Question

6. By what number should
-12/35
be divided to get 3/7 ?
​

Answer 1

Answer:

$\tt \huge \frac{ - 4}{5}$

Step-by-step explanation:

$\tt \huge \: let \: the \: no. \: be \: x \\ \\ \tt \huge \frac{ - 12}{35} \div x = \frac{3}{7} \\ \\ \tt \huge \: x = \frac{3}{7} \times \frac{ - 35}{12} = \frac{ - 4}{5}$

Answer 2

★ How to do :-

Here, we are given with a number which should be divided. We are also given with the answer obtained while dividing those both. But, we aren't given with the second number that should be divided by first number. We are asked to find the value of the same. Here, we use a concept known as shifting the numbers from one hand side to the other. While doing this process, the sign of the particular number changes. We can also verify the statement by substituting the value of second number. So, let's solve!!

$\:$

➤ Solution :-

Let, the other number be x.

${\tt \leadsto \dfrac{(-12)}{35} \div x = \dfrac{3}{7}}$

${\tt \leadsto x = \dfrac{(-12)}{35} \div \dfrac{3}{7}}$

Take the reciprocal of second fraction and multiply both the fractions.

${\tt \leadsto x = \dfrac{(-12)}{35} \div \dfrac{7}{3}}$

Write both numerators and denominators in a common fraction.

${\tt \leadsto x = \dfrac{\cancel{(-12)} \times 7}{35 \times \cancel{3}} = \dfrac{(-4) \times 7}{35 \times 1}}$

Again cancel the remaining two numbers in numerator and denominator.

${\tt \leadsto x = \dfrac{(-4) \times \cancel{7}}{\cancel{35} \times 1} = \dfrac{(-4) \times 1}{5 \times 1}}$

Simplify the fraction to get the answer.

${\tt \leadsto x = \dfrac{(-4)}{5}}$

$\:$

${\red{\underline{\boxed{\bf So, \: the \: other \: number \: is \: \: \dfrac{(-4)}{5}}}}}$

━━━━━━━━━━━━━━━━━━━━━━━

Verification :-

${\tt \leadsto \dfrac{(-12)}{35} \div x = \dfrac{3}{7}}$

Substitute the value of x.

${\tt \leadsto \dfrac{(-12)}{35} \div \dfrac{(-4)}{5} = \dfrac{3}{7}}$

Take the reciprocal of second fraction and multiply both fractions.

${\tt \leadsto \dfrac{(-12)}{35} \times \dfrac{5}{(-4)} = \dfrac{3}{7}}$

Write both numerators and denominators in the common fraction.

${\tt \leadsto \dfrac{\cancel{(-12)} \times 5}{35 \times \cancel{(-4)}} = \dfrac{3}{7}}$

Write the resulting fraction.

${\tt \leadsto \dfrac{3 \times 5}{35 \times 1} = \dfrac{3}{7}}$

Now, multiply teh numbers on numerator and denominator.

${\tt \leadsto \dfrac{15}{35} = \dfrac{3}{7}}$

Write the fraction on LHS in lowest form by cancellation method.

${\tt \leadsto \dfrac{3}{7} = \dfrac{3}{7}}$

So,

${\sf \leadsto LHS = RHS}$

$\:$

Hence verified !!