Question

the numbers 2,3,4 and 5 occur(2k-3), (2+5k),(5k-7) and (k+2) times respectively. the average of the numbers is 2.85.Later on, the number 2 was replaced by 6 in all the places , what is the average of the new number

Answer 1

Answer:

[(k) + (2k+3) + (3k-5) + (5k+1)]/4 = 63

=> 11k -1 = 252

=> k = 253/11 = 23

Step-by-step explanation:

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Answer 2

Solution :-

→ Number 2 occurs = (2 + 5k) times .

So,

→ Total sum = 2 * (2 + 5k) = (4 + 10k)

and,

→ Number 3 occurs = (5k - 7) times .

So,

→ Total sum = 3 * (5k - 7) = (15k - 21)

and,

→ Number 4 occurs = (2k - 3) times .

So,

→ Total sum = 4 * (2k - 3) = (8k - 12)

and,

→ Number 5 occurs = (k + 2) times .

So,

→ Total sum = 5 * (k + 2) = (5k + 10)

then,

→ Sum of all numbers = Average * Total numbers

→ (4 + 10k) + (15k - 21) + (8k - 12) + (5k + 10) = 2.85 * (2 + 5k + 5k - 7 + 2k - 3 + k + 2)

→ 38k - 19 = 2.85 * (13k - 6)

→ 19(2k - 1) = 2.85 * (13k - 6)

→ 2k - 1 = 0.15(13k - 6)

→ 2k - 1 = 1.95k - 0.9

→ 2k - 1.95k = 1 - 0.9

→ 0.05k = 0.1

→ k = 2 .

therefore,

→ Number 2 occurs = (2 + 5k) = 2 + 5 * 2 = 12 times .

now, given that, number 2 is replaced by 6 in all the places .

then,

→ Total numbers = Same as before = 13k - 6 = 13 * 2 - 6 = 20

→ Total sum = 6 * 12 + (15k - 21) + (8k - 12) + (5k + 10) = 72 + (28k - 23) = 28k + 49 = 28 * 2 + 49 = 56 + 49 = 105

hence,

→ New Average = Total sum / Total numbers

→ New Average = 105/20

→ New Average = 5.25 (Ans.)

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