Quantitative Aptitude - Arithmetic Ability Questions

Let S be the sample space.
Here n(S)= 23 = 8
Let E be the event of getting atmost two heads. Then,
n(E) = {(H,T,T), (T,H,T), (T,T,H), (H,H,T), (T,H,H), (H,T,H)}
Required probability = n(E)/n(S) = 6/8 = 3/4.

As these three strategies P, Q and R are independent so these will cut cost one after the other.
If initial cost is Rs 100, then
20% cost is cut after initializing strategy P, then cost will remain 80% = 80
further 30% cost is cut after strategy Q, then cost will remain 70% of 80 = 56
further 10% cost is cut after strategy R, then cost will remain 90% of 56 = 50.4

Thus final cost remains 50.4 % of the original cost. Hence net saving is 100 - 50.4 = 49.6 %.

Given digits are 0, 4, 2, 6

Required 4 digit number should be greater than 6000.

So, first digit must be 6 only and the remaining three places can be filled by one of all the four digits.

This can be done by

1x4x4x4 = 64

Greater than 6000 means 6000 should not be there.

Hence, 64 - 1 = 63.

Given,721+722+723+724

721(1 + 7 + 49 + 343)
721(400)

Now, 400 is divisible by 25. The remainder will be 0

There are 5 slots.

                   __ __ __ __ __

The first slot must be a four. There are 4 ways to put a four in the first slot.

There are 4 ways to put a five in the second slot, and there are 4 ways to put a six in the third slot. etc.

(4)(4)(4)(4)(4) = 1024

Therefore there are 1024 different ways to produce the desired hand of cards.

Total number of letters in the word ABYSMAL are 7

Number of ways these 7 letters can be arranged are 7! ways

But the letter is repeated and this can be arranged in 2! ways

Total number of ways arranging ABYSMAL = 7!/2! = 5040/2 = 2520 ways.

Cost of 4 mangoes = Rs. 6 

Cost of 1 mango = Rs. 6/4 = Rs. 1.5

Now, Selling Price of 4 mangoes = Rs. 4

Selling Price of 1 mango = Rs. 4/4 = Rs. 1.0

• We observe that, S.P < C.P

Hence, the shopkeeper incurs a loss.

Required loss% = loss/C.P x 100

Loss = 1.5 - 1 = 0.5

C.P = 1.5

Loss% = (0.5/1.5) x 100

 = (5/15) x 100

 = 1/3 x 100

 = 100/3

 = 33.333%.

L.C.M. of 252, 308 and 198 = 2772.

So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.