Quantitative Aptitude - Arithmetic Ability Questions

Ans. 
5 = 3 + (13 + 1)
15 = 5 + (23 + 2)
45 = 15 + (33 + 3)
113 = 45 + (43 + 4)
? = 113 + (53 + 5), i.e. ? = 243

Let x and y be the respective km's travelled by man via taxi and by his own car.
Given x + y = 90 => x = 90 - y
But according to the question,
7x + 6y = 595
7(90-y) + 6y = 595
=> 630 - 7y + 6y = 595
=> y = 630 - 595 = 35
=> x = 90 - 35 = 55
Therefore, the distance travelled by taxi is 55 kms.

Volume of the wall = (2400 x 800 x 60)  cu.cm  

Volume of bricks   = 90% of the volume of the wall 

= [(90/100) x 2400 x 800 x 60] cu.cm  

Volume of 1 brick = (24 x 12 x 8)  cu.cm 

Number of bricks = [(90/100) x (2400 x 800 x 60) ]/ (24 x 12 x 8) = 45000

Let the total sale be Rs. x.

Then, 2.5%. of x = 12.50

=> (25/100 * 1/100 * x) =  125/10

=> x = 500.

Assume A be the cost price.

Therefore,
35×A×10100 - 25×A×5100 = 1500

=>  A = Rs 37,500.

Population afer 3 years = 64000 * (1 +  5/(2 *100) )^3 = (64000 * 41/40 * 41/40 * 41/40 ) = 68921..

Let the number of sides be n. 

The number of diagonals is given by ⁿC₂ - n  

^{Therefore, n}C₂ - n = 44, n>0 

ⁿC₂ - n = 44

n² - 3n - 88 = 0 

n² -11n + 8n - 88 = 0  

As n>0, n will not be -8. Therefore, n=11.

From the information provided we know that,

Principal + 8% p.a. interest on principal for n years = 180 …….. (1)

Principal + 4% p.a. interest on principal for n years = 120 ……… (2)

Subtracting equation (2) from equation (1), we get

4% p.a. interest on principal for n years = Rs.60.

Now, we can substitute this value in equation (2),

i.e Principal + 60 = 120

= Principal = Rs.60.

We know that SI = , where p is the principal, n the number of years and r the rate percent of  interest.

In equation (2), p = Rs.60, r = 4% p.a. and the simple interest = Rs.60.

Therefore, 60 =(60*n*4)/100

=> n = 100/4 = 25 years.