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When the price of oranges is lowered by 40%, 4 more oranges

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farukqmul
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When the price of oranges is lowered by 40%, 4 more oranges [#permalink] New post   Updated on: 08 Jul 2012, 06:22
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When the price of oranges is lowered by 40%, 4 more oranges can be purchased for $12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price?

(A) 8
(B) 12
(C) 16
(D) 20
(E) 24
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B

Originally posted by farukqmul on 08 Jul 2012, 06:19.
Last edited by Bunuel on 08 Jul 2012, 06:22, edited 1 time in total.
Edited the question.
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Bunuel
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Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] New post   08 Jul 2012, 06:33
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farukqmul wrote:
When the price of oranges is lowered by 40%, 4 more oranges can be purchased for $12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price?

(A) 8
(B) 12
(C) 16
(D) 20
(E) 24


Say \(x\) is the original price of an orange, then:

\(xn=12\);
and
\(0.6x*(n+4)=12\) --> \(x(n+4)=20\) --> \(xn+4x=20\) --> \(12+4x=20\) --> \(x=2\).

So, for 24 dollars 24/2=12 oranges can be purchased at the original price of $2.

Answer: B.
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Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] New post   15 Jul 2012, 03:22
Bunuel wrote:
farukqmul wrote:
When the price of oranges is lowered by 40%, 4 more oranges can be purchased for $12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price?

(A) 8
(B) 12
(C) 16
(D) 20
(E) 24


Say \(x\) is the original price of an orange, then:

\(xn=12\);
and
\(0.6x*(n+4)=12\) --> \(x(n+4)=20\) --> \(xn+4x=20\) --> \(12+4x=20\) --> \(x=2\).

So, for 24 dollars 24/2=12 oranges can be purchased at the original price of $2.

Answer: B.

are there any other ways?
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Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] New post   22 Oct 2016, 12:32
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Set up:

12/p = x
12/0.60 = x+4

Manipulate and plug the first equation into the second --> you'll find p = $2

Thus $24/$2 per orange = 12 oranges
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Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] New post   22 Oct 2016, 18:22
farukqmul wrote:
When the price of oranges is lowered by 40%, 4 more oranges can be purchased for $12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price?

(A) 8
(B) 12
(C) 16
(D) 20
(E) 24


let r=number of oranges purchased originally
[12/(r+4)]/(12/r)=.6
r/(r+4)=.6
r=6 oranges
$12/6=$2 per orange
$24/$2=12 oranges
B
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Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] New post   21 Apr 2017, 01:28
Here is what i did in this question =>

Let the price of each orange be $x
Hence for $12 => we can buy => 12/x oranges

Now new price => x(1-40/100) => $0.6x
Hence for $12 we could now buy => 12/0.6x


As per the given information => 12/x+4=12/0.6x
Hence 12+4x/x=20/x
=> 12+4x=20
=>x=2

So the price of each orange is $2

Now for 24 dollars => we can buy => 1/2*24=12 oranges

SMASH THAT B.
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Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] New post   30 Jun 2018, 19:53
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12/(.6x)=12/x+4
20=12+4x
x=2
24/2

Answer B
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Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] New post   04 Jul 2018, 19:25
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farukqmul wrote:
When the price of oranges is lowered by 40%, 4 more oranges can be purchased for $12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price?

(A) 8
(B) 12
(C) 16
(D) 20
(E) 24


We use the equation: price per item x no. of items = total cost. Here, we let p = the original price of an orange and q = the original number of oranges purchased. We can create the equation for the original total cost:

pq = 12

q = 12/p

After the orange’s price is lowered, we have that 0.6p = the new (reduced) price of an orange and (q + 4) = the new number of oranges that can be purchased at the reduced price. Our new equation for total cost is:

(0.6p)(q + 4) = 12

0.6pq + 2.4p = 12

Substituting for q, we have:

0.6p(12/p) + 2.4p = 12

7.2 + 2.4p = 12

2.4p = 4.8

p = 2

That is, each orange is $2. So for $24, we can buy 24/2 = 12 oranges.

Answer: B
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Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] New post   09 Sep 2019, 11:02
Expert Reply
farukqmul wrote:
When the price of oranges is lowered by 40%, 4 more oranges can be purchased for $12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price?

(A) 8
(B) 12
(C) 16
(D) 20
(E) 24


Let P denote the price per orange (before discount) and Q denote the number of oranges that can be bought for $12 (before discount). We can create the equations:

P * Q = 12

Q = 12/P

and

0.6P * (Q + 4) = 12

0.6PQ + 2.4P = 12

Substituting, we have:

0.6P * 12/P + 2.4P = 12

7.2 + 2.4P = 12

2.4P = 2.8

P = 2, so for 24 dollars, 12 oranges can be purchased.

Answer: B
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Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] New post   24 Sep 2020, 09:45
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farukqmul wrote:
When the price of oranges is lowered by 40%, 4 more oranges can be purchased for $12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price?

(A) 8
(B) 12
(C) 16
(D) 20
(E) 24


The price of oranges is lowered by 40%
Let x = the ORIGINAL price per orange (in dollars)
So, 0.6x = the REDUCED price per orange

4 more oranges can be purchased for $12 than can be purchased for the original price.
Let's start with a "word equation"

(number of oranges that can be purchased for $12 at the REDUCED price) = (number of oranges that can be purchased for $12 at the ORIGINAL price) + 4
Substitute values to get: 12/0.6x = 12/x + 4
Multiply both sides of the equation by x to get: 12/0.6 = 12 + 4x
Simplify: 20 = 12 + 4x
Subtract 12 from both sides: 8 = 4x
Solve: x = 2
In other words, the ORIGINAL price per orange is $2

How many oranges can be purchased for 24 dollars at the ORIGINAL price?
24/2 = 12 oranges

Answer: B
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Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] New post   24 Sep 2020, 11:48
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Let x be the original price per orange.
Suppose, we buy y oranges, then,
xy =12

Now, after reduction:
0.6x (y+4) =12
0.6xy + 2.4x = 12
0.6*12 + 2.4x =12
x=2

So, total number of oranges which can be bought = 24/2 = 12
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Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] New post   19 Oct 2021, 23:47
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farukqmul wrote:
When the price of oranges is lowered by 40%, 4 more oranges can be purchased for $12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price?

(A) 8
(B) 12
(C) 16
(D) 20
(E) 24


Here is how one can do it without any variables:

We know, Revenue = Price * Quantity
So if price becomes 60% i.e. 3/5, the quantity will become 5/3 to retain the same total amount of $12.

New quantity of oranges : Old quantity of oranges = 5:3
The difference of 2 on ratio scale is actually 4 oranges so multiplier is 2.
Hence we can now purchase 10 oranges in $12 where initially we could purchase only 6 oranges for $12.
So initially we could purchase 12 oranges for $24.

Answer (B)

Check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... of-ratios/
to understand the ratios and multiplier.
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Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] New post   02 May 2023, 10:17
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Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink]
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