Quantitative Aptitude - Arithmetic Ability Questions

The above given sequence follows a pattern that,

$$\begin{gathered} 1 \\ 22 = 4 \\ 33 = 27 \\ 44 = 256 \\ 55 = 3125 \\ 66 = 46656 \\ and so on... \end{gathered}$$

Therefore, the next two numbers in the given sequence is 256 & 3125.

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.

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

Time = (100 x 81) / (450 x 4.5) = 4 years.

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.

L.C.M. of 6, 9, 15 and 18 is 90.

Let required number be 90k + 4, which is multiple of 7.

Least value of k for which (90k + 4) is divisible by 7 is k = 4.

=>Required number = (90 x 4) + 4   = 364.