1. Two numbers differ by 8. If one-half the larger is added to one-fourth the smaller, the resulting sum is 22.Find the numbers.
2. Ten years ago a man was three times as old as his son. In 12 more years the son is three-fifths as old
as his father. How old are they now?
3. A and B working togrther can complete a piece of work in 30 days. After they have worked together for 18 days, A has to leave, and B finishes to work alone for 20 more days. Find the time it takes for each to finish the work alone.
1.
smaller number --- x
larger number --- x+8
(1/2)(x+8) + (1/4)x = 22
etc
2.
son's age 10 years ago --- x
man's age 10 years ago ---- 3x
in 12 years, son's age = x+22
in 12 years, man's age = 3x+22
x=22 = (3/5)(3x+22)
etc
3.
Let job to be done be 1 unit
let A's rate be 1/x, where x is the number of days
let B's rate be 1/y, where y is the number of days
combined rate = 1/x + 1/y = (x+y)/(xy)
1 /( (x+y)/(xy) ) = 30
xy/(x+y) = 30
xy = 30(x + y) **
They work together for 18 days, so when A leaves 12 days of work is left
or 12(x+y)/(xy) of the job is to be done by B
12(x+y)/(xy) ÷ (1/y) = 20
12(x+y)(y)/(xy) = 20
3(x+y)/x = 5
5x = 3(x+y) ***
*** ÷ **
5x/xy = 3(x+y)/(30(x+y))
5/y = 1/10
y = 50
sub in ***
5x = 3(x+50)
5x = 3x + 150
2x = 150
x = 75
Working alone, A would take 75 days, and B would work 50 days.
Check my arithmetic and algebra.
1. Let's say the two numbers are x and y, with x being the larger number and y being the smaller number. According to the problem, we have two pieces of information:
1) x - y = 8 (The two numbers differ by 8)
2) (1/2)x + (1/4)y = 22 (One-half the larger number added to one-fourth the smaller number equals 22)
To solve this system of equations, we can use substitution or elimination method. Let's use the substitution method:
From equation 1, we can express y in terms of x: y = x - 8
Substituting this value of y into equation 2, we get:
(1/2)x + (1/4)(x - 8) = 22
Multiplying through by 4 to get rid of the fractions, we have:
2x + x - 8 = 88
3x - 8 = 88
3x = 96
x = 32
Substituting the value of x back into equation 1, we find:
32 - y = 8
y = 32 - 8
y = 24
So, the two numbers are 32 and 24.
2. Let's assign the present ages of the man and his son as x and y, respectively.
According to the problem, we have two pieces of information:
1) x - 10 = 3(y - 10) (Ten years ago, the man was three times as old as his son)
2) y + 12 = (3/5)(x + 12) (In 12 more years, the son's age will be three-fifths the father's age)
To solve this system of equations, let's simplify equation 1:
x - 10 = 3y - 30
x = 3y - 20
Substituting the value of x from equation 1 into equation 2, we have:
y + 12 = (3/5)((3y - 20) + 12)
y + 12 = (3/5)(3y - 8)
5(y + 12) = 9y - 24
5y + 60 = 9y - 24
4y = 84
y = 21
Substituting the value of y back into equation 1 to find x:
x = 3(21) - 20
x = 43
So, the man is currently 43 years old, and his son is currently 21 years old.
3. Let's assume A's efficiency is a and B's efficiency is b, measured as a fraction of the work completed per day.
According to the problem, we have two pieces of information:
1) (a + b) = 1/30 (A and B working together can complete the work in 30 days)
2) 18a + 20b = 1 (After they worked together for 18 days, A left, and B finished the work alone for 20 more days)
We can use these two equations to solve for a and b.
To solve this system of equations, let's express b in terms of a from equation 1:
b = 1/30 - a
Substituting the value of b into equation 2, we have:
18a + 20(1/30 - a) = 1
18a + 20/30 - 20a = 1
(18a - 20a) + 2/30 = 1
-2a + 1/15 = 1
-2a = 1 - 1/15
-2a = 14/15
a = (14/15)(-1/2)
a = -7/30
Substituting the value of a back into equation 1, we find:
-7/30 + b = 1/30
b = 1/30 + 7/30
b = 8/30
b = 4/15
So, A takes 30/7 days to finish the work alone, and B takes 15/4 days to finish the work alone.