Question

a two digit number is 4 times the summer of its digits and twice the product of the digits . find the number .​

Answer 1

Answer:Given : A two-digit number is 4 times the sum of its digits and twice the product of the digits.

Solution:

Let the digit in the unit's place be x and the digit at the tens place be y.

Number = 10y + x

The number obtained by reversing the order of the digits is = 10x + y

 

ATQ :

Condition : 1

10y + x = 4(x + y)

10y + x = 4x + 4y

4x + 4y - 10y – x = 0

3x – 6y = 0

3(x – 2y) = 0

x – 2y = 0

x = 2y ……………(1)

Condition : 2

10y + x = 2xy…………..(2)

On Substituting the value of x in equation (2) we obtain :  

10y + 2y = 2 × (2y) × y

12y = 4y²

4y2 – 12y = 0

4y(y – 3) = 0

y(y – 3) = 0

y = 0 or y = 3

On putting y = 0 in eq (1)  we obtain :

x = 2y

x = 2 × 0

x = 0

On putting y = 3 in eq (1)  we obtain :

x = 2y

x = 2 × 3

x = 6

x = 0 and y = 0 pair of solution does not give a two digit number.

From x = 6 and y = 3 we obtain a number :  

Number = 10y + x = 10 × 3 + 6 =  30 + 6 = 36

Hence, the number is 36.

Hope this answer will help you…

Step-by-step explanation:

Answer 2

GivEn:

• A two digit number is 4 times the sum of its digits and twice the product of the digits.

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To find:

• Find the number.

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SoluTion:

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☯ Let ones place digit be x

☯ Let tens place digit be y

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Therefore,

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The number is = 10y + x

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${\underline{\bf{\bigstar\;As\;per\;given\; Question\;:}}}$

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A two digit number is 4 times the sum of its digits and twice the product of the digits.

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★ 4 times the sum of digit,

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$:\implies\sf 10y + x = 4(x + y)$

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$:\implies\sf 10y + x = 4x + 4y$

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$:\implies\sf 10y - 4y = 4x - x$

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$:\implies\sf 6y = 3x$

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$:\implies\sf x = \dfrac{ \cancel{6}y}{ \cancel{3}}$

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$:\implies\bf \red{x = 2y}\;\;\;\;\;\;\;\;\;\bigg\lgroup\bf eq.\;(1)\bigg\rgroup$

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And

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★ Twice the product of the digits,

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$:\implies\sf 10y + x = 2xy$

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$\;\;\;\small\sf \underline{Dividing\;both\;sides\;by\;xy\;:}$

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$:\implies\sf \dfrac{10y}{xy} + \dfrac{x}{xy} = \dfrac{2xy}{xy}$

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$:\implies\sf \dfrac{10 \cancel{y}}{x \cancel{y}} + \dfrac{ \cancel{x}}{ \cancel{x}y} = \dfrac{2 \cancel{xy}}{ \cancel{xy}}$

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$:\implies\bf \blue{ \dfrac{10}{x} + \dfrac{1}{y} = 2}\;\;\;\;\;\;\;\;\;\bigg\lgroup\bf eq.\;(2)\bigg\rgroup$

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${\underline{\sf{\bigstar\;Now,\;Putting\;value\:of\;eq.(1)\;in\;eq.(2)\;:}}}$

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$:\implies\sf \dfrac{10}{2y} + \dfrac{1}{y} = 2$

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$:\implies\sf \dfrac{ \cancel{10}}{ \cancel{2}y} + \dfrac{1}{y} = 2$

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$:\implies\sf \dfrac{5}{y} + \dfrac{1}{y} = 2$

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$:\implies\sf \dfrac{5 + 1}{y} = 2$

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$:\implies\sf \dfrac{6}{y} = 2$

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$:\implies\sf y = \cancel{ \dfrac{6}{2}}$

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$:\implies{\underline{\boxed{\bf{\pink{y = 3}}}}}\;\bigstar$

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${\underline{\sf{\bigstar\;Now,\;Putting\;value\:of\;y\;in\;eq.(1)\;:}}}$

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$:\implies\sf x = 2 \times 3$

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$:\implies{\underline{\boxed{\bf{\purple{x = 6}}}}}\;\bigstar$

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$\therefore$ Hence, The required number is,

$:\implies\sf (10y + x) = 10 \times 3 + 6 = \bf{36}.$