m−n)

1 + 1
1+PPmn 1+PPmn

Pm + Pn ⇒ Pm+Pn = 1
Pm+Pn Pn+Pm Pm+Pn

22

28. A man starts climbing a 11 m high wall at 5 pm. In each
minute he climbs up 1 m but slips down 50 cm. At what time
will he climb the wall?
a) 5:30 pm
b) 5:21 pm
c) 5:25 pm
d) 5:27 pm
Solution:
1st Method:
Man climbs 1 m and slips down 50 cm (0.5 m) in one minute
i.e. he climbs (1 – 0.5 = 0.5 m) in one minute.
But in the last minute he will be climbing 1 m as he gets on the
top so no slip.
Time taken to climb 11 meter = [(10/0.5) +1] = 21 minutes.
He climbs the wall at 5:21 pm.
2nd Method (short-cut):
Reqd. time = [(ht. of pole - slipped distance) /(climbed
distance - slipped distance)] × t;
= [(x – z)/(y – z)] × t;
Required time,
= {(11 – 0.5)/(1 – 0.5)} × 1
= 21 minutes.

23

29. A monkey climbs a 60 m high pole. In first minute he climbs 6
m and slips down 3 m in the next minute. How much time is
required by it to reach the top?
a) 35minutes
b) 33 minutes
c) 37 minutes
d) 40 minutes
Solution:
Monkey climbs 6 m in 1st minute and slips down 3 m in next
minute
i.e. Monkey climbs 3 meter in 2 minute then he climbs in one
minute, = 3/2 m.
But in the last minute he climbs 6 m as he gets on the top so
there is no slip.
Time required = 2 × (54/3) + (6/6) = 37 minutes.

30. Ram Lal is a renowned packager of fruits in Varanasi. He
packs 70 mangoes or 56 guavas every day working 7 hours a
day. His wife also helps him. She packs 30 mangoes or 24
guavas working 6 hours per day. Ram Lal has to pack 3300
mangoes and 2400 guavas with help of his wife. They work
alternately, each day 10 hours. His wife started packaging firs
day and works every alternate day. Similarly, Ram Lal started
his work second day and worked alternatively till the

24

completion of the work. In how many days the work will
finished?
a) 82
b) 85
c) 84
d) 80
Solution:
Mango Packing rate of Ram Lal = 70/7 = 10 per hour
Guavas Packing rate of Ram Lal = 56/7 = 8 per hour.
Mango Packing rate of Ram Lal's Wife = 30/6 = 5 per hour
Guavas Packing rate of Ram Lal's Wife = 24 /6 = 4 per hour.
On comparing,
8 Guavas = 10 Mangoes
1 Guavas = 10/8 Mangoes
2400 Guavas = (10 × 2400) /8 = 3000 Mangoes.
This means that they need to pack (6300 + 3000) Mangoes.
Combined Mango Packing rate of both = 10 + 5 = 15 per hour.
We will take 2 days = 1-time unit.
Working 10 hours a day they will pack 150 Mangoes in 1-time
unit.
Total time taken to pack 6300 Mangoes = 6300 /150 = 42-time
unit.
So, total time taken = 2 × 42 = 84 days.

25