 # 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 16th May to 30th May, 2017.

Created On: Apr 25, 2017 18:23 IST 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 16th May to 30th 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 x2–x–1 = 0 and Sn = αnn, for all integers n ≥ 1. Then for every integer n≥2

(A) Sn+Sn–1 = Sn+1

(B) Sn–Sn–1 = Sn

(C) Sn–1 = Sn+1

(D) Sn+Sn–1 = 2Sn+1

Ans : (A)

Sol:

We have,

x2–x–1 = 0

So,

α+β =1

Also,

α2–α–1 = 0

α2 = α + 1                           …(1)

And, β2–β–1=0

β2 = β + 1                         …(2)

Now,

Sn +Sn–1 = αnn + αn-1n-1  = (αnn–1)+( βnn–1) = αn–1 (α+1) + βn–1 (β+1)

From (1) and (2)

Sn +Sn–1 = αn–12) + βn–12) = αn+1  + βn+1 = Sn+1

So, Sn +Sn–1 = Sn+1

Q. The number of digits in 20301 (given log102 = 0.3010) is

(A) 602

(B) 301

(C) 392

(D) 391

Ans: (C)

Sol:

log 20301 = 301 × log 20 = 301 × 1.3010 = 391.6010,

So, the number of digits in 20301 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 a3sin(B–C) + b3 sin(C–A) + c3sin(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 .

a3sin(B–C) + b3 sin(C–A) + c3sin(A–B) = 0