Aptitude and Reasoning Questions

Total number of cases=36

Number of favourable cases(sum 5 or 6)=10

P(getting total of 5 or 6)=10/36=5/18

 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

Let age of retirement = S
so according to the given condition 25000(x-20)
=10000 x 3 + 12000 + 14000 + 16000 + 18000 + 20000 + 22000 + 24000 + 26000 + 28000 + 30000 + 30000(S-33)
= 30000 + 210000 + 30000S-990000
= 30000S - 750000
or 25000S - 500000 = 30000x - 750000
or 5000x = 250000
or S = 50 yrs.

F = Rs. 8100

R = 5%

T = 3 months = 1/4 years

BD=FTR100=8100×14×5100=Rs.101.25

TD=FTR100+TR=8100×14×5100+14×5=Rs.100

Therefore BD - TD = 101.25-100 = Rs.1.25

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

Clearly, If 60 % is equally divided then remaining 40% of total profit will be divided between 2 persons A and B in ratio of (profit of A) : (profit of B)= 12,500 : 8500 = 125 : 85 = 25:17
now difference in amount = 25-17 = 8 this difference = 240
therefore 40% of total profit = 240*(25+17)/(25-17) = 1260
∴ 100% profit = 1260 / 40 × 100 = 3150

C.I. =Rs[4000x(1+10/100)²-4000]

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

Sum

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

= Rs.1750.

To find the minimum distance, we have to get the LCM of 75, 80, 85

Now, LCM of 75, 80, 85 = 5 x 15 x 16 x 17 = 20400

Hence, the minimum distance each should walk so that thay can cover the distance in complete steps = 20400 cms = 20400/100 = 204 mts.