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In the Land of OZ only one or two-letter words are used.

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by swerve » Mon May 07, 2018 11:16 am

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Answers

A

B

C

D

E

Stats

Difficulty—

In the Land of OZ, only one or two-letter words are used. The local language has 66 different letters. The parliament decided to forbid the use of the seventh letter. How many words have the people of OZ lost because of the prohibition?

A. 65
B. 66
C. 67
D. 131
E. 132

The OA is E.

Please, can anyone explain this PS question for me? I need help. I tried to solve it but I can't get the correct answer. Thanks.
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by Jay@ManhattanReview » Mon May 07, 2018 8:04 pm
swerve wrote:In the Land of OZ, only one or two-letter words are used. The local language has 66 different letters. The parliament decided to forbid the use of the seventh letter. How many words have the people of OZ lost because of the prohibition?

A. 65
B. 66
C. 67
D. 131
E. 132

The OA is E.

Please, can anyone explain this PS question for me? I need help. I tried to solve it but I can't get the correct answer. Thanks.
Before the prohibition:

The number of 1-letter words = 66;
The number of 2-letter words = 66*66 = 66^2

Total number of words = 66 + 66^2

After the prohibition:

The number of 1-letter words = 65;
The number of 2-letter words = 65*65 = 65^2

Total number of words = 65 + 65^2

The number of words lost = [66 + 66^2] - [65 + 65^2] = 1 + [66^2 - 65^2] = 1 + (66 - 65)(66 + 65) = 1 + 1*(131) = 1 + 131 = 132.

The correct answer: E

Hope this helps!

-Jay
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by Brent@GMATPrepNow » Tue May 08, 2018 8:37 am
Jay@ManhattanReview wrote:
swerve wrote:In the Land of OZ, only one or two-letter words are used. The local language has 66 different letters. The parliament decided to forbid the use of the seventh letter. How many words have the people of OZ lost because of the prohibition?

A. 65
B. 66
C. 67
D. 131
E. 132
Jay's approach (above) is exactly how I would have solved it.
The only way my approach differs is when we get to:

The number of words lost = [66 + 66²] - [65 + 65²]
At this point, we might notice that each of the 5 answer choices has a DIFFERENT UNITS DIGIT.
So, we can save some time by focusing solely on the UNITS DIGITS
For example, notice that 66² = ---6 [we need not concern ourselves with the other digits. We need only recognize that (66)(66) = some number with 6 as its units digit ]
Likewise 65² = (65)(65) = ----5

So, we get:
The number of words lost = [66 + 66²] - [65 + 65²]
= [66 + ---6] - [65 + ----5]
= (------2) - (----0)
= ------2

Since only answer choice E has a units digit of 2, the correct answer must be E

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by regor60 » Tue May 08, 2018 9:01 am
Or you can focus on the words no longer permitted.

One letter word = 1 word lost

Two letter word: call first letter the forbidden letter. Then there are 65 other letters except for the forbidden one that can't be paired so 65 choices. Multiply by 2 since the forbidden letter can be swapped to the second position = 130 lost words.

Finally, recognize you can have a two letter word with both letters being forbidden = 1 lost word

Total lost words 132, E
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by Scott@TargetTestPrep » Wed May 09, 2018 3:51 pm
swerve wrote:In the Land of OZ, only one or two-letter words are used. The local language has 66 different letters. The parliament decided to forbid the use of the seventh letter. How many words have the people of OZ lost because of the prohibition?

A. 65
B. 66
C. 67
D. 131
E. 132
We are given that in the Land of OZ only one- or two-letter words are used. If all 66 letters can be used, then we have:

1-letter words = 66

2-letter words = 66^2

Thus, there are 66 + 66^2 words if all 66 letters can be used.

When the seventh letter is taken away, we have:

1-letter words = 65

2-letter words = 65^2

Thus, there are 65 + 65^2 words when the seventh letter is taken away.

The number of lost words is:

(66 + 66^2) - (65 + 65^2)

66 + 66^2 - 65 - 65^2

1 + 66^2 - 65^2

Noting that the expression 66^2 - 65^2 is a difference of squares, we have:

1 + (66 - 65)(66 + 65)

1 + (1)(131) = 132

Answer: E

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