the product of the ages of a set of quadruplets is 16 times the sum of their ages. how old are they?
explain this please
2007-09-11 12:59:00 · 20 answers · asked by Tyler J 1 in Science & Mathematics ➔ Mathematics
Assume that the age of each child is x therefore:
The product of their ages would be:
x*x*x*x = x^4
The sum of their ages would be:
x + x + x + x = 4x
And you know that:
x^4 = 16*4x
Divide both sides of the equation by x:
x^4/x = 16*4x/x --> x^3 = 64
Therefore, x equals the qubic root of 64:
x = 64^(1/3) = 4
Hope this helps :-)
2007-09-11 13:01:40 · answer #1 · answered by Anthony P - Greece 2 · 0⤊ 4⤋
If you multiply the 4 ages it ends up being 16 times bigger than if you add up the 4 ages.
So if the quadruplets are 1 year old, then the sum is 4 (1+1+1+1.) The product is also 4 (1x1x1x1)
Therefore we know 1 is not the correct answer.
16(4x )= x ^4
I don't know why people are giving you the answer, that is not the same as explaining it.
2007-09-11 20:08:37 · answer #2 · answered by Lost Poet 6 · 0⤊ 1⤋
Since they are quadruplets they have the same age
A x B x C x D = 16 x (A + B + C + D)
A ^ 4 = 16 x 4 x A
A^4 = 64 A
A ^ 4 / (A) = 64
A ^3 = 64 therefore A = 4
Each of them is 4 years old
2007-09-11 20:09:08 · answer #3 · answered by Rintje 1 · 0⤊ 2⤋
Let the age of one of the quadruplets be x, then every one else is x years old as well.
What is the product of their ages? Write it out.
so it is (is meaning =) to 16 times the sum of the ages.
Set up the equation, set it equal to zero and factor and you'll have the right answer.
Doug
2007-09-11 20:05:26 · answer #4 · answered by douglas 2 · 0⤊ 1⤋
4 babies born in same days => quadruplets
a,b,c,d => age of each person to find
abcd = 16(a+b+c+d)
born in same days => same age
a=b=c=d
aaaa = 16(a+a+a+a)
a^4 = 16 (4a)
a^4 = 64a
a^3 = 64 (third root both sides)
a = third root (4^3)
a = 4
Answer : Each child has 4 years
Verification
4 x 4 x 4 x 4 = 16(4 + 4 + 4 + 4)
256 = 16 (16)
256 = 256
2007-09-11 20:27:17 · answer #5 · answered by frank 7 · 0⤊ 1⤋
Since they are quadruplets (born at the same time), their ages are the same. Let's say each one of them is X years old.
Then you can say:
16(x+x+x+x)=x * x * x * x
(notice how I translated words into this formula)
Simplify this and you get:
16(4x) = x^4
64x = x^4
64 = x^3 (divide both sides by x)
x=third root of (64)
x=4
4 years old
2007-09-11 20:05:52 · answer #6 · answered by tkquestion 7 · 2⤊ 1⤋
4.
The product of the their age is 4 to the power of 4 = 256.
The sum of their ages is 16 (4+4+4+4).
256/16 = 16
2007-09-11 20:05:54 · answer #7 · answered by SoulDawg 4 UGA 6 · 1⤊ 2⤋
They are 4 years old.
4+4+4+4= 16 (sum of their ages)
4*4*4*4= 256 (product of their ages)
256/ 16= 16 (the sum of their ages! =] )
2007-09-11 20:09:05 · answer #8 · answered by swimaholic 2 · 0⤊ 2⤋
Let x be the babies' age.
x*x*x*x = 16(x + x + x + x)
x^4 = 16*4x
x^4 = 64x
x^4 - 64x = 0
x(x^3 - 64) = 0
either x = 0 or x^3 = 64
either x = 0 or x = cube root of 64 = 4
The babies can either be 0 (they've just been born) or 4.
2007-09-12 10:40:40 · answer #9 · answered by ... 6 · 0⤊ 1⤋
They are the same age = x
So x^4 = 16*4x
x^4-64x=0
x(x^3-64)
x^3 = 64
x = 4 = age of each quad
2007-09-11 20:07:55 · answer #10 · answered by ironduke8159 7 · 0⤊ 1⤋