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D01-11

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D01-11  [#permalink]

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New post 15 Sep 2014, 23:11
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A
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E

Difficulty:

  25% (medium)

Question Stats:

75% (01:19) correct 25% (01:16) wrong based on 164 sessions

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If \(S\) is the sum of the digits of a given number, \(T\) is the sum of the digit of \(S\), and \(G\) is the sum of digits in \(T\). For example \(S\) of 987 is \(9+8+7 = 24\), \(T\) of \(S\) is \(2+4 = 6\) and \(G\) of 6 is 6. Therefore \(G\) of 987 is 6. Which of the following has the greatest \(G\)?

A. 94123
B. 91964
C. 64678
D. 62355
E. 45689

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Re D01-11  [#permalink]

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New post 15 Sep 2014, 23:11
Official Solution:

If \(S\) is the sum of the digits of a given number, \(T\) is the sum of the digit of \(S\), and \(G\) is the sum of digits in \(T\). For example \(S\) of 987 is \(9+8+7 = 24\), \(T\) of \(S\) is \(2+4 = 6\) and \(G\) of 6 is 6. Therefore \(G\) of 987 is 6. Which of the following has the greatest \(G\)?

A. 94123
B. 91964
C. 64678
D. 62355
E. 45689


The explanation follows as under: \(G\) of 45689 is 5 whereas the rest have \(G\) smaller than 5.

A. \(S = 9+4+1+2+3 = 19\), \(T = 1+9 = 10\) and \(G = 1+0 = 1\).

B. \(S = 29\), \(T = 11\) and \(G = 2\).

C. \(S = 31\), \(T = 4\), and \(G = 4\).

D. \(S = 21\), \(T = 3\), and \(G = 3\).

E. \(S = 32\), \(T = 5\), and \(G = 5\).

Therefore E is the highest.


Answer: E
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Re: D01-11  [#permalink]

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New post 04 Dec 2014, 10:05
Is there a shortcut to do this? It's not hard to add up all the options manually..
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Re: D01-11  [#permalink]

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New post 05 Dec 2014, 05:41
E
But took 2.15sec
Solved by adding each Digits in mind and noted down
Finding Out just S is Enough
Greater S value >>Mean Greater T >> Greater G!
:-D
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Re: D01-11  [#permalink]

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New post 05 Dec 2014, 14:56
kanusha wrote:
E
Greater S value >>Mean Greater T >> Greater G!
:-D


it doesn't work
look
S in B) is greater than S in D), however T is greater in D) than in B)
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Re: D01-11  [#permalink]

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New post 20 Feb 2016, 13:39
this requires just the ability to process basic arithmetic quickly in your head.
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Re: D01-11  [#permalink]

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New post 05 Oct 2016, 04:50
another way is to cancel out the number in your head first cut 9 then 5 etc.. then u will be left with e last value with maximum weightage
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Re D01-11  [#permalink]

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New post 16 Jul 2017, 11:28
I think this is a poor-quality question and I agree with explanation. Because there is no shorter way to solve this question.
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Re: D01-11  [#permalink]

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New post 18 Jul 2017, 08:53
seeing the question stem I understand that its simply adding the individual digits of each number. so option E yields 32 which is the highest.
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Re: D01-11  [#permalink]

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New post 21 Aug 2017, 14:28
rayzha wrote:
Is there a shortcut to do this? It's not hard to add up all the options manually..


Yes, these type of problems are called digit sum. The shortest way is to find out the remainder when divided by 9.
This should give the answer.

Cheers!
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Re: D01-11  [#permalink]

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New post 16 Oct 2017, 02:49
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Hello umg,

I too had the same concern. I did some digging on this and found some interesting properties. You too may find them useful.

Sum of individual sums of a number is called Digit sum or Beejank (in Vedic maths). One shortcut to quickly arrive at the digit sum is to simply ignore 9's. Let me take the first option - 94123. The Digit sum is - 9 + 4 + 1 + 2 + 3 = 1 (Simply strike out the digit 9 or the numbers that add up to 9. You can try it on others options and find that this holds true.

While I was at it, I also discovered that digit-sum/Beejank can be used to quickly check if the addition/multiplications done are correct. There are a lot of links that explain this approach. One such link is - http://www.quickermaths.com/checking-of ... out-nines/

umg wrote:
I think this is a poor-quality question and I agree with explanation. Because there is no shorter way to solve this question.

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D01-11  [#permalink]

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New post 25 Nov 2017, 22:04
I solved the question using simple addition. I don't know the probability of this question showing up on the GMAT exam. but thanks to susheelh and his method:

susheelh wrote:

While I was at it, I also discovered that digit-sum/Beejank can be used to quickly check if the addition/multiplications done are correct. There are a lot of links that explain this approach. One such link is - http://www.quickermaths.com/checking-of ... -out-nines


I could solve the question in a minute or a few seconds less.
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D01-11  [#permalink]

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New post 21 Aug 2018, 09:45
A. 94123
B. 91964
C. 64678
D. 62355
E. 45689


Hi,Only way to solve is to add up?
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D01-11 &nbs [#permalink] 21 Aug 2018, 09:45
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