Model Question Paper 1

Additive inverse of $\frac{5}{7}$ is $\frac{-5}{7}$.

According to the given condition,

$$\begin{aligned}\Rightarrow \frac{5}{7} + \Big(\frac{-5}{7}\Big)\\[1em] \Rightarrow \frac{5 - 5}{7}\\[1em] \Rightarrow \frac{0}{7}\\[1em] \Rightarrow 0\end{aligned}$$

Hence, Option 2 is the correct option.

Let the other rational number be $x$.

$$\begin{aligned}\Rightarrow \frac{4}{5} \times x = 1\\[1em] \Rightarrow x = 1 \times \frac{5}{4}\\[1em] \Rightarrow x = \frac{5}{4}\end{aligned}$$

Hence, Option 3 is the correct option.

As we know, $\Big(\frac{a}{b}\Big)^m = \frac{a^m}{b^m}, a^m \times a^n = a^{m+n}$ and $a^n = a^m \Rightarrow n = m$.

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{4}{9}\right)^x \times \left(\frac{3}{2}\right)^{-1} = \frac{8}{27}\\[1em] \Rightarrow \left(\frac{2^2}{3^2}\right)^x \times \frac{2}{3} = \frac{2^3}{3^3}\\[1em] \Rightarrow \left(\frac{2}{3}\right)^{2x} \times \left(\frac{2}{3}\right)^1 = \left(\frac{2}{3}\right)^3\\[1em] \Rightarrow \left(\frac{2}{3}\right)^{2x + 1} = \left(\frac{2}{3}\right)^3\\[1em] \Rightarrow 2x + 1 = 3\\[1em] \Rightarrow 2x = 3 - 1\\[1em] \Rightarrow 2x = 2\\[1em] \Rightarrow x = 1\end{aligned}$$

Hence, $\bm{x = 1}$.

L.C.M. of 2 and 3 = 6

Converting the given rational numbers into like rational numbers,

$$\begin{aligned}\frac{-1}{2} = \frac{-1 \times 3}{2 \times 3} = \frac{-3}{6}\\[1em] \frac{2}{3} = \frac{2 \times 2}{3 \times 2} = \frac{4}{6}\end{aligned}$$

The rational numbers lying between $\frac{-3}{6}$ and $\frac{4}{6}$ are:

$\frac{-2}{6}, \frac{-1}{6}, \frac{0}{6}, \frac{1}{6}, \frac{2}{6}, \frac{3}{6}$

Taking any two of them, we get 0 and $\frac{2}{6} = \frac{1}{3}$.

Hence, two rational numbers between $\mathbf{\frac{-1}{2}}$ and $\mathbf{\frac{2}{3}}$ are 0 and $\mathbf{\frac{1}{3}}$.

L.C.M. of 5 and 3 = 15

Converting the given rational numbers into like rational numbers,

$$\begin{aligned}\frac{-4}{5} = \frac{-4 \times 3}{5 \times 3} = \frac{-12}{15}\\[1em] \frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15}\end{aligned}$$

Since $\frac{-12}{15} \lt \frac{-11}{15} \lt \frac{-10}{15} \lt ..... \lt \frac{-2}{15} \lt \frac{10}{15}$,

Hence, ten rational numbers between $\mathbf{\frac{-4}{5}}$ and $\mathbf{\frac{2}{3}}$ are $\mathbf{\frac{-11}{15}, \frac{-10}{15}, \frac{-9}{15}, \ldots, \frac{-2}{15}}$.

Finding prime factors of 50653 :

$$\begin{array}{l|r} 37 & 50653 \\ \hline 37 & 1369 \\ \hline 37 & 37 \\ \hline & 1 \\ \end{array}$$

Thus,

$$\begin{aligned}\Rightarrow \sqrt[3]{50653}\\[1em] \Rightarrow \sqrt[3]{37 \times 37 \times 37}\\[1em] \Rightarrow 37\end{aligned}$$

Hence, $\mathbf{\sqrt[3]{50653} = 37}$.

Finding prime factors of 3645 :

$$\begin{array}{l|r} 3 & 3645 \\ \hline 3 & 1215 \\ \hline 3 & 405 \\ \hline 3 & 135 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$$

3645 = (3 × 3 × 3) × (3 × 3 × 3) × 5

Since 5 is left after grouping the prime factors in triplets, so 3645 should be divided by 5.

3645 ÷ 5 = 729 = (3 × 3 × 3) × (3 × 3 × 3), which is a perfect cube.

Hence, the smallest number by which 3645 should be divided to obtain a perfect cube is 5.

Solving L.H.S.,

$$\begin{aligned}\Rightarrow p \times (q + r)\\[1em] \Rightarrow \frac{-3}{5} \times \Big(\frac{1}{2} + \frac{-7}{9}\Big)\\[1em] \Rightarrow \frac{-3}{5} \times \Big(\frac{9 - 14}{18}\Big)\\[1em] \Rightarrow \frac{-3}{5} \times \frac{-5}{18}\\[1em] \Rightarrow \frac{15}{90}\\[1em] \Rightarrow \frac{1}{6}\end{aligned}$$

Solving R.H.S.,

$$\begin{aligned}\Rightarrow p \times q + p \times r\\[1em] \Rightarrow \Big(\frac{-3}{5} \times \frac{1}{2}\Big) + \Big(\frac{-3}{5} \times \frac{-7}{9}\Big)\\[1em] \Rightarrow \frac{-3}{10} + \frac{21}{45}\\[1em] \Rightarrow \frac{-3}{10} + \frac{7}{15}\\[1em] \Rightarrow \frac{-9 + 14}{30}\\[1em] \Rightarrow \frac{5}{30}\\[1em] \Rightarrow \frac{1}{6}\end{aligned}$$

Since L.H.S. = R.H.S. = $\frac{1}{6}$,

Hence, $\bm{p \times (q + r) = p \times q + p \times r}$ is verified.

Finding prime factors of 7056 :

$$\begin{array}{l|r} 2 & 7056 \\ \hline 2 & 3528 \\ \hline 2 & 1764 \\ \hline 2 & 882 \\ \hline 3 & 441 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

7056 = (2 × 2) × (2 × 2) × (3 × 3) × (7 × 7)

$$\begin{aligned}\Rightarrow \sqrt{7056}\\[1em] \Rightarrow \sqrt{(2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (7 \times 7)}\\[1em] \Rightarrow 2 \times 2 \times 3 \times 7\\[1em] \Rightarrow 84\end{aligned}$$

Hence, $\mathbf{\sqrt{7056} = 84}$.

Since the remainder is 436, so 59000 is not a perfect square.

Here, (242)² < 59000 < (243)²

So, the next perfect square after 59000 is (243)².

(243)² = 59049

∴ Required number = 59049 - 59000 = 49

Hence, the least number which must be added to 59000 to make it a perfect square is 49.