Quantitative Aptitude - Arithmetic Ability Questions

 P1+20100n>2P⇔65n>2

 65×65×65×65>2

so, answer is 4 years

Total number of cases=p+q

Favourable cases=p

Probability of that event is=p/(p+q)

Given compound interest for 3 years = Rs. 1513.2

and simple interest for 5 years = Rs. 2400

Now, we know that  C.I = P1+R100n - 1

=> 1513.2 = P1+R1003 - 1 ...........(A)

And S.I = PTR/100

=> 2400 = P5R/100 ..................(B)

By solving (A) & (B), we get

R = 5%.

Except 437 all others are prime numbers.

Select a number which ocurs on two dice out of six numbers (1, 2, 3, 4, 5, 6). This can be done in C16, ways.

Now select two distinct number out of remaining 5 numbers which can be done in C25 ways. Thus these 4 numbers can be arranged in 4!/2! ways.

So, the number of ways in which two dice show the same face and the remaining two show different faces is 

 C16×C25×4!2!=720

 =>  n(E) = 720

 ∴PE=72064=59

50 men can build a tank in 40 days 

Assume 1 man does 1 unit of work in 1 day

Then the total work is 50×40 = 2000 units

50 men work in the first 10 days and completes 50×10 = 500 units of work

45 men work in the next 10 days and completes 45×10 = 450 units of work

40 men work in the next 10 days and completes 40×10 = 400 units of work

35 men work in the next 10 days and completes 35×10 = 350 units of work

So far 500 + 450 + 400 + 350 = 1700 units of work is completed and

Remaining work is 2000 - 1700 = 300 units

30 men work in the next 10 days. In each day, they does 30 units of work.

Therefore, additional days required = 300/30 =10

Thus, total 10+10+10+10+10 = 50 days required.

You can find the integers which when divided by 5 have a remainder 2 by adding 2 to all multiples of 5. So we have n = 7 , 12, 17, 22 etc.

From this series we can see that n does not have to be odd.

Also n + 1 can be a prime because, for example, 12 + 1 = 13

And (n + 2) / 7 has a remainder 2 in some cases but not all.

Remember the question asks us for what MUST be true, and we see that none of the statements are true in all cases. However, adding 3 to any of the values of n will always give a multiple of 5.

let inner radius be r meters. 

$$\begin{gathered} 2πr=440⇒r=440×744=70m \\ radius of outer circle =70+14m \end{gathered}$$  

Therefore, Radius of outer circle = 70 + 14 = 84m