u Search

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

Report Error

View Answer Report Error Discuss

Filed Under: Volume and Surface Area

Q:

The volume of a wall, 5 times as high as it is broad and 8 times as long as it is high, is 12.8 cu. meters. Find the breadth of the wall.

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

Answer & Explanation Answer: A) 40cm

Explanation:

Let the breadth of the wall be x metres.

Then, Height = 5x metres and Length = 40x metres.

x * 5x * 40x = 12.8

=> $x^{3} = 12.8 / 200 = 128 / 2000 = 64 / 1000$

=> x = 4/10 m

=> x = (4/10)*100 = 40 cm

Report Error

View Answer Report Error Discuss

Filed Under: Volume and Surface Area

Q:

Find the surface area of a cuboid 16 m long, 14 m broad and 7 m high

A) 868 sq.cm	B) 920 sq.cm
C) 727 sq.cm	D) 900 sq.cm

Answer & Explanation Answer: A) 868 sq.cm

Explanation:

Surface area = [2 (16 x 14 + 14 x 7 + 16 x 7)] sq.cm

= (2 x 434)sq.cm = 868 sq.cm

Report Error

View Answer Report Error Discuss

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

Report Error

View Answer Report Error Discuss

Filed Under: Probability

Q:

What is the probability of getting at least one six in a single throw of three unbiased dice?

A) 1/36	B) 91/256
C) 13/256	D) 43/256

Answer & Explanation Answer: B) 91/256

Explanation:

Find the number of cases in which none of the digits show a '6'.

i.e. all three dice show a number other than '6', 5×5×5=125 cases.

Total possible outcomes when three dice are thrown = 216.

The number of outcomes in which at least one die shows a '6' = Total possible outcomes when three dice are thrown - Number of outcomes in which none of them show '6'.

=216−125=91

The required probability = 91/256

Report Error

View Answer Report Error Discuss

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

Report Error

View Answer Report Error Discuss

Filed Under: Probability

Q:

Two cards are drawn from a pack of well shuffled cards. Find the probability that one is a club and other in King

A) 1/52	B) 1/26
C) 1/13	D) 1/2

Answer & Explanation Answer: B) 1/26

Explanation:

Let X be the event that cards are in a club which is not king and other is the king of club.
Let Y be the event that one is any club card and other is a non-club king.
Hence, required probability:
=P(A)+P(B)

= $[\frac{12 C_{1} * 1 C_{1}}{52 C_{2}}] + [\frac{13 C_{1} * 3 C_{1}}{52 C_{2}}]$

= $[12 * \frac{2}{52 * 51}] + [13 * 3 * \frac{2}{52 * 51}]$ = $\frac{24 + 78}{52 * 51}$ = $\frac{1}{26}$

Report Error

View Answer Report Error Discuss

Filed Under: Probability

Q:

The probability of success of three students X, Y and Z in the one examination are 1/5, 1/4 and 1/3 respectively. Find the probability of success of at least two.

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

Answer & Explanation Answer: C) 1/6

Explanation:

P(X) = $\frac{1}{5}$ , P(Y) = $\frac{1}{4}$ , P(Z) = $\frac{1}{3}$

Required probability:

= [ P(A)P(B){1−P(C)} ] + [ {1−P(A)}P(B)P(C) ] + [ P(A)P(C){1−P(B)} ] + P(A)P(B)P(C)

= $(\frac{1}{4} * \frac{1}{3} * \frac{4}{5}) + (\frac{3}{4} * \frac{1}{3} * \frac{1}{5}) + (\frac{2}{3} * \frac{1}{4} * \frac{1}{5}) + (\frac{1}{4} * \frac{1}{3} * \frac{1}{5})$

= $\frac{4}{60} + \frac{3}{60} + \frac{2}{60} + \frac{1}{60}$ = $\frac{10}{60}$ = $\frac{1}{6}$

Report Error

View Answer Report Error Discuss

Filed Under: Probability

Searching for "u"

Quick Links