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REY [17]
2 years ago
6

15. Jim had 103 red and blue marbles. After giving of his blue marbles and 15 of his red marbles

Mathematics
1 answer:
sesenic [268]2 years ago
4 0

This question above is incomplete

Complete Question

Jim had 103 red and blue marbles. After giving 2/5 of his blue marbles and 15 of his red marbles to Samantha, Jim had 3/7 as many red marbles as blue marbles. How many blue marbles did he have originally?

Answer:

70 Blue marbles

Step-by-step explanation:

Let red marbles = R

Blue marbles = B

Step 1

Jim had 103 red and blue marbles.

R + B = 103.......Equation 1

R = 103 - B

Step 2

After giving 2/5 of his blue marbles and 15 of his red marbles to Samantha, Jim had 3/7 as many red marbles as blue marbles

2/5 of B to Samantha

Jim has = B - 2/5B = 3/5B left

He also gave 15 red marbles to Samantha

= R - 15

The ratio of what Jim has left

= Red: Blue

= 3:7

= 3/7

Hence,

R - 15/(3/5)B = 3/7

Cross Multiply

7(R - 15) = 3(3/5B)

7R - 105 = 3(3B/5)

7R - 105 = 9B/5

Cross Multiply

5(7R - 105) = 9B

35R - 525 = 9B............ Equation 2

From Equation 1, we substitute 103 - B for R in Equation 2

35(103 - B) - 525 = 9B

3605 - 35B - 525 = 9B

Collect like terms

3605 - 525 = 9B + 35B

3080 = 44B

B = 3080/44

B = 70

Therefore, Jim originally had 70 Blue marbles.

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