Some marbles in a bag are red and the rest are blue. If one red
marble is removed, then one-•‐seventh of the remaining marbles are red.
If two blue marbles are removed instead of one red, then one fifth of
the remaining marbles are red. How many marbles were in the bag
originally?
Answers
Answer:
Step-by-step explanation:
Given Some marbles in a bag are red and the rest are blue. Assuming that one of the red marble is removed, then one-seventh of the remaining marbles are also red . If two blue marbles are removed instead of one red, then one fifth of the remaining marbles are red. How many marbles were in the bag originally?
Let m and n be the number of red and blue marbles inside a bag.
According to the question 1 red marble is removed, so it will be m + n – 1 marbles and also m – 1 red marbles left.
So we get m – 1 / m + n – 1 = 1/7 (since 1/7 th of remaining)
Now when 2 blue marbles are taken out there are m red marbles
So total marbles left in the bag will be m / m + n – 2 = 1/5 (since 1/5 th are remaining)
From the above equations we get
7m – 7 = m + n – 1 = 6m – n – 6 = 0
5m = m + n – 2 = 4m – n + 2 = 0
Subtracting both equations we get
2m – 8 = 0
Or m = 4
Now 6m – n – 6 = 0
6(4) – n – 6 = 0
24 – n – 6 = 0
Or n = 18
Therefore total number of marbles will be m + n = 4 + 18 = 22
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