Quantitative Aptitude - Arithmetic Ability Questions

Since the compound interest is taken here,

 PW1+51002=1764

=> PW = 1600

The year 2004 is a leap year. So, it has 2 odd days.

But, Feb 2004 not included because we are calculating from March 2004 to March 2005.

So it has 1 odd day only.

The day on 6th March, 2005 will be 1 day beyond the day on 6th March, 2004.

Given that, 6th March, 2005 is Monday.

6th March, 2004 is Sunday (1 day before to 6th March, 2005).

we can treat every two consecutive terms as one.
So, we will have a total of 100 terms of the nature:
(2 + 5) + (7 + 6) + (12 + 7).... => 7, 13, 19,....

We know the sum of n terms  nn+12

Now, a= 7, d=6 and n=100
Hence the sum of the given series is
S= 100/2 x[2 x 7 + 99 x 6]
=> 50[608]
=> 30,400.

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

$$\begin{gathered} 0.01+0.0064=0.01+6410000 \\ =0.01+8100 =0.01+0.08=0.3 \end{gathered}$$

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.