Rational Numbers Test - 7

We have, \(\frac{\left(-18 \frac{1}{3} \times 2 \frac{8}{11}\right)-\left(4 \frac{5}{7} \times 2 \frac{1}{3}\right)}{\left|\frac{3}{5}+\left(\frac{-9}{10}\right)\right|+\left|-\left(\frac{-3}{5}\right)\right|}\)

\(=\frac{\left(\frac{-55}{3} \times \frac{30}{11}\right)-\left(\frac{33}{7} \times \frac{7}{3}\right)}{\left|\frac{3}{5}-\frac{9}{10}\right|+\left|\frac{3}{5}\right|}=\frac{-50-11}{\left|\frac{6-9}{10}\right|+\frac{3}{5}}\)

\(=\frac{-61}{\frac{3}{10}+\frac{3}{5}}=\frac{-61}{\frac{3+6}{10}}\)

\(=\frac{-61 \times 10}{9}=\frac{-610}{9}\)

\(=-67 \frac{7}{9}\)

Since, denominator is 0, it is not a rational number.

\(\frac{1}{4}\) can be written as \(\frac{1 \times 4}{4 \times 4}=\frac{4 \times(-1)}{16 \times(-1)}=\frac{-4}{-16}\)

\(\therefore\) \(\frac{1}{4}\) and \(\frac{-4}{-16}\) are equivalent rational numbers.

\(\frac{\sqrt{2}}{3}\) is not a rational number.

The ascending order of \(\frac{4}{7}, \frac{1}{3}, \frac{2}{5}\) and \(\frac{5}{9}\) is \(\frac{1}{3}<\frac{2}{5}<\frac{5}{9}<\frac{4}{7}\).

The two middle numbers are \(\frac{2}{5}\) and \(\frac{5}{9}\).

\(\therefore\) Average \(=\frac{\frac{2}{5}+\frac{5}{9}}{2}=\frac{43}{90}\)

Let the total area of field be 1. Part of field in which the farmer grows tomatoes \(=1-\left(\frac{2}{3}+\frac{1}{4}\right)=1-\left(\frac{8+3}{12}\right)=1-\frac{11}{12}=\frac{1}{12}\)

\(\frac{-3}{6}, \frac{-4}{3}, \frac{-9}{4}, \frac{-13}{4}\)

First we need to make the denominator same by taking L.C.M (6, 3, 4, 4) = 12

Now, \(\frac{-3}{6}=\frac{-3 \times 2}{6 \times 2}=\frac{-6}{12} ; \frac{-4}{3}=\frac{-4 \times 4}{3 \times 4}=\frac{-16}{12}\); \(\frac{-9}{4}=\frac{-9 \times 3}{4 \times 3}=\frac{-27}{12} ; \frac{-13}{4}=\frac{-13 \times 3}{4 \times 3}=\frac{-39}{12}\)

Since, \(-6>-16>-27>-39\)

\(\therefore\) \(\frac{-3}{6}, \frac{-4}{3}, \frac{-9}{4}, \frac{-13}{4}\) are in descending order.

Let us consider a fraction 3/5 and a rational number = −4/7. By definition, 3/5 is a rational number.

So, p is true.

By definition, −4/7 is not a fraction, since, −4 is not a natural number.

So, q is false.

Total quantity of rice \(=87 \frac{1}{2} k g=\frac{175}{2} k g\)

So, each friend got \(=\frac{175}{2} \div 3=\frac{175}{2} \times \frac{1}{3}=\frac{175}{6}=29 \frac{1}{6} k g\)

Since, every integer having a denominator 1 can be expressed in P/q form, p is true. Since, a rational number with denominator other than 1 is not an integer (e.g., 3/5 ), q is false.

  L.C.M. of 5, 4 and 20 is 20.

\(\frac{4}{-5} \times \frac{4}{4}=\frac{16}{-20} ; \frac{3}{4}=\frac{3 \times 5}{4 \times 5}=\frac{15}{20}\)\(-2<\frac{4}{-5}<\frac{-11}{20}<\frac{3}{4}\) or \(\frac{3}{4}>\frac{-11}{20}>\frac{4}{-5}>-2\) is the required descending order. Rewrite equivalent fractions of the given fractions and then compare then. Arrange them in descending order.

Sum of \(\frac{5}{12}\) and \(\frac{-17}{24}\) is \(\frac{5}{12}+\left(\frac{-17}{24}\right)=\frac{5}{12}-\frac{17}{24}=\frac{10-17}{24}=\frac{-7}{24}\)

Product of \(\frac{2}{5}\) and \(\frac{7}{4}\) is \(\frac{2}{5} \times \frac{7}{4}=\frac{7}{10}\)

According to the question, we have

\(\frac{-7}{24} \div \frac{7}{10}=\frac{-7}{24} \times \frac{10}{7}=\frac{-5}{12}\)

Delegation from India = \(\frac{3}{20} \times 1200=180\)

Delegation from Britain \(\frac{1}{4} \times 1200=300\)

Delegation from Germany \(=\frac{3}{10} \times 1200=360\)

Delegation from America \(=1200-(180+300+360)=1200-840=360\)

Distance between home and school \(=\frac{6}{7} k m\) And, distance between school and landmark \(=\frac{5}{6} k m\)

\(\therefore\) Distance between home and landmark \(=\left(\frac{6}{7}-\frac{5}{6}\right)=\frac{36-35}{42}=\frac{1}{42} k m\)

Weight of each mango = \(\frac{1}{10}\)kg

Wight of box = \(\frac{3}{5}\) kg

Number of mangoes \(=\frac{3}{5} \div \frac{1}{10}=\frac{3}{5} \times 10=6\)