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.

Let Speed of boat in still water = b

Let Speed of still water = w

Then we know that,

Speed of Upstream = U = boat - water

Speed of Downstream = D = boat + water

Given, U + D = 82

b - w + b + w = 82

2b = 82

=> b = 41 kmph

From the given data,

41 - w = 105/3 = 35

w = 6 kmph

Now,

b + w = 126/t

=> 41 + 6 = 126/t

=> t = 126/47  = 2.68 hrs.

The cow can graze an area of 2826 sq.m i.e it is able to cover area of circle of 2826 sq.m

Therefore, πr2 =2826
r2 =(2826×7/22) = 899.18 = 900
r = 30 m.

Therefore, the radius of the circle implies the length of the rope = 30 mts.

Let the actual distance travelled be x km.
Then x/8=(x+20)/12
=> 12x = 8x + 160
=> 4x = 160
=> x = 40 km.

Let l, b the sides of the base, so that the area of the base = lb
remains constant.
Let ‘h’ be the height of the solid.
Volume of the solid, V = l× b× h.
Since the volume is to be increased by 50%

Since each number to be divisible by 5, we must have 5 0r 0 at the units place. But in given digits we have only 5.

So, there is one way of doing it.

Tens place can be filled by any of the remaining 5 numbers.So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

Required number of numbers = (1 x 5 x 4) = 20.

$$\begin{gathered} 10+25+108+154+225 \\ 10+25+108+154+15 \\ 10+25+108+13 \\ 10+25+11 \\ 10+6 \\ 16 \\ =4 \end{gathered}$$

N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)

= H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4