Question

Out of four numbers, the average of first three
is 15 and that of the last three is 16. If the last
number is 19, the first is​

Answer 1

Answer:

let the numbers be a, b, c, 19

• a+b+c/3= 15
• a+b+c= 45 ------ (1)
• also, b+c+ 19/3= 16
• b+c+19=48
• b+c= 29
• substituting (b+c ) in eqn (1), we get
• a+ 29= 45
• a= 16

so, the first number is 16

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Answer 2

Answer:

Sum of first 3 numbers (number 1,2,3}= $15 \times 3 = 45$

Sum of last 3 numbers(number 2,3,4) =$16 \times 3 = 48$

Now according to the question 4th number.= 19

So according to second equation,

$no.2 + no.3 + 19 = 48 \\ transposing \: 19 \\no.2 + no.3 = \implies \: 48 - 19 = 39 \\ \implies\: no.2 + no.3 \: = 39$

Putting the value of 2 and 3 in first equation

$no.1 + no.2 + no.3 \\ putting \:the \: value \: of \: 2 \: and \: 3 \\ no.1 + 39 = 45 \\ transposing \:39\\ no.1 = 45 - 39 = 6 \\ so \: value \: of \: 1 = 6$