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# In a history class of a humanities course, there are N students

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DS Forum Moderator
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In a history class of a humanities course, there are N students  [#permalink]

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25 Mar 2018, 23:19
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00:00

Difficulty:

45% (medium)

Question Stats:

63% (02:10) correct 38% (02:00) wrong based on 48 sessions

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In a history class of a humanities course, there are N students present on a particular day. If N is a two digit number, then what is the value of N?

(1) If 2 more students had been present in the class, they could have been evenly divided in groups of 5 each.

(2) If 2 less students had been present in the class, they could have been evenly divided in groups of 7 each.
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Re: In a history class of a humanities course, there are N students  [#permalink]

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26 Mar 2018, 00:38
amanvermagmat wrote:
In a history class of a humanities course, there are N students present on a particular day. If N is a two digit number, then what is the value of N?

(1) If 2 more students had been present in the class, they could have been evenly divided in groups of 5 each.

(2) If 2 less students had been present in the class, they could have been evenly divided in groups of 7 each.

We'll translate our statements into equations so we understand what we need to do.
This is a Precise approach.

We need to solve for N.
(1) This tells us that N + 2 = 5k, that is, it is divisible by 5. Selecting k = 1 or k = 2 give different values of N.
Insufficient.

(2) This tells us that N - 2 = 7p, that is, it is divisible by 7. Similarly to the above, this is insufficient.
Insufficient.

Combined:
We know have two equations: N+2=5k and N-2=7p
Subtracting the second from the first gives 4=5k-7p. We know that both k and p are integers so we can cycle through the different options.
We'll cycle through values of 'p' as there are less of them.
If p=1 --> 5k=11. Impossible.
If p=2 we add 7 to the above --> 5k=11+7=18. Impossible.
p=3 --> 5k = 18+7 = 25 --> k=5. In this case N = 23
p=4 --> 5k = 25+7 = 32. Impossible.
p=5 --> 5k = 32+7 = 39. Impossible.
p=6 --> 5k = 39+7 = 46. Impossible.
p = 7--> 5k = 46+7 = 53. Impossible.
p = 8 --> 5k = 53+7 = 60 ---> k = 12. In this case N = 58.
Then (Combined) is still insufficient.

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Re: In a history class of a humanities course, there are N students  [#permalink]

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26 Mar 2018, 04:01
1
amanvermagmat wrote:
In a history class of a humanities course, there are N students present on a particular day. If N is a two digit number, then what is the value of N?

(1) If 2 more students had been present in the class, they could have been evenly divided in groups of 5 each.

(2) If 2 less students had been present in the class, they could have been evenly divided in groups of 7 each.

it is clear that A & B are not sufficient.

solving them together:

Considering statement 1 :
we have all the digits from 1 to 99 that has unit digit of 3 or 8 ,, only then adding two will be evenly divisible by 5.

now, Considering statement 2, keeping in mind that the number must be ending with 3 or 8 :
7 14 21 28 35 42 49 56 63 70 77 84 91 98

9 16 23 30 37 44 51 58 65 72 79 86 93 100

more than one solution hence, not sufficient.

kudoss, if it helped.
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Re: In a history class of a humanities course, there are N students  [#permalink]

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26 Mar 2018, 14:24
Statement 1
$$N + 2 = 5k$$ --> $$N = 5k - 2$$ where $$k$$ is an integer
Make a quick list of possible values of N:
{13, 18, 23, 28 etc}
Not sufficient

Statement 2
$$N - 2 = 7p$$ --> $$N = 7p + 2$$ where $$p$$ is an integer
Make a quick list of possible values of N:
{16, 23, 30, etc}
Not sufficient

Combine 1 & 2
Per statement 1, N can be {13, 18, 23, 28 etc}
Per statement 2, N can be {16, 23, 30, etc}
Note that the first term in common is 23. The next term in common will be: 23 + LCM(5,7) = 23 + 35 = 58
More than 2 values of N are possible.
Not sufficient

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Re: In a history class of a humanities course, there are N students  [#permalink]

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31 Mar 2018, 05:28
amanvermagmat wrote:
In a history class of a humanities course, there are N students present on a particular day. If N is a two digit number, then what is the value of N?

(1) If 2 more students had been present in the class, they could have been evenly divided in groups of 5 each.

(2) If 2 less students had been present in the class, they could have been evenly divided in groups of 7 each.

St 1: N + 2 = 5k , Or N = 5k - 2 -------- (I) Insufficient as we have no info about N and k

N can be 3, 8, 13, 18, 23 etc

St 2: N - 2 = 7k , Or N = 7k + 2 -------(II) Insufficient as we have no info about N and k

N can be 9, 16, 23, 30 etc

We need to check to for some other values with formula : (LCM of 5 & 7) + (first common value)

35k + 23 : - common values are 23, 58, 93 etc.

Insufficient

(E)

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Re: In a history class of a humanities course, there are N students &nbs [#permalink] 31 Mar 2018, 05:28
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