# UPSEE 2017 Solved Mathematics Sample Paper Set-III

Find UPSEE 2017 Solved Mathematics Sample Paper Set-III pdf in this article. Solving this paper will help in time management and knowing difficulty level of UPSEE 2017 engineering exam

Find **UPSEE 2017 Solved Mathematics Sample Paper Set-III**. This sample paper will be very helpful for those engineering aspirants who are going to appear for UPSEE-2017 Exam. Solving sample papers and previous year question papers are must do activities to do good in any entrance examination. Questions in this paper are prepared with great expertise and every question of this paper is important for UPSEE 2017 Examination. Also, many a times question is repeated in UPSEE/UPTU examination. So, practicing more questions will certainly give you an extra edge over other students.

This paper consists of 50 questions. Each correct answer will give 4 marks. There is no negative marking in this paper. Questions given in this paper cover complete syllabus of UPSEE. All questions are very important as they are developed with great expertise.

**About Exam:**

UPSEE is a state level entrance examination for various disciplines. Many prestigious government and private college is associated with this university. UPSEE engineering entrance exam paper is easier than JEE Main and Advanced. It provides good opportunity for those who cannot make to IITs or NITs.

**Few questions from the sample paper are given below:**

**Q. **If two sets *A *and *B *are having 39 elements incommon, then the number of elements common to each of the sets *A *× *B *and *B *× *A *are

(A) 2^{39}

(B) 39^{2}

(C) 78

(D) 351

**Correct Option: (B) **

**Sol.**

If in two sets *A *and *B **n* elements are common, then the number of elements common to each of the sets *A *× *B *and *B *× *A *are *n*^{2}.

So, the number of elements common to each set is 39 × 39 = (39)^{2}.

**Q. **The nth term of the series5+ 7+13 + 31 + 85 +....is

(A) 3^{n}^{-}^{1 }+ 2

(B) 3 + 3^{n}^{-}^{1 }

(C) 4 + 3^{n}^{-}^{1}

(D) None of these

**Correct Option: (C)**

**Sol**.

We have,

*S*_{n}* *= 5 + 7 + 13 + 31+85+....+*T _{n }*

_{-}

_{1}+

*T*...(i)

_{n}S* _{n}* = 5 + 7 + 13 + 31 + ....+

*T*

_{n}_{-}

_{2}+T

_{n}_{-}

_{1+}T

*...(ii)*

_{n}On subtracting Eq. (ii) from Eq. (i), we get

0=5 + 2 + 6 + 18 + ....to *n* terms] -*T _{n}*

Þ *T*_{n} = 5 + [2 + 6 + 18 + ....to (*n* −1) terms)]

Now,

2, 6, 18... is in GP with first term 2 and common ration as 3.

**Q. **The last two digits of the number 3^{400} are

(A) 81

(B) 43

(C) 29

(D) 01

**Correct Option: (D)**

**Sol**.

3^{400 }= (3^{4})^{100}

= (81)^{100}

= (1 + 80)^{100}

= ^{loo}*C*_{0} + ^{loo}*C*_{1}80 + ^{100}*C*_{2}(80)^{2} + ..... + ^{1oo}*C*_{100}(80)^{100}

Since the last two digits of each term in the binomial expansion is 00 except the first term which has last two digits as 01.

So, last two digits of the number 3^{400 }are 01.

**Download Complete Solved Sample Paper**

**Also, Get:**

**UPSEE 2017 Solved Chemistry Sample Paper Set-II**

## Comments