mr. lee had a bag of red beans and a bag of black beans. 20% of the total number of beans were black, and there were 180 ounces more red beans than black beans. he transferred some red beans to the bag containing black beans, so that, the bag now contained 30% of the total number of beans. How many ounces of beans were there in the bag of mixed beans
X oz. of black beans.
(X+180) oz. of red beans.
Total = X + (X+180) = 2x + 180.
X/(2x+180) = 0.20
x = 0.4x + 36
0.6x = 36
X = 60 Oz. of Black beans.
X+180 = 60+180 = 240 Oz. of red beans.
Total = 60 + 240 = 300 oz. of beans.
0.3 * 300 = 90 oz. of mixed beans.
r = b+180
b = .2(r+b)
so,
b = 60
r = 240
Total beans = 300 oz
After the transfer, 30% = 90 oz
So, that means that 30 oz of red beans were transferred, added to the 60 oz of black beans already there.
pawg
Well, well, well, looks like Mr. Lee had a colorful bean situation going on! Let's break it down and solve the mystery of the mixed beans.
Let's say the total number of beans Mr. Lee had is "x." According to the information given, 20% of the total beans were black beans. So, we can say that 0.2x represents the number of black beans.
Now, it's time to use our detective skills! We know that there were 180 ounces more red beans than black beans. So, we can write an equation:
0.2x + 180 = 0.8x
Why 0.8x? Well, that's because we subtract the transferred red beans from the total red beans (0.2x) to get the remaining red beans (0.8x).
Solving this equation, we find that x = 600. So, the total number of beans Mr. Lee had is 600.
Now, Mr. Lee, being the bean enthusiast he is, started transferring some red beans to the black bean bag. After the transfer, the bag now contains 30% of the total number of beans. Let's call the number of red beans transferred "y."
The equation for the number of red beans after the transfer becomes:
0.8x - y = 0.3(x - y)
Simplifying this equation, we find that:
0.8x - y = 0.3x - 0.3y
0.5x = 0.7y
Substituting the value of x which we found earlier, we get:
0.5(600) = 0.7y
300 = 0.7y
Dividing both sides by 0.7, we find that y = 428.57 (approximately).
Since we can't have a fraction of a bean (that would be quite the sight!), we round that value down to the nearest whole number, which is 428. So, Mr. Lee transferred 428 red beans to the black bean bag.
Now, let's find the total number of beans in the mixed bag. We add up the number of black beans (0.2x) and the remaining red beans (0.8x - y):
Total number of beans = 0.2x + 0.8x - y
Plugging in the numbers, we find:
Total number of beans = 0.2(600) + 0.8(600) - 428
Total number of beans = 120 + 480 - 428
Total number of beans = 172 ounces
So, there were 172 ounces of beans in the bag of mixed beans. Case closed!
To find the number of ounces of beans in the bag of mixed beans, we need to break down the problem step-by-step:
Let's represent the number of red beans as R and the number of black beans as B.
1) We're given that 20% of the total number of beans were black. This can be represented as B = 0.2(R + B).
2) It's also stated that there were 180 ounces more red beans than black beans. This can be expressed as R = B + 180.
3) Then, the problem states that some red beans were transferred to the bag containing black beans, resulting in the bag now containing 30% of the total number of beans. This can be represented as:
(B + R) × 0.30 = B.
Now, let's solve these equations to find the values of B and R:
From equation 1:
0.2(R + B) = B.
0.2R + 0.2B = B.
0.2R = 0.8B.
From equation 2:
R = B + 180.
Let's substitute the value of R from equation 2 into equation 1:
0.2(B + 180) = 0.8B.
0.2B + 36 = 0.8B.
0.8B - 0.2B = 36.
0.6B = 36.
B = 36 / 0.6
B = 60.
Now that we know B, we can substitute it back into equation 2 to find R:
R = B + 180.
R = 60 + 180.
R = 240.
Finally, to find the total number of ounces of beans in the bag of mixed beans:
(B + R) × 0.30 = B.
(60 + 240) × 0.30 = B.
300 × 0.30 = B.
B = 90.
Therefore, there were 90 ounces of beans in the bag of mixed beans.