SAT (Stage-2) Mock Test - 4

Let c, w and n be the numbers of questions correct, wrong, and not attempted, respectively.
c + w + n = 50 … (i)
c - w/3 - n/6 = 32
6c - 2w - n = 32 × 6 = 192 … (ii)
(i) + (ii)
7c - w = 242
w = 7c - 242
The minimum value of w is 3, when c = 35.
Alternative method: You can solve this question by option also. Because the net score is an integer (32), the number of wrong questions must be a multiple of 3 and also the number of unattempted questions a multiple of 6. So by manipulation and with the help of options you can do this question faster.

For roots to be equal, b² - 4ac = 0
Now, b² is always positive.
Thus, 4ac must be positive for the above equation to be 0.
This is only possible if both a and c are positive or both are negative.
Thus, a and c must have the same sign.

Let the radius of one circle be r₁ cm. Then, the radius of the other circle is (12 - r₁) cm.
Area of circle = π(radius of circle)²
Given: Sum of areas = 74π
πr₁² + π(12 - r₁)² = 74π
On solving, r₁ = 7 or 5 cm
Then, radius of the other circle is 5 cm or 7 cm.
Difference = 2 cm

The diameters are 8 cm and 4 cm. So, the radii are 4 cm and 2 cm.
Volume of sphere = ⁴/₃π(R³ - r³) = ⁴/₃π(4³ - 2³) = ²²⁴/₃π
Since the metal is melted, Volume of sphere = Volume of cone
Diameter of cone = 8 cm
So, radius of cone = 4 cm
Volume of cone = ¹/₃πr²h
i.e. ¹/3π4²h = ²²⁴/3π
h = 14 cm

Two hemispheres of radius r will join to form a sphere.
Surface area of a sphere = 4πr²

The readings in the class interval 310 - 330 are 310, 310, 320, 319, 318, 316.
As there are 6 readings in this class, its frequency is 6.