Question

in a bag there are blue and red sweets the ratio of blue sweets to red sweets is 5:3 what fraction of the sweets are blue

Answer 1

Answer:

Let the common ratio between blue and red sweets be x

Then,

Number of blue sweets = 2x

Number of green sweets = 7x

If the common ratio between the green and red sweets be y

Then,

Number of green sweets = 3y

Number of red sweets = y

Here the number of green sweets must be equal

Therefore,

7x=3y7x=3y

Also the sum of all of them is 140

Thus,

2x+3y+y < 1402x+3y+y<140

\implies 2x+4y < 140⟹2x+4y<140

\implies x+2y < 70⟹x+2y<70

\implies x+2\times\frac{7x}{3} < 70⟹x+2×

3

7x

<70

\implies 3x+14x < 210⟹3x+14x<210

\implies 17x < 210⟹17x<210

\implies x < \frac{210}{17}⟹x<

17

210

\implies x < 12.36⟹x<12.36

x should be a whole number

Therefore, x=12x=12

And the sum of all the sweets will be 136

Therefore,

y=\frac{7\times 12}{3}y=

3

7×12

\implies y=28⟹y=28

Thus the greatest possible number of red sweets is 28

Hope this helps.

Answer 2

Answer:

1 In a bag there are blue sweets and red sweets. The ratio of blue sweets to red sweets is 5:3. What fraction of the sweets are blue? 5 parts Blue parts in Total. 8.

Step-by-step explanation:

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