Answer
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Hint: First of all, we will assume E’s score, then we will have A’s and D’s score in terms of E. Then using their scores, we will have B’s score. Now, adding off their scores and dividing it 5, equate the resultant to 36 and we will get the required answer.
Complete step-by-step answer:
Let us suppose that E scored \[x\] runs. ………(1)
Now, since we are given that D scored 5 more than E.
Therefore, D scored \[x + 5\]. …………(2)
Now, we are also given that E scored 8 fewer than A.
Therefore, since E scored $x$ runs and A scored 8 more than E.
Therefore, A scored \[x + 8\] runs. ………….(3)
Now, we are also given that B scored as many as D and E combined.
By (1) and (2), we have that B scored \[x + x + 5\] runs which is equal to $2x + 5$ runs. ………(4)
Now, we are also given that and B and C scored 107 between them.
Therefore, the sum B’s and C’s score is 107.
Now, using (4) in this, we will have:-
$ \Rightarrow 2x + 5 + C's$ score is equal to 107.
So, C’s score will be $107 - 2x - 5 = 102 - 2x$ runs. ………(5)
Now we have the average of everyone’s score as 36 runs.
So, let sum of everyone’s score using (1), (2), (3), (4) and (5):-
Sum of scores is $x + 8 + 2x + 5 + 102 - 2x + x + 5 + x$.
On simplifying it, we have: Sum of scores = $3x + 120$.
So, using the formula of average now which is: $Average = \dfrac{{Sum}}{{Frequency}}$.
Hence, $36 = \dfrac{{3x + 120}}{5}$
Cross multiplying it to get:-
$3x + 120 = 36 \times 5 = 180$
Taking 120 from LHS to RHS, we will get:-
$3x = 60$
Taking 3 from LHS to RHS, we will get:-
$x = 20$
Hence, E scored 20 runs.
Note: The students must keep in mind that if they will just randomly take everyone’s score equal to some variable, there will be a whole lot of variables and equations to solve, which may create a lot of confusion and more room for mistakes. So, you must try to assume as few variables as possible.
Complete step-by-step answer:
Let us suppose that E scored \[x\] runs. ………(1)
Now, since we are given that D scored 5 more than E.
Therefore, D scored \[x + 5\]. …………(2)
Now, we are also given that E scored 8 fewer than A.
Therefore, since E scored $x$ runs and A scored 8 more than E.
Therefore, A scored \[x + 8\] runs. ………….(3)
Now, we are also given that B scored as many as D and E combined.
By (1) and (2), we have that B scored \[x + x + 5\] runs which is equal to $2x + 5$ runs. ………(4)
Now, we are also given that and B and C scored 107 between them.
Therefore, the sum B’s and C’s score is 107.
Now, using (4) in this, we will have:-
$ \Rightarrow 2x + 5 + C's$ score is equal to 107.
So, C’s score will be $107 - 2x - 5 = 102 - 2x$ runs. ………(5)
Now we have the average of everyone’s score as 36 runs.
So, let sum of everyone’s score using (1), (2), (3), (4) and (5):-
Sum of scores is $x + 8 + 2x + 5 + 102 - 2x + x + 5 + x$.
On simplifying it, we have: Sum of scores = $3x + 120$.
So, using the formula of average now which is: $Average = \dfrac{{Sum}}{{Frequency}}$.
Hence, $36 = \dfrac{{3x + 120}}{5}$
Cross multiplying it to get:-
$3x + 120 = 36 \times 5 = 180$
Taking 120 from LHS to RHS, we will get:-
$3x = 60$
Taking 3 from LHS to RHS, we will get:-
$x = 20$
Hence, E scored 20 runs.
Note: The students must keep in mind that if they will just randomly take everyone’s score equal to some variable, there will be a whole lot of variables and equations to solve, which may create a lot of confusion and more room for mistakes. So, you must try to assume as few variables as possible.
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