Quantitative Aptitude - Arithmetic Ability Questions

Given (2a + 3b)/(5a - 3b) = 23/5
=> 10a + 15b = 115a - 69b
=> 105a = 84b => 5a = 4b
a/b = 4/5 => a=4 & b=5
=> bxa = 5x4 = 20.

As Trains are moving in same direction,

Relative Speed = 60-20 = 40 kmph

--> 40×518=1009 m/sec

Length of Train= Speed * Time

Length = 1009×6

$$\begin{gathered} ⇒2003=6623 mts \\ \end{gathered}$$ 

Total number of events=6 x 6 x 6=216
Let A be the event of getting a total of atleast 6 and B denoted event of getting a total of less than 6 i.e.,3,4,5.

So,B={(1,1,1),(1,1,2),(1,2,1),(2,1,1),(1,1,3),(1,3,1),(3,1,1),(1,2,2),(2,1,2),(2,2,1)}

Favourable number of cases=10

Therefore,P(B)=10/216

=> P(A) = 1 - P(B)

             = 1 - (10/216) = 103/108

Average speed = total distance / total time

Total distance covered = 6 miles; total time = 45 minutes = 0.75 hours

Average speed = 6/ 0.75 = 8 miles/hour

Let the sum be Rs. x.

I  gives, S.I. = Rs. 7000 and T = 7 years.

II gives, Sum + S.I. for 5 years = 2 x Sum Sum = S.I. for 5 years.

Now, S.I. for 7 years = Rs. 7000.

therefore, S.I. for 1 year =  Rs. 1000.

Thus, I and II both are needed to get the answer.