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 August 12, 2026 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.
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
Made by
Davneet Singh
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