TNPSC Question Paper-2013 Questions and answers

Principal P = Rs.6000

Time,  N = 1(1/2) = 3/2 years

Rate,  R = 6(2/3)% = 20/3 %

SI = \( \large\frac{PNR}{100} = \frac{6000 \times 3}{2 \times 100}\times \frac{20}{3} = Rs.600 \)

Amount, A = P + I = Rs.6000 + 600

=Rs.6,600

Area of an equilateral triangle=\( \large\frac{\sqrt{3}a^{2}}{4} \)sq.units

= \( 4\sqrt{3} = \frac{\sqrt{3}}{4}a^{2} \)

=\( a^{2} = 16 \)

= 4

So perimeter 3a = 3 X 4 = 12 m. 

Let the radius sphere A be r1, and the radius of sphere B be r2

Also given, r2 = \( \large\frac{r1}{4}B \)

Ratio of volume between spheres A and B = \( \large\frac{4 \pi r1^{3}}{3} : \frac{4 \pi r2^{3}}{3} \)

=\( r1^{3} : \left( \frac{r1}{4}\right)^{3} \)

=\( 64r1^{3} : r1^{3}\)

= 64 : 1

Let the radius of the sphere be r.

Given, Surface area = volume

\( 4 \pi r^{2}  = \frac{4}{3} \pi r^{2}\)

R = 3 units

Represent man by A and son by B

A’s 1 day work = \( \large\frac{1}{5} \)

(A+B)'S 1 day work = \( \large\frac{1}{3} \)

Son alone can do the work in \( \large\frac{15}{2} = 7\frac{1}{2}days \)

Total number of students studying in 7 institutions = 27,300

Total number of students studying in institutions M and S

 =\( \large\frac{17}{100}\times 27300 + \frac{14}{100} \times 27300 \)

=(17 + 14)273

=31 X 273 = 8463

Total marks in 7 subjects = 7 X 75 = 525

Total marks in 6 subjects = 6 X 72 = 432

Marks obtained in science subject = 525 - 432=93