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Caleb spends $72.50 on 50 hamburgers for the marching band. If single

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Bunuel
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Caleb spends $72.50 on 50 hamburgers for the marching band. If single [#permalink] New post   30 Jun 2015, 04:25
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Caleb spends $72.50 on 50 hamburgers for the marching band. If single burgers cost $1.00 each and double burgers cost $1.50 each, how many double burgers did he buy?

A. 5
B. 10
C. 20
D. 40
E. 45

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E
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Re: Caleb spends $72.50 on 50 hamburgers for the marching band. If single [#permalink] New post   30 Jun 2015, 04:38
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Caleb spends $72.50 on 50 hamburgers for the marching band. If single burgers cost $1.00 each and double burgers cost $1.50 each, how many double burgers did he buy?

A. 5
B. 10
C. 20
D. 40
E. 45

Solution -

Lets say, Single hamburgers x and Double hamburgers y
Given that,
x+y=50 and 1x+1.5y=72.50. By solving the equations y=45. ANS E.

Thanks
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Re: Caleb spends $72.50 on 50 hamburgers for the marching band. If single [#permalink] New post   30 Jun 2015, 10:10
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Bunuel wrote:
Caleb spends $72.50 on 50 hamburgers for the marching band. If single burgers cost $1.00 each and double burgers cost $1.50 each, how many double burgers did he buy?

A. 5
B. 10
C. 20
D. 40
E. 45

Kudos for a correct solution.


No. of Single Burger = S
No. of Double Burger = D

$1*S + $1.5*D = 72.50-----------Equation (1)

Total Burgers = S + D = 50-----------Equation (2)

Solving the two equations Simultaneously [Subtract Eq(1) from Eq(2)]

($1*S + $1.5*D) - (S + D) = 72.5 - 50

0.5*D = 22.5

i.e. D = 45
i.e. S = 50-45 = 5

Answer: Option E
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Re: Caleb spends $72.50 on 50 hamburgers for the marching band. If single [#permalink] New post   30 Jun 2015, 10:16
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suppose no. of single burger is x , and no. of double burger is y

So , x+ y = 50 ------1
and x +1.5 y = 72.5 ------2

Substract eqn 1 from 2

we get , 0.5 y= 22.5
so y =45

answer is E
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Re: Caleb spends $72.50 on 50 hamburgers for the marching band. If single [#permalink] New post   30 Jun 2015, 10:20
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x+1.5y = 72.5
x+y = 50
------------------
0.5y = 22.5
y = 45, Ans is E.
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Re: Caleb spends $72.50 on 50 hamburgers for the marching band. If single [#permalink] New post   30 Jun 2015, 10:29
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Let # of single burger=x , # of double =y
So we have x+y=50 ---1
We also have x*1+y*1.5 =72.5 =>2x+3y=145 ----2
With equation 1 &2 we can find out y =45

Hence answer is E
Thanks,
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Re: Caleb spends $72.50 on 50 hamburgers for the marching band. If single [#permalink] New post   06 Jul 2015, 01:29
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Bunuel wrote:
Caleb spends $72.50 on 50 hamburgers for the marching band. If single burgers cost $1.00 each and double burgers cost $1.50 each, how many double burgers did he buy?

A. 5
B. 10
C. 20
D. 40
E. 45

Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

Let s = the number of single burgers purchased
Let d = the number of double burgers purchased

Caleb bought 50 burgers: s + d = 50

Caleb spent $72.50 in all: s + 1.5d = 72.50

Combine the two equations by subtracting equation 1 from equation 2.

s + 1.5d = 72.50
-(s + d = 50)
______________
0.5d = 22.5

d = 45.

Answer: E.
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Re: Caleb spends $72.50 on 50 hamburgers for the marching band. If single [#permalink] New post   07 Jul 2015, 05:20
Bunuel wrote:
Caleb spends $72.50 on 50 hamburgers for the marching band. If single burgers cost $1.00 each and double burgers cost $1.50 each, how many double burgers did he buy?

A. 5
B. 10
C. 20
D. 40
E. 45

Kudos for a correct solution.


single burger= x
double burger =y

(1)x+ (1.5)y= 72.5

Although there is another equation x+y=50. I ignore it. I looked into the answer choice which I can multiply in 1.5 and its tenth digit 0.5 so when subtracting from 72.5 give me integer number. Two choices give 5 & 45. Logically stares with 45

x +(1.5) 45= 72.5..........x= 72.5- 67.5 = x=5 so 5+45=50


answer E
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Re: Caleb spends $72.50 on 50 hamburgers for the marching band. If single [#permalink] New post   10 Oct 2016, 15:47
Bunuel wrote:
Bunuel wrote:
Caleb spends $72.50 on 50 hamburgers for the marching band. If single burgers cost $1.00 each and double burgers cost $1.50 each, how many double burgers did he buy?

A. 5
B. 10
C. 20
D. 40
E. 45

Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

Let s = the number of single burgers purchased
Let d = the number of double burgers purchased

Caleb bought 50 burgers: s + d = 50

Caleb spent $72.50 in all: s + 1.5d = 72.50

Combine the two equations by subtracting equation 1 from equation 2.

s + 1.5d = 72.50
-(s + d = 50)
______________
0.5d = 22.5

d = 45.

Answer: E.



Well, I didn't solve it this way, but I believe that this method works as well.
$1x+ $1.5x= $72.5
So, $2.5x= $72.5
So, x= 29
So 1*29= 29 single burgers and
1.5* 29= 43.5 double burgers
It's not 45, but it's not either any of the other choices, so choice E is the closest.
What do you think Bunuel?
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Re: Caleb spends $72.50 on 50 hamburgers for the marching band. If single [#permalink] New post   10 Oct 2016, 19:53
Bunuel can you please tell why the above method I'd wring silver by jennisa

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Re: Caleb spends $72.50 on 50 hamburgers for the marching band. If single [#permalink] New post   10 Oct 2016, 23:54
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ashishahujasham wrote:
Bunuel can you please tell why the above method I'd wring silver by jennisa

Sent from my A1601 using GMAT Club Forum mobile app


No, that method is not correct. jennisa assumes that the number of single and double burgers is the same, hence she denotes both of them by x (in the correct solution we denoted those with s and d). Next, when she multiplies x (the # of burgers) by $1.5, she gets the amount of money but she assumes it's the number of double burgers, which is wrong.
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Re: Caleb spends $72.50 on 50 hamburgers for the marching band. If single [#permalink] New post   16 Jul 2017, 23:52
Though many people have solved by using the Algebraic method, I guess this can also be solved by looking at the answer choice.

Since the Double burger cost is $1.5, single burger is $1 and total is $72.5, the number of Double burger HAS to be odd. If not, the total cost will not end with 0.5. This Eliminates B,C,D. Answer is either A or E. A quick check will confirm that E is the correct answer.

Bunuel wrote:
Caleb spends $72.50 on 50 hamburgers for the marching band. If single burgers cost $1.00 each and double burgers cost $1.50 each, how many double burgers did he buy?

A. 5
B. 10
C. 20
D. 40
E. 45

Kudos for a correct solution.
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Re: Caleb spends $72.50 on 50 hamburgers for the marching band. If single [#permalink] New post   20 Jul 2017, 16:33
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Bunuel wrote:
Caleb spends $72.50 on 50 hamburgers for the marching band. If single burgers cost $1.00 each and double burgers cost $1.50 each, how many double burgers did he buy?

A. 5
B. 10
C. 20
D. 40
E. 45


We can let x = the number of single burgers and y = the number of double burgers, and thus we can create the following two equations:

x + y = 50

and

x + 1.5y = 72.50

If we subtract the first equation from the second, we have:

0.5y = 22.50

Multiply both sides by 2, we have:

y = 45

Answer: E
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Re: Caleb spends $72.50 on 50 hamburgers for the marching band. If single [#permalink] New post   14 Jan 2020, 10:56
Bunuel wrote:
Caleb spends $72.50 on 50 hamburgers for the marching band. If single burgers cost $1.00 each and double burgers cost $1.50 each, how many double burgers did he buy?

A. 5
B. 10
C. 20
D. 40
E. 45

Kudos for a correct solution.


\(s + d = 50\)------------------->(I)

Now, \(1*s + \frac{3d}{2} = \frac{725}{10}\)

Or, \(\frac{10s}{10} + \frac{15d}{10} = \frac{725}{10}\)

Or, \(10s + 15d = 725\)

Or, \(2s + 3d = 145\)------------------->(II)

(I) *2 = > \(2s + 2d = 100\)------------------->(III)


Now, (II) - (III) => \(d = 45\), Hence Answer must be (E) 45
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Re: Caleb spends $72.50 on 50 hamburgers for the marching band. If single [#permalink]
14 Jan 2020, 10:56
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