Quantitative Aptitude - Arithmetic Ability Questions

4 persons can be selected from 9 in 9C4 ways =126

Fvaourable events =4C2*5C2 =60

So,required probability = 60/126 = 10/21

Let the age of Manisha and Sudeshna is 5x and 6x years respectively.

According to the given data,

5x+8/6x+8 = 7/8

42x + 56 = 40x + 64

x = 4

Required Difference of their ages = (6 × 4)  –  (5 × 4)

= 4 years.

10% profit at half plot = 600000/2 x 10/100 = Rs. 30,000

15% profit at remaining half plot = 600000/2 x 15/100 = Rs. 45,000

Now, total profit = 30000 + 45000 = 75000

Profit % = 75000/600000 x 100 = 12.5 %

The mother completes the job in x hours.
So, the daughter will take 2x hours to complete the same job.

In an hour, the mother will complete 1/x of the total job.
In an hour, the daughter will complete 1/2x of the total job.

So, if the mother and daughter work together, in an hour they will complete 1/x + 1/2x of the job.
i.e., in an hour they will complete 3/2x of the job.

The question states that they complete the entire task in 6 hours if they work together.
i.e., they complete 1/6 th of the task in an hour.

Equating the two information, we get 3/2x = 1/6
By solving for x, we get 2x = 18 or x = 9.

The mother takes 9 hours to complete the job.

Let M, N and O worked together for x days.

From the given data,
M alone worked for 8 days
M,N,O worked for x days
N, O worked for 1 day

But given that
M alone can complete the work in 18 days
N alone can complete the work in 36 days
O alone can complete the work in 54 days

The total work can be the LCM of 18, 6, 54 = 108 units

M's 1 day work = 108/18 = 6 units
N's 1 day work = 108/36 = 3 units
O's 1 day work = 108/54 = 2 units

Now, the equation is
8 x 6 + 11x + 5 x 1 = 108
48 + 11x + 5 = 108
11x = 103 - 48
11x = 55
x = 5 days.

Hence, all M,N and O together worked for 5 days.

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).

we can treat every two consecutive terms as one.
So, we will have a total of 100 terms of the nature:
(2 + 5) + (7 + 6) + (12 + 7).... => 7, 13, 19,....

We know the sum of n terms  nn+12

Now, a= 7, d=6 and n=100
Hence the sum of the given series is
S= 100/2 x[2 x 7 + 99 x 6]
=> 50[608]
=> 30,400.

On 31st December, 2005 it was Saturday.

Number of Odd days from the year 2006 to the year 2009 = (1+1+2+1)=5 days.

 On 31st december 2009, It was Thursday

Thus, on 1st Jan, 2010 it is Friday.