Quantitative Aptitude - Arithmetic Ability Questions

S.I. for 1 year =  Rs. (854 - 815) = Rs. 39.

S.I. for 3 years = Rs.(39 x 3) = Rs. 117.

Principal = Rs. (815 - 117) = Rs. 698

Principal = Rs. 16000; Time = 9 months =3 quarters;
Rate = 20% per annum = 5% per quarter.
Amount = Rs. [16000 x (1+(5/100))3] = Rs. 18522.

CJ. = Rs. (18522 - 16000) = Rs. 2522

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E = event of getting at most two heads.

Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.

P(E) =n(E)/n(S)=7/8.

In two throws of a die, n(S) = (6 x 6) = 36.

Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.

P(E) =n(E)/n(S)=4/36=1/9.

If the period between the two months is divisible by 7, then that two months will have the same calender .

 
(a). Oct + Nov  = 31 + 30 = 61 (not divisible by 7)

(b). Apr + May + Jun + Jul + Aug + Sep + Oct = 30 + 31 + 30 + 31 + 31 + 30 + 31  = 214 (not divisible by 7)

(c). Jun + July + Aug + Sep = 30 + 31 + 31 + 30 = 122 (not divisible by 7)

(d). Apr + May + June = 30 + 31 + 30 = 91 (divisible by 7)

Hence, April and July months will have the same calendar.

B covers 28 meters in 7sec. So, B's time over the course = (7/28)*100 =250 sec.

Whereas A's time over the course = 250 -7 = 243 sec.

i.e A's time over the course is 4min , 3 sec.

Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.
Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).
⇒ [(140/100) × 5x],[(150/100) × 7x] and [(175/100) × 8x]
⇒ 7x, 21x/2 and 14x.

⇒ The required ratio =7x : 21x/2 : 14x
⇒ 14x : 21x : 28x
⇒ 2 : 3 : 4