Quantitative Aptitude Test - 2 - PDF Flipbook
Quantitative Aptitude Test - 2
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GATE
EEE
Quantitative
Aptitude
Test-02Solutions
QUANTITATIVE APTITUDE
1. One year ago the ratio between Samir and Ashok's age was 4: 3
one year hence the ratio of their age will be 5: 4. What is the
sum of their present ages in years?
a) 12 years
b) 15 years
c) 16 years
d) Cannot be determined
Answer: (c)
Solution:
One year ago be 4x and 3x years
4 +2 = 5/4
3 +2
⇒ 4(4x + 2) = 5(3x + 2)
⇒ x = 2.
Present ages = (4x + 1 + 3x + 1) = 16 Years
2. The population of a new city is 5 million and is growing at 20%
annually. How many years would it take to double at this growth
rate?
a) 3 - 4 years
b) 4 - 5 years
c) 5 - 6 years
d) 6 - 7 years
Answer: (a)
1
Solution:
Another Method
Hence it takes 3 to 4 years
3. The average of 25 results is 18. The average of first 12 of those
is 14 and the average of last 12 is 17. What is the 13th result?
a) 74
b) 75
c) 69
d) 78
Answer: d
Solution:
Sum of 1st 12 results = 12*14
Sum of last 12 results = 12*17
13th result = x (let)
Now, 12*14 + 12*17 + x = 25*18
Or, x = 78.
4. How many leap years does 100 years have?
a) 25
b) 24
c) 4
d) 26
Answer: (b)
Solution:
2
Given year is divided by 4, and the quotient gives the number of
leap years.
Here,
100/4 = 25.
But, as 100 is not a leap year ⇒ 25 – 1 = 24 leap years.
5. A function f(x) is linear and has a value of 29 at x = –2 and 39 at
x = 3. Find its value at x = 5.
a) 59
b) 45
c) 43
d) 35
Answer: (c)
Solution:
Let f(x) = ax + b ………. (i)
at x = – 2, f(– 2) = a(– 2) + b = 29
∴ – 2a + b = 29 ……... (ii)
at x = 3, f(3) = a(3) + b = 9
∴ 3a + b = 39 ……… (iii)
By solving equation (ii) and (iii)
– 2a + b = 29 ………. (ii)
3a + b = 39 ………. (iii)
(–) (–) (–)
+5a = +10
a=2
Substitute a = 2, in the equation (ii)
3
– 2(2) + b = 29
– 4 + b = 29
b = 33
∴ f(x) = 2x + 33
at x = 5, f(5) = 2 × 5 + 33 = 43
6. A garrison of 500 persons had provisions for 27 days. After 3
days a reinforcement of 300 persons arrived. For how many
more days will the remaining food last now?
a) 12 days
b) 16 days
c) 14 days
d) 15 days
Answer: (d)
Solution:
Given that fort had provision for 500 persons for 27 days
Hence, after 3 days, the remaining food is sufficient for 500
persons for 24 days
Remaining persons after 3 days = 500 + 300 = 800
Assume that after 10 days, the remaining food is sufficient for
800 persons for x days
More men, Less days (Indirect Proportion)
Men 500 : 800} :: x : 24
⟹ 500 × 24 = 800x
⟹ 5 × 24 = 8x
⟹ x = 5 × 3 = 15
4
7. A clock is set at 5 am. If the clock loses 16 minutes in 24 hours,
what will be the true time when the clock indicates 10 pm on 4th
day?
a) 9.30 pm
b) 10 pm
c) 10.30 pm
d) 11 pm
Answer: (d)
Solution:
Time from 5 am to 10 pm on the 4th day = 3 days 17 hours = 3
× 24 + 17 = 89 hours
Given that clock loses 16 minutes in 24 hours
⇒ >23 hour 44 minutes of the given clock = 24 hours in a
normal clock
⟹ 236404 hours of the given clock = 24 hours in a normal clock
⟹ 231115 hours of the given clock = 24 hours in a normal clock
⟹ 356 hours of the given clock = 24 hours in a normal clock
15
⟹ 89 hours of the given clock = 24 × 15 × 89 hours in a
356
normal clock = 24 × 15
356
= 6 × 15 = 90 hours
So, the correct time is 90 hours after 5 am = 3 days 18 hours
after 5 am = 11 pm on the 4th day
5
8. An electronic device makes a beep after every 60 sec. Another
device makes a beep after every 62 sec. They beeped together at
10 a.m. The time when they will next make a beep together at
the earliest,
a) 9: 31 am
b) 10: 31 am
c) 11: 31 am
d) 12: 31 am
Answer: (b)
Solution:
L.C.M. of 60 and 62 seconds is 1860 seconds
1860/60 = 31 minutes
They will beep together at 10:31 a.m.
9. Four cards are randomly selected from a pack of 52 cards. If the
first two cards are kings, what is the probability that the third
card is a king?
a) 4/52
b) 2/50
c) (1/52) × (1/52)
d) (1/52) × (1/51) × (1/50)
Answer: (b)
Solution:
Probability of an event A, denoted as p(A), is defined as
P(A) = Number of cases favorable to A
Number of possible outcomes
6
→ Number of cases favorable to A = two kings only from a
pack of 52
→ Number of possible outcomes = 50
The probability that the third card is a king p(A) = 2
50
10. The area of a square is 4096 sq cm. Find the ratio of the
breadth and the length of a rectangle whose length is twice the
side of the square and breadth is 24 cm less than the side of the
square.
a) 18 : 5
b) 7 : 16
c) 5 : 14
d) 5 : 16
Answer: (d)
Solution:
Let the length and the breadth of the rectangle be 1 cm and b cm
respectively. Let the side of the square be a cm.
a2 = 4096 = 212
a = (212)1/2 = 26 = 64
L = 2a and b = a – 24
b : 1 = a – 24 : 2a = 40 : 128 = 5 : 16
11. What is the maximum value of m such that 7m divides into 14!
evenly?
a) 1
b) 2
c) 3
7
d) 4
Answer: (b)
Solution:
The term 14! equals the product of the numbers 1, 2, 3, 4, 5, 6,
7, 8, 9, 10, 11, 12, 13 and 14.
Only two of these numbers are divisible by 7.
The numbers are 7 and 14.
Hence, 14! can be expressed as the product of k×7×14, where k
is not divisible by 7.
Now, since there are two 7s in 14!, the numbers 7 and 72 divide
14!
73 and further powers of 7 leave a remainder when divided into
14!.
Hence, the maximum value of m is 2.
12. A, B, C enter into a partnership investing Rs. 35,000, Rs.
45,000 and Rs. 55,000 respectively. The respective shares of A,
B, C in annual profit of Rs. 40,500 are:
a) Rs. 10,500, Rs. 13,500, Rs. 16,500
b) Rs. 11,500, Rs. 13,000, Rs. 16,000
c) Rs. 11,000, Rs. 14,000, Rs. 15,500
d) Rs. 11,500, Rs. 12,500, Rs. 16,500
Answer: (a)
Solution:
A: B: C = 35000:45000: 55000 = 7:9:11
A's share = 40500 * 7/27 = Rs. 10500
8
B's share = 40500 * 9/27 = Rs. 13500
C's share = 40500 * 11/27 = Rs. 16500
13. A student took five papers in an examination, where the full
marks were the same for each paper. His marks in these papers
were in the proportion of 6: 7 : 8 : 9 : 10. In all papers together,
the candidate obtained 60% of the total marks then, the number
of papers in which he got more than 50% marks is
a) 1
b) 3
c) 4
d) 5
Answer: (c)
Solution:
Let the marks obtained in five subjects be 6x, 7x, 8x, 9x and
10x.
Total marks obtained = 40x
Max. Marks of the five subjects = 40x/0.6
[40x is 60% of total marks]
Max. Marks in each subject = 40x/0.6*5 = 13.33x
Hence, % of each subject = 6x*100/13.33 = 45.01% or,
7x*100/13.33 = 52.51
In same way other percentage are 60.01%, 67.52%, 75.01%.
Hence, number of subjects in which he gets more than 50%
marks = 4.
9
14. (x% of y) + (y% of x) is equivalent to_____
a) 2% of xy
b) 2% of (xy 1100)
c) xy% of 100
d) 100% of xy
Answer: (a)
Solution:
(x% of y) + (y% of x) = × + ×
100 100
= + = 2
100 100
= 2% of xy
15. Three flags each of different colours are available for a military
exercise, using these flags different codes can be generated by
waving
I. Single flag of different colours.
II. Any two flags in a different sequence of colours.
III. Three flags in a different sequence of colours.
The maximum number of codes that can be generated is.
a) 6
b) 9
c) 15
d) 18
Answer: (c)
Solution:
This type of question becomes very easy when we assume three
colour are red(R) blue (B) and Green (G).
10
We can choose any colour.
Now according to the statement 1 i.e.., codes can be generated
by waving single flag of different colours, then number of ways
are three i.e.., R.B.G from statement III three flags in different
sequence of colours, then number of ways are six i.e.., RBG,
BGR, GBR, RGB, BRG, GRB.
Hence total number of ways by changing flag = 3 + 6 + 6 = 15
16. The sum of the digits of a two-digit number is 12. If the new
number formed by reversing the digits is greater than the
original number by 54, find the original number.
a) 39
b) 57
c) 66
d) 93
Answer: (a)
Solution:
The new number formed by reversing the digits it greater than
the original number is possible in options A and B only.
Option ‘a’:
Sum of two digits in the number = 3 + 9 = 12
After reversing the two digits’ number = 93
The difference between the new number formed and original
number = 93 – 39 = 54
∴ option ‘A’ is correct
11
Option ‘b’:
Original number = 57
After reversing, the number is formed = 75
The difference between these two numbers = 75 – 57 = 18
∴ It is not.
17. The Manager of a company accepts only one employees leave
request for a particular day. If five employees namely Roshan,
Mahesh, Sripad, Laxmipriya and Shreyan applied for the leave
on the occasion of Diwali. What is the probability that Laxmi
priya’s leave request will be approved?
a) 1
b) 1/5
c) 5
d) 4/5
Answer: (b)
Solution:
Number of applicants = 5
On a day, only 1 leave is approved.
Now favorable events = 1 of 5 applicants is approved
Probability that Laxmi priya's leave is granted = 1/5.
18. Mahesh marks an article 15% above the cost price of Rs. 540.
What must be his discount percentage if he sells it at Rs.
496.80?
a) 18%
b) 21%
12
c) 20%
d) 19%
Answer: (c)
Solution:
CP = Rs. 540, MP = 540 + 15% of 540 = Rs. 621
SP = Rs. 496.80, Discount = 621 – 496.80 = 124.20
Discount % = 124.2/621 * 100 = 20%
19. Raman and Manoj attempted to solve a quadratic equation.
Raman made a mistake in writing down the constant term. He
ended up with the roots (4, 3). Manoj made a mistake in writing
down the coefficient of x. He got the roots as (3, 2). What will
be the exact roots of the original quadratic equation?
a) (6, 1)
b) (–3, –4)
c) (4, 3)
d) (–4, –3)
Answers: (a)
Solution:
Since Raman made a mistake in constant term, so his product of
roots will be wrong but his sum of roots will be correct.
Hence, the sum of roots is 4 + 3 = 7.
Manoj made a mistake in coefficient of ‘x’, so his sum of roots
will be wrong but product of roots will be correct.
Hence, the product of roots is 3 × 2 = 6.
Hence, the required equation is x2 – 7x + 6 = 0.
13
The roots of this equations are 6 and 1.
20. Shantanu started cycling along the boundaries of a square field
ABCD from corner point A. After 1/2 h, he reached the corner
point C, diagonally opposite to point A. If his speed was 16
km/h. find the area of the field?
a) 8 sq km
b) 9 sq km
c) 19 sq km
d) None of the above
Answer: (d)
Solution:
Distance = (AB + BC) = 2L = 16 × 1/2 = 8 km
∴ L = 8/2 = 4 km
∴ Area of the field = 4 × 4 = 16 sq km.
21. The sum of three numbers is 98. If the ratio of the first to the
second is 2:3. And that of the second to the third is 5:8, then the
second number is:
a) 20
b) 30
c) 48
d) 58
Answer: (b)
Explanation:
Let the three parts A, B, C.
Then, A: B = 2:3 and B: C = 5:8 = (5 * 3/5): (8 * 3/5) = 3:24/5
14
A: B: C = 2:3:24/5 = 10:15:24
⇒ B = 98 * 15/49 = 30.
22. The following sequence of numbers is arranged in increasing
order: 1, x, x, x, y, y, 9,16,18 given that the mean and median are
equal, and are also equal to twice the mode, the value of y is
a) 5
b) 6
c) 7
d) 8
Answer: (d)
Solution:
Given increasing order of number sequence is
1, x, x, x, y, y, 9, 16, 18
Total number of numbers = 9
Mean = Median = 2 Mode (given)
Mean = sum of numbers
Total no.of numbers
= 1 + + + + + + 9 + 1 6 + 18
9
= 1 + 3 + 2 + 9 + 1 6 + 18
9
= 3 + 2 + 44 ………. (i)
9
Median ⟹ As number of terms are odd (i.e.) 9 so, median will
be middle of the sequence = y
Mode = Number showing maximum frequency of repetition = x
(three times)
From the given condition
15
y = 2x ……… (ii)
By solving equation (i) and (ii), we get
x = 4, y = 8
∴ The value of y is 8
23. 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%
Answer: (d)
Solution:
Amount of Rs. 100 for 1 year when compounded half-yearly
= Rs. �100 × �1 + 1300�2�
= Rs. 106.09
∴ Effective rate = (106.09 – 100) % = 6.09%
24. If the radius of a right circular cone is increased by 50% its
volume increases by
a) 75%
b) 100%
c) 125%
d) 237.5%
Answer: (c)
16
Solution:
Volume of a right circular cone
V = 1 πR2H
3
R = radius of a cone
H = height of the cone
V ∝ R2
V1 = R2
V2 = (1.5)2R2 with increasing
V2 = 2.25 R2 = 2.25 V1
∴ Volume increases = V2 − V1 × 100
V1
= 2.25V1− V1 × 100
V1
= V1(1.25) × 100
V1
= 1.25 × 100
= 125%
25. Find the value of �75983 ×75983−45983 ×45983�
30000
a) 121966
b) 121766
c) 123966
d) 121866
Answer: (a)
Solution:
Given expression = (75983)2− (45983)2 = � 2− 2)�,
(75983 − 45983 ) ( −
where a = 75983, b = 45983
17
= ( + )( − ) = (a + b)
( − )
= (75983 + 45983)
= 121966
26. Two dice are thrown simultaneously. The probability that the
product of the numbers appearing on the top faces of the dice is
a perfect square is
a) 1/9
b) 2/9
c) 1/3
d) 4/9
Answer: (b)
Solution:
Total chances = 6 × 6 = 36 events
Product of numbers on 2 dice have to prefect square = Favorable
chances = (1, 1), (2, 2), (1, 4), (3, 3), (4, 1), (5, 5), (4, 4), (6, 6) =
8
Probability = Favourable chances = 8 = 2
Total chances 36 9
∴ Option (b) is correct
27. The value of 1 2 + 1 3 + 1 1 is
(216)− 3 (256)− 4 5
(32)−
a) 102
b) 105
c) 107
d) 109
Answer: (a)
18
Solution:
1 2 + 1 3 + 1 1
(216)− 3 4 5
(256)− (32)−
= 1 2 + 1 3 + 1 1
3 4 5
(63)− (44)− (25)−
= 1 −2 + 1 −3 + 1 −1
3 4 5
63 × 64 × 25 ×
= 1 + 1 + 1
(6)−2 (4)−3 (2)−1
= (62 + 43 + 21)
= 36 + 64 + 2) = 102
28. A runs twice as fast as B and B runs thrice as fast as C. The
distance covered by C in 72 minutes, will be covered by A in:
a) 18 minutes
b) 24 minutes
c) 16 minutes
d) 12 minutes
Answer: (d)
Solution:
Ratio of the speed of A, B and C = 6:3:1
Then, ratio of time taken;
= 1/6:1/3:1
= 1:2:6;
Hence, time taken by
A = 72/6
A = 12 minutes.
19
29. If A travels to his school from his house at the speed of 3 km/h,
then he reaches the school 5 minutes late. If he travels at the
speed of 4 km/h, he reaches the school 5 minutes earlier than
school time. The distance of his school from his house is:
a) 1 km
b) 2 km
c) 3 km
d) 4 km
Answer: (b)
Solution:
Let the distance between school and home be x km.
The difference of time when A goes school to school with these
two different speeds is 10 min = 10/60 hour.
Now,
(x/3) – (x/4) = 10/60
Or, x/12 = 1/6
Or, x = 12/6 = 2 km.
30. The charges per hour of internet surfing is increased by 25%
then find the percentage decrease in the time period of surfing
user (a net savvy) who can afford only 10% increase in
expenditure:
a) 22%
b) 12%
c) 15%
d) 9.09%
20
Answer: (b)
Solution:
Time * Rate = total charges
100 *100 = 10000
X *125 = 110
[25% increase in rate, user can afford only 10% increase]
X = (110/125) *100 = 88%
Thus, decrease in time = 12%.
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