A shopkeeper has 50 kg sugar he sells one seventh of it at rupees 21 per kg three-seventh of it at rupees 35 per kg and the remaining at rupees 28 per kg how much money did he collected by selling the sugar
Answers
Solution
1/7+3/7+x = 50
4/7+x = 50
x = 50-4/7
= (350-4)/7
= 346/7
The remaining at rupees 28 per kg,
346(28)/7 = 1384Rs
He collected by selling the sugar
1384+21+35 = 1440Rs
Given,
The total amount of sugar = 50 kg
1/7th sold at Rs. 21 per kg, 3/7th sold at Rs. 35 per kg and remaining at Rs. 28 per kg
To find,
The money he collected by selling the sugar.
Solution,
The money he collected by selling the sugar will be Rs. 1500.
We can easily solve this problem by following the given steps.
According to the question,
1/7th of 50 kg is sold at Rs. 21 per kg
50/7 kg = Rs. (50×21/7)
50/7 kg = Rs. (50×3)
50/7 kg = Rs. 150
3/7th of 50 kg is sold at Rs. 35 per kg
3×50/7 kg = 150/7 kg
150/7 kg = Rs. (150×35/7)
150/7 kg = Rs. (150×5)
150/7 kg = Rs. 750
The remaining part of 50 kg sugar will be:
(1)- (1/7+3/7)
(1) - (4/7)
Taking the LCM of 1 and 7,
7-4/7
3/7
So, 3/7th of 50 kg is sold at Rs. 28 per kg.
3×50/7 kg = 150/7 kg
150/7 kg = Rs. (150×28/7)
150/7 kg = Rs. (150×4)
150/7 kg = Rs. 600
So, the total amount he has will be Rs. (150+250+600).
Total amount = Rs. 1500
Hence, the money he collected by selling the sugar is Rs. 1500.