## Discussion

If *a* = 3*b*, *b*^{2} = 2*c*, 9*c* = *d*, then

(A) | 1/2 |

(B) | ... |

(C) | ... |

(D) | ... |

(E) | ... |

(F) | ... |

The solution is

I am in 7th grade and I am about to take the SAT and want to be prepared. While looking at this I had trouble solving and have not seen stuff like this before. With this in mind the work afterwards did not help me under stand it better. Thank you for you assistance.

This is substitution, while common substitution gives you things like x=5 or n=8 for you to work with, a similar thing is happening right here, however now they give you x=2y, y=4z and z=2.

Substituting y=4(2) so y=8, x=2(8) so x=16.

Given is:

a = 3b

b^2=2c

9c=d

And we start out with:

a^2 / d

plug in 9c for d

(a^2) / (9c)

plug in 3b for a

((3b)^2) / (9c)

now we simply the numerator (3b)^2 becomes 9b^2

Now we have (9)b^2 / (9)c

Divide both numerator and denominator by 9

(1)b^2 / (1)c ergo b^2 / c

Now we plug 2c for b^2 as given and we'll get

2c / c ergo 2c / 1c

Now divide both numerator and denominator by c

2 / 1 = 2 and that's our answer.

Niels

Greetings from Holland

Substituting y=4(2) so y=8, x=2(8) so x=16.

Given is:

a = 3b

b^2=2c

9c=d

And we start out with:

a^2 / d

plug in 9c for d

(a^2) / (9c)

plug in 3b for a

((3b)^2) / (9c)

now we simply the numerator (3b)^2 becomes 9b^2

Now we have (9)b^2 / (9)c

Divide both numerator and denominator by 9

(1)b^2 / (1)c ergo b^2 / c

Now we plug 2c for b^2 as given and we'll get

2c / c ergo 2c / 1c

Now divide both numerator and denominator by c

2 / 1 = 2 and that's our answer.

Niels

Greetings from Holland

Posted: 01/16/2012 15:49

How did they get 2c/c?

(1)b^2 / (1)c ergo b^2 / c

Now we plug 2c for b^2 as given and we'll get

2c / c ergo 2c / 1c

The question includes equations, one of them is b^2 = 2c which in turn means everywhere we end up with b^2 in our working equation we can replace it with 2c. We can also replace 2c for b^2 if that's more convenient since their equal.

Assuming you got the part where we end up with:

b^2 / c

All we do in the next step is the replacement from b^2 to 2c:

2c / c

Niels

Now we plug 2c for b^2 as given and we'll get

2c / c ergo 2c / 1c

The question includes equations, one of them is b^2 = 2c which in turn means everywhere we end up with b^2 in our working equation we can replace it with 2c. We can also replace 2c for b^2 if that's more convenient since their equal.

Assuming you got the part where we end up with:

b^2 / c

All we do in the next step is the replacement from b^2 to 2c:

2c / c

Niels

Posted: 02/26/2012 18:08

Why Dose the c come to top

Can you explain the problem again because aren't you suppose you suppose to put (9c)^2 since its a^2 / b^2?

Posted: 05/24/2012 06:55

Please show your step(s) that take you to a^2 / b^2

a^2 / d >> sub. a for 3b

(3b)^2 / d >> sub. d for 9c

(3b)^2 / 9c >> simplify numerator

(9)b^2 / 9c >> divide numerator and denominator by 9

(1)b^2 / (1)c or just b^2 / c

b^2 / c >> sub. b^2 for 2c

2c / c >> divide num. and den. by c

2(1) / (1) or just 2 / 1

2 / 1 = 2

Niels

(3b)^2 / d >> sub. d for 9c

(3b)^2 / 9c >> simplify numerator

(9)b^2 / 9c >> divide numerator and denominator by 9

(1)b^2 / (1)c or just b^2 / c

b^2 / c >> sub. b^2 for 2c

2c / c >> divide num. and den. by c

2(1) / (1) or just 2 / 1

2 / 1 = 2

Niels

so for this problem a can be either 3b or b^2 and b^2 can be either 2c or 9c? i've never sloved this problem..

Posted: 12/04/2012 15:46

The answer is (B).

Posted: 07/29/2013 16:31

if a=3b,b^2=2c,9c=d then a^2/d= ?????

english please?

english please?

A^2 would translate to (3b)^2 so 9b^2. But b^2 is 2C so now you have 18c. Divided by d and 9c is d. So 18c/9c is 2.

Posted: 10/14/2013 16:13

I'm so confused on how they got that solution

Who is using Arcadia prep in 2015? Am I the only one? Cos no new posts are being made