Question 28 (Choice 2) - CBSE Class 10 Sample Paper for 2023 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

Last updated at May 29, 2023 by Teachoo

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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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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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