Question paper, Set 4

H.C.F of $45$ and $75$ is $15$.

$\displaystyle \frac{45}{75} = \frac{15 \times 3}{15 \times 5} = \frac{3}{5}$

Simplest form $= 3 : 5$

Answer: $3 : 5$

Sum of the terms of the ratio $= 1 + 3 + 5 = 9$

$\displaystyle \text{1st share} = \frac{1}{9} \times 4500 = 500$

$\displaystyle \text{2nd share} = \frac{3}{9} \times 4500 = 1500$

$\displaystyle \text{3rd share} = \frac{5}{9} \times 4500 = 2500$

Answer: ₹500, ₹1500, ₹2500

$\text{Product of extremes} = 4 \times 27 = 108$

$\text{Product of means} = 9 \times 12 = 108$

They are equal, so $4 : 9 :: 12 : 27$. Yes.

Answer: Yes

$\text{Mean proportional} = \sqrt{9 \times 16} = \sqrt{144} = 12$

Answer: $12$

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

$\displaystyle \frac{x+y}{y} = \frac{9 + 4}{4} = \frac{13}{4}$

Answer: $\displaystyle \frac{13}{4}$

Shadow (m) | Height (m)

$6$ | $9$

$10$ | $y$

Direct proportion: $x_1 y_2 = x_2 y_1$

$\displaystyle 6 \times y = 10 \times 9 \Rightarrow y = \frac{10 \times 9}{6} = 15$

Answer: $15$ m

Total food $= 600 \times 25 = 15{,}000$ man-days

Food used in 5 days $= 600 \times 5 = 3000$ man-days

$\text{Remaining food} = 15{,}000 - 3000 = 12{,}000$

Men it can feed for 16 days $\displaystyle = \frac{12{,}000}{16} = 750$

$\text{More men} = 750 - 600 = 150$

Answer: $150$ more men

$0.065 = (0.065 \times 100)\% = 6.5\%$

Answer: $6.5\%$

$\displaystyle \left(\frac{7}{25} \times 100\right)\% = 28\%$

Answer: $28\%$

Final value after an decrease of $6\%$ $\displaystyle = \left(\frac{100 - R}{100}\right)P$

$\displaystyle = \frac{100 - 6}{100} \times 1350 = \frac{94}{100} \times 1350 = 1269$

Answer: ₹1269

Population $n$ years ago $\displaystyle = \frac{P}{\left(1 + \frac{R}{100}\right)^n}$

$\displaystyle = \frac{13{,}310}{\left(1 + \frac{10}{100}\right)^3} = \frac{13{,}310}{\left(\frac{11}{10}\right)^3}$

$\displaystyle = 13{,}310 \times \frac{10}{11} \times \frac{10}{11} \times \frac{10}{11} = 10{,}000$

Answer: $10{,}000$

$\displaystyle \text{C.P} = \frac{100}{100 - \text{Loss\%}} \times \text{S.P}$

$\displaystyle = \frac{100}{100 - 15} \times 612 = \frac{100}{85} \times 612 = 720$

Answer: ₹720

$\text{M.P} = 600 + 30\% \text{ of } 600 = 780$

$\displaystyle \text{S.P} = \frac{100 - 10}{100} \times 780 = 702$

$\text{Profit} = 702 - 600 = 102$

$\displaystyle \text{Profit \%} = \left(\frac{102}{600} \times 100\right)\% = 17\%$

Answer: M.P ₹780; S.P ₹702; profit $17\%$

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

$\displaystyle 12x = 15y \Rightarrow \frac{y}{x} = \frac{12}{15} = \frac{4}{5}$

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

$\displaystyle \text{Loss \%} = \left(1 - \frac{y}{x}\right) \times 100 = 20\%$

Answer: Loss of $20\%$

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

$\displaystyle = \frac{48 + 52 + 61 + 39}{4} = \frac{200}{4} = 50$

Answer: $50$

Let $AB = x$. It covers $\displaystyle \frac{x}{4}$ in 3 h.

$\displaystyle S = \frac{D}{T} \Rightarrow 45 = \frac{x/4}{3}$

$\Rightarrow x = 45 \times 3 \times 4 = 540 \text{ km}$

Answer: $540$ km

$\displaystyle \frac{T_1}{T_2} = \frac{1}{3}$ ($T_1$ for the car, $T_2$ for the bus)

$s_1 : s_2 = T_2 : T_1 = 3 : 1$

$\displaystyle \text{Speed of bus} = \frac{75}{3} = 25 \text{ km/h}$

Answer: $25$ km/h

To cross a pole, the train travels its own length.

$\displaystyle S = \frac{D}{T} = \frac{150}{9} = 16\frac{2}{3} \text{ m/s}$

$\displaystyle = 16\frac{2}{3} \times \frac{18}{5} = 60 \text{ km/h}$

Answer: $60$ km/h

$\text{Amount} = P + \text{S.I} \Rightarrow 1143 = P + 243$

$P = 1143 - 243 = 900$

$\displaystyle \text{S.I} = \frac{P \times T \times R}{100} \Rightarrow 243 = \frac{900 \times T \times 6}{100}$

$\displaystyle T = \frac{243 \times 100}{900 \times 6} = 4\frac{1}{2} \text{ years}$

Answer: Principal ₹900; time $\displaystyle 4\frac{1}{2}$ years

$\displaystyle A = P\left(1 + \frac{R}{100}\right)^n$

$\displaystyle 28{,}090 = 25{,}000\left(1 + \frac{R}{100}\right)^2$

$\displaystyle \Rightarrow \left(1 + \frac{R}{100}\right)^2 = \frac{28{,}090}{25{,}000} = \frac{2809}{2500} = \left(\frac{53}{50}\right)^2$

$\displaystyle \Rightarrow 1 + \frac{R}{100} = \frac{53}{50} \Rightarrow \frac{R}{100} = \frac{3}{50}$

$\Rightarrow R = 6\%$

Answer: $6\%$ per annum