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BITSAT Mathematics

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BITSAT Mathematics

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BITSAT 2019 – Entrance Examination, conducted in online Mode, has: a duration of 3 Hours and consists of 4 parts, namely – Physics, Chemistry, English Proficiency & Logical Reasoning and Mathematics/ Biology. The 4 sections are not separately timed and there is no break in between the sections. There is a Negative marking in BITSAT 2019 exam and 1 mark is deducted for each wrong answer. The below sections gives the detailed information about BITSAT Physics part.


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Parts Subject No of Questions Duration
Part I Physics 40 3 hours
(Without break)
Part II Chemistry 40
Part III (a) English Proficiency and 15
(b) Logical Reasoing 10
Part IV Mathematics or Bilogy
(For B.Pharm candidates)
45
Total 150


The BITSAT Mathematics section, has 45 objective questions. Below mentioned are the different categories of expected questions in the BITSAT Mathematics Section.


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[Click Here] for BITSAT 2019 Exam Syllabus


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1. The ages of the two persons differ by 20 years. If 5 year ago, the older one be 5 times as old as the younger one, then their present ages, in year are:

    A. 25, 5
    B. 30, 10
    C. 35, 15
    D. 50, 30


Answer: Option B

Exaplanation:
Let the age be x and y years now.

Then, x−y=20 ——– (i)

and (x−5)=5(y−5) ——- (ii)

On solving both equations we get:

x=30 and y=10


2. A man arranges to pay off a debt of Rs 3600 by 40 annual installments which are in A.P. When 30 of the installments are paid he dies leaving one-third of the debt unpaid.
The value of the \({8}^{th}\) installment is:

    A. Rs 35
    B. Rs 50
    C. Rs 65
    D. Rs 70


Answer: Option C

Exaplanation:

Let the first installment be ‘a’ and the common difference between any two consecutive installments be ‘d’

Using the formula for the sum of an A.P.
S = \(\frac{n}{2}\) [2a+(n−1)d]

We have,
3600 ⇒ 180And 2400 ⇒ 160 = \(\frac{40}{2}\) [2a + (40 − 1)d]
= 2a + 39d = \(\frac{30}{2}\)[2a + (30 − 1)d] = 2a + 29d(1)(2)
On solving both the equations we get:
d = 2 and a = 51
Note: To check out the calculations, check the comment by Anita.
Value of 8th installment;
= 51 + (8 − 1)2
= Rs. 65


3. A student was asked to divide a number by 6 and add 12 to the quotient. He, however first added 12 to the number and then divided it by 6, getting 112 as the answer.

    A. 114
    B. 118
    C. 122
    D. 124


Answer: Option C

Exaplanation:
Let the number be x, then operations undertook by the student:
= (\(\frac{x+126}{6}\)) = 112
⇒ x = 660
Correct answer:
= \(\frac{660}{6}\) + 12
= 122


4. A man arranges to pay off a debt of Rs 3600 by 40 annual instalments which are in A.P. When 30 of the instalments are paid he dies leaving one-third of the debt unpaid. The value of the \({8}^{th}\) instalment is:

    A. Rs 35
    B. Rs 50
    C. Rs 65
    D. Rs 70


Answer: Option C

Exaplanation:

Let the first instalment be ‘a’ and the common difference between any two consecutive instalments be ‘d’

Using the formula for the sum of an A.P.

S = \(\frac{n}{2}\) [2a+(n−1)d]

We have,

3600 = \(\frac{40}{2}\) [2a+(40−1)d]

⇒ 180 = 2a+39d——– (i)

2400 = \(\frac{30}{2}\) [2a+(30−1)d]

⇒ 160 = 2a+29d——– (ii)

On solving both the equations we get:

d = 2 and a = 51
Value of \({8}^{th}\) instalment = 51 + (8 − 1)2

= Rs 65

5. The sum of A and B’s age is 43 years. 11 year hence, A’s age will be 7/6 times B’s age then. Find B’s present age.

    A. 22 years
    B. 20 years
    C. 24 years
    D. 19 years


Answer: Option D

Exaplanation:

Let A’s age be x and B’s age be y.

x + y = 43 ——– (i)

(x + 11) = \(\frac{7}{6}\)(y+11)

⇒ 6x − 7y = 11 ——– (ii)

On solving both the equations, we get:

y = 19 years

1. If 2ycosθ = xsinθ and 2xsecθ – ycosecθ = 3, then value of \({x}^{2}\) + \({y}^{2}\) is

    A. 1
    B. 2
    C. 4
    D. None of these


Answer: Option D

Exaplanation:

Given:

2ycosθ = xsinθ ; 2xsecθ – ycosecθ = 3


2xsecθ – ycosecθ = 3
2x/cosθ – y/sinθ = 3
2xsinθ – ycosθ = 3sinθ cosθ ..(i)


2ycosθ = xsinθ
ycosθ = xsinθ/2


Substitute ycosθ in (i),
2xsinθ – (xsinθ/2) = 3sinθ cosθ
(4xsinθ – xsinθ)/2 = 3sinθ cosθ
4xsinθ – xsinθ = 6sinθ cosθ
3xsinθ = 6sinθ cosθ
x = 2cosθ


ycosθ = 2cosθsinθ/2
y = sinθ
x2 + y2 = 4cos2θ + sin2θ
= 4(1) (wkt, sin2θ + cos2θ = 1 )
= 4.


2. If cot A + cosec A = and A is an acute angle, then the value of cos A is

    A. \(\frac{4}{5}\)
    B. 1
    C. \(\frac{1}{2}\)
    D. \(\frac{1}{\sqrt{3}}\)


Answer: Option A

Exaplanation:
Here, cot A + cosec A = 3
⇒ \(\frac{cosA}{sinA}\) + \(\frac{1}{sinA}\) = 3
⇒ cos A + 1 = 3sin A
Squaring both sides.
⇒ \({(cos A + 1)}^{2}\) = \({(9sin}^{2}\) A

⇒ \({(cos A)}^{2}\) + 2cos A + 1 = 9 \((1- {cos}^{2}A)\)

⇒ 10 \({cos}^{2}\)A + 2 cos A – 8 = 0

⇒ 5 \({cos}^{2}\) A + cosA – 4 = 0

⇒ (5 cos A – 4) (cos A + 1) = 0

(∵ A is an acute angle)

∴ cos A = \(\frac{4}{5}\)


3. The value of sin (45° + θ) – cos (45° – θ) is

    A. 1
    B. 0
    C. 2 cos θ
    D. 2 sin θ


Answer: Option B

Exaplanation:
sin(45° + θ) – cos(45° – θ)
= sin{90° – (45° – θ)} – cos (45° – θ)
= cos (45° – θ) – cos (45° – θ)
{∵sin(90 – A) = cos A }
= 0


4. If \(\frac{2 sin θ – cos θ}{}\) = 1, then the value of cot θ is

    A. \(\frac{1}{2}\)
    B. \(\frac{1}{3}\)
    C. 3
    D. 2


Answer: Option A


Exaplanation:

Dividing numerator and denominator by sin θ

⇒ \(\frac{\frac{2Sinθ – Cosθ}{Sinθ}}{\frac{ Cosθ + Sinθ }{Sinθ}}\) = 1

⇒ \(\frac{2 – cot θ}{1 + cot θ}\) = 1

⇒ 2 – cot θ = 1 + cot θ

⇒ cot θ = \(\frac{1}{2}\)


5. Minimum value of 12 \({cos}^{2}\) a + 3 \({sec}^{2}\) a is

    A. 14
    B. 10
    C. 11
    D. 12


Answer: Option D


Exaplanation:

Let A = 12 \({cos}^{2}\) a + 3 \({sec}^{2}\) a

⇒ A = 3(4\({cos}^{2}\) a + \({sec}^{2}\) a)

⇒ A = 3[ \({( 2 cos α – sec α)}^{2}\) + 4]

⇒ A = [ \({( 2 cos α – sec α)}^{2}\) + 12]

For the minimum value of A, 2008α – sec α = 0

2 cos α = sec α

Hence, the minimum value of the expression = 12.

1. There are 8 positive numbers and 6 negative numbers. 4 numbers are chosen at random and multiplied. Find the probability that the product is a positive number ?

    A. \(\frac{1001}{505}\)
    B. \(\frac{101}{505}\)
    C. \(\frac{505}{1001}\)
    D. \(\frac{450}{965}\)


Answer: Option C

Exaplanation:

14C4 = \(\frac{14*13*12*11}{4*3*2*1}\) 1001
4 No positive + 4 no negative + (2 no positive * 2 no negative)
= 6C4 + 8C4 + (6C2 × 8C2) = 15 + 70 + 15*28 = 505
P = \(\frac{505}{1001}\)


2. A five digit number is formed with the digit 1,2,3,4 and 5 without repetition. Find the chance that the number is divisible by 5?

    A. \(\frac{1}{4}\)
    B. \(\frac{1}{5}\)
    C. \(\frac{1}{3}\)
    D. \(\frac{1}{2}\)


Answer: Option B

Exaplanation:
5! = 5*4*3*2*1 = 120
4! = 4*3*2*1 = 24
P = \(\frac{24}{120}\) = \(\frac{1}{5}\)


3. A bag contains 6 red, 5 blue and 2 green balls. If 2 balls are picked at random, what is the probability that both are green?

    A. \(\frac{1}{78}\)
    B. \(\frac{2}{75}\)
    C. \(\frac{4}{89}\)
    D. None of these


Answer: Option A

Exaplanation:
13C2 = \(\frac{13*12}{2}\) = 78
2C2 = 1
P = \(\frac{1}{78}\)


4. If You toss a coin and roll a die. What is the probability of getting a head and a 3 on the die?

    A. \(\frac{1}{10}\)
    B. \(\frac{1}{8}\)
    C. \(\frac{1}{12}\)
    D. \(\frac{2}{3}\)


Answer: Option C

Exaplanation:
\(\frac{1}{2}\) * \(\frac{1}{6}\) = \(\frac{1}{12}\)


5. From a pack of 52 cards, 3 cards are drawn. What is the probability that one is ace, one is queen and one is jack?

    A. \(\frac{1}{5520}\)
    B. \(\frac{15}{4525}\)
    C. \(\frac{13}{5005}\)
    D. \(\frac{16}{5525}\)


Answer: Option C

Exaplanation:

= \(\frac{4C1 × 4C1 × 4C1}{52C3}\)

= \(\frac{4 × 4 × 4}{22100}\)

= \(\frac{16}{5525 }\)

1. Sampling error can be reduced by

    A. Non-probability sampling
    B. Increasing the population size
    C. Decreasing the sample size
    D. Increasing the sample size


Answer: Option D


2. Neymann Pearson lemma provides always

    A. an unbiased test
    B. a most powerful test
    C. an admissible test
    D. a minimax test


Answer: Option B


3. If there is linear trend present in the population, then which of the following methods is the most efficient sampling technique?

    A. Cluster sampling
    B. Systematic sampling
    C. Stratified sampling
    D. Simple random sampling


Answer: Option B


4. When the population consists of heterogeneity, which sampling procedure is preferred?

    A. Stratified Random Sampling
    B. Simple Random Sampling
    C. Systematic Sampling
    D. Double Sampling


Answer: Option A


5. The Wishart distribution is a multivariate generalization of

    A. Normal distribution
    B. t-distribution
    C. Chi-square distribution
    D. F-distribution


Answer: Option C

1. Which of the following vector combinations will result in the greatest displacement?

    A. \vec{e} + \vec{d}
    B.\vec{e} + \vec{d}\vec{a} – \vec{c}
    C.\vec{a} – \vec{c}
    D.

\vec{e} – \vec{b}

Answer – Option B

Explanation –


2. A vector equally inclined to axes is

    A. [0, 8]
    B. [-12, 8]
    C. [0, 12]
    D. [8, 12]


Answer – Option C

Explanation –


3.The magnitude of the vector,
V =(5,-3,2)
is


    A. 4
    B. 10
    C. 38
    D.

20

Answer – Option B

Explanation –


4. A vector field which has a vanishing divergence is called as ____________

    A. Solenoidal field
    B. Rotational field
    C. Hemispheroidal field
    D.

Irrotational field


Answer – Option A

Explanation –


5. Chose the curl of f⃗ (x,y,z)=x2i^+xyzj^–zk^ at the point (2, 1, -2)

    A. 2i^+2k^
    B. -2i^-2jk^
    C. -4i^-4j^+2k^
    D.

-2i^-2j^

Answer – Option D

Explanation –

1. The distance of point (2, 5, 7) from the x-axis is

    A. 2
    B. \(\sqrt{74}\)
    C. 5
    D. \(\sqrt{53}\)


Answer: Option B


Exaplanation:
As distance of point (2, 5, 7) from the x-axis is
\(\sqrt{{5}^{2} + {7}^{2}}\) = \(\sqrt{25 + 49}\) = \(\sqrt{74}\)


2. P is a point on the line segment joining the points (3, 5, -1) and (6, 3, -2). If y-coordinate of point P is 2, then its x-coordinate will be

    A. 2
    B. \(\frac{17}{3}\)
    C. \(\frac{15}{2}\)
    D. -5


Answer: Option C


Exaplanation:

∴ \(\frac{3k + 5}{k + 1}\) = 2

⇒ 3k + 5 = 2k + 2 ⇒ k = -3

∴ x – coordinate is

\(\frac{6k – 3}{k – 1}\) = \(\frac{-18 + 3}{-3 + 1}\) = \(\frac{15}{2}\)


3. PDirection ratios of a line are 2, 3, -6. Then direction cosines of a line making obtuse angle with the y-axis are

    A. \(\frac{2}{7}\), \(\frac{-3}{7}\), \(\frac{-6}{7}\)
    B. \(\frac{-2}{7}\), \(\frac{3}{7}\), \(\frac{-6}{7}\)
    C. \(\frac{-2}{7}\), \(\frac{-3}{7}\), \(\frac{6}{7}\)
    D. \(\frac{-2}{7}\), \(\frac{-3}{7}\), \(\frac{-6}{7}\)


Answer: Option C


Exaplanation:

as direction cosines of a line whose direction ratio are \(\frac{2}{7}\), \(\frac{3}{7}\), \(\frac{-6}{7}\)
As angle with the y-axis is obtuse,
∴ cos β < 0,
Therefore direction ratios are \(\frac{-2}{7}\), \(\frac{-3}{7}\), \(\frac{6}{7}\)


4. The equations of y-axis in space are

    A. x = 0, y = 0
    B. x = 0, z = 0
    C. y = 0, z = 0
    D. y = 0


Answer: Option B


Exaplanation:

As on the y-axis, x-coordinate and z-coordinate are zeroes.


5.The line joining the points (0, 5, 4) and (1, 3, 6) meets XY-plane at the point ________ .

    A. (-2, -7, 0)
    B. (-2, -9, -0)
    C. (-2, 9, -0)
    D. (-2, 9, 0)


Answer: Option D

Exaplanation:
(-2, 9, 0), as line is\(\frac{x−1}{1}\) = \(\frac{y−3}{-2}\) = \(\frac{z−6}{2}\) = λ
General point on line is (λ + 1, -2λ + 3, 2λ + 6)
If it meets AT-plane, then 2λ + 6 = 0
⇒ λ = – 3
∴ Point is (-2, 9, 0)

1. Find \({lt}_{(x,y,z)→(0,0,0)}\) \(\frac{sin(x).sin(y)}{x.z}\)

    A.
    B. 5
    C.1
    D. Does Not Exist


Answer: Option D

Exaplanation:

Put x = t : y = at : z = t

= \({lt}_{t→0}\)\(\frac{sin(t).sin(at)}{t}^{2}\)

\({lt}_{t→0}\) \(\frac{sin(t)}{t}\) x (a) x \({lt}_{t→0}\) \(\frac{sin(at)}{at}\)

= (1) * (a) * (1) = a


2. Find \({lt}_{(x,y,z)→(2,2,4)}\) \(\frac{{x}^{2} + {y}^{2} – {z}^{2} + 2xy}{x + y – z}\)

    A.
    B. 123
    C.9098
    D. 8


Answer: Option D

Exaplanation:

Simplifying the expression yields

\({lt}_{(x,y,z)→(0,0,0)}\) \(\frac{{x + y}^{2} – {z}^{2}}{(x + y)−z}\)

\({lt}_{(x,y,z)→(0,0,0)}\) \(\frac{(x + y + z).(x + y − z)}{x + y − z}\)

\({lt}_{(x,y,z)→(0,0,0)}\) (x+y+z) = 2 + 2 + 4

= 8


3. Find \({lt}_{(x,y,z,w)→(3,1,1,11)}\) \(\frac{{x}^{4} + {y}^{2} + {z}^{2} + 2{x}^{2}y + 2yz + 2{x}^{2}z – {w}^{2}} {{x}^{2} + y + z – w}\)

    A.
    B. 123
    C.9098
    D. 8


Answer: Option D

Exaplanation:

Simplifying the expression we have

\({lt}_{(x,y,z,w)→(3,1,1,11)}\) \(\frac{{x}^{2} + y+ {z}^{2} – {w}^{2} }{{x}^{2} + y + z – w}\)

\({lt}_{(x,y,z,w)→(3,1,1,11)}\) \(\frac{({x}^{2} + y+ z – w) .({x}^{2} + y+ z – w) }{{x}^{2} + y + z – w}\)

\({lt}_{(x,y,z,w)→(3,1,1,11)}\)\(({x}^{2} + y+ z + w)\) = (32+1+1+11)

= 9+1+1+11=22


4. Given that limit exist find \({lt}_{(x,y,z)→(-9,-9,-9)}\) \(\frac{tan((x+9)(y+11)(z+7))}{(x+9)(y+10)}\)

    A. 2
    B. 1
    C. 4
    D. 3


Answer: Option C

Exaplanation:

We can parameterize the curve by
x = y = z = t

\({lt}_{(t)→(-9)}\)\(\frac{tan((t + 9)(t + 11)(t + 7))}{(t + 9)(t + 10)}\)

\({lt}_{(t)→(-9)}\)\(\frac{tan((t + 9)(t + 11)(t + 7))}{(t + 9)(t + 11)(t + 7)}\) x \({lt}_{(t)→(-9)}\)\(\frac{tan((t + 11)(t + 7))}{(t + 10)}\)

= \(\frac{(−9+11)(−9+7)}{(−9+10)}\) = \(\frac{(2)(2)}{(1)}\)

= 4

5.Given that limit exist find \({lt}_{(x,y,z)→(0,0,0,)}\)\(\frac{(Cos(\frac{x }{2} – x) tan(y).cot( \frac{x }{2}) – z)}{sin(x). sin(y). sin(z)}\)

    A. 99
    B. 0
    C. 1
    D. 100


Answer: Option C

Exaplanation:
Put x = y = z = t

\({lt}_{(x,y,z)→(0)}\)\(\frac{(Cos(\frac{x }{2} – t) tan(y).cot( \frac{x }{2}) – t)}{sin(x). sin(y). sin(z)}\)

\({lt}_{(x,y,z)→(0)}\) \(\frac{(sin(t)) ({tan}^{2}(t))}{{sin}^{3}(t)}\)

\({lt}_{(x,y,z)→(0)}\) \(\frac{{tan}^{2}(t)}{sin(t)}\) = \({lt}_{(x,y,z)→(0)} \) \(\frac{1}{{cos}^{2}(t)}\)

= \(\frac{1}{{cos}^{2}(t)}\) = \(\frac{1}{1}\) = 1