Solveeit Logo

Question

Question: The average temperatures of Tuesday, Wednesday and Thursday was \({{42}^{\circ }}C\). The average te...

The average temperatures of Tuesday, Wednesday and Thursday was 42C{{42}^{\circ }}C. The average temperature of Wednesday, Thursday and Friday was 47C{{47}^{\circ }}C. If the temperature of Tuesday was 43C{{43}^{\circ }}C, then the temperature of Friday was?
(A)58C (B)50C (C)53C (D)49C \begin{aligned} & \left( A \right){{58}^{\circ }}C \\\ & \left( B \right){{50}^{\circ }}C \\\ & \left( C \right){{53}^{\circ }}C \\\ & \left( D \right){{49}^{\circ }}C \\\ \end{aligned}

Explanation

Solution

Assuming temperatures of individual days as variables, we will form separate linear equations in these variables. Since, Wednesday and Thursday appear in both the average temperature measure, we can take the average temperature of Wednesday and Thursday to be a single variable. Solving these discrete linear equations will give us the temperature of Friday.

Complete step by step solution:
Let us first assign some terms that we are going to use later in our problem.
Let the temperature on Tuesday be ‘x’, then it has been given to us in the problem as 43C{{43}^{\circ }}C.
Let the sum of temperatures of Wednesday and Thursday be equal to ‘y’. And,
Let the temperature on Friday be ‘z’, then we have to find this temperature ‘z’.

According to our first statement: The average temperatures of Tuesday, Wednesday and Thursday was 42C{{42}^{\circ }}C. Mathematically, this could be written as:
x+y3=42C\Rightarrow \dfrac{x+y}{3}={{42}^{\circ }}C

Using the value of ‘x’ given to us in the problem, we get:
43+y3=42 43+y=126 y=83C \begin{aligned} & \Rightarrow \dfrac{43+y}{3}=42 \\\ & \Rightarrow 43+y=126 \\\ & \therefore y={{83}^{\circ }}C \\\ \end{aligned}

Now, according to our second statement: The average temperature of Wednesday, Thursday and Friday was 47C{{47}^{\circ }}C. Writing this as a mathematical expression, we get the equation:
y+z3=47C\Rightarrow \dfrac{y+z}{3}={{47}^{\circ }}C

Using the value of ‘y’ calculated above, we get:
83+z3=47 83+z=141 z=58C \begin{aligned} & \Rightarrow \dfrac{83+z}{3}=47 \\\ & \Rightarrow 83+z=141 \\\ & \therefore z={{58}^{\circ }}C \\\ \end{aligned}

Thus, the final value of ‘z’ comes out to be 58C{{58}^{\circ }}C.
Hence, the temperature on Friday was 58C{{58}^{\circ }}C.

So, the correct answer is “Option A”.

Note: While solving this problem, we did not consider the temperatures of Wednesday and Thursday as different entities. This was done to reduce the number of variables we have in our problem. Linear equations in less variables are comparatively easier to solve then equations containing more variables.