Quantitative Aptitude (QA) Test

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Quantitative Aptitude (QA) Test - 38

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Question 1
3 / -1

Najib and Adil together can do a piece of work in 4 days. Adil and Eoin together can do the same work in 6 days. Najib and Adil together worked for 2 days and then Eoin did the work alone for 3 days. After that Eoin left and Najib and Adil together completed the rest of the work in 1 day. In how many days could all the three together have completed the whole work?

A
3 days

B
4 days

C
5 days

D
6 days

Solution

Let Najib, Adil, and Eoin be able to do the work alone in 1/N, 1/A, 1/E days.

Since Najib and Adil together do one-fourth work in a day. In three days, they do three-fourth of the work. So, Eoin does one-fourth of the work in three days. Hence Eoin does the whole work alone in 12 days. 1/E =12.

Hence

N+ A = 0.25.......... (1)

A+ E =1/6............ (2)

E = 1/12................. (3)

Solving the three equations we have

1/N= 6(1/A)=12

Hence amount of work done by the three of them together in a day = 1/12+1/6+1/12=1/3

Time taken by three of them together to complete the whole work = 3 days.
Question 2
3 / -1

Mistry can fill a large tank from a river using a bucket in 60 minutes. Saha is 25% more efficient than Mistry and Datta is 20% more efficient than Saha. Mistry and Saha are allotted for filling up the tank. Datta, being a wicked kid, starts emptying the tank. Assuming filling the tank requires equal time and effort as that for emptying, what is the time required to fill the tank?

A
60 minutes

B
15 minutes

C
120 minutes

D
80 minutes

Solution

Given, Mistry can fill the tank in 60 minutes

So Saha can fill the tank in (60 X 100)/125 minutes = 48 minutes.

So Datta can empty (or fill) the tank in (48 X 100)/120 = 40 minutes.

Now, amount of the tank filled in 1 minute = (1/60+ 1/48 - 1/40) part of the tank = 1/80 part of the tank.

Therefore, the tank will be filled in 80 minutes.
Question 3
3 / -1

If a, b, c are in GP and log_ca, log_bc, log_ab, are in AP, then the common difference of AP is

A
3

B
3/2

C
1/2

D
2/3

Solution

Let log_ca,=x and log_bc, log_ab,=y and z respectively

So xyz=1 and 2y=x+z(AP)

Also a,b,c are in GP so b²=ac

Or, log_bb²= log_ba+log_bc

Or, 2=(1/z)+y

From above we have 2(2-1/z)=x+z

Or, x=4-(2/z)-z

Putting the value of x

We get (4-2/z-z)(2-1/z)z=1

Or, 2z²-7z+2=0 (i)

Now,

Common difference = z-y = z-(2-1/z)=(z²-2z+1)/z

Using (i) we get (2z²-4z+2)/2z= 3z/2z = 3/2
Question 4
3 / -1

From a collection of 12 bulbs of which 6 are defective 3 bulbs are chosen at random for three sockets in a room. Find the probability that the room is lighted if one bulb is sufficient to light the room.

A
10/11

B
10/13

C
10/12

D
10/14

Solution

Total no. of possible outcomes = ¹²C₃ = 220. Number of ways of selecting three defective bulbs from 6 defective bulbs = ⁶C₃=20. Probability that the room is not lighted = 20/220=1/11. Probability that the room is lighted= 1-1/11=10/11.
Question 5
3 / -1

A man borrows a certain sum of money and pays it back in 2 yrs in two equal installments. If C.I. is reckoned at 5% per annum and he pays back annually Rs. 441, what sum did he borrow?

A
Rs.820

B
Rs.800

C
Rs.882

D
Rs.850

Solution

Money borrowed that becomes 441 in 1 year = 441 / (1 + 5/100) = 441 x 20 /21 = 420

Money borrowed that becomes 441 in 2 years = 441 / (1 + 5/100)² = 441 x (20/21) x (20/21) = 400

Total Money = 420 + 400 = 820
Question 6
3 / -1

If M is the mid-point of QR, find the length of QR.

A
15 units

B
10 units

C
12 units

D
10√7 units

Solution

By Appollonius theorem,

6² + 8² = 2 * (5)² + 2 * (QM)²

Therefore, QM = 5 units

M is the midpoint of QR.

So, length of QR = 10 units
Question 7
3 / -1

A
2008 - 1/2008

B
2007 - 1/2007

C
2007 - 1/2008

D
2008 - 1/2007

Solution
Question 8
3 / -1

The sum of the areas of two circles which touch each other externally is 153π. If the sum of their radii is 15, find the ratio of the larger to the smaller radius

A
4 : 1

B
2 : 1

C
3 : 1

D
None of these

Solution
Question 9
3 / -1

A merchant bought some goods at a discount of 20% of the list price. He wants to mark them at such a price that he can give a discount of 20% of the marked price and still make a profit of 20% of the selling price. The per cent of the list price of which he should mark them is:

A
20

B
100

C
125

D
80

Solution

Let the list price be P

Merchant bought it at a discount of 20% i.e. cost price = 0.8P

Profit = 20% of selling price

Selling price = cost price + profit

Therefore, Selling price = cost price + 20% of selling price

SP = CP + 0.2 SP

0.8 SP = CP

Cost price = 0.8 × selling price

Hence,

Selling price = P

He gives 20% on marked price

Therefore,

Selling price = 0.8 × marked price

Marked price = P/0.8 = 1.25 P = 1.25 list price

Hence, option 3
Question 10
3 / -1

Find the orthocenter of the triangle formed by the points (0,0), (8,0) and (4,6).

A
(4, 8)

B
(4, 8/3)

C
(42/13, 28/13)

D
(-4, 8/3)

Solution

As we know that orthocenter is the point of the intersection of the perpendiculars drawn from the vertices to the opposite sides of the triangle.

In the above figure AD is perpendicular to BC and BE is perpendicular to AC.

O is the orthocenter of the triangle ABC.

Now we have to find out the equation of AD and BE.

Slope of BC =-(6/4) = -(3/2)

So the slope of AD will be (2/3) and it is passing through (0,0) hence the equation of AD is , 2x - 3y = 0 ....(1)

Similarly slope of BE is - (2/3) and it is passing through (8,0), hence the equation of BE is, 3y = -2( x - 8)

⇒ 3y = -2x + 16

⇒ 2x + 3y = 16 .....(2)

Now the intersection of (1) and (2) will give us the co-ordinates of orthocenter.

On solving (1) and (2) we will get (x, y) = (4, 8/3)