for no. 3:

Assume that the no. of right exercises = x and no. of wrong exercises = y

Therefore

x+y=50 ⇒ y=50-x

and

5x-3y=130 ......(1) substitute y=50-x in (1) you get 5x-3(50-x)=130

or

5x-150+3x=130

8x=280 x= 35 , y= 15

so the no. of right exercises = 35

the no. of wrong exercises = 15

For no. (4)

Assume that the age of the first = x now

the age of the second= y now

therefore x+y=30 y=30 - x

before 8 years (x-8)(y-8)=48..........(1) but y=30 - x

(x-8)(30-x-8)=48 so (x-8)(22-x)=48

22x-x^2-176+8x=48 ⇒ x^2-30x+224=0 ⇒ (x-16)(x-14)=0

x=16 and y=14 or

x=14 and y=16

So the older brother's nowadays age = 16

Best Wishes

Riad Zaidan

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2) Find two positive consecutive integers whose squares' sum equals 481.

3) A student gets 5 points for every exercise he gets right, and looses 3 points for each he gets wrong. After doing 50 exercises he had 130 points. How many did he got right?

4) The sum of two brothers ages equals 30 years now. 8 years ago, the product of the ages was 48. What's the older brother's nowadays age?

Enjoy.

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