Quantitative Aptitude Test - 1 - PDF Flipbook
Quantitative Aptitude Test - 1
302 Views
92 Downloads
PDF 4,878,894 Bytes
GATE
CSE
Quantitative
Aptitude
Test-01Solutions
QUANTITATIVE APTITUDE
1. 10 years ago, chandravati’s mother was 4 times older than her
daughter, after 10 years the mother will be twice older than the
daughter. The present age of chandravati is:
a) 5 years
b) 10 years
c) 20 years
d) 30 years
Answer: (c)
Solution:
Chandravati’s age 10 years ago be x years
Mother’s age 10 years ago = (4x) years
2(x + 20) = (4x + 20)
⇒ x = 10
Present age of Chandravati = (x + 10)
= 20 years
2. There are two candidate’s P and Q in an election. During the
campaign, 40% of the voters promised to vote for P, and rest for
Q' However, on the day of election 15% of the voters went back
on their promise to vote for P and instead voted for Q, 25% of
the voters went back on their promise to vote for Q and instead
voted for P. Suppose, P lost by 2votes, then what was the total
number of voters?
a) 100
b) 10
1
c) 90
d) 95
Answer: (a)
Solution:
P:Q
40 : 60 15% of 40 = 8 × 40 = 6
100
−6 : +6 25% of 60 = 25 × 60 = 15
34 66 100
15% of 40 = 25 × 60 = 15
100
The different between P and Q is 2
∴ The total no. of voters is 100
3. In a boat there are 8 men whose average weight is increased by
1 kg when 1 man of 60 kg is replaced by a new man. What is
weight of new comer?
a) 70
b) 66
c) 68
d) 69
Answer: (c)
Solution:
Member in group * age increased = difference of replacement
Or, 8*1 = new comer – man going out
Or, new comer = 8 + 60;
Or, new comer = 68 years.
2
4. What was the day of the week on, 16th July, 1776?
a) Tuesday
b) Wednesday
c) Monday
d) Saturday
Answer: (a)
Solution:
16th July, 1776 = (1775 years + Period from 1st Jan, 1776 to
16th July, 1776)
Counting of odd days:
1600 years have 0 odd day.
100 years have 5 odd days.
75 years = (18 leap years + 57 ordinary years)
= [(18 x 2) + (57 x 1)]
= 93 (13 weeks + 2 days) = 2 odd days
1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day.
Jan Feb Mar Apr May Jun Jul
31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days = (28 weeks +
2days)
Total number of odd days = (0 + 2) = 2.
Required day was 'Tuesday'.
5. If Log(P): (1/2) Log(Q): (1/3) Log (R) then which of the
following options is TRUE?
a) P2 = Q2R2
b) Q2 = PR
3
c) Q2 = R3P
d) R = P2Q2
Answer: (b)
Solution:
Given P = Q1/2 = R1/3
�Q1�2�4 = P4
⇒ Q2 = P4
Q2 = P. P3
(∴ P3 = �R1�3�3 = R)
Q2 = P. R
6. If 40 men can make 30 boxes in 20 days, how many more men
are needed to make 60 boxes in 25 days?
a) 28
b) 24
c) 22
d) 26
Answer: (b)
Solution:
Given that 40 men can make 30 boxes in 20 days
Let x more men are needed to make 60 boxes in 25 days
Then (40 + x) men can make 60 boxes in 25 days
More boxes, more men (direct proportion)
More days, less men (indirect proportion)
Hence we can write as-
4
boxes 30 : 60
::40 : (40 + x)
days 25 : 20
⟹ 30 × 25 × (40 + x) = 60 × 20 × 40
⟹ 25 × (40 + x) = 2 × 20 × 40
⟹ 5 × (40 + x) = 2 × 4 × 40
⟹ (40 + x) = 2 × 4 × 8 = 64
⟹ x = 64 – 40 = 24
7. At what time between 3 o'clock and 4 o'clock, both the needles
of a clock will coincide each other?
a) 16 2/11 minutes past 3
b) 16 4/11 minutes past 3
c) 15 4/11 minutes past 3
d) 15 2/11 minutes past 3
Answer: (b)
Solution:
The two hands of a clock will be together between H and (H +
1) o' clock at (60H/11) minutes past H o' clock.
Here H = 3.
Hands will be together at 60H minutes past 3
11
= 60 × 3 minutes past 3
11
= 180 minutes past 3
11
= 16 4 minutes past 3
11
5
8. The H.C.F. and L.C.M. of two numbers are 12 and 5040
respectively if one of the numbers is 144, find the other number?
a) 110
b) 180
c) 360
d) 420
Answer: (d)
Solution:
Product of 2 numbers = product of their HCF and LCM
144 * x = 12 * 5040
x = (12*5040)/144
x = 420
9. If (1.001)1259 : 3.52 and (1.001)2062: 7.85 then (1.001)3321 =
a) 2.23
b) 4.33
c) 11.37
d) 27.64
Answer: (d)
Solution:
(1.001)1259 = 3.25 and (0.001)2062 = 7.85
(1.001)3321 = (1.001)1259 × (0.001)2062
= 3.25 × 7.85
= 27.63 ≈ 27.64
6
10. A wire in the form of a circle of radius 3.5 m is bent in the
form of a rectangle, whose length and breadth are in the ratio of
6 : 5. What is the area of the rectangle?
a) 60 cm2
b) 30 cm2
c) 45 cm2
d) 15 cm2
Answer: (b)
Solution:
The circumference of the circle is equal to the perimeter of the
rectangle.
2 (l + b) = 2 r
Let l = 6x and b = 5x
2(6x + 5x) = 2 × 22 × 305
7
⟹x=1
Therefore, l = 6 cm and b = 5 cm
Area of the rectangle = 6 × 5 = 30 cm2
11. How many natural numbers below 660 are divisible by 5 and
11 but not by 3?
a) 8
b) 9
c) 10
d) 11
Answer: (a)
7
Solution:
If the number is divisible by 5 and 11 it must be divisible by 55.
The numbers are less than 660.
Hence, dividing 659 by 55 gives the number of multiples of 55
= 11 (ignoring fraction part).
The 11 multiples of 55 which are less than 560, but of these 11
multiples some can be multiples of 3.
The numbers of such, multiples are the quotient of 11 by 3.
Quotient of 113 = 3.
Out of 11 multiples of 55, 3 are multiples of 3.
Hence, numbers less than 660 and divisible by 5 and 11 but not
by 3 = 11 − 3 = 8
12. X and Y started a business with capitals Rs. 20000 and Rs.
25000. After few months Z joined them with a capital of Rs.
30000. If the share of Z in the annual profit of Rs. 50000 is Rs.
14000, then after how many months from the beginning did Z
join?
a) 7
b) 6
c) 3
d) None of these
Answer: (d)
Solution:
Investments of X, Y and Z respectively are Rs. 20000, Rs.
25000 and Rs. 30000
8
Let investment period of Z be x months.
Ratio of annual investments of X, Y and Z is
(20000 * 12): (25000 * 12) : (30000 * x) = 240: 300: 30x = 8:
10: x
The share of Z in the annual profit of Rs. 50000 is Rs. 14000.
⇒ [x/ (18 + x)] 50000 = 14000
⇒ [x/ (18 + x)] 25 = 7
⇒ 25x = 7x + (18 * 7)
⇒ x = 7 months.
Z joined the business after (12 – 7) months. i.e., 5 months.
13. The price of Maruti car rises by 30 percent while the sales of
the car come down by 20%. What is the percentage change in
the total revenue?
a) – 4%
b) – 2%
c) + 4%
d) + 2%
Answer: (c)
Solution:
Let initial price of Maruti Car be Rs. 100.
As price increases 30%, price of car will become,
(100 + 30% of 100) = Rs. 130.
Due to increase in price, sales is down by 20%. It means, it is
going make 20% less revenue as expected after increment of
price.
9
So, New revenue = (130 – 20% of 130)
= Rs. 104.
The initial revenue was Rs. 100 which becomes Rs. 104 at the
end. It means there is 4% increment in the total revenue.
Mind Calculation Method:
100 = 30%↑ (price effect)
⇒ 130 = 20 %↓ (sales effects)
⇒ 104.
Hence, 4% rises.
14. The cost function for a product in a firm is given by 5q2, where
q is the amount of production. The firm can sell the product at a
market price of 50 per unit. The number of units to be produced
by the firm such that the profit is maximized is
a) 5
b) 10
c) 15
d) 25
Answer: (a)
Solution:
The cost function for a product in a firm = 5q2
q = The amount of production
= The number of units to be produced by the firm
Cost price (C.P) = 5q2
Selling price (S. P) = 50q
Profit = S. P – C. P
10
q S.P = (50q) C.P = 5q2 Profit
1 50(1) = 20 5(1)2 = 5 45
2 50(2) = 100 5(2)2 = 20 80
3 50(3) = 150 5(3)2 = 45 105
4 50(4) = 200 5(4)2 = 80 120
5 50(6) = 250 5(5)2 = 125 125
6 50(7) = 300 5(6)2 = 180 120
7 50(8) = 350 5(7)2 = 245 105
8 50(8) = 400 5(8)2 = 320 80
Profit = 50q – 5q2 is maximum at q = 5
The profit is maximized at q = 5
Alternative method
Selling price = 50q
Cost price = 5q2
Profit (y) = 50q – 5q2
yˈ = 50 – 10q = 0
⟹ 10q = 50
⟹q=5
yˈˈ = –10 < 0
⟹ yˈˈ(5) < 0
∴ y is maximum at q = 5
Maximum profit = y(5)
= 20(5) – 5(s)2
= 250 – 125
= 125
11
15. A mixed doubles tennis game is to be played two teams (each
consists of one male and one female) There are four married
couples. No team is to consist a husband and his wife. What is
the maximum number of games that can be played?
a) 12
b) 48
c) 36
d) 42
Answer: (d)
Solution:
Married couples = MF MFMFMF
AB CD EF GH
Possible teams = AD CB EB GB
AF CF ED GD
AH CH EH GF S
Since one male can be paired with 3 other females,
Total teams = 4 × 3 = 12.
Team AD can play only with CB, CF, CH, EB, EH, GB, GF
(7 teams)
Team AD cannot play with AF, AH, ED and GD
The same will apply with all teams,
so number of total matches = 12 × 7 = 84.
But every match includes 2 teams,
so the actual number of matches = 84/2 = 42.
12
16. A number is as much as greater than 75 as it is smaller than
117. The number is:
a) 91
b) 93
c) 89
d) 96
Answer: (d)
Solution:
From given options
→ option A ⟹ 91 – 75 = 16
→ option B ⟹ 93 – 75 = 18
→ option C ⟹ 89 – 75 = 14
→ option D ⟹ 96 – 75 = 21
The difference of given options with 75 is greatest in option (d)
so, it is much greater than 75 and it also smaller than 117
Therefore, option d is correct
Alternative method:
From options, maximum value from 75 and nearer smaller than
117 is 96 only.
17. The first 8 alphabets are written down at random. What is the
probability that the letters b, c, d, e always come together?
a) 1/7
b) 8!
c) 7!
d) 1/14
13
Answer: (d)
Solution:
The 8 letters can be written in 8! Ways.
n(S) = 8!
Let E be the event that the letters b, c, d, e always come together
when the first 8 alphabets are written down.
Now the letters a, (bcde), f, g and h can be arranged in 5! Ways.
The letters b, c, d and e can be arranged themselves in 4! Ways.
n(E) = 5! × 4!
Now, the required P(E) = n(E)/n(S)
= 5! × 4!/8!
= 1/14
Hence the answer is 1/14.
18. Mohit sold an article for Rs. 18000. Had he offered a discount
of 10% on the selling price, he would have earned a profit of
8%. What is the cost price of the article?
a) Rs. 15000
b) Rs. 16200
c) Rs. 14700
d) Rs. 15900
Answer: (a)
Solution:
Let the CP be Rs. x.
Had he offered 10% discount, profit = 8%
14
Profit = 8/100 x and hence his SP = x + 8/100 x = Rs. 1.08x =
18000 – 10/100(18000) = 18000 – 1800 = Rs. 16200
⇒ 1.08x = 16200
⇒ x = 15000
19. All the page numbers from a book are added, beginning at
page 1. However, one-page number was mistakenly added
twice. The sum obtained was 1000. Which page number was
added twice?
a) 44
b) 45
c) 10
d) 12
Answers: (c)
Solution :
Let the total pages be ‘n’.
So we have {n(n + 1)}/2 = 100
Since one page was added twice, so 1000 is not the actual sum
but it is an increased sum.
We have n(n + 1) = 2000.
Now by hit and trial we can say that the value of n = 44 i.e.
initially there were 44 pages and their sum was
(44 × 45)/2 = 990
Since the given sum is 1000, so we can say that the page number
10 was added twice.
15
20. In a 20 m race, A runs at a speed of 2 m/s. If A Gives B a start
of 10 m and still beats him by 5 s, what will be the speed of B?
a) 117/21 m/s
b) 317/21 m/s
c) 21 m/s
d) 2 m/s
Answer: (a)
Solution:
Time taken by A to cover 200 m = 200/2 = 100 s
∴ B covers (200 – 10) = 190 m in (100 + 5) = 105 s
∴ B’s speed = 190/105 = 38/21 = 117/21 m/s
21. The salaries of A, B, C are in the ratio 2:3:5. If the increments
of 15%, 10% and 20% are allowed respectively in their salaries,
then what will be the new ratio of their salaries?
a) 3:3:10
b) 10:11:20
c) 23:33:60
d) Cannot be determined
Answer: (c)
Solution:
Let A = 2k, B = 3k and C = 5k
A's new salary = 115/100 of 2k = 23/10 k
B's new salary = 110/100 of 3k = 33/10 k
C's new salary = 120/100 of 5k = 6k
New ratio = 23k/10: 33k/10: 6k = 23:33:60
16
22. If y: 5x2 + 3, then the tangent at x = 0, y = 3
a) passes through x = 0, y = 0
b) has a slope of + 1
c) is parallel to the x- axis
d) has a slope of – 1
Answer: (c)
Solution:
y = 5x2 + 3
dy = 10x
dx
x=0
then dy = 0
dx
∴ There is no slope, it is parallel to the x-axis
23. The least number of complete years in which a sum of money
put out at 20% compound interest will be more than doubled is:
a) 3
b) 4
c) 5
d) 6
Answer: (b)
Solution:
P �1 + 12000� > 2P ⟹ �65� > 2.
Now, �65 × 6 × 6 × 56� > 2
5 5
So, n = 4 years
17
24. A foundry has a fixed daily cost of Rs 50,000 whenever it
operates and a variable cost of Rs. 800Q, where Q is the daily
production in tonnes. What is the cost of production in Rs per
tonne for a daily production of 100 tonnes?
a) 1300
b) 1400
c) 1200
d) 1600
Answer: (a)
Solution:
Total cost for producing Q tones = 50000 + 800 Q
Cost of production per ton = 50000 + 800 Q = 50000 + 800
Q Q
Cost of production per ton for a daily production if 100 tonnes
= 50000 + 800 = 1300
Q
25. Find the value of 3− 3 Where a= 343, by b = 113
2+ + 2
a) 220
b) 225
c) 230
d) 235
Answer: (c)
Solution:
Given expression = 3− 3
2+ + 2
Where
18
a = 343, by b = 113
= (a – b) = (343 – 113) = 230.
26. In any given year, the probability of an earthquake greater than
Magnitude 6 occurring in the Garhwal Himalayas is 0.04. The
average time between successive occurrences of such
earthquakes is ____ year?
a) 22 years
b) 25 years
c) 24 years
d) 23 years
Answer: (b)
Solution:
Average Time Interval (T) = 1 (P) = 1 = 25 years
Probabilit 0.04
27. �2116�−32 ÷ �217�−34 = ?
a) 3/4
b) 2/3
c) 4/9
d) 1/8
Answer: (c)
Solution:
�2116�−32 ÷ �217�−43 = 2 ÷ 4 = 2
(216)3 (27)3 (216)3
4
(27)3
= 2 = 6�3 × 23� = 62 = 36 = 4
3�3 × 43� 34 81 9
�63�3
(33)43
19
28. The speed of A and B are in the ratio 3:4. A takes 20 minutes
more than B to reach a destination. Time in which A reach the
destination?
a) 1(1/3) hours
b) 2 hours
c) 2(2/3) hours
d) 1(2/3) hours
Answer: (a)
Solution:
Ratio of speed = 3:4;
Ratio of time taken = 4:3 (As Speed ∝ (1/Time), When distance
remains constant.)
Let time taken by A and B be 4x and 3x hour respectively.
Then,
4x – 3x = 20/60;
Or, x = 1/3;
Hence, time taken by A = 4x
hours = 4*1/3 = 1(1/3) hours.
29. Two guns are fired from the same place at an interval of 6
minutes. A person approaching the place observes that 5
minutes 52 seconds have elapsed between the hearings of the
sound of the two guns. If the velocity of the sound is 330 m/sec,
the man was approaching that place at what speed (in km/h)?
a) 24
b) 27
20
c) 30
d) 36
Answer: (b)
Solution:
Difference of time = 6 min - 5 mins. 52 secs = 8 secs.
Distance covered by man in 5 mins. 52 secs = distance covered
by sound in 8 secs = 330*8 = 2640 m.
Hence, speed of man = 2640/5 min. 52 secs;
= 2640 m/352 secs = 2640*18/352*5 kmph
= 27 kmph.
30. Two forest officials in their respective division were involved
in the harvesting of tendu leave. One division had an average
output of 21 tons from a hectares and other division, which had
12 hectares of land less, dedicated to tendu leaves, and got 25
tons of tendu from a hectare. As a result, the second division
harvested 300 tons of tendu leaves more than the first. How
many tons of tendu leaves did the first division harvest.
a) 3150
b) 3450
c) 3500
d) 3600
Answer: (a)
Solution:
Let First division have x hectare of the land then second division
will have (x – 12) hectare of land.
21
Now according to the question,
25 *(x – 12) = 21x + 300
Or, x = 150 hectare.
Hence, the first division harvest 3150 tons tendu leaves.
22
Data Loading...