If one needs to score 34% to pass a test, the pass marks out of 40 would be:

 

52.5% of 800 + 30.5% of 2800 = ? + 87.30

420 + 854 - 87.30 = ?

1274 - 87.30 = ?

? = 1186.70

Let the original price of rice be Rs. x

Let rice a man can buy for Rs. 500 at rs. x/kg be = R kgs

From given data, for Rs. 500

x ---- R kgs

(x + 25x/100) ---- (R - 5) kgs

=> Rx = (R - 5)(x + 25x/100)

=> Rx = (R - 5)(125x/100)

=> 100Rx = 125Rx - 625x

=> R = 25

So at rate of Rs. x/kg, man get 25 kgs for Rs. 500

=> x = 500/25 = Rs. 20

Let C's salary be Rs. x

Then B's salary = 80x/100

=> A's salary = (50/100) * (80x/100) = 40x/100

But given total earnings in a month = 22000

=> x + 80x/100 + 40x/100 = 22000

=> x + 120x/100 = 22000

=> x = 10000

=> A's salary = 40 x 10000/100 = Rs. 4000 

Let number of pages in a book = p

According to questions

p - [3p/7 + (2/5 x 4p/7)] = 36

= p - 23p/35 = 36

=> 12p = 36 × 35

p = 105

Let marks obtained by S = M

R's marks = 0.80M

Q's marks = 1.25 (0.80M)

P's marks = (0.90) (1.25) (0.80M)

= 0.9M

But 0.9M = 360

M = ?

M = 400

Hence % of marks obtained by S

= 400/500 x 100 = 80%.

Let number of boys = B

Number of girls = G

From the given data,

1B/3 + 1G/2 = 300

But given B/3 = 100

=> B = 300

G/2 = 300 - 100

G = 200 x 2

=> G = 400

Therefore, total number of students in the college = B + G = 700.

Required % is given by 

=> 12100-12x100

= 1200/88

= 13.636%

Let no. of boys be 'B' and no. of girls be 'G'

Here from given data,

B + G = 1650 .....(1)

B - G = 400 .....(2)

From (1) & (2), we get

2B = 2050

=> B = 1025

=> G = 1650 - 1025 = 625

Hence, the % of girls = 625 x 100/1650

=> 38% (approx).