Problem 1 :
The denominator of a fraction is 4 more than twice the numerator. When both the numerator and denominator are decreased by 6, then the denominator becomes 12 times the numerator and determine the fraction.
Problem 2 :
The sum of a two digit number and the number formed by interchanging the digit is 132. If 12 is added to the number, then new number becomes 5 times the sum of the digits. Find the number.
Problem 3 :
Ten years ago, father was twelve times as old as his son and ten years hence, he will be twice as old as his son will be. Find their present ages.
Problem 4 :
A man starts his job with a certain monthly salary and earns a fixed increment every year. If his monthly salary be $ 1500 after 4 years of service and $ 1800 after 10 years of service, what is his starting salary and what is his annual increment.
1. Answer :
Let x be the numerator of the fraction.
Then the denominator is (2x + 4).
So, the required fraction is
= x / (2x + 4) -----(1)
Given : If the numerator and denominator are decreased by 6, then the denominator becomes 12 times the numerator and determine the fraction.
Then, we have
(x - 6) / (2x + 4 - 6) = x / 12x
(x - 6) / (2x - 2) = 1 / 12
Simplify.
12(x - 6) = 2x - 2
12x - 72 = 2x - 2
10x = 70
x = 7
2x + 4 = 2(7) + 4
2x + 4 = 18
(1)-----> x / (2x + 4) = 7 / 18
So, the required fraction is 7/18.
2. Answer :
Let xy be the two digit number.
Given : The sum of two digit number and the number formed by interchanging the digit is 132.
xy + yx = 132
10x + y + 10y + x = 132
11x + 11y = 132
Divide each side by 11.
x + y = 12
Subtract x from each side.
y = 12 - x -----(1)
Given : If 12 is added to the number, then new number becomes 5 times the sum of the digits.
xy + 12 = 5(x + y)
10x + y + 12 = 5x + 5y
5x - 4y + 12 = 0
Substitute (12 - x) for y.
5x - 4(12 - x) + 12 = 0
5x - 48 + 4x + 12 = 0
9x - 36 = 0
9x = 36
x = 4
Substitute 4 for x in (1).
(1)-----> y = 12 - 4
y = 8
xy = 48
So, the required two digit number is 48.
3. Answer :
Let f and s be the present ages of father and son.
Given : Ten years ago, father was twelve times as old as his son.
Then, we have
f - 10 = 12(s - 10)
Simplify.
f - 10 = 12s - 120
f = 12s - 110 -----(1)
Given : Ten years hence, father will be twice as old as his son will be
Then, we have
f + 10 = 2(s + 10)
Simplify.
f + 10 = 2s + 20
f = 2s + 10 -----(2)
From (1) and (2), we get
12s - 110 = 2s + 10
10s = 120
s = 12
Substitute 12 for s in (2).
f = 2(12) + 10
f = 24 + 10
f = 34
So, the present ages of father and son are 34 years and 12 years respectively.
4. Answer :
The man gets fixed increment every year.
So, his monthly salary each year is growing linearly.
Then, we can have the following linear equation for the information given above.
y = mx + b -----(1)
y -----> Monthly salary
x -----> Number of years
m -----> Yearly increment
b -----> Starting monthly salary
So, we have to solve for m and b.
Given : Monthly salary is $ 1500 after 4 years of service.
Substitute 1500 for y and 4 for x in (1).
(1)-----> 1500 = 4m + b
or
4m + b = 1500 -----(2)
Given : Monthly salary is $ 1800 after 10 years of service.
Substitute 1800 for y and 10 for x in (2).
(1)-----> 1800 = 10m + b
or
10m + b = 1800 -----(3)
Solving (2) and (3), we get
m = 50
b = 1300
So, his starting monthly salary is $1300 and his annual increment is $50.
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