Let number of notes of each denomination be x.

Then x + 5x + 10x = 480

 16x = 480

Therefore x = 30.

Hence, total number of notes = 3x = 90.

Let the number of students in rooms A and B be x and y respectively.

Then, x - 10 = y + 10     \(\Rightarrow\) x - y = 20 .... (i)

     and x + 20 = 2(y - 20)     \(\Rightarrow\) x - 2y = -60 .... (ii)

Solving (i) and (ii) we get: x = 100 , y = 80.

therefore  The required answer A = 100.

Let the cost of a chair and that of a table be Rs. x and Rs. y respectively.

Then, 10x = 4y   or   y =\(\frac{5}{2}x.\)

Therefore  15x + 2y = 4000

 15x + \(2\times\frac{5}{2}x = 4000\)

 20x = 4000

Therefore x = 200.

So, y = \(\left(\frac{5}{2}\times200\right)=500\)

Hence, the cost of 12 chairs and 3 tables = 12x + 3y

= Rs. (2400 + 1500)

    = Rs. 3900.

2ab = (a² + b²) - (a - b)²

   = 29 - 9 = 20

   ab = 10.

Let the price of a saree and a shirt be Rs. x and Rs. y respectively.

Then, 2x + 4y = 1600 .... (i)

    and x + 6y = 1600 .... (ii)

Divide equation (i) by 2, we get the below equation.

=> x +  2y =  800. --- (iii)

Now subtract (iii) from (ii)

Therefore, y = 200.

Now apply value of y in (iii)

=>  x + 2 x 200 = 800

=>  x + 400 = 800

Therefore x = 400

Solving (i) and (ii) we get x = 400, y = 200.

Therefore Cost of 12 shirts = Rs. (12 x 200) = Rs. 2400.

Let C's share = Rs. x

Then, B's share = Rs. \(\frac{x}{4}\)  ,   A's share = Rs. \(\left(\frac{2}{3}\times\frac{x}{4}\right)= Rs.\frac{x}{6}
\)

Therefore \(\frac{x}{6} + \frac{x}{4}+x = 1360
\)

\(\Rightarrow\frac{17x}{12}=1360\)

\(\Rightarrow x = \frac{1360\times12}{17} = Rs .960\)

Hence, B's share = Rs. \(\left(\frac{960}{4}\right)= Rs. 240.\)

Let savings in N.S.C and P.P.F. be Rs. x and Rs. (150000 - x) respectively. Then,

\(\frac{1}{3}x=\frac{1}{2}\left(150000-x\right)\)

\(\Rightarrow\frac{x}{3}+\frac{x}{2}=75000\)

\(\Rightarrow\frac{5x}{6}=75000\)

\(\Rightarrow x=\frac{75000\times6}{5}=90000\)      

Savings in Public Provident Fund = Rs. (150000 - 90000) = Rs. 60000

Let the total number of shots be x. Then,

Shots fired by A = \(\frac{5}{8}x\)

Shots fired by B = \(\frac{3}{8}x\)

Killing shots by A =\(\frac{1}{3} of \frac{5}{8}x = \frac{5}{24}x\)

Shots missed by B = \(\frac{1}{2} of \frac{3}{8}x = \frac{3}{16}x\)

Therefore \(\frac{3x}{16} = 27 or x =\left(\frac{27\times16}{3}\right)=144. \)

Birds killed by A = \(\frac{5x}{24} = \left(\frac{5}{24}\times144\right) = 30.\)

Original share of 1 person = \(\frac{1}{8}\)

New share of 1 person = \(\frac{1}{7}\)

Increase = \(\left(\frac{1}{7}- \frac{1}{8}\right)= \frac{1}{56}\)

Therefore Required fraction = \(\frac{(\frac{1}{56})}{(\frac{1}{8})} = \left(\frac{1}{56}\times\frac{8}{1}\right) = \frac{1}{7} \)

Let the capacity of 1 bucket = x.

Then, the capacity of tank = 25x.

New capacity of bucket = \(\frac{2}{5}x\)

 Required number of buckets = \(\frac{25x}{(\frac{2x}{5})}\)

= \(\left(25x\times\frac{5}{2x}\right)\)

= \(\frac{125}{2}\)

=62.5