Question

The mean of 5 number is 21 if one of the number is excluded then the mean of remaining numbers is is 22.5 the excluded number is


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Answer 1

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★ Question :-

Find the number of cakes of soap each measuring 9 cm x 5 cm x 2 cm that can be placed in a big cuboidal box measuring 1.5 m x 0.9 m x 0.5 m.

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★ Solution :-

Given,

» Dimensions of box = 1.5 m × 0.9 m × 0.5 m

= 150 m × 90 m × 50 m

» Dimensions of each soap cake

= 9 cm × 5 cm × 2 cm

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Question :

The mean of 5 number is 21 if one of the number is excluded then the mean of remaining numbers is is 22.5 the excluded number is?

Given :

Mean of 5 numbers = 21

If one number is excluded then mean = 22.5

To find :

Excluded number

Solution :

Mean of 5 numbers = 21

We know,

$: \implies\underline{\boxed{\sf{Mean = \dfrac{Sum \ of\ numbers}{Total \ number} }}} \\\\ \\ : \implies\sf 21 = \dfrac{Sum \ of \ 5 \ numbers}{5} \\\\ \\ : \implies\sf Sum \ of \ 5\ numbers = 21 \times 5 \\\\ \\ : \implies\bold{Sum \ of \ 5 \ numbers = 105} \dots\dots \ (i)$

Now mean of 4 numbers = 22.5

$: \implies\sf 22.5 = \dfrac{Sum \ of \ 4 \ numbers}{4} \\\\ \\ : \implies\sf Sum \ of \ 4 \ numbers = 4 \times 22.5 \\\\ \\ : \implies\bold{Sum \ of \ 4 \ numbers = 90} \dots\dots \ (ii)$

Now excluded number :

$: \implies\sf Excluded \ number = Sum \ of \ 5 \ numbers - Sum \ of \ 4 \ numbers \\\\ \\ : \implies\sf Excluded \ number = 105 - 90 \\\\ \\ : \implies\underline{\boxed{\sf{Excluded \ number = 15 }}} \ \star$

$\therefore\underline{\boxed{\textsf{Excluded number is = {\textbf{15}}}}} \qquad\qquad \bigg\lgroup\bold{Required \ answer} \bigg\rgroup$

Answer 2

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${\fbox{\fbox{\huge\sf\purple{QuEsTiOn}}}}$

Find the number of cakes of soap each measuring 9 cm x 5 cm x 2 cm that can be placed in a big cuboidal box measuring 1.5 m x 0.9 m x 0.5 m.

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${\fbox{\fbox{\huge\sf\purple{AnSwER}}}}$

Given,

» Dimensions of box = 1.5 m × 0.9 m × 0.5 m

= 150 m × 90 m × 50 m

» Dimensions of each soap cake

= 9 cm × 5 cm × 2 cm

__________________________

Question :

The mean of 5 number is 21 if one of the number is excluded then the mean of remaining numbers is is 22.5 the excluded number is?

Given :

Mean of 5 numbers = 21

If one number is excluded then mean = 22.5

To find :

Excluded number

Solution :

Mean of 5 numbers = 21

We know,

$: \implies\underline{\boxed{\sf{Mean = \dfrac{Sum \ of\ numbers}{Total \ number} }}} \\\\ \\ : \implies\sf 21 = \dfrac{Sum \ of \ 5 \ numbers}{5} \\\\ \\ : \implies\sf Sum \ of \ 5\ numbers = 21 \times 5 \\\\ \\ : \implies\bold{Sum \ of \ 5 \ numbers = 105} \dots\dots \ (i)$

Now mean of 4 numbers = 22.5

$: \implies\sf 22.5 = \dfrac{Sum \ of \ 4 \ numbers}{4} \\\\ \\ : \implies\sf Sum \ of \ 4 \ numbers = 4 \times 22.5 \\\\ \\ : \implies\bold{Sum \ of \ 4 \ numbers = 90} \dots\dots \ (ii)$

Now excluded number :

$: \implies\sf Excluded \ number = Sum \ of \ 5 \ numbers - Sum \ of \ 4 \ numbers \\\\ \\ : \implies\sf Excluded \ number = 105 - 90 \\\\ \\ : \implies\underline{\boxed{\sf{Excluded \ number = 15 }}} \ \star$

$\therefore\underline{\boxed{\textsf{Excluded number is = {\textbf{15}}}}} \qquad\qquad \bigg\lgroup\bold{Required \ answer} \bigg\rgroup$