Quantitative Aptitude - Arithmetic Ability Questions

Let there are L men

job requires LK man hours.

job completed in first 4 hrs = L x 4 = 4L

job completed in next 4 hrs = 4 x L/2 = 2L

job completed in next 4 hrs = 4 x L/4 = L

job completed in last 4 hrs = 4 x L/8 = L/2

4L + 2L + L + L/2 = KL

K = 7+1/2 = 7.5 hours.

Given,
Compound rate, R = 10% per annum
Time = 2 years
C.I = Rs. 420
Let P be the required principal.
A = (P+C.I)
Amount, A = P1 + r100n

(P+C.I) = P1 + 101002

(P+420) = P[11/10][11/10]

P-1.21P = -420

0.21P = 420

Hence, P = 420/0.21 = Rs. 2000

Birds speeds in mtrs/sec is 'x' and '2x' .
While flying from Engine to end, relative speed = (x+10) m/sec
from end to engine, flying speed = (2x - 10) mtr/sec
so
1000/(x+10) + 1000/(2x-10) = 187.5 secs
solving it, we get
so x = 8.728 m/sec and 2x= 17.456 m/sec

x = 31.4208 km/hr and 2x = 62.8416 km/hr.

Let the principle be Rs. P
As the amount double itself the interest is Rs. P too
So P = P x r x 15/100
=> r = 100/15 = 20/3 % = 6.66 %.

Let the required Price be Rs. X .Then, Lest toys , Less cost  (Direct Proportion) 

Therefore 6 : 5  :: 264.37 : x

=> 6 * x = 5 * 264.37

=> X= 220.308

Therefore,  Approximate price of 5 toys = Rs. 220

Walking at 3/4th of usual rate implies that time taken would be 4/3th of the usual time. In other words, the time taken is 1/3rd more than his usual time

so 1/3rd of the usual time = 15min
or usual time = 3 x 15 = 45min = 45/60 hrs = 3/4 hrs.

P = Rs. 900, R = 3 ½ % = 7/2 %, T = 2 years. 

Therefore, 

S.I. = PTR/100

S.I. = Rs. (900 x 7/2 x 2/100) = Rs. 63

Now, P = Rs. 200, S.I. = Rs. 63, R = 5 %

Time = ((100 x 63) / (200 x 5) ) years = 6.3 years.

Days in 4 years => 

Let the first year is Normal year i.e, its not Leap year. A Leap Years occurs once for every 4 years.

4 years => 365 + 365 + 365 + 366(Leap year)

4 years => 730 + 731 = 1461

Therefore, Number of Days in 4 Years = 1461 Days.