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ISI MSQE 2004-14 ME-I SAMPLE PAPER SOLUTION

ISI: Year 2004

1.(a) 2. (b) 3. (c) 4. (a) 5. (c) 6. (d) 7. (b)

8. (c) 9. (c) 10. (a) 11. (d) 12. (c) 13. (c) 14.(d)

15. (c) 16. (a) 17. (b) 18. (c) 19. (b) 20. (a) 21. (b)

22. (d) 23. (d) 24. (a) 25. (c) 26. (c) 27. (a)

28. (d) 29. (d) 30. (b)

ISI: Year 2005

1. (a) 2. (c) 3. (a) 4. (c) 5. (a) 6. (c)

7. (a) 8. (c) 9. (d) 10. (d) 11. (c) 12. (d)

ISI: Year 2006

1. (a) 2. (a) 3. (d) 4. (c) 5. (d) 6. (a) 7. (c) 8. (d)

9. (b) 10. (d) 11. (c) 12. (c) 13. (a) 14. (a) 15. (d)

ISI: Year 2007

1. (a) 2. (b) 3. (a) 4. (d) 5. (d) 6. (c) 7. (b)

8. (a) 9. (c) 10. (d) 11. (b) 12. (b) 13. (d) 14. (a)

15. (c) 16. (a) 17. (b) 18. (d) 19. (a) 20. (b) 21. (b)

22. (d) 23. (c) 24. (d) 25. (b) 26. (d) 27. (c) 28. (d)

29. (d) 30. (a)

ISI: Year 2008

1. (d) 2. (b) 3. (c) 4. (b) 5. (c) 6. (c) 7. (b)

8. (d) 9. (d) 10. (c) 11. (c) 12. (b) 13. (b) 14. (c)

15. (b) 16. (c) 17. (a) 18. (b) 19. (a) 20. (d) 21. (d)

22. (c) 23. (a) 24. (a) 25. (b) 26. (a) 27. (c) 28. (d)

29. (d) 30. (c)

ISI: Year 2009

1. (b) 2. (d) 3. (b) 4. (a) 5. (d) 6. (c) 7. (c) 8. (c)

9. (c) 10. (c) 11. (b) 12. (b) 13. (c) 14. (b) 15. (d) 16. (b)

17. (b) 18. (a) 19. (c) 20. (a) 21. (c) 22. (a) 23. (b) 24. (d)

25. (b) 26. (a) 27. (b) 28. (b) 29. (a) 30. (d)

ISI: Year 2010

1. (a) 2. (a) 3. (a) 4. (d) 5. (a) 6. (c) 7. (b) 8. (b) 9. (c) 10. (a)

11. (c) 12. (c) 13. (b) 14. (b) 15. (c) 16. (c) 17. (b) 18. (c) 19. (b) 20. (d)

21. (d) 22. (b) 23. (d) 24. (c) 25. (a) 26. (b) 27. (c) 28. (d) 29. (b) 30. (c)

ISI: Year 2011

1. (d) 2. (d) 3. (b) 4. (d) 5. (c) 6. (c) 7. (d) 8. (a) 9. (c) 10. (b)

11.(d) 12. (c) 13. (a) 14. (d) 15. (c) 16. (a) 17. (d) 18. (b) 19. (d) 20. (b)

21.(c) 22. (c) 23. (a) 24. (a) 25. (b) 26. (a) 27. (b) 28. (c) 29. (c) 30. (b)

ISI: Year 2012

1. (c) 2. (c) 3. (c) 4. (b) 5. (c) 6. (a) 7. (d) 8. (a) 9. (a) 10. (c)

11.(a) 12. (b) 13. (a) 14. (c) 15. (a) 16. (a) 17. (d) 18. (b) 19. (c) 20. (d)

21.(c) 22. (d) 23. (a) 24. (d) 25. (c) 26. (a) 27. (d) 28. (b) 29. (b) 30. (a)

ISI: Year 2013

1. (d) 2. (c) 3. (a) 4. (b) 5. (a) 6. (c) 7. (a) 8. (c) 9. (a) 10. (a)

11.(c) 12. (b) 13. (d) 14. (d) 15. (d) 16. (c) 17. (c) 18. (a) 19. (c) 20. (d)

21.(c) 22. (d) 23. (a) 24. (c) 25. (c) 26. (a) 27. (b) 28. (d) 29. (b) 30. (d)

ISI: Year 2014

1. (d) 2. (b) 3. (b) 4. (c) 5. (b) 6. (a) 7. (a) 8. (a) 9. (c) 10. (b)

11.(c) 12. (a) 13. (d) 14. (a) 15. (b) 16. (a) 17. (b) 18. (c) 19. (a) 20. (a)

21.(b) 22. (b) 23. (a) 24. (b) 25. (c) 26. (b) 27. (d) 28. (d) 29. (d) 30. (b)

Solution to Sample Questions for ME I (Mathematics), 2013

1. (d) (

)

( (

) )

2. ( )

( )

( )( )

.

3. (a) By AM GM inequality , (

) , where i = 1,2,3,...,n.

Multiplying all these we have (1+a1)(1+a2)....(1+an) , since a1a2...an=1 .

4. (b) P[X=0]=P[X=1] , where XBin(n,p) , 0 Number of Head] = 2

43 , where Number of Head>Number of Tail = Number of Tail >

Number of Head. So, answer is

10. (a) f (x,y) = m when y = 0. f(x+k,y) = f(x,y)=m when y=0

f (x,y+k)= f(x,y)+kx = f(x,0) + kx= m+kx , when y=0

So, f (x,k) = m+kx , f (x,y)=m+xy .

10. (b) ( ) (

) (

) . Solve these three equations to find n .

11. (d) (a+b+c)2 0 a2+b2+c2 + 2(ab+bc+ca) 0 (ab+bc+ca) -

.

12. (d) f(x) is not differentiable at x = 4 and 5 .

16. (c)A1 = { 2,4,6,....} , A2= {3,6,9,....}, so A1A2 ={6,12,18,...}

={6k , kN} = A5.

17. (c) *

( )+ *

(

)+

(

)

.

18.(a) For Binomial distribution E(K+1)= ( )( )

=

+1 , where p =

. So, the value of the given sum is = n2n-1

+ 2n.

19. (c) Perform these elementary row operations to the given matrix to reduce it into row-

reduced matrix form : (i) R3' = R3 R2 + R4 ,

(ii) R2 is interchanged by R4,

(iii) R4' = R4 - R3 ,

Then R4 of the given matrix will vanish. So, rank is 3.

20. (d) The two integers with product is maximum is of the form (

)

( ).

21. (c) S = {a2+a

4+a

6+.......}+{ab+a

2b

2+a

3b

3+.....} =

.

22. (d) The necessary condition to exist maxima and minima at any point x=a is that f '(a)=0 .

But here is no real a for which f '(x)=0 . [Since x2 4x + 8 =0 doesnt have any real

solution]

25. (c) On integration, we have log(f(x))= x + c f(x) =kex f(x)= kex + m.

Given f(0)=e2

, f(1)=e3 gives k = e

2 .

27. (b) Compute A2, then put the value of A2, A, I in the given form to get the answer.

28. (d) The number of permutations is =

29. (b) Mean deviation about mean cant exceed the standard deviation. To prove this

statement use Cauchy-Schwarz inequality & choose ai = xi - and bi = 1.

Solution to Sample Questions for ME II (Economics) 2013

Q.1. Agents utility maximization problem is the following :

s.t. ( ) ( )

&

(a) Solving the above problem we get :

( ) (

( )

)

Hence , saving = w -

.

(b) Clearly, Savings doesnt change in rate of interest rate r .

Q.2. (a) If the price of a MD is Rs. 20 and the marginal cost is Rs. 15 per MD , vendors profit

maximization problem is the following :

s.t.

Thus, each vendor would want to sell 100 MD a day .

(b)Given competitive behavior, free entry-exit from the industry and constant returns to scale

technology , we have zero profit condition.

That is , price equals marginal cost. Thus, demand is

d(15) = 4400 120(15) = 2600

Since each vendor sells 100 units and demand is 2600 units , there are 26 vendors selling MD in

the market .

(c) If number of vendors are 20 and each vendor produces 100 units , price is given by

p =

(d) The maximum price that a vendor is willing to pay for the permit is equal to the profit

that a vendor gets if he operates i.e. 5100=500.

Q.3. The production possibility frontier of the two inputs is given by

Since final product can be sold at the end of the day at a per unit price of Rs. 1.

The firms profit maximization problem is :

s.t.

&

Thus, firm will hire the worker, produces

.

Q.4. Since monopolists sale to the government is positive, his marginal revenue at the point of

sale in the private market must be Rs. 100 . Now price in the private market is Rs. 150. We can

compute the price elasticity of demand in the following way :

TR(x) = p(x) . x

Differentiating TR(x) w.r.t. x , we get ,

MR(x) = p(x) + x ( )

= p(x) + ( )

( )

( )

= p(x)(

)

Now substituting p(x) = 150 and MR(x) = 100 in the above we get elasticity, = - 3 .

Q.5. (a) Let f (K,L) denotes the production function.

( ) ( ) ( )

( )

( ) .

Thus, production function satisfies constant returns to scale.

(b) Profit maximization problem of the competitive producer is

s.t. L 0 , K 0

The above problem is equivalent to

( ) ( )

s.t. L 0 , K 0

Clearly, when 0 w , demand function for labor is not defined.

(c) Also, the demand function for capital is defined when price of capital service is zero

provided w >

Q.7. E (Planned Expenditure) = C* + c(Y T) (I

* bi) + G

* + NX(other than investment) . Let

NX = Export Import = Export mY , where m = marginal propensity to import.

E = (C* + G

* I

* cT + Export) + cY mY ........(For the sake of simplicity export component is

assumed to be autonomous)

In equilibrium E = Y

Here Y( 1 (c m) ) = Autonomous component of expenditure

( ( ))

Assuming interest rate to be constant so when I* is positive output will decrease by (

( )) , where = Change in imports.

So, the given statement is TRUE.

Q.8. Given that income elasticity of demand for all goods are positive, i.e. (

) (

) . This

implies that (

) Hence, goods are normal.