Quantitative Aptitude - Arithmetic Ability Questions

Let the total Profit be Z.
Ramu's profit share is 2/3 of profit (i.e 2Z/3)
Ravi's profit is (Z-2Z/3) =Z/3
Hence the profit ratio is , Ravi : Ramu = Z/3 : 2Z/3 = 1:2

Let total Capital be Rs.X and Ramu has contributed for Y months.Since Ramu's profit share is 2/3, his invest share will be 2/3 in capital.

Ravi's invest for 16 months / Ramu's invest for Y months = Ravi's profit share / Ramu's profit share
i.e. (X/6 x 16) / (2X/3 x Y) = 1/2

Solving the above equation, we get Y = 8.
So Ramu's money has been used for 8 months.

Sum =Rs.(50 x 100/2x5)

= Rs. 500.

Amount

=[Rs.500x(1+5/100)²]

=Rs(500x21/20x21/20).

=Rs. 551.25

C.I

= Rs. (551.25 - 500)

= Rs. 51.25

Can be easily solved with following formula

a2-b2=a-b×a+b

= (25.732 - 15.732 ) x (25.732 + 15.732)

=414.64

Let the leak will empty the tank in x hrs 

Then, 1/9 - 1/x = 1/13

x = 29.25 hrs

6,400 gents, 2400 ladies in that company.
So total 8,800employees.
for 8,800 employees, we want 2400 ladies.
for 12,100employees we want how many ladies ?
=> (12,100/8,800) x 2400 = 3300.
So we want 3300 - 2400 = 900 more ladies.

Total work is given by L.C.M of 72, 48, 36

Total work = 144 units

Efficieny of A = 144/72 = 2 units/day

Efficieny of B = 144/48 = 3 units/day

Efficieny of C = 144/36 = 4 units/day

According to the given data,

2 x p/2 + 3 x p/2 + 2 x (p+6)/3 + 3 x (p+6)/3 + 4 x (p+6)/3 = 144 x (100 - 125/3) x 1/100

3p + 4.5p + 2p + 3p + 4p = 84 x 3 - 54

p = 198/16.5

p = 12 days.

Now, efficency of D = (144 x 125/3 x 1/100)/10 = 6 unit/day

(C+D) in p days = (4 + 6) x 12 = 120 unit

Remained part of work = (144-120)/144 = 1/6.

P=4s

Perimeter of a square p=4s ,where s=length of its side 

=> p=4(14)=56 feets 

Area  of a square A = s x s

=> A=196

The perimeter is 56 feet, and the area is 196 square feet.