Question paper, Set 6

$2$ hours $= 2 \times 60$ minutes $= 120$ minutes

$\displaystyle \frac{40 \text{ min}}{120 \text{ min}} = \frac{40}{120} = \frac{1}{3}$

Ratio $= 1 : 3$

Answer: $1 : 3$

Sum of the terms of the ratio $= 3 + 4 + 5 = 12$

$\displaystyle \text{1st share} = \frac{3}{12} \times 7200 = 1800$

$\displaystyle \text{2nd share} = \frac{4}{12} \times 7200 = 2400$

$\displaystyle \text{3rd share} = \frac{5}{12} \times 7200 = 3000$

Answer: ₹1800, ₹2400, ₹3000

Product of means $=$ product of extremes.

$36 \times x = 12 \times 15$

$\displaystyle x = \frac{12 \times 15}{36} = 5$

Answer: $x = 5$

$\text{Mean proportional} = \sqrt{25 \times 36} = \sqrt{900} = 30$

Answer: $30$

Apply componendo and dividendo to $\displaystyle \frac{a+b}{a-b} = \frac{9}{5}$:

$\displaystyle \frac{(a+b) + (a-b)}{(a+b) - (a-b)} = \frac{9 + 5}{9 - 5}$

$\displaystyle \Rightarrow \frac{2a}{2b} = \frac{14}{4} \Rightarrow \frac{a}{b} = \frac{7}{2}$

$a : b = 7 : 2$

Answer: $7 : 2$

Length (m) | Cost (₹)

$6$ | $348$

$4$ | $y$

Direct proportion: $x_1 y_2 = x_2 y_1$

$\displaystyle 6 \times y = 4 \times 348 \Rightarrow y = \frac{4 \times 348}{6} = 232$

Answer: $232$ rupees

Food left after 10 days is enough for 2400 people for $40 - 10 = 30$ days.

Now there are $2400 + 600 = 3000$ people.

People and days are in inverse proportion: $x_1 y_1 = x_2 y_2$

$2400 \times 30 = 3000 \times y$

$\displaystyle y = \frac{72{,}000}{3000} = 24 \text{ days}$

Answer: $24$ days

$\displaystyle \frac{9}{25} = \left(\frac{9}{25} \times 100\right)\% = 36\%$

Answer: $36\%$

$\displaystyle 12\% \text{ of } 450 = \frac{12}{100} \times 450 = 54$

Answer: $54$

$\text{Increase} = 736 - 640 = 96$

$\displaystyle \% \text{ of increase} = \left(\frac{96}{640} \times 100\right)\% = 15\%$

Answer: $15\%$

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

$\displaystyle = 64{,}000\left(1 + \frac{5}{100}\right)^3 = 64{,}000 \times \left(\frac{21}{20}\right)^3$

$= 74{,}088$

Answer: $74{,}088$

Here $\text{S.P} < \text{C.P}$, so there is a loss.

$\text{Loss} = 1250 - 1100 = 150$

$\displaystyle \text{Loss \%} = \left(\frac{150}{1250} \times 100\right)\% = 12\%$

Answer: $12\%$ loss

$\text{Discount} = 15\% \text{ of } 2400 = 360$

$\text{S.P} = 2400 - 360 = 2040$

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

$\displaystyle = \frac{100}{120} \times 2040 = 1700$

$\text{Profit} = 2040 - 1700 = 340$

Answer: Discount ₹360; S.P ₹2040; C.P ₹1700; profit ₹340

$\displaystyle \text{C.P} \text{ of one} = \frac{100}{12}, \quad \text{S.P} \text{ of one} = \frac{100}{10}$

$\displaystyle \text{Profit \%} = \frac{10 - \frac{25}{3}}{\frac{25}{3}} \times 100$

$\displaystyle = \frac{\frac{5}{3}}{\frac{25}{3}} \times 100 = 20\%$

Answer: $20\%$

$x_1 + x_2 + \cdots + x_n = n\bar{x}$

$\text{Sum} = 15 \times 42 = 630$

Answer: $630$

$\displaystyle T = \frac{D}{S} = \frac{150}{72} = \frac{25}{12} \text{ h}$

$= 2$ h $5$ min

$9.40 + 2$ h $5$ min $=$ 11.45 am

Answer: 11.45 am

$s_1 : s_2 = T_2 : T_1$

$\displaystyle 5 : 7 = T_2 : 14 \Rightarrow T_2 = \frac{5 \times 14}{7} = 10 \text{ hours}$

Answer: $10$ hours

$\displaystyle \text{Speed} = 72 \times \frac{5}{18} = 20 \text{ m/s}$

$\text{Distance in } 25 \text{ s} = 20 \times 25 = 500 \text{ m}$

This distance $=$ length of train $+$ length of bridge.

$\text{Bridge} = 500 - 180 = 320 \text{ m}$

Answer: $320$ m

$\displaystyle \text{Amount} = P + \frac{P \times T \times R}{100} = P\left(1 + \frac{T \times R}{100}\right)$

$\displaystyle 1298 = P\left(1 + \frac{4 \times \frac{9}{2}}{100}\right) = P\left(\frac{59}{50}\right)$

$\displaystyle P = 1298 \times \frac{50}{59} = 1100$

Answer: ₹1100

$\displaystyle A = P\left(1 + \frac{R}{100k}\right)^{nk}$, with $k = 4$ (quarterly)

$\displaystyle = 40{,}000\left(1 + \frac{8}{100 \times 4}\right)^{\frac{1}{2} \times 4} = 40{,}000 \times \left(\frac{51}{50}\right)^2$

$= 41{,}616$

$\text{C.I} = 41{,}616 - 40{,}000 = 1616$

Answer: Amount ₹41,616; C.I ₹1616