# BITSAT 2017 Solved Mathematics Question Paper Set-I

Read this article to get download link of BITSAT 2017 Mathematics Solved Question Paper. This paper will be very helpful for the coming BITSAT examination. Birla Institute of Technology and Science Admission Test(BITSAT) is a computer-based admission test conducted by BITS. This year BITSAT will be held between 16^{th} May to 30^{th} May, 2017.

Birla Institute of Technology and Science Admission Test(BITSAT) is a computer-based admission test conducted by BITS. This year BITSAT will be held between **16 ^{th} May to 30^{th} May, 2017**. The Mathematics section of BITSAT contains 50 questions.

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

**Q.** The letters of the word COCHIN are permuted and all the permutations are arranged in alphabetical order as in English dictionary. The number of words that appear before the word COCHIN is

(A) 360

(B) 192

(C) 96

(D) 48

**Ans: (C)**

**Sol. **

In dictionary the words starting from CC, CH, CI and CN will appear first

Now, fixing CC the remaining 4 letters O, H, I and N can be arranged in 4! ways.

Similarly the words starting from CH, CI and CN will also be arranged in 4! ways.

So, number of words that appear before the word COCHIN is (4!)4 = 96

**Q. **Let α,β be the roots of x^{2}–x–1 = 0 and S_{n} = α^{n}+β^{n}, for all integers n ≥ 1. Then for every integer n≥2

(A) S_{n}+S_{n–1} = S_{n+1}

(B) S_{n}–S_{n–1} = S_{n}

(C) S_{n–1} = S_{n+1}

(D) S_{n}+S_{n–1} = 2S_{n+1}

**Ans : (A)**

**Sol: **

We have,

x^{2}–x–1 = 0

So,

α+β =1

Also,

α^{2}–α–1 = 0

α^{2} = α + 1 …(1)

And, β^{2}–β–1=0

β^{2} = β + 1 …(2)

Now,

S_{n} +S_{n–1} = α^{n}+β^{n} + α^{n-1}+β^{n-1} = (α^{n}+α^{n}–1)+( β^{n}+β^{n–1}) = α^{n–1} (α+1) + β^{n–1} (β+1)

From (1) and (2)

S_{n} +S_{n–1} = α^{n–1} (α^{2}) + β^{n–1} (β^{2}) = α^{n+1} + β^{n+1} = S_{n+1}

So, S_{n} +S_{n–1} = S_{n+1}

**Q. **The number of digits in 20^{301} (given log_{10}2 = 0.3010) is

(A) 602

(B) 301

(C) 392

(D) 391

**Ans: (C)**

**Sol: **

log 20^{301} = 301 × log 20 = 301 × 1.3010 = 391.6010,

So, the number of digits in 20^{301 }is 392 digits.

**Q. **In a ΔABC, a,b,c are the side of the triangle opposite to the angles A,B,C respectively. Then the value of a^{3}sin(B–C) + b^{3} sin(C–A) + c^{3}sin(A–B) is equal to

(A) 0

(B) 1

(C) 3

(D) 2

**Ans : (A)**

**Sol.**

By using sine law and 3A formula, the value of .

a^{3}sin(B–C) + b^{3} sin(C–A) + c^{3}sin(A–B) = 0

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