kitchen table math, the sequel: help desk

Friday, June 11, 2010

help desk

If 20 percent of x equals 80 percent of y, which of the following expresses y in terms of x?

(A) y = 16% of x
(B) y = 25% of x
(C) y = 60% of x
(D) y = 100% of x
(E) y = 400% of x

The Official SAT Study Guide 2006
p. 550

I absolutely cannot do this question, and I don't know why.


Crimson Wife said...

the answer therefore is E

Catherine Johnson said...

That's not the book's answer...

Catherine Johnson said...

I did the math the same way you did & got y = .25x.

I also drew a bar model, for pete's sake.

I keep getting y is 25% of x.

That's not Barron's answer.

Crimson Wife said...

Oops, I solved for x rather than y. Should be 0.2/0.8 or 25%.

That's what I get for trying to be a show-off, LOL!

Crimson Wife said...

What does Barron's claim the answer is?

Catherine Johnson said...

right - I saw that

you just turned the numbers around; you have the same answer I do

Barron's says:

(C) y = 60% of x

Crimson Wife said...

If I'd thought the problem through like I'm always telling my DD to do, it should've been obvious that x > y and therefore E is not the correct answer.

Bostonian said...

I did the problem and agree that the answer is (B). Books sometimes have typos -- don't overthink this.

Catherine Johnson said...

Is the answer definitely B??

Because if it isn't, I'm in trouble.

SteveH said...

It's B. Trust the math ... and yourself.


Where could you go wrong?

Catherine Johnson said...

Steve, you're talking to a person who spent a quarter of a century thinking 7x6 was 43.

ChemProf said...

I also agree it is b. If it were c, you'd have;

.2x = .8y = .8(.6x) = .42x

which can only be true if x = 0. Sometimes, if you aren't sure about the book's answer, try plugging it back in and see what you get.

James said...

agreed; it's 25% The book's just wrong. Maybe it's one of those experiential learning books where you just got to experience the joy of "real math," and not being limited by "right answers." That makes the comments section a group project . . .

Hainish said...

I deduced B right away, but didn't work out any actual equations, even in my head. Does this mean that I have good mathematical intuition, that I would fail a present-day algebra course, or both?

Crimson Wife said...

Hey Singapore Math enthusiasts- how would I explain the below problem to my DD using the bar model method? She had trouble following along when I showed her how to do it using algebra. It's from the "Intensive Practice" 3A book BTW.

1 mango and 1 papaya together cost $4.70. Krystle bought 2 mangoes and 5 papayas for $13.00. How much did the 2 mangoes cost?

Here's how I did it:

let m = the cost of the mango and let p = the cost of the papaya

m + p = $4.70 so
m = $4.70 - p

5p + 2m = $13.00
5p + 2($4.70 - p) = $13.00
5p + $9.40 - 2p = $13.00
3p = $3.60
p = $1.20

m = $4.70 - $1.20 = $3.50
2m = $7.00

The IP book confirms this is the answer, but there's no worked solution showing how to do it using the bar models. Can anybody help?

bky said...

Let M represent a bar for a mango, and P for papya. Then we have

$13 { MM PPPPP


$13 { MP MP PPP

the pair {MP MP represents the cost: 2x4.70 = 9.40

so the 3 papyas cost $(13 - 9.40) = $3.60. Hence one papya costs $1.20. So the 5 papyas coast $(5x1.20) = $6.00, leaving $7 for the two mangos.

This uses the bar diagram only as a book-keeping device. An adult would probably solve this problem the same way using arithmetic rather than algebra, but without the diagram. But for someone working through Singagpore 3, I would say the diagram and multi-step arithmetic should do the trick. This kind of problem requires actual problem solving for a child: what do I know, what can I calculate from that, how does it get me to where I want to be?

Glen said...

"1 mango and 1 papaya together cost $4.70. Krystle bought 2 mangoes and 5 papayas for $13.00. How much did the 2 mangoes cost?"

Isn't this a perfect example of what so many of us are dreaming of for US education? The reform math folks accuse us of being traditionalists, which means (they insist) that we don't care about understanding, just mindless calculation drill.

Yet, I never had anything like this problem back when I was in 3rd grade. The math in my era was a lot better than reform math, but I don't dream of going back to it; I dream of going forward to something like this.

This problem represents what I think of as conceptual understanding: looking past the irrelevant surface details (fruits) and seeing the underlying abstract relationships. I don't know what reform math'ers mean by conceptual understanding, because their insistence on "real world relevance" ("go cut pictures out of real magazines") shows that they are looking at the surface details, not at the "underlying concepts".

Catherine Johnson said...


Actually, now that I've read the thread here, I'm wondering whether I've contradicted myself...