Question

the mean (3x-1),(2x-3),(x+7),(1x+1) and (x+3)is 19. find the value of x​

Answer 1

Given :-

• Mean of 5 digits is 19

• Digits are (3x-1),(2x-3),(X+7),(X+1) and (X+3)

To Find:-

• Value of X .

Solution :-

Formula -

$\boxed{\sf{\implies Mean = \dfrac{ Sum\ of\ digits}{number\ of\ digits } }}$

Using the above mentioned formula

$\sf{\implies Mean = \dfrac{ Sum\ of\ digits}{number\ of\ digits } }$

$\sf{\implies 19 = \dfrac{ (3x-1)+(2x-3)+(x+7)+(1x+1)+ (x+3)}{5} }$

$\sf{\implies 19 = \dfrac{ 3x-1+2x-3+x+7+x+1+ x+3}{5} }$

$\sf{\implies 19 = \dfrac{ (8x + 11 - 4 }{5} }$

$\sf{\implies 19 \times 5= 8x + 7 }$

$\sf{\implies 95 - 7 = 8x }$

$\sf{\implies 88 = 8x }$

$\sf{\implies X = \dfrac{88}{8} \rightarrow 11 }$

$\boxed{\sf{\implies X = 11 }}$

Answer 2

✯✯ QUESTION ✯✯

The mean (3x-1),(2x-3),(x+7),(1x+1) and (x+3)is 19. find the value of x..

━━━━━━━━━━━━━━━━━━━━

✰✰ ANSWER ✰✰

• (3x-1) (2x-3) (x+7) (1x+1) and (x+3)
• Mean = 19
• Total No. of Observations = 5

➥Using Formula : -

$\implies\tt{\small{\boxed{\bold{\bold{\blue{\sf{Mean=\dfrac{Sum\:of\:all\:observations}{Total\:No.\:of\:Observations}}}}}}}}$

➥Putting Values : -

$\implies\tt{19=\dfrac{3x-1+2x-3+x+7+1x+1+x+3}{5}}$

$\implies\tt{19=\dfrac{3x+2x+x+x+x-1-3+7+1+3}{5}}$

$\implies\tt{19=\dfrac{8x-1-3+7+1+3}{5}}$

$\implies\tt{19=\dfrac{8x+7}{5}}$

➥Now ,

$\implies\tt{1(8x+7)=19\times{5}}$

$\implies\tt{8x+7=95}$

$\implies\tt{8x=95-7}$

$\implies\tt{8x=88}$

$\implies\tt{x=\cancel\dfrac{88}{8}}$

$\red\longmapsto\:\large\underline{\boxed{\bf\green{x}\orange{=}\purple{11}}}$

☛So , The value of x is 11...