Question

The sum of the digits of a 2-digit number is 12. The number is 6 times the unit digit.
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Answer 1

Given Question:

The sum of the digits of a 2-digit number is 12. The number is 6 times the unit digit.

Required Solution:

${\boxed {\huge {\bf {\pink { Answer \leadsto \purple{48} }}}}}$

How?

${\underline {\underline {\rm { Given: } }}}$

•The sum of the digits of a 2-digit number is 12.

•The number is 6 times the units digit.

${\underline {\underline {\rm { To \: Find:} }}}$

•The original number.

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${\underline {\underline {\rm { Calculation:} }}}$

Let us assume the number at tens place be a and number at unit place be b.

So, the original number will become = $\sf { 10a + b }$

Now according to the question:

• $\sf { a + b = 12 }$

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➝ $\sf { b = 12 - a }$................(i) [/tex]

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Also,

•$\sf { 10a + b = 6b }................(ii)$

Substituting values of b from equation (i)

➝ $\sf { 10a + b = 6b }$

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➝ $\sf { 10a + 12-a = 6(12-a) }$

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➝ $\sf { 9a + 12 = 72-6a }$

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➝ $\sf { 9a + 6a = 72-12 }$

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➝ $\sf { 15a = 60 }$

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➝ $\sf { a = \dfrac{60}{15} }$

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➝ $\sf { a = 4 }$ $\red { \bigstar }$

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$\therefore$ From equation (i),

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➝ $\sf { b = 12- a}$

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➝ $\sf { b = 12- 4}$

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➝ $\sf { b = 8}$$\red { \bigstar }$

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Hence, the original number = $\sf { 10a + b }$

➝ $\sf { 10(4) +8}$

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➝ $\sf { 40 +8}$

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➝ $\sf { 48}$ $\green { \bigstar }$

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$\therefore$ The original number is 48.

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Answer 2

Answer:

yes may sagot na.............