Quantitative Aptitude - Arithmetic Ability Questions

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

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.

Here n(S) = 6 x 6 = 36

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

=> n(E)=20

Required Probability n(P) = n(E)/n(S) = 20/36 = 5/9.

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

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.

Here total parts of milk and water in the solution is 6+2 = 8 parts.

1part = 640/8 = 80

old mixture contains 6parts of milk and 2 parts of water(6:2).

To get new mixture containing half milk and half water, add 4parts of water to the old mixture then 6:(2+4) to make the ratio same.

i.e, 4 x 80 = 320 ml.

The Multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60.

The Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60.

The Common Multiples of 4 & 6 upto 60 numbers are given as 12, 24, 36, 48, 60.

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%