# Quantitative Aptitude Test - 5 - PDF Flipbook

Quantitative Aptitude Test - 5

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Quantitative

Aptitude

Test-05Solutions

QUANTITATIVE APTITUDE

1. Ajay’s age is 125% of what it was 10 years ago, but 83⅓% of

what it will be after ten years. What is his present age?

a) 45 years

b) 50 years

c) 55 years

d) 60 years

Answer: (b)

Solution:

Let the present age be x years.

Then, 125% of (x – 10) = x

And 831/3% of (x + 10) = x

Therefore, 125% of (x – 10) = 831/3% of (x + 10)

5x/12 = 250/12

x = 50 years

2. The smallest angle of a triangle is equal to two thirds of the

smallest angle of a quadrilateral. The ratio between the angles of

the quadrilateral is 3:4:5:6. The largest angle of the triangle is

twice its smallest angle. What is the sum, in degrees, of the

second largest angle of the triangle and the largest angle of the

quadrilateral?

a) 180o

b) 110o

c) 130o

d) 150o

1

Answer: (a)

The ratio of angles of quadrilateral = 3 : 4 : 5: 6

Sum of angles of quadrilateral = 3600

18 k = 360o ⟹ k = 20o

So smallest angle of quadrilateral = 3k = 60o

Smallest angle of triangle = 2 (smallest angle of quadrilateral)

3

= 23(60o) = 40o

Largest angle of triangle = 2 × (smallest angle of triangle)

= 2 × 40o = 80o

∴ second largest angle of triangle + largest angle of quadrilateral

= 180 – (40 + 80) + 6k = 60o + 120o

= 180o

3. 3 years ago the average of a family of 5 members was 17 years.

A baby having been born, the average age of the family is the

same today. The present age of the baby is:

a) 1 years

b) 3/2 years

c) 2 years

d) 3 years

Answer: (c)

Solution:

Let age of the baby is x.

3 years ago total age of the family = 5 × 17 = 85 years.

Total age of the 5 member at present time,

2

= 85 + 3 × 5 = 100 years

Total age of the family at present time including baby,

= 100 + X.

The average of the family including baby at present time,

= 17 years.

(100 + X)/6 = 17

100 + X = 102

X = 102 – 100 = 2 years.

4. The second day of a month is Sunday, what will be the last day

of the next month which has 31 days?

a) Friday

b) Saturday

c) Sunday

d) Can't be determined

Answer: (d)

Solution:

We cannot find out the answer because the number of days of

the current month is not given.

5. One percent of the people of country X are taller than 6 ft. Two

percent of the people of country Y are taller than 6 ft. There are

thrice as many people in country X as in country y. Taking both

countries together, what is the percentage of people taller than 6

ft?

a) 3.0

b) 2.5

3

c) 1.5

d) 1.25

Answer: (d)

Solution:

y = 3x

XY

Population 3P P

Taller than 6ft 0.03P 0.02P

Hence percentage of people taller than 6ft

= 0.03P + 0.02P × 100 = 0.05P × 100 = 1.25%

3P + P 4P

6. If 10 programmers, working 16 hours a day, can finish the

coding of a project in 25 days, then how many days will be

taken by 12 software engineers, working 5 hours a day to finish

the coding of the project? It is known that 3 software engineers

code as much as 2 programmers in the same time.

a) 25 days

b) 50 days

c) 75 days

d) 100 days

Answer: (d)

Solution:

Number Hrs Days

Programmer 10 16 25

Engineers 12 5x

4

Also, 3E = 2P i.e. 12 E = 8P ⇒ 10 P × 16 × 25 = 12 E × 5 × x.

⇒10 P × 16 × 25 = 8P × 5 × x. Number of days = x = 100 days.

Hence, answer is 100 days.

7. At what time between 6 and 7 will the hands be perpendicular?

a) 481/11 minutes past 6 and 164/11 minutes past 6

b) 48 minutes past 6 and 163/11 minutes past 6

c) 491/11 minutes past 6 and 164/11 minutes past 6

d) 482/11 minutes past 6 and 163/11 minutes past 6

Answer: (c)

Solution:

The traditional way.

At 5 o' clock, the hands are 30 minute spaces apart

Hence minute hand needs to gain 15 more minute spaces or 45

more minute spaces so that the hands will be in right angles

(90Â° between them)

We know that 55 min spaces are gained by minute hand (with

respect to hour hand) in 60 min

Hence time taken for gaining 15 minute spaces by minute hand

= 60 ∙ 15 = 12 ∙ 15 = 180 = 16141 minutes

55 11 11

Hence time taken for gaining 45-minute spaces by minute hand

= 60 ∙ 45 = 12 ∙ 45 = 540 = 49111 minutes

55 11 11

Hence the hands will be perpendicular at 16141 minutes past 6

and 49111 minutes past 6

5

8. Which greatest possible length can be used to measure exactly

15-meter 75 cm, 11-meter 25 cm and 7-meter 65 cm?

a) 255 cm

b) 265 cm

c) 275 cm

d) 285 cm

Answer: (a)

Solution:

Convert first all terms into cm.

I.e. 1575 cm, 1125cm, 765cm.

Now whenever we need to calculate this type of question, we

need to find the HCF.

HCF of above terms is 255.

9. 10% of the population in a town is HIV+. A new diagnostic kit

for HIV detection is available; this kit correctly identifies HIV–

individuals 95% of the time, and HIV individuals 89% of the

time. A particular patient is tested using this kit and is found to

be positive. The probability that the individual is actually

positive is___.

Answer: 0.48 to 0.49

The probability of selecting HIV+ individual in the population =

10%

The probability of selecting HIV+ individual in the population =

90%

6

The probability HIV detection kit correctly identifies HIV+

patient = 95%

The probability HIV detection kit correctly identifies HIV–

patient = 89%

The required probability = �11000×19050� = 95 = 0.486

�11000×19050�+�19000×11010� 194

10. The parameter of a square is equal to the perimeter of a

rectangle of length 16 cm and breadth 14 cm. Find the

circumference of a semicircle whose diameter is equal to the

side of the square. (Round off your answer to two decimal

places)

a) 77.14 cm

b) 47.14 cm

c) 84.92 cm

d) 23.57cm

Answer: (d)

Solution:

Let the side of the square be a cm.

Parameter of the rectangle = 2(16 + 14) = 60 cm Parameter of

the square = 60 cm

i.e. 4a = 60

A = 15

Diameter of the semicircle = 15 cm

Circumference of the semicircle = 1/2(∏)(15) = 1/2(22/7)(15)

= 330/14 = 23.57 cm to two decimal places

7

11. pqr is a three-digit natural number so that pqr = p! + q! + r!.

What is the value of (q + r)p?

a) 1296

b) 3125

c) 19683

d) 9

Answer: (d)

Solution:

100p + 10q + r = p! + q! + r!

Now, 6! = 720 and 7! = 5040.

If 7 is one of digits, then the sum of the factorials becomes four

digits’ number or more.

Hence the numbers 7, 8, 9 can be neglected.

Consider 6! = 720.

But 7 cannot be there in hundred's place.

Hence, we can neglect 6 also.

Now, 5! = 120, 4! = 24,

3! = 6, 2! = 2 and 1! = 1.

To get a three-digit number, 5 has to be present in the number.

But 5 cannot be in hundreds place as then the number greater

than 500 which cannot be obtained as the sum of factorial.

Also, maximum possible number is 5! + 4! + 3! = 150

Also 'p' cannot be zero as it is a three-digit number.

Hence p = 1.

Then different possible cases are 154, 153, 152, 125, 135, 145

8

From this only 145 satisfies the condition

Thus, (4 + 5) = 9.

12. A, B and C enter into partnership by making investments in the

ratio 3:5:7. After a year, C invests another Rs.337600 while A

withdraw Rs.45600. The ratio of investments then changes to

24:59:167. How much did A invest initially?

a) Rs.45600

b) Rs.96000

c) Rs.141600

d) None of these

Answer: (c)

Solution:

Let initial investments be 3x, 5x and 7x rupees

= (3x – 45600):5x:(7x + 337600)

= 24:59:167

3x − 45600/5x = 24/ or x = 47200

Initial investment of A = Rs. (47200 × 3) = Rs.141600

13. The cost of packaging of the mangoes is 40% the cost of fresh

mangoes themselves. The cost of mangoes increased by 30% but

the cost of packaging decreased by 50%, then the percentage

change of the cost of packed mangoes, if the cost of packed

mangoes is equal to the sum of the cost of fresh mangoes and

cost of packaging:

a) 14.17%

b) 7.14%

9

c) 8.87%

d) 6.66%

Answer: (b)

Solution:

Cost of fresh mangoes + Cost of packaging = Total cost.

Let initial Cost of fresh, mangoes = 100.

Then, packaging cost = 40.

Thus, Initial total cost = 100 + 40 = 140

After increasing in cost of fresh mangoes 30%,

Cost of fresh mangoes = 130

And cost of packing go down by 50 % so,

Cost of packing = 20.

Now Total cost = 130 + 20 = 150.

Increased cost = 150 – 140 = 10.

% increased = (10 × 100) /140 = 7.14%.

14. How many four-digit numbers can be formed with the 10 digits

0, 1, 2, .... 9 if no number can start with 0 and if repetitions are

not allowed?

a) 3268

b) 4536

c) 3657

d) 5894

Answer: (b)

10

Solution:

⟹ Four-digit number without having repeated any digit are of

four kinds so, there are 10P4 numbers of this kind

⟹ Those numbers that does contain zero in the first digit so,

there are 9P3 numbers of this kind.

∴ Hence, the total number of four-digit numbers are without

start with ‘0’ and repetitions are not allowed

= 10P4 – 9P3 = 10! − 9!

(10−4)! (9−3)!

= 10 × 9 × 8 × 7 – 9 × 8 × 7

= 5040 – 504 = 4536

15. In how many ways can you rearrange the word JUMBLE such

that the rearranged word starts with a vowel?

a) 120

b) 240

c) 360

d) 60

Answer: (b)

Solution:

JUMBLE is a six-lettered word. Since the rearranged word has

to start with a vowel, the first letter can be either U or E. The

balance 5 letters can be arranged in 5P5 or 5! Ways. Total

number of words

= 2 × 5! = 240.

11

16. The given question is followed by two statements; select the

most appropriate option that solves the question Capacity of a

solution tank A is 70% of the capacity of tank B. How many

gallons of solution are in tank A and tank B?

Statements:

Tank A is 80% full and tank B is 40% full

Tank A if full contains 14,000 gallons of solution.

a) Statement I alone is sufficient

b) Statement II alone is sufficient

c) Either statement I or II alone is sufficient.

d) Both the statements I and II together are sufficient

Answer: (d)

Solution:

Capacity of tank A = 70% of the capacity of tank B

Statements:

I. Tank A is 80% full and tank B is 40% full

II. Tank A if full contains 14,000 gallons of solution

Tank A 100% = 1400

1% = 14 gallons

∴ Solution in the tank A = 80% = 80 × 14 = 1120 gallons

Capacity of tank A = 70% of capacity of tank B

1400 gallons = 70%

12

1% = 20 gallons

Now, solution in the tank B = 40%

= 40 × 20 = 800 gallons

∴ Both the statements I and II together are sufficient.

17. Out of sixty students, there are 14 who are taking Economics

and 29 who are taking Calculus. What is the probability that a

randomly chosen student from this group is taking only the

Calculus class?

a) 8/15

b) 7/15

c) 1/15

d) 4/15

Answer: (b)

Solution:

Given total students in the class = 60

Students who are taking Economics = 24 and

Students who are taking Calculus = 32

Students who are taking both subjects = 60 – (24 + 32)

= 60 – 56 = 4

Students who are taking calculus only = 32 – 4 = 28

probability that a randomly chosen student from this group is

taking only the Calculus class = 28/60 = 7/15.

13

18. A sells a bicycle to B at a profit of 20%. B sells it to C at a

profit of 25%. If C pays Rs. 225 for it, the cost price of the

bicycle for A is:

a) Rs. 110

b) Rs. 120

c) Rs. 125

d) Rs. 150

Answer: (b)

Solution:

125% of 120% of A = 225

125/100 × 120/100 × A = 225

A = 225 × 2/3 = 150.

19. Using the factoring method, solve the quadratic equation: x2 +

4x + 4 = 0

a) 0 and 1

b) 1 and 2

c) 2

d) – 2

Answer: (d)

Solution:

x2 + 4x + 4 = 0

x2 + 2x + 2x + 4 = 0

x(x + 2) + 2(x + 2) = 0

(x + 2)(x + 2) = 0

(x + 2)2 = 0 ⇒ x = – 2

14

20. A and B run a km race. If A Gives B a start of 50 m, A wins by

14 secs and if A Gives B a start of 22 secs, B wins by 20 m. The

time taken by A to run a km is

a) 125 sec

b) 120 sec

c) 105 sec

d) 100 sec

Answer: (d)

Solution:

Let time taken by A to run 1000 m = a

and time taken by B to run 1000 m = b

If A gives B a start of 50 m, A wins by 14 sec

distance travelled by A = 1000

distance travelled by B = (1000 – 50) = 950

Time taken by B to run 950 m - Time taken by A to run 1000 m

= 14 sec

950 b – a = 14

1000

95 b – a = 14 ----- (Equation 1)

100

If A gives B a start of 22 sec, B wins by 20 m

i.e., A starts 22 seconds after B starts from the same starting

point.

i.e., here B runs for b seconds where A runs for (b – 22) seconds

⇒ Distance Covered by B in b seconds – Distance Covered by

A in (b – 22) seconds = 20

15

⟹ 1000 – 1000 × (b – 22) = 20

1000 × (b – 22) = 980

1000b – 22000 = 980a

100b – 2200 = 98a

50b – 1100 = 49a

50b – 49a = 1100 ------ (Equation 2)

(Equation 1) × 49

⟹ 951×0049b – 49a = 14 × 49

⟹ 46.55b – 49a = 686 ------- (Equation 3)

(Equation 2) – (Equation 3) ⟹ 3.45b = 414

⟹ b = 414 = 120 seconds

3.45

Substituting this value of b in Equation 1

⟹ 95 × 120 − a = 14

100

⟹ 114 – a = 14

⟹ a = 114 – 14 = 100

i.e. The time taken by A to run a km = 100 seconds

21. If 2/3 of A = 75% of B = 0.6 of C, then A: B: C is

a) 2:3:3

b) 3:4:5

c) 4:5:6

d) 9:8:10

Answer: (d)

16

Solution:

According to the question,

Or, (2A/3) = (75B/100) = (C × 6/10);

Above relation gives;

A × 2/3 = B × 3/4;

→ A/B = 9/8;

And, B × 3/4 = C × 3/5;

→ B/C = 8:10;

Thus, A:B:C = 9:8:10.

22. In a process, the number of cycles to failure decreases

exponentially with an increase in load. At a load of 80 units, it

takes 100 cycles for failure. When the load is halved, it takes

10000 cycles for failure. The load for which the failure will

happen in 5000 cycles is___.

a) 40.00

b) 46.02

c) 60.01

d) 92.02

Answer: (b)

Solution:

Load failure cycles

80 100

40 10000

There is one failure 5000 cycles load must be between 80 and 40

∴ 46.02

17

23. A sum of money lent at compound interest for 2 years at 20%

per annum would fetch Rs.482 more, if the interest was payable

half yearly than if it was payable annually. The sum is

a) 10000

b) 20000

c) 40000

d) 50000

Answer: (b)

Solution:

Let sum = Rs.x

C.I. when compounded half yearly = � �1 + 11000�4 − � = 4641

10000

C.I when compounded annually = � �12000�2 − � = 11

25

4641 x – 11 x = 482

10000 25

⟹ x = 20000

24. Two finance companies, P and Q, declared fixed annual rates

of interest on the amounts invested with them. The rates of

interest offered by these companies may differ from year to

year. Year-wise annual rates of interest offered by these

companies are shown by the line graph provided below.

18

If the amounts invested in the companies, P and Q, in 2006 are

in the ratio 8:9, then the amounts received after one year as

interests from company’s P and Q would be in the ratio:

a) 2:3

b) 3:4

c) 6:7

d) 4:3

Answer: (d)

Solution:

The amounts invested in the companies of, P and Q in 2006 = 8

:9

The rate of invested of company ‘P’ in 2006 = 6%

The rate of invested of company ‘Q’ in 2006 = 4%

The amounts received after one year by P and Q companies in

2006 year

PQ

6% of 8 : 4% of 9

6 × 8 : 4 × 9

100 100

4 :3

25. If x = 65, Find the value of x2+y2

y x2−y2

a) 61/11

b) 51/9

c) 41/11

d) 65/11

19

Answer: (a)

Solution:

= 2+ 2 22+1

22−1

2− 2

= = =� �2+ 1 �65�2+ 1 61

�65�2− 1 25

� �2− 1 11

25

= 61 × 25 = 61

25 11 11

26. A square pyramid has a base perimeter x, and the slant height

is half of the perimeter. What is the lateral surface area of the

pyramid?

a) x2

b) 0.75 x2

c) 0.50 x2

d) 0.25 x2

Answer: (d)

Solution:

Base perimeter of square pyramid = x = p

Slant height = = l

2

Lateral surface area of pyramid

20

= 1 × Base perimeter (p) × Slant height (l)

2

= 1 × p × l

2 2

= 1 × x ×

2 2

= 2 = 0.25 x2

4

27. If x= (8 + 3√7), what is the value of �√x − √1x�?

a) √12

b) 4

c) 2

d) √14

Answer: (d)

Solution:

(a – b)2 = a2 – 2ab + b2

a2 – b2 = (a – b)(a + b)

�√ − √1 �2= x – 2 + 1

= x + 1 – 2

= (8 + 3√7) + (8 1 − 2

+ 3√7)

= (8 + 3√7) + �8 + (8 − 3√7) − 2

3√7�(8 − 3√7)

= (8 + 3√7) + (8 − 3√7) − 2

82 − �3√7�2

= (8 + 3√7) + (8 − 3√7) − 2

64 − 63

= (8 + 3√7) + (8 − 3√7) − 2

1

21

= 8 + 3√7 + 8 – 3√7 – 2 = 14

We got that �√ − √1 �2 = 14

Hence, �√ − √1 � = √14

28. A runs 7/4 times as fast as B. If A Gives B a start of 300 m,

how far must the winning post be if both A and B have to end

the race at same time?

a) 1400 m

b) 700 m

c) 350 m

d) 210 m

Answer: (b)

Solution:

. A B Reason (ST = D)

Speed 7 4 Given

Time 4 7 Since, Speed ∝ 1/time

Distance 4 7 Distance ∝ time

Now,

7x – 4x = 300 (A runs 7x m and B runs 4x)

x = 100

7x = 7 × 100 = 700 m

Winning post is 700 m away, As A runs 700 m to complete the

race.

22

29. The wheel of an engine of 300 cm in circumference makes 10

revolutions in 6 seconds. What is the speed of the wheel (in

km/h)?

a) 18

b) 20

c) 27

d) 36

Answer: (a)

Solution:

Circumference = One revolution

Distance covered in 10 revolutions = 300 × 10 = 3000 cm = 30

m

30 meters are covered in 6 seconds.

Speed of the wheel = 30 /6 = 5 m/s

5 m/s = (5 × 18)/5 = 18 km/h. [To convert m/s to km/h, we use

to multiply by factor 18/5]

30. Two persons having different productivity of labour, working

together can reap a field in 2 days. If one-third of the field was

reaped by the first man and rest by the other one working

alternatively took 4 days. How long did it take for the faster

person to reap the whole field working alone?

a) 3

b) 6

c) 8

d) 12

23

Answer: (a)

Solution:

Total efficiency of two persons = 50% [As they complete work

in 2 days.]

First Person completes work = 1/3 = 33.33% [In 2 days]

Rest work will be completed by Second man = 2/3 = 66.66% [In

2 days]

So, efficiency of second person is greater.

Efficiency of second person = 66.66/2 = 33.33% per day

Then, Second person will complete whole work in,

= 100/33.33 = 3 days.

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