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The average of the five numbers is 6.8. If one of the numbers is multi

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The average of the five numbers is 6.8. If one of the numbers is multi  [#permalink]

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08 Apr 2015, 05:16
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The average of the five numbers is 6.8. If one of the numbers is multiplied by 3, the average of the numbers increases to 9.2. Which of the five numbers is multiplied by 3?

(A) 1.5
(B) 3.0
(C) 3.9
(D) 4.0
(E) 6.0

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The average of the five numbers is 6.8. If one of the numbers is multi  [#permalink]

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08 Apr 2015, 05:54
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1
Or we can just do some very easy algebra.
$$a_1 + a_2 + a_3 + a_4 + a_5 = 34$$(1)
lets say first number was multiplied by 3
$$3*a_1 + a_2 + a_3 + a_4 + a_5 = 46$$ (2)
subtract (2) from (1), $$2*a_1 = 12, a1 = 6$$
Option E
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Re: The average of the five numbers is 6.8. If one of the numbers is multi  [#permalink]

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08 Apr 2015, 05:23
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Initial avg is 6.8
So the summation of the nos is 6.8*5 = 34

Now the new avg is 9.2
So summation will be 9.2*5 = 46.
So change in summation is 12.
This comes when a number is multiplied by 3.
In the original sum we already have that number added once. To bring additional 12 increment by multiplying by 3, we get a total increase of 18. Thus, 6 is multiplied by 3.

We can also do trial and error on all nos and see the knw tht gives 12 increase.

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The average of the five numbers is 6.8. If one of the numbers is multi  [#permalink]

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08 Apr 2015, 09:29
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Bunuel wrote:
The average of the five numbers is 6.8. If one of the numbers is multiplied by 3, the average of the numbers increases to 9.2. Which of the five numbers is multiplied by 3?

(A) 1.5
(B) 3.0
(C) 3.9
(D) 4.0
(E) 6.0

Kudos for a correct solution.

$$x_1 + x_2 + x_3 + x_4 + x_5 = 5*(6.8) = 34$$

$$3x_1 +x_2 + x_3 + x_4 + x_5 = 5*(9.2) = 46$$

Subtract the first from the second:

$$2x_1 = 12$$

$$x_1 = 6.$$

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Re: The average of the five numbers is 6.8. If one of the numbers is multi  [#permalink]

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08 Apr 2015, 13:03
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Bunuel wrote:
The average of the five numbers is 6.8. If one of the numbers is multiplied by 3, the average of the numbers increases to 9.2. Which of the five numbers is multiplied by 3?

(A) 1.5
(B) 3.0
(C) 3.9
(D) 4.0
(E) 6.0

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Let X = {x1,x2,x3,x4,x5}

Sum(X) = 6.8*5 = 34

Let y be the no. that is multiplied by 3

Thus, Sum(X) - y +y*3 = 9.2*5

=> 34 +2y = 46 => 2y = 12 => y = 6

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Re: The average of the five numbers is 6.8. If one of the numbers is multi  [#permalink]

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08 Apr 2015, 14:43
2
Equation 1: a+b+c+d+e=6.8*5=34
Equation 2: 3a+b+c+d+e=9.2*5=46
Deducting equation 1 from 2 we get : 2a=46-34=12=>a=6

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Re: The average of the five numbers is 6.8. If one of the numbers is multi  [#permalink]

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09 Apr 2015, 03:28
1

a+b+c+d+e = 6.8 * 5
3a+b+c+d+e = 9.2*5

by subtracting the two equations we get
2a = 12 thus a = 6
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Re: The average of the five numbers is 6.8. If one of the numbers is multi  [#permalink]

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13 Apr 2015, 06:13
Bunuel wrote:
The average of the five numbers is 6.8. If one of the numbers is multiplied by 3, the average of the numbers increases to 9.2. Which of the five numbers is multiplied by 3?

(A) 1.5
(B) 3.0
(C) 3.9
(D) 4.0
(E) 6.0

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

You can do this problem in a few different ways, but perhaps the best way is Algebra! No matter how you choose the address the question you will need to determine the magnitude of the increase. Since “sum (total) = average * # of terms” You can take the average of 6.8 times the five terms and get a beginning total of 34. The new total is 9.2 times 5 which equals 46. So the increase is 12.

In order to create an equation you need to ask yourself “what happened to cause that increase of 12?” The question stem tells you that one of the numbers was multiplied by 3. So when one of the numbers (we can call that number “x”) was multiplied by 3 the total increased by 12.

The equation formed from this information is simply “3x = x + 12.” The “3x” is because the number is multiplied by 3 and the “x + 12” is because you had the x to start with (there were five numbers right? and x was one of them) and you added 12 because of the increase to the sum.

So if “3x = x + 12” then x = 6. So the correct answer is E.

This question can be done based on knowledge of number properties and can even be done by working directly with the answer choices. However, neither of these methods is as reliable for most students as the algebra is. I have worked with the question for years and I can tell you that more people choose D than choose the correct answer. Yet very few of the people who get this wrong used algebra. Those who use algebra generally seem to get this question right.

Make sure that you are very comfortable with algebra, after all, bringing your “A Game” is essential to your success on the Quant section!
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Re: The average of the five numbers is 6.8. If one of the numbers is multi  [#permalink]

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08 Aug 2017, 00:07
Excellent Question.
Here is what I did on this question =>
Sum (initial) => 5*6.8 => 34
Sum (final) => 5*9.2 => 46

Now difference => 12

Let p be the number multiplied by 3.

p => 3p

Thus we added 2p to the list.

Therefore => 2p = 12

Thus p = 6

Hence D.

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Re: The average of the five numbers is 6.8. If one of the numbers is multi  [#permalink]

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19 Nov 2018, 14:00
Bunuel wrote:
The average of the five numbers is 6.8. If one of the numbers is multiplied by 3, the average of the numbers increases to 9.2. Which of the five numbers is multiplied by 3?

(A) 1.5
(B) 3.0
(C) 3.9
(D) 4.0
(E) 6.0

Kudos for a correct solution.

this is pretty good question

x1+x2+x3+x4+x5=5∗(6.8)=34x1+x2+x3+x4+x5=5∗(6.8)=34

3x1+x2+x3+x4+x5=5∗(9.2)=463x1+x2+x3+x4+x5=5∗(9.2)=46

Subtract the first from the second:

-After subtracting above average we will get 12
till this step i understood

then i divided by 3 i got answer 4 and not 6

where i got wrong??
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The average of the five numbers is 6.8. If one of the numbers is multi  [#permalink]

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20 Nov 2018, 05:13
gmatdordie wrote:
Bunuel wrote:
The average of the five numbers is 6.8. If one of the numbers is multiplied by 3, the average of the numbers increases to 9.2. Which of the five numbers is multiplied by 3?

(A) 1.5
(B) 3.0
(C) 3.9
(D) 4.0
(E) 6.0

Kudos for a correct solution.

this is pretty good question

x1+x2+x3+x4+x5=5∗(6.8)=34x1+x2+x3+x4+x5=5∗(6.8)=34

3x1+x2+x3+x4+x5=5∗(9.2)=463x1+x2+x3+x4+x5=5∗(9.2)=46

Subtract the first from the second:

-After subtracting above average we will get 12
till this step i understood

then i divided by 3 i got answer 4 and not 6

where i got wrong??

you need to divide by 2 , not 3

3x1 - x1 = 2x1 = 12 ; x1 = 6
and assume better variables. That was really hard to look at
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Re: The average of the five numbers is 6.8. If one of the numbers is multi  [#permalink]

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20 Nov 2018, 05:36
Bunuel wrote:
The average of the five numbers is 6.8. If one of the numbers is multiplied by 3, the average of the numbers increases to 9.2. Which of the five numbers is multiplied by 3?

(A) 1.5
(B) 3.0
(C) 3.9
(D) 4.0
(E) 6.0

Kudos for a correct solution.

When you increase a number (by multiplying it by 3), the avg goes from 6.8 to 9.2 i.e. an increase of 2.4 in all 5 numbers.
This is a total increase of 2.4*5 = 12

This 12 represents twice the original number because when we multiply a number by 3, we add twice the number to it. So original number is 12/2 = 6

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Re: The average of the five numbers is 6.8. If one of the numbers is multi  [#permalink]

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24 Aug 2019, 05:30
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Bunuel wrote:
The average of the five numbers is 6.8. If one of the numbers is multiplied by 3, the average of the numbers increases to 9.2. Which of the five numbers is multiplied by 3?

(A) 1.5
(B) 3.0
(C) 3.9
(D) 4.0
(E) 6.0

Kudos for a correct solution.

The sum of the original 5 numbers is 6.8 x 5 = 34.

If we let k = the number that is multiplied by a factor of 3, then the new sum is 34 - k + 3k = 34 + 2k. We can create the equation:

(34 + 2k)/5 = 9.2

34 + 2k = 46

2k = 12

k = 6

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The average of the five numbers is 6.8. If one of the numbers is multi  [#permalink]

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24 Aug 2019, 20:03
Bunuel wrote:
The average of the five numbers is 6.8. If one of the numbers is multiplied by 3, the average of the numbers increases to 9.2. Which of the five numbers is multiplied by 3?

(A) 1.5
(B) 3.0
(C) 3.9
(D) 4.0
(E) 6.0

let t=original total
n=original number
eq1: t=5*6.8
eq2: t+2n=5*9.2
subtracting eq1 from eq2,
n=6.0
E
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Re: The average of the five numbers is 6.8. If one of the numbers is multi  [#permalink]

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24 Aug 2019, 20:47
a1+....=34=(5*6.8)
3a1+....=46=(5*9.2)
3a1-a1=12
2a1=12
a1=6
Option E

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Re: The average of the five numbers is 6.8. If one of the numbers is multi  [#permalink]

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24 Aug 2019, 21:13

x+y=6.8*5 ====> x+y=34
x+3y=9.2*5 ====> x+3y=46

solving the two equations for y we get y=6.
y is the number that is multiplied by 3.
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Re: The average of the five numbers is 6.8. If one of the numbers is multi  [#permalink]

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24 Aug 2019, 22:13
Bunuel wrote:
The average of the five numbers is 6.8. If one of the numbers is multiplied by 3, the average of the numbers increases to 9.2. Which of the five numbers is multiplied by 3?

(A) 1.5
(B) 3.0
(C) 3.9
(D) 4.0
(E) 6.0

Kudos for a correct solution.

Given:
1. The average of the five numbers is 6.8.
2. One of the numbers is multiplied by 3, the average of the numbers increases to 9.2.

Asked: Which of the five numbers is multiplied by 3?

Total of 5 numbers = 6.8*5 = 34
Let the number multiplied by 3 be x
34 + 2x = 9.2*5 = 46
x =6

IMO E
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Re: The average of the five numbers is 6.8. If one of the numbers is multi  [#permalink]

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24 Aug 2019, 23:33
Let Z be the sum of the four number that is not multiplied by 3
Let Y be the number that is multiplied by 3

Z + Y = 5*6.8 = 34.....(1)
Z + 3Y = 5*9.2 = 46...(2)

Subtract (1) from (2)

2Y = 12
Y = 6

Hence, E.
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Re: The average of the five numbers is 6.8. If one of the numbers is multi  [#permalink]

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25 Aug 2019, 11:05
Bunuel wrote:
The average of the five numbers is 6.8. If one of the numbers is multiplied by 3, the average of the numbers increases to 9.2. Which of the five numbers is multiplied by 3?

(A) 1.5
(B) 3.0
(C) 3.9
(D) 4.0
(E) 6.0

Kudos for a correct solution.

sum of 5 nos; 34
3x+sum of 4 no ; 46
subtract ; 2x=12 ; x=6
IMOE
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Re: The average of the five numbers is 6.8. If one of the numbers is multi   [#permalink] 25 Aug 2019, 11:05
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