# My rice problem

• January 30th 2011, 07:10 PM
nimanamjouyan
My rice problem
A rice company decides to combine a firstrate rice that costs 5000 per kilo with a secondrate type that costs 4000 per kilo. The result is a new type of rice which costs 4600 per kilo. How many kilograms of the first rate rice are there in 40 kilo of the new type of rice?

A.24
B.32
C.30
D.33

Please explain the solution clearly.With many thanks.
• January 30th 2011, 07:34 PM
Soroban
Hello, nimanamjouyan!

Quote:

A rice company decides to combine a first-rate rice that costs 5000 per kg
with a second-rate type costs 4000 per kg.
The result is a mixture of rice which costs 4600 per kg.
How many kilograms of the first rate rice are there in 40 kg of the mixture?

. . $(A)\;24 \qquad (B)\;32 \qquad (C)\;30 \qquad (D)\;33$

Let $\,x$ = kg of first-rate rice.
Then $40-x$ = kg of second-rate rice.

We have $\,x$ kg of first-rate rice at 5000/kg.
. . Its value is: . $5000x$

We have $40-x$ kg of second-rate rice at 4000/kg.
. . Its value is: . $4000(40-x)$

The total value of the mixture is:
. . $5000x + 4000(40-x) \:=\:1000x + 160,\!000$ .[1]

But we know that mixture will be 40 kg of rice at 4600/kg.
. . Its value will be: . $40 \times 4600 \:=\:184,\!000$ .[2]

We just described the value of the mixture in two ways.

There is our equation! . $\hdots\;\;1000x + 160,\!000 \:=\:184,\!000$

• January 30th 2011, 08:05 PM
Ithaka
Quote:

Originally Posted by nimanamjouyan
A rice company decides to combine a firstrate rice that costs 5000 per kilo with a secondrate type that costs 4000 per kilo. The result is a new type of rice which costs 4600 per kilo. How many kilograms of the first rate rice are there in 40 kilo of the new type of rice?

A.24
B.32
C.30
D.33

Please explain the solution clearly.With many thanks.

Let x= mass of the first rate rice in 1 Kg of mixture
y= mass of second rate rice in 1 Kg of mixture.

(x and y both measured in Kg)

Form 2 equations connecting x and y and solve them.
• January 30th 2011, 10:30 PM
nimanamjouyan
I have tried forming two equations like that but I couldn't get an answer.
• January 31st 2011, 05:29 AM
HallsofIvy
Well, you saw Soroban form the equations. Do you understand how he did that?
• January 31st 2011, 06:08 AM
bjhopper
a simplified solution

mark a cost line in relative costs per kg

40 46 50
delta 46-40 = 6 parts of 50
delta 50-46 = 4 parts of 40
makes 10 parts of 46

to make 40 kg of 46 cost requires .6 x 40 = 24 kg of 50 and .4 x40 = 16 kg of 40

bjh
• January 31st 2011, 06:49 AM
Ithaka
Quote:

Originally Posted by nimanamjouyan
I have tried forming two equations like that but I couldn't get an answer.

x+y=1
5000x+4000y=4600

(it is an alternate solution to the one provided above by Soroban, hence x and y have different meanings, leading to different equations)
• January 31st 2011, 10:36 AM
Quote:

Originally Posted by nimanamjouyan
A rice company decides to combine a firstrate rice that costs 5000 per kilo with a secondrate type that costs 4000 per kilo. The result is a new type of rice which costs 4600 per kilo. How many kilograms of the first rate rice are there in 40 kilo of the new type of rice?

A.24
B.32
C.30
D.33

Please explain the solution clearly.With many thanks.

You have 2 different weights for which you need to solve for only one.

To find 2 unknowns, you need 2 clues.
By formulating the clues as a pair of equations, you can solve the equations.
Only one weight is needed, so we can eliminate the 2nd weight from our equations.

The clues are

(1) the sum of the weights is 40 kg
(2) the sum of the prices is 4600(40)=184000

Price is (weight in kg)(price per kg)

Formulating these, let $x=$weight of first rate rice, $y=$ weight of second rate rice.

$x+y=40$

$x(5000)+y(4000)=184000$

Therefore, y can be eliminated by substituting it for 40-x.

$y=40-x\Rightarrow\ 5000x+4000(40-x)=4600$

allows us to find x.

Alternatively, multiply both sides of $x+y=40$ by $4000$

and subtract the resulting pair of equations to find x.
This also eliminates y.

viz....

$4000x+4000y=160000$

$5000x+4000y=184000$

Subtracting these

$5000x+4000y-(4000x+4000y)=184000-160000$

$5000x-4000x+4000y-4000y=24,000$

$1000x=24,000$