MAT PO CAT

# Quantitative Aptitude

## Sample Paper 2

Question 1. A club consists of members whose ages are in A.P., the common difference being 3 months. If the youngest member of the club is just 7 years old and the sum of the ages of all the members is 250 years, then the numbers of members in the club are:

1. 15
2. 20
3. 25
4. 30

Answer: This can be calculated by using the SUM formula for AP.
S = (n/2)[2a + (n – 1)d]
Or, 250 = (n/2)[2 xx 7 + (n – 1) 0.25]
Or, 500 = n[14 + (n – 1)0.25]
Or, n(14 + 0.25n – 0.25) = 500
Or, 0.25n^2 + 14n – 0.25n = 500
Or, n^2 + 56n – n = 2000
Or, n^2 + 55n – 2000 = 0
Or, n^2 + 80n – 25n – 2000 = 0
Or, n(n + 80) – 25(n + 80) = 0
Or, (n – 25)(n + 80) = 0
So, n = 25 OR n = - 80
Discarding the negative value, we have n = 25
So, option C is the right answer.

Question 2. A metal cube of edge 12 cm is melted and formed into three smaller cubes. If the edges of two smaller cubes are 6cm and 8cm, find the edge of the third smaller cube.

1. 10cm
2. 14cm
3. 12cm
4. 16cm

Answer: The edge of the third cube will always be less than 12 cm. So, option ‘A’ is the right answer.
You can cross check the following: 12^3 = 6^2 + 8^3 + 10^3
Or, 1728 = 216 + 512 + 1000

Question 3. A well has to be dug out that is to be 22.5 m deep and of diameter 7m. Find the cost of plastering the inner curved surface at Rs. 3 per sq. meter.

1. Rs. 1465
2. Rs. 1485
3. Rs. 1475
4. Rs. 1495

Answer: In this case, you need to multiply the diameter and depth by Pi. Since, Pi = 22/7 and 22 (numerator in Pi) is divisible by 11 so, product should be divisible by 4. Only one option, i.e. ‘B’ is divisible by 11 and hence, ‘B’ is the right answer.

Conventional method:

Curved surface area of cylinder = π xx\ d \xx\ h
= (22)/7 xx 7 xx 22.5 = 495 sq cm
Now, cost = Rate xx CSA = 3 xx 495 = Rs. 1485

Question 4. Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 meters per second into a cylindrical tank, the radius of whose base is 60 cm. By how much will the level of water rise in 30 minutes?

1. 2m
2. 4m
3. 3m
4. 5m

Answer: Volume of water which flows through the pipe in 30 minutes will be same as volume of water filled in the cylindrical tank in 30 minutes.
Volume of water through pipe in 1 second = π r^2\h
= π 1^2 xx 600 = 600 π cubic cm
Volume of water in 30 minutes = 600 xx 60 xx 30 second = 1080000 π cubic cm
Or, 60^2 xx\ h = 1080000
Or, h = (1080000)/(3600) = 300 cm = 3 m
Hence, C is the right answer.

Question 5. A ladder 15 m long reaches a window which is 9 m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 12m high. Find the width of the street.

1. 19m
2. 21m
3. 20m
4. 22m

Answer: The ladder becomes hypotenuse of the right angled triangle for which values of perpendicular and hypotenuse are known. Here; ratio of 15/9 = 5:3. This shows that the base of the right angled triangle should be part of Pythagoras Triplet. Thus, three sides should be in ratio 5:3:4.
Base = 12
When ladder is turned to the other side of the street, perpendicular becomes 12. Hence, base = 9 (to make Pythagoras Triplet.)
So, width of street = 12 + 9 = 21 m
Hence, B is the right answer.

Question 6. The horizontal distance between two trees of different heights is 60m. The angle of depression of the top of the first tree when seen from the top of the second tree is 45°. If the height of the second tree is 80 m, find the height of the first tree.

1. 20m
2. 22m
3. 18m
4. 16m

Answer: Tan 45° = p/b = 1
This means that difference between heights of two trees = 60 m
So, height of first tree = 80 – 60 = 20 m
Hence, A is the right answer.

Question 7. An aero plane flying at a height of 300 metres above the ground passes vertically above another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 60° and 45°,respectively.Thenthe height of the lower plane from the ground, in metres, is:

1. 100√3
2. 50
3. 100/√3
4. 150(√3 + 1)

Answer: In case of first aeroplane; perpendicular p = 300 m and θ = 60°
We know; tan 60° = p/b
Or, sqrt3 = (300)/b
Or, b = (300)/sqrt3 = 100sqrt3 m

In case of second aeroplane; b = 100sqrt3 and θ = 45°
We know; tan 45°= p/b
Or, 1 = p/(100sqrt3)
Or, p = 100sqrt3 m
Hence, A is the correct answer.

Question 8. A person standing on the bank of river observes that the angle of elevation of the top of a tree on the opposite bank of the river is 60° and when he retires 40 metres away from the tree the angle of elevation becomes 30°.The breadth of the river is:

1. 40m
2. 20m
3. 30m
4. 60m

Answer: Here; AC = width of river
In ΔABC; tan 60° = (AB)/(BC)
Or, sqrt3 = (AB)/(BC)
Or, AB = BCsqrt3 ……. (1)

In ΔABD; tan 30° = (AB)/(BD)
Or, 1/sqrt3 = (AB)/(BD)
Or, AB = BDsqrt3 ..... (2)

From equations (1) and (2);
BCsqrt3 = (BD)/sqrt3
Or, 3BC = BD
As per question; BD = BC + 40
So, 3BC = BC + 40
Or, 2BC = 40
Or, BC = 20 m
Hence, B is the right answer.

Question 9. A circular grassy plot of land, 42m in diameter, has a path 3.5 m wide running around it on the outside. The cost of gravelling the path at Rs.4 per square metre is:

1. Rs.1002
2. Rs.3002
3. Rs.2002
4. Rs.1802

Answer: As 11 is a factor of Pi so 11 shall be a factor of area of circle. By this logic, 11 shall be a factor of final cost. Here, only one option is divisible by 11. So, option C is the right answer.

Conventional Method:

Area of path = Area of bigger circle – Area of smaller circle
= π R^2 - π r^2
= π (R^2 - r^2
= π (24.5^2 - 21^2
= π (24.5 + 21)(24.5 – 21)
= π xx 45.5 xx 3.5
(22)/7 xx 45.5 xx 3.5 = 500.5 sq m
Now, cost = Area xx Rate = 500.5 xx 4 = Rs 2002

Question 10. A plot of land in the form of a rectangle has a dimension 240m X 180m.A drain outlet 10m wide is drug all around it(on the outside) and the earth dug all around it is spread over the plot, increasing its surface level by 25cm.The depth of the drain outlet is:

1. 1.225m
2. 1.229m
3. 1.227m
4. 1.223m

Answer: Area of plot = 240 xx 180 = 43200 sq m
Area covered by plot and drain = (240 + 20) xx (180 + 20) = 260 xx 200 = 52000 sq m
Area of drain = 52000 – 43200 = 8800 sq m
Volume of soil spread over the plot = 43200 xx 0.25 = 10800
Depth of drain = volume/area
= (10800)/(8800) = 1.227 m
Option C is the right answer.