Quantitative Aptitude - Arithmetic Ability Questions

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

Money collected = (59.29 x 100) paise = 5929 paise.

  Number of members = 5929 = 77.

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

Let the two-digit number be 10a + b
a + b = 12 --- (1)
If a>b, a - b = 6
If b>a, b - a = 6
If a - b = 6, adding it to equation (1), we get
2a = 18 => a =9
so b = 12 - a = 3
Number would be 93.
if b - a = 6, adding it to the equation (1), we get
2b = 18 => b = 9
a = 12 - b = 3.
Number would be 39.
There fore, Number would be 39 or 93.

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

The distance is constant in this case.

Let the time taken for travel with a speed of 10 kmph be 't'.

Now the speed of 15 kmph is 3/2 times the speed of 10 kmph.

Therefore, time taken with the speed of 15 kmph will be 2t/3 (speed is inversely proportional to time)

Extra time taken = t - 2t/3 = t/3

=> 1pm - 11am = 2hrs

=> t/3 = 2h

=> t = 6 hrs.

Now, Distance = speed x time = 10 x 6 = 60 kms

Time he takes to reach at noon = 6 - 1 = 5 hrs

Now, Speed = 60/5 = 12 kmph.

C.I. =Rs[4000x(1+10/100)²-4000]

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

Sum

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

= Rs.1750.

1 boy's 1 day work = 1/70

1 girl's 1 day work = 1/140

(5 boys + 10 girls)'s 1 day work = (5/70 + 10/140) = (1/14 + 1/14) = 1/7

5 boys and 10 girls will complete the work in 7 days.