# Aptitude model placement papers with solution - set 7

## Aptitude model placement papers with solution - set 7

Q1. What is the angle between the hour and minute hand at 3:40?

A. 135
B. 130
C. 120
D. 125

Solution:

When minute hand is behind the hour hand, the angle between the minute hand and hour hand at M minutes past H hours is given by:

=> 30 * (H-M/5) + M/2

When minute hand is ahead of hour hand, the formula becomes:

=> 30* (M/5-H) – M/2

Applying the second formula as here the minutes hand is ahead of the hour hand.

=> 30* (40/5-3) – 40/2

=> 30* (8-3) – 20

=> 30*5 – 20

=> 150-20

=> 130

Thus the angle formed is 130 degrees.

Q2. 15th august 2010 was which day of the week?

A. Thursday
B. Sunday
C. Wednesday
D. Friday

Solution

15th August 2010 can be written as 2009 + days from 1st January 2010 to 15th August 2010.

=> Total number of odd days in 400 years = 0

Hence, total number of odd days in 2000 years = 0 (as 2000 is a perfect multiple of 400)

Odd in days in the period 2001-2009:
7 normal years + 2 leap yeas

=> (7*1) + (2*2) = 11

=> Odd days will be 11- (7*1) = 4

Days from January 1 to August 15 in 2010: 31+28+31+30+31+30+31+15

= 227 days.

= 32 weeks and 3 days, this gives additional 3 odd days.

=> Total odd days= 3+4=7

=> 7 odd days=1 week= 0 odd days

=> 0 odd days= Sunday

Thus, 15th August 2010 was a Sunday.

Q3. A boatman rows 96 km downstream in 8 hours with a stream speed of 4kmph. How much time will he take to cover 8km upstream?

A. 4 hours
B. 6 hour
C. 2 hours
D. 1 hours

Solution

Speed = distance/time

Speed downstream: 96 / 8 km/hr = 12kmph

Speed of stream = 4kmph

Effective speed of boat = (12-4) kmph
= 8kmph

Distance to be travelled upstream= 8 km

Speed upstream = boat speed-current speed

= 8-4 kmph

= 4 kmph

Time taken = distance/speed = 8/ 4 hours
=2 hours

Thus it will take 2 hours to go upstream.

Q4. What is the true discount if a banker’s discount on a bill due 4 months hence is 420 at 15%?

A. 400
B. 410
C. 390
D. 380

Solution

Using the formula:

True Discount = (Banker’s Discount * 100 * 100)+ Total Rate

=> True Discount = 420*100/100+15
= 400

Thus the true discount is Rs 100.

Q5. India has a run rate of 3.2 in the first 10 overs. What should be the run rate in remaining 40 overs to win a game of 282 runs?

A. 6.25
B. 5.5
C. 7.25
D. 8

Solution:

Runs scored in first 10 overs:

=> Rate*Overs= 3.2*10 = 32 runs

=> Required runs = 282-32 = 250

=> Left overs= 40

Required rate = runs/overs

= 250/40

= 6.25

=> 6.25 runs per over is needed to win the game.

Q6. While calculating the edge of a square, a worker makes an error of 2% in excess. What % error does he make in calculating area? (%)

A. 4.04
B. 4
C. 40..4
D. 4.004

Solution

Given Error = 2% while measuring the side of a square.

If the correct value of the side of square is 100, the measured value:

=> 100 + 2% *100

= 100+2=102

The area of square with edge 100 = side*side

=> 100*100

=> 10000

The area of square with side 102 = 102*102= 10404

Error in area calculation = 10404-1000 = 404

% error= (404/10000)*100
= 4.04%

Q7. The sum of ages of 5 children born at three year interval each is 50. How old is the youngest child?

A. 6
B. 5
C. 4
D. 3

Solution

Let the age of youngest child be X

According to the question,

X + (X+3) + (X+6) + (X+9) + (X+12) = 50

5X + 30= 50

5X = 50-30

5X = 20

X = 20/5 = 4

Age of youngest child is 4 years.

Q8. A pipe is 30 m long and is 45% longer than another pipe. Find the length of the other pipe.

A. 20.68
B. 20
C. 20.12
D. 20.5

Solution

Let length of other pipe be X
According to question,
30 = 45/100 X + X
30 = 0.45X + X
30 = 1.45 X
X= 30/1.45
X = 20.68m

Thus the length of the other pipe is 20.68 metres.

Q9. A man cycles at a rate of 4kmph. He cycles for 45 minutes and then takes a 15 minute break. Determine the total time taken to cover 12 kms by him.

A. 4 hours 35 minutes
B. 4 hours
C. 3 hours
D. 3 hours 45 minutes

ANSWER: D. 3 hours 45 minutes

Solution

Speed of man = 4kmph

Time for which he cycled = 45 minutes.

Distance covered = speed*time

=> 4*45/60

=> 3km

Total time to cover 3 km = 45 minutes + 15 minutes = 60 minutes = 1 hour

Thus 3 km is covered in 1 hour

Required distance to be covered = 12 km

Beginning from origin point, he covers 9 km in 3 hours taking 15 minutes rest after every 45 minutes. To complete 12 kms he goes another 3 km in 45 minutes.

Thus, total time he takes is 3 hours 45 minutes.

Q10. In a race, A beats B by 15 metres and C by 29 metres. If B and C run over the course together, B wins by 15 metres. What is the length of the course?

A. 225m
B. 120m
C. 220m
D. 160m

Solution:

Let X be the length of the course

according to the question,
A beats B by 15 metres and C by 29 metres

Applying this,
If A runs X metres, B runs (X-15) and C runs ( X-29) metres.

B and C together run, B runs X metres and C runs (X-15)

=> If B runs 1 metre, C runs X-15/X

=> B runs X-15 m, C runs (X-15/x) * (X-15)

APPLYING THIS,
X-29 = (X-15)(X-15) / X

Solving,
X2 -29X= X2 – 30 X + 225
X = 225

Thus, length of the course is 225 metres.