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(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.

(b):Unit digit will be 1. For the second last digit we will multiply the last digit of power and the second last digit of the number = 4 * 4 = 6.

Hence the last two digits are 61.

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

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

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

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.

(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

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 $=Rs.8448\times \frac{16}{33}=Rs.4096$ $$

B’s share = Rs. 4352

Now amount of A & B at 21 =

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

(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

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(b):A = price of one apple

B = price of one tomato

Given that:

3A + 4B = 5A + 3B

B = 2A

So, one tomato can be purchased in the same amount as that used to purchase two apples.

Therefore instead of 18 tomatoes one will be able to purchase 36 apples.

(a) :Since the winner gets 60 % of the total votes, the loser must have got 40% of the total votes. Let the total number of votes be T. Then

0.6T – 0.4T = 150

0.2T = 150 T = 750

We need to find the votes polled by the losing party, that is

40 % of 750 = 0.4 * 750 = 300 votes

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.

(c):Discount = 3%

Price = 650.

S.P = .97 * 650 = 630.5

(b):Total time required to empty it =

$\frac{1}{\left(\frac{1}{3}-\frac{1}{3.5}\right)}=21hours.$

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Let L be the vertical distance between the centers of two adjacent balls.

$L=\sqrt[]{\left[{(2r)}^2-r^2\right]}=r\sqrt[]{3}$

Also, if the box can hold n sphere then (n – 1) L + 2r

≤ 9r [ the last ball height = 2r]

$(n-1)\sqrt[]{3}r\le 7r$

$n\le \frac{7}{\sqrt[]{3}}+1$

n ≤ 5.04

Since, n is a natural number.

n = 5

So, the box can hold maximum of 5 spheres.

9B + 3 + 5G + 4 = 200  9B + 5G = 193

The positive integer solutions are (B, G) = (2, 35) (7, 25) (12, 17) (17, 8)

Only one of the (B, G) gives G > B and 9B + 3 > 100, i.e. (B, G) = (12, 17)

17 * 5 = 85