Question

The average of three prime numbers lying between 47 and 74 is 191/3. The greatest possible sum between any two of these 3 prime numbers is
$\begin{align}
  & \left( a \right)120 \\
 & \left( b \right)138 \\
 & \left( c \right)132 \\
 & \left( d \right)136 \\
\end{align}$

Answer 1

Hint: To solve the question given above, first we will assume that the numbers are a, b and c. Then we will take their arrangement and equate it to $\dfrac{191}{3}$. From there we will get the sum of a, b and c. Now, we will find out all the prime numbers lying between 47 and 74. Then we will select three of them such that their sum is equal to the sum of a, b and c. Then we will find out the sum of maximum of (a, b and c) and second maximum of (a, b and c).

Complete step by step solution:
It is given in the question that the average of three prime numbers lying between 47 and 74 is $\dfrac{191}{3}$. Let us assume that these three prime numbers are a, b and c. Now, we have to calculate their average. The average of any data is given by the
formula: $Average=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+........{{x}_{n}}}{n}$
where ${{x}_{1}},{{x}_{2}},{{x}_{3}}..........{{x}_{n}}$ are the entities and n is the total number of entities.
Thus, we have:
$\begin{align}
  & Average=\dfrac{a+b+c}{3}=\dfrac{191}{3} \\
 & \Rightarrow \dfrac{a+b+c}{3}=\dfrac{191}{3} \\
 & \Rightarrow a+b+c=191 \\
\end{align}$
Now, from the equation above, we can say that the sum of prime numbers should be 191.
Now, we will find all prime numbers lying between 47 and 74. These are 53, 59, 61, 67, 71 and 73. Now, we have to select three numbers from the above prime numbers such that their sum is 191. Here, we will get two cases:
Case 1: a=53, b=67, c=71.Here we can see that the sum of numbers is 191. Now, we have to find the largest possible sum between two numbers. Thus, we have:
$\begin{align}
  & a+b=53+67=120 \\
 & b+c=67+71=138 \\
 & a+c=53+71=124 \\
\end{align}$
Now, we can see that the greatest sum in this case is 138.
Case 2: a=59, b=61, c=71. Here we can see that the sum of numbers is 191. Now, we have to find the largest possible sum between two numbers. Thus, we have:
$\begin{align}
  & a+b=59+61=120 \\
 & b+c=61+71=132 \\
 & a+c=59+71=130 \\
\end{align}$
Here, the greatest sum is 120.
From the above two cases, we can see that the greatest sum is 138.
Hence, option (b) is correct.

Note: In the solution of the above question, we have checked the sum using hit and trial method i.e. we have taken any three numbers and check if their sum is 191. The alternative method to do this is as follows: The last digit of 191 is 1. So, we have to find those one digit numbers (1, 3, 7 and 9). Which when selected their sum will give 11. There are only two cases: (1, 1, 9) and (7, 3, 1). a, b and c should have the last digit as the digits in these cases. Thus a=59, b=61 and c=71 or a=53, b=67 and c=71.