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ISC Class 12 Mathematics Solved Question Paper 2015

Short Answer Type

21.

If z = x + iy, w = $\frac{2 - iz}{2 z - i} and |w| = 1$ , find the locus of z and illustrate it in the Argand plane.

22.

Find the volume of a parallelopiped whose eidges are reprsented by the vectors:

$\vec{a} = 2 \hat{i} - 3 \hat{j} - 4 \hat{k}, \vec{b} = \hat{i} + 2 \hat{j} - \hat{k}, and \vec{c} = 3 \hat{i} + \hat{j} + 2 \hat{k}$ .

23.

Find the equation of the plane passing through the intersection of the planes:

x + y +z + 1 = 0and 2x - 3y + 5z - 2 = 0 and the point (- 1, 2 , 1)

24.

Find the shortest distance between the lines $r_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k} + λ (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) and r_{2} = 2 \hat{i} + 4 \hat{j} + 5 \hat{k} + μ (4 \hat{i} + 6 \hat{j} + 8 \hat{k})$

25.

Mr. Nirav borrowed Rs. 50,000 from the bank for 5 years. The rate of interest is 9 % per annum compounded monthly. Find the payment he makes monthly if he pays back at the beginning of each month.

26.
A dietician wishes to mix two kinds of food X and Y in such a way that the mixture contains atleast 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:

Food Vitamin A Vitamin B Vitamin C

X 1 unit 2 units 3 units

Y 2 units 2 units 1 unit

One kg of food X costs Rs.24 and one kg of food Y costs Rs.36. Using Linear Programming, find the least cost of the total mixture which will contain the required vitamins.

Let the mixture contain x kgs of food X and y kgs of food Y. Clearly, x, y ≥ 0. Let us construct the following table from the given data:

Food	Amount	Vitamin A	Vitamin B	Vitamin C	Cost Rs
X	x	x	2x	3x	16x
Y	y	2y	2y	y	20y
Total	x + y	x + 2y	2x + 2y	3x + y	16x + 20y
Requirements		Atlest 10	Atlest 12	Atlest 8

We need to minimize z = 24x + 36y,

Subject to constraints

$x + 2 y \geq 10$

$2 x + 2 y \geq 12$

$3 x + y \geq 8$

$x \geq 0, y \geq 0$

Plotting the constraints:

Line = $x + 2 y \geq 10$ , If x = 0 $\to 2 y = 10 \to y = 5$

At (0, 0), in the inequality, we have 0 + 0 = 0 which is not $\geq$ 0, so area associated with this inequality is unbounded and away from origin.

Line x + y = 6

If $x = 0 \to y = 6 and vice versa$

Line 3x + y = 8

If x = 0 $\to y = 8 and if y = 0 \to 3 x = 8 \to x = \frac{8}{3}$

Finding the feasible region:

Since x and y are $\geq$ 0, the feasible region is in the first quadrant.

On solving equations x + 2y = 10 and x + y = 6, we get

Substituting for $x = 6 - y \to 6 - y + 2 y = 10 \to y = 4$

Therefore, x = 6 - 4 = 2

On solving equations x + 2y = 10 and 3x + y = 8, we get

Substituting for x = 10-2y $\to 3 \times (10 - 2 y) + y = 8 \to 30 - 6 y + y = 8$

$\to 5 y = 22 \to y = \frac{22}{5}$

Therefore, x = $10 - 2 \times \frac{22}{5} = \frac{50 - 44}{5} = \frac{6}{5}$

On solving equations x+y = 6 and 3x+y = 8, we get,

Substituting for y = 6 - x $\to 3 x + 6 - x = 8 \to 2 x = 2 \to x = 1$

Therefore, y = 6 -1 = 5

The values of Z at the corner points are calculated as follows:

x	y	Z(cost)
10	0	240
0	8	288
1	5	204
2	4	192

192 is a minimum cost.

27.

A bill for Rs. 7650 was drawn on 8^th March, 2013, at 7 months. It was discounted on 18^thMay, 2013 and the holder of bill received Rs. 7497. What is the rate of interest charged by bank?