Quantitative Aptitude - Arithmetic Ability Questions

n(S) = C252 = 1326

 Let  A = event of getting both red cards

and B = event of getting both queens

then A∩B = event of getting two red queens

n(A) = C226 = 325,   n(B) = C24 = 6

 n(A∩B)=C22=1

 ∴ PA=3251326, PB = 61326

 PA∩B=11326

∴P ( both red or both queens) = PA∪B

= PA+PB-PA∩B=3251326+1221-11326=55221

A completed their work in 8 X 2 = 16 days

B completed their work in 8 X 3 = 24 days

A and B one day work1/16 + 1/245/48

A and B completed their work in 48/5 days = 9.6 days

Let A’s investment be= 7a
Let B’s investment be= 6a
Let C’s investment be= 5a
Their investment in 2ndyear:= (125% of 7a): (120% of 6a) : (115% of 5a) = 175 : 144 : 115

With slight modifications, the basic formula can be made to deal with compounding at intervals other than annually.

Since the compounding is done at six-monthly intervals, 4 per cent (half of 8 per cent) will be added to the value on each occasion.

Hence we use r = 0.04. Further, there will be ten additions of interest during the five years, and so n = 10. The formula now gives:

V = P(1 + r)10 = 5,000 x (1.04)10 = 7,401.22

Thus the value in this instance will be £7,401.22.

In a case such as this, the 8 per cent is called a nominal annual

rate, and we are actually referring to 4 per cent per six months.

Let the distance he covered each way = d kms

According to the question,

d/45 - d/50 = 1

=> d = 450 kms.

Hence, the total distance he covered in his way = d + d = 2 d = 2 x 450 = 900 kms.

Let rate = R% and time = R years.

Then, (1200 x R x R) / 100 = 432

=> R = 6.

Let 'Y' be the youngest player.

The last song can be sung by any of the remaining 3 players. The first 3 players can sing the song in (3!) ways.

The required number of ways = 3(3!) = 4320.