Quantitative Aptitude - Arithmetic Ability Questions

Here by trial and error method, we can obseve that

404404 = 404 x 1001; 415415 = 415 x 1001, etc.

So, any number of this form is divisible by 1001.

Let the three parts M, N, O.
Then, M:N = 2:3 and N:O = 5:8
= (5 x 3/5) : (8 x 3/5) = 3:24/5

M:N:O = 2:3:24/5

= 10:15:24 => N = 98 x 15/49 = 30.

Let original length = x and original breadth = y. 

Decrease in area = xy -[(80/100x) * (90/100y)]  

=[(xy-(18/25)xy)]

=(7/25)xy

Decrease % =[(7/25)xy * (1/xy) * 100] =28%

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.

Given numbers are 585 546 514 404 206 369
Then interchanging the 1st and 3rd digits of the given numbers, we get
585 645 415 404 602 963
Now, arranging them in ascending order
963 645 602 585 415 404
Then the last second number is 415.

20 x 55 x 66 = L x 65 x 70
Here L = 72,600/4550 = 15.95 =~ 16
Reduction is 20-16 = 4

which is 20% reduction.

1. To find the interest, subtract the principal from the balance.

       $618 - $600 = $18

2.Use the simple interest formula and solve for r.

       I = Prt

      18= 600 x r x (1/2)

       r= 0.06 =6%

1523+2323=19-423+19+423  

 The remainder on dividing by 19 will be -423+423= 0.