This post contains solutions of Indian Statistical Institute, ISI 2015 BStat – BMath Objective Problems. Try to solve them.

This is a work in progress. Candidates, please submit objective problems in the comments section (even if you partially remember them). If you do not remember the options, that is fine too. We can work with the questions.

(these problems and solutions are suggested by students; they will cross-checked soon)

- For positive x, find the maximum value of .
- ANS: 10

- Find the integer solutions of
- ANS:

- ; Find the number of solutions to this equation.
- ANS: 4

- ABCD…I is a 9 digit number (all digits distinct). A+B+C = 9, D, E, F, G are consecutive odd numbers in decreasing order. Last three-digit are three even numbers in decreasing order. Find A.
- ANS: 8

- Consider functions which map from set A to set B. Set A has 10 elements. Set B has two elements, one even and one odd. Suppose , if a is mapped to b then we write f(a) = b. Suppose the function f has the property = even. How many such functions are there?
- , f(t) > 1 for all t, then
**g(x) = c has a unique solution for all x,**- g(x) = c has no solution for some c
- g(x) has a minima or maxima at x = 0

- A=(0, 8) and B= (12, 0) be two points. From any point P sum of the distances from the given points . How many lattice points P= (x, y) , x>0, y> 0 (both x, y are integer) are there such that d_p is minimized.
- 2n + 1 integers are there. Sum of any n numbers is always less than the rest n+1 numbers.
- all integers must be equal;
- all are positive;
- all integers are negative;
- such a set does not exist;

- Find the angle between the tangents to the hyperbolae and xy=1 at their points of intersection.
- Ans:

- There are two circles which are represented by the equations and . Points P and Q lie on circle 1 and circle 2 respectively. Point P lies above x-axis and point Q lies below x-axis. PQ is joined such that it is a common tangent to both the circles. Find the length of PQ.
- ANS:

- The lengths of two sides of a triangle are 2 cm and 3 cm. Find the maximum possible area of the triangle
- ANS: 3

- A set is called multiplicatively closed
*if the product of any two elements of the set is also an element of the set. Consider the following two sets:*S1 = {a + ib : a, b are integers}

S2 = {a + wb : a, b are integers; w is a non real cube root of unity}*Then:**(1)***Both S1 and S2 are multiplicatively closed.**

(2) Neither is closed.

(3) Only S1 is closed.

(4) Only S2 is closed. - The rate of increase of a population is directly proportional to the current population. If it takes 50 years for the population to double, find the time it takes to triple.(1) 50 log (3, base 2)

(2) 50 log (3/2) - A function y=f(x) is defined in (0, 2) U (4, 6) such that dy/dx = 1

Then:

(1) f(x) = x

(2) f(x) is always increasing

(3) f(5.5) – f(4.5) = f(1.5) – f(0.5) - How many points P of integral coordinates are there on Y axis which are there such that triangles AOP and BOP are both right angled and such that A is (10,0) and B is (7,0) and O is origin ?A) infinite

B) 2

C) 4 - How many surjective functions are there from A to B such that A has 6 distinct elements and B has 3 elements.A)52

B)120

C)200

D) - What is the least area of the triangle bounded by a tangent to the ellipse , X axis and Y axis ?A) ab

B) - How many complex numbers z are there such that (z conjugate= )A) 2

B) 4

C)6

D)8 - How many factors are there which are either perfect squares, perfect cubes or both of A)252

B)

C)

D) - How many triples of prime numbers are there such that a+b+c=30A) 2

B)

8) Most probably- does not exist. I have a doubt about the question. Is it ‘Sum of any n numbers is always less than the rest n+1 numbers’ or ‘ Sum of any n numbers is always less than the sum of rest n+1 numbers’?

Its the former one, all numbers are equal.

Not Necessarily. Take the case of five equal negative integers.

Thanks for pointing out the obvo, Krishna. My poor sentence formation above might imply something else, but actually, I dropped this question.

All are positive Take the case of 3,3,3,4,4. any two numbers’ sum is less than that of the remaining

Thanks for your thoughts, I agree with you. Tbh, I feel that all are positive would be the ‘perfectest’ solution.

It is option (a).if all numbers are equal then it satisfy the question.

7) distance is minimized when, the three points are collinear. The equation of the line is 2x+3y=24, for x,y>0, only solutions are (3,6), (6,4) and (9,2). Though the question is not clear!

5) It is very easy. Just simple counting. 2^9

Debjit, do you remember any of the objective problems. If you do, then please let us know

Actually, I am in class 12 now. I will give the exam next year. I am just solving the problems.

Oh, ok good

4) If we consider this number- 1089753642 then it is satisfying all the conditions. There are more such examples.

Q16 should be 540. There are 3^6 total functions out of which 3*2^6-3 are not onto whence number of onto functions is 540. Was 540 a possible option there?

2) Most probably no solutions over integers

R.H.S.= 3(mod 4)

X=0,1,2,3(mod 4)

=>X^2=0,1(mod 4)

so, R.H.S.=0,1,2(mod 4)

so, no case is matching, hence no integer solution exists.

Sorry for the Typo. Here is the correct one-

R.H.S.= 3(mod 4)

Now,

X=0,1,2,3(mod 4)

=>X^2=0,1(mod 4)

so, L.H.S.=0,1,2(mod 4)

so, no case is matching, hence no integer solution exists.

In problem number 4, I’m pretty sure that the question implied that that the rest 3 were descending evens, not consecutive descending evens. And I believe it was a nine-digit number?

Yeah…It should be….otherwise there will be a lot of counter examples!

Yes, and I had obtained the solution as 6.

nope.it was a ten digit number.and i got 8

I remembered another problem, find the angle between the tangents to the hyperbolae x^2 – y^2 =1 and xy=1 at their points of intersection. Answer is of course pi/2

After reading this question only once and seeing the answer, I believe, there should me one nice solution in projective geometry. May be not!

Spandan, were you able to solve problem 6?

Yes, I was. It said f(t)>1 for all t, so I simply put f(t)=2 for all t!

Subjective problems:

1. Find all n such that 5^n+1 is divisible by 7.

2. There are a set of three unit circles each of which is tangent to the other two. A triangle is formed by drawing tangents to two circle at a time.Find the lengths of the sides of the triangle.

I will post more if I remember.

Thanks. We have found all the subjective problems. If you remember any objective problem then that will be great

There are two circles which are represented by the equations x^2 + y^2 -9=0 and x^2+y^2-20x+96=0. Points P and Q lie on circle 1 and circle 2 respectively.Point P lies above x-axis and point Q lies below x-axis. PQ is joined such that it is a common tangent to both the circles. Find the length of PQ.

ANSWER: 5* root(3)

One more question (don’t remember options):

Consider the circle x^2 + y^2 = 9. Take a point P on it, and above the X axis. Another circle is

x^2 + y^2 -20x + 96 = 0. Take a point Q on this circle and below the axis. Then find the length of PQ such that the line is tangent to both the circles.

The lengths of two sides of a triangle are 2 cm and 3 cm. Find the maximum possible area of the triangle.

I believe the answer is 3.

A set is called multiplicatively closed

if the product of any two elements of the set is also an element of the set. Consider the following two sets:S1 = {a + ib : a, b are integers}S2 = {a + wb : a, b are integers; w is a non real root of unity}

Then:(1) Both S1 and S2 are multiplicatively closed.

(2) Neither is closed.

(3) Only S1 is closed.

(4) Only S2 is closed.

Sorry for the typo. w is a non real CUBE root of unity.

Dear Aronyo, I believe the answer is option 3, isn’t it?

Spandan, that is what I thought at first. But if you take (a1 + wb1) and (a2 + wb2) and multiply them, (a1 + wb1)(a2 + wb2) = a1a2 + (a1b2 + a2b1)w + b1b2(w^2), then w^2 can be written as (-1 -w). Putting w^2 = -1-w, the expression becomes linear with respect to w, and it reduces to a + wb form. So option 1 is the answer.

The rate of increase of a population is directly proportional to the current population. If it takes 50 years for the population to double, find the time it takes to triple.

(1) 50 log (3, base 2)

(2) 50 log (3/2)

Thank you Aronyo. Due to your effort we now have many more questions. Hopefully with the help of rest of the students, we will be able to reconstruct the entire paper

Always happy to help, sir 🙂 It is the least I can do to help future students.

I will be posting more if I remember.

A function y=f(x) is defined in (0, 2) U (4, 6) such that dy/dx = 1

Then:

(1) f(x) = x

(2) f(x) is always increasing

(3) f(5.5) – f(4.5) = f(1.5) – f(0.5)

I believe option 3 is correct.

I agree, I too put option 3.

I heard that there was a condition in question 4 that, A>B>C.

Yes you are right.

Hello All

Most probably the 8th question should have two answers simultaneously:

1) All numbers are equal

2)All numbers are positive (as if they are negative it does not satisfy conditions of problem)

However out of confusion I marked only the option :- all numbers are positive

Can anyone please clarify.

Well, I assumed that the numbers were in AP, thus, we can show that all the numbers can never be negative.

Also, if the numbers are all positive, then its correct for all cases when they are equal and for some other cases as well, when all are positive.

Hence, though both options both options are correct, your choice, in a way, is ‘more correct’ I think.

Anyway, if I were one of those ‘grey emirates’ @ ISI, then I would rather have this question as a bonus tbh.

I am extremely grateful to you for your thoughts on the question. How much do you think will be the cutoff in subjective. And will I get any partial credit for not pointing out n=3k where k is only odd positive number.

I read in the main forum, that you had solved 7 problems in subjective, I had a conversation with sir earlier, and according to him, more than 6 that should be more than enough. Congrats!

‘Correct’.if the number in AP then it satisfy the question. let a,a+b,a+2b….a+2nb are in AP then sum of any n numbers is less than the sum of n+1 numbers.For this I think the option is ‘all numbers is positive’.It is more appropriate than the other option

Three such Positive numbers remind us of the triangle inequality, isn’t it?

More problems from ISI objective :

1. How many points P of integral coordinates are there on Y axis which are there such that triangles AOP and BOP are both right angled and such that A is (10,0) and B is (7,0) and O is origin ?

A) infinite

B) 2

C) 4

Sorry i dont remember D option

2. How many surjective functions are there from A to B such that A has 6 distinct elements and B has 3 elements.

A)52

B)120

C)200

D)

3. What is the least area of the triangle bounded by a tangent to the ellipse x^2/a^2 + y^2/b^2 , X axis and Y axis ?

A) ab

B) (a^2+b^2-ab)

4. How many complex numbers z are there such that (z conjugate= z^2)

A) 2

B) 4

C)6

D)8

5. How many factors are there which are either perfect squares, perfect cubes or both of 2^14*3^9^5^5

A)252

B)

C)

D)

6. How many triples of prime numbers a<b<c are there such that a+b+c=30

A) 2

B)

Q3 a^2+b^2/2 is the answer. I believe it was C

Well, I had put the answer as ab!

Can’t. Take the case of x^2/16+y^2/9=1. The tangents touch the axes at (5,0) and (0,5). The area is 0.5*5*5 = 25/2 = 16+9/2

Krishnan, I appreciate your method, but please be patient enough to go through mine as well-

Equation of tangent to a regular ellipse-

y= mx + sqrt.([am]^2 + b^2]

Thus, it will meet the axes @ points having magnitude of sqrt.([am]^2 + b^2] and {sqrt.([am]^2 + b^2]}/m.

Thus the area would be min. of 1/2 * [am]^2 + b^2]/m.

Minimizing this function, we get ab!

(P.S- In your problem, you had considered a specific tangent, giving you the area 12.5. 12 is possible by my method, try it and see.)

Thanks again for your suggestions!

Yes. You are right Spandan. I missed the part that said Least Area.

1 and 4 are B right?

Oh and Q1 is how many equidistant points?

I am extremely sorry for any stupid typos from my part . I will try remembering more problems

Thanks for the contribution Biswarup. With the help of students like you, we are able to reconstruct most of the objective paper.

Thank you , Sir

Certainly I will try my best to remember the other problems as well.

x^2+y^2=2015 no.s of integer soln (x,y)

ans:ZERO

That is what I thought too. But then I remember a question in TOMATO which was x^2+y^2=2007 having more than two integral solutions. Can anyone please explain that?

No bro if u have book test of mathematics at 10+2 level u can clearly see that its ans is zero

objective 2012 q.no-2, no….. its none what that asked in current year oftion.

Simplifying

x2 + y2 = 2015

Solving

x2 + y2 = 2015

Solving for variable ‘x’.

Move all terms containing x to the left, all other terms to the right.

Add ‘-1y2’ to each side of the equation.

x2 + y2 + -1y2 = 2015 + -1y2

Combine like terms: y2 + -1y2 = 0

x2 + 0 = 2015 + -1y2

x2 = 2015 + -1y2

Simplifying

x2 = 2015 + -1y2

Reorder the terms:

-2015 + x2 + y2 = 2015 + -1y2 + -2015 + y2

Reorder the terms:

-2015 + x2 + y2 = 2015 + -2015 + -1y2 + y2

Combine like terms: 2015 + -2015 = 0

-2015 + x2 + y2 = 0 + -1y2 + y2

-2015 + x2 + y2 = -1y2 + y2

Combine like terms: -1y2 + y2 = 0

-2015 + x2 + y2 = 0

The solution to this equation could not be determined.

In general

x^2+y^2=N

It will have integer solns only if N contains prime of the form 4k+1 type.

If all primes are 4k+1 type then +ve solns=>No of factors of N,(be careful in case of prefect square) it will be -1 then

If it contains some 4k+3 prime with odd power then 0 solns

If it contains even power of 4k+3 primes.then +ve solns = no of factors of 4k+1 form only.

Integral soln=4*positive solns

Is your interview at ISI Kolkata over? What kind of questions were asked? I am having my interview for bmath on 22nd June. Any suggestions??

I have correctly attempted 23 objectives and 5 subjectives in this years entrance for bmath. What are my chances for being selected for interview ??

Congrats! You are in mate I think, of course, considering that all your attempts are right.

Thanks a lot for replying bro…..

You are welcome.

Most welcome.

Q16 should be 540. There are 3^6 total functions possible out which 3*2^6-3 are not onto whence 540 functions. Was 540 a possible option there?

When will result of isi bstat be out

But they are-

check out http://www.isical.ac.in/~admission/

spandan . how many did you solve in objective paper

29 questions.

look out at question 20, I am getting answer as D option i.e., 2

the cases are 2,5,23 and 2,17,11.

and also question 18 should get answer 2 as solutions must be cis(2*3.14/3) and -cis(3.14/3)

question no. 17 answer should be ab

Qno.16 – answer 120

Q.no 15 answer should be infinite

Qno – 14 answer option is D

check and correct me for these 2015 isi solutions.

Actually in question 8 all the options include “must be” condition that is satisfied only if all integers are equal

Let A = { ( a,b,c ) : a,b,c are prime numbers, a<b<c, a+b+c= 30} . The number of elements in A is ?

(A) 0

(B) 1

(C) 2

(D) 3

Can someone help…my ans is coming wrong ?