Total of 1018 chocolates were distributed in a class such that each st...
Solution: Let x, y and z be the numbers of students who got 9, 10 and 12 chocolates respectively. 9x + 10y+ 12z = 1018
total number of students = 24 + 43 + 31 =98 Hence, option 4.
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Total of 1018 chocolates were distributed in a class such that each st...
Problem Analysis:
Let's assume the number of students getting 9 chocolates is a, the number of students getting 10 chocolates is b, and the number of students getting 12 chocolates is c. We need to find the total number of students in the class.
Given:
- Each student gets either 9 or 10 or 12 chocolates.
- Most students got 10 chocolates.
- The number of students getting 10 chocolates is the same as the sum of the number of students getting 12 chocolates and half the number of students getting the least number of chocolates.
Key Observations:
- Most students got 10 chocolates, so b > a and b > c.
- The number of students getting 10 chocolates is the same as the sum of the number of students getting 12 chocolates and half the number of students getting the least number of chocolates, i.e., b = c + (0.5)a.
Solution:
To find the total number of students, we need to determine the values of a, b, and c.
From the given information, we have the equation b = c + (0.5)a.
We also know that the total number of chocolates distributed is 1018. Since each student gets either 9 or 10 or 12 chocolates, we can write the equation:
9a + 10b + 12c = 1018
We have two equations and three variables. To solve this, we need another equation. Let's use the fact that most students got 10 chocolates, i.e., b > a and b > c.
Since each student gets either 9 or 10 or 12 chocolates, the minimum number of chocolates a student can get is 9. So, let's assume a = 0 and consider cases where a > 0.
Case 1: a = 1
Substituting a = 1 in the equation b = c + (0.5)a, we get b = c + 0.5.
We can substitute these values in the equation 9a + 10b + 12c = 1018 and solve for c.
9(1) + 10(c + 0.5) + 12c = 1018
9 + 10c + 5 + 12c = 1018
22c = 1004
c = 45.6
Since c should be an integer, the case a = 1 is not valid.
Case 2: a = 2
Substituting a = 2 in the equation b = c + (0.5)a, we get b = c + 1.
We can substitute these values in the equation 9a + 10b + 12c = 1018 and solve for c.
9(2) + 10(c + 1) + 12c = 1018
18 + 10c + 10 + 12c = 1018
22c = 990
c = 45
Since c is an integer, the case a = 2 is valid. From the equation b = c + 1, we get b = 46.
The total number of students in the class is a + b + c = 2 + 46 + 45
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