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Mark has a bag full of red and blue marbles. For every 7 red
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22 Jan 2020, 00:47
22 Jan 2020, 00:47
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Question Stats:
80% (01:06) correct
20% (00:00) wrong based on 10 sessions
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Mark has a bag full of red and blue marbles. For every 7 red marbles in the bag, there are 5 blue marbles. Mark removes half of the red marbles from the bag. What fraction of the marbles does Mark remove from the bag?
A. \(\frac{7}{24}\)
B. \(\frac{2}{20}\)
C. \(\frac{14}{48}\)
D. \(\frac{7}{20}\)
E. \(\frac{13}{20}\)
A. \(\frac{7}{24}\)
B. \(\frac{2}{20}\)
C. \(\frac{14}{48}\)
D. \(\frac{7}{20}\)
E. \(\frac{13}{20}\)
Re: Mark has a bag full of red and blue marbles. For every 7 red
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22 Jan 2020, 02:04
22 Jan 2020, 02:04
1
According to the question,
Red marbles = 7k
Blue marbles = 5k
Total marbles = 12k
If half of the red marbles are removed then fraction of the marbles removed = \(\frac{3.5k}{12k}\)
It is hard to cut the marbles in half so k will take an even value.
if k = 2 then \(\frac{3.5k}{12k} = \frac{7}{24}\)
if k = 4 then \(\frac{3.5k}{12k} = \frac{14}{48} = \frac{7}{24} \) in the lowest form.
OA,A
Red marbles = 7k
Blue marbles = 5k
Total marbles = 12k
If half of the red marbles are removed then fraction of the marbles removed = \(\frac{3.5k}{12k}\)
It is hard to cut the marbles in half so k will take an even value.
if k = 2 then \(\frac{3.5k}{12k} = \frac{7}{24}\)
if k = 4 then \(\frac{3.5k}{12k} = \frac{14}{48} = \frac{7}{24} \) in the lowest form.
OA,A
huda wrote:
Mark has a bag full of red and blue marbles. For every 7 red marbles in the bag, there are 5 blue marbles. Mark removes half of the red marbles from the bag. What fraction of the marbles does Mark remove from the bag?
A. \(\frac{7}{24}\)
B. \(\frac{2}{20}\)
C. \(\frac{14}{48}\)
D. \(\frac{7}{20}\)
E. \(\frac{13}{20}\)
A. \(\frac{7}{24}\)
B. \(\frac{2}{20}\)
C. \(\frac{14}{48}\)
D. \(\frac{7}{20}\)
E. \(\frac{13}{20}\)
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Re: Mark has a bag full of red and blue marbles. For every 7 red
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23 Jan 2020, 01:23
23 Jan 2020, 01:23
1
Question)
Mark has a bag full of red and blue marbles. For every 7 red marbles in the bag, there are 5 blue marbles. Mark removes half of the red marbles from the bag. What fraction of the marbles does Mark remove from the bag?
Solution:
For every 7 red marbles in the bag, there are 5 blue marbles
=> Ratio of number of red marbles to number of blue marbles = 7 : 5
Let the number of red and blue marbles be 14k and 10k (we take an even number since we need to take half the number of red marbles)
[Note: We can multiply a ratio with any number keeping it unchanged. Observe that since the ratio has to be 7 : 5, 14k : 10k also equals 7 : 5]
Thus, total number of marbles = 24k
Number of marbles removed by Mark = 1/2 * 14k = 7k
=> Required fraction of total marbles that Mark removed = 7k/24k = 7/24
Answer A
Mark has a bag full of red and blue marbles. For every 7 red marbles in the bag, there are 5 blue marbles. Mark removes half of the red marbles from the bag. What fraction of the marbles does Mark remove from the bag?
Solution:
For every 7 red marbles in the bag, there are 5 blue marbles
=> Ratio of number of red marbles to number of blue marbles = 7 : 5
Let the number of red and blue marbles be 14k and 10k (we take an even number since we need to take half the number of red marbles)
[Note: We can multiply a ratio with any number keeping it unchanged. Observe that since the ratio has to be 7 : 5, 14k : 10k also equals 7 : 5]
Thus, total number of marbles = 24k
Number of marbles removed by Mark = 1/2 * 14k = 7k
=> Required fraction of total marbles that Mark removed = 7k/24k = 7/24
Answer A
