
If \[5\] spiders can catch \[5\] files in \[5\] minutes, how many flies can \[100\] spiders catch in \[100\] minutes?
(a) \[100\]
(b) \[500\]
(c) \[1000\]
(d) \[2000\]
Answer
490.2k+ views
Hint: In this problem we are going to find the number of flies that can be caught by \[100\] spiders in \[100\]minutes. Read the given question carefully. First find the number of flies that can be caught by a spider in \[1\] minute then in \[100\] minutes. By multiplying it with the \[100\] spiders we can solve this.
Complete step-by-step answer:
We are given the problem,
There are \[5\] spiders can catch \[5\] files in \[5\] minutes, then find the number of flies can \[100\] spiders catch in \[100\] minutes.
If \[5\] spiders can catch \[5\] files in \[5\] minutes, then
We can determine one spider can catch reciprocal of the given,
\[ \Rightarrow 1\] spiders can catch \[\dfrac{1}{5}\] flies in \[1\] minute.
In \[100\] minutes number of flies catch by \[1\] spider \[ = \dfrac{1}{5} \times 100 = 20\]
Therefore, In \[100\] minutes the number of flies caught by \[1\] spider is \[20\].
Now, in \[100\] minutes number of flies catch by \[100\] spider \[ = 20 \times 100 = 2000\]
In \[100\] minutes the number of flies caught by \[100\] spider is \[2000\].
Hence option (d) \[2000\] is correct.
So, the correct answer is “Option d”.
Note: From the step by step solution we can see that if the number of spider’s flies catch also increases and if the time increases flies catch also increases.
From this we can say that the number of flies caught is directly proportional to the number of spiders and times.
Two quantities are said to be directly proportional, if on the increase (or decrease) of the one, the other also increases (or decreases) to the same extent.
Hence this question can also be solved in the direct proportion method.
Let the number of flies be \[x\].
More spiders, More flies (Direct Proportional)
More time, More flies (Direct Proportional)
\[\therefore 5 \times 5 \times x = 100 \times 100 \times 5\]
\[x = \dfrac{{100 \times 100 \times 5}}{{5 \times 5}}\]
\[x = 2000\].
From this we can understand that a problem can be solved in more than one way. Try to understand the question clearly and find the best solution thus the answer can be easily achieved.
Complete step-by-step answer:
We are given the problem,
There are \[5\] spiders can catch \[5\] files in \[5\] minutes, then find the number of flies can \[100\] spiders catch in \[100\] minutes.
If \[5\] spiders can catch \[5\] files in \[5\] minutes, then
We can determine one spider can catch reciprocal of the given,
\[ \Rightarrow 1\] spiders can catch \[\dfrac{1}{5}\] flies in \[1\] minute.
In \[100\] minutes number of flies catch by \[1\] spider \[ = \dfrac{1}{5} \times 100 = 20\]
Therefore, In \[100\] minutes the number of flies caught by \[1\] spider is \[20\].
Now, in \[100\] minutes number of flies catch by \[100\] spider \[ = 20 \times 100 = 2000\]
In \[100\] minutes the number of flies caught by \[100\] spider is \[2000\].
Hence option (d) \[2000\] is correct.
So, the correct answer is “Option d”.
Note: From the step by step solution we can see that if the number of spider’s flies catch also increases and if the time increases flies catch also increases.
From this we can say that the number of flies caught is directly proportional to the number of spiders and times.
Two quantities are said to be directly proportional, if on the increase (or decrease) of the one, the other also increases (or decreases) to the same extent.
Hence this question can also be solved in the direct proportion method.
Let the number of flies be \[x\].
More spiders, More flies (Direct Proportional)
More time, More flies (Direct Proportional)
\[\therefore 5 \times 5 \times x = 100 \times 100 \times 5\]
\[x = \dfrac{{100 \times 100 \times 5}}{{5 \times 5}}\]
\[x = 2000\].
From this we can understand that a problem can be solved in more than one way. Try to understand the question clearly and find the best solution thus the answer can be easily achieved.
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