Question paper, Set 2

$2$ m $= 2 \times 100$ cm $= 200$ cm

$\displaystyle \frac{45 \text{ cm}}{200 \text{ cm}} = \frac{45}{200} = \frac{9}{40}$

Ratio $= 9 : 40$

Answer: $9 : 40$

Sum of the terms of the ratio $= 2 + 3 + 4 = 9$

$\displaystyle \text{1st share} = \frac{2}{9} \times 3600 = 800$

$\displaystyle \text{2nd share} = \frac{3}{9} \times 3600 = 1200$

$\displaystyle \text{3rd share} = \frac{4}{9} \times 3600 = 1600$

Answer: ₹800, ₹1200, ₹1600

Let the fourth proportional be $x$. Then $6 : 14 :: 21 : x$.

$6 \times x = 14 \times 21$

$\displaystyle x = \frac{14 \times 21}{6} = 49$

Answer: $49$

$\text{Mean proportional} = \sqrt{4 \times 49} = \sqrt{196} = 14$

Answer: $14$

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

$\displaystyle \frac{a+b}{a-b} = \frac{7 + 3}{7 - 3} = \frac{10}{4} = \frac{5}{2}$

Answer: $\displaystyle \frac{5}{2}$

People | Rice (kg)

$25$ | $4$

$60$ | $y$

Direct proportion: $x_1 y_2 = x_2 y_1$

$\displaystyle 25 \times y = 60 \times 4 \Rightarrow y = \frac{60 \times 4}{25} = 9.6$

Answer: $9.6$ kg

Let $x$ soldiers leave. Then $900 - x$ remain.

No. of soldiers and days are in inverse proportion: $x_1 y_1 = x_2 y_2$

$900 \times 28 = (900 - x) \times 36$

$\displaystyle 900 - x = \frac{900 \times 28}{36} = 700$

$x = 900 - 700 = 200$

Answer: $200$ soldiers

$\displaystyle \frac{3}{8} = \left(\frac{3}{8} \times 100\right)\% = 37.5\%$

Answer: $37.5\%$

$\displaystyle 35\% \text{ of } 260 = \frac{35}{100} \times 260 = 91$

Answer: $91$

Final value after an increase of $12\%$ $\displaystyle = \left(\frac{100 + R}{100}\right)P$

$\displaystyle = \frac{100 + 12}{100} \times 750 = \frac{112}{100} \times 750 = 840$

Answer: ₹840

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

$\displaystyle = 25{,}000\left(1 + \frac{4}{100}\right)^2 = 25{,}000 \times \left(\frac{26}{25}\right)^2$

$= 27{,}040$

Answer: $27{,}040$

Here $\text{S.P} > \text{C.P}$, so there is a profit.

$\text{Profit} = 552 - 480 = 72$

$\displaystyle \text{Profit \%} = \left(\frac{72}{480} \times 100\right)\% = 15\%$

Answer: $15\%$ profit

Let the label price (M.P) be ₹$x$.

$\displaystyle \text{C.P} = x - 20\% \text{ of } x = \frac{80}{100}x$

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

$\displaystyle \Rightarrow 7200 = \frac{120}{100} \times \frac{80}{100}x$

$\displaystyle \Rightarrow x = \frac{7200 \times 100 \times 100}{120 \times 80} = 7500$

Answer: ₹7500

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

$\displaystyle 21x = 18y \Rightarrow \frac{y}{x} = \frac{21}{18} = \frac{7}{6}$

$y > x$, so there is a gain.

$\displaystyle \text{Gain \%} = \left(\frac{y}{x} - 1\right) \times 100 = 16\frac{2}{3}\%$

Answer: Gain of $\displaystyle 16\frac{2}{3}\%$

The numbers are $2, 3, 5, 7, 11, 13$.

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

$\displaystyle = \frac{2 + 3 + 5 + 7 + 11 + 13}{6} = \frac{41}{6} = 6\frac{5}{6}$

Answer: $\displaystyle \frac{41}{6} = 6\frac{5}{6}$

Time $\displaystyle = 4 + \frac{30}{60} = \frac{9}{2}$ hours

$\displaystyle D = S \times T = 65 \times \frac{9}{2} = 292.5 \text{ km}$

Answer: $292.5$ km

$\displaystyle s_1 = \frac{d}{T_1}$, $\displaystyle s_2 = \frac{d}{T_2} \Rightarrow \frac{s_1}{s_2} = \frac{T_2}{T_1}$

$T_1 : T_2 = s_2 : s_1 = 56 : 42 = 4 : 3$

Answer: $4 : 3$

Distance $=$ length of train $+$ length of platform $= 240 + 360 = 600$ m

$\displaystyle \text{Speed} = 54 \times \frac{5}{18} = 15 \text{ m/s}$

$\displaystyle T = \frac{D}{S} = \frac{600}{15} = 40 \text{ s}$

Answer: $40$ s

$\displaystyle P + (\text{S.I for } 4\frac{1}{2} \text{ years}) = 1470$

$P + (\text{S.I for } 3 \text{ years}) = 1380$ ... (1)

Subtracting: S.I for $\displaystyle 1\frac{1}{2}$ years $= 90$

$\displaystyle \text{S.I for } 3 \text{ years} = \frac{3}{\frac{3}{2}} \times 90 = 180$

$(1) \Rightarrow P = 1380 - 180 = 1200$

$\displaystyle 180 = \frac{1200 \times 3 \times R}{100} \Rightarrow R = \frac{180 \times 100}{1200 \times 3} = 5\%$

Answer: Sum ₹1200; rate $5\%$ per annum

$\displaystyle A = P\left(1 + \frac{R}{100k}\right)^{nk}$, with $k = 2$ (half-yearly)

$\displaystyle = 16{,}000\left(1 + \frac{10}{100 \times 2}\right)^{1 \times 2} = 16{,}000 \times \left(\frac{21}{20}\right)^2$

$= 17{,}640$

$\text{C.I} = 17{,}640 - 16{,}000 = 1640$

Answer: Amount ₹17,640; C.I ₹1640