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Re: A certain marching band has fewer than 50 members. During performances [#permalink]
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tdrowberry wrote:
Sorry, but this is a problem I just do not understand. How could we be sure the lowest combo of the two formulas would be the correct one that seven would also be apart of? Why is there an LCM if we don't use it?



1) Why would the combo suit the third formation too?
Finally the number of people are the same in all three formations, they are same set of people. Since you know only possibility of number people is 23, you can place 23 in rows of 7 to get remainder.

2) Why LCM?
LCM would help you get the next numbers. Here we know that number of people is less than 50 and there is only one possibility. But say the number was between 50 and 100. Then next would be 30+23 and 30+30+23 or 53 and 83. So here you would have two possibilities and if you put them in rows of 6..
53 is 7*7+4, so 4 people in last row.
83 is 7*11+6, so 6 people in last row.
Two answers here, but we had our single solution when it was less than 50.
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Re: A certain marching band has fewer than 50 members. During performances [#permalink]
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let rows in A be x 5x+3
and rows in B be y 6y+5
5x+3 = 6y+5
2=5x-6y
possible when x is 4 and y is 3
we get total as 23
so in formation C we will have 3 rows of 7 each and 2 in last
option B


hugogva wrote:
A certain marching band has fewer than 50 members. During performances, the members march in different formations. When they march in Formation A, each row has 5 members, except the last row, which has 3 members. When they march in Formation B, each row has 6 members, except the last row, which has 5 members. If they march in Formation C, each row has 7 members, except the last row. How many band members are in the last row in Formation C?

A) 1
B) 2
C) 3
D) 4
E) 6
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Re: A certain marching band has fewer than 50 members. During performances [#permalink]
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Let
#of people in the band = n
#of rows in formation A = x
#of rows in formations B = y

Then, A) n = 5x + 3
B) n = 6y+ 5
Or, 5x + 3 = 6y + 5
Now no. Of rows must be an interger so ,
5x = 6y+ 2
X = (6y + 2) / 5

For y = 3 ; x will be integer i.e X= 4

#of people in the band will be n = 5.4 + 3 = 23

Now from the options, only 2 satisfies
23 = 7.3 + 2

Ans : 2

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Re: A certain marching band has fewer than 50 members. During performances [#permalink]
First Column
Formation A -> Write out multiples of 5 under 50.
Formation B -> Write out multiples of 6 under 50.
Formation C -> Write out multiples of 7 under 50.

Second Column
Formation A -> Add 3 to every multiple of 5
Formation B -> Add 5 to every multiple of 6

Formation A -> 20 + 3 = 23
Formation B -> 18 + 5 = 23
This is the only common number, thus there are 23 people in the band.

Formation C -> 7 people per row means 21 + ___ -> this number must be 2.

Option B
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Re: A certain marching band has fewer than 50 members. During performances [#permalink]
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