Class 12th

(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

(d):The three different Maths books can be arranged in 3! ways. Now there are four places created for the Chemistry books such that they are not together.

The four Chemistry books can be arranged in the available four places in 4! ways.

In the first group, one question can be selected or can be rejected; so three questions can be dealt with in 2 * 2 * 2 ways, but this includes the case when all three questions have been left; so they can be selected in 23 – 1 = 7 ways. Similarly four questions of the second group can be selected in 24 – 1 = 15 ways. Thus all seven questions can be selected in 15 * 7 = 105 ways; but this includes the case when all questions have been solved; hence leaving that case, total number of ways required is 105 – 1 = 104

Every number between 1000 and 2000, which is divisible by five and which can be formed by the given digits, must contains 5 in unit's place and 1 in thousand's place. Thus we are left with four digits out of which we are to place two between 1 and 5, which can be done in 4P2 = 12 ways. Hence, 12 numbers can be formed.