Quantitative Aptitude - Arithmetic Ability Questions

P.W = [100 x Amount] / [100+(R x T)]

=[100 x 930] / [100+(8x3)] = Rs. 750 

T.D = Amount - P.W = 930 - 750 = Rs.180

Given total sum = Rs. 427
And given that 3 times A’s share, 4 times B’s share and 7 times C’s share are all equal.
=> 3A = 4B = 7C
But given
=> A + B + C = 427
Now express A and B in terms of C i.e
=> (7C/3) + (7C/4) + C = 427
=> C = 84

Increase in volume = Volume of the cube = (15 x 15 x 15) cm3.

\Rise in water level = volume/area = (15 x 15 x 15)/(20 x 15) cm = 11.25 cm.

Using, Crn=Cn-rn we get 

n – 10 = 12

or, n = 12 + 10 = 22

Current ages of Yuvaraj = 4A and Ganguly = 3A
So that 4A + 6 = 26
A = 5
Current age of Ganguly = 3A
=> 3 x 5 = 15 years

Let,

no. of rice bowl = x
no. of dal bowl = y
no. of meat bowl = z

Then, x + y + z = 65 ....(1)

Also given that, 2x = 3y = 4z

substituting value of y & z in terms of x i.e. y = 2x/3 and z = x/2 in eq (1),
We get x = 30.

As 1 rice bowl is shared between 2 guests,
Therefore, there are 60 guests.

Length of the wire fencing = perimeter = 2(90 + 50) = 280 metres

Two poles will be kept 5 metres apart. Also remember that the poles will be placed along the perimeter of the rectangular plot, not in a single straight line which is very important.

Hence number of poles required = 280 / 5 = 56

Let the C.P of one item is Rs. P
and that of other is Rs. (7500 - P)

According to the data given
C.P = S.P
=> Px(116/100) + (7500-P)x(86/100) = 7500
=> 30P = 105000
=> P = 3500

Required difference between selling prices
= Rs. [(3500/100) x 116] - [(4000/100) x 86]
= 4060-3440
= Rs. 620