Mix Test

Result

Mix Test - 3

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following values of x and y satisfy the equation 3x + 4y = 12?

A
x = 4, y = 0

B
x = 0, y = 3

C
x = , y = 1

D
All of the above

Solution

3x + 4y = 12
(4, 0) 3(4) + 4(0) = 12 = RHS
(0, 3) 3(0) + 4(3) = 12 = RHS
+ 4(1) = 12 = RHS
Hence, all the values of x and y satisfy the given equation.
Question 2
1 / -0

If |x²- 25| = 25 - x², then which of the following conditions describes the value of x?

A
x > 5 or x < - 5

B
- 5 < x < 5

C
x = ± 5

D
None of these

Solution

For x² - 25 ≥ 0,
Or x² ≥ 25
Or x ≥ 5 or x ≤ - 5,
We have Ix² - 25I = x² - 25
Thus, the equation becomes
x² - 25 = 25 - x²
Or x = 5 or - 5
For x² - 25 < 0
Or x² < 25
Or -5 < x < 5
Ix² - 25I = -(x² - 25) = 25 - x²
Thus, we have 25 - x² = 25 - x² which shall be true for all -5 < x < 5
Hence, -5 ≤ x ≤ 5
Thus, answer option 4 is correct.
Question 3
1 / -0

If the roots of ax² + b = 0 are real and distinct, then

A
ab > 0

B
a = 0

C
ab < 0

D
a < 0 and b > 0

Solution

Roots of the equation ax² + b = 0 are real and distinct when, D > 0
0² - 4ab > 0 ab < 0
Question 4
1 / -0

In the figure, what is the perimeter of triangle OPQ?

A
4 + 2

B
8 + 4

C
6 + 2

D
6 + 2

Solution

The length of OQ = √(4 + 4) = √8 = √(4 2) = 2√2
The length of OP = √(16 + 16) = √32 = √(16 2) = 4√2
The length of PQ = √{(2 + 4)² + (2 – 4)²} = √(36 + 4) = √40 = √(4 10) = 2√10
So, the perimeter of OPQ = 2√2 + 4√2 + 2√10
= 6√2 + 2√10
Question 5
1 / -0

A chord of a circle of radius 14 cm subtends an angle of 90° at the centre of the circle. Find the area of the corresponding minor sector.

A
22 cm²

B
308 cm²

C
154 cm²

D
462 cm²

Solution

If is the angle subtended by the chord at the centre of a circle and r is the radius of the circle, then the area of minor sector of the circle is .
_{Thus, area of the minor sector of angle 90°}₌ × 14 × 14 cm² = 154 cm²
Question 6
1 / -0

The solution set of 3(2x + 4) > 4x + 10 is

A
x = - 1

B
x > - 1

C
x < 1

D
x - 1

Solution

3(2x + 4) > 4x + 10
6x + 12 > 4x + 10
2x + 12 > 10
2x > - 2
x > - 1
Question 7
1 / -0

The vertices of a triangle ABC are (2, 1), (4, 3) and (2, 5). D, E, F are the mid-points of the sides. Then, the area of the triangle DEF (in square units) is

A
1

B
1.5

C
3

D
4

Solution

The mid points are D (3, 2), E (3, 4), F (2, 3).Area =
= [3 (4 – 3) – 2 (3 – 2) + 1 (9 – 8)] = 1 sq. unit
Question 8
1 / -0

A right circular cone and a hemisphere have equal radii of bases and equal volumes. Find the ratio of their heights.

A
3 : 2

B
3 : 1

C
2 : 1

D
1 : 1

Solution

Volume of cone = = Volume of a hemisphere =
Height of hemisphere = Radius of its base = r
We have to find out h/r.
By the equation formed above, we can clearly see that h/r = 2 : 1.
Question 9
1 / -0

If the largest 4 digit number is substracted from the smallest 6 digit number, then the balance is

A
101112

B
90000

C
90001

D
90011

Solution

Smallest 6 digit number = 100000
Largest 4 digit number = 9999
Hence, required balance = 100000 - 9999 = 90001
Question 10
1 / -0

The lengths of the sides of a rectangular park are in the ratio 3 : 2. If its area is 3750 sq. metres, then the cost of fencing it at the rate of Rs. 5 per metre is

A
Rs. 3125

B
Rs. 3750

C
Rs. 1875

D
Rs. 1250

Solution

Let the lengths of the sides of the rectangle be 3x and 2x m.
Area = 6x² m² = 3750 m²
x² = 625
x = 25
Perimeter = 2 × (3x + 2x) m = 2 × 5x m = 250 m
Cost of fencing = 250 m × Rs. 5/m = Rs. 1250
Question 11
1 / -0

If two parallel lines are intersected by a transversal, then

A
any pair of alternate angles will be equal

B
any pair of corresponding angles will be equal

C
any pair of co-exterior angles will be equal

D
All of the above

Solution

Two parallel straight lines are intersected by a transversal.
All of the given options are correct.
Question 12
1 / -0

By selling 12 articles, a shopkeeper gains a profit equal to the cost price of 2 articles. What is his profit percentage?

A
16.66%

B
17.14%

C
18.28%

D
20.33%

Solution

Assume CP of one article as Re 1.
So, by selling 12 articles (worth Rs. 12), he can make a profit of Rs. 2.
So, profit% = = 16.66%
Question 13
1 / -0

Find the least number of oranges needed to satisfy the thirst of 23 football players if one glass of juice requires 5 oranges and each player drinks 2 glasses of juice.

A
762

B
672

C
230

D
267

Solution

Total number of football players = 23
Each player drinks 2 glasses of juice.
∴ Number of glasses of juice required = 23 × 2 = 46
Number of oranges needed for 1 glass juice = 5
Number of oranges needed for 46 glasses of juice = 46 × 5 = 230
Question 14
1 / -0

Directions: Which of the following options will replace the question mark (?) in the equation given below?

+ + + ………… up to 50 terms = ?

A

B

C

D

Solution

– + – + ……….. + – = 1 – =
Question 15
1 / -0

What are the values of x, y and z in the given figure?

A
x = 118°, y = 32°, z = 30°

B
x = 128°, y = 30°, z = 32°

C
x = 95°, y = 32°, z = 53°

D
x = 118°, y = 32°, z = 35°

Solution

x + 30° + 32° = 180°
x = 180° - 62° = 118°
x = 118°
y = 32°(Vertically opposite angles)
z = 30°(Vertically opposite angles)