Anuj had some chocolates, and he divided them into two lots A and B. He sold the first lot at the rate of ₹2 for 3 chocolates and the second lot at the rate of ₹1 per chocolate, and got a total of ₹400. If he had sold the first lot at the rate of ₹1 per chocolate, and the second lot at the rate of ₹4 for 5 chocolates, his total collection would have been ₹460. Find the total number of chocolates he had.

 

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Transcript

Question 28 (Choice 2) Anuj had some chocolates, and he divided them into two lots A and B. He sold the first lot at the rate of ₹2 for 3 chocolates and the second lot at the rate of ₹1 per chocolate, and got a total of ₹400. If he had sold the first lot at the rate of ₹1 per chocolate, and the second lot at the rate of ₹4 for 5 chocolates, his total collection would have been ₹460. Find the total number of chocolates he had. Let Number of chocolates in first lot = x Number of chocolates in second lot = y Given that He sold the first lot at the rate of ₹2 for 3 chocolates and the second lot at the rate of ₹1 per chocolate, and got a total of ₹400. Let Number of chocolates in first lot = x Number of chocolates in second lot = y Given that He sold the first lot at the rate of ₹2 for 3 chocolates and the second lot at the rate of ₹1 per chocolate, and got a total of ₹400. Number of chocolates in 1st lot × 𝟐/𝟑 + Number of chocolates in 2nd lot × 1 = 400 x × 2/3 + y × 1 = 400 2𝑥/3 + y = 400 Multiplying by 3 both sides 3(2𝑥/3 + y) = 3 × 400 2x + 3y = 1200 Also, given that he had sold the first lot at the rate of ₹1 per chocolate, and the second lot at the rate of ₹4 for 5 chocolates, his total collection would have been ₹460 Number of chocolates in 1st lot × 1 + Number of chocolates in 2nd lot × 𝟒/𝟓 = 460 x × 1 + y × 4/5 = 460 x + 4𝑦/5 = 460 Multiplying by 5 both sides 5(x + 4𝑦/5) = 5 × 460 5x + 4y = 2300 Now, our equations are 2x + 3y = 1200 …(1) 5x + 4y = 2300 …(2) From (1) 2x + 3y = 1200 2x = 1200 − 3y x = 𝟏/𝟐(1200 – 3y) Putting value of x in (2) 5x + 4y = 2300 5 × 𝟏/𝟐(1200 – 3y) + 4y = 2300 Multiplying by 2 both sides 2 × 5 × 𝟏/𝟐(1200 – 3y) + 2 × 4y = 2 × 2300 5(1200 − 3y) + 8y = 4600 6000 − 15y + 8y = 4600 6000 – 7y = 4600 6000 – 4600 = 7y 1400 = 7y 7y = 1400 y = 1400/7 y = 200 Putting y = 200 in (1) 2x + 3y = 1200 2x + 3(200) = 1200 2x + 600 = 1200 2x = 1200 − 600 2x = 600 x = 600/2 x = 300 Thus, x = 300, y = 200 Now, we need to find Total Number of Chocolates Total Number of Chocolates = x + y = 300 + 200 = 500

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.