Quantitative Aptitude - Arithmetic Ability Questions

Given a jeep travels a certain distance taking 6 hrs in forward journey
During the return journey, it takes 4 hrs with an increased speed 12 km/hr
Let 'x' be the distance.
Now, speed is given by distance/time.
Here the difference between both speeds = 12

 x4 - x6 = 12

 3x - 2x12 = 12

=> x = 144 km

Therefore, the distance travelled by the jeep in forward or return journey is x = 144 km.

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.

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.

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.

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).

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.

Here distance d = 800 mts

speed s = 63 - 3 = 60 kmph = 60 x 5/18 m/s

time t = 800x1860x5= 48 sec.