Problem 1 :
If we add 1 to the numerator and subtract 1 from the
denominator, a fraction reduces to 1, it becomes 1/2 if we only add 1 to the
denominator. What is the fraction?
Solution :
Let x/y be the required fraction.
(x + 1)/(y – 1) = 1
(x + 1) = (y - 1)
x – y = -1 - 1
x – y = -2 -----(1)
x/(y + 1) = 1/2
2 x = (y + 1)
2x – y = 1 -----(2)
(1) - (2) :
x = 3
Substitute 3 for x in (1).
(1)-----> 3 - y = -2
-y = -5
y = 5
Then,
x/y = 3/5
So, the required fraction is 3/5.
Problem 2 :
Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu ?
Solution :
Let 'x' be the present age of Nuri.
Let 'y' be the present age of Sonu.
Five years ago, Nuri was thrice as old as sonu.
(x – 5) = 3(y – 5)
x – 5 = 3y – 15
x – 3y = -10 -----(1)
Ten years later, Nuri will be twice as old as Sonu.
(x + 10) = 2(y + 10)
x + 10 = 2y + 20
x – 2y = 10 -----(2)
(1) - (2) :
y = 20
Substitute y = 20 in (2).
(2)-----> x - 2(20) = 10
x - 40 = 10
x = 50
So, the present ages of Nuri and Sonu 50 years and 20 years respectively.
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