Quantitative Aptitude - Arithmetic Ability Questions

Let the third number be 100.

Then, the first number = 100 - 20 = 80 and second number = 100 - 30 = 70.

Difference between the first and second number = 80-70 = 10

 Required percentage = 1080 x 100% = 100/8 = 12.5%

speed = 12 km/h = 12×518=103m/s 

distance covered = 20×2×227×50=440007m 

time taken  = distance /speed =  440007×310sec=4400×37×160min=2207min 

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.

Then, E= {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3,4),(3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1),6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

n(E) = 27.

P(E) = n(E)/n(S) = 27/36 = 3/4.

Let 'B' alone can do the work in 'x' days

6/30 + 18/x = 1

=> x = 22.5

1/30 + 1/22.5 = 7/90 

=> 90/7 = 12 6/7 days

Assume x soldiers join the fort. 1200 soldiers have provision for 1200 (days for which provisions last them)(rate of consumption of each soldier)
= (1200)(30)(3) kg.

Also provisions available for (1200 + x) soldiers is (1200 + x)(25)(2.5) k

As the same provisions are available
=> (1200)(30)(3) = (1200 + x)(25)(2.5)

x = ([(1200)(30)(3)] / (25)(2.5)) - 1200 => x = 528.

Given total rupees = 20 Rs
No. of one rupee stamps = 3
Now, remaining money = Rs. 17
With that he buys only 2 and 5 rupee stamps
Let number of Rs. 5 stamps = K
Let number of Rs. 2 stamps = L
5K + 2L = 17
K = 3, L = 1 (possible)
L = 6, K = 1 (possible)
=> But given that they are different in number so, K is not equal to 3
one rupee stamps = 3
2 two stamps = 6
5 rupee stamps = 1
Total number of stamps = 10.

 = 

   =         where  

B ---> speed of boat in still water

R---->speed of current

    ( By componendo and dividendo)

Let length = x meters, then breadth = 0.6x 

Given that  perimeter = 800 meters 

=> 2[ x + 0.6x] = 800 

=> x = 250 m 

Length = 250m and breadth = 0.6 x  250 = 150m 

Area = 250 x 150 = 37500 sq.m