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Q:

Which among the following validators validates controls if controls contain data?

A) CompareValidator	B) RangeValidator
C) RequiredFieldValidator	D) None

Answer & Explanation Answer: C) RequiredFieldValidator

Explanation:

RequiredFieldValidator is used to validate the controls if they contain data

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Filed Under: .NET
Job Role: Software Architect

Q:

Server controls contain attributes whereas HTML controls have properties only.

A) TRUE

B) FALSE

Answer & Explanation Answer: B) FALSE

Explanation:

Server controls contain properties whereas HTML controls have attributes.

Workspace

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Subject: Web Technology
Job Role: Software Architect

Q:

A cube of edge 15 cm is immersed completely in a rectangular vessel containing water . If the dimensions of the base of vessel are 20 cm x 15 cm, find the rise in water level

A) 11.25cm	B) 11.56cm
C) 11.66cm	D) 10.86cm

Answer & Explanation Answer: A) 11.25cm

Explanation:

Increase in volume = Volume of the cube = (15 x 15 x 15) cm3.

\Rise in water level = volume/area = (15 x 15 x 15)/(20 x 15) cm = 11.25 cm.

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Filed Under: Quantitative Aptitude - Arithmetic Ability

Q:

A bag contains 7 green and 5 black balls. Three balls are drawn one after the other. The probability of all three balls being green, if the balls drawn are not replaced will be:

A) 123/897	B) 23/67
C) 7/44	D) 12/45

Answer & Explanation Answer: C) 7/44

Explanation:

Here, n(E) = $7 C_{1} \times 5 C_{1} \times 5 C_{1}$

And, n(S) = $12 C_{1} * 11 C_{1} * 10 C_{1}$

P(S) = $\frac{7 * 6 * 5}{12 * 11 * 10}$ = 7/44

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Filed Under: Probability

Q:

A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?

A) 1/6	B) 1/3
C) 1/2	D) 1/4

Answer & Explanation Answer: C) 1/2

Explanation:

P(odd) = P (even) = $\frac{1}{2}$ 1(because there are 50 odd and 50 even numbers)

Sum or the three numbers can be odd only under the following 4 scenarios:

Odd + Odd + Odd = $\frac{1}{2} * \frac{1}{2} * \frac{1}{2}$ = $\frac{1}{8}$

Odd + Even + Even = $\frac{1}{2} * \frac{1}{2} * \frac{1}{2}$ = $\frac{1}{8}$

Even + Odd + Even = $\frac{1}{2} * \frac{1}{2} * \frac{1}{2}$ = $\frac{1}{8}$

Even + Even + Odd = $\frac{1}{2} * \frac{1}{2} * \frac{1}{2}$ = $\frac{1}{8}$

Other combinations of odd and even will give even numbers.

Adding up the 4 scenarios above:

= $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$ + $\frac{1}{8}$ = $\frac{4}{8}$ = $\frac{1}{2}$

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Filed Under: Probability

Q:

A bag contains 12 white and 18 black balls. Two balls are drawn in succession without replacement. What is the probability that first is white and second is black?

A) 6/145	B) 36/145
C) 18/95	D) 12/145

Answer & Explanation Answer: B) 36/145

Explanation:

The probability that first ball is white:
=C112C130=1230=25

Since, the ball is not replaced; hence the number of balls left in bag is 29.
Hence the probability the second ball is black:
=C118C129=1829

Required probability =(25)×(1829)

=36/145

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Filed Under: Quantitative Aptitude - Arithmetic Ability

Q:

A bag contains 6 white and 4 black balls .2 balls are drawn at random. Find the probability that they are of same colour.

A) 1/2	B) 7/15
C) 8/15	D) 1/9

Answer & Explanation Answer: B) 7/15

Explanation:

Let S be the sample space

Then n(S) = no of ways of drawing 2 balls out of (6+4) = $10 C_{2}$ 10 = $\frac{10 * 9}{2 * 1}$ =45

Let E = event of getting both balls of same colour

Then,n(E) = no of ways (2 balls out of six) or (2 balls out of 4)

= $6 C_{2} + 4 C_{2}$ = $\frac{6 * 5}{2 * 1} + \frac{4 * 3}{2 * 1}$ = 15+6 = 21

Therefore, P(E) = n(E)/n(S) = 21/45 = 7/15

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Filed Under: Probability

Q:

Find the probability that one ball is white when two balls are drawn at random from a basket that contains 9 red, 7 white and 4 black balls.

A) 18/95	B) 18/190
C) 1/2	D) 91/190

Answer & Explanation Answer: D) 91/190

Explanation:

Total number of elementary events = $20 C_{2}$ ways =190

There are 7 white balls out of which one white can be drawn in 7 $C_{1}$ ways and one ball from remaining 13 balls can be drawn in $13 C_{1}$ ways.

So,favourable events = $7 C_{1} * 13 C_{1}$

Therfore,required probability = ( $7 C_{1} * 13 C_{1}$ )/ $20 C_{2}$ =91/90

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Filed Under: Probability

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