Question

$\huge\blue {\underline { \overline{ \bf \mid\: \: \: \: \: QUESTION \: \: \: \: \mid}}}$



If x is the average (arithmetic mean) of m and 9, y is the average of 2m and 15, and z is the average of 3m and 18, what is the average of x, y, and z in terms of m ?​

Answer 1

$\huge \text{Given Question - }$

If x is the average (arithmetic mean) of m and 9, y is the average of 2m and 15, and z is the average of 3m and 18, what is the average of x, y, and z in terms of m ?

LET'S UNDERSTAND IT !

First understand what is Given in the Question and what we have to find

Given -

• x is the average of m and 9
• y is the average of 2m and 15
• z is the average of 3m and 18

To Find -

• the average of x, y, and z in terms of m

LET'S SOLVE IT !

x is the average of m and 9 which mean

$\sf \dashrightarrow \sf{{x = \dfrac{ m + 9}{2} }..(1)}$

y is the average of 2m and 15 which mean

$\sf \dashrightarrow \sf{{y = \dfrac{ 2m + 15}{2} }..(2)}$

z is the average of 3m and 18

$\sf \dashrightarrow \sf{{z = \dfrac{ 3m + 18}{2} }..(3)}$

Now we have 3 equations and if we add all three equation we get :-

$\sf \dashrightarrow \sf{{2x + 2y + 2z = m + 2m + 3m + 9 + 15 + 18}}$

$\sf \dashrightarrow \sf{{2(x + y + z )= 6m + 42}}$

$\sf \dashrightarrow \sf{{x + y + z = \dfrac{6m + 42}{2} }}$

$\sf \dashrightarrow \sf{{x + y + z = 3m + 21 }}$

now we have to find average of x, y, and z

$\sf \dashrightarrow \sf{{x + y + z = \dfrac{3m + 21}{3} }}$

$\sf \dashrightarrow \sf{{x + y + z = m + 7 }}$

hence , The average of x, y, and z in terms of m will be m + 7 $\blue{\bigstar}$

Answer 2

The average of x, y, and z in terms of m will be m + 7