Quantitative Aptitude - Arithmetic Ability Questions

B runs 35 m in 7 sec.

 B covers 200 m in (7/35*200) = 40sec

B's time over the course = 40 sec.

 A's time over the course (40 - 7) sec = 33 sec.

T = (100 x I)/ (P x r)

Let the distance travelled by Tilak in first case or second case = d kms

Now, from the given data,

d/20 = d/22 + 36 min

=> d/20 = d/22 + 3/5 hrs

=> d = 132 km.

Hence, the total distance travelled by him = d + d = 132 + 132 = 264 kms.

As A1 speaks always after A2, they can speak only in  1st  to 9th places and 

A2 can speak in 2nd to 10 the places only when A1 speaks in 1st place 

A2 can speak in 9 places the remaining 

 A3, A4, A5,...A10  has no restriction. So, they can speak in 9.8! ways. i.e

when A2 speaks in the first place, the number of ways they can speak is 9.8!.

When A2 speaks in second place, the number of ways they can speak is  8.8!.

When A2 speaks in third place, the number of ways they can speak is  7.8!. When A2 speaks in the ninth place, the number of ways they can speak is 1.8!

Therefore,Total Number of ways they can  speak = (9+8+7+6+5+4+3+2+1) 8! = 92(9+1)8! = 10!/2

Prime Numbers :: Numbers which are divisible by only 1 and itself are Prime Numbers.

It's answer will be 91.

Because 91 can be divisible by 7,13,91,1.

It is quite clear that prime number should be divisible only by itself and by 1.

Given Dileep and Narender started a business.

Dileep invested 8000 for 12 months

Let Narender invested 'p' for 6 months

At the end of the year, their profit ratio is 1 : 1.

Hence,

8000 x 12 : p x 6

p : 16000

p/16000 = 1/1

=> p = 16,000

Hence, Narender invested amount is Rs. 16,000.

15! - 14! - 13!

= (15 × 14 × 13!) - (14 × 13!) - (13!)

= 13! (15 × 14 - 14 - 1)

= 13! (15 × 14 - 15)

= 13! x 15 (14 - 1)

= 15 × 13 × 13!