Given: A local cricket team played $20$ matches in one season. It won $25$ % of them. To do: To find the number of matches did they win.Solution:Number of matches played by cricket team $=20$Percentage of won matches $=25$%Total matches won by them $=25$ % of $20$$=\frac{25}{100}\times20$$=5$Hence, they won $5$ matches. 

Solution:(i) Five rational numbers between -1 and 0We can write -1, 0 as $\frac{-6}{6}$, $\frac{0}{6}$So 5 rational numbers between $\frac{ -6}{6}$ and $\frac{0}{6}$ are$\frac{-5}{6}$, $\frac{-4}{6}$, $\frac{-3}{6}$, $\frac{-2}{6}$, $\frac{-1}{6}$(ii) Five rational numbers between  -2 and -1We can write -2, -1 as $\frac{-12}{6}$, $\frac{-6}{6}$So 5 rational numbers between $\frac{-12}{6}$ and $\frac{-6}{6}$ are$\frac{-11}{6}$, $\frac{-10}{6}$, ... Read More

Given: (i) $\frac{-3}{5}, \ \frac{-6}{10}, \ \frac{-9}{15}, \ \frac{-12}{20}$........(ii) $\frac{-1}{4}, \ \frac{-2}{8}, \ \frac{-3}{12}$..... (iii) $\frac{-1}{6}, \ \frac{2}{-12}, \ \frac{3}{-18}, \ \frac{4}{-24}$...... (iv) $\frac{-2}{3}, \ \frac{2}{-3}, \ \frac{4}{-6}, \ \frac{6}{-9}$.....To do: To write four more rational numbers in each of the given patterns.Solution:(i) $\frac{-3}{5}, \ \frac{-6}{10}, \ \frac{-9}{15}, \ \frac{-12}{20}$........​This given pattern of ... Read More

Given: Rational numbers:$(i)$. $\frac{-2}{7}$$(ii)$. $\frac{5}{-3}$$(iii)$. $\frac{4}{9}$To do: To write the rational numbers equivalent to the above given rational numbers.Solution:$(i).\ -\frac{2}{7}$Multiplying the given rational number by the same numerator and denominator, the product obtained will be equal to the given rational number.Now we will multiply the given rational number by $\frac{2}{2}, \ ... Read More

Given: Rational numbers:$(i)$. $\frac{3}{4}$$(ii)$. $\frac{-5}{8}$$(iii)$. $\frac{-7}{4}$$(iv)$. $\frac{7}{8}$To do: To draw the number line and represent the given rational numbers on it.Solution: Here given rational numbers are represented on the number lines as below:$(i)$. Representation of $\frac{3}{4}$ on the number line:Let us draw a number line. As we know that $\frac{3}{4}$ ... Read More

Given :Meena pays an interest of Rs. 45 for one year at the rate of 9% p.a.To find :We have to find the sum borrowed.Solution :Let the sum she borrowed be Rs. PTherefore, $Simple\ Interest = \frac{PTR}{100}$$45 = \frac{P\times1\times 9}{100}$$9P = 45\times100$$P = \frac{45\times100}{9}$$P = 5\times100$$P = 500$Therefore, Meena borrowed ... Read More

Given: $P =Rs.\ 56000$, $SI =Rs.\ 280$, $t =2 years$.To do: To find the rate $(R)$.Solution:We know that,$R=\frac{(100\times SI)}{(P\times T)}$$=\frac{(100\times 280)}{(56000\times 2)}$$=\frac{(1\times 28)}{(56\times 2)}$$=\frac{(1\times 14)}{(56\times 1)}$$=\frac{(1\times 1)}{(4\times 1)}$$=\frac{(1)}{(4)}$$=0.25$ %

$(a)$ Here, Principal $(P)=Rs.\ 1, 200$Rate $(R)=12 \% p.a.$,  Time $(T)=3\ years$Simple Interest$=\frac{P\times R\times T}{100}$$=\frac{1200\times12\times3}{100}$$= Rs.\ 432$Now, $Amount=Principal+Simple Interest$$=1200+432$$=Rs.\ 1, 632$$(b)$ Here, Principal $(P)=Rs.\ 7, 500$,  Rate $(R)=5 \% p.a.$,  Time $(T)=3\ years$$Simple Interest=\frac{P\times R\times T}{100}=\frac{7500\times5\times3}{100}$$=Rs.\ 1125$Now, $Amount=Principal+Simple\ Interest$$=7, 500+1, 125$$=Rs.\ 8625$Read More

Given: Amina buys a book for $Rs.\ 275$ and sells it at a loss of $15 \%$. To do: To find that how much she sells it for.Solution:The cost of a book$=Rs.\ 275$Loss percent$=15$ %Loss$=$Loss % of C.P.$=15$ % of $Rs.\ 275$$= \frac{15}{100}\times275$$=Rs.\ 41.25$Therefore, $S.P.=C.P.-Loss$$=275-41.25=Rs.\ 233.75$Hence, Amina sells a book for $Rs.\ ... Read More

Solution:$(i)$. Given ratio$=10:3:12$Total part$=10+3+12=25$Part of Carbon$=\frac{3}{25}$Percentage of Carbon part in chalk$=\frac{3}{25}\times100=12$ % $(ii)$. Quantity of Carbon in chalk stick$=3\ g$Let the weight of chalk be $x\ g$.Then, $12$ % of $x=3$$\Rightarrow\ \frac{12}{100}\times x=3$$\Rightarrow x=\frac{3\times100}{12}=25\ g$Hence the weight of the chalk stick is $25\ g$.Read More