MCQ questions (1 Marks) for SA2 mathematics
1. There are three non-collinear points. The number of circles passing through them are:
(a) 2 (b) 1
(c) 3 (d) 4
Answer:(b) 1
There can only be one and only one circle that can pass through any three collinear points.
2. In the given figure, OM | to the chord AB of the circle with centre O. If OA = 13 cm and AB = 24 cm, then OM equals:
(a) 3 cm (b) 4 cm
(c) 5 cm (d) √4.7
Answer:(c) 5 cm
Given that ∠OMA = 90° .
ΔAOM and ΔBOM are congruent and hence AM=AB and therefore,
AM = (1/2)24 = 12 cm.
Using Pythagoras theorem, we have
OM = √AO² - AM²
=> OM = √13² - 12²
=> OM = √169 - 144 = √25 = 5 cm
3. In the given figure, a circle is centred at O, the value of x is:
(a) 55° (b) 70°
(c) 110° (d) 125°
Answer:(c) 110°
OC = OB (radius of the circle)
=> ∠OCB = ∠OBC
=> ∠OCB = 35°
Again, OC = OA (radius of the circle)
=> ∠OCA = ∠OAC
=> ∠OCA = 20°
Therefore, ∠ACB = ∠OCB + ∠OCA = 35 +20 = 55°
And ∠AOB = 2 x 55 = 110 °
(Angle subtended at the centre of the circle by an arc is double of the angle subtended at any point on the circle )
4. In figure, O is the centre of the circle and .∠OBA = 60. Then ∠ACB equals
(a) 60° (b) 45°
(c) 30° (d) 90°
Answer:(c) 30°
Given that,
∠OBA = 60 °
and OA=OB (radius of circle )
But ∠OAB = 60 ° (because in a triangle angles opposite to opposites sides are equal)
Again, in triangle OAB, we have
∠OAB + ∠OBA + ∠AOB = 180°
=> 60° + 60° + ∠AOB = 180°
=> ∠AOB = 180° - 120° = 60°
Therefore, ∠ACB = (1/2)(∠AOB)
=> ∠ACB = (1/2)(60°) = 30°
5. If ∠C = ∠D = 50, then four points A, B, C, D :
(a) Are con cyclic (b) Do not lie on same circle
(c) Are collinear (d) A, B, C and A, B, C lie on different circles.
Answer: (a) Are con cyclic
Since ∠C= ∠D = 50, Since angles on the same side of the line segment are equal therefore they are concyclic.
6. The distance of chord length 16 cm from the centre of the circle of radius 10 cm is
(a) 6 cm (b) 8 cm
(c) 10 cm (d) 12 cm
Answer: The distance of the chord = √(10)² - 8²
= √(10)² - 8²
= √100 - 64
= √36
= 6 cm
7. The distance of a chord 8 cm long chord from the centre of a circle of radius of 5 cm is :
(a) 50° (b) 60°
(c) 65° (d) 70°
Answer: 3 cm
Distance of the chord from the centre of the circle = √5² - 4²
= √25 -16
= √9
= 3 cm
8. Area of an isosceles right triangle is 8 cm². Its hypotenuse is:
(a) √32 cm (b) 4 cm
(c) 4√3 cm (d) 2√6 cm
Answer:√32 cm
Area of triangle = (1/2)(b)(h)
8 = (1/2)(b)(b) (Since its an isosceles triangle)
=> 16 = (b)(b)
=> b = 4 cm
Again the hypotenuse = √4² + 4² = √2.4² = √32 cm
9. Side of an equilateral triangle is 4 cm. Its area is :
(i) 4√3 (ii) √3/4
(iii) √3 (iv) 2 √3
Answer: The area of an equilateral triangle = (√3/4)(side)² = (√3/4)(4)² = 4 √3 cm²
10.A square and an equilateral triangle have equal perimeters. If the diagonal of the square is 12√2, then the area of the triangle (in cm² ) is :
(a) 24√ 2 (b) 25√3
(c) 48√ 3 (d) 64√3
Answer: (d) 64√3
Relationship between diagonal of the square and the side of the square is
Diagonal of a square of side S = √2 S where S is the side of the square.
=> 12√2 = √2 S
=> S = 12 cm
And therefore the perimeter of the square = 4x side = 4x12 = 48 cm
Again, the perimeter of an equilateral triangle with side a = 3a
But given that the perimeter of the given square is equal to the perimeter of the square.
=> 3a = 48
=> a = 48/3 = 16 cm
And therefore area of the equilateral triangle = (√ 3/4)a² = (√ 3/4)(16)² = 64√ 3 cm²
11. The area of the ΔABC in which AB=BC= 4 cm and ∠B =90° is :
(a) 16 cm² (b) 8 cm²
(c) 4 cm² (d) 12 cm²
Answer: (b) 8 cm²
From the condition given in the question it is clear that ΔABC is a right angle triangle with base = height of the triangle.
Therefore, area of the triangle = (1/2) (4)(4) = 8 cm²
12.The area of an equilateral is 16√3 m². Its perimeter , in meters, is :
(a) 12 m (b) 48 m
(c) 24 m (d) 306m
Answer:(c) 24 m
The area of an equilateral triangle with side a is (√3/4) (a)².
Therefore, (√3/4) (a)².= 16√3
=> (1/4)a² = 16
=> a² = 64
=> a = 8
Therefore the perimeter of the square is 3(8) = 24 m
13. The perimeter of a triangle is 36 cm and its sides are in the ratio a:b:c = 3:4:5 then a,b,c are respectively:
(A) 9 cm, 15 cm, 12cm (B) 15cm, 12cm, 9 cm
(C) 12cm, 9cm, 15cm (D) 9cm, 12cm, 15cm
Answer: (D) 9cm, 12cm, 15cm
Let the three sides be 3x, 4x and 5x.
Then according to the conditions given in the questions, we have
3x + 4x + 5x =36
12x = 36
x = 3 cm
Thus the three sides are 3 x 3, 4 x3 and 5 x 3 = 9 cm,12 cm and 15 cm
14. In an equilateral triangle of side 'a', the length of the altitude is:
(a) (√3a/4) (b) (√3a/2)
(c) (√3a) (d) (2√3a)
Answer: (b) (√3a/2)
In a equilateral triangle the median and the altitude are the same.
Thus. in the figure
in Δ APC , using the Pythagoras theorem, we have
AP² + (a/2)² = a² (or the altitude)
=> AP² = a² - a²/4
=> AP² = (3/4)a²
=> AP = (√3a/2)
15. The volume of sphere of diameter r is:
(a) (4/3)πr³ (b) (8/3)πr³
(c) (1/6)πr³ (d) (1/3)πr³
Answer:(c) (1/6)πr³
Given that diameter of sphere = r
therefore the radius of sphere = (r/2)
And the volume of sphere = (4/3)π(r/2)³ = (1/6)πr³
16. The curved surface area of a right circular cylinder of height 14 cm is 88 cm². Diameter of the cylinder is:
(a) 1 cm (b) 2 cm
(c) 3 cm (d) 3 cm
Answer: (b) 2 cm
The formula for curved surface for a right circular cylinder is 2πrh
and its is given here that the surface area = 88 cm²
=> 2πrh = 88
=> 2(22/7)(r)(14) = 88
=> 2r = 88x7 / 22x14
=> diameter = 2 cm
17. A conical tent is 9 cm high and the radius of its base is 12 cm. The slant height of the cone would be
(a) 13 cm (b) 14 cm
(c) 15 cm (d) 16 cm
Answer:(c) 15 cm
The slant height, height and the radius of the base of a cone forms hypotenuse, height and base of a right angled triangle respectively.
Then, Slant height = √ 12² + 9²
= √144 +81
= √225
= 15 cm
18. If half of the perimeter of a square is 12 cm. Then area and side of the square are:
(a) 6cm², 16cm (b) 36 cm², 6cm
(c) 16 cm², 62 (d) 36 cm², 18 cm
Answer: (b) 36 cm², 6cm
Half of the perimeter of the square = 2s = 12 cm
Therefore the side of the square s = 6 cm
Therefore the area of the square = 6 x 6 = 36 cm²
19. In a cylinder, if radius is halved and height is doubled , the volume will be:
(a) Same (b) Doubled
(c)Halved (d) Four times
Answer:(c)Halved
The volume of the cylinder = πr²h
Again, if the radius is halved and the height doubled the the volume becomes = π(r/2)²(2h)
= π(r²/4))(2h)
= (1/2)πr²h
And therefore halved.
20. The ratio of the curved surface areas of two cones if their diameters of the bases are equal and slant heights are in the ratio 4:3 is:
(a) 3:4 (b) 4:3
(c) 12:15 (d) none of these
Answer: (b) 4:3
Let the slant heights be 4x and 3x.
Again there bases are the same, therefore the radius are the same.
=> R = r
Therefore the ratio of curved surface areas = (2πR²H)/(2πr²h)
= H/h (As the radius are the same for both the tents)
= 4/3
21. In a cylinder, radius is doubled and height is halved, curved surface area will be :
(a) halved (b) doubled
(c) same (d) four times
Answer:(c) same
Initial curved surface area = 2πrh
Final curved surface area after halving the height and doubling the radius is
= 2π(2r)(h/2) = 2πrh
22. The mean of first five prime numbers is:
(a) 5.6 (b) 5.2
(c) 5.8 (d) 4.5
Answer: (a) 5.6
First five prime numbers are 2,3,5,7,11
Therefore the mean = (2+3+5+7+11)/5 = 28/5 = 5.6
23 . Class-mark of the interval 70-80 is:
(a) 65 (b) 85
(c) 75 (d) 150
Answer:(c) 75
Class-mark = (70+80)/2 = 150/2 = 75
24. If the mean of the data 2,x+1,9,x-2 is 4, then the value of 'x' is:
(a) 2 (b) 3
(c) 4 (d) 5
Answer: (b) 3
Mean of the given numbers is {2 + (x+1) + 9 +(x-2) } /4 = 4
=> {2 + x + 1 + 9 + x - 2} = 16
=> {10+ 2x} = 16
=> 2x = 6
x = 3
25. The median of the data 70,50,20,30,40,50,40,50,80 is:
(a) 45 (b) 49.5
(c) 50 (d) 56
Answer:(c) 50
Arranging in the increasing order we get,
20,30,40,40,50,50,50,70,80
There are 9 values in the questions and there the median lies at (n+1)/2th =5th value. It is 50 in these case.
26. The mdeian of the following data is 5,9,8,6,3,5,7,12,15,is
(a) 4 (b) 6
(c) 5 (d) 7
Answer: (d) 7
Arranging the given 9 values in increasing order we get 3,5,5,6,7,8,9,12,15
The median is defined as (n+1)/2th values when we arrange the data in ascending or descending order, if n is odd.
therefore the mean is (9+1)/2 = 5th value which is 7 .
27. Mode of the data :
4,6,9,6,4,2,4,8,6,4,3,4,6 is:
(a) 6 (b) 4
(c) 3 (d) 2
Answer:
Mode of a given set of values is defined as the value the repeats the most number of times in the given set.
Here it is easily seen that the mode is 4 as it is repeated the most number of times.
28. A fair coin is tossed 100 times and the Head occurs 58 times and tail 42 times. The experimental probability of getting a Head is:
(a) 1/2 (b) 21/50
(c) 29/50 (d) 42/58
Answer:(c) 29/50
Number of times a Head occurs = 58
And number of times the coin was tossed = 100
Therefore probability of getting a Head = 58/100 = 29/50
29. Which of the following cannot be a probability of an event?
(a) 0 (b) 1
(c) -1/6 (d) 2/3
Answer:(c) -1/6
The probability of a event always lies between 0 and 1.
Therefore, -1/6 does not represent a probability of an event.
30. In a survey, , out of 200 students, it is observed that 125 students like mathematics. What is the probability that the students who do not like mathematics is :
(a) 5/8 (b) 7/8
(c) 3/8 (d) /
Answer:(c) 3/8
Number of students out of 200 that does not like mathematics is 200 - 125 = 75
Therefore, the required probability is 75/200 = 3/8
1. There are three non-collinear points. The number of circles passing through them are:
(a) 2 (b) 1
(c) 3 (d) 4
Answer:(b) 1
There can only be one and only one circle that can pass through any three collinear points.
2. In the given figure, OM | to the chord AB of the circle with centre O. If OA = 13 cm and AB = 24 cm, then OM equals:(a) 3 cm (b) 4 cm
(c) 5 cm (d) √4.7
Answer:(c) 5 cm
Given that ∠OMA = 90° .
ΔAOM and ΔBOM are congruent and hence AM=AB and therefore,
AM = (1/2)24 = 12 cm.
Using Pythagoras theorem, we have
OM = √AO² - AM²
=> OM = √13² - 12²
=> OM = √169 - 144 = √25 = 5 cm
3. In the given figure, a circle is centred at O, the value of x is:
(a) 55° (b) 70°(c) 110° (d) 125°
Answer:(c) 110°
OC = OB (radius of the circle)
=> ∠OCB = ∠OBC
=> ∠OCB = 35°
Again, OC = OA (radius of the circle)
=> ∠OCA = ∠OAC
=> ∠OCA = 20°
Therefore, ∠ACB = ∠OCB + ∠OCA = 35 +20 = 55°
And ∠AOB = 2 x 55 = 110 °
(Angle subtended at the centre of the circle by an arc is double of the angle subtended at any point on the circle )
4. In figure, O is the centre of the circle and .∠OBA = 60. Then ∠ACB equals
(a) 60° (b) 45°
(c) 30° (d) 90°
Answer:(c) 30°
Given that,
∠OBA = 60 °
and OA=OB (radius of circle )
But ∠OAB = 60 ° (because in a triangle angles opposite to opposites sides are equal)
Again, in triangle OAB, we have
∠OAB + ∠OBA + ∠AOB = 180°
=> 60° + 60° + ∠AOB = 180°
=> ∠AOB = 180° - 120° = 60°
Therefore, ∠ACB = (1/2)(∠AOB)
=> ∠ACB = (1/2)(60°) = 30°
5. If ∠C = ∠D = 50, then four points A, B, C, D :
(c) Are collinear (d) A, B, C and A, B, C lie on different circles.
Answer: (a) Are con cyclic
Since ∠C= ∠D = 50, Since angles on the same side of the line segment are equal therefore they are concyclic.
6. The distance of chord length 16 cm from the centre of the circle of radius 10 cm is
(a) 6 cm (b) 8 cm
(c) 10 cm (d) 12 cm
Answer: The distance of the chord = √(10)² - 8²
= √(10)² - 8²
= √100 - 64
= √36
= 6 cm
7. The distance of a chord 8 cm long chord from the centre of a circle of radius of 5 cm is :
(a) 50° (b) 60°
(c) 65° (d) 70°
Answer: 3 cm
Distance of the chord from the centre of the circle = √5² - 4²
= √25 -16
= √9
= 3 cm
8. Area of an isosceles right triangle is 8 cm². Its hypotenuse is:
(a) √32 cm (b) 4 cm
(c) 4√3 cm (d) 2√6 cm
Answer:√32 cm
Area of triangle = (1/2)(b)(h)
8 = (1/2)(b)(b) (Since its an isosceles triangle)
=> 16 = (b)(b)
=> b = 4 cm
Again the hypotenuse = √4² + 4² = √2.4² = √32 cm
9. Side of an equilateral triangle is 4 cm. Its area is :
(i) 4√3 (ii) √3/4
(iii) √3 (iv) 2 √3
Answer: The area of an equilateral triangle = (√3/4)(side)² = (√3/4)(4)² = 4 √3 cm²
10.A square and an equilateral triangle have equal perimeters. If the diagonal of the square is 12√2, then the area of the triangle (in cm² ) is :
(a) 24√ 2 (b) 25√3
(c) 48√ 3 (d) 64√3
Answer: (d) 64√3
Relationship between diagonal of the square and the side of the square is
Diagonal of a square of side S = √2 S where S is the side of the square.
=> 12√2 = √2 S
=> S = 12 cm
And therefore the perimeter of the square = 4x side = 4x12 = 48 cm
Again, the perimeter of an equilateral triangle with side a = 3a
But given that the perimeter of the given square is equal to the perimeter of the square.
=> 3a = 48
=> a = 48/3 = 16 cm
And therefore area of the equilateral triangle = (√ 3/4)a² = (√ 3/4)(16)² = 64√ 3 cm²
11. The area of the ΔABC in which AB=BC= 4 cm and ∠B =90° is :
(a) 16 cm² (b) 8 cm²
(c) 4 cm² (d) 12 cm²
Answer: (b) 8 cm²
From the condition given in the question it is clear that ΔABC is a right angle triangle with base = height of the triangle.
Therefore, area of the triangle = (1/2) (4)(4) = 8 cm²
12.The area of an equilateral is 16√3 m². Its perimeter , in meters, is :
(a) 12 m (b) 48 m
(c) 24 m (d) 306m
Answer:(c) 24 m
The area of an equilateral triangle with side a is (√3/4) (a)².
Therefore, (√3/4) (a)².= 16√3
=> (1/4)a² = 16
=> a² = 64
=> a = 8
Therefore the perimeter of the square is 3(8) = 24 m
13. The perimeter of a triangle is 36 cm and its sides are in the ratio a:b:c = 3:4:5 then a,b,c are respectively:
(A) 9 cm, 15 cm, 12cm (B) 15cm, 12cm, 9 cm
(C) 12cm, 9cm, 15cm (D) 9cm, 12cm, 15cm
Answer: (D) 9cm, 12cm, 15cm
Let the three sides be 3x, 4x and 5x.
Then according to the conditions given in the questions, we have
3x + 4x + 5x =36
12x = 36
x = 3 cm
Thus the three sides are 3 x 3, 4 x3 and 5 x 3 = 9 cm,12 cm and 15 cm
14. In an equilateral triangle of side 'a', the length of the altitude is:
(a) (√3a/4) (b) (√3a/2)
(c) (√3a) (d) (2√3a)
Answer: (b) (√3a/2)
In a equilateral triangle the median and the altitude are the same.
Thus. in the figure
AP² + (a/2)² = a² (or the altitude)
=> AP² = a² - a²/4
=> AP² = (3/4)a²
=> AP = (√3a/2)
15. The volume of sphere of diameter r is:
(a) (4/3)πr³ (b) (8/3)πr³
(c) (1/6)πr³ (d) (1/3)πr³
Answer:(c) (1/6)πr³
Given that diameter of sphere = r
therefore the radius of sphere = (r/2)
And the volume of sphere = (4/3)π(r/2)³ = (1/6)πr³
16. The curved surface area of a right circular cylinder of height 14 cm is 88 cm². Diameter of the cylinder is:
(a) 1 cm (b) 2 cm
(c) 3 cm (d) 3 cm
Answer: (b) 2 cm
The formula for curved surface for a right circular cylinder is 2πrh
and its is given here that the surface area = 88 cm²
=> 2πrh = 88
=> 2(22/7)(r)(14) = 88
=> 2r = 88x7 / 22x14
=> diameter = 2 cm
17. A conical tent is 9 cm high and the radius of its base is 12 cm. The slant height of the cone would be
(a) 13 cm (b) 14 cm
(c) 15 cm (d) 16 cm
Answer:(c) 15 cm
The slant height, height and the radius of the base of a cone forms hypotenuse, height and base of a right angled triangle respectively.
Then, Slant height = √ 12² + 9²
= √144 +81
= √225
= 15 cm
18. If half of the perimeter of a square is 12 cm. Then area and side of the square are:
(a) 6cm², 16cm (b) 36 cm², 6cm
(c) 16 cm², 62 (d) 36 cm², 18 cm
Answer: (b) 36 cm², 6cm
Half of the perimeter of the square = 2s = 12 cm
Therefore the side of the square s = 6 cm
Therefore the area of the square = 6 x 6 = 36 cm²
19. In a cylinder, if radius is halved and height is doubled , the volume will be:
(a) Same (b) Doubled
(c)Halved (d) Four times
Answer:(c)Halved
The volume of the cylinder = πr²h
Again, if the radius is halved and the height doubled the the volume becomes = π(r/2)²(2h)
= π(r²/4))(2h)
= (1/2)πr²h
And therefore halved.
20. The ratio of the curved surface areas of two cones if their diameters of the bases are equal and slant heights are in the ratio 4:3 is:
(a) 3:4 (b) 4:3
(c) 12:15 (d) none of these
Answer: (b) 4:3
Let the slant heights be 4x and 3x.
Again there bases are the same, therefore the radius are the same.
=> R = r
Therefore the ratio of curved surface areas = (2πR²H)/(2πr²h)
= H/h (As the radius are the same for both the tents)
= 4/3
21. In a cylinder, radius is doubled and height is halved, curved surface area will be :
(a) halved (b) doubled
(c) same (d) four times
Answer:(c) same
Initial curved surface area = 2πrh
Final curved surface area after halving the height and doubling the radius is
= 2π(2r)(h/2) = 2πrh
22. The mean of first five prime numbers is:
(a) 5.6 (b) 5.2
(c) 5.8 (d) 4.5
Answer: (a) 5.6
First five prime numbers are 2,3,5,7,11
Therefore the mean = (2+3+5+7+11)/5 = 28/5 = 5.6
23 . Class-mark of the interval 70-80 is:
(a) 65 (b) 85
(c) 75 (d) 150
Answer:(c) 75
Class-mark = (70+80)/2 = 150/2 = 75
24. If the mean of the data 2,x+1,9,x-2 is 4, then the value of 'x' is:
(a) 2 (b) 3
(c) 4 (d) 5
Answer: (b) 3
Mean of the given numbers is {2 + (x+1) + 9 +(x-2) } /4 = 4
=> {2 + x + 1 + 9 + x - 2} = 16
=> {10+ 2x} = 16
=> 2x = 6
x = 3
25. The median of the data 70,50,20,30,40,50,40,50,80 is:
(a) 45 (b) 49.5
(c) 50 (d) 56
Answer:(c) 50
Arranging in the increasing order we get,
20,30,40,40,50,50,50,70,80
There are 9 values in the questions and there the median lies at (n+1)/2th =5th value. It is 50 in these case.
26. The mdeian of the following data is 5,9,8,6,3,5,7,12,15,is
(a) 4 (b) 6
(c) 5 (d) 7
Answer: (d) 7
Arranging the given 9 values in increasing order we get 3,5,5,6,7,8,9,12,15
The median is defined as (n+1)/2th values when we arrange the data in ascending or descending order, if n is odd.
therefore the mean is (9+1)/2 = 5th value which is 7 .
27. Mode of the data :
4,6,9,6,4,2,4,8,6,4,3,4,6 is:
(a) 6 (b) 4
(c) 3 (d) 2
Answer:
Mode of a given set of values is defined as the value the repeats the most number of times in the given set.
Here it is easily seen that the mode is 4 as it is repeated the most number of times.
28. A fair coin is tossed 100 times and the Head occurs 58 times and tail 42 times. The experimental probability of getting a Head is:
(a) 1/2 (b) 21/50
(c) 29/50 (d) 42/58
Answer:(c) 29/50
Number of times a Head occurs = 58
And number of times the coin was tossed = 100
Therefore probability of getting a Head = 58/100 = 29/50
29. Which of the following cannot be a probability of an event?
(a) 0 (b) 1
(c) -1/6 (d) 2/3
Answer:(c) -1/6
The probability of a event always lies between 0 and 1.
Therefore, -1/6 does not represent a probability of an event.
30. In a survey, , out of 200 students, it is observed that 125 students like mathematics. What is the probability that the students who do not like mathematics is :
(a) 5/8 (b) 7/8
(c) 3/8 (d) /
Answer:(c) 3/8
Number of students out of 200 that does not like mathematics is 200 - 125 = 75
Therefore, the required probability is 75/200 = 3/8
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