Quantitative Aptitude - Arithmetic Ability Questions

729 = 9 x 9 x 9

=> Cube root of 729 = 9

Now, required square of 9 = 9 x 9 = 81.

Let the number of correct answers be X.

Number of incorrect answers = (60 – X).

Therefore, 4x – (60 – x) = 130 

=> 5x = 190  => x = 38

Total number of exams written by 100 students = 48 + 45 + 38 = 131
Now let us say x members are writing only 1 exam, y members are writing only 2 exams, z members are writing only 3 exams.
Therefore, x + 2y + 3z = 131 also x + y + z = 100.
Given that z = 5. So x + 2y = 116 and x + y = 95.
Solving we get y = 21.
So 21 members are writing exactly 2 exams.

Let S be the sample space.
Here n(S)= 23 = 8
Let E be the event of getting atmost two heads. Then,
n(E) = {(H,T,T), (T,H,T), (T,T,H), (H,H,T), (T,H,H), (H,T,H)}
Required probability = n(E)/n(S) = 6/8 = 3/4.

It takes four of those little squares -- each one 1/2 inch on a side -- to cover one square inch. That's because each one is 1/2 x 1/2, or 1/4 of a square inch.
In a rectangle that is 48 inches by 48 inches, there are 2304 square inches.(1 ft = 12 inches)
Since it takes 4 little squares to cover 1 sq inch, then it would need 2304 x 4 squares (1/2 inch on a side each) to cover a rectangle 4 feet by 4 feet.

That is 9,216 of those squares.

After servicing, the distance covered by car in 6 hours = 60 * 6 = 360 km Without servicing, speed of car = 50 kmph => Required time = Distance/speed = 360/50 = 7.2 hrs.

When two dice are thrown simultaneously, the probability is n(S) = 6x6 = 36

Required, the sum of the two numbers that turn up is less than 12

That can be done as n(E)

= { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5) }

= 35

Hence, required probability = n(E)/n(S) = 35/36.