From equation (1)

x = 25 -y
Substituting in equation (2)
50 (25 - y) + 100y = 2000

1250 - 50y + 100y = 2000
50y = 2000 - 1250 = 750

y = 75050 = 15
x = 25 - 15 = 10
Hence, Rubina received ten `50 notes and fifteen `100 notes.
Solution through the elimination method:
In the equations, coefficients of x are 1 and 50 respectively. So,

Equation (1)
× 50 : 50x + 50y = 1250
Equation (2)
× 1 : 50x + 100y = 2000
same sign, so subtract
(-) (-) (-)
-50y = -750
or
y = 750-- = 15
50Substitute y in equation (1)
x + 15 = 25x
= 25 - 15 = 10

Hence Rubina received ten D50 notes and fifteen D100 rupee notes.
Example-9. In a competitive exam, 3 marks are awarded for every correct answer and forevery wrong answer, 1 mark will be deducted. Madhu scored 40 marks in this exam. Had 4marks been awarded for each correct answer and 2 marks deducted for each incorrect answer,Madhu would have scored 50 marks. If Madhu has attempted all questions, how many questionswere there in the test?
Solution : Let the number of correct answers be x
and the number of wrong answers be y.
When 3 marks are given for each correct answer and 1 mark deducted for each wronganswer, his score is 40 marks.
So 3x - y = 40
(1)
His score would have been 50 marks if 4 marks were given for each correct answer and 2marks deducted for each wrong answer.
Thus, 4x - 2y = 50
(2)


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