Question paper, Set 5

H.C.F of $56$ and $72$ is $8$.

$\displaystyle \frac{56}{72} = \frac{8 \times 7}{8 \times 9} = \frac{7}{9}$

Simplest form $= 7 : 9$

Answer: $7 : 9$

Sum of the terms of the ratio $= 6 + 7 = 13$

$\displaystyle \text{First part} = \frac{6}{13} \times 2275 = 1050$

$\displaystyle \text{Second part} = \frac{7}{13} \times 2275 = 1225$

Answer: ₹1050 and ₹1225

Let 1 g of metal A be mixed with $x$ g of metal B; the alloy is $(1 + x)$ g.

$20w + x(6w) = (1 + x)16w$

$\Rightarrow 20w - 16w = x(16w) - x(6w)$

$\displaystyle \Rightarrow 4w = x(10w) \Rightarrow x = \frac{4}{10} = \frac{2}{5}$

$\displaystyle \text{Required ratio} = 1 : x = 1 : \frac{2}{5} = 5 : 2$

Answer: $5 : 2$

Let the fourth proportional be $x$. Then $4 : 10 :: 18 : x$.

$4 \times x = 10 \times 18$

$\displaystyle x = \frac{10 \times 18}{4} = 45$

Answer: $45$

$\displaystyle \frac{25}{x} = \frac{x}{64}$

$x^2 = 25 \times 64 = 1600$

$x = \sqrt{1600} = 40$

Answer: $x = 40$

8 litres → 120 km

1 litre → $\displaystyle \frac{120}{8} = 15$ km

Answer: $15$ km

Remaining pumps $= 15 - 6 = 9$

Pumps | Time (min)

$15$ | $36$

$9$ | $y$

Inverse proportion: $x_1 y_1 = x_2 y_2$

$\displaystyle 9 \times y = 15 \times 36 \Rightarrow y = \frac{15 \times 36}{9} = 60$

Answer: $60$ minutes

$\displaystyle 64\% = \frac{64}{100} = \frac{16}{25}$

Answer: $\displaystyle \frac{16}{25}$

$1$ day $= 24$ hours

$\displaystyle \left(\frac{9}{24} \times 100\right)\% = 37.5\%$

Answer: $37.5\%$

(i) $\displaystyle \left[\frac{\text{Greater} - \text{smaller}}{\text{smaller}} \times 100\right]\%$

$\displaystyle = \left(\frac{75 - 60}{60} \times 100\right)\% = \left(\frac{15}{60} \times 100\right)\% = 25\%$

(ii) $\displaystyle \left[\frac{\text{Greater} - \text{smaller}}{\text{Greater}} \times 100\right]\%$

$\displaystyle = \left(\frac{15}{75} \times 100\right)\% = 20\%$

Answer: (i) $25\%$ (ii) $20\%$

Let $q = x$; then $p = 5x$.

$\displaystyle \text{Required \%} = \left(\frac{5x - x}{5x} \times 100\right)\% = \left(\frac{4}{5} \times 100\right)\% = 80\%$

Answer: $80\%$

$\text{C.P} \text{ of } 400 \text{ quintals} = 400 \times 3500 = 14{,}00{,}000$

$\displaystyle \text{Profit} = 6\% \text{ of } 14{,}00{,}000 = \frac{6}{100} \times 14{,}00{,}000 = 84{,}000$

$\displaystyle \text{S.P} = \frac{100 + 6}{100} \times 14{,}00{,}000 = 14{,}84{,}000$

Answer: Profit ₹84,000; S.P ₹14,84,000

Let the C.P be ₹$x$.

$x + 15\% \text{ of } x = 690$

$\displaystyle \Rightarrow \frac{100x + 15x}{100} = 690$

$\displaystyle \Rightarrow x = \frac{690 \times 100}{115} = 600$

Answer: ₹600

When two articles have the same S.P, one at $a\%$ profit and the other at $a\%$ loss, there is always a loss of $\displaystyle \left(\frac{a}{10}\right)^2\%$.

$\displaystyle \text{Loss \%} = \left(\frac{15}{10}\right)^2\% = 2.25\%$

Check: C.P's are $3400$ and $4600$, total $8000$; S.P total $7820$; loss $180$ on $8000$.

Answer: Loss of $2.25\%$

The numbers are $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$.

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

$\displaystyle = \frac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10}{10} = \frac{55}{10} = 5.5$

Answer: $5.5$

Let the average of the first 16 innings be $P$.

$\displaystyle \frac{16P + 85}{17} = P + 3$

$\Rightarrow 16P + 85 = 17P + 51$

$\Rightarrow P = 85 - 51 = 34$

$\text{New average} = P + 3 = 37$

Answer: $37$

$1$ km/h $\displaystyle = \frac{5}{18}$ m/s

$\displaystyle 162 \text{ km/h} = 162 \times \frac{5}{18} = 45 \text{ m/s}$

Answer: $45$ m/s

For three equal distances: average speed $\displaystyle = \frac{3s_1 s_2 s_3}{s_1 s_2 + s_2 s_3 + s_3 s_1}$

$\displaystyle = \frac{3 \times 10 \times 20 \times 60}{10 \times 20 + 20 \times 60 + 60 \times 10}$

$\displaystyle = \frac{36{,}000}{2000} = 18 \text{ km/h}$

Answer: $18$ km/h

$P = 1600$, $\displaystyle R = 4\frac{1}{2}\%$, $T = 2$ years

$\displaystyle \text{S.I} = \frac{P \times T \times R}{100} = \frac{1600 \times 2 \times \frac{9}{2}}{100} = 144$

Answer: ₹144

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

$\displaystyle = 50{,}000\left(1 + \frac{20}{100}\right) \left(1 - \frac{10}{100}\right) \left(1 + \frac{5}{100}\right)$

$\displaystyle = 50{,}000 \times \frac{6}{5} \times \frac{9}{10} \times \frac{21}{20}$

$= 56{,}700$

Answer: $56{,}700$