
A bag of salt weighs kg. If kg of salt is to be put in such bags, how many bags will be required?
Answer
479.4k+ views
Hint: Here, with the help of unitary method, we will find that 1 kg of salt can be put in how many bags. Then, we will multiply both sides by to find the required number of bags which can carry this quantity of salt.
Complete step-by-step answer:
According to the question, a bag of salt weighs kg. Hence,
Weight of 1 bag
Now, by unitary method we can say that,
salt bags
According to the questions, kg of salt is to be put in such bags.
Now,
Since, bags
Therefore, of salt will be put into :
bags
Hence, 18 bags are required to store kg of salt.
Note: Here, we have used a unitary method to solve this question. Unitary method is a technique in which we find the value of a single unit (say 1kg salt) and then, we find the required value by just multiplying it on both the sides of this single unit. Also, all the quantities were provided in mixed fraction hence, we were required to convert them to an improper fraction to solve this question. The method to convert a mixed fraction involved the multiplication of the whole number by the denominator of the given fraction and then, writing that number as a numerator by adding the existing numerator in the product. Hence, these two steps were essential to solve this question.
Complete step-by-step answer:
According to the question, a bag of salt weighs
Weight of 1 bag
Now, by unitary method we can say that,
According to the questions,
Now,
Since,
Therefore,
Hence, 18 bags are required to store
Note: Here, we have used a unitary method to solve this question. Unitary method is a technique in which we find the value of a single unit (say 1kg salt) and then, we find the required value by just multiplying it on both the sides of this single unit. Also, all the quantities were provided in mixed fraction hence, we were required to convert them to an improper fraction to solve this question. The method to convert a mixed fraction involved the multiplication of the whole number by the denominator of the given fraction and then, writing that number as a numerator by adding the existing numerator in the product. Hence, these two steps were essential to solve this question.
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