Quantitative Aptitude - Arithmetic Ability Questions

(100-x)%  of 88% of 8400 = 6896

=> 100-x10088100x8400 = 6896

=> x = 6.709%

Let the distance in one direction = k kms

Speed in still water = 4.5 kmph

Speed of river = 1.5

Hence, speed in upstream = 4.5 - 1.5 = 3 kmph

Speed in downstream = 4.5 + 1.5 = 6 kmph

Time taken by Rajesh to row upwards = k/3 hrs

Time taken by Rajesh to row downwards = k/6 hrs

Now, required Average speed =$$\begin{gathered} Total distanceTotal speed = 2kk3 + k6 \\ = 2k x 186k + 3k = 4 kmph \end{gathered}$$ 

Therefore, the average speed of the whole journey = 4kmph.

Let A be event of selecting a number divisible by 3. B be the event of selecting a number divisible by 7.

  

 A = { 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 }, so n(A)=10
 B = { 7, 14, 21, 28 }, n(B)= 4 

 

Since A and B are not mutually exclusive So :

Possible outcomes = (RY, YP, PR)

2C1 4C1 + 4C1 6C1 + 6C1 2C1

Required probability = (2C1 4C1 + 4C1 6C1 + 6C1 2C1)/(12C2)

= 8 + 24 + 12/66

= 44/66

= 2/3.

From the given data,

Virat : Rohit = 3 : 2

Rohit : Dhoni = 3 : 2

Ratio of runs scored by Virat, Rohit and Dhoni respectively

= 3 x 3 : 2 x 3 : 2 x 2

9 : 6 : 4

Runs scored by Virat = 9/19 x 285 = 135.

Given,

mean of 100 observations is 40

=> Total of 100 observations = 40 x 100 = 4000

84 is misread as 48

=> Difference = 84 - 48 = 36

=> Now, new total of 100 observations = 4000 + 36 = 4036

Correct Mean = 4036/100 = 40.36

Here it is asked that 3 is what percent of 5

Let 3 is K percent of 5.

=> K x 5/100 = 3

=> K = 300/5 = 60

Hence, 3 is 60% of 5.

(A+B)'s 15 days work = 120×15=34

Remaining work =  1/4

Now, 1/4 work is done by A in 10 days.

Whole work will be done by A in (10 x 4) = 40 days.