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Problem Solving Strategy - Kaplan Premier 2011 - Page 375

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by anuptvm » Sun Jul 04, 2010 11:21 am
The question is as follows:

7^b + 7^b+7^b+7^b+7^b+7^b +7^b =

a. 7^b
b. 7^b+7
c.7^7b
d. 8^b
e. 49^b

The strategy described here is to substitute for b with 0 and 1 and see if the equation can be solved.

picking b=1:
= 7^1+7^1+7^1+7^1+7^1+7^1+7^1
= 7+7+7+7+7+7+7
= 49
Plugging b=1 into the answer stems

a. 7
b. 49 (I'm not sure if this is correct isn't this 7^8)
c. 7^7
d. 8
e. 49

The explanation given is that since there are 2 answers with the possible value 49 we need to substitute b=0 and try again.

picking b=0:
= 7^0+7^0+7^0+7^0+7^0+7^0+7^0
= 1+1+1+1+1+1+1
=7

Plugging b=0 into the answer stems

a. 1
b. 7 (I'm not sure if this is correct isn't this 7^7)
c. 1
d. 1
e. 1

And the answer is stated to be B.

Could someone explain how this has been worked out.

Thanks for your help

Anup
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by selango » Sun Jul 04, 2010 11:27 am
I think option B is (7^b)*7

Can you check it?
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by kvcpk » Sun Jul 04, 2010 11:32 am
Hi Anup,

as per the problem I see here, There is no correct answer in the list of options..

I do not have a copy of that book. Are you sure that there is no typo in your post?

There are 7 7^b terms.
So answer should be 7^(b+1) or 7*7^b
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by anuptvm » Sun Jul 04, 2010 11:49 am
kvcpk wrote:Hi Anup,

as per the problem I see here, There is no correct answer in the list of options..

I do not have a copy of that book. Are you sure that there is no typo in your post?

There are 7 7^b terms.
So answer should be 7^(b+1) or 7*7^b
Thanks guys for jumping onto this immediately.

First of Anand can you explain how you get to the answer B which is 7^(b+7) (this is the answer that Kaplan has quoted as the correct one)

Kvcpk - There isn't any typo there are seven 7^b items that are being summed.

Thanks
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by GMATGuruNY » Sun Jul 04, 2010 1:42 pm
anuptvm wrote:The question is as follows:

7^b + 7^b+7^b+7^b+7^b+7^b +7^b =

a. 7^b
b. 7^b+7
c.7^7b
d. 8^b
e. 49^b

The strategy described here is to substitute for b with 0 and 1 and see if the equation can be solved.

picking b=1:
= 7^1+7^1+7^1+7^1+7^1+7^1+7^1
= 7+7+7+7+7+7+7
= 49
Plugging b=1 into the answer stems

a. 7
b. 49 (I'm not sure if this is correct isn't this 7^8)
c. 7^7
d. 8
e. 49

The explanation given is that since there are 2 answers with the possible value 49 we need to substitute b=0 and try again.

picking b=0:
= 7^0+7^0+7^0+7^0+7^0+7^0+7^0
= 1+1+1+1+1+1+1
=7

Plugging b=0 into the answer stems

a. 1
b. 7 (I'm not sure if this is correct isn't this 7^7)
c. 1
d. 1
e. 1

And the answer is stated to be B.

Could someone explain how this has been worked out.

Thanks for your help

Anup
Hi, Anup!

The other posters are correct. If you're not misreading answer choice B, then it has a typo; the correct answer should be either 7^b * 7 or 7^(b+1).

Solving algebraically:

7^b + 7^b+7^b+7^b+7^b+7^b +7^b = 7(7^b) = 7^(b+1).

Solving by plugging in, if b=2:

7^b + 7^b+7^b+7^b+7^b+7^b +7^b
=7^2 + 7^2+7^2+7^2+7^2+7^2 +7^2
=49 + 49 + 49 + 49 + 49 + 49 + 49
=7 * 49 = 343.

Plugging b=2 into answer choice b:
7^(b+7) = 7^(2+7) = 7^9 = 823,543.

So if answer choice B is intended to be the correct answer, it needs to read:

7^b * 7 = 7^2 * 7 = 49 * 7 = 343
or
7^(b+1) = 7^(2+1)=7^3=343.

Hope this helps!
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by anuptvm » Sun Jul 04, 2010 7:28 pm
GMATGuruNY wrote: Hi, Anup!

The other posters are correct. If you're not misreading answer choice B, then it has a typo; the correct answer should be either 7^b * 7 or 7^(b+1).

Solving algebraically:

7^b + 7^b+7^b+7^b+7^b+7^b +7^b = 7(7^b) = 7^(b+1).

Solving by plugging in, if b=2:

7^b + 7^b+7^b+7^b+7^b+7^b +7^b
=7^2 + 7^2+7^2+7^2+7^2+7^2 +7^2
=49 + 49 + 49 + 49 + 49 + 49 + 49
=7 * 49 = 343.

Plugging b=2 into answer choice b:
7^(b+7) = 7^(2+7) = 7^9 = 823,543.

So if answer choice B is intended to be the correct answer, it needs to read:

7^b * 7 = 7^2 * 7 = 49 * 7 = 343
or
7^(b+1) = 7^(2+1)=7^3=343.

Hope this helps!
Mitch,

Thank you for explaining this, I kind of figured that out too, for a moment I was stumped about my algebra :D.

However, Kaplan uses this example to explain the technique of Picking numbers to solve such equations (specifically the number 0 and 1).

I guess they have a bad example here. Do you have any thoughts on using the numbers 0 and 1 to substitute in equations?

Thanks
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