SAT (Stage-1) Mock Test

Result

SAT (Stage-1) Mock Test - 63

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Weekly Quiz Competition

Question 1
1 / -0

In the given figure, PQO is an equilateral triangle and O is the centre of the circle. Points P and Q lie on the circle. If the diameter of the circle is 16 cm, then what is the area of the shaded region?

A

B

C

D

Solution
Question 2
1 / -0

Let x and y be real numbers such that (x² – y²)(x² – 2xy + y²) = 3 and x – y = 1. What is the value of xy?

A
2

B
1 + √2

C
1 – √2

D
1

Solution

(x² – y²)(x² – 2xy + y²) = 3
Or, (x – y)(x + y)(x² – 2xy + y²) = 3
Or, (x – y)(x + y)(x – y)² = 3
x + y = 3
We have, x + y = 3 and x – y = 1
Adding two, we get 2x = 4
i.e. x = 2
and y = 1
Value of xy = 2
Question 3
1 / -0

The sum of the first n terms of an AP is zero. If the first term and the common difference of the AP are 4 and – 2, then what is the value of n?

A
3

B
6

C
5

D
4

Solution

Here, S = 0, a = 4 and d = –2.

=> 10n – 2n²= 0
=> 2n(n – 5) = 0
n = 5 or n = 0 (n cannot be 0)
n = 5
Question 4
1 / -0

In the given figure, O is the centre, AB is the diameter and ∠ACL = ∠LCO = 20°. What is the measure of ∠DCA?

A
40°

B
50°

C
60°

D
70°

Solution

The angle between the radius of a circle and the tangent drawn to the circle is equal to 90°.

Thus, in the given figure, ∠DCO = 90°
Also, ∠DCO = ∠DCA + ∠ACL + ∠LCO
90° = ∠DCA + 20° + 20°
⇒ ∠DCA = 90° - 40°
∠DCA = 50°
Question 5
1 / -0

Area of the triangle formed by the points (1, -3), (3, 3) and (5, -3) is

A
8 sq. units

B
10 sq. units

C
12 sq. units

D
14 sq. units

Solution
Question 6
1 / -0

Which of the following options is the x–coordinate of the point dividing the line segment joining the points (m, n) and (n, m) in the ratio n : m internally?

A

B
0

C

D
1

Solution
Question 7
1 / -0

Mr. Problem wants to buy pens as well as pencils, and he has Rs. 190 with him. Every pen costs Rs. 15 and every pencil costs Rs. 8. What is the maximum number of items he can buy, if he has to exhaust all the money he has?

A
15

B
18

C
20

D
22

Solution

Let the number of pen and that of pencil be x and y, respectively.

Then, 15x + 8y = 190

Maximum positive integral values of x and y satisfying the equation are given below:

x y

10 5

2 20

Maximum number of pens and pencils together is 20 pencils and 2 pens.
i.e. For full money division, he can buy 2 pens and 20 pencils.
2 × 15 + 20 × 8 = 30 + 160 = 190
Total items = 20 + 2 = 22
Question 8
1 / -0

A covered wooden box has the inner measures as 115 cm, 75 cm and 35 cm and the thickness of the wood is 2.5 cm. Find the volume of the wood.

A
80,000 cm³

B
82,125 cm³

C
84,150 cm³

D
85,000 cm³

Solution

Thickness of the wood = 2.5 cm

Now, external dimensions of the wooden box are as follows.

Length = 120 cm, width = 80 cm and height = 40 cm

Volume of the wood = (120 cm × 80 cm × 40 cm) - (115 cm × 75 cm ^_× 35 cm)

= (384,000 - 301,875) cm³ = 82,125 cm³
Question 9
1 / -0

By which of the following expressions should the polynomial 6x² - 13x + 10 be divided so that the quotient is 3x - 2 and the remainder is 4?

A
(2x + 5)

B
(2x + 3)

C
(2x - 5)

D
(2x - 3)

Solution
Question 10
1 / -0

If one of the zeroes of the cubic polynomial x³+ ax² + bx + c is –1, then the product of the other two zeroes is

A
b – a + 1

B
b – a – 1

C
a – b

D
b – a

Solution

Let p(x) = x³ + ax² + bx + c

Let the zeroes be z₁, z₂and –1.

z₁z₂(-1) = -c

Or, z₁z₂ = c = Product of the other two zeroes ... (i)

Now, as one zero is –1, p(-1) = 0

Or, (-1)³ + a(-1)² + b(-1) + c = 0

Or, -1 + a – b + c = 0

Or, b + 1 – a = c … (ii)

Equating the left hand sides of equations (i) and (ii),

z₁z₂ = b + 1 – a
Question 11
1 / -0

Which of the following gives the H.C.F. of 121, 49, 63 and 841?

A
1

B
11

C
14

D
18

Solution

These four numbers have only 1 as the H.C.F.

Factors of 63 = 1 × 3 × 3 × 7

Factors of 121 = 1 × 11 × 11

Factors of 49 = 1 × 7 × 7

Factors of 841 = 1 × 29 × 29

So, the H.C.F. of 49, 63,121 and 841 = 1
Question 12
1 / -0

The mean of 12, 10, p, p + 2 and 11 is 9. What is the value of p?

A
8.5

B
7.5

C
6.0

D
5.0

Solution
Question 13
1 / -0

In the given figure, DE || BC, AD/DB = 3/1 and ar(∆ADE) = 9 units². What is the area of ∆ABC?

A
10 units²

B
11 units²

C
12 units²

D
16 units²

Solution

AD = 3k units

DB = 1k unit

Now, AB = AD + BD = 4k units

Now, using the property of similar triangles,
Question 14
1 / -0

In ΔABC, DE is parallel to BC and it divides the triangle into two equal parts. The ratio of the area of ΔADE to that of ΔABC is

A
1 : 1

B
1 : 2

C
2 : 1

D
2 : 3

Solution

DE ∥ BC and D , E is the mid point of AB and AC SO, DE = BC/2

⇒ Area of ∆ABC = 2(Area of ∆ADE)
Question 15
1 / -0

In the given figure, X/U/UY = XV/VZ. Which of the following relationships is correct?

A
UV = YZ

B
∠XYZ = ∠XZY

C
∠XUV = ∠XYZ

D
XV = VZ

Solution

UV/UY = XV/VZ (Given)

UV || YZ (By Basic Proportionality Theorem)

Therefore, ∠XUY = ∠XYZ (Corresponding angles)