Question
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The correct answer is:
Related Questions to study
A team of runners is needed to run a 
-mile relay race. If each runner must run 
mile. Find the number of runners they need to run the race.
The fractions can also be simplified before multiplying by factoring out common factors in the numerator and denominator.
A team of runners is needed to run a -mile relay race. If each runner must run mile. Find the number of runners they need to run the race.
The fractions can also be simplified before multiplying by factoring out common factors in the numerator and denominator.
Multiply 
.
The fractions can also be simplified before multiplying by factoring out common factors in the numerator and denominator.
Multiply .
The fractions can also be simplified before multiplying by factoring out common factors in the numerator and denominator.
Simplify 
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The fractions can also be simplified before multiplying by factoring out common factors in the numerator and denominator.
Simplify =
The fractions can also be simplified before multiplying by factoring out common factors in the numerator and denominator.
Simplify 
The fractions can also be simplified before multiplying by factoring out common factors in the numerator and denominator.
Simplify
The fractions can also be simplified before multiplying by factoring out common factors in the numerator and denominator.
Simplify 
The fractions can also be simplified before multiplying by factoring out common factors in the numerator and denominator.
Simplify
The fractions can also be simplified before multiplying by factoring out common factors in the numerator and denominator.
Simplify 
The factions can also be simplified first before multiplying by factoring out common factors in the numerator and the denominator.
Simplify
The factions can also be simplified first before multiplying by factoring out common factors in the numerator and the denominator.
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Another approach to the question could be that first we can cut both the 3s, i.e., the 3 in the numerator of the first fraction and the 3 in the denominator of the second fraction. Also, we can reduce the 8 in the denominator of the first fraction and the 2 in the numerator in the second fraction. The 8 gets reduced to 4.
Hence, 
Thus, option (a) is the correct option.
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Another approach to the question could be that first we can cut both the 3s, i.e., the 3 in the numerator of the first fraction and the 3 in the denominator of the second fraction. Also, we can reduce the 8 in the denominator of the first fraction and the 2 in the numerator in the second fraction. The 8 gets reduced to 4.
Hence,
Thus, option (a) is the correct option.
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In the question, another approach could be that we can cut the 2 in the denominator of the first fraction by the 4 in the numerator of the second fraction and reduce the 4 to 2. Then, we get 
Thus, we get option (c) as the correct option.
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In the question, another approach could be that we can cut the 2 in the denominator of the first fraction by the 4 in the numerator of the second fraction and reduce the 4 to 2. Then, we get
Thus, we get option (c) as the correct option.
George has 
pan of brownies. He eats of them. The fraction of brownies George ate. is
George has pan of brownies. He eats of them. The fraction of brownies George ate. is
Multiply .
In the question, another approach could be that we can reduce the first fraction 
Then we get, 
. Here, the 3 in the denominator of the first fraction gets cut by the 3 in the numerator of the second fraction. Hence, we get the product as 
.
Thus, we get option (a) as the correct option.
Multiply .
In the question, another approach could be that we can reduce the first fraction
Then we get, . Here, the 3 in the denominator of the first fraction gets cut by the 3 in the numerator of the second fraction. Hence, we get the product as .
Thus, we get option (a) as the correct option.
Multiply 
.
In the question, another approach could be that we can reduce 
into by dividing both the numerator and denominator by 2 which is the HCF of both the numerator and the denominator. Then we simply get 
Thus, the correct option is option (a)
Multiply .
In the question, another approach could be that we can reduce into by dividing both the numerator and denominator by 2 which is the HCF of both the numerator and the denominator. Then we simply get
Thus, the correct option is option (a)
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In the question, another approach could be that we can cut both the threes and the sevens, i.e., 3 in the numerator of the first fraction and the 3 in the denominator of the second fraction and 7 in the denominator of the first fraction and 7 in the numerator of the second fraction.. That way we do not have to reduce the fraction later into its simplest form.
Hence, 
Thus, we get option (d) as the correct option.
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In the question, another approach could be that we can cut both the threes and the sevens, i.e., 3 in the numerator of the first fraction and the 3 in the denominator of the second fraction and 7 in the denominator of the first fraction and 7 in the numerator of the second fraction.. That way we do not have to reduce the fraction later into its simplest form.
Hence,
Thus, we get option (d) as the correct option.
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In the question, another approach could be that we can cut both the sixes, i.e., 6 in the numerator of the first fraction and the 6 in the denominator of the second fraction. That way we do not have to reduce the fraction later into its simplest form.
Thus, we get option (d) as the correct option.
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In the question, another approach could be that we can cut both the sixes, i.e., 6 in the numerator of the first fraction and the 6 in the denominator of the second fraction. That way we do not have to reduce the fraction later into its simplest form.
Thus, we get option (d) as the correct option.
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