# This is a work in progress. Please suggest improvements in the comment section.

# Problems

# Objective Section (Answer Key)

1. A | 2. A | 3. C | 4. A | 5. C | 6. B |

7. C | 8. | 9. C | 10. B | 11. A | 12. D |

13. C | 14. C | 15. D | 16. C | 17. B | 18. B |

19. B | 20. B | 21. A | 22. A | 23. B | 24. B |

25. C | 26. B | 27. B | 28. B | 29. B | 30. B |

# Subjective Section (Sequential Hints)

# Faculty Team (for this section)

#### Ashani Dasgupta

Founder – Faculty at Cheenta.

Pursuing Ph.D. at University of Wisconsin, Milwaukee. USA

Research Interest: Geometric Group Theory

#### Srijit Mukherjee

Faculty – Admin at Cheenta

Pursuing B.Stat at Indian Statistical Institute, India

#### Writabrata Bhattacharya

Faculty at Cheenta

Pursuing B.Sc. Math at Chennai Mathematical Institute, India

Research Interest – Algebraic Geometry

#### Ishan Sengupta

Associate Faculty at Cheenta

Pursuing B.Stat from Indian Statistical Institute, India

In the third question the area of the resultant polygon is 2/3rd of the area of the triangle

Area = 2/3 x 9√3/2

= 3√3

The answer is option d

Ankush double checking (possibly you are right), isn’t area of the triangle rt3/4 *(3^2)? 2/3rd of that is ..

Okay I made a mistake at such a silly thing. Thanks for correcting it sir

16th Question the right answer should be 33 , it’s not such an easy case of PHP as it seems like, for that we have to calculate

Number of Student who visited 5 days- 5C5=1

Number of Student who visited 4 days- 5C4=5

Number of Student who visited 3 days- 5C3=10

Number of Student who visited 2 days- 5C2=10

Number of Student who visited 1 day – 5C1=5

Number of Student who visited 0 day – 5C0=1

Total 32 cases, now applying Pigeon Hole Principle we need 33 students

Yes. Fixed it! Thank you.

For objective #16, if we represent the days visited as a YES/NO tuple of (a1,a2,…,a5), then for each student, 2^5=32 such tuples are possible. So, there are 32 boxes and so the minimum number of students needed to guarantee the condition (in question) has to be 33. So answer should be (C)33.

Yes. We just fixed that option. Thank you.

26 should be option A.

Are the options of 18 correct

Check that n=17 works for Problem 26

Yes, the options of 18 are correct.

18’s correct option should be A.Let’s assume we sent two children together, one children alone and both adults separately then we found that the boat has to cross the river 9 times

In ques. 29 graph of x^2 and e^x intersect at 2 points so ans. Should be 2

Trying drawing the graphs using GeoGebra and check how many times they intersect.

In ques. 19, for adults it would be 4 times(2 for each) and for 2 children 2 times and as it is written no adult so ‘max’… So for last 1 children, 2 times so total 4+2+2=8.. Sir please tell options are true.

Remember, someone needs to row the boat back. It cannot come back empty to take more passengers.

23

option B

as 79, 78, 77, 74, 73, 69

are not possible hence 100 – 6 = 94

Yes that is right. That you for the input.

total numbers are 101 not 100

10th question

isn’t that 1 is a finite number hence option D should also be correct

No because D restricts the angles to certain values. But there are solutions for all angles.

in 10th question 1 is also a finite number

hence option D should also be correct

No because D restricts the angles to certain values. But there are solutions for all angles.

they have told m are integer’s’ and n is not equal to zero which includes all real numbers

they haven’t specifically said that values n are also are integers

I hope that works out. However, usually, when mathematicians say integers m and n not 0, they usually mean both m and n are integers and n is not 0. Anyway, maybe your answer is correct. We have posted unofficial options.

Please provide the other subjective answers

for the 11th march to 15th march question,why should we consider not swimming on any dates as a possibility ?The empty set should not be considered,since we are asked to figure out whether they ‘swim’ on the same dates.Not swimming at all shouldn’t be an option.

A child can be absent on all days and still be a student of the class.

I gave ISI BMath 2019 and scored 93 in UGA, And in UGB I attempted total 6 questions,3 questions correctly with 2 being partially answered and one being fully wrong, I am OBC too, Will I be called for the interview?

That seems possible.

but the one i did wrong was the first question…. is it something dangerous…. i mean about the impression to the checker?

How is everyone’s marks in UGA 2019? (acc. to official answer key) How much are you scoring?

99

What is the expected cutoff of ISI BMath 2019? (Like 2UGA+9UGB >= How much??)

Correct option of question no. 19 should be D. We can get 3 also there from the determinant.

HOW DID YOU GET THE SOLUTION FOR Q.26 ?

I COULD NOT FIND ANY SOLUTION SO I THINK ANSWER WILL BE 0 i.e (A).

IF YOU GET 1 SOLUTION STISFYING THE CONDITION THEN FOR WHAT VALUE OF n IS IT FOR?

n = 17

respected sir, the solutions of all the subjective questions are not uploaded.