Question

There are 4 Different Natural Numbers. If 4th Number is added to the Average of First 3 Numbers then we get 36, 34, 32 and 30.
Find the Numbers.​

Answer 1

AnswEr :

⋆ Let the Numbers be a, b, c and d.

• According to the Question Now :

⇝ Average of First 3 No. + 4th = 36

⇝ (a + b + c) /3 + d = 36

⇝ a + b + c + 3d /3 = 36

⇝ a + b + c + 3d = 36 × 3

⇝ a + b + c + 3d = 108 ⠀⠀⠀⠀— eq.( I )

⇝ Average of First 3 No. + 4th = 34

⇝ (b + c + d) /3 + a = 34

⇝ b + c + d + 3a /3 = 34

⇝ b + c + d + 3a = 34 × 3

⇝ b + c + d + 3a = 102⠀⠀⠀⠀— eq.( II )

⇝ Average of First 3 No. + 4th = 32

⇝ (a + d + c) /3 + c = 32

⇝ a + d + c + 3b /3 = 32

⇝ a + d + c + 3b = 32 × 3

⇝ a + d + c + 3b = 96⠀⠀⠀⠀⠀— eq.( III )

⇝ Average of First 3 No. + 4th = 30

⇝ (a + b + d) /3 + c = 30

⇝ a + b + d + 3c /3 = 30

⇝ a + b + d + 3c = 30 × 3

⇝ a + b + d + 3c = 90 ⠀⠀⠀⠀⠀— eq.( IV )

━━━━━━━━━━━━━━━━━━━━━━━━

• Adding eq. ( I ), ( II ), ( III ) and, ( IV ) :

⇝ a + b + c + 3d = 108

⇝ b + c + d + 3a = 102

⇝ a + d + c + 3b = 96

⇝ a + b + d + 3c = 90

_________________________

⇝ 6a + 6b + 6c + 6d = 396

⇝ 6(a + b + c + d) = 396

• Dividing Both term by 6

⇝ a + b + c + d = 66 ⠀⠀⠀⠀⠀⠀— eq.( V )

━━━━━━━━━━━━━━━━━━━━━━━━

• Subtracting eq.( V ) from ( I ) :

↦ a + b + c + 3d = 108

↦ a + b + c + d ⠀= 66

⠀-⠀ -⠀⠀-⠀⠀-⠀⠀⠀-

________________________

↦ 2d = 42

• Dividing Both term by 2

↦ d = 21

_________________________________

• Subtracting eq.( V ) from ( II ) :

↦ 3a + b + c + d = 102

↦ a + b + c + d ⠀= 66

⠀-⠀ -⠀⠀-⠀⠀-⠀⠀⠀-

________________________

↦ 2a = 36

• Dividing Both term by 2

↦ a = 18

_________________________________

• Subtracting eq.( V ) from ( III ) :

↦ a + 3b + c + d = 96

↦ a + b + c + d ⠀= 66

⠀-⠀ -⠀⠀-⠀⠀-⠀⠀⠀-

________________________

↦ 2b = 30

• Dividing Both term by 2

↦ b = 15

_________________________________

• Subtracting eq.( V ) from ( IV ) :

↦ a + b + 3c + d = 90

↦ a + b + c + d ⠀= 66

⠀-⠀ -⠀⠀-⠀⠀-⠀⠀⠀-

________________________

↦ 2c = 24

• Dividing Both term by 2

↦ c = 12

∴ Natural Numbers are 18, 15, 12 and 21.

Answer 2

${\huge\bf{\mid{\overline{\underline{correct\:Question}}}\mid}}$

There are 4 Different Natural Numbers. If 4th Number is added to the Average of First 3 Numbers then we get 36, 34, 32 and 30 ??

First lets understand Question ,,, it has been said that , when we add any three numbers average out of four to the fourth number , we will get average as 36, 34, 32 and 30 respectively ...

${\huge\bf{\mid{\overline{\underline{Solution}}}\mid}}$ ::----------

Let the four numbers be a, b, c and d ....

A/q, now ,

$\frac{a + b + c}{3} + d = 36 \\ \\ \frac{a + c + d}{3} + b = 34 \\ \\ \frac{a + b + d}{3} + c = 32 \\ \\ \frac{b + c + d}{3} + a = 30$

or, we can say that ,,,,

a+b+c+3d = 36×3 = 108 --------- Equation(1)

a+c+d+3b = 34×3 = 102 --------- Equation(2)

a+b+d+3c = 32×3 = 96 --------- Equation(3)

b+c+d+3a = 30×3 = 90 --------- Equation(4)

$\textsf{Adding all these Equation now we get,,}$

→ 6a + 6b +6c + 6d = 108+102+96+90 = 396

Taking 6 common now ,,

→ 6(a+b+c+d) = 396

→ (a+b+c+d) = 396/6 = 66 --------- Equation(5)

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$\textbf{</strong><strong>Now</strong><strong> </strong><strong>subtracting</strong><strong> </strong><strong>Equation</strong><strong> </strong><strong>(</strong><strong>5</strong><strong>)</strong><strong> </strong><strong>from</strong><strong> </strong><strong>Equation</strong><strong> </strong><strong>(</strong><strong>1</strong><strong>)</strong><strong> </strong><strong>we</strong><strong> </strong><strong>get</strong><strong>,</strong><strong>,</strong><strong>}$

→ (a+b+c+3d) - (a+b+c+d) = 108 - 66

→ 2d = 42

→ $\red{\bold{d = 21}}$

______________________________________

$\textbf{Now subtracting Equation (5) from Equation (2) we get,,}$

→ (a+c+d+3b) - (a+b+c+d) = 102 - 66

→ 2b = 36

→ $\red{\bold{b = 18}}$

_______________________________________

$\textbf{Now subtracting Equation (5) from Equation (3) we get,,}$

→ (a+b+d+3c) - (a+b+c+d) = 96-66

→ 2c = 30

→ $\red{\bold{c = 15}}$

_______________________________________

$\textbf{Now subtracting Equation (5) from Equation (4) we get,,}$

→ (b+c+d+3a) - (a+b+c+d) = 90-66

→ 2a = 24

→ $\red{\bold{a = 12}}$

_______________________________________

$\bold{\pink{Hence\:our\:4\:Required\:Numbers\:Are:-12,\:15,\:18\:and\:21}}$

$\large\underline\textbf{Hope it Helps you .}$