Quantitative Aptitude - Arithmetic Ability Questions

Let the place be at a distance of 'd' kms

From the given data,

$$\begin{gathered} d5 - 1 + d5 + 1 = 7560 kms \\ d4 + d6 = 54 \\ \end{gathered}$$

5d/12 = 5/4 => d = 3 kms.

Hence, the place is 3 kms far.

Let the rquired number be 'P'

According to the question,

$$\begin{gathered} 100 x P100 = 20 \\ => P = 20 x 100100 = 20 \end{gathered}$$

Hence, 100 percent of the number 20 is 20.

Now, you can also try for

20 percent of what number is 100

$$\begin{gathered} 20 x P100 = 100 \\ => P = 100 x 10020 = 500 \\ => P = 500 \end{gathered}$$

Hence, 20 percent of the number 500 is 100.

I
= ptr

Therefore, the total amount the owner received on January 17, 2001 was

$1295.57 + $10.47 =
$1306.04

In case B, odd times even will give even

Perimeter = Distance covered in 8 min. = (12000/60) * 8 = 1600m 

Let length = 3x metres and breadth = 2x metres.  

Then, 2(3x + 2x) = 1600 or x = 160.  

Length = 480 m and Breadth = 320 m.

Area = (480 x 320) = 153600 

Let 'M' be the maximum marks in the examination.

Therefore, Madhu got 32% of M = 32M/100 = 0.32M
And Kumar got 42% of M = 42M/100 = 0.42M.
In terms of the maximum marks Kumar got 0.42M - 0.32M = 0.1M more than Madhu. ---- (1)

The problem however, states that Kumar got 16 marks more than the cut-off mark and Madhu got 8 marks less than the cut-off mark. Therefore, Kumar has got 16 + 8 = 24 marks more than Madhu. ---- (2)

Now, Equating (1) and (2), we get
0.1M = 24 => M = 24/0.1 = 240
'M' is the maximum mark and is equal to 240 marks.
We know that Madhu got 32% of the maximum marks.
Therefore, Madhu got 32 x 240/100 = 76.8 marks.
We also know that Madhu got 8 marks less than the cut-off marks.

Therefore, the cut-off marks will be 8 marks more than what Madhu got
= 76.8 + 8 = 84.8.

If the queen of spades and the four of diamonds must be in hand,we have 50 cards remaining out of which we are choosing 3.

So, 50C3