Question

10. The sum of the digits of a two digit number is 12 and product of the digits is 27, one-third of
that number is :
(a)31
(b) 13
(c) 43
(d) 34​

Answer 1

$\underline{\underline{\huge{\gray{\tt{\textbf Answer :-}}}}}$

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$\sf\large\bold{\orange{\underline{\blue{ Given :-}}}}$

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• Sum of the digits of a two digit number is 12

• Product of the digits is 27

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$\sf\large\bold{\orange{\underline{\blue{ To \: Find :-}}}}$

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• One-third of that number

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$\sf\large\bold{\orange{\underline{\blue{ Solution :-}}}}$

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• Let the ten's digit be "x"

• Let the one's digit be "y"

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$\underline{\bold{\texttt{Original number :}}}$

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➠ 10x + y

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$\rm{ \large{\pink{\underline\blue{According \: to \: the \ question :}}}}$

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Given that , sum of the digits of a two digit number is 12

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So ,

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➜ x + y = 12 ⚊⚊⚊⚊ ⓵

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➜ x = 12 - y ⚊⚊⚊⚊ ⓶

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Also given that , product of the digits is 27

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So,

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➜ x × y = 27 ⚊⚊⚊⚊ ⓷

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⟮ Putting x = 12 - y from ⓶ to ⓷ ⟯

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➜ x × y = 27

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➜ (12 - y) × y = 27

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➜ 12y - y² = 27

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➜ y² - 12y + 27 = 0

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➜ y² - 3y - 9y + 27 = 0

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➜ y(y - 3)-9(y - 3) = 0

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➜ (y - 3)(y - 9) = 0

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• y = 3 ⚊⚊⚊⚊ ⓸
• y = 9 ⚊⚊⚊⚊ ⓹

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Case I [ y = 3 ]

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⟮ Putting y = 3 from ⓸ to ⓶ ⟯

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➜ x = 12 - y

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➜ x = 12 - 3

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➜ x = 9

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• Hence the one's digit is 3 & ten's digit is 9

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∴ The original number is 93

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Case II [ y = 9 ]

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⟮ Putting y = 9 from ⓹ to ⓶ ⟯

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➜ x = 12 - y

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➜ x = 12 - 9

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➜ x = 3

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• Hence the one's digit is 9 & ten's digit is 3

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∴ The original number is 39

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From Case I & Case II we got that the original number is either 93 or 39

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So ,

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$\underline{\bold{\texttt{One third of 93 :}}}$

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➜ $\sf \dfrac { 1 } { 3 } \times 93$

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➨ 31

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• Hence one third of 93 is 31

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$\underline{\bold{\texttt{One third of 39 :}}}$

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➜ $\sf \dfrac { 1 } { 3 } \times 39$

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➨ 13

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• Hence one third of 39 is 13

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∴ Both a) 31 & b) 13 are correct options