A fruit vendor has 900 apples and 945 oranges. he packs them into baskets. each basket contains only one of the two fruits but in equal number. find the number of fruits to be put in each basket in order to have minimum no of baskets.
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Solution:-
The fruit vendor has 900 apples and 945 oranges.
He packs them in the baskets in such a way that each basket has only one kind of the two fruits, but in equal number.
Therefore, we have to find HCF of 900 and 945.
Prime factors of 900 = 2 × 2 × 3 × 3 × 5 × 5
Prime factors of 945 = 3 × 3 × 3 × 5 × 7
So, the HCF of 900 and 945 is 3 × 3 × 5 = 45
So, if he packs 45 of a kind of fruit in each basket, then he can pack all his fruits in minimum number of baskets.
The fruit vendor has 900 apples and 945 oranges.
He packs them in the baskets in such a way that each basket has only one kind of the two fruits, but in equal number.
Therefore, we have to find HCF of 900 and 945.
Prime factors of 900 = 2 × 2 × 3 × 3 × 5 × 5
Prime factors of 945 = 3 × 3 × 3 × 5 × 7
So, the HCF of 900 and 945 is 3 × 3 × 5 = 45
So, if he packs 45 of a kind of fruit in each basket, then he can pack all his fruits in minimum number of baskets.
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Answer:
we want to find hcf of 900 and 945
Step-by-step explanation:
900=3×3×2×2×5×5
945=3×3×5×7×3
hcf=3×3×5= 45 maximum no. of fruits can packed =45
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