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2/3/2011 Subjective Test Paper - Math - Meritnati… Math 2004 Set 1 Close Subjective Test (i) The question paper consists of three sections A, B and C Section A is compulsory for all students. In addition to Section A, every student has to attempt either Section B OR Section C. (ii) For Section A Question numbers 1 to 8 are of 3 marks each. Question numbers 9 to 15 are of 4 marks each. Question numbers 16 to 18 are of 6 marks each. (iii) For Section B/Section C Question numbers 19 to 22 are of 3 marks each. Question numbers 23 to 25 are of 4 marks each. Question number 26 is of 6 marks. (iv) All questions are compulsory. (v) Internal choices have been provided in some questions. You have to attempt only one of the choices in such questions. (vi) Use of calculator is not permitted. However, you may ask for logarithmic and statistical tables, if required. Question 1 ( 3.0 marks) If , prove that , n ∈ N. Solution: It is given that We will prove the required result by the principal of mathematical induction. For n = 1, we have Thus, the result is true for n = 1. Now let the result be true for n = k, where k ∈ N. Thus, … (1) Now, we will prove the result for n = k + 1. That is, we have to prove that meritnation.com/…/zWWUBora1Ru4ms… 1/28 2/3/2011 Subjective Test Paper - Math - Meritnati… Thus, Therefore, the result is true for n = k + 1 also. Hence, by principal of mathematical induction, the result is true for all n ∈ N. That is, ,n∈N Question 2 ( 3.0 marks) Using the properties of determinants, prove that Solution: We have: By applying C 1 → C 1 − C 2 and C 2 → C 2 − C 3 , we obtain Expanding the determinant along R1 , we obtain Thus, Question 3 ( 3.0 marks) From a bag containing 20 tickets, numbered from 1 to 20, two tickets are drawn at random. Find the probability that (i) Both the tickets have prime numbers on them (ii) On one there is prime number and on the other there is a multiple of 4 Solution: It is given that from a bag containing 20 tickets numbered from 1 to 20, two tickets are drawn at random. Out of 20 tickets, two tickets can be drawn in 20 C ways. meritnation.com/…/zWWUBora1Ru4ms… 2/28 2/3/2011 Subjective Test Paper - Math - Meritnati… Out of 20 tickets, two tickets can be drawn in 20 C 2 ways. Therefore, total number of elementary events = 20 C 2 = 190 (i) The prime numbers amongst the numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19 Out of these 8 prime numbers, two numbers can be selected in 8 C 2 ways. Thus, number of outcomes favourable to the event “both the tickets have prime numbers on them” = 8C 2 = 28 Therefore, required probability = (ii) The numbers which are multiples of 4 from amongst the numbers from 1 to 20 are 4, 8, 12, 16, 20 Out of these 5 numbers, one can be selected in 5 C 1 ways. Out of 8 prime numbers from 1 to 20, one prime number can be selected in 8 C 1 ways. Thus, the number of favourable outcomes for the event “on one there is prime number and on the other there is a multiple of 4” = 5 C 1 × 8 C 1 = 5 × 8 = 40 Therefore, required probability = Question 4 ( 3.0 marks) Two dice are tossed once. Find the probability of getting an even number on the first die or a total of 8. Or From a lot of 30 bulbs, which includes 6 defective bulbs, a sample of 3 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs. Solution: Let S be the sample space associated with the experiment of throwing two dice. Then, n(S) = 36 ∴Total number of elementary events = 36 Let A and B be two events given by: A = Getting an even number on the first die B = Getting a total of 8 Then, A ∩ B = Getting an even number on the first die and a total of 8 And A ∪ B = Getting an even number on the first die or a total of 8 We have: A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} B = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)} A ∩ B = {(2, 6), (4, 4), (6, 2)} ∴P(A) = , P(B) = , and P(A ∩ B) = We know that, P(A ∪ B) = P(A) + P(B) − P(A ∩ B) Thus, probability of getting an even number on the first die or a total of 8 when two dice are thrown is . Or It is given that from a lot of 30 bulbs, which includes 6 defective bulbs, a sample of 3 bulbs is drawn at random with replacement. Hence, number of non-defective bulbs in the lot = 30 − 6 = 24 meritnation.com/…/zWWUBora1Ru4ms… 3/28 2/3/2011 Subjective Test Paper - Math - Meritnati… Let X denote the number of defective bulbs in the sample. Clearly, X can assume the values 0, 1, 2, 3 such that P(X = 0) = Probability of getting no defective bulbs P(X = 1) = Probability of getting one defective bulb P(X = 2) = Probability of getting 2 defective bulbs P(X = 3) = Probability of getting 3 defective bulbs Hence, the probability distribution of X is as follows: X 0 1 2 3 P(X) Question 5 ( 3.0 marks) Evaluate: Solution: The given integral is Put x − a = t. Then, dx = dt. Therefore, we have Question 6 ( 3.0 marks) Evaluate: Solution: meritnation.com/…/zWWUBora1Ru4ms… 4/28 2/3/2011 Subjective Test Paper - Math - Meritnati… The given integral is Put ex = t. Then, ex dx = dt. Therefore, Question 7 ( 3.0 marks) Solve the differential equation: Solution: The given differential equation is We can observe that the functions y + x and y − x are homogeneous functions of degree 1, so the given differential equation is homogeneous. Let y = vx Differentiating both sides of this relationship with respect to x, we obtain Putting these values in equation (1), we obtain Integrating both sides, we obtain meritnation.com/…/zWWUBora1Ru4ms… 5/28 2/3/2011 Subjective Test Paper - Math - Meritnati… This is the required solution of the given differential equation. Question 8 ( 3.0 marks) Solve the differential equation: , given that , when x = 0. Solution: The given differential equation is: Integrating both sides with respect to x, we obtain Again, integrating both sides of the above relationship with respect to x, we obtain Now, it is given that , when x = 0. Therefore, from equation (1), we obtain , when x = 0 From equation (2), we obtain Putting the values of C1 and C2 in equation (2), we obtain meritnation.com/…/zWWUBora1Ru4ms… 6/28 2/3/2011 Subjective Test Paper - Math - Meritnati… This is the required solution of the given differential equation. Question 9 ( 4.0 marks) For each x in a Boolean Algebra B, prove that Solution: We have: x′ = x′+ 0 = x′ + x.y ( it is given that x.y = 0) = (x′ + x).(x′ + y) (By distributivity of ‘+’ over ‘.’) = 1. (x′ + y) (Since x + x′ = x′ + x = 1) = (x + y). (x′ + y) ( it is given that x + y = 1) = (y + x). (y + x’) (By commutativity of ‘+’) = y + (x. x′) (By distributivity of ‘+’ over ‘.’) = y + 0 (Since x. x′ = 0) =y Therefore, Question 10 ( 4.0 marks) Evaluate: Or Evaluate: Solution: We have: meritnation.com/…/zWWUBora1Ru4ms… 7/28 2/3/2011 Subjective Test Paper - Math - Meritnati… Or We have: Question 11 ( 4.0 marks) Differentiate with respect to x: Solution: Let Differentiating with respect to x, we obtain meritnation.com/…/zWWUBora1Ru4ms… 8/28 2/3/2011 Subjective Test Paper - Math - Meritnati… Question 12 ( 4.0 marks) Differentiate with respect to x from first principles. Solution: Let . Then, Question 13 ( 4.0 marks) The volume of a spherical balloon is increasing at the rate of 25 cm3 /sec. Find the rate of change of its surface area at the instant when its radius is 5 cm. Solution: Let r cm, V cm3 , and S cm2 be the radius, volume, and surface area of the spherical balloon. Then, , where r is a function of time t It is given that: meritnation.com/…/zWWUBora1Ru4ms… 9/28 2/3/2011 Subjective Test Paper - Math - Meritnati… When r = 5, Thus, the rate of change of surface area of the spherical balloon at the instant when its radius is 5 cm is 10 cm2 /sec. Question 14 ( 4.0 marks) Evaluate: Solution: The given integral is Put x2 = y This gives 2x dx = dy Therefore, we have … (1) Let , where A and B are integers ⇒ 1 = A(3 + y) + B(1 + y) … (2) Putting y = −1 in the above equation, we obtain 1 = A(3 − 1) + B(1 − 1) ⇒ Putting y = −3 in equation (2), we obtain 1 = A(3 − 3) + B(1 − 3) ⇒ Therefore, we have meritnation.com/…/zWWUBora1Ru4ms… 10/28 2/3/2011 Subjective Test Paper - Math - Meritnati… Now, from equation (1), we obtain Question 15 ( 4.0 marks) Prove that: Or Prove that: Solution: Adding equations (1) and (2), we obtain Or The given integral is: meritnation.com/…/zWWUBora1Ru4ms… 11/28 2/3/2011 Subjective Test Paper - Math - Meritnati… Question 16 ( 6.0 marks) Using matrix method, solve the following system of linear equations: x+y−z=1 3x + y − 2z = 3 x − y − z = −1 Or If , prove that A2 − 4A − 5I = O. Hence, find A−1 . Solution: The given system of equations can be written in the form AX = B, where Thus, A is non-singular and so its inverse exists. meritnation.com/…/zWWUBora1Ru4ms… 12/28 2/3/2011 Subjective Test Paper - Math - Meritnati… Hence, x = 2, y = 1, and z = 2 is the solution of the given system of equations. Or It is given that We have to prove that A2 − 4A − 5I = O Hence, A2 − 4A − 5I = O Now, multiplying both sides of the above equation by A−1 , we obtain A − 4I − 5A−1 = O (Since AA1 = 1 and I A−1 = A−1 ) ⇒ 5A−1 = A − 4I meritnation.com/…/zWWUBora1Ru4ms… 13/28 2/3/2011 Subjective Test Paper - Math - Meritnati… Question 17 ( 6.0 marks) Show that the maximum volume of the cylinder which can be inscribed in a sphere of radius cm is 500π cm3 . Solution: Let r cm and h cm be the radius and height of the cylinder respectively. From the given figure, height of cylinder, Let V cm3 be the volume of the sphere. If r = 0, then V = 0. This is not possible since maximum volume of the cylinder is a positive quantity. Therefore, we will take r2 = 50 Now, for r2 = 50, meritnation.com/…/zWWUBora1Ru4ms… 14/28 2/3/2011 Subjective Test Paper - Math - Meritnati… Therefore, the volume is maximum when Now, height of cylinder = h cm = Thus, required volume of the cylinder = (πr2 h) cm3 = (π × 50 × 10) cm2 = 500π cm3 Question 18 ( 6.0 marks) Find the area bounded by the circle x2 + y2 = 16 and the line y = x in the first quadrant. Solution: Equation of the circle is x2 + y2 = 16 and the line y = x. Solving these two equations, we obtain x = y = Hence, the line intersects the circle at the point in the first quadrant. The area enclosed between the circle x2 + y2 = 16 and the line y = x in the first quadrant is as shown by the shaded area OABCO. ar (OABCO) = ar (OACO) + ar (ABCA) Thus, the required area bounded the given circle and the given line in the first quadrant is 2π square units. Section B Question 19 ( 3.0 marks) If the vectors are coplanar, then find the value of λ. meritnation.com/…/zWWUBora1Ru4ms… 15/28 2/3/2011 Subjective Test Paper - Math - Meritnati… Solution: Three vectors are said to be coplanar if one of the given vectors is expressible as a linear combination of the other two. Hence, the vectors will be coplanar, if From equation (2), it can be observed that the value of λ can be found just by finding the values of x and y. For this, equations (1) and (3) have to be solved. On solving equations (1) and (3), we obtain x= and y = Putting these values in equation (2), we obtain λ= Thus, the value of λ is 2. Question 20 ( 3.0 marks) Prove that: Or The volume of the parallelopiped whose edges are is 546 cubic units. Find the value of λ. Solution: We have: Therefore, Or We know that if the vectors represent the coterminous edges of a parallelopiped, then the volume (V) of the parallelopiped is given by the scalar triple product of i.e., The vectors associated with the coterminous edges of parallelopiped are and . Therefore, its volume is given by: meritnation.com/…/zWWUBora1Ru4ms… 16/28 2/3/2011 Subjective Test Paper - Math - Meritnati… . Therefore, its volume is given by: It is given that volume of the parallelopiped is 546 cubic units. Hence, the values of λ are −87 or 95. Question 21 ( 3.0 marks) The Cartesian equations of a line are 3x + 1 = 6y − 2 = 1 − z. Find the fixed point through which it passes, its direction ratios, and also its vector equation. Solution: We know that the general equation of a line passing through the point (x1 , y1 , z1 ) with directional ratios proportional to a, b, c is given by, The given Cartesian equations of the line are 3x + 1 = 6y − 2 = 1 − z. This can be written as This shows that the given line passes through and has direction ratios proportional to i.e., 2, 1 −6. In vector form, this means that the line passes through the point having position vector and is parallel to the vector . Therefore, its vector equation is given by: Question 22 ( 3.0 marks) Find the equation of the plane passing through the points (0, −1, 0), (1, 1, 1) and (3, 3, 0). Solution: It is known that the equation of the plane passing through three points (x1 , y1 , z1 ), (x2 , y2 , z2 ), and (x3 , y3 , z3 ) is given by the formula: meritnation.com/…/zWWUBora1Ru4ms… 17/28 2/3/2011 Subjective Test Paper - Math - Meritnati… Since the plane passes through the points (0, −1, 0), (1, 1, 1), and (3, 3, 0), This is the equation of the required plane. Question 23 ( 4.0 marks) Prove that the plane x + 2y − z = 4 intersects the sphere x2 + y2 + z2 − x + z − 2 = 0 in a circle of radius unity. Also, find the centre of the circle. Solution: Let C be the centre of the sphere and let M be the foot of perpendicular from C on the plane. Let P be any point on the circle. Then, CP = Radius of the sphere The given equation of the sphere is x2 + y2 + z2 − x + z − 2 = 0 We know that the coordinate of the centre and radius of a sphere are at and respectively. Comparing the equation x2 + y2 + z2 − x + z − 2 = 0 with , we obtain Hence, the coordinate of the centre (C) and radius (CP) of the sphere x2 + y2 + z2 − x + z − 2 = 0 are and respectively. Equation of the normal line CM to the given plane x + 2y − z = 4 through centre C of the sphere is Any point on line (1) is of the form . If this point lies on the given plane x + 2y − z = 4, then meritnation.com/…/zWWUBora1Ru4ms… 18/28 2/3/2011 Subjective Test Paper - Math - Meritnati… Therefore, coordinates of the centre of the circle = Now, CM = Length of the perpendicular from C on the plane x + 2y − z = 4 Thus, by Pythagoras Theorem, we obtain Hence, the given plane x + 2y − z = 4 intersects the sphere x2 + y2 + z2 − x + z − 2 = 0 in a circle of radius unity. Question 24 ( 4.0 marks) Three forces and act along OA, OB and OC, where O is the orthocenter of ∆ABC. If the forces be in equilibrium, prove that P: Q: R = a: b: c. Solution: Using angle sum property for quadrilateral OFAE, we obtain ∠FAE + ∠OFA + ∠EOF + ∠OEA = 360° ⇒ ∠FAE + 90° + ∠EOF + 90° = 360° (Since O is the orthocentre, OF ⊥ AB and OE ⊥ AC) ⇒ ∠EOF = 180° − ∠FAE = 180° − ∠A Now, ∠BOC = ∠EOF (Vertically opposite angles) ⇒ ∠BOC = 180° − ∠A Similarly, ∠COA = ∠DOF = 180° − ∠B ∠AOB = ∠DOE = 180° − ∠C Since the forces P, Q, and R are in equilibrium, applying Lami’s theorem, we obtain Also, by sine formula, meritnation.com/…/zWWUBora1Ru4ms… 19/28 2/3/2011 Subjective Test Paper - Math - Meritnati… Thus, from relationships (1) and (2), we obtain Thus, P: Q: R = a: b: c Question 25 ( 4.0 marks) The resultant of two like parallel forces and acting at A and B, 8 cm apart, is 40 N. If the resultant passes through C, where AC = 3 cm, find the magnitude of the forces. Solution: Let the two forces be of magnitudes P and Q acting at points A and B respectively. Let the resultant force act at a point C on AB. Then, AB = 8 cm and P + Q = 40 N (Given) Now, we have However, P + Q = 40 N ⇒ P = (40 − 15) N = 25 N Thus, the required magnitudes of the tw o forces are 25 N and 15 N. Question 26 ( 6.0 marks) A particle moving in a straight line with uniform acceleration describes successive equal distance in times t 1 , t 2 and t 3 . Prove that . Or A stone is projected at an angle α with the horizontal. Given its velocity when it attains half of the maximum height (it can attain) is times of the velocity at the maximum height. Prove that Solution: Let u be the initial velocity and a be the uniform acceleration. Let the successive equal distances be AB = BC = CD = s (say) Clearly, the particle describes distance AB = s in time t 1 , distance AC = 2 s in time t 1 + t 2 , and distance AD = 3 s in time t1 + t2 + t3 . Therefore, by using , we obtain AB = … (1) meritnation.com/…/zWWUBora1Ru4ms… 20/28 2/3/2011 Subjective Test Paper - Math - Meritnati… AC = … (2) AD = … (3) Subtracting equation (1) from equation (2), we obtain Subtracting equation (2) from (3), we obtain Also, from (1), we obtain … (6) Or Let P and Q be the points on the path of projectile such that AP is the maximum height attained by the projectile and BQ = AP. Let the velocities at the points P and Q be v1 and v2 respectively. We know that if a projectile is projected at velocity u with angle of projection α, then velocity (v) of a projectile at any height y is given by the formula: We also know that maximum height (H) attained by the projectile is Therefore, for point P, we have meritnation.com/…/zWWUBora1Ru4ms… 21/28 2/3/2011 Subjective Test Paper - Math - Meritnati… For point Q, we have It is given that: Section C Question 27 ( 3.0 marks) The true discount and the banker’s gain on a certain bill of exchange due after a certain period of time are respectively Rs 700 and Rs 17.50. Find the face of the bill. Out of current syllabus Solution: Let the face value of the bill be Rs S and let the bill be due after n years. Let the rate of interest be r per annum. Then, True discount = and Banker’s gain = Now, it is given that: True discount = Rs 700 and Banker’s gain = Rs 17.50 and Putting , we obtain meritnation.com/…/zWWUBora1Ru4ms… 22/28 2/3/2011 Subjective Test Paper - Math - Meritnati… Hence, the face value of the bill is Rs 28,700. Question 28 ( 3.0 marks) A bill of Rs 1,000 drawn on May 7, 2003 for six months was discounted on August 29, 2003 for a cash payment of Rs 988. Find the rate of interest charged by the bank. Out of current syllabus Solution: Date of drawing the bill is May 7, 2003. Term of the bill = 6 months ∴Due date of the bill is November 10, 2003. Date of discounting the bill is August 29, 2003. ∴Number of days from the date of discounting (i.e., August 29, 2003) to due date (i.e., November 10, 2003) = 2 + 30 + 31 + 10 = 73 (For Aug) (For Sept) (For Oct) (For Nov) Thus, we have S = 1000, Let the interest be r per rupee per annum. Then, Banker’s discount = Srt = = 200r … (1) It is given that the bill was discounted for a cash payment of Rs 988. ∴Banker’s discount = Rs (1000 − 988) = Rs 12 … (2) From (1) and (2), we obtain 200r = 12 Hence, rate of interest charged by the bank = 100r% = Question 29 ( 3.0 marks) A company has two plants to manufacture T.V.s. The first plant manufactures 70% of the T.V.s. and the rest are manufactured by the other plant. 80% of the T.V.s. manufactured by the first plant is rated of standard quality while that of the second plant, only 70% are of standard quality. If a T.V. chosen at random is found to be of standard quality, find the probability that it was produced by the first plant. Solution: Let E 1 be the event of choosing plant I, E 2 be the event of choosing plant II, and A be the event of choosing a T.V., which is of standard quality. Then, we have P(E 1 ) = P(E 2 ) = P(A/ E 1 ) = P(choosing a T.V. of standard quality from plant I) = P(A/ E 2 ) = P(choosing a T.V. of standard quality from plant II) = meritnation.com/…/zWWUBora1Ru4ms… 23/28 2/3/2011 Subjective Test Paper - Math - Meritnati… Now, probability of choosing a T.V. from plant I, given that it is of standard quality, is P(E 1 /A). Using Bayes’ theorem, we have P(E 1 /A) = Question 30 ( 3.0 marks) A pair of dice is thrown 7 times. If getting the total 7 is considered a success, find the probability of (i) no success (ii) at least 6 successes. Or If the probability that an individual suffers from reaction by an injection is , find the probability that out of 5000 individuals given that injection (i) exactly 5 will suffer from reaction (ii) no one will suffer from reaction. Given that Out of current syllabus Solution: Let p denote the probability of getting a total of 7 in a single throw of a pair of dice. Then, [Since the sum can be 7 in any of the ways: (1, 6), (6, 1), (2, 5), (5, 2), (3, 4) and (4, 3)] Let X denote the number of successes in 7 throws of a pair of dice. Then X is a binomial variate with parameters n = 7 and . Therefore, we have i. Probability of no successes = ii. Probability of at least 6 successes = P(X ≥ 6) = P(X = 6) + P(X = 7) Or meritnation.com/…/zWWUBora1Ru4ms… 24/28 2/3/2011 Subjective Test Paper - Math - Meritnati… Or Let p be the probability that an individual suffers from reaction by the injection and n be the total number of individuals who took the injection. It is given that n = 5000 and Since p is very small and n is large, we shall use Poisson’s distribution. We have m = np ⇒ m = 5000 × 0.001 = 5 Let X denote the number of individuals who suffer from reaction by the injection. Then, X is a Poisson variate such that i. The probability that exactly 5 individuals will suffer from a reaction is given by: ii. The probability that no individual will suffer from a reaction is given by: Question 31 ( 4.0 marks) X and Y entered into a joint business with their capitals in the ratio 3 : 2. At the end of 3 months, X took out one-third of his capital, but after another 3 months Y put in a sum equal to what X had taken out. If at the end of the year Y gets Rs 11,000 more than w hat X got as profit, find (i) the total profit in the business (ii) the profit share of X in the business Out of current syllabus Solution: It is given that X and Y entered into a joint business with their capitals in the ratio 3 : 2. Therefore, let the capitals of X and Y be Rs 3x and Rs 2x respectively. Now, Effective capital of X = Rs 3x for 3 months + Rs 2x for 9 months = (Rs 3x × 3 + Rs 2x × 9) for one month = Rs 27x for one month ∴X’s average investment = Rs Effective capital of Y = Rs 2x for 6 months + Rs 3x for 6 months = (Rs 2x × 6 + Rs 3x × 6) for one month = Rs 30x for one month ∴Y’s average investment = Rs It is given that the profit is divided in the ratio of average investments. ∴Profits of X and Y are in the ratio = 9 : 10 Let the profits of X and Y be of Rs 9p and 10p respectively. Since Y’s profit is Rs 11000 more than X’s profit, 10p − 9p = 11000 ⇒ p = 11000 Hence, X’s profit = Rs (9 × 11000) = Rs 99000 Y’s profit = Rs (10 × 11000) = Rs 110000 meritnation.com/…/zWWUBora1Ru4ms… 25/28 2/3/2011 Subjective Test Paper - Math - Meritnati… Y’s profit = Rs (10 × 11000) = Rs 110000 Thus, total profit in the business = Rs (99000 + 110000) = Rs 209000 Question 32 ( 4.0 marks) A television set is available for Rs 20, 000 cash or Rs 5,000 as cash down payment followed by 6 equal annual instalments, the first to be paid one year after the date of purchase. If the rate of interest under the instalment plan is 10% per annum, determine the amount of instalment. [Given that (1.1)−6 = 0.5644] Out of current syllabus Solution: We have: Value of the television set = Rs 20,000 Cash down payment = Rs 5,000 ∴Balance = Rs 20,000 − Rs 5,000 = Rs 15,000 This balance is to be paid in 6 equal instalments. Let Rs R be the amount of each instalment. Then, the sum of Rs 15,000 is the present value of an annuity R for 6 years at 10% per annum. Thus, we have P = Rs 15,000, n = 6, and i = Hence, the required amount of each instalment is Rs 3443.5. Question 33 ( 4.0 marks) If the cost function of an article manufactured by a company is given by C(x) = , where x stands for the output. Find the output at which: (i) The marginal cost is minimum (ii) The average cost is minimum Out of current syllabus Solution: We have: C(x) = ⇒ MC = 300 − 20x + x2 meritnation.com/…/zWWUBora1Ru4ms… 26/28 2/3/2011 Subjective Test Paper - Math - Meritnati… For, x = 10, we have >0 Thus, MC is minimum when x = 10 Hence, MC is minimum when the output is of 10 units. Now we will calculate the average cost (AC). Now, for x = 15, w e have >0 Thus, AC is minimum when the output is 15 units. Question 34 ( 6.0 marks) A diet for a sick person must contain at least 4000 units of vitamins, 50 units of minerals and 1400 calories. Two foods X and Y are available at a cost of Rs 4 and Rs 3 per unit respectively. One unit of the food X contains 200 units of vitamins, 1 unit of minerals and 40 calories, whereas one unit of food Y contains 100 units of vitamins, 2 units of minerals and 40 calories. Find what combination of X and Y should be used to have least cost, satisfying the requirements. Solution: Let x units of food X and y units of food Y are used. Since food X costs Rs 4 per unit and food Y costs Rs 3 per unit, the total cost of x units of food X and y units of food Y is Rs (4x + 3y). Let Z denote the total cost. Then, Z = 4x + 3y Since each unit of food X contains 200 units of vitamins, x units of food X will contain 200x units of vitamins. Also, each unit of food Y contains 100 units of vitamins. Therefore, y units of food Y will contain 100y units of vitamins. However, the minimum requirement of vitamins is 4000 units. Therefore, we have 200x + 100y ≥ 4000 Or, 2x + y ≥ 40 Similarly, the total units of minerals in x units of food X and y units of food Y is x + 2y. However, the minimum requirement of minerals is 50 units. Therefore, we have x + 2y ≥ 50 Finally, the total calories in x units of food X and y units of food Y is 40x + 40y. However, the minimum requirement of calories is 1400. Therefore, we have 40x + 40y ≥ 1400 i.e., x + y ≥ 35 Clearly, x ≥ 0, y ≥ 0 Thus, the mathematical formulation of the problem is Minimize Z = 4x + 3y subject to 2x + y ≥ 40 x + 2y ≥ 50 x + y ≥ 35 x ≥ 0, y ≥ 0 The feasible region determined by the system of constraints is as follows: meritnation.com/…/zWWUBora1Ru4ms… 27/28 2/3/2011 Subjective Test Paper - Math - Meritnati… The coordinates of the vertices (corner points) of shaded feasible region are A (50, 0), B (20, 15), C (5, 30), and D (0, 40). The values of Z at these corner points are as follows: Corner points Z = 4x + 3y A (50, 0) 200 B (20, 15) 125 C (5, 30) 110 → Minimum D (0, 40) 120 It is observed that the feasible region is unbounded. It is clear that if we draw the inequality 4x + 3y < 110, there are no points in common with the feasible region. Therefore, Z is minimum at x = 5 and y = 30. Thus, the cost will be minimum (w hich is Rs 110) when 5 units of food X and 30 units of food Y are used. meritnation.com/…/zWWUBora1Ru4ms… 28/28

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PREVIOUS YEAR PAPERS CBSE BOARD EXAM AIEEE BITSAT ISAT VITEEE IIT-JEE STUDY MATERIAL PHYSICS CLASS XI XII SAMPLE PAPERS KEY SOLUTIONS ANSWERS QUESTIONS

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