Aptitude and Reasoning Questions

In the figure, there are 21 columns containing 2 cubes each and 4 columns containing 1 cube each. 

∴Total number of cubes = (21 x 2) + (4 x 1) = 42 + 4 = 46.

Here in 2nd column if you add the individual digits

 we get  3 + 7 + 0 =10 , 2 + 2 + 4 = 8 , 7 + 3 + 0 = 10

7   10   3  --> 7+3 =10

6   8     2  --> 6+2 =8

x   10   1  --> x+1 =10

Here we can observe that by adding 1st and 3rd column elements we get 2nd column elements

 --> x = 9

Suppose the merchant will take advantage of the cash discount of 4% of $20 000 = $800 by paying the bill within 30 days from the date of invoice. He needs to borrow $20 000 = $800 = $19 200. He would borrow this money on day 30 and repay it on day 100 (the day the original invoice is due) resulting in a 70-day loan. The interest he should be willing to pay on borrowed money should not exceed the cash discount $800.

r=I/pt=21.73%

The highest simple interest rate at which the merchant can afford to borrow money is 21.73%. This is a break-even rate. If he can borrow money, say at a rate of 15%, he should do so. He would borrow $19 200 for 70 days at 15%. Maturity value of the loan is $19 200(1+0.15(70/365))=$19 752.33

savings would be $20 000 − $19 752.33 = $247.67

Draw the two chords as shown in the figure. Let O be the center of the circle. Draw OC
perpendicular to both chords. That divides the two chords in half.
So CD = 10 and AB = 14. Draw radii OA and OD, both equal to radius r.
We are given that BC = 5, the distance between the two chords. Let
OB = x.

We use the Pythagorean theorem on right triangle ABO

AO² = AB² + OB²
r² = 14² + x²

We use the Pythagorean theorem on right triangle DCO

DO² = CD² + OC²

We see that OC = OB+BC = x+5, so

r² = 10² + (x+5)²

So we have a system of two equations:

r² = 14² + x²
r² = 10² + (x+5)²

Since both left sides equal r², set the right sides
equal to each other.

14² + x² = 10² + (x+5)²
196 + x² = 100 + x² + 10x + 25
196 = 125 + 10x
71 = 10x
7.1 = x

r² = 14² + x²
r² = 196 + (7.1)²
r² = 196 + 50.41
r² = 246.41
r = √246.41
r = 15.69745202 cm

In all other groups, the third and second letters are consecutive and first letter is three steps ahead of the second.

Let initial quantity of milk & water in the mixture are 5x & 4x
2 litres water is added.
Hence,
New quantity of milk & water will be 5x & (4x+2)
ATQ,
5x/(4x+2) = 10/9
x/(4x+2) = 2/9 9x = 8x + 4
x=4
Hence,
New amount of water in the mixture = 4x+2 = (4*4) + 2 = 18 litres.

Given Prabhas is twice as good a workman as Rana.

Prabhas finishes the work in 3 hrs

=> Rana finishes the work in 6 hrs.

Number of hours required together they could finish the work

= 3 x 6 / 3 + 6

= 18/9

= 2 hrs.