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A bag contains some red and blue balls. If Jane removes 1 red ball from the bag, then 1/7 of the remaining balls in the bag are red. If, instead of 1 red ball, Jane removes 2 blue balls from the bag, then 1/5 of the remaining balls in the bag are red. How many balls are there in the bag at first?
post-22501
Besides using the Model Method, these 2 methods could also be used.

Units and Parts Method :

Scenario 1 : If Jane removes 1 red ball from the bag,
red ——- 1u
blue ——- 6u

Scenario 2 ; If Jane removes 2 blue balls from the bag,
red ——- 1p
blue ——- 4p

1u + 1 ——- 1p
6u ——- 4p + 2
6u ——- 4(1u + 1) + 2
1u ——- 6/2 = 3
7u + 1 = 7 x 3 + 1 = 22

Scenario 2 ; If Jane removes 2 blue balls from the bag,
red ——- 1p
blue ——- 4p
1u + 1 ——- 1p
6u ——- 4p + 2
6u ——- 4(1u + 1) + 2
1u ——- 6/2 = 3
7u + 1 = 7 x 3 + 1 = 22

The Listing Method :

Scenario 1 : If Jane removes 1 red ball from the bag : 1 + 7 = 8, 15, 22, …
Scenario 2 ; If Jane removes 2 blue balls from the bag ; 2 + 5 = 7, 12, 17, 22, …

Ans : 22 balls.

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