Question paper, Set 9

H.C.F of $66$ and $121$ is $11$.

$\displaystyle \frac{66}{121} = \frac{11 \times 6}{11 \times 11} = \frac{6}{11}$

Simplest form $= 6 : 11$

Answer: $6 : 11$

Sum of the terms of the ratio $= 7 + 11 = 18$

$\displaystyle \text{First part} = \frac{7}{18} \times 3240 = 1260$

$\displaystyle \text{Second part} = \frac{11}{18} \times 3240 = 1980$

Answer: ₹1260 and ₹1980

$\displaystyle \frac{a^2}{l \times b} = \frac{3}{2}$

$\displaystyle \Rightarrow \frac{12 \times 12}{l \times 8} = \frac{3}{2}$

$\displaystyle \Rightarrow l = \frac{144 \times 2}{8 \times 3} = 12 \text{ m}$

$\displaystyle \text{Required ratio} = \frac{4a}{2(l + b)} = \frac{4 \times 12}{2(12 + 8)} = \frac{48}{40} = \frac{6}{5}$

Ratio $= 6 : 5$

Answer: $6 : 5$

$l : 24 = 7 : 4$

$\displaystyle \frac{l}{24} = \frac{7}{4}$

$\displaystyle l = \frac{7 \times 24}{4} = 42 \text{ cm}$

Answer: $42$ cm

Let the third proportional be $c$. Then $12 : 18 :: 18 : c$.

$12 \times c = 18 \times 18$

$\displaystyle c = \frac{324}{12} = 27$

Answer: $27$

7 days → ₹4200

1 day → ₹$\displaystyle \frac{4200}{7} = 600$

Answer: ₹600

$\displaystyle A \propto \frac{B}{C} \Rightarrow A = \frac{kB}{C}$ ... (1)

$\displaystyle 15 = \frac{k(10)}{4} \Rightarrow k = 6$

$\displaystyle (1) \Rightarrow A = \frac{6B}{C} = \frac{6 \times 18}{6} = 18$

Answer: $A = 18$

$\displaystyle \frac{5}{16} = \left(\frac{5}{16} \times 100\right)\% = 31.25\%$

Answer: $31.25\%$

Let the number be $x$.

$20\% \text{ of } 90 = 30\% \text{ of } x + 6$

$\displaystyle \Rightarrow 18 = \frac{30x}{100} + 6$

$\displaystyle \Rightarrow 18 - 6 = \frac{30x}{100} \Rightarrow x = \frac{12 \times 100}{30} = 40$

Answer: $40$

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

$\displaystyle = \left(\frac{63 - 56}{56} \times 100\right)\% = \left(\frac{7}{56} \times 100\right)\% = 12.5\%$

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

$\displaystyle = \left(\frac{7}{63} \times 100\right)\% = 11\frac{1}{9}\%$

Answer: (i) $12.5\%$ (ii) $\displaystyle 11\frac{1}{9}\%$

Increase in consumption $\displaystyle = \left[\frac{R}{100 - R} \times 100\right]\%$

$\displaystyle = \left(\frac{20}{80} \times 100\right)\% = 25\%$

Answer: $25\%$

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

$\displaystyle = \frac{100}{90} \times 21{,}600 = 24{,}000$

For a gain of $15\%$: $\displaystyle \text{S.P} = \frac{100 + \text{Profit\%}}{100} \times \text{C.P}$

$\displaystyle = \frac{115}{100} \times 24{,}000 = 27{,}600$

Answer: C.P ₹24,000; S.P ₹27,600

When a profit is made, $\text{C.P} = \text{M.P} - \text{Discount} - \text{Profit}$

$\text{C.P} = 1500 - 150 - 270 = 1080$

Answer: ₹1080

Gain $\displaystyle \% = \left(\frac{\text{Error}}{\text{True value} - \text{Error}} \times 100\right)\%$

Error $= 1000 - 800 = 200$ g

$\displaystyle \text{Gain \%} = \left(\frac{200}{1000 - 200} \times 100\right)\% = \left(\frac{200}{800} \times 100\right)\% = 25\%$

Answer: $25\%$

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

$\displaystyle = \frac{125 + 140 + 155 + 180 + 200}{5} = \frac{800}{5} = 160$

Answer: $160$

$x_1 + \cdots + x_{12} = 40(12) = 480 \quad (1)$

$\displaystyle \frac{x_1 + \cdots + x_{12} + x_{13}}{13} = 40 + 3 = 43$

$480 + x_{13} = 43(13) = 559$

$x_{13} = 559 - 480 = 79$

Answer: $79$ runs

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

$\displaystyle 126 \text{ km/h} = 126 \times \frac{5}{18} = 35 \text{ m/s}$

Answer: $35$ m/s

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

$\displaystyle = \frac{2 \times 50 \times 75}{50 + 75} = \frac{7500}{125} = 60 \text{ km/h}$

Answer: $60$ km/h

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

$\displaystyle R = \frac{270 \times 100}{1800 \times 3} = 5\%$

Answer: $5\%$ per annum

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

$\displaystyle = 12{,}500\left(1 + \frac{8}{100}\right)^2 = 12{,}500 \times \left(\frac{27}{25}\right)^2$

$= 14{,}580$

$\text{C.I} = A - P = 14{,}580 - 12{,}500 = 2080$

Answer: Amount ₹14,580; C.I ₹2080