Respuesta :
Answer:
The greatest possible number of red sweets is 28.
Explanation:
Let g represent the number of green sweets. We know that the ratio of blue to green is 2:7; this means the number of blue sweets is 2/7 the number of green sweets, or 2/7g.
We also know that the number of green sweets to the number of red sweets is 3:1; this means the number of red sweets to the number of green sweets is 1:3, or that the number of red sweets is 1/3 the number of green sweets:
1/3g
We know that there are fewer than 140 total sweets; this gives us the inequality
g + 2/7g + 1/3g < 140
We must find a common denominator; 21 will work:
21/21g + 6/21g + 7/21g < 140
34/21g < 140
Divide both sides by 34/21:
(34/21g)÷(34/21) < 140÷(34/21)
g < 140÷(34/21)
To divide fractions, flip the second one and multiply:
g < 140×(21/34)
g < 2940/34
This means the number of red sweets will be 1/3 of this:
r < 2940/34 × 1/3
r < 2940/102
102 will go into 2940 28 times, so 28 is the largest number of red sweets.
Greatest possible number of red sweets in the box will be 28.
It's given in the question,
- Ratio of the number of blue (b) and green sweets (g) = 2 : 7
[tex]\frac{b}{g}=\frac{2}{7}[/tex] ⇒ [tex]b=\frac{2}{7}g[/tex] ------(1)
- Ratio of the number of green (g) and red sweets (r) = 3 : 1
[tex]\frac{g}{r}=\frac{3}{1}[/tex] ⇒ [tex]r=\frac{1}{3}g[/tex] -------(2)
Since, total number of blue, green and red sweets in the box are less than 140,
[tex]b+g+r< 140[/tex]
Substitute the values of 'b' and 'r' from equations (1) and (2),
[tex]\frac{2}{7}g+g+\frac{1}{3}g< 140[/tex]
[tex]\frac{6g+21g+7g}{21}< 140[/tex]
[tex]\frac{34}{21}g< 140[/tex]
[tex]g< \frac{140\times 21}{34}[/tex]
Since, [tex]g=3r[/tex],
[tex]3r< \frac{140\times 21}{34}[/tex]
[tex]r< \frac{140\times 21}{34\times 3}[/tex]
[tex]r<28.82[/tex]
Therefore, greatest possible number of red sweets will be 28.
Learn more,
https://brainly.com/question/24669471
