Quantitative Aptitude - Arithmetic Ability Questions

C.I. =Rs[4000x(1+10/100)²-4000]

Rs.(4000x11/10x11/10-4000) = Rs.940.

Sum

=Rs. [420 *100 /3 * 8]

= Rs.1750.

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%

If earlier he is getting Rs. 1  per hr and working for 1 hr, income = Rs. 1
Now he is getting Rs 1.3333 per hr and working for 0.66667 hrs

Then income for 0.66667 hrs = Rs. 0.8889.

If payment per hr is increased by 33.33 %  and working hours dereased by 33.33%, then 11.11 % decrease in income.

Let the present age of Raj and Rajitha be 4x and 9x.
Then, according to question,
9x = 2(9x-12)-3 => 9x = 27 => x = 3
Raj's present age = 4x = 12yrs,
After 5 years Raj's age will be 12 + 5 = 17 yrs.

PW = 100×Amount100+R×T 

 Since 8 does not occur in 1000, we have to count the number of times 8 occurs when we list the integers from 1 to 999. Any number between 1 and 999 is of the form xyz, where 0≤x,y,z≤9.

Let us first count the numbers in which 8 occurs exactly once.

Since 8 can occur atone place in 3C1ways. There are 3 x 9 x 9 such numbers.

Next, 8 can occur in exactly two places in 3C2×9 such numbers. Lastly, 8 can occur in all three digits in one number only.

Hence, the number of times 8 occur is 

 1×3×92+2×3×9+3×1 = 300

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%

Area to be plastered= [2(l + b) x h] + (l x b)

= [2(25 + 12) x 6] + (25 x 12)  

= (444 + 300)  

= 744 

Cost of plastering = Rs. 744 x (75/100)= Rs. 558