Quantitative Aptitude - Arithmetic Ability Questions

Let the sum be Rs. x. Then,   

x2+x2×100100+252×2-x×100100+252×1 = 40  

(x/2)+(2x/5)-(8x/9) = 40  

x=3600.  

Therfore, amount of bill=Rs. 3600.

The best approach is to pick a number that satisfies the rules that x and y are greater than 0. So for example you could choose 25 for y.

The value of x before doubling is then 25 - 50/25 = 23

The value after doubling y will be 50 - 50/50 = 49, which is more than double.

Step 1 :2x + 5y

Step 2 :3(2x + 5y) = 6x + 15y

Step 3 :(6x + 15y) - (x +y) = 6x + 15y - x - y (watch that last minus sign!)

This gives 5x + 14y

   1 1
   8 A 9
+ 4 C 6
-  6 B 2
   7 2 3
We may represent the given sum, as shown below:

1 + A + C - B = 12
A + C - B = 11

Now,giving the maximum values to A and C, i.e.

A = 9 and C = 9, we get B = 7.

Given  65×359×113

⇒2×35×5×79×113 

⇒25×35×59×79×113

Thus, there are (5 + 5 + 9 + 9 + 3)= 31 prime numbers.

we can treat every two consecutive terms as one.
So, we will have a total of 100 terms of the nature:
(2 + 5) + (7 + 6) + (12 + 7).... => 7, 13, 19,....

We know the sum of n terms  nn+12

Now, a= 7, d=6 and n=100
Hence the sum of the given series is
S= 100/2 x[2 x 7 + 99 x 6]
=> 50[608]
=> 30,400.