Quantitative Aptitude - Arithmetic Ability Questions

Let the number of days be 'p' 

As the work is same, we know that

M1 x D1 x H1 = M2 x D2 x H2

Where M = Men, D = Days, H = Hours per day 

Here M1 = 9, D1 = 15, H1 = 7 

M2 = 6, D2 = p, H2 = 9 

=> 9 x 15 x 7 = 6 x p x 9 

=> p = 35/2 = 17.5 days.

(300 – 101)/11 = 199/11 = 18 1/11

18 Numbers.

Ratio of length and breadth is 6 : 5
∴ Length = 6x
Breadth = 5x
∴ Area = l x b = 30x2
∴ Perimeter = 2 (l + b) = 2(6x + 5x) =22x
∴ (30*x*x)/(22x)= 60/11
∴ X = 4
∴ Length = 6x = 6 x 4 = 24 units

Let Rs. K invested in each scheme

Two years C.I on 20% = 20 + 20 + 20x20/100 = 44%

Two years C.I on 15% = 15 + 15 + 15x15/100 = 32.25%

Now,

(P x 44/100) - (P x 32.25/100) = 528.75

=> 11.75 P = 52875

=> P = Rs. 4500

Hence, total invested money = P + P = 4500 + 4500 = Rs. 9000.

A=P(1+r)^t

M = 10 days

The ratio of efficiency of M & N are 3 : 2

Hence, the time rquired for N alone = 15 days

=> Required time taken t=by both to complete the work = M x N / M + N 

= 10 x 15/ 10 + 15

= 150/25

= 6 days.

Suppose that the original price of the car = Rs. x 

Then new price of the car

=> (x) + (x ×25/100) = Rs. 5x/4 

To restore the original price, the new price must be decreased by

5x/4 − x = x/4

So required percentage =(x/4)/(5x/4) × 100%

= 20%