SOLUTION: A woman is 35 years old and her son is 12 years old . How many years ago was the product of their ages 174

Algebra ->  College  -> Linear Algebra -> SOLUTION: A woman is 35 years old and her son is 12 years old . How many years ago was the product of their ages 174      Log On


   

Question 1158871: A woman is 35 years old and her son is 12 years old . How many years ago was the product of their ages 174
Found 4 solutions by jim_thompson5910, greenestamps, josgarithmetic, MathTherapy:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Method 1:

Find the prime factorization of 174
First note that it is an even number, so
174/2 = 87
174 = 2*87

Then let's factor 87. This is a multiple of 3 because the digits add to 8+7 = 15 which is a multiple of 3
87/3 = 29
87 = 3*29

Therefore,
174 = 2*87
174 = 2*3*29

The mother is 35 and 29 is one of the factors. There is no way to have the son be this old as we're rewinding into the past. The mother can only attain this age. This must mean that 35-29 = 6 years is the answer.

Six years ago, her son was 12-6 = 6 years old

The product of their ages six years ago was
29*6 = 174
which confirms our answer

================================================================


Method 2:

x = number of years into the past
35-x = woman's age x years ago
12-x = son's age x years ago
(35-x)(12-x) = product of their ages x years ago
(35-x)(12-x) = 35(12-x)-x(12-x) ... distributive property
(35-x)(12-x) = 420-35x-12x+x^2 ... distributive property again
(35-x)(12-x) = 420-47x+x^2
(35-x)(12-x) = x^2-47x+420
note: you can use the FOIL rule or the box method to expand out (35-x)(12-x)

Set this equal to the desired product (174) and get everything to one side
x^2-47x+420 = 174
x^2-47x+420-174 = 0
x^2-47x+246 = 0

Use the quadratic formula to solve for x
x+=+%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29 or x+=+%28-b-sqrt%28b%5E2-4ac%29%29%2F%282a%29

x+=+%28-%28-47%29%2Bsqrt%28%28-47%29%5E2-4%281%29%28246%29%29%29%2F%282%281%29%29 or x+=+%28-%28-47%29-sqrt%28%28-47%29%5E2-4%281%29%28246%29%29%29%2F%282%281%29%29

x+=+%2847%2Bsqrt%281225%29%29%2F%282%29 or x+=+%2847-sqrt%281225%29%29%2F%282%29

x+=+%2847%2B35%29%2F%282%29 or x+=+%2847-35%29%2F%282%29

x+=+%2882%29%2F%282%29 or x+=+%2812%29%2F%282%29

x+=+41 or x+=+6

Since the woman is 35 years old, it is not possible to go back 41 years into the past before she was born. A similar situation happens with the son as well. So we ignore x = 41 as a solution. The only practical solution is x = 6.

This is the same result as we got with method 1.

Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


The difference between their ages is 23; it always will be.

So we need to find two numbers whose difference is 23 and whose product is 174.

174 = 2*87 = 2*3*29

The numbers we are looking for are 29 and 6.

Their current ages are 23 and 12; it was 6 years ago that the product of their ages was 174.

Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
174=2%2A87=3%2A58=6%2A29

If x years ago, product was 174, then %2835-x%29%2812-x%29=174
%2835-x%29%2812-x%29=29%2A6, and these two factors differ by 23, on each side of this equation.

35-x=29
x=35-29
highlight%28x=6%29--------------six years ago

Answer by MathTherapy(10556) About Me  (Show Source):
You can put this solution on YOUR website!

A woman is 35 years old and her son is 12 years old . How many years ago was the product of their ages 174
Let the number of years in the past that the woman's and her son's ages multiplied to 174, be y

Then we get: (35 - y)(12 - y) = 174

matrix%281%2C3%2C+420+-+47y+%2B+y%5E2%2C+%22=%22%2C+174%29

matrix%281%2C3%2C+y%5E2+-+47y+%2B+246%2C+%22=%22%2C+0%29

(y - 6)(y - 41) = 0 ------ Factoring the quadratic

y - 6 = 0        OR                   y - 41_______y = 41 (ignore) 

Number of years ago when woman's and son's ages multiplied to 174 or, highlight_green%28matrix%281%2C3%2C+y%2C+%22=%22%2C+6%29%29