Question paper, Set 7

$4$ cm $= 4 \times 10$ mm $= 40$ mm

$\displaystyle \frac{36 \text{ mm}}{40 \text{ mm}} = \frac{36}{40} = \frac{9}{10}$

Ratio $= 9 : 10$

Answer: $9 : 10$

Sum of the terms of the ratio $= 5 + 7 = 12$

$\displaystyle \text{First part} = \frac{5}{12} \times 1980 = 825$

$\displaystyle \text{Second part} = \frac{7}{12} \times 1980 = 1155$

Answer: ₹825 and ₹1155

Let Divya's old salary $= 4k$ and Hari's old salary $= 7k$.

$\displaystyle \frac{4k + 2000}{7k + 2000} = \frac{14}{23}$

$\Rightarrow (4k + 2000)23 = (7k + 2000)14$

$\Rightarrow 92k + 46{,}000 = 98k + 28{,}000$

$\Rightarrow 46{,}000 - 28{,}000 = 98k - 92k$

$\Rightarrow 18{,}000 = 6k \Rightarrow k = 3000$

Old salaries: Divya $= 4k = 12{,}000$, Hari $= 7k = 21{,}000$

Answer: Divya: ₹12,000, Hari: ₹21,000

$\text{Product of extremes} = 5 \times 25 = 125$

$\text{Product of means} = 8 \times 15 = 120$

They are not equal, so the numbers are not in proportion. No.

Answer: No

$\text{Mean proportional} = \sqrt{27 \times 75}$

$= \sqrt{(9 \times 3) \times (3 \times 25)}$

$= \sqrt{9 \times 9 \times 25} = 3 \times 3 \times 5 = 45$

Answer: $45$

6 kg → ₹270

1 kg → ₹$\displaystyle \frac{270}{6} = 45$

Answer: ₹45

New cost of a fan $= 1200 + 240 = 1440$

Cost of a fan (₹) | Fans

$1200$ | $30$

$1440$ | $y$

Inverse proportion: $x_1 y_1 = x_2 y_2$

$\displaystyle 1440 \times y = 1200 \times 30 \Rightarrow y = \frac{1200 \times 30}{1440} = 25$

Answer: $25$ fans

$\displaystyle 7 : 8 = \frac{7}{8}$

$\displaystyle \frac{7}{8} = \left(\frac{7}{8} \times 100\right)\% = 87.5\%$

Answer: $87.5\%$

Let the number be $x$.

$\displaystyle x = \frac{25}{100}x + 54$

$\displaystyle \Rightarrow \frac{100x - 25x}{100} = 54$

$\displaystyle \Rightarrow \frac{75x}{100} = 54 \Rightarrow x = \frac{54 \times 100}{75} = 72$

Answer: $72$

$\displaystyle \text{Decrease} = 8\% \text{ of } 1250 = \frac{8}{100} \times 1250 = 100$

$\displaystyle \text{Students now} = \frac{100 - 8}{100} \times 1250 = 1150$

Answer: 100 left; 1150 now

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

$\displaystyle = \left(\frac{25}{100 - 25} \times 100\right)\% = \left(\frac{25}{75} \times 100\right)\% = 33\frac{1}{3}\%$

Answer: $\displaystyle 33\frac{1}{3}\%$

Let her buy $x$ oranges at each price, $2x$ in all.

$\displaystyle \text{C.P} \text{ of } x \text{ at } 10 \text{ for } 15 = \frac{15x}{10} = \frac{3}{2}x$

$\displaystyle \text{C.P} \text{ of } x \text{ at } 15 \text{ for } 15 = \frac{15x}{15} = x$

$\displaystyle \text{Total } \text{C.P} = \frac{3}{2}x + x = \frac{5}{2}x$

$\displaystyle \text{S.P} \text{ of } 2x = 2x \times \frac{12}{8} = 3x$

$\displaystyle \text{Gain \%} = \frac{3x - \frac{5}{2}x}{\frac{5}{2}x} \times 100 = 20\%$

Answer: Gain of $20\%$

$\text{Discount} = \text{M.P} - \text{S.P} = 800 - 680 = 120$

$\displaystyle \text{Discount \%} = \left(\frac{120}{800} \times 100\right)\% = 15\%$

Answer: $15\%$

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

Error $= 1000 - 950 = 50$ ml

$\displaystyle \text{Gain \%} = \left(\frac{50}{1000 - 50} \times 100\right)\% = \left(\frac{50}{950} \times 100\right)\% = 5\frac{5}{19}\%$

Answer: $\displaystyle 5\frac{5}{19}\%$

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

$\displaystyle = \frac{3.5 + 4.2 + 5.8 + 6.5}{4} = \frac{20}{4} = 5$

Answer: $5$

Let the average be $P$ and the new person's weight $N$. Sum before $= 10P$.

$\displaystyle \frac{10P - 58 + N}{10} = P + 1.5$

$\Rightarrow 10P - 58 + N = 10P + 15$

$\Rightarrow N = 58 + 15 = 73 \text{ kg}$

Answer: $73$ kg

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

$\displaystyle 12 \text{ m/s} = 12 \times \frac{18}{5} = 43.2 \text{ km/h}$

Answer: $43.2$ km/h

Average speed $\displaystyle = \frac{\text{Total distance}}{\text{Total time}}$

$\displaystyle = \frac{150 + 90}{3 + 2} = \frac{240}{5} = 48 \text{ km/h}$

Answer: $48$ km/h

$T = 9$ months $\displaystyle = \frac{9}{12} = \frac{3}{4}$ year

$\displaystyle \text{S.I} = \frac{P \times T \times R}{100} = \frac{840 \times \frac{3}{4} \times 5}{100} = 31.5$

Answer: ₹31.5

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

$\displaystyle = 2{,}00{,}000\left(1 + \frac{5}{100}\right) \left(1 + \frac{4}{100}\right) \left(1 + \frac{2.5}{100}\right)$

$\displaystyle = 2{,}00{,}000 \times \frac{21}{20} \times \frac{26}{25} \times \frac{41}{40}$

$= 2{,}23{,}860$

Answer: $2{,}23{,}860$