Need help with quadratic equations and quadrilaterals.?

Question

1) A sum of Rs 120 was distributed among a certain number of boys. If there had been 4 boys less, each would have received a rupee more. Find the no. of boys. ( Ans - 24).

2) Find two consecutive natural numbers, the sum of whose reciprocals is 11/30. (Ans - 5 and 6).

3) In parallelogram ABCD, E is the mid point of side AB and CE bisects angle BCD. Prove that -
i) AE = AD
ii) DE bisects angle ADC
iii) Angle DEC is a right angle.

Answer 1

1.
let x be the number of students.
then each student will get Rs.120/4
=>
120/x + 1 = 120/(x-4)
=>(120 + x)(x-4) = 120x
=>120x + x^2 -480 -4x = 120x
=>x^2 - 4x -480 = 0
=>x^2 -24x+20x-480=0
=>x(x-24)+20(x-24)=0
=>(x+20)(x-24)
x=-20 or x=24
x=20 is the answer

2.
let x,x+1 are the numbers
then
1/x+1/(x+1) = 11/30
(x+1 + x)/x(x+1) = 11/30
(2x+1)*30 = 11x(x+1)
60x+30 = 11x^2+11x
=>11x^2-49x-30=0
11x^2-55x+6x-30=0
11x(x-5)+6(x-5)=0
(11x+6)(x-5)=0
x=-6/11, or 5

Answer 2

1)
let no of stu be x
decreased stu. x-4
money = 120
atq
120/c-4 - 120/x = 1
120x -120x +480 = x^2 - 4x
x^2 -4x -480
=>x^2 -24x+20x-480=0
=>x(x-24)+20(x-24)=0
=>(x+20)(x-24)
x=-20 or x=24
x=24 is the answer