Question paper, Set 3

$2$ kg $= 2 \times 1000$ g $= 2000$ g

$\displaystyle \frac{750 \text{ g}}{2000 \text{ g}} = \frac{750}{2000} = \frac{3}{8}$

Ratio $= 3 : 8$

Answer: $3 : 8$

Sum of the terms of the ratio $= 4 + 5 = 9$

$\displaystyle \text{First part} = \frac{4}{9} \times 1350 = 600$

$\displaystyle \text{Second part} = \frac{5}{9} \times 1350 = 750$

Answer: ₹600 and ₹750

Let the numbers be $2k$, $3k$, $4k$.

$(2k)^2 + (3k)^2 + (4k)^2 = 725$

$\Rightarrow 4k^2 + 9k^2 + 16k^2 = 725$

$\displaystyle \Rightarrow 29k^2 = 725 \Rightarrow k^2 = \frac{725}{29} = 25$

$\Rightarrow k = 5$

Numbers: $2 \times 5 = 10$, $3 \times 5 = 15$, $4 \times 5 = 20$

$\text{Sum} = 9k = 9 \times 5 = 45$

Answer: $10, 15, 20$; sum $= 45$

$8 : 20 = x : 45$

$\displaystyle \frac{8}{20} = \frac{x}{45}$

$\displaystyle x = \frac{8 \times 45}{20} = 18$

Answer: $18$

Let the third proportional be $c$. Then $9 : 15 :: 15 : c$.

$9 \times c = 15 \times 15$

$\displaystyle c = \frac{225}{9} = 25$

Answer: $25$

15 pens → ₹210

1 pen → ₹$\displaystyle \frac{210}{15} = 14$

Answer: ₹14

Days | Hours per day

$10$ | $6$

$12$ | $y$

Inverse proportion: $x_1 y_1 = x_2 y_2$

$\displaystyle 12 \times y = 10 \times 6 \Rightarrow y = \frac{10 \times 6}{12} = 5$

Answer: $5$ hours

$\displaystyle 3 : 5 = \frac{3}{5}$

$\displaystyle \frac{3}{5} = \left(\frac{3}{5} \times 100\right)\% = 60\%$

Answer: $60\%$

Let the number be $x$.

$45\% \text{ of } x = 60\% \text{ of } x - 21$

$\displaystyle \Rightarrow \frac{45x}{100} = \frac{60x}{100} - 21$

$\displaystyle \Rightarrow 21 = \frac{15x}{100} \Rightarrow x = \frac{21 \times 100}{15} = 140$

Answer: $140$

$\text{Increase} = 63{,}000 - 56{,}000 = 7000$

$\displaystyle \% \text{ of increase} = \left(\frac{7000}{56{,}000} \times 100\right)\% = 12.5\%$

Answer: Increase ₹7000; $12.5\%$

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

$\displaystyle = \left(\frac{10}{110} \times 100\right)\% = 9\frac{1}{11}\%$

Answer: $\displaystyle 9\frac{1}{11}\%$

$\text{C.P} = 150 \times 2400 + 15{,}000 = 3{,}75{,}000$

$\text{S.P} = 27 \times 100 \times 150 = 4{,}05{,}000$

$\text{S.P} > \text{C.P}, \text{ Profit} = 4{,}05{,}000 - 3{,}75{,}000 = 30{,}000$

$\displaystyle \text{Profit \%} = \left(\frac{30{,}000}{3{,}75{,}000} \times 100\right)\% = 8\%$

Answer: Profit ₹30,000; $8\%$

$\text{M.P} = 540 + 25\% \text{ of } 540$

$\displaystyle = 540 + \frac{25}{100} \times 540 = 540 + 135 = 675$

Answer: ₹675

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

Error $= 1000 - 960 = 40$ g

$\displaystyle \text{Gain \%} = \left(\frac{40}{1000 - 40} \times 100\right)\% = \left(\frac{40}{960} \times 100\right)\% = 4\frac{1}{6}\%$

Answer: $\displaystyle 4\frac{1}{6}\%$

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

$\text{Sum} = 12 \times 18.5 = 222$

Answer: $222$

$x_1 + \cdots + x_{5} = 24(5) = 120$, and the youngest is $8$.

The other 4 members: $x_1 + \cdots + x_{4} = 120 - 8 = 112$

8 years ago each of them was 8 years younger.

$\displaystyle \text{Required average} = \frac{112 - 8(4)}{4} = \frac{112 - 32}{4} = \frac{80}{4} = 20$

Answer: $20$ years

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

$\displaystyle 25 \text{ m/s} = 25 \times \frac{18}{5} = 90 \text{ km/h}$

Answer: $90$ km/h

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

$\displaystyle = \frac{2 \times 90 \times 60}{90 + 60} = \frac{10{,}800}{150} = 72 \text{ km/h}$

Answer: $72$ km/h

$P = 1500$, $R = 8\%$, $\displaystyle T = 2\frac{1}{2} = \frac{5}{2}$ years

$\displaystyle \text{S.I} = \frac{P \times T \times R}{100} = \frac{1500 \times \frac{5}{2} \times 8}{100} = 300$

$\text{Amount} = P + \text{S.I} = 1500 + 300 = 1800$

Answer: ₹1800

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

$\displaystyle = 15{,}625\left(1 + \frac{4}{100}\right)^3 = 15{,}625 \times \left(\frac{26}{25}\right)^3$

$= 17{,}576$

$\text{C.I} = A - P = 17{,}576 - 15{,}625 = 1951$

Answer: Amount ₹17,576; C.I ₹1951