SAT (Stage-1) Mock Test

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SAT (Stage-1) Mock Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

If sinα and cosα are the roots of equation ax² + bx + 1 = 0, then which of the following is true?

A

B

C
a² + b² = 2a

D
a(a + 2) = b²

Solution
Question 2
1 / -0

What is the equation of median on side AC of ΔABC whose vertices are A(-3, -2), B(-1, 1) and C(2, -2)?

A
y + 6x = -5

B
x - y = 2

C
x + y = 2

D
y - x = 3

Solution

Let the mid-point of AC be 'D'.
Question 3
1 / -0

In parallelogram ABCD, AB = 16 cm. The altitudes corresponding to the sides DC and AD are 8 cm and 10 cm, respectively. Find AD.

A
12.8 cm

B
14.0 cm

C
18.0 cm

D
3.4 cm

Solution

Area of parallelogram = CD × AE = AD × FC

AB × AE = AD × FC (∵ AB = CD)
16 × 8 = AD × 10
AD = 12.8 cm
Question 4
1 / -0

ΔLMN ~ ΔXYZ and ar(ΔXYZ) = 121 cm². If LM/XY = 7/11, then ar(∆LMN) is equal to

A
49 cm²

B
36 cm²

C
25 cm²

D
16 cm²

Solution
Question 5
1 / -0

In the given figure, O is the centre of the circle and PQ is a tangent to the circle. What is the measure of ∠OBQ?

A
25°

B
20°

C
15°

D
10°

Solution

In ΔAPQ,

55° + 25° + ∠QAP = 180°

∠QAP = 180° - 80° = 100°

Now, ∠QAB + ∠QAP = 180°

∠QAB = 80°

So, ∠QOB = 2 ∠QAB = 160°

Further, in ΔQOB,

∠QOB + 2∠OBQ = 180°

∠OBQ = (180° - 160°)/2

∠OBQ = 10°
Question 6
1 / -0

In the given figure, ABC is a right-angled triangle and BD ⊥AC. Which of the following relationships is true?

A
AB : BC : : AD : DC

B
AB : BC : : AD² : DC²

C

D

Solution
Question 7
1 / -0

If volume of a hemisphere is 144π m³, then what is its total surface area?

A

B

C
72π m²

D
108π m²

Solution
Question 8
1 / -0

The lateral surface area and height of a prism whose base is an equilateral triangle, are 54 m² and 2√3 m, respectively. What is the area of the base of the prism?

A

B

C

D

Solution
Question 9
1 / -0

A cylindrical vessel, whose radius is 6 cm, contains water. If a solid sphere of radius 3 cm is dipped into the vessel completely, then what is the rise in the water level in the vessel?

A
3 cm

B
2 cm

C
1 cm

D
0.5 cm

Solution

When solid sphere is completely inserted in the cylinder, water level rises in cylinder.

Let the new height be 'H'.

Total volume of cylinder with height 'H' = Total volume of initial cylinder + Volume of sphere

36 H = 36 h + 36
36 (H - h) = 36
H - h = 1 cm
Rise in the water level in the vessel = 1 cm
Question 10
1 / -0

If cosec (α + β) = sec (α - β) = 2/√3, then what is the value of cot 2α?

A
0

B
1/√3

C
1

D
√3

Solution

sin (α + β) = sin (60°), cos (α - β) = cos (30°) (Minimum value of α and β; α > 0, β > 0)

α + β = 60°

α - β = 30°

Adding these equations,

2α = 90°

cot (2α) = cot (90°) = 0

Question 11

1 / -0

What is the median of marks of 20 students in the following frequency distribution?

Marks	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50
Number of students	3	4	5	6	2

15.5

26

25.5

35.5 Solution

Class interval	Frequency	Commulative frequency
0 - 10	3	3
10 - 20	4	7
20 - 30	5	12
30 - 40	6	18
40 - 50	2	20

Median = l + (((n/2) - cf)/f) × h
Here, n = 20
10th item lies in (20 - 30) class.
l = 20
h = 10
f = 5
cf = 7

Median = 20 + 6 = 26

Question 12
1 / -0

If the mean of the following frequency distribution is 14, then what is the value of P?

Marks 0 - 10 10 - 20 20 - 30

Number of students 3 P 2

A
4

B
5

C
6

D
7

Solution

Class interval x_i f_i x_i f_i

0 - 10 5 3 15

10 - 20 15 P 15P

20 - 30 25 2 50

Total x 5 + P 65 + 15P

⇒ 65 + 15P = 70 + 14P

⇒ P = 5
Question 13
1 / -0

Four persons A, B, C and D marked four identical articles at Rs. 5000 each. A sold the article at a discount of 60%, B sold the article at two successive discounts of 60% and 20%, C sold the article at two successive discounts of 40% each and D sold the article at a discount of 70%. What was the descending order of selling prices for A, B, C and D?

A
A, C, B, D

B
C, B, A, D

C
D, A, B, C

D
D, B, A, C

Solution

A sold at a discount of 60%.

S.P. for A = 40% of 5000 = 2000

B sold at a discount of 60% and 20%.

Total discount = 60% + 20% - (60 × 20)/100 = 68%

S.P. for B = (100 - 68)% of 5000 = 1600

C sold at a discount of 40% and 40%.

Total discount = 40% + 40% - (40 × 40)/100 = 64%

S.P. for C = (100 - 64)% of 5000 = 1800

D sold at a discount of 70%.

S.P. for D = 30% of 5000 = 1500

S.P. comparison: A > C > B > D
Question 14
1 / -0

A man can row a boat 20 km upstream in 2 hours and 40 km downstream in the same time. What is the ratio of speed of the stream and that of the boat in still water?

A
1 : 3

B
2 : 1

C
3 : 1

D
1 : 2

Solution

Let the speed of boat in still water = B and the speed of the stream = S

(B - S)2 = 20

B - S = 10 ...Eqn (1)

And, (B + S)2 = 40

B + S = 20 ... Eqn (2)

Adding (1) and (2),

2B = 30

B = 15

S = 20 - 15 = 5

Ratio of the stream and the boat = 5/15 = 1/3 i.e. 1 : 3
Question 15
1 / -0

The number of terms between 30 and 530 which are divisible by 11 are

A
46

B
40

C
36

D
None of these

Solution

The terms are 33, 44, …, 528.
This is an AP with a common difference 11.
Let 'n' be the number of terms.
528 = 33 + (n - 1)11
495 = 11(n - 1)

n - 1 = 495/11 = 45
n = 46

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Class interval	x_i	f_i	x_i f_i
0 - 10	5	3	15
10 - 20	15	P	15P
20 - 30	25	2	50
Total	x	5 + P	65 + 15P

SAT (Stage-1) Mock Test - 50

Result

SAT (Stage-1) Mock Test - 50

Score

-

Rank

-

SHARING IS CARING

Question 1

a2 + b2 = 2a

a(a + 2) = b2

Solution

Question 2

y + 6x = -5

x - y = 2

x + y = 2

y - x = 3

Solution

Question 3

12.8 cm

14.0 cm

18.0 cm

3.4 cm

Solution

Question 4

49 cm2

36 cm2

25 cm2

16 cm2

Solution

Question 5

25°

20°

15°

10°

Solution

Question 6

AB : BC : : AD : DC

AB : BC : : AD2 : DC2

Solution

Question 7

72π m2

108π m2

Solution

Question 8

Solution

Question 9

3 cm

2 cm

1 cm

0.5 cm

Solution

Question 10

0

1/√3

1

√3

Solution

Question 11

15.5

26

25.5

35.5

Solution

Question 12

4

5

6

7

Solution

Question 13

A, C, B, D

C, B, A, D

D, A, B, C

D, B, A, C

Solution

Question 14

1 : 3

2 : 1

3 : 1

a² + b² = 2a

a(a + 2) = b²

49 cm²

36 cm²

25 cm²

16 cm²

AB : BC : : AD² : DC²

72π m²

108π m²