SAT (Stage-2) Mock Test

Result

SAT (Stage-2) Mock Test - 37

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Question 1
1 / -0

A test has 50 questions. A student scores 1 mark for a correct answer, -1/3 for a wrong answer and -1/6 for not attempting a question. If the net score of a student is 32, the number of questions answered wrongly by that student cannot be less than

A
6

B
12

C
3

D
9

Solution

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.
Question 2
1 / -0

A
0

B
1

C
-1

D
√-1

Solution

Alternate Method:
Go with option checking.
Question 3
1 / -0

If the zeros of the quadratic polynomial ax²+ bx + c (c ^_≠ 0) are equal, then

A
c and a have opposite signs

B
c and b have opposite signs

C
c and a have the same sign

D
c and b have the same sign

Solution

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.
Question 4
1 / -0

A
one integral root

B
two real roots

C
infinitely many roots

D
no root

Solution

Solving the equation, x = 3 is the solution. But since at x = 3 the equation becomes undefined, it is not a valid solution. So, the equation has no roots.
Question 5
1 / -0

The sum of n terms of the series 1 + (1 + 3) + (1 + 3 + 5) + ... is

A
n²

B

C

D
None of the above

Solution

n^th term of the given series is:
a_n = 1 + 3 + 5 + ... + (2n - 1) = n²
Question 6
1 / -0

A

B

C

D
None of these

Solution
Question 7
1 / -0

A

B

C
p/q

D

Solution
Question 8
1 / -0

Two circles touch each other externally such that the shortest distance between their centres is 12 cm and the sum of their areas is 74π cm². What is the difference between their radii?

A
4 cm

B
2 cm

C
1 cm

D
0 cm

Solution

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
Question 9
1 / -0

A metallic spherical shell of internal and external diameters respectively 4 cm and 8 cm is melted and recast into a cone of base diameter 8 cm. The height of the cone is

A
12 cm

B
14 cm

C
15 cm

D
18 cm

Solution

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
Question 10
1 / -0

If two solid hemispheres of the same radius 'r' are joined together along their bases, then the curved surface area of this new solid would be

A
4πr²

B
6πr²

C
3πr²

D
8πr²

Solution

Two hemispheres of radius r will join to form a sphere.
Surface area of a sphere = 4πr²
Question 11
1 / -0

In the given figure, BC is a diameter of the circle and ^_∠BAO= 60°. Then, ∠ADC is equal to

A
30°

B
45°

C
60°

D
120°

Solution

Angle ABO is equal to 60° since OA and OB being the radii of the circle are of equal length.
Now, angle ADC = angle ABO = 60° (because they are angles on the same chord)
Question 12
1 / -0

In the figure given below, if PQR is a tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to

A
20°

B
40°

C
35°

D
45°

Solution

Angle AQP = Angle BQR = 70° (alternate angles)
Angle AQB = 180° - (70° + 70°) = 40°
Question 13
1 / -0

If point C(a, a) is the circumcentre of a triangle formed by vertices D(2, 3), E(3, 2) and F(1, 1), then what is the value of a?

A
2

B
3/2

C
7/6

D
11/6

Solution

As C is the circumcentre of triangle DEF, the distance from C to the points D, E and F will be the same.
i.e. CF = CE
Question 14
1 / -0

If the point P(6, 3) lies on the line segment joining the points A(4, 2) and B(8, 4), then

A

B
AP = AB

C

D

Solution
Question 15
1 / -0

A grouped frequency table with class intervals of equal size, using 250 - 270 (270 not included in this interval) as one of the class intervals, is constructed for the following data:

268, 220, 368, 258, 242, 310, 272, 342, 310, 290, 300, 320, 319, 304, 402, 318, 406, 292, 354, 278, 210, 240, 330, 316, 406, 215, 258, 236

The frequency of the class 310 - 330 is

A
4

B
5

C
6

D
7

Solution

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.