Question

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?​

Answer 1

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

Reference link will be

https://brainly.in/question/10830846