<p>If a, b, c, and f are four nonzero numbers then all of the following proportions are equivalent EXCEPT
(A) a/f = b/c
(B) f/c=b/a
(C) c/a = f/b
(D) a/c = b/f
(E) af/bc = 1/1</p>
<p>Please and thank you!</p>
<p>If a, b, c, and f are four nonzero numbers then all of the following proportions are equivalent EXCEPT
(A) a/f = b/c
(B) f/c=b/a
(C) c/a = f/b
(D) a/c = b/f
(E) af/bc = 1/1</p>
<p>Please and thank you!</p>
<p>The answer would be A.
The way I would approach this problem is take each of the option out of fraction form (multiply the denominators of the fractions on both sides), so:
A)ac=bf
B)af=bc
C)cb=af
D)af=bc
E)af=bc</p>
<p>From this you can clearly see that B-E are the same, but A is different.</p>
<p>Ohhhh, I see that now. Thank you!!</p>
<p>Here’s another one where I got stumped:</p>
<p>If a and b are positive integers and [a^(1/2) b^(1/3)]^6 = 432, what is the value of ab?
(A) 6
(B) 12
(C) 18
(D) 24
(E) 36</p>
<p>[a^(1/2)]^6 = a^3
[b^(1/3)]^6 = b^2</p>
<p>(a^3)(b^2) = 432
factorize it …432 >> (27)(16) = (3^3)(4^2)
ANS = 12 … (B)</p>
<p>Feel free to add more difficult sums, I needa test myself before the exam tho :)</p>
<p>Thank you again! Okay, I’ll post up more </p>
<p>Short-answer:
A school ordered $600 worth of lightbulbs. Some of the lightbulbs cost $1 each and the others cost $2 each. If twice as many $1 bulbs as $2 bulbs were ordered, how many lightbulbs were ordered altogether?</p>
<p>I got 360, but apparently, it’s wrong. :/</p>
<p>You welcome
Ummm …well, let the number of bulbs costing $2 be “x” …so the price is “2x”
so, the number of bulbs costing $1 will be “2x” …and the price remains “2x” cuz its multiplied by 1.
2x + 2x = 600
x = 150
If I didn’t mess up, the answer would be 450.</p>
<p>Here’s another way to think: Since there are twice as many $1 lightbulbs as there are $2 lightbulbs, the total prices for the $1 and $2 lightbulbs must be the same:
$600/$2 = 300</p>
<p>Then, remembering that half of these were actually two $1 lightbulbs:
300 * 1.5 = 450</p>
<p>I’m afraid it might have made more sense in my head than in my explanation. You could always use:</p>
<p>2(x) + 1(2x) = 600
Solve for x ( = 150)
Find 3x (total # of lightbulbs = 450)</p>
<p>450 is correct! I misunderstood the question and thought there were more $2 bulbs than $1 bulbs. “Twice as many ___ as ___” confuses me along with some other similar phrases. :/</p>