Quantitative Aptitude - Arithmetic Ability Questions

Here n(S) = 6 x 6 = 36

E={(1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,1),(5,4),(6,3) ,(6,6),(1,3),(2,2),(2,6),(3,1),(3,5), (4,4),(5,3),(6,2)}

=> n(E)=20

Required Probability n(P) = n(E)/n(S) = 20/36 = 5/9.

We know that,

$$\begin{gathered} i2 = -1 \\ i3 = i2 x i = -1 x i = -i \\ i4 = i2 x i2 = -1 x -1 = 1 \\ Hence, i233 =i4 x 58 + 1 = i232 x i1 = 1 x i = i \end{gathered}$$

Principal = Rs.1000;

Amount = Rs.1331;

Rate = Rs.10%p.a.

Let the time be n years.Then

[1000(1+10/100)Λn;]

= 1331.

= (1331/1000)

= (11/10)³

Therefore n = 3 years.

i=j/m

Maturity value
= PV(1 + i)^n

Term
= 5 years - 21 months=
3.25 years

Price paid
= FV(1+ i )^-n

Probability of getting a tail when a single coin is tossed =12
Probability of getting 4 when a die is thrown =16

Required probability  =(12)×(16)
= 1/12

Maximum earning will be only when he will won on the maximum yielding table.

     A ----> 10:1

     B ----> 20:1

     C ----> 30:1

 i.e, he won on B and C but lost on A

 20 x 200 + 30 x 200 -1 x 200 = 9800

 minimum earning will be when he won on  table A and B and lose on that table 3.

Therefore, 10 x 200 + 20 x 200 - 1 x 200  =  6000-200 = 5800

Therefore, Difference= 9800 - 5800 = 4000

As given Kapil's 8th birthday was celebrated 2 years ago,

Kapil’s present age = 8 + 2 = 10 years.

Father's age is twice of Kapil’s age after 10 years.
=> F + 10 = 2 (K + 10) = 2 (10+ 10)
F = 30years,

Avanthi's age =1/6(F) = 30/6 = 5 years.