Quantitative Aptitude - Arithmetic Ability Questions

Let the middle digit be x.

Then, 2x = 10 or x = 5. So, the number is either 253 or 352

Since the number increases on reversing the digits, so the hundred's digit is smaller than the unit's digit.

Hence, required number = 253.

Given Expression = 18800470÷20 = 40÷20 = 2

Here, n(E) = 7C1×5C1×5C1

And,  n(S) = 12C1*11C1*10C1

P(S) = 7*6*512*11*10 = 7/44

In a circle of n different persons, the total number of arrangements possible = (n - 1)!
Total number of arrangements = n(S) = (15 – 1)! = 14 !
Taking three persons as a unit, total persons = 13 (in 4 units)
Therefore no. of ways for these 13 persons to around the circular table = (13 - 1)! = 12!
In any given unit, 3 particular person can sit in 3!. Hence total number of ways that any three person can sit =
n(E) = 12! X 3!
Therefore P (E) = probability of three persons sitting together = n(E) / n(S) = 12! X 3 ! / 14!
= 12!x3x2 / 14x13x12! = 6/14x13 = 3/91

Let the first number be K and second number be L
Then, 2K + 3L = 100.....(1)
3K + 2L = 120.......(2)
solving (1)&(2), we get
K = 32 and L = 12

Hence the largest number is 32.

For an income of Re. 1 in 9% stock at 96, investment = Rs. 96/9 = Rs.32/3

For an income Re. 1 in 12% stock at 120, investment = Rs. 120/12 = Rs. 10.

Ratio of investments =(32/3) : 10 = 32 : 30 = 16 : 15.

Time = (100 x 81)/(450 x 4.5) years

= 4 years.

A Committee of 5 persons is to be formed from 6 gentlemen and 4 ladies by taking. 

(i) 1 lady out of 4 and 4 gentlemen out of 6 

(ii) 2 ladies out of 4 and 3 gentlemen out of 6 

(iii) 3 ladies out of 4 and 2 gentlemen out of 6 

(iv) 4 ladies out of 4 and 1 gentlemen out of 6 

In case I the number of ways = C14×C46 = 4 x 15 = 60 

In case II the number of ways = C24×C36 = 6 x 20 = 120 

In case III the number of ways = C34×C26 = 4 x 15 = 60

In case IV the number of ways = C44×C16 = 1 x 6 = 6 

Hence, the required number of ways = 60 + 120 + 60 + 6 = 246