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 |

There is no time limit for individual parts of the test. The candidate can go back and change any of his/her answers among the

If a candidate answers all the

These extra questions will be from

The

[Click Here] for BITSAT 2019 Exam Syllabus

**Note**:

Each correct answer fetches**3 marks**, while each incorrect answer has a penalty of **1 mark (-1mark)**. No marks are awarded for questions not attempted. While the candidate can skip a question, the computer will not allow the candidate to choose more than one option as correct answer. There will be **150 questions** in all.

Each correct answer fetches

**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: **

**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. **

**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: **

**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.
**

**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

**Answer**: Option D

**Exaplanation**:

Given:

2ycosÎ¸ = xsinÎ¸ ; 2xsecÎ¸ – ycosecÎ¸ = 3

2xsecÎ¸ – ycosecÎ¸ = 3

2x/cosÎ¸ – y/sinÎ¸ = 3

2xsinÎ¸ – ycosÎ¸ = 3sinÎ¸ cosÎ¸ ..(i)

2ycosÎ¸ = xsinÎ¸

ycosÎ¸ = xsinÎ¸/2

Substitute ycosÎ¸ in (i),

2xsinÎ¸ – (xsinÎ¸/2) = 3sinÎ¸ cosÎ¸

(4xsinÎ¸ – xsinÎ¸)/2 = 3sinÎ¸ cosÎ¸

4xsinÎ¸ – xsinÎ¸ = 6sinÎ¸ cosÎ¸

3xsinÎ¸ = 6sinÎ¸ cosÎ¸

x = 2cosÎ¸

ycosÎ¸ = 2cosÎ¸sinÎ¸/2

y = sinÎ¸

x2 + y2 = 4cos2Î¸ + sin2Î¸

= 4(1) (wkt, sin2Î¸ + cos2Î¸ = 1 )

= 4.

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

**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**

**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**

**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**

**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.

**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? **

**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? **

**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?**

**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?**

**Answer**: Option C

**Exaplanation**:

= \(\frac{4C1 × 4C1 × 4C1}{52C3}\)

= \(\frac{4 × 4 × 4}{22100}\)

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

**Answer**: Option D

**2. Neymann Pearson lemma provides always**

**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? **

**Answer**: Option B

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

**Answer**: Option A

**5. The Wishart distribution is a multivariate generalization of **

**Answer**: Option C

\vec{e} – \vec{b}

**Answer –** Option B

**Explanation –**

**2. A vector equally inclined to axes is **

** Answer –** Option C

**Explanation –**

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

20

**Answer –** Option B

**Explanation –**

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

Irrotational field

** Answer –** Option A

**Explanation –**

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

-2i^-2j^

** Answer –** Option D

**Explanation –**

**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 **

**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 **

**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**

**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 ________ .**

**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)

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

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

**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 \({lt}_{(x,y,z)→(-9,-9,-9)}\) \(\frac{tan((x+9)(y+11)(z+7))}{(x+9)(y+10)}\) **

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

**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