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Q:

Find the quadratic equations whose roots are the reciprocals of the roots of $2 x^{2} + 5 x + 3$ ?

A) $3 x^{2} + 5 x + 2 = 0$

B) $5 x^{2} + 3 x + 2 = 0$

C) $3 x^{2} - 5 x - 2 = 0$

D) None

A) A	B) B
C) C	D) D

Answer & Explanation Answer: A) A

Explanation:

The quadratic equation whose roots are reciprocal of $2 x^{2} + 5 x + 3 = 0$ can be obtained by replacing x by 1/x.

Hence, 2(1/x)(1/x)+ 5(1/x) + 3 = 0

=> $3 x^{2} + 5 x + 2 = 0$

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Filed Under: Mathematical Operations
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Q:

The average age of seven persons sitting in a row facing east is 26 years. If the average age of the first three persons is 19 years and the average age of the last three persons is 32 years, then find the age of the person sitting in the middle of the row ?

A) 32 yrs	B) 29 yrs
C) 24 yrs	D) 27 yrs

Answer & Explanation Answer: B) 29 yrs

Explanation:

Total age seven persons = (26 x 7)years
Total age of the first three persons and the last three persons are (19 x 3) years and (32 x 3) years respectively.
Age of the person sitting in the middle of the row = 26 x 7 - 19 x 3 - 32 x 3 = 182 - 57 - 96 = 29 years.

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Filed Under: Problems on Ages
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Q:

I. $a^{2} - 7 a + 12 = 0$ ,

II. $b^{2} - 3 b + 2 = 0$

Solve both the equations to find the values of a and b ?

A) if a > b	B) if a < b
C) if the relationship between a and b cannot be established.	D) if a ≤ b

Answer & Explanation Answer: A) if a > b

Explanation:

I. (a - 3)(a - 4) = 0

=> a = 3, 4

II. (b - 2)(b - 1) = 0

=> b = 1, 2

=> a > b

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Filed Under: Statement and Conclusions
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Q:

A boat takes 19 hours for travelling downstream from point A to point B and coming back to a point C which is at midway between A and B. If the velocity of the stream is 4 kmph and the speed of the boat in still water is 14 kmph, what is the distance between A and B ?

A) 180 km	B) 160 km
C) 140 km	D) 120 km

Answer & Explanation Answer: A) 180 km

Explanation:

Speed in downstream = (14 + 4) km/hr = 18 km/hr;

Speed in upstream = (14 – 4) km/hr = 10 km/hr.

Let the distance between A and B be x km. Then,

x/18 + (x/2)/10 = 19 ⇔ x/18 + x/20 = 19 ⇒ x = 180 km.

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Filed Under: Boats and Streams
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Q:

What is the maximum velocity with which a body of mass 'm' must enter a vertical loop of radius 'R'so that it can complete the loop?

A) 0 B) gR C) $\sqrt{2} g R$ D) $\sqrt{5 g R}$

A) A	B) B
C) C	D) D

Answer & Explanation Answer: D) D

Explanation:

To complete the vertical loop, the minimum speed required at the lowest point = $\sqrt{5 g R}$ .

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Filed Under: Physics
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Q:

The Internet was launched in 1969 and was originally called ?

A) AARPNET	B) CERNET
C) ARPANET	D) CERN

Answer & Explanation Answer: C) ARPANET

Explanation:

The Internet was originally called Advanced Research Project Agency Network (ARPANET).

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Filed Under: Web Technology
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Q:

Solve the equation ${(\frac{1}{2})}^{2 x + 1} = 1$ ?

A) -1/2	B) 1/2
C) 1	D) -1

Answer & Explanation Answer: A) -1/2

Explanation:

Rewrite equation as ${(\frac{1}{2})}^{2 x + 1} = {(\frac{1}{2})}^{0}$

Leads to 2x + 1 = 0

Solve for x : x = -1/2

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Filed Under: Logarithms
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Q:

A box contains 4 different black balls, 3 different red balls and 5 different blue balls. In how many ways can the balls be selected if every selection must have at least 1 black ball and one red ball ?

A) $2^{4} - 1$

B) $2^{4} (2^{5} - 1)$

C) $(2^{4} - 1) (2^{3} - 1) 2^{5}$

D) None

A) A	B) B
C) C	D) D

Answer & Explanation Answer: C) C

Explanation:

It is explicitly given that all the 4 black balls are different, all the 3 red balls are different and all the 5 blue balls are different. Hence this is a case where all are distinct objects.

Initially let's find out the number of ways in which we can select the black balls. Note that at least 1 black ball must be included in each selection.

Hence, we can select 1 black ball from 4 black balls
or 2 black balls from 4 black balls.
or 3 black balls from 4 black balls.
or 4 black balls from 4 black balls.

Hence, number of ways in which we can select the black balls

= 4C1 + 4C2 + 4C3 + 4C4
= $2^{4} - 1$ ........(A)

Now let's find out the number of ways in which we can select the red balls. Note that at least 1 red ball must be included in each selection.

Hence, we can select 1 red ball from 3 red balls
or 2 red balls from 3 red balls
or 3 red balls from 3 red balls

Hence, number of ways in which we can select the red balls
= 3C1 + 3C2 + 3C3
= $2^{3} - 1$ ........(B)

Hence, we can select 0 blue ball from 5 blue balls (i.e, do not select any blue ball. In this case, only black and red balls will be there)
or 1 blue ball from 5 blue balls
or 2 blue balls from 5 blue balls
or 3 blue balls from 5 blue balls
or 4 blue balls from 5 blue balls
or 5 blue balls from 5 blue balls.

Hence, number of ways in which we can select the blue balls
= 5C0 + 5C1 + 5C2 + … + 5C5
= $2^{5}$ ..............(C)

From (A), (B) and (C), required number of ways
= $2^{5} (2^{4} - 1) (2^{3} - 1)$

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Filed Under: Permutations and Combinations
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