Question

The sum of two numbers is 90. If 50 exceeds three seventh of the smaller number by the same amount by which the larger number exceed to 16. Find the number.

Answer 1

$\huge{\blue{\bold{\underline{\underline{Answer :}}}}}$

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$\large\underline{\green{\bf Given :-}}$

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• The sum of two numbers is 90

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• 50 exceeds three seventh of the smaller number by the same amount by which the larger number exceed to 16

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$\large\underline{\red{\bf To \: Find :-}}$

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• The numbers

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$\large\underline{\orange{\bf Solution :-}}$

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• Let the larger number be "x"

• Let the smaller number be "y"

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$\purple{\underline \bold{According \: to \: the \ question :}}$

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The sum of two numbers is 90

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➜ x + y = 90

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➜ x = 90 - y ⚊⚊⚊⚊ ⓵

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Also given that , 50 exceeds three seventh of the smaller number by the same amount by which the larger number exceed to 16

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• Let the number which exceeds and is common in both case be "k"

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So,

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➜ $\sf 50 = \dfrac { 3y } { 7 } + k$

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➜ $\sf k + \dfrac { 3y } { 7 } = 50$ ⚊⚊⚊⚊ ⓶

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➜ 16 + k = x ⚊⚊⚊⚊ ⓷

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⟮ Putting x = 90 - y from ⓵ to ⓷ ⟯

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➜ 16 + k = x

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➜ 16 + k = 90 - y

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➜ k + y = 90 - 16

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➜ k + y = 74 ⚊⚊⚊⚊ ⓸

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⟮ Subtracting equation ⓶ from ⓸ ⟯

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➜ $\sf k + y - (k + \dfrac { 3y } { 7 }) = 74 - 50$

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➜ $\sf k + y - k - \dfrac { 3y } { 7 } = 24$

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➜ $\sf \dfrac { 7y - 3y } { 7 } = 24$

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➜ $\sf \dfrac { 4y } { 7 } = 24$

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➜ 4y = 24 × 7

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➜ 4y = 168

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➜ $\sf y = \dfrac { 168 } { 4 }$

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➨ y = 42

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• Hence the smaller number is 42

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⟮ Putting y = 42 in ⓵ ⟯

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➜ x = 90 - y

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➜ x = 90 - 42

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➨ x = 48

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• Hence the larger number is 48

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⟮ Putting y = 42 in ⓸ ⟯

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➜ k + y = 74

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➜ k + 42 = 74

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➜ k = 74 - 42

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➨ k = 32

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• Hence the common exceeding number is 32

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∴ The two numbers are 48 & 42

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