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In order to find the highest sum, we have to be able to combine the highest number with the lowest, and so on. For example, if 7 is the highest number, we place it as a numerator while 1 is the denominator, giving us 7/1. We can repeat this process, while putting the 3 highest with the 3 lowest. 7/1, 6/2, 5/3. This will give us: 7+3+ 1 2/3= 10 +1 2/3= 11 2/3. The largest possible sum is 11 2/3.
So here's how I solved the problem: So we need to get the highest possible sum out of the three that means you should try to make the biggest fractions out of the numbers (1, 2, 3, 4, 5, 6, 7) so we would take the three smallest numbers and three biggest numbers and put them together to make it so that a number is paired with another that is either as small compare to how big the number is or how big the number is compare to how smaller the number is. The numbers would be: (1,2,3,5,6,7) 7/1, 6/2, and 5/3 and when adding the fractions you get 11 4/6 which is the biggest sum because it is the sum of the biggest possible fractions
To start off, I picked six numbers out of the seven provided. I chose the six largest numbers, as the problem said to "determine the largest possible sum that can be obtained." I picked the numbers 2, 3, 4, 5, 6, and 7. The fractions that I made with these numbers are: 2/5 3/6 4/7 I chose these fractions, as when converted to percentages, they are around 50%. This is about as high as it would get with any other fractions made out of the provided numbers. Then I found the lowest common multiple of 5, 6, and 7, as they were the denominators. The answer was 210. The new fractions are: 84/210 105/210 120/210 Then, I added up the fractions. 84/210 + 105/210 + 120/210 = 309/210 = 1 33/70. The largest possible sum is 1 33/70.
11 2/3 is the largest sum possible. The largest fraction you can make with number 1,2,3,4,5,6, and 7 is 7/1, with the value of 7. The second largest you can make is 6/2 which is equivalent to 3. Then we have numbers 3,4,5 left. Out of these 5/3 creates the biggest value, which is 1 2/3. There fore with these three fractions, you are bound to have the highest possible sum, which is 11 2/3
Sine I need to select six numbers from the 7 numbers given and make 3 fractions, I decided to take the 3 biggest numbers and use them as numerators and pair them with the 3 smallest numbers to give me improper fractions. Then I paired up the largest with the smallest and so on: 7/1 6/2 5/3. Next I added them together: 7/1+6/2+5/3=42/6+18/6+10/6= 70/6 = 11 4/6 = 11 2/3 So the highest possible sum is 11 2/3, or 11.67
In order to find the highest sum, we have to be able to combine the highest number with the lowest, and so on. For example, if 7 is the highest number, we place it as a numerator while 1 is the denominator, giving us 7/1. We can repeat this process, while putting the 3 highest with the 3 lowest. 7/1, 6/2, 5/3. This will give us:
ReplyDelete7+3+ 1 2/3= 10 +1 2/3= 11 2/3.
The largest possible sum is 11 2/3.
So here's how I solved the problem:
ReplyDeleteSo we need to get the highest possible sum out of the three that means you should try to make the biggest fractions out of the numbers (1, 2, 3, 4, 5, 6, 7) so we would take the three smallest numbers and three biggest numbers and put them together to make it so that a number is paired with another that is either as small compare to how big the number is or how big the number is compare to how smaller the number is. The numbers would be:
(1,2,3,5,6,7)
7/1, 6/2, and 5/3
and when adding the fractions you get 11 4/6 which is the biggest sum because it is the sum of the biggest possible fractions
To start off, I picked six numbers out of the seven provided. I chose the six largest numbers, as the problem said to "determine the largest possible sum that can be obtained." I picked the numbers 2, 3, 4, 5, 6, and 7. The fractions that I made with these numbers are:
ReplyDelete2/5
3/6
4/7
I chose these fractions, as when converted to percentages, they are around 50%. This is about as high as it would get with any other fractions made out of the provided numbers. Then I found the lowest common multiple of 5, 6, and 7, as they were the denominators. The answer was 210. The new fractions are:
84/210
105/210
120/210
Then, I added up the fractions.
84/210 + 105/210 + 120/210 = 309/210 = 1 33/70.
The largest possible sum is 1 33/70.
The largest sum of the fractions would be 11.6666666667. As 5/2 + 6/1 + 4/3 = 11.6666666667
ReplyDeleteUsing guess and check, I found that 7/1, 6/2 and 5/3 equal 11 2/3, which is the highest I could find
ReplyDelete11 2/3 is the largest sum possible.
ReplyDeleteThe largest fraction you can make with number 1,2,3,4,5,6, and 7 is 7/1, with the value of 7.
The second largest you can make is 6/2 which is equivalent to 3.
Then we have numbers 3,4,5 left.
Out of these 5/3 creates the biggest value, which is 1 2/3. There fore with these three fractions, you are bound to have the highest possible sum, which is 11 2/3
My work was done hard copy. This was the answer I obtained...
ReplyDeleteLargest sum possible - 11 2/3
Sine I need to select six numbers from the 7 numbers given and make 3 fractions, I decided to take the 3 biggest numbers and use them as numerators and pair them with the 3 smallest numbers to give me improper fractions. Then I paired up the largest with the smallest and so on: 7/1 6/2 5/3.
ReplyDeleteNext I added them together:
7/1+6/2+5/3=42/6+18/6+10/6= 70/6 = 11 4/6 = 11 2/3
So the highest possible sum is 11 2/3, or 11.67
I did it hard copy and got 11 2/3
ReplyDeleteI did it hard copy but the 3 fractions to create the biggest sum are 7/1, 6/2, and 5/3. These add up to 140/12, 11 2/3 (11 6/12), or 11.67.
ReplyDelete11 2/3 is correct! Does anyone know what this answer is as an improper fraction?
ReplyDeleteI did mine hard copy. I got 11 2/3. I obtained this answer by putting the largest numbers over the smallest
ReplyDelete