Set A consists of three consecutive positive multiples of 3, and set B

Question

Set A consists of three consecutive positive multiples of 3, and set B consists of five consecutive positive multiples of 5. If the sum of the integers in set A is equal to the sum of the integers in set B, what is the least number that could be a member of set A?

(A) 69
(B) 72
(C) 75
(D) 78
(E) 81

(A) 69
(B) 72
(C) 75
(D) 78
(E) 81

Source: Advanced Quant Manhattan Prep 2020

chetan2u · Accepted Answer

Let Set A be { 3x-3, 3x, 3x+3 }, so sum will be 9x, and Set B be { 5y-10, 5y-5, 5y, 5y+5, 5y+10 }, so sum will be 25y.

Given that 9x=25y, so the least value of x will be 25 =>  9*25=25*9

SO x=25, and the least number =  3x-3=3*25-3=72

B

prateek25 · Answer

The three numbers in set A should have a sum divisible by 5.

A. 69 + 72 + 75 = 216- rejected 
B. 72 + 75 + 78= 225 - keep
C. 75 + 78 + 81= 234- rejected 
D 78 + 81 + 84 = 243 - rejected 
E 81 + 84 + 87 = 252- rejected

Ans 72, B

Posted from my mobile device

Kinshook · Answer

Given: Set A consists of three consecutive positive multiples of 3, and set B consists of five consecutive positive multiples of 5. 
Asked: If the sum of the integers in set A is equal to the sum of the integers in set B, what is the least number that could be a member of set A?

A = {3k, 3k+3, 3k+6}; k>0; k is a positive integer
B = {5m, 5m+5, 5m + 10, 5m+15, 5m+20}; m>0; m is a positive integer

3k + 3k + 3 + 3k + 6 = 5m + 5m+5 + 5m + 10 + 5m+15 + 5m+20
9k + 9 = 25m + 50
9k = 25m + 41
 k = \frac{(25m+41)}{9} 
When m = 7;  k = \frac{(175 + 41 = 216)}{9} = 24 
 k_{min} = 24 
 3k_{min} = 72

IMO B

ManishaPrepMinds · Answer

The other two approaches mentioned are the perfect way to solve !!

We can also use POE also to get the answer:

Given: 
 Set A consists of three consecutive positive multiples of 3
 set B consists of five consecutive positive multiples of 5
 the sum of the integers in set A = sum of the integers in set B

Infer: Set A and Set B are evenly spaced sets 
 Thus, Mean of the set = Median of the set 
Sum = number of items* Mean
Sum of Set A = 3 * Mean of Set A
Sum of Set B = 5 *Mean of Set B
=> 3 * Mean of Set A = 5 *Mean of Set B
=>  Mean of Set A is multiple of 5
=> Median of Set A is multiple of 5 
Question is asking for :
the least number that could be a member of set A

(A) 69  Not Possible as the median is 72, not a multiple of 5 
(B) 72 {72, 75, 78 }   Possible, as the median is 75, a multiple of 5. Let's hold it 
(C) 75  Not Possible as the median is 78, not a multiple of 5 
(D) 78   Not Possible as the median is 81, not a multiple of 5 
(E) 81  Not Possible as the median is 84, not a multiple of 5

As we are left with only answer choice B, that's the right answer

ScottTargetTestPrep · Answer

We can let a = median of set A and b = median of set B. Then the sum of the 3 integers in set A would be 3a and the sum of the 5 integers in set B would be 5b. Since they are equal, we have:

3a = 5b

We see that a must be a multiple of 5 since 3 is not a multiple of 5. Since a is a multiple of 5, then (a - 3), the smallest member in set A, must be 3 less than a multiple of 5. Looking at the answer choices, we see that only 72 satisfies the condition because it is 3 less than 75 (which is a multiple of 5). Therefore, 72 could be the least number in set A.

Answer: B

Crytiocanalyst · Answer

3*3x=5*5y 
since the number chosen in 5 multiples were 5y-10, 5y-5, 5y, 5y+5, 5y+10
and 3 multiples are 3x-3 ,3x,3x+3 
=>9x=25y 
=>therefore x=25 and y=9 
and the least number to satisfy the same is 75-3=72
Threfore IMO B

Scuven · Answer

If SUM of B must be equal to SUM of A, then the sum of all the A items MUST BE a multiple of 5.

We thus can go reversal by plugging- in the answers in set A and see whether the sum is a multiple of 5.

IF 72 is the lowest term of A, then set A is (72, 72+3, 72+(2*3)) = 72,75,78

You don't have to calculate the sum, just see if the unit digit of the sum is a 5 or a zero.

Do the same for the other answers and you see that B is the only pattern of which sum gives a number multiple of 5

KabxW · Answer

Here is a 30 second solution to this problem.

Take a look at the options. For the sum to equal we need the sum of three numbers in Set A to be divisible by 5.

So the units digit can be 5 or 0. Nothing else.

Check the options. A+B+C = Nope
B+C+D = Yuppp
C+D+E = Nope

No other combination will be possible as they are consecutive.
Out of B,C,D the smallest is (B) 72
There's your answer.

Kudos please, if I helped

Budhaditya_Saha · Answer

Posted from my mobile device

Why did you take m=9 and not m=1? Because its the least value that has been asked for. I put m=1 and got k=33. Where am I wrong?

DmitryFarberMPrep · Answer

Actually, m can't be 1 or 9. I believe Kinshook meant to test m=7, since 25(7) = 175. We need a value that produces a multiple of 9 in the numerator, and m=7 is the lowest one that produces that result.

If you test m=1, you shouldn't get 33. k = (25*1 + 41)/9  = 67/9 = 7r4 (not an integer).

findingmyself · Answer

Given that Set A consists of three consecutive positive multiples of 3, and set B consists of five consecutive positive multiples of 5.

A = 3x, 3x+3, 3x+6  
  B = 5y, 5y+5, 5y + 10, 5y+15, 5y+20

Since A=B, Summing them together,   
 
 9x + 9 = 25y + 50

Taking 9 out of the brackets 
 9(x + 1) = 25(y + 2)

This can be true when (x+1)=25 and (y+2)=9 (Since 25*9=9*25) 
 x+1=25-->x=24 
 3x= smallest=3*24=72

weirdapoet · Answer

Here is my approach: 
A = sum of multiples of 3
B = sum of multiples of 5 => the last digit of B could be 0 or 5 
Then, plug numbers in those answers to set A, quickly calculate the last digit which is matched the condition above so we have the final answer is B.

CodeBXT · Answer

Hello  chetan2u ,

Your explanation is correct to understand how to get the answer. But the use of 3x-3,3x,3x+3 for set A and 5y-10, 5y-5 does not satisfy the requirement that both Set A and Set B should consist of three consecutive POSITIVE multiples.

Just a little suggestion.

Cheers

Set A consists of three consecutive positive multiples of 3, and set B

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