Question

Rohit spent ⅓ of his money on a wallet and ¼ of the remainder on a belt. what fraction of his money did he spend on the belt??​

Answer 1

Answer:

1/6

Step-by-step explanation:

1. The money Rohit spent on the belt

= one-fourth of the money left after he spent on his wallet

= 1/4 of (3/3-1/3)

=1/4×2/3

=1/6

HENCE, he spent 1/6th fraction of his money on the belt

Answer 2

Answer:

$\frac{1}{6}$ fraction of his money did he spend on the belt.

Step-by-step explanation:

Given,Rohit spent ⅓ of his money on a wallet and ¼ of the remainder on a belt.

This is a problem of Fraction.

We know in fraction, total amount always $= 1$

He spent $\frac{1}{3}$

of his money on a wallet.

So now remaining money

$= (1 - \frac{1}{3} ) \\ = \frac{3 - 1}{3} \\ = \frac{2}{3}$

It is also given that he spend $\frac{1}{4}$

of the remainder on belt.

So,he spend on belt

$= \frac{1}{4} \times \frac{2}{3} \\ = \frac{1}{2 \times 3} \\ = \frac{1}{6}$

This is a problem of part of fraction of Algebra.

Some important Algebra formulas:

${(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} \\ {(x + y)}^{3} = {x}^{3} + 3 {x}^{2} y + 3x {y}^{2} + {y}^{3} \\ {(x - y)}^{3} = {x}^{3} - 3 {x}^{2} y + 3x {y}^{2} - {y}^{3} \\ {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y) \\ {x}^{3} - {y}^{3} = {(x - y)}^{3} + 3xy(x - y) \\ {x}^{2} - {y}^{2} = (x + y)(x - y) \\ {x}^{2} + {y}^{2} = {(x - y)}^{2} + 2xy \\ {x}^{2} - {y}^{2} = {(x + y)}^{2} - 2xy \\ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} ) \\ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )$

Know more about fraction,

1.https://brainly.in/question/9833636

2.https://brainly.in/question/16383044

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