Class 12th

(b):Rise of milk-level = volume of 3 spheres / Base area of cylinder

^{$=\frac{\{\frac{4}{3}\pi {(5)}^3*3\}}{\{\pi {(5)}^{2}cm\}}=20cm$}

After x hours $\frac{x}{12}$ $\frac{}{}$ of the first candle is burnt out = $\frac{x}{8}$ $\frac{}{}$ of the second is burnt out.

Remaining parts are $\left(1-\frac{x}{12}\right)and\left(1-\frac{x}{8}\right)$ $$

$Now,\left(1-\frac{x}{12}\right)=2\left(1-\frac{x}{8}\right)orx=6hrs.$

(c) :As per the problem:

Initial amount of rum = 30 litres and water = 18 litres

New ratio of rum to water = 1 : 2

Therefore, if x is the amount of water to be added, then

$\frac{30}{(18+x)}=\frac{1}{2}$

18 + x = 60

x = 42 litres

(a): $Asgiven\frac{p-3}{q}=-1\Rightarrow p+q=3$ $$ . (1)

$$ $Also\frac{p}{q+2}=\frac{1}{4}\Rightarrow 4p-q=2$  . (2)

(c):Let X & Y be their ages at the time of marriage.

X – Y = 8. (1)

(X – 8) = 2 (Y – 8). (2)

Solve to get X = 24, Y = 16.

Let the solutions added be 2, 4 and 1 litre, respectively. Then quantity of ethanol in the solution is

$2*\frac{1}{3}+4*\frac{2}{5}+1*\frac{3}{7}$

$\begin{array}{l}=\frac{2}{3}+\frac{8}{5}+\frac{3}{7}\\ =\frac{70+168+45}{105}\\ =\frac{283}{105}\end{array}$

So, the quantity of methanol

$\begin{array}{l}=7-\frac{283}{105}\\ =\frac{735-283}{105}\\ =\frac{452}{105}\end{array}$

So the ratio of ethanol to methanol in the resultant solution is 283: 452.

(a):Two digit numbers where sum is 10 are 19, 28, 37, 46, 55, 64, 73, 82 and 91. Out of the given number only 37 + 36 = 73. 37 is the original number and 73 is the number obtained by reversing the digits.

Let us assume that each receives 100 when 21.

100 = P1 (1.0625)3 …. (1)

100 = P2 (1.0625)2 …. (2)

From (1) & (2), P1 : P2 = 16 : 17

Now A's share $=Rs.8448*\frac{16}{33}=Rs.4096$ $$

B's share = Rs. 4352

Now amount of A & B at 21 =

(4096) $$ ${\left(\frac{17}{16}\right)}^3=Rs.4913$ .

(c):x * y = 0.8x (y + 4) or y = 16 y + 4 = 20.

Old price = $\frac{}{}$ = $\frac{}{}$ = Rs. 0.625/sugar

New price = $\frac{}{}$ = Rs. 0.5/sugar