Aptitude model placement papers with solution - set 7

ANSWER: B. 130

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.

ANSWER: B. Sunday

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.

ANSWER: C. 2 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.

ANSWER: A. 400

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.

ANSWER: A. 6.25

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.

ANSWER: A. 4.04

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%

ANSWER: C. 4

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.

ANSWER: A. 20.68

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.

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.

ANSWER: A. 225m

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.