At a birthday party, every child gets $2$ chocolates, every mother gets $1$ chocolate, while no father gets a chocolate. In total $69$ persons get $70$ chocolates. If the number of children is half of the number of mothers and fathers put together, then how many fathers are there?
Fathers Calculation: Birthday Party Chocolates
This problem requires setting up and solving a system of equations based on the given constraints about people and chocolates.
Defining Variables and Equations
Let $C$ represent the number of children, $M$ the number of mothers, and $F$ the number of fathers.
- Total Persons: The total number of people is 69.
$C + M + F = 69$ - Total Chocolates: Each child gets 2 chocolates, each mother gets 1, and fathers get none. Total chocolates are 70.
$2C + M = 70$ - Children-Parents Relation: The number of children equals half the sum of mothers and fathers.
$C = \frac{1}{2}(M + F)$
This implies $2C = M + F$.
Step-by-Step Solution
-
Find the number of children ($C$). Use the equation $2C = M + F$ and substitute it into the total persons equation $C + M + F = 69$.
Replace $M + F$ with $2C$:
$C + (2C) = 69$Combine terms and solve for $C$:
$3C = 69$
$C = \frac{69}{3}$
$C = 23$There are 23 children.
-
Find the number of mothers ($M$). Use the total chocolates equation $2C + M = 70$ and the value $C = 23$.
Substitute $C=23$:
$2(23) + M = 70$
$46 + M = 70$Solve for $M$:
$M = 70 - 46$
$M = 24$There are 24 mothers.
-
Find the number of fathers ($F$). Use the total persons equation $C + M + F = 69$ and the values $C = 23$ and $M = 24$.
Substitute $C=23$ and $M=24$:
$23 + 24 + F = 69$
$47 + F = 69$Solve for $F$:
$F = 69 - 47$
$F = 22$There are 22 fathers.
Verification
Confirm the solution satisfies all conditions:
- Total Persons: $23 + 24 + 22 = 69$ (Correct)
- Total Chocolates: $2(23) + 1(24) = 46 + 24 = 70$ (Correct)
- Children relation: $23 = \frac{1}{2}(24 + 22) = \frac{1}{2}(46)$ (Correct)
The number of fathers is 22.
Important Questions from Algebra (Notes)
- What is the remainder when 2023²⁰²⁴ + 2025²⁰²⁴ is divided by 2024?
In an examination, a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer. If she/he attempts all 60 questions and secures 130 marks, the number of questions she/he attempts wrongly, are?
Match List-I with List-II
List-1 List-II (A) If $\begin{bmatrix}\lambda-1 & 0 \\ 0 & \lambda-1 \end{bmatrix} $, then $\lambda$ is (I) 0 (B) If A=$ \begin{bmatrix}1 & 2 \\2 & 4 \end{bmatrix} $, then $\Delta$ is (II) 1 (C) If A = $ \begin{bmatrix}1 & 0 \\0 & \frac{1}{2} \end{bmatrix} $, then $|A^{-1}|$ is (III) -2 (D) If $ \begin{bmatrix}a+1 & 1 \\1 & 2 \end{bmatrix} = \begin{bmatrix}-1 & 1 \\1 & 2 \end{bmatrix} $, then a is (IV) 2 Choose the correct answer from the options given below:
- If (x - 1) is a factor of $2x^2 - 5x + k = 0$, then the value of k is:
- If $x = (2+\sqrt{3})^{\frac{1}{3}} + (2+\sqrt{3})^{-\frac{1}{3}}$ and $x^3-3x + k = 0$, then the value of k is:
