A farmer put up a watermelon stand. On the first day, he sold 650 kilograms of watermelons. On the second day, he sold 350 kilograms less than on the first day. On the third day, he sold 6 times less kilograms than on the second day. How many kilograms of watermelons did the farmer sell over the three days?
Answers
Given : A farmer put up a watermelon stand.
On the first day, he sold 650 kilograms of watermelons.
On the second day, he sold 350 kilograms less than on the first day.
On the third day, he sold 6 times less kilograms than on the second day.
To Find : How many kilograms of watermelons did the farmer sell over the three days?
Solution:
Sold on first day = 650 kg
Sold on 2nd day = 650 - 350 = 300 kg
Sold on 3rd day = 6 times less than on the second day.
First understand meaning of 6 times less than 300 here : ( As many students gets confused over this term )
x times less than a Means (1/x) time of a
x times more than a means x times of a
Hence 6 times less than 300 kg = (1/6) * 300 = 50 kg
Sold on 3rd day = 50 kg
Total Sold = 650 + 300 + 50 = 1000 kg
the farmer sold watermelon over the three days = 1000 kg
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⇒On the first day, he sold 650 kilograms of watermelons.
⇒On the second day, he sold 350 kilograms less than on the first day.
⇒On the third day, he sold 6 times less kilograms than on the second day.
⇒Sold on first day = 650 kg
⇒Sold on 2nd day = 650 - 350 = 300 kg
⇒Sold on 3rd day = 6 times less than on the second day.
⇒Hence 6 times less than 300 kg = (1/6) * 300 = 50 kg
⇒Sold on 3rd day = 50 kg
⇒Total Sold = 650 + 300 + 50 = 1000 kg