Question

$\large\underbrace{\bf{ \: \: \: \: \: \: \: \: \:Question\: \: \: \: \: \: \: \: \: }}$

Solve all the questions​

Answer 1

$\large\underline{\sf{Solution-6}}$

Given that, Cost of 9 toys is Rs 333.

Let assume that cost of 16 toys be Rs x.

Now, more toys, so more cost. So, it means, number of toys and cost are in direct variation.

So,

$\begin{array}{|c|c|c|c} \hline \rm Number \: of \: toys&\rm 9&\rm 16 \\ \hline\rm Cost \: of \: toy \: (in \: Rs)&\rm 333&\rm x \\ \hline \end{array}$

So, using Law of direct variation, we have

$\rm \: \dfrac{9}{333} = \dfrac{16}{x}$

$\rm \: \dfrac{1}{27} = \dfrac{16}{x}$

$\bf\implies \:x \: = \: Rs \: 432$

$\large\underline{\sf{Solution-7}}$

Given that, length of 85.5 kg of rod is 22.5 m

Let assume that length of rod whose weight is 22.8 kg is x m.

Now, less weight, so less length. It means, Length of the rod and weight of the rod is in direct variation.

So,

$\begin{array}{|c|c|c|c} \hline \rm Weight \: of \: rod \: (in \: kg)&\rm 85.5&\rm 22.8 \\ \hline\rm Length \: of \: rod \: (in \: m)&\rm 22.5&\rm x \\ \hline \end{array}$

Now, By using law of direct variation, we have

$\rm \: \dfrac{85.5}{22.5} = \dfrac{22.8}{x}$

$\rm \: x = \dfrac{22.8 \times 22.5}{85.5}$

$\bf\implies \:x = 6 \: m$

$\large\underline{\sf{Solution-8}}$

Given table is

$\begin{array}{|c|c|c|c|} \hline \rm \: \: \: x \: \: \: &\rm 9&\rm a&\rm 81 \rm \\ \hline\rm y&\rm 27&\rm 9&\rm b \rm \\ \hline \end{array}$

Since, it is given that x and y varies inversely.

$\rm \: 9 \times 27 = 9 \times a = 81 \times b$

Taking first and second member, we get

$\rm \: 9 \times 27 = 9 \times a$

$\bf\implies \:a = 27$

On taking first and third member, we get

$\rm \: 9 \times 27 = 81 \times b$

$\bf\implies \:b = 3$

So, complete table is

$\begin{array}{|c|c|c|c|} \hline \rm \: \: \: x \: \: \: &\rm 9&\rm 27&\rm 81 \rm \\ \hline\rm y&\rm 27&\rm 9&\rm 3 \rm \\ \hline \end{array}$

$\large\underline{\sf{Solution-9}}$

Given that, weight of 12 m of rod is 42 kg.

Let assume that weight of rod whose length is 6 m is x kg.

Now, less length, so less weight. It means, Length of the rod and weigh of the rod is in direct variation.

So,

$\begin{array}{|c|c|c|c} \hline \rm Weight \: of \: rod \: (in \: kg)&\rm 42&\rm x \\ \hline\rm Length \: of \: rod \: (in \: m)&\rm 12&\rm 6 \\ \hline \end{array}$

So, By using law of direct variation, we have

$\rm \: \dfrac{42}{12} = \dfrac{x}{6}$

$\rm \: \dfrac{7}{2} = \dfrac{x}{6}$

$\bf\implies \:x \: = \: 21 \: kg$

$\large\underline{\sf{Solution-10}}$

Given that, cost of 15 oranges is Rs 70.

Let assume that cost of 39 oranges be Rs x.

So,

$\begin{array}{|c|c|c|c} \hline \rm Number \: of \: oranges&\rm 15&\rm 39 \\ \hline\rm Cost \: of \: oranges \: (in \: Rs)&\rm 70&\rm x \\ \hline \end{array}$

So, by using Law of direct variation,

$\rm \: \dfrac{15}{70} = \dfrac{39}{x}$

$\rm \: \dfrac{3}{14} = \dfrac{39}{x}$

$\rm \: \dfrac{1}{14} = \dfrac{13}{x}$

$\bf\implies \:x \: = \: Rs \: 182$

Answer 2

$\huge\red{A}\pink{N}\orange{S} \green{W}\blue{E}\gray{R} =$

Solution−6

Given that, Cost of 9 toys is Rs 333.

Let assume that cost of 16 toys be Rs x.

Now, more toys, so more cost. So, it means, number of toys and cost are in direct variation.

So,

$\begin{gathered}\begin{array}{|c|c|c|c} \hline \rm Number \: of \: toys&\rm 9&\rm 16 \\ \hline\rm Cost \: of \: toy \: (in \: Rs)&\rm 333&\rm x \\ \hline \end{array} \\\end{gathered} NumberoftoysCostoftoy(inRs)933316x$

So, using Law of direct variation, we have

$\begin{gathered}\rm \: \dfrac{9}{333} = \dfrac{16}{x} \\ \end{gathered} 3339 = x16$

$\begin{gathered}\rm \: \dfrac{1}{27} = \dfrac{16}{x} \\ \end{gathered} </p><p>271= x16$

$\begin{gathered}\bf\implies \:x \: = \: Rs \: 432 \\ \end{gathered} ⟹x=Rs432$

$\large\underline{\sf{Solution-7}} Solution−7$

Given that, length of 85.5 kg of rod is 22.5 m

Let assume that length of rod whose weight is 22.8 kg is x m.

Now, less weight, so less length. It means, Length of the rod and weight of the rod is in direct variation.

So,

$Weightofrod(inkg)</p><p>Lengthofrod(inm)</p><p></p><p> </p><p>85.5</p><p>22.5</p><p></p><p> </p><p>22.8</p><p>x$

Now, By using law of direct variation, we have

$\begin{gathered}\rm \: \dfrac{85.5}{22.5} = \dfrac{22.8}{x} \\ \end{gathered} 22.585.5= x22.8$

$\begin{gathered}\rm \: x = \dfrac{22.8 \times 22.5}{85.5} \\ \end{gathered} x= 85.522.8×22.5$

$\begin{gathered}\bf\implies \:x = 6 \: m \\ \end{gathered} ⟹x=6m</p><p>$

$\large\underline{\sf{Solution-9}}Solution−9$

Given that, weight of 12 m of rod is 42 kg.

Let assume that weight of rod whose length is 6 m is x kg.

Now, less length, so less weight. It means, Length of the rod and weigh of the rod is in direct variation.

So,

$\begin{gathered}\begin{array}{|c|c|c|c} \hline \rm Weight \: of \: rod \: (in \: kg)&\rm 42&\rm x \\ \hline\rm Length \: of \: rod \: (in \: m)&\rm 12&\rm 6 \\ \hline \end{array} \\\end{gathered} Weightofrod(inkg)Lengthofrod(inm)4212x6$

So, By using law of direct variation, we have

$\begin{gathered}\rm \: \dfrac{42}{12} = \dfrac{x}{6} \\ \end{gathered} 1242= 6x$

$\begin{gathered}\rm \: \dfrac{7}{2} = \dfrac{x}{6} \\ \end{gathered} 27= 6x$

$\begin{gathered}\bf\implies \:x \: = \: 21 \: kg\\ \end{gathered} ⟹x=21kg$

$\large\underline{\sf{Solution-10}} Solution−10$

Given that, cost of 15 oranges is Rs 70.

Let assume that cost of 39 oranges be Rs x.

So,

$\begin{gathered}\begin{array}{|c|c|c|c} \hline \rm Number \: of \: oranges&\rm 15&\rm 39 \\ \hline\rm Cost \: of \: oranges \: (in \: Rs)&\rm 70&\rm x \\ \hline \end{array} \\\end{gathered} NumberoforangesCostoforanges(inRs)157039x$

So, by using Law of direct variation,

$\begin{gathered}\rm \: \dfrac{15}{70} = \dfrac{39}{x} \\ \end{gathered} 7015= x39$

$\begin{gathered}\rm \: \dfrac{3}{14} = \dfrac{39}{x} \\ \end{gathered} 143= x39$

$\begin{gathered}\rm \: \dfrac{1}{14} = \dfrac{13}{x} \\ \end{gathered} 141 = x13$

$\begin{gathered}\bf\implies \:x \: = \: Rs \: 182 \\ \end{gathered} ⟹x=Rs182$