Math, asked by drksdeepa, 8 days ago

The product of two rational numbers is 63 / 40 if one of the number is -7/5 find the other number

Answers

Answered by AditDip
2

Answer:

-9/8 is the answer

Step-by-step explanation:

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Answered by MasterDhruva
30

How to do :-

Here, we are given with a rational number that should be multiplied by an other rational number. We are also given with the answer obtained while multiplying those both rational numbers. But, we are not given with the second number that the first number on LHS to be multiplied with. We are asked to find the same. We use some other concepts to find out the answer of this question. The concept is named as transposition method. This method is always used when we need to find the value of any non-given value. By using this concept, the sign of the particular number changes. So, let's solve!!

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Solution :-

Let the other number be x,

{\tt \leadsto \dfrac{(-7)}{5} \times x = \dfrac{63}{40}}

Shift the number on LHS to RHS.

{\tt \leadsto x = \dfrac{63}{40} \div \dfrac{(-7)}{5}}

Take the reciprocal of second fraction and multiply both fractions.

{\tt \leadsto x = \dfrac{63}{40} \times \dfrac{5}{(-7)}}

Write the fractions in lowest form by cancellation method.

{\tt \leadsto x = \dfrac{\cancel{63} \times 5}{40 \times \cancel{(-7)}} = \dfrac{9 \times 5}{40 \times (-1)}}

Multiply the remaining numbers.

{\tt \leadsto x = \cancel \dfrac{45}{(-40)} = \dfrac{9}{(-8)}}

Write the obtained fraction in the form of mixed fraction.

{\tt \leadsto \dfrac{9}{(-8)} = -1 \dfrac{1}{8}}

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{\red{\underline{\boxed{\bf So, \: the \: other \: number \: is \: \: -1 \dfrac{1}{8}}}}}

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Verification :-

{\tt \leadsto \dfrac{(-7)}{5} \times x = \dfrac{63}{40}}

Substitute the value of x.

{\tt \leadsto \dfrac{(-7)}{5} \times \dfrac{(-9)}{8} = \dfrac{63}{40}}

Write both numerators and denominators in a common fraction.

{\tt \leadsto \dfrac{(-7) \times (-9)}{5 \times 8} = \dfrac{63}{40}}

Multiply the numerators and denominators on LHS.

{\tt \leadsto \dfrac{63}{40} = \dfrac{63}{40}}

So,

{\sf \leadsto LHS = RHS}

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Hence verified !!

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