SAT (Stage-1) Mock Test

Result

SAT (Stage-1) Mock Test - 83

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Weekly Quiz Competition

Question 1
1 / -0

If the sum of the first 5 terms of an AP is 160 and its first term is 6, find the common difference.

A
11

B
12

C
13

D
16

Solution
Question 2
1 / -0

In the given figure, AB and AC are equal chords and BCDE is a cyclic quadrilateral. If∠DEX = 80° and ∠ACD = 30°, what is the measure of ∠CAB?

A
100°

B
90°

C
80°

D
70°

Solution

Given, ∠DEX = 80°

So, ∠DCB = 80°(Cyclic quadrilateral)

And, ∠DCA = 30°

∠DCA + ∠ACB = 80°

30°+ ∠ACB = 80°

∠ACB = 50°

Since, AB = AC

∠ACB = ∠ABC = 50°

∠BAC = 180° - (50° + 50°)

= 80°
Question 3
1 / -0

If AB and AC are two tangents to a circle with centre O, which of the following relationships is true?

A
AB + OB = AC + OC

B
AC || OB

C
AB = OC

D
AB + OB = AC

Solution

The lengths of the tangents drawn from an external point to a circle are equal.

Thus, in the given figure, OB and OC are the radii of the circle.
⇒ OB = OC ...(i)
Also, AB = AC ...(ii) [∵ AB and AC are tangents to the circle]
By adding equations (i) and (ii), we get
OB + AB = OC + AC
AB + OB = AC + OC
Question 4
1 / -0

If one of the diagonals of a square has vertices (1, 2) and (4, 6), what is the area of the square?

A
2.5 sq. units

B
5.0 sq. units

C
12.5 sq. units

D
25.0 sq. units

Solution
Question 5
1 / -0

The pair of equations 2x + 4y + 10 = 0 and -6x - 12y + 2 = 0 has

A
a unique solution

B
exactly two solutions

C
infinitely many solutions

D
no solution

Solution

From the given equations, we can conclude that:

Hence, the lines are parallel and have no solution.
Question 6
1 / -0

If the system of equations 2kx + 6y = 1 and 6x + 2ky = 2p - 1 is inconsistent, which of the following cannot be the value of p?

A
3

B
1

C
2

D
- 3

Solution

For inconsistent,

i.e. 4k² = 36

or, k = ± 3

So, 2p - 1 ≠ ±1

2p ≠ 0 or, 2p ≠ 2

p ≠ 0 or, p ≠ 1
Question 7
1 / -0

On rationalising the denominator of , we get . What are the values of a, b and c respectively?

A
2, 3, 6

B
2, 6, 3

C
3, 2, 6

D
6, 2, 3

Solution
Question 8
1 / -0

The probability that a leap year selected at random will contain 53 Sundays is

A
1/7

B
2/7

C
7/366

D
26/183

Solution

A leap year has 366 days.
366/7 = 52, remainder = 2
⇒ There are 52 weeks.
⇒ 52 Sundays, 2 days extra;
For 53 Sundays, either one of these should be a Sunday.
So, P (53 Sundays) = P (1^st day is Sunday) + P (2^nd day is Sunday)
= (1/7) + (1/7) = 2/7
Question 9
1 / -0

If α, β are the roots of the equation 2x² - 7x + 5 = 0, then α³ + β³is equal to

A
273/8

B
7

C
133/8

D
None of these

Solution
Question 10
1 / -0

Which of the following equations can be obtained after completing the square of 3x² - 11x - 4 = 0?

A

B

C

D

Solution
Question 11
1 / -0

Work efficiency of two persons is in the ratio 2 : 3. Work assigned to them is in the ratio 4 : 5. Find the ratio of the time taken by them, if they start at the same time.

A
4 : 5

B
6 : 5

C
5 : 6

D
2 : 3

Solution

According to the given condition,
Let work efficiency of A be 2x and that of B be 3x.
Work done by A is 4y and that by B be 5y.
Ratio of time taken by them is:
4y/2x : 5y/3x = 2 : 5/3 = 6 : 5
Question 12
1 / -0

For a grouped data, if l = 45, h = 15, f₁ = 35, f₀= 30 and mode = 54.375, then what is the value of f₂?

Given:
l = Lower limit of the modal class
h = Size of the class interval
f₁ = Frequency of the modal class
f₀ = Frequency of the class preceding the modal class
f₂ = Frequency of the class succeeding the modal class

A
22.5

B
32

C
48

D
38.4

Solution
Question 13
1 / -0

Look at the pie chart that shows the number of students who like different colours and answer the question below.

One student is selected randomly from the group. Find the probability that the student likes orange or pink or peach colour.

A
51/100

B
15/49

C
16/49

D
49/100

Solution

Number of students who like orange or pink or peach colour = 16 + 15 + 18 = 49
Total number of students = 16 + 15 + 17 + 19 + 18 + 15 = 100
Probability = 49/100
Question 14
1 / -0

What is the total surface area of a box with length, breadth and height as 16 cm, 8 cm and 6 cm, respectively?

A
680 cm²

B
136 cm²

C
272 cm²

D
544 cm²

Solution

Length = = 16 cm
Breadth = b = 8 cm
Height = h = 6 cm
Total surface area = 2(b + bh + h)
= 2(16 ×8 + 8 ×6 + 6 ×16) cm²
= 2(128 + 48 + 96) cm²
= 2 ×272 cm² = 544 cm²
Question 15
1 / -0

What is the maximum number of cubes of side 5 cm that can be obtained from a cube of side 20 cm?

A
24 cubes

B
60 cubes

C
64 cubes

D
84 cubes

Solution

Side of the bigger cube (Side₁) = 20 cm
Volume of the bigger cube (Volume₁) = (20)³= 8000 cm³
Side of the smaller cube (Side₂) = 5 cm
Volume of the smaller cube (Volume₂) = (5)³ = 125 cm³

Number of smaller cubes with side 5 cm =