Question paper, Set 1

H.C.F of $36$ and $48$ is $12$.

$\displaystyle \frac{36}{48} = \frac{12 \times 3}{12 \times 4} = \frac{3}{4}$

Simplest form $= 3 : 4$

Answer: $3 : 4$

Sum of the terms of the ratio $= 3 + 4 = 7$

$\displaystyle \text{First part} = \frac{3}{7} \times 840 = 360$

$\displaystyle \text{Second part} = \frac{4}{7} \times 840 = 480$

Answer: ₹360 and ₹480

Let Kiran's old salary $= 3k$ and Arun's old salary $= 5k$.

$\displaystyle \frac{3k + 3000}{5k + 3000} = \frac{15}{23}$

$\Rightarrow (3k + 3000)23 = (5k + 3000)15$

$\Rightarrow 69k + 69{,}000 = 75k + 45{,}000$

$\Rightarrow 69{,}000 - 45{,}000 = 75k - 69k$

$\Rightarrow 24{,}000 = 6k \Rightarrow k = 4000$

$\text{Arun's new salary} = 5k + 3000 = 20{,}000 + 3000 = 23{,}000$

Answer: ₹23,000

Product of means $=$ product of extremes.

$21 \times x = 7 \times 36$

$\displaystyle x = \frac{7 \times 36}{21} = 12$

Answer: $x = 12$

$\text{Mean proportional} = \sqrt{12 \times 75}$

$= \sqrt{(4 \times 3) \times (3 \times 25)}$

$= \sqrt{4 \times 9 \times 25} = 2 \times 3 \times 5 = 30$

Answer: $30$

9 m → ₹585

1 m → ₹$\displaystyle \frac{585}{9} = 65$

Answer: ₹65

Days | Men

$35$ | $48$

$28$ | $y$

Inverse proportion: $x_1 y_1 = x_2 y_2$

$\displaystyle 28 \times y = 35 \times 48 \Rightarrow y = \frac{35 \times 48}{28} = 60$

Answer: $60$ men

$\displaystyle 35\% = \frac{35}{100} = \frac{7}{20}$

Answer: $\displaystyle \frac{7}{20}$

$3$ kg $= 3 \times 1000$ g $= 3000$ g

$\displaystyle \left(\frac{450}{3000} \times 100\right)\% = 15\%$

Answer: $15\%$

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

$\displaystyle = \left(\frac{96 - 80}{80} \times 100\right)\% = \left(\frac{16}{80} \times 100\right)\% = 20\%$

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

$\displaystyle = \left(\frac{16}{96} \times 100\right)\% = 16\frac{2}{3}\%$

Answer: (i) $20\%$ (ii) $\displaystyle 16\frac{2}{3}\%$

If $A$ is $R\%$ more than $B$, then $B$ is less than $A$ by $\displaystyle \left[\frac{R}{100 + R} \times 100\right]\%$

$\displaystyle = \left(\frac{25}{100 + 25} \times 100\right)\% = \left(\frac{25}{125} \times 100\right)\% = 20\%$

Answer: $20\%$

$\text{C.P} \text{ of } 150 \text{ eggs} = 150 \times 6 = 900$

$\displaystyle 12 \text{ eggs} \to 78, \quad 1 \text{ egg} \to \frac{78}{12} = 6.5$

Remaining eggs $= 150 - 6 = 144$

$\text{S.P} = 144 \times 6.5 = 936$

$\text{S.P} > \text{C.P} \Rightarrow \text{Profit} = 936 - 900 = 36$

$\displaystyle \text{Profit \%} = \left(\frac{36}{900} \times 100\right)\% = 4\%$

Answer: Profit ₹36; $4\%$

$\displaystyle \text{Discount} = 12\% \text{ of } 1250 = \frac{12}{100} \times 1250 = 150$

$\text{S.P} = \text{M.P} - \text{Discount} = 1250 - 150 = 1100$

Answer: Discount ₹150; S.P ₹1100

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

$\displaystyle \text{Loss \%} = \left(\frac{10}{10}\right)^2\% = 1\%$

Check: C.P's are $900$ and $1100$, total $2000$; S.P total $1980$; loss $20$ on $2000$.

Answer: Loss of $1\%$

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

$\displaystyle = \frac{14 + 22 + 27 + 31 + 36}{5} = \frac{130}{5} = 26$

Answer: $26$

$\displaystyle \frac{s_1 + s_2 + \cdots + s_{24}}{24} = 40 \Rightarrow s_1 + \cdots + s_{24} = 40(24) \quad (1)$

$\displaystyle \frac{s_1 + \cdots + s_{24} + T}{25} = 40 + 1 \Rightarrow s_1 + \cdots + s_{24} + T = 41(25) \quad (2)$

$(2) - (1): T = 41(25) - 40(24)$

$= 1025 - 960 = 65$

Answer: $65$ kg

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

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

Answer: $20$ m/s

For equal distances at speeds $s_1$ and $s_2$: average speed $\displaystyle = \frac{2s_1 s_2}{s_1 + s_2}$

$\displaystyle = \frac{2 \times 30 \times 20}{30 + 20} = \frac{1200}{50} = 24 \text{ km/h}$

Answer: $24$ km/h

$P = 2400$, $R = 7\%$, $T = 3$ years

$\displaystyle \text{S.I} = \frac{P \times T \times R}{100} = \frac{2400 \times 3 \times 7}{100} = 504$

Answer: ₹504

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

$\displaystyle = 8000\left(1 + \frac{5}{100}\right)^2 = 8000 \times \left(\frac{21}{20}\right)^2$

$= 8820$

$\text{C.I} = A - P = 8820 - 8000 = 820$

Answer: Amount ₹8820; C.I ₹820