Quantitative Aptitude - Arithmetic Ability Questions

A number is divisible by 9 only if the sum of the digits of the number is divisible by 9.

Here 86236 = 8 + 6 + 2 + 3 + 6 = 25

We must add 2 to 25 to become 27 which is divisible by 9.

Given total students in the class = 60
Students who are taking Economics = 24 and
Students who are taking Calculus = 32
Students who are taking both subjects = 60-(24 + 32) = 60 - 56 = 4
Students who are taking calculus only = 32 - 4 = 28
probability that a randomly chosen student from this group is taking only the Calculus class = 28/60 = 7/15.

The year 2004 is a leap year. So, it has 2 odd days.

But, Feb 2004 not included because we are calculating from March 2004 to March 2005.

So it has 1 odd day only.

The day on 6th March, 2005 will be 1 day beyond the day on 6th March, 2004.

Given that, 6th March, 2005 is Monday.

6th March, 2004 is Sunday (1 day before to 6th March, 2005).

Number of white marbles = 4

Number of Black marbles = 5

Total number of marbles = 9

Number of ways, two marbles picked randomly = 9C2

Now, the required probability of picked marbles are to be of same color = 4C2/9C2  +  5C2/9C2

= 1/6 + 5/18

= 4/9.

Ans. B.
According to the question ,
Marked Price = 1600 + Cost price - (i)
MP – 500 = 125/100 * CP - (ii)
MP = 5/4 CP + 500 - (ii)
4MP = 5CP + 2000 -(ii)
Solving (i) and (ii)
Cost Price = 4400
Now , the selling price
= 130/100 * 4400
= 5720

Let the present age's of mother be 'M' and that of her son be 'S'.

According to the given data in the question,

M - 5 = 3(S - 5)

F - 3S = -10  ...............(1)

Then given ,

M + 5 : S + 5 = 11 : 5 

=> 5M - 11S = 30 ..............(2)

From (1) & (2), we get

S = 20

Then by substituting S = 20 in (1) or (2), we get

M = 50.

Hence, the present age of mother is M = 50 years.

From the given information, the circular arrangement is

Here F is second to the right of B and the remaning all are wrong.

Given A,B,C can complete a work in 15,20 and 30 respectively.

The total work is given by the LCM of 15, 20, 30 i.e, 60.

A's 1 day work = 60/15 = 4 units

B's 1 day work = 60/20 = 3 units

C's 1 day work = 60/30 = 2 units

(A + B + C) worked for 2 days = (4 + 3 + 2) 2 = 18 units

 Let B + C worked for x days = (3 + 2) x = 5x units

C worked for 2 days = 2 x 2 = 4 units

Then, 18 + 5x + 4 = 60

22 + 5x = 60

5x = 38

x = 7.6

Therefore, total number of days taken to complete the work = 2 + 7.6 + 2 = 11.6 = 11 3/5 days.