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Q:

A rectangular block 6 cm by 12 cm by 15 cm is cut up into an exact number of equal cubes. Find the least possible number of cubes.

A) 30	B) 40
C) 10	D) 20

Answer & Explanation Answer: B) 40

Explanation:

Volume of the block = (6 x 12 x 15) cu.cm = 1080 cu.cm

Side of the largest cube = H.C.F. of 6 cm, 12 cm, 15 cm

= 3 cm.

Volume of this cube = (3 x 3 x 3) cu.cm = 27 cu.cm

Number of cubes = $\frac{1080}{27}$ = 40.

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Filed Under: Volume and Surface Area

Q:

Find the number of bricks, each measuring 24 cm x 12 cm x 8 cm, required to construct a wall 24 m long, 8m high and 60 cm thick, if 10% of the wall is filled with mortar?

A) 35000	B) 45000
C) 55000	D) 65000

Answer & Explanation Answer: B) 45000

Explanation:

Volume of the wall = (2400 x 800 x 60) cu.cm

Volume of bricks = 90% of the volume of the wall

= [(90/100) x 2400 x 800 x 60] cu.cm

Volume of 1 brick = (24 x 12 x 8) cu.cm

Number of bricks = [(90/100) x (2400 x 800 x 60) ]/ (24 x 12 x 8) = 45000

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Filed Under: Volume and Surface Area

Q:

I forgot the last digit of a 7-digit telephone number. If 1 randomly dial the final 3 digits after correctly dialing the first four, then what is the chance of dialing the correct number?

A) 1/999	B) 1/1001
C) 1/1000	D) 4/1000

Answer & Explanation Answer: C) 1/1000

Explanation:

It is given that last three digits are randomly dialled. then each of the digit can be selected out of 10 digits in 10 ways.
Hence required probability = ${(\frac{1}{10})}^{3}$ = 1/1000

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Filed Under: Probability

Q:

In a charity show tickets numbered consecutively from 101 through 350 are placed in a box. What is the probability that a ticket selected at random (blindly) will have a number with a hundredth digit of 2?

A) 0.25	B) 0.5
C) 0.75	D) 0.40

Answer & Explanation Answer: D) 0.40

Explanation:

250 numbers between 101 and 350 i.e. n(S)=250

n(E)=100th digits of 2 = 299−199 = 100

P(E)= n(E)/n(S) = 100/250 = 0.40

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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 number X is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3. What is the probability that |X|<2

A) 3/7	B) 3/4
C) 4/5	D) 5/7

Answer & Explanation Answer: A) 3/7

Explanation:

X can take 7 values.
To get |X|+2) take X={−1,0,1}

=> P(|X|<2) = Favourable CasesTotal Cases = 3/7

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Filed Under: Probability

Q:

Two dice are thrown together .What is the probability that the sum of the number on the two faces is divided by 4 or 6.

A) 7/18	B) 14/35
C) 8/18	D) 7/35

Answer & Explanation Answer: A) 7/18

Explanation:

Clearly, n(S) = 6 x 6 = 36
Let E be the event that the sum of the numbers on the two faces is divided by 4 or 6.
Then,E = {(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2),(6,6)}
n(E) = 14.
Hence, P(E) = n(E)/n(S) = 14/36 = 7/18

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Filed Under: Probability

Q:

A bag contains 50 tickets numbered 1,2,3,4......50 of which five are drawn at random and arranged in ascending order of magnitude.Find the probability that third drawn ticket is equal to 30.

A) 551/15134	B) 1/2
C) 552/15379	D) 1/9

Answer & Explanation Answer: A) 551/15134

Explanation:

Total number of elementary events = $50 C_{5}$
Given,third ticket =30

=> first and second should come from tickets numbered 1 to 29 = $29 C_{2}$ ways and remaining two in $20 C_{2}$ ways.

Therfore,favourable number of events = $29 C_{2} * 20 C_{2}$

Hence,required probability = $29 C_{2} * 20 C_{2}$ $/ 50 C_{5}$ =551 / 15134

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Filed Under: Probability

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