Algebra Honors (Period 6 & 7)

Puzzle Problems 9-7

This includes digit, age, and fraction problems

Example 1: Digit problem
The sum of the digits in a 2 digit number is 12. The new number obtained when the digits are reversed is 36 more than the original number. Find the original number.

Let t = the ten's digit of the original number
Let u = the unit's digit of the original number

then the quantity relationship is t + u = 12

The value expression of the original number is 10t +u so when the digits are reversed the value would be 10u+ t

SO, the value equation becomes
10u + t = 10t + u + 36 or
9u - 9t = 36 Now, divide by 9

u - t = 4
So the two equations are
u - t = 4
u + t = 12

2u = 16
u = 8
If u = 8 then t= 4
and the original number is {48}

EXAMPLE 2: Age Problem

Mini is 4 years older than Ronald
5 years ago she was twice as old as he was. Find their ages now.

Let m = Mimi's age now
Let r = ronald's age now
The two equations created from this information are the following:
m = r + 4
and
m-5 = 2(r-5)

m -5 = 2r - 10 -->I would use substitution with this one so
(r+4) - 5 = 2r - 10
r-1 = 2r - 10
9 = r
Since Ronald is 9 years old
Mimi is 13 years old

Example 3: Fraction Problem

The numerator of a fraction is 3 less than the denominator
If the numerator and the denominator are each increased by 1, the value of the resulting fraction is 3/4 What was the original fraction?

Let n = the numerator
Let d = the denominator

n/d is the fraction

n = d -3
which can be translated to n - d = -3

and
n+1=3
d+1 4

4(n+1) = 3(d+1)
4n + 4 = 3d+ 3

4n-3d = -1
n - d = -3

Using the addition/subtraction method and multiplying the 2nd equation by (-4)

4n-3d = -1
-4n - (-4)d = -3(-4)

4n-3d = -1
-4n+4d = 12
d = 1

Since d = 11
n = 8
and the fraction is 8/11
{8/11}

We then turned to Page 447 and completed several even problems # 8-16 even.
Here are a few of the problems worked out. TUrn to our textbook to read the actual wording of each problem.

#8. Let t = ten's digit
Let u = unit's digit

t = u-5

t + u = (1/3)(10t + u)
3t + 3u = 10t + u
-7t + 2u = 0

Using substitution

-7(u-5) + 2u + 0
-7u + 35 + 2u = 0
-5u = -35
u = 7

Since u = 7 t = 2 and the number is 27
{27}

10.
Let t = ten's digit
Let u = unit's digit

The original number would be 10t + u and the reverse would be 10u + t

so 10u + t = 10t + u -54
9u -9t = -54
divide that by 9
u - t = -6
so
u - t = -6
u + t = 8
2u =2
u = 1
so the number must be 71
{71}

12.
Cecelia is 24 years younger than Joe
Let c = Cecelia's age NOW
:et j = Joe's age now
c + 24 = j
c - 6 = (1/2)(j - 6)

2c - 12 = j- 6

2c - j = 6
c - j = -24

Becomes
2c - j = 6
- c + j = 24
c = 30

So Cecelia's age is 30 years old and Joe is 54 years old