Quantitative Aptitude - Arithmetic Ability Questions

Clearly, Rate = 5% p.a .,

Time = 3 years

S.I =Rs.1200.

So,Principal

=Rs.(100 x 1200/3x5)

=Rs.8000.

Amount

=Rs.[8000 x (1+5/100)³]

=Rs(8000x21/20x21/20x21/20)

= Rs.9261

C.I

=Rs.(9261-8000)

=Rs.1261.

S.I=PNR/100

Length of train A = 48 x 9 x 5/18 = 120 mts

Length of train B = 48 x 24 x 5/18 - 120

=> 320 - 120 = 200 mts.

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

let E= Event of geting a sum 9={(3,6),(4,5),(5,4),(6,3)}

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

In a family with 2 children there are four possibilities:
1) the first child is a boy and the second child is a boy (bb)
2) the first child is a boy and the second child is a girl (bg)
3) the first child is a girl and the second child is a boy (gb)
4) the first child is a girl and the second child is a girl (gg)

But already given that one child is boy. So we have three possibilities of (bb)(bg)(gb).
n(E)= both are boys=BB=1
n(S)= 3
Required probability P = n(E)/n(S) = 1/3.

As the work is same
M1D1 = M2D2

Here Let the number of men employed was M
=> 11M = 7(M+4)
=> 11M - 7M = 28
=> M = 7

Let the rate of interest be R

(Interest at rate (R+4)%)–(Interest at R%)

= 600(PT(R+4)/100) –(PTR/100)

= 6003PR +12P –3PR = 60000

12P = 60000

P = 5000

Given expression = 0.2×0.2×0.010.1×0.1×0.02=0.04+0.010.01+0.02=0.050.03=53