Question paper, Set 8

H.C.F of $84$ and $60$ is $12$.

$\displaystyle \frac{84}{60} = \frac{12 \times 7}{12 \times 5} = \frac{7}{5}$

Simplest form $= 7 : 5$

Answer: $7 : 5$

Sum of the terms of the ratio $= 2 + 5 + 4 = 11$

$\displaystyle \text{1st share} = \frac{2}{11} \times 5280 = 960$

$\displaystyle \text{2nd share} = \frac{5}{11} \times 5280 = 2400$

$\displaystyle \text{3rd share} = \frac{4}{11} \times 5280 = 1920$

Answer: ₹960, ₹2400, ₹1920

$6 : 21 = x : 28$

$\displaystyle \frac{6}{21} = \frac{x}{28}$

$\displaystyle x = \frac{6 \times 28}{21} = 8$

Answer: $8$

$\text{Mean proportional} = \sqrt{5 \times 45} = \sqrt{225} = 15$

Answer: $15$

By dividendo, if $\displaystyle \frac{a}{b} = \frac{c}{d}$ then $\displaystyle \frac{a-b}{b} = \frac{c-d}{d}$.

$\displaystyle \frac{p-q}{q} = \frac{11 - 6}{6} = \frac{5}{6}$

Answer: $\displaystyle \frac{5}{6}$

Cost (₹) | Bananas

$90$ | $36$

$35$ | $y$

Direct proportion: $x_1 y_2 = x_2 y_1$

$\displaystyle 90 \times y = 35 \times 36 \Rightarrow y = \frac{35 \times 36}{90} = 14$

Answer: $14$ bananas

$\displaystyle x \propto \frac{1}{y} \Rightarrow x = \frac{k}{y} \Rightarrow xy = k$

$\text{(i) } k = 12 \times 15 = 180$

$\displaystyle \text{(ii) } x = \frac{k}{y} = \frac{180}{20} = 9$

$\displaystyle \text{(iii) } y = \frac{k}{x} = \frac{180}{30} = 6$

Answer: (i) $180$ (ii) $9$ (iii) $6$

$\displaystyle 17.5\% = \frac{17.5}{100} = \frac{175}{1000} = \frac{7}{40}$

Answer: $\displaystyle \frac{7}{40}$

$\displaystyle \left(\frac{18}{72} \times 100\right)\% = 25\%$

Answer: $25\%$

$\text{Decrease} = 1500 - 1320 = 180$

$\displaystyle \% \text{ of decrease} = \left(\frac{180}{1500} \times 100\right)\% = 12\%$

Answer: $12\%$

$\displaystyle A = P\left(1 - \frac{R}{100}\right)^n$ with $P = 50{,}000$, $R = 6$, $n = 2$

$\displaystyle = 50{,}000\left(1 - \frac{6}{100}\right)^2 = 50{,}000 \times \left(\frac{47}{50}\right)^2 = 44{,}180$

Answer: $44{,}180$

Let the C.P be ₹$x$.

$\text{Profit} = 565 - x, \quad \text{Loss} = x - 435$

$565 - x = x - 435 \Rightarrow 1000 = 2x \Rightarrow x = 500$

Answer: ₹500

$\displaystyle \text{C.P} = \frac{100}{100 + 20} \times 1800 = 1500$

$\displaystyle \text{S.P} = \frac{100 - 10}{100} \times 1800 = 1620$

$\text{Profit} = 1620 - 1500 = 120$

$\displaystyle \text{Profit \%} = \left(\frac{120}{1500} \times 100\right)\% = 8\%$

Answer: C.P ₹1500; S.P ₹1620; profit $8\%$

Let C.P of each $= x$ and S.P of each $= y$.

$\displaystyle 27x = 30y \Rightarrow \frac{y}{x} = \frac{27}{30} = \frac{9}{10}$

$y < x$, so there is a loss.

$\displaystyle \text{Loss \%} = \left(1 - \frac{y}{x}\right) \times 100 = 10\%$

Answer: Loss of $10\%$

The numbers are $1, 3, 5, 7, 9, 11, 13$.

Average $\displaystyle = \frac{\text{Sum of observations}}{\text{No. of observations}}$

$\displaystyle = \frac{1 + 3 + 5 + 7 + 9 + 11 + 13}{7} = \frac{49}{7} = 7$

Answer: $7$

$\displaystyle S = \frac{D}{T} = \frac{2.04 \text{ km}}{6 \text{ s}} = \frac{2.04 \times 1000}{6} \text{ m/s}$

$= 340 \text{ m/s}$

Answer: $340$ m/s

$s_1 : s_2 = T_2 : T_1$

$\displaystyle 15 : 20 = T_2 : 6 \Rightarrow T_2 = \frac{15 \times 6}{20} = 4.5 \text{ hours}$

Answer: $4.5$ hours

$\displaystyle S = \frac{D}{T} = \frac{900 \text{ m}}{6 \times 60 \text{ s}} = 2.5 \text{ m/s}$

$\displaystyle = 2.5 \times \frac{18}{5} = 9 \text{ km/h}$

Answer: $9$ km/h

$\displaystyle R = 6\frac{1}{4}\% = \frac{25}{4}\%$, $\displaystyle T = 3\frac{1}{2} = \frac{7}{2}$ years

$\displaystyle \text{S.I} = \frac{P \times T \times R}{100} = \frac{480 \times \frac{7}{2} \times \frac{25}{4}}{100}$

$= 105$

$\text{Amount} = P + \text{S.I} = 480 + 105 = 585$

Answer: S.I ₹105; amount ₹585

$\displaystyle A = P\left(1 + \frac{R_1}{100}\right)\left(1 + \frac{R_2}{100}\right)\left(1 + \frac{R_3}{100}\right)$

$\displaystyle = 50{,}000\left(1 - \frac{4}{100}\right) \left(1 + \frac{15}{100}\right) \left(1 + \frac{10}{100}\right)$

$\displaystyle = 50{,}000 \times \frac{24}{25} \times \frac{23}{20} \times \frac{11}{10}$

$= 60{,}720$

$\text{Net profit} = 60{,}720 - 50{,}000 = 10{,}720$

Answer: ₹10,720