Quantitative Aptitude - Arithmetic Ability Questions

Let 4 girls be one unit and now there are 6 units in all.

They can be arranged in 6! ways.

In each of these arrangements 4 girls can be arranged in 4! ways. 

Total number of arrangements in which girls are always together = 6! x 4!= 720 x 24 = 17280

Average =   (76 + 65 + 82 + 67 + 85 )/ 5 = 375/5   =  75.

Distance covered in one revolution = 88×10001000= 88m. 

2πR2=88 ⇒2×227×R=88 ⇒R=88×744

Let the total profit be Rs. x

Amount paid to B as salary = (120 x 12)

                                       = Rs.1440

Share of each = 

Interest paid by B = 

Total money received by  A = 

                                       = Rs.

Total money received by B

W1:A1 W2:A2 ..... WN:AN

  2 : 3                   4 : 5                    5 : 7 

W1W1+A1 = 25 W2W2+A2=49 WNWN+AN = 512

= 72/180

= 80/180

= 75/180  

 =>     5   :   3

Therefore, the ratio is  5: 3

Given that 

6 men and 8 boys can do a piece of work in 10 days 

26 men and 48 boys can do the same in 2 days

As the work done is equal,

10(6M + 8B) = 2(26M + 48B)

60M + 80B = 52M + 96B

=> M = 2B

=> B = M/2 ……(1)

Now Put (1) in 15M + 20B

=> 15M + 10M = 25M

Now, 6M + 8B in 10 days

=> (6M + 4M) 10 = 100M

Then D(25M) = 100M

=> D = 4 days.

The mathematical formula for calculating compound interest depends on several factors. These factors include the amount of money deposited called the principal, the annual interest rate (in decimal form), the number of times the money is compounded per year, and the number of years the money is left in the bank.

 FV=p1+rnnt

FV = Future value of the Deposit

p = Principal or Amount of Money deposited

r = Annual Interest Rate (in decimal form )

n = No of times compounded per year

t = time in years

FV=40001+0.0644(5)= 5387.42

The toys are different; The boxes are identical 

If none of the boxes is to remain empty, then we can pack the toys in one of the following ways 

a. 2, 2, 1 

b. 3, 1, 1 

Case a. Number of ways of achieving the first option 2 - 2 - 1 

Two toys out of the 5 can be selected in 5C2 ways. Another 2 out of the remaining 3 can be selected in 3C2 ways and the last toy can be selected in 1C1 way. 

However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2 

Therefore, total number of ways of achieving the 2 - 2 - 1 option is ways 5C2*3C2= 15 ways

Case b. Number of ways of achieving the second option 3 - 1 - 1

Three toys out of the 5 can be selected in 5C3 ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.

Therefore, total number of ways of getting the 3 - 1 - 1 option is 5C3 = 10 = 10 ways.

Total ways in which the 5 toys can be packed in 3 identical boxes

= number of ways of achieving Case a + number of ways of achieving Case b= 15 + 10 = 25 ways.