SAT math problem?

Question

1. If m^15 = 48 over y and m ^13 = 2 over 6y and m > 0, what is the value of m?

2. If 1 over 4 of an even positive # and 5 over 6 of the next larger even # have a sum of 32, what is the average of the 2 #'s?

I posted this question before and I got this answer: Just wondering if anyone could explain how to get the 91 over 3.

(1/4)(x)+(5/6)(x+2)=32

(13/12)x=91/3

x=28
x+2=30

Avg: 29

Thank you. I appreciate any steps.reasoning shown :].

Answer 1

1)m^15=48/y ----- m^13=2/6y
ym^15=48 ------- ym^13=1/3
y=48/m^15 ------ y=1/3m^13
48/m^15 = 1/3m^13
144m^13=m^15
144 = m²
m=12 or -12
m=12 (m>0)

2) let x = the even pos #, so x+2 = the next even number
(1/4)(x) + (5/6)(x+2) = 32
(1/4)(x) + (5/6)(x)+(5/3) = 32
(3/12)(x) + (10/12)(x) = 96/3 - 5/3
(13/12)(x) = 91/3
13x=364
x=28
x+2=30
(28+30)/2 = 29

Answer 2

If m^15 = 48 over y
=> y = 48/m^15


and m ^13 = 2 over 6y
=> y = 2/(6m^13)

=> 2/(6m^13) = 48/m^15
=> m^15/m^13 = 6*48/2
=> m^2 = 144
=> m = 12

2. If 1 over 4 of an even positive # and 5 over 6 of the next larger even # have a sum of 32, what is the average of the 2 #'s?

let no. = x
=> next larger even # = x+2

1x/4 + 5/6(x+2)= 32
3x + 10x + 20 =384...............X by LCM 12
13 = 364
x = 28

=> 2 nos = 28 & 30
=> average of the 2 #'s = 29

QED

Answer 3

Hello

First one

m^15 = 48/y and m^13 = 2/6y

Solve each for y and we have y = 48/m^15 and y = 1/(3m^13) now set they equal giving us

48/m^15 = 1/(3m^13) then multiply both sides by 3m^15 and we have 144 = m^2 so m = +/- 12 but since y>0 we use y = 12

Second one
Let x = first even number then the next one is x+2 so

1/4(x) + 5/6(x+2) = 32

Multiply by 12 giving us 3x + 10(x+2) = 384 so 13x + 20 = 384 or 3x = 364 so x = 28 now x + 2 = 30 and the average is 29


Hope this helps!

Answer 4

Hi,

Solving m^15 = 48/y for y gives y = 48/m^15.

Solving m^13 = 2/(6y) for y gives y = 1/(3m^13)

Since both of these expressions equal y, they equal each other.

48................1
--------..=..---------
m^15.......3m^13

Cross multiplying gives:

144m^13 = m^15

144 = m²

m = 12 <== 1st answer

#2

1x/4 + 5(x+2)/6 = 32

Multiplying by 12 to eliminate fractions gives:

3x + 10x + 20 = 384
13x = 364
x = 28
x + 2 = 30

Their average is (28 + 30)/2 = 29 <== answer


I hope that helps!! :-)

Answer 5

1. use the fact that p^c divided by p^d = p^[c-d], so,

m^15 divided by m^13 = 48/y divided by 2/6y,so,

m^2 = 48/y * 6y/2,so,

m^2 = [48*6]*2, so,

m^2 = 144, so,

m = 12.

2. x/4 +5/6 of [x+2] =32, so, multiply thro` by 12, the L.C.M. of
4 and 6, so,
3x + 10[x+2] = 384, so,


3x + 10x + 20 = 384, so,

13x +20 = 384, so,

13x = 364,so,

x = 364/13