Quantitative Aptitude - Arithmetic Ability Questions

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%

Here in 100P2, P says that permutations and is defined as in how many ways 2 objects can be selected from 100 and can be arranged.

That can be done as,

100 P2 = 100!/(100 - 2)!

= 100 x 99 x 98!/98!

= 100 x 99 

= 9900.

It won't requires that much solving.

Here given the increase in weekly collection by Rs. 1,05,000

That is this increase denotes for all 7 days.

=> Gross Collection increase per day = Rs. 1,05,0007 = Rs. 15,000

Hence, the gross collection increase per day = Rs. 15,000.

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.

We know that,

$$\begin{gathered} i2 = -1 \\ i3 = i2 x i = -1 x i = -i \\ i4 = i2 x i2 = -1 x -1 = 1 \\ Hence, i233 =i4 x 58 + 1 = i232 x i1 = 1 x i = i \end{gathered}$$

Given year, 2013 when divided by 4 leaves a remainder of 1.

NOTE: When the remainder is 1, 11 is substracted from the given year to get the result.

So, 2013 - 11 =2002

Speed of lorry = 36012 = 30 kmph

Speed of van =250100 x 30 = 75 kmph

Speed of bike = 35 x 75 = 45 kmph

Therefore, now required average speed of bike and van  = 75 + 452= 60 kmph.

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