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Under Graduate Entrance Exams IIT JEE IEEE Entrance Exam # Mathematics Permutations and Combinations Prepare / Learn

Q3. Different calendars for the month of February are made so as to serve for all the coming years. The number of such calendars is

A. 7

B. 2

C. 14

D. 8

Answer : Option C
Explaination / Solution:

The month of February has either 28 days (non-Leap Year) or 29 days(Leap Year).

Thus there are 2 possibility for the number of days in February.

Also the first day of the month of February can be any of the 7 days in a week.

So total number of February calendars = 2 * 7 =14

Under Graduate Entrance Exams IIT JEE IEEE Entrance Exam # Mathematics Three Dimensional Geometry Prepare / Learn

Q4. In the Cartesian form two lines

\frac{x - x_{1}}{a_{1}} = \frac{y - y_{1}}{b_{1}} = \frac{z - z_{1}}{c_{1}}

and

\frac{x - x_{2}}{a_{2}} = \frac{y - y_{2}}{b_{2}} = \frac{z - z_{2}}{c_{2}}

are coplanar if

Answer : Option D
Explaination / Solution:

In the Cartesian form two lines

\frac{x - x_{1}}{a_{1}} = \frac{y - y_{1}}{b_{1}} = \frac{z - z_{1}}{c_{1}}

and

\frac{x - x_{2}}{a_{2}} = \frac{y - y_{2}}{b_{2}} = \frac{z - z_{2}}{c_{2}}

are coplanar if

Under Graduate Entrance Exams IIT JEE IEEE Entrance Exam # Mathematics Differential Equations Prepare / Learn

Q5. Forming a differential equation representing the given family of curves by eliminating arbitrary constants a and b from

\frac{x}{a} + \frac{y}{b} = 1

yields the differential equation

A. y″ = y

B. y'' = 0

C. y′′=y3

D. y″ = 2y

Answer : Option B
Explaination / Solution:

Under Graduate Entrance Exams IIT JEE IEEE Entrance Exam # Mathematics Complex Numbers and Quadratic Equations Prepare / Learn

Q6. The value of

{(\frac{1 + ω}{ω^{2}})}^{3}

A. 1

B. 0

C. None of these

D. -1

Answer : Option D
Explaination / Solution:

Under Graduate Entrance Exams IIT JEE IEEE Entrance Exam # Mathematics Binomial Theorem Prepare / Learn

Q7.

If the coefficients of 2nd ,3rd and 4th terms in the expansion of $(1 + x)^{n}$ are in A.P. then

A. n2−5n+14=0

B. n2−9n+14=0

C. n2−9n+18=0

D. n2−6n+12=0

Answer : Option B
Explaination / Solution:

Under Graduate Entrance Exams IIT JEE IEEE Entrance Exam # Mathematics Maths Sets Prepare / Learn

Q8. If Q = { x : = 1/y , where y ∈ N } , then

A. 1/2

\notin

B. 0

\in

C. 2

\in

D. 1 ∈ Q

\in

Answer : Option D
Explaination / Solution:

N is set of natural number,so

x = 1/y

when y=1 then x=1

1 $\in$ Q

Under Graduate Entrance Exams IIT JEE IEEE Entrance Exam # Mathematics Limits and Derivatives Prepare / Learn

Q9.

\frac{d}{d x} (\sin^{- 1} (1 - x))

is equal to

A. −12x−x2√

B. 1x2−2x√

C. 12x−x2√

Answer : Option D
Explaination / Solution:

Under Graduate Entrance Exams IIT JEE IEEE Entrance Exam # Mathematics Linear Programming Prepare / Learn

Q10. A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding/cutting machine for at the most 12 hours. The profit from the sale of a lamp is Rs 5 and that from a shade is Rs 3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximize his profit?

A. 5 Pedestal lamps and 5 wooden shades; Maximum profit = Rs 38

B. 4 Pedestal lamps and 5 wooden shades; Maximum profit = Rs 36

C. 5 Pedestal lamps and 4 wooden shades; Maximum profit = Rs 35

D. 4 Pedestal lamps and 4 wooden shades; Maximum profit = Rs 32

Answer : Option D
Explaination / Solution:

Let number of pedestal lamps manufactured = x
And number of wooden shades manufactured = y
Therefore , the above L.P.P. is given as :
Maximise , Z = 5x +3y , subject to the constraints : 2x +y ≤ 12 and. 3x +2y ≤ 20 , x, y ≥ 0.

Corner points	Z =5x +3 y
O( 0 , 0 )	0
D(6,0 )	30
A(0,10)	30
B(4,4)	32…………………(Max.)

Here Z = 32 is maximum.
i.e 30 packages of screws A and 20 packages of screws B; Maximum profit = Rs 410.
i.e. 4 Pedestal lamps and 4 wooden shades; Maximum profit = Rs 32 .

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