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Danny spends $360 buying his favorite dolls. If he buys only

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Danny spends $360 buying his favorite dolls. If he buys only  [#permalink]

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New post 29 Jul 2014, 06:55
1
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A
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D
E

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Question Stats:

71% (02:54) correct 29% (02:46) wrong based on 141 sessions

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Danny spends $360 buying his favorite dolls. If he buys only small LemonHead dolls, which are $1 cheaper than the large LemonHead dolls, he could buy 5 more dolls than if he were to buy only large LemonHead dolls. How much does a large LemonHead doll cost?

A. $5
B. $6
C. $7.2
D. $8
E. $9
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Re: Danny spends $360 buying his favorite dolls. If he buys only  [#permalink]

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New post 29 Jul 2014, 07:25
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goodyear2013 wrote:
Danny spends $360 buying his favorite dolls. If he buys only small LemonHead dolls, which are $1 cheaper than the large LemonHead dolls, he could buy 5 more dolls than if he were to buy only large LemonHead dolls. How much does a large LemonHead doll cost?

A. $5
B. $6
C. $7.2
D. $8
E. $9


Let p $ be the price of large lemonhead doll and n be the no. dolls danny can buy

so we have n*p=360

also if price of 1 dollar less then danny can buy 5 more small head dolls so

(p-1)*(n+5)=360=pn..

pn +5p-n-5=pn or 5p= 5+n or \(5p=5+\frac{360}{p}\)

5p^2-360-5p=0

or p^2-p-72=0

p^2-9p+8p-72=0

or p =9 or -8 (price cannot be negative so p =9

note that the problem can be solved by plugging as well..

Starting with D...because C is not great option to start with here

if p=8 then danny can buy 45 large head dolls and p-1=7 $, he can buy \(\frac{360}{7}\neq{Integer}\)So drop this option

Look at p=9 so Large head dolls=40 and for p-1=$8, then he can buy 45 dolls which is exactly what q. stem says..
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Danny spends $360 buying his favorite dolls. If he buys only  [#permalink]

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New post 29 Jul 2014, 08:58
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goodyear2013 wrote:
Danny spends $360 buying his favorite dolls. If he buys only small LemonHead dolls, which are $1 cheaper than the large LemonHead dolls, he could buy 5 more dolls than if he were to buy only large LemonHead dolls. How much does a large LemonHead doll cost?

A. $5
B. $6
C. $7.2
D. $8
E. $9


I'd plug the options for this question. Usually we start with C but since C is not an integer let's skip it for a while and plug D first.

If large doll costs $8, then he can buy 360/8 = 45 large dolls and 360/7 = 50.something small dolls. The difference is more than 5. Discard.

Now, increase the price of the large doll to decrease the difference.

If large doll costs $9, then he can buy 360/9 = 40 large dolls and 360/8 = 45 small dolls. The difference is 5, which is what we wanted.

Answer: E.
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Re: Danny spends $360 buying his favorite dolls. If he buys only  [#permalink]

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New post 31 Jul 2014, 23:16
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Lets say price of big = x

So quantity obtained \(= \frac{360}{x}\) (Total = 360)

So, price of small = x-1

& Qty of small \(= \frac{360}{x} + 5\)

\((x-1)(\frac{360}{x} + 5) = 360\)

\(x^2 - x - 72 = 0\)

x = 9

Answer = E
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Re: Danny spends $360 buying his favorite dolls. If he buys only  [#permalink]

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New post 15 Apr 2015, 03:28
X: Price of small dolls; Y: Price of big dolls.
Y = X +1

360(1/x - 1/y) = 5
<=>x.y = 72
We can make trial: 9x8 = 72.
Choose (E) y = 9.
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Re: Danny spends $360 buying his favorite dolls. If he buys only  [#permalink]

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New post 30 Jan 2016, 18:06
I didn't want to complicate with the algebra...thus, I just plugged in numbers:
6 is way lower than needed, thus we can right away eliminate A, B - because we need higher price.
C is out because it yields a non-integer number when we check for the price of the small doll (360/6.2 = not integer)
D is out, because 360/7 = non-integer.
thus, we are remained with E.
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Re: Danny spends $360 buying his favorite dolls. If he buys only  [#permalink]

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New post 23 Oct 2016, 19:56
S = small doll
L = large doll
D = # of dolls

Use the following formulas:

360/L-1 = D+5
360/L = D

You'll eventually get it down to the following:

5L^2 -5L = 360
(L^2 -L) = 72
L^2 -L -72 = 0
(L+8)(L-9) = 0
L = -8 or 9
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Re: Danny spends $360 buying his favorite dolls. If he buys only  [#permalink]

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New post 18 Mar 2019, 09:32
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goodyear2013 wrote:
Danny spends $360 buying his favorite dolls. If he buys only small LemonHead dolls, which are $1 cheaper than the large LemonHead dolls, he could buy 5 more dolls than if he were to buy only large LemonHead dolls. How much does a large LemonHead doll cost?

A. $5
B. $6
C. $7.2
D. $8
E. $9


Let L = cost of 1 LARGE LemonHead doll
So L-1 = cost of 1 SMALL LemonHead doll

If he buys only small LemonHead dolls, which are $1 cheaper than the large LemonHead dolls, he could buy 5 more dolls than if he were to buy only large LemonHead dolls.
Let's first write a "word equation": (# of LARGE dolls for $360) = (# of SMALL dolls for $360) - 5
Substitute values to get: $360/L = $360/(L-1) - 5

ASIDE: At this point, it might be faster to just TEST THE ANSWER CHOICES to see which one satisfies the above equation.
Or we can just solve the equation for L. Let's do that.

Multiply both sides by L to get: 360 = 360L/(L-1) - 5L
Multiply both sides by (L-1) to get: 360(L-1) = 360L - 5L(L-1)
Expand to get: 360L - 360 = 360L - 5L² + 5L
Subtract 360L from both sides to get: -360 = -5L² + 5L
Rearrange to get: 5L² - 5L - 360 = 0
Divide both sided by 5 to get: L² - L - 72 = 0
Factor to get: (L - 9)(L + 8) = 0
So, EITHER L = 9 OR L = -8
Since L cannot be negative, it must be the case that L = 9

Answer: E

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Brent
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Re: Danny spends $360 buying his favorite dolls. If he buys only  [#permalink]

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New post 19 Mar 2019, 19:11
goodyear2013 wrote:
Danny spends $360 buying his favorite dolls. If he buys only small LemonHead dolls, which are $1 cheaper than the large LemonHead dolls, he could buy 5 more dolls than if he were to buy only large LemonHead dolls. How much does a large LemonHead doll cost?

A. $5
B. $6
C. $7.2
D. $8
E. $9


We can let the cost of a large doll = d and thus the cost of a small doll = d - 1. We can let q = the number of large dolls that can be purchased with $360, and thus q + 5 = the number of small dolls that can be purchased with $360.

We can create the equations:

360 = (d - 1)(q + 5)

360 = dq - q + 5d - 5

and

360 = dq

360/d = q

Substituting, we have:

360 = d(360/d) - 360/d + 5d - 5

360 = 360 - 360/d + 5d - 5

0 = -360/d + 5d - 5

Multiplying the equation by d, we have:

0 = -360 + 5d^2 - 5d

d^2 - d - 72 = 0

(d - 9)(d + 8) = 0

d = 9 or d = -8

Since d can’t be negative, d = 9.

Answer: E
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Re: Danny spends $360 buying his favorite dolls. If he buys only   [#permalink] 19 Mar 2019, 19:11
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