On a piece of paper is a list of six 2-digit prime numbers. They have a mean and median of 39. Their mode is 31, and the smallest number is 13. What are the six numbers? Explain how you deduced your answer.
Answers
Answer:
13,31,31,47,53,59 are the 6 numbers
Step-by-step explanation:
Mode is 31 so we know that 31 will be repeated, set of numbers 13, 31, 31, x, x, x.
Median is 39, total numbers is even so 2 middle numbers will be median. (31+x)/2 = 39
(31+x)=78
x= 78-31
x= 47
set of numbers is now 13, 31, 31, 47, x, x
mean is 39 so =
(13+31+31+47+x+x)/6 = 39
13+31+31+47+x+x= 234
13+31+31+x+x= 187
13+x+x= 125
x+x= 112
2x= 112
x= 56
+3, -3 on x value equals to last two numbers
Final answer= 13, 31, 31, 47, 53, 59
Solution :-
Let us assume that, the given six 2 - digits prime numbers are a, b , c , d , e and f ( in ascending order. )
conclusion :-
- smallest number = a = 13 .
- Mode = value that comes more number of times => conclusion 31 comes more than 1 time .
- Median = (3rd term + 4th term)/2 = 39 as total numbers are even number .
- Mean = (13 + b + c + d + e + f) / 6 = 39 => (b + c + d + e + f) = 39 * 6 - 13 = 221 .
→ (b + c + d + e + f) = 221 (from mean)
→ c + d = 78 (from median)
we get,
→ b + e + f = 221 - 78 = 143 .
since c and d both can't be 31 (mode) as 31*2 ≠ 78 .
- b , e or f = 31 .
- sum of other 2 = 112 .
- 112/2 = 56 => Prime numbers between 50 and 60 => 53 and 59 => as 53 + 59 = 112 .
as numbers are in ascending order , therefore,
→ e = 53
→ f = 59 .
→ b = 31 .
now, one number from c or d is also 31 , (which is c in ascending order)
- d = 78 - 31 = 47 .
hence,
- a = 13 .
- b = 31 .
- c = 31 .
- d = 47 .
- e = 53 .
- f = 59 .
Learn more :-
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