A number is divided into two parts so that the difference of the parts is 40. If four-fifth of the smaller part exceeds one-third of the greater part by 24. Find the number
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A number is divided into two parts so that the difference of the parts is 40. If four-fifth of the smaller part exceeds one-third of the greater part by 24. Find the number
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[tex]\Large{\red{\underline{\underline{\bf{Solution:}}}}} [/tex]
[tex]⟹4/5th
[/tex]
[tex]=1/3th
[/tex]
[tex]\sf{\implies \dfrac{4x}{5}=\dfrac{x + 40}{3} + 24}[/tex]
[tex]\sf{\implies \dfrac{4x}{5}-\dfrac{x + 40}{3} = 24}[/tex]
[tex]\sf{\implies \dfrac{12x-5x-200}{15}=24}[/tex]
[tex]\sf{\implies \dfrac{7x-200}{15}=24}[/tex]
⟹7x−200=360
⟹7x=360+200
⟹7x=560
⟹x=80
[tex]\sf\small\underline\orange{Hence,\:the\: smaller\: number\:(x)=80:-}[/tex]
[tex]\sf\small\underline\orange{Hence,\:the\: greater\: number\:(x+40)=120} [/tex]
Answer:
Let the required number be 'x', then four fifths of the number = 4/5 x
And three fourths of the number = 3/4x
It is given that 4/5x is greater than3/4 x by 4
⇒ 4/5x- 3/4x = 4
16x- 15x / 20 = 4
= x/20 = 4
x= 80
hence the required number is 80.
Step-by-step explanation:
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