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Hovkial
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A man bought some oranges at the rate of 5 per dollar. He bought the s 08 Sep 2019, 11:05
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D
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95% (hard)

Question Stats:

46% (02:52) correct 54% (03:28) wrong based on 201 sessions

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A man bought some oranges of one type at the rate of 5 per dollar. He bought the same number of oranges of another type at the rate of 4 per dollar. He mixed both types of oranges and then sold the mix at 9 oranges for $2. After calculations, the man discovered that he made a loss of $3. What was the total number of oranges that the man bought?

(A) 90

(B) 180

(C) 540

(D) 1080

(E) 1620
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D
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A man bought some oranges at the rate of 5 per dollar. He bought the s 08 Sep 2019, 19:46
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Hovkial
A man bought some oranges of one type at the rate of 5 per dollar. He bought the same number of oranges of another type at the rate of 4 per dollar. He mixed both types of oranges and then sold the mix at 9 oranges for $2. After calculations, the man discovered that he made a loss of $3. What was the total number of oranges that the man bought?

(A) 90

(B) 180

(C) 540

(D) 1080

(E) 1620
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Let the number of each be x...
Cost of oranges purchased =\(x*\frac{1}{5}+x*\frac{1}{4}=\frac{9x}{20}\)
Selling price of these 2x oranges = \(2x*\frac{2}{9}=\frac{4x}{9}\)..

Now it is given that there is a loss of 3, so \(CP-SP=3...............\frac{9x}{20}-\frac{4x}{9}=3...............\frac{81x-80x}{180}=3...........x=3*180=540...........2x=2*540=1080\)

D
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A man bought some oranges at the rate of 5 per dollar. He bought the s 08 Sep 2019, 23:05
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Assume total number of apples of each type= LCM(5,4,9)=180

36........45
........x.....
1...........1

x= (45+36)/2=40.5

SP of the mixture of two types= 40$/180 apples

if he sells 180 apples, there is 0.5$ loss.
There is 3$ loss, if he sells= 180*3/0.5=1080 Apples


Hovkial
A man bought some oranges of one type at the rate of 5 per dollar. He bought the same number of oranges of another type at the rate of 4 per dollar. He mixed both types of oranges and then sold the mix at 9 oranges for $2. After calculations, the man discovered that he made a loss of $3. What was the total number of oranges that the man bought?

(A) 90

(B) 180

(C) 540

(D) 1080

(E) 1620
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A man bought some oranges at the rate of 5 per dollar. He bought the s 09 Sep 2019, 00:55
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cost of oranges x/5+x/4 ; 9x/20
SP of oranges ; 2x/9 * 2 ; 4x/9
so
CP-SP = 3 ; 9x/20-4x/9 = 3
solve for x = 540
2*x ; 1080
IMO D


Hovkial
A man bought some oranges of one type at the rate of 5 per dollar. He bought the same number of oranges of another type at the rate of 4 per dollar. He mixed both types of oranges and then sold the mix at 9 oranges for $2. After calculations, the man discovered that he made a loss of $3. What was the total number of oranges that the man bought?

(A) 90

(B) 180

(C) 540

(D) 1080

(E) 1620
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A man bought some oranges at the rate of 5 per dollar. He bought the s 09 Sep 2019, 04:04
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Hovkial
A man bought some oranges of one type at the rate of 5 per dollar. He bought the same number of oranges of another type at the rate of 4 per dollar. He mixed both types of oranges and then sold the mix at 9 oranges for $2. After calculations, the man discovered that he made a loss of $3. What was the total number of oranges that the man bought?

(A) 90

(B) 180

(C) 540

(D) 1080

(E) 1620
Show more

Given:
1. A man bought some oranges of one type at the rate of 5 per dollar.
2. He bought the same number of oranges of another type at the rate of 4 per dollar.
3. He mixed both types of oranges and then sold the mix at 9 oranges for $2.
4. After calculations, the man discovered that he made a loss of $3.

Asked: What was the total number of oranges that the man bought?

Let the number of oranges the man bought be x.

1. A man bought x of one type at the rate of 5 per dollar.
2. He bought x oranges of another type at the rate of 4 per dollar.
He bought 2x oranges for x/5 + x/4 = $9x/20

3. He mixed both types of oranges and then sold the mix at 9 oranges for $2.
He sold oranges 2x oranges for = 2x *2/9 = $4x/9

4. After calculations, the man discovered that he made a loss of $3.
9x/20 - 4x/9 = 3
(81 - 80) x/180 = 3
x = 180 * 3 = 540
2x = 1080

IMO D
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A man bought some oranges at the rate of 5 per dollar. He bought the s 09 Sep 2019, 04:06
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nick1816
Assume total number of apples of each type= LCM(5,4,9)=180

36........45
........x.....
1...........1

x= (45+36)/2=40.5

SP of the mixture of two types= 40$/180 apples

if he sells 180 apples, there is 0.5$ loss.
There is 3$ loss, if he sells= 180*3/0.5=1080 Apples


Hovkial
A man bought some oranges of one type at the rate of 5 per dollar. He bought the same number of oranges of another type at the rate of 4 per dollar. He mixed both types of oranges and then sold the mix at 9 oranges for $2. After calculations, the man discovered that he made a loss of $3. What was the total number of oranges that the man bought?

(A) 90

(B) 180

(C) 540

(D) 1080

(E) 1620
Show more


nick1816
You replaced ORANGES with APPLES
The solution is fine.
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A man bought some oranges at the rate of 5 per dollar. He bought the s 12 Sep 2019, 11:31
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Hovkial
A man bought some oranges of one type at the rate of 5 per dollar. He bought the same number of oranges of another type at the rate of 4 per dollar. He mixed both types of oranges and then sold the mix at 9 oranges for $2. After calculations, the man discovered that he made a loss of $3. What was the total number of oranges that the man bought?

(A) 90

(B) 180

(C) 540

(D) 1080

(E) 1620
Show more

We let x = the number of oranges of each type he bought. We can create the equation:

x/5 + x/4 = (2x)(2/9) + 3

Multiplying the equation by 180, we have:

36x + 45x = 80x + 540

81x = 80x + 540

x = 540

Therefore, he bought 2(540) = 1080 oranges.

Answer: D
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A man bought some oranges at the rate of 5 per dollar. He bought the s 17 Dec 2025, 08:46
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Bought same number orange of each type and sold after mixing.....it means when sells one orange of a type ....next one sold must be of da other type......
So.....cost price of one type = 1 / 5 $ = 20 cents
Cost price of da other type = 1 / 4 $ = 25 cents
So...when sells 2 consecutive orange....their cost price = 20 + 25 = 45 cents

And.... selling price of 1 orange = 2 / 9 $ = 200 / 9 cents
So.... selling price of 2 orange = 400 / 9 cents

Loss from selling 2 orange = 45 - [ 400 / 9 ] = 5 / 9 cent

So... 5 / 9 cent loss from selling = 2 orange
1 cent loss from selling = 2 / [ 5 / 9 ] = 18 / 5 orange
So....3$ or 300 cent loss from selling = [ 18 × 300 ] / 5 = 1080 orange........

So....he bought 1080 orange......

! nah id win!
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