Category: Class 10 Maths Objective Questions

[CLASS 10] Mathematics MCQs on Surface Area & Volumes – Frustum of a Cone

Mathematics Problems for Class 10 on “Surface Area & Volumes – Frustum of a Cone”.

1. The frustum of a cone has its circular ends radius as 15 cm and 30 cm where the height is 10cm. Find the volume of the frustum.
a) 21544.4 cm2
b) 16485 cm2
c) 16570 cm2
d) 16572.5 cm2
Answer: b
Clarification: r = 15 cm, R = 30 cm, h = 10 cm
The volume of a Frustum = (frac {1}{3})πh(r2 + rR + r2)
= (frac {1}{3}) × 3.14 × 10 (225 + 450 + 900)
= 16485 cm3

2. Find the slant height if the frustum of height 10cm and having its diameter as 20 cm and 40 cm.
a) 10.4 cm
b) 14.14 cm
c) 15.50 cm
d) 89.4 cm
Answer: b
Clarification: h = 10 cm, R = 20 cm, r = 10 cm
The slant height of a cone s2 = h2 + (R – r)2
s2 = √(100 + (20 – 10)2)
s = 14.14 cm

3. Find the Lateral surface area of the frustum, if the slant height is 15cm and having radii as 14 cm and 21 cm.
a) 1648.5 cm2
b) 1115 cm2
c) 2105.4 cm2
d) 1237.2 cm2
Answer: a
Clarification: s = 15 cm, R = 42 cm, r = 28 cm
Lateral Surface area of Frustum (L) = π(R + r)s
L = 3.14(21 + 14 )15
= 1648.5 cm2

4. Find the total surface area of a frustum whose slant height is 10 cm and having radii as 10 cm and 20 cm.
a) 1923.21 m3
b) 2512 cm2
c) 2184.21 m3
d) 2842.21 m3
Answer: b
Clarification: s = 10 cm, R = 20 cm, r = 10 cm
Total surface area of a cone = π(R + r)s + πR2 + πr2
= 3.14(20 + 10)10 + (3.14 × 202) + (3.14 × 102)
= 942 + 1256 + 314
= 2512 cm2

5. Find the difference between the radii of a Frustum, if the slant height and height are 10 cm and 8 cm.
a) 4 cm
b) 1 cm
c) 2 cm
d) 6 cm
Answer: d
Clarification: The slant height of a cone = h2 + (R – r)2
102 = 82 + (R – r)2
(R – r)2 = 100 – 64
(R – r) = 6 cm

6. The frustum of a cone has its circular ends radius as 5 cm and 15 cm where the height is given as 8 cm. Find the volume of the frustum.
a) 2152.76 cm3
b) 1254 cm3
c) 2721.33 cm3
d) 1421.76 cm3
Answer: c
Clarification: r = 5 cm, R = 15 cm and h = 8 cm
The volume of a Frustum = (frac {1}{3}) πh(r2 + rR + r2)
= (frac {1}{3}) × 3.14 × 8(25 + 75 + 225)
= 2721.33 cm3

7. Find the Lateral surface area of the frustum, if the slant height is 12 cm and having radii as 15 cm and 35 cm.
a) 1984 cm2
b) 1804 cm2
c) 1984 cm2
d) 1884 cm2
Answer: d
Clarification: s = 12 cm, R = 35 cm, r = 15 cm
Lateral Surface area of frustum = π(r + R)s
L = 3.14 × (15 + 35) × 12
= 1884 cm2

8. The difference between the radii is 12 cm and slant height is given as 15 cm. Find the height of the frustum.
a) 5.4 cm
b) 1.2 cm
c) 3 cm
d) 9 cm
Answer: d
Clarification: Given R – r = 12 cm and slant height = 15 cm
The slant height of a cone s2 = h2 + (R – r)2
225 = h2 + 144
h2 = 225 – 144
h2 = 81
h = 9 cm

9. Find the total surface area of a frustum whose slant height is 18 cm and having radii as 8 cm and 16 cm.
a) 2361.28 cm2
b) 2205.6 cm2
c) 1628.23 cm2
d) 2604 cm2
Answer: a
Clarification: s = 18cm, R = 16 cm, r = 8 cm
Total surface area of a cone = π(R + r)s + πR2 + πr2
= 3.14(16 + 8)18 + (3.14 × 256) + (3.14 × 64)
= 1356.48 + 803.84 + 200.96
= 2361.28 cm2

10. Find the slant height if the Frustum of height 22 cm has its diameter as 34 and 50cm.
a) 83.40 cm
b) 43.90 cm
c) 23.40 cm
d) 73.48 cm
Answer: c
Clarification: h = 22 cm, R = 25 cm, r = 17 cm
The slant height of a cone (s2) = h2 + (R – r)2
s2 = 484 + (25 – 17)2
s2 = 548
s = 23.40 cm

[CLASS 10] Mathematics MCQs on Quadratic Equation – Determination of Types of Roots

Mathematics MCQs for Schools on “Quadratic Equation – Determination of Types of Roots”.

1. For the equation x2 + 5x – 1, which of the following statements is correct?
a) The roots of the equation are equal
b) The discriminant of the equation is negative
c) The roots of the equation are real, distinct and irrational
d) The discriminant is equal to zero
Answer: c
Clarification: Roots are real. ∴ b2 – 4ac ≥ 0
52 – 4(1)(1)
25 – 4 = 21 which is greater than 0. Hence, the discriminant of the equation is greater than zero, so roots are real.

2. If the roots of the equation ax2 + bx + c are real and equal, what will be the relation between a, b, c?
a) b = ±(sqrt {ac})
b) b = ±(sqrt {4c})
c) b = ±(sqrt {-4ac})
d) b = ±(sqrt {4ac})
Answer: d
Clarification: Roots are real and equal. ∴ b2 – 4ac = 0
b2 = 4ac
b = ±(sqrt {4ac})

3. What will be the value of k, so that the roots of the equation are x2 + 2kx + 9 are imaginary?
a) -5 < k < 5
b) -3 < k < 3
c) 3 < k < -3
d) -5 < k < 3
Answer: b
Clarification: Roots are imaginary. ∴ b2 – 4ac < 0
(2k)2 – 4(9)(1) < 0
4k2 – 36 < 0
k2 – 9 < 0
k2 < 9
k < ±3
-3 < k < 3

4. What will be the nature of the roots of the quadratic equation 5x2 – 11x + 13?
a) Imaginary
b) Real
c) Irrational
d) Equal
Answer: a
Clarification: To check the nature of the roots, the discriminant must be either equal to zero, less than zero or greater than zero.
Discriminant = b2 – 4ac = – 112 – 4 × 5 × 13 = 121 – 260 = – 139
Since discriminant is less than zero, the roots of the equation are imaginary.

5. What will be the nature of the roots of the quadratic equation x2 + 10x + 25?
a) Imaginary
b) Real
c) Irrational
d) Equal
Answer: d
Clarification: To check the nature of the roots, the discriminant must be either equal to zero, less than zero or greater than zero.
Discriminant = b2 – 4ac = 102 – 4 × 25 × 1 = 100 – 100 = 0
Since discriminant is equal to zero, the roots of the equation are equal.

6. What will be the nature of the roots of the quadratic equation 2x2 + 10x + 9?
a) Imaginary
b) Real
c) Irrational
d) Equal
Answer: b
Clarification: To check the nature of the roots, the discriminant must be either equal to zero, less than zero or greater than zero.
Discriminant = b2 – 4ac = 102 – 4 × 2 × 9 = 100 – 72 = 28
Since discriminant is greater than zero, the roots of the equation are real and distinct.

7. The equation 9x2 – 2x + 5 is not true for any real value of x.
a) False
b) True
Answer: b
Clarification: To check the nature of the roots, the discriminant must be either equal to zero, less than zero or greater than zero.
Discriminant = b2 – 4ac = -22 – 4 × 9 × 5 = 4 – 180 = -176
Since discriminant is less than zero, the roots of the equation are imaginary. Hence, for any real value of x the equation is not true.

8. The value of p for which the equation 8x2 + 9px + 15 has equal roots is (frac {4sqrt {30}}{9}).
a) True
b) False
Answer: a
Clarification: Roots are equal. ∴ b2 – 4ac = 0
(9p)2 – 4(8)(15) = 0
81p2 – 480 = 0
81p2 = 480
p = ± (sqrt {frac {480}{81}}) = ± (frac {4sqrt {30}}{9})

9. What will be the value of a, for which the equation 5x2 + ax + 5 and x2 – 12x + a will have real roots?
a) a = 37
b) 10 < a < 36
c) 36 < a < 10
d) a = 9
Answer: b
Clarification: The roots of both the equations are real.
Discriminant of 5x2 + ax + 5 : b2 – 4ac = a2 – 4 × 5 × 5 = a2 – 100
Since, roots are real; discriminant will be greater than 0.
a2 ≥ 100
a ≥ ±10
Now, discriminant of x2 – 12x + a : b2 – 4ac = -122 – 4 × 1 × a = 144 – 4a
Since, roots are real; discriminant will be greater than 0.
144 ≥ 4a
a ≤ (frac {144}{4}) = 36
For both the equations to have real roots the value of a must lie between 36 and 10.

10. What will be the value of k, if the roots of the equation (k – 4)x2 – 2kx + (k + 5) = 0 are equal?
a) 18
b) 19
c) 20
d) 21
Answer: c
Clarification: Roots are equal. ∴ b2 – 4ac = 0
-(2k)2 – 4(k – 4)(k + 5) = 0
4k2 – 4(k2 – 4k + 5k – 20) = 0
4k2 – 4(k2 + k – 20) = 0
4k2 – 4k2 – 4k + 80 = 0
-4k = – 80
k = (frac {-80}{-4}) = 20

To practice Mathematics MCQs for Schools,

[CLASS 10] Mathematics MCQs on Trigonometric Ratios of Specific Angles

Mathematics Multiple Choice Questions & Answers on “Trigonometric Ratios of Specific Angles – 1”.

1. What is the value of sec θ when θ is 45°?
a) √3
b) 1
c) 0
d) √2
Answer: d
Clarification: The value of cos 45° is (frac {1}{sqrt 2}) and secant is inverse to cos.
So, sec θ = (frac {1}{cos , theta } = frac {1}{1/sqrt 2})
= √2

2. What is the value of sin 0° + cos 0°?
a) 0
b) 2
c) 1
d) ∞
Answer: c
Clarification: The value of sin 0° is 0 and the value of cos 0° is 1.
sin 0° + cos 0° = 0 + 1
= 1

3. Evaluate cos 30° sin 60° + cos 60° sin 30°.
a) 2
b) 0
c) 1
d) ∞
Answer: c
Clarification: cos 30° sin 60° + cos 60° sin 30° = (frac {sqrt 3}{2} . frac {sqrt {3}}{2} + frac {1}{2} . frac {1}{2})
= (frac {3 + 1}{4})
= (frac {4}{4})
= 1

4. What is the value of cos (30° + 60°)?
a) 1
b) (frac {1}{4})
c) (frac {3}{4})
d) 0
Answer: a
Clarification: cos (30° + 60°) = cos (90°) = 1
The value of cos (90°) is 1

5. 60° lies in the_____ quadrant.
a) first
b) second
c) third
d) fourth
Answer: a
Clarification: A plane is divided into four quadrants. The first quadrant ranges from 0° to 90°. So, 60° lies between 0° to 90° which is in the first quadrant of a plane.

6. What is the value of (frac {2 cot , ⁡60^{circ }}{1 + cot , ⁡⁡45^{circ }})?
a) Cot 30°
b) Cot 45°
c) Cot 60°
d) Tan 60°
Answer: c
Clarification: (frac {2 cot , ⁡⁡60^{circ }}{1 + cot , ⁡⁡45^{circ }} = frac {2 (frac {1}{sqrt 3})}{1 + 1})
= (frac {frac {2}{sqrt {3}}}{2})
= (frac {1}{sqrt 3})
= Cot 60°

7. Cosec 0° is _____
a) Not defined
b) 1
c) 0
d) 2
Answer: a
Clarification: Cosec 0° = (frac {1}{Sin , ⁡0^{circ } } = frac {1}{0})
Any number divided with 0 is always not defined.

8. The value of sin 90° + cos 0° + √2 cos 45° is_____
a) 2
b) 1
c) 4
d) 3
Answer: d
Clarification: sin 90° + cos 0° + √2 cos 45° = 1 + 1 + √2 ((frac {1}{sqrt {2}}))
= 1 + 1 + 1
= 3

9. Cosec 90° is _____
a) 0
b) 2
c) 1
d) √3
Answer: c
Clarification: Cosec 90° = (frac {1}{Sin , ⁡90^{circ } })
= (frac {1}{1})
= 1

10. (frac {Cos , ⁡0^{circ }}{1 – Sin , ⁡30^{circ }}) is _____
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: (frac {Cos , ⁡0^{circ }}{1 – Sin , ⁡30^{circ }} = frac {1}{1 – 1/2})
= (frac {1}{1/2})
= 2

[CLASS 10] Mathematics MCQs on Statistics – Mean of Grouped Data

Mathematics Multiple Choice Questions & Answers on “Statistics – Mean of Grouped Data”.

1. What is statistics?
a) Statistics is the collection of data
b) Statistics is the collection, classification and interpretation of data
c) Statistics is the classification of data
d) Statistics is the interpretation of data
Answer: b
Clarification: Statistics is derived from the language and is a branch of mathematics that deals with the collection, classification and interpretation of data.

2. Who is the pioneer that contributed to the development of statistics?
a) Albert Einstein
b) Lewis Capaldi
c) Ronald Fisher
d) Harold Fisher
Answer: c
Clarification: Sir Ronald Fisher contributed to the development of new theories in statistics. Statistics deals with the collection, classification and interpretation of data.

3. What is Range?
a) Largest value – Smallest value
b) Smallest value – Largest value
c) Mid value – Average value
d) Average value – Mid value
Answer: a
Clarification: Range is defined as the difference between the largest value and the smallest value of the variable in a distribution.
R = L – S     where L = Largest value, S = Smallest value.

4. What is the formula for the arithmetic mean?
a) (frac {Number , of , the , observations}{Sum , of , observations})
b) (frac {Sum , of , the , observations}{Number , of , observations})
c) (frac {Product , of , the , observations}{Number , of , observations})
d) (frac {Sum , of , the , observations}{Product , of , the , observations})
Answer: b
Clarification: Arithmetic mean can also be called the average of given variables in a distribution. The formula for arithmetic mean is the ratio of the sum of the observations to the number of observations.
Arithmetic mean = (frac {Sum , of , the , observations}{Number , of , observations})

5. Mean of the data can be represented as x = (frac {sum f i xi}{sum fi}).
a) False
b) True
Answer: b
Clarification: Let f1, f2 …. fn are the frequencies of respective observations x1, x2 …. xn. Then the mean of the data can be written as x = (frac {f1 x1+f2x2+⋯+fnxn}{f1+f2+⋯+fn})
x = (frac {sum f i xi}{sum fi})

6. What is the arithmetic mean of the observations 2, 8.2, 3, 9, 11.2, 4?
a) 5.5
b) 8.45
c) 6.23
d) 7.1
Answer: c
Clarification: Arithmetic mean = (frac {Sum , of , the , observations}{Number , of , observations})
= (frac {2+8.2+3+9+11.2+4}{6})
= 6.23

7. What is the mean of 142, 143, 145, 158, 139?
a) 135.4
b) 145.4
c) 0.23
d) 135.5
Answer: b
Clarification: Arithmetic mean = (frac {Sum , of , the , observations}{Number , of , observations})
= (frac {142+143+145+158+139}{5})
= 145.4

8. Find the sum of the observations if the mean is 143 and the number of observations is 8.
a) 1148
b) 1344
c) 1244
d) 1144
Answer: d
Clarification: Arithmetic mean = (frac {Sum , of , the , observations}{Number , of , observations})
143 = (frac {Sum , of , the , observations}{8})
Sum of the observations = 143 × 8
= 1144

9. Find the sum of the observations if the mean is 23 and the number of the observations is 11?
a) 283
b) 256
c) 293
d) 253
Answer: d
Clarification: Arithmetic mean = (frac {Sum , of , the , observations}{Number , of , observations})
23 = (frac {Sum , of , the , observations}{11})
Sum of the observations = 23 × 11
= 253

10. Find the number of observations if the sum of the observations is 37.4 and the mean of the observations is 6.23?
a) 9
b) 7
c) 6
d) 4
Answer: c
Clarification: Arithmetic mean = (frac {Sum , of , the , observations}{Number , of , observations})
6.23 = (frac {37.4}{Number , of , observations})
Number of the observations = (frac {37.4}{6.23})
= 6

[CLASS 10] Mathematics MCQs on Solution of Quadratic Equation by Squaring Method

Mathematics Multiple Choice Questions & Answers on “Solution of Quadratic Equation by Squaring Method”.

1. What are the roots of the equation x2-9x-10?
a) 9, 1
b) -9, -1
c) -10, 1
d) 10, -1
Answer: d
Clarification: x2-9x-10=0
Shifting -10 to RHS
x2-9x=10
Adding (frac {b^2}{4}) on both sides, where b=-9
x2 – 9x + (frac {-9^2}{4} = frac {-9^2}{4}) + 10
x2 – 9x + (frac {81}{4} = frac {81}{4}) + 10
x2 – 9x + (frac {81}{4} = frac {121}{4})
((x-frac {9}{2}))2=((frac {11}{2}))2
x – (frac {9}{2}) = ± (frac {11}{2})
x = (frac {11}{2}+frac {9}{2}=frac {20}{2}) = 10 and x = (frac {-11}{2} + frac {9}{2} = frac {-2}{2}) = -1
The roots of the equation are 10 and -1.

2. How many methods are there to find the roots of a quadratic equation?
a) 1
b) 2
c) 3
d) 4
Answer: d
Clarification: There are four methods of solving quadratic equations namely, factorization, completing the square method, using quadratic formula and using square roots.

3. A metro travels at the speed of 396 km at a uniform speed. If the speed had been 5 km/hr. more, it would have taken 2 hours less for the same journey. What is the speed of the train?
a) 32 km/hr
b) 30 km/hr
c) 20.06 km/hr
d) 29.06 km/hr
Answer: d
Clarification: Let the speed of the metro train be x km/hr
Time taken to cover 396 km at x km/hr = (frac {396}{x}) hrs
Time taken to cover 396 km at x+5 km/hr = (frac {396}{x+5}) hrs
∴ (frac {396}{x}-frac {396}{(x+5)})=2
(frac {x+5-x}{x(x+5)}=frac {2}{396})
(frac {5}{x^2+5x}=frac {1}{198})
990=x2+5x
x2+5x=990
Adding (frac {b^2}{4}) on both sides, where b=5
x2 + 5x + (frac {5^2}{4}=frac {5^2}{4}) + 990
x2 + 5x + (frac {25}{4}=frac {25}{4}) + 990
((x + frac {5}{2}))2=((frac {sqrt {3985}}{2}))2
x + (frac {5}{2}) = ±(frac {sqrt {3985}}{2})
x = (frac {63.12}{2} -frac {5}{2}=frac {58.12}{2}) = 29.06 and x = (frac {-63.12}{2} -frac {5}{2} = frac {-68.12}{2}) = -34.06
Since, speed cannot be negative hence x=29.06 km/hr

4. The sum of areas of two squares is 625m2. If the difference in their perimeter is 20m then, what will be the sides of squares?
a) 10, 15
b) 15.28, 20.28
c) 13, 67.34
d) 35.67, 46.78
Answer: b
Clarification: Let the sides of two squares be x and y.
Their areas will be x2 and y2 respectively.
Sum of their areas is 625 m2
x2+y2=625     (1)
Their perimeters will be 4x and 4y
Difference in their perimeters is 20 m
4x-4y=20
x-y=5
x=5+y     (2)
Substituting (2) in (1) we get,
(y+5)2+y2=625
y2+10y+25+y2=625
2y2+10y+25=625
2y2+10y-620=0
y2+5y=310
Adding (frac {b^2}{4}) on both sides, where b=5
y2 + 5y + (frac {5^2}{4}=frac {5^2}{4}) + 310
y2 + 5y + (frac {25}{4}=frac {25}{4}) + 310
y2 + 5y + (frac {25}{4}=frac {1265}{4})
((y+frac {5}{2}))2 = ((frac {sqrt {1265}}{2}))2
y+(frac {5}{2}) = ±(frac {sqrt {1265}}{2})=±(frac {35.56}{2})
y = (frac {35.56}{2}-frac {5}{2}=frac {30.56}{2}) = 15.28 and y = (frac {-35.56}{2}-frac {5}{2}=frac {-40.56}{2}) = -20.28
Since, length cannot be negative hence, y=15.28
Now, x=5+y=5+15.28=20.28

5. The diagonal of rectangular field is 20 more than the shorter side. If the longer side is 10 more than shorter side what will be the area of the rectangular field?
a) 1000
b) 1100
c) 1200
d) 1500
Answer: c
Clarification: Let the length of shorter side be x.
Length of longer side = x+10
Length of diagonal = x+20

Here, ∆BDC forms a right-angled triangle. Hence by Pythagoras Theorem,
BD2=BC2+DC2
(x+20)2=x2+(x+10)2
x2+40x+400=x2+x2+20x+100
40x+400=x2+20x+100
x2-20x-300=0
x2-20x=300
Adding (frac {b^2}{4}) on both sides, where b=-20
x2 – 20x + (frac {-20^2}{4}=frac {-20^2}{4}) + 300
x2 – 20x + (frac {400}{4}=frac {400}{4}) + 300
x2 – 20x + (frac {400}{4}=frac {1600}{4})
((x-frac {20}{2}))2 = ((frac {40}{2}))2
x – (frac {20}{2}) = ±(frac {40}{2})
x = (frac {40}{2}+frac {20}{2}=frac {60}{2}) = 30 and x = (frac {-40}{2}+frac {20}{2}=frac {-20}{2}) = -10
Since length cannot be negative, hence x = 30
The length of the other side is x + 10 = 30 + 10 = 40
Area of rectangle = 40 × 30 = 1200 units

6. A rectangular field is 50 m long and 20 m wide. There is a path of equal width all around it having an area of 121 m2. What will be the width of the path?
a) 10.205 m
b) 11.205 m
c) 12.56 m
d) 13.76 m
Answer: b
Clarification:

Let the width of the path be x metres.
Length of the field including the path = 50+2x m
Breadth of the field including the path = 14+2x m
Area of the field including the path = (50+2x)(14+2x) m2
Area of the field excluding the path = 50×14=700 m2
∴ area of the path = [(50+2x)(14+2x)]-700
∴ [(50+2x)(14+2x)]-700=121
700+100x+28x+4x2-700=121
4x2+128x-121=0
x2+42x=(frac {121}{4})
Adding (frac {b^2}{4}) on both sides, where b=42
x2 + 42x + (frac {42^2}{4}=frac {42^2}{4}+frac {121}{4})
x2 + 42x + (frac {1764}{4}=frac {1764}{4}+frac {121}{4})
x2 + 42x + (frac {1764}{4}=frac {1885}{4})
((x+frac {21}{2}))2=((frac {sqrt {1885}}{2}))2
x + (frac {21}{2}) = ±(frac {sqrt {1885}}{2}) = ±(frac {43.41}{2})
x = (frac {43.41}{2}-frac {21}{2}=frac {22.41}{2}) = 11.205 and x = (frac {-43.41}{2}-frac {21}{2}=frac {-64.41}{2}) = -32.20
Since, the length cannot be negative; therefore, x=11.205 m

7. The area of right angled triangle is 500 cm2. If the base of the triangle is 10 less than the altitude of the triangle, what are the dimensions of the triangle?
a) base = 32 cm, altitude = 42 cm
b) base = 42 cm, altitude = 62 cm
c) base = 76 cm, altitude = 42 cm
d) base = 32 cm, altitude = 55 cm
Answer: a
Clarification: Let the altitude of the triangle be x cm.
The base will be x-10 cm.
The area of triangle is 500 cm2
(frac {1}{2}) × base × altitude = 672
(frac {1}{2}) × (x-10) × x = 672
x2-10x=1344
Adding (frac {b^2}{4}) on both sides, where b=-10
x2 – 10x + (frac {-10^2}{4}=frac {-10^2}{4}) + 1344
x2 – 10x + (frac {100}{4}=frac {100}{4}) + 1344
x2 – 10x + (frac {100}{4}=frac {5476}{4})
((x-frac {10}{2}))2=((frac {74}{2}))2
x-(frac {10}{2})=±(frac {74}{2})
x = (frac {74}{2}+frac {10}{2}=frac {84}{2}) = 42 and x = (frac {-74}{2}+frac {10}{2}=frac {-64}{2}) = -32
Since, length cannot be negative, hence x=42 cm
Base will be x-10=42-10=32 cm

8. The perimeter of rectangle is 24 cm and area of rectangle is 35cm2. What are the dimensions of the rectangle?
a) l=9, b=5
b) l=12, b=5
c) l=5, b=7
d) l=5, b=5
Answer: c
Clarification: Let the length of the rectangle be l cm and breadth be b cm.
Perimeter of rectangle = 24
2(l+b)=24
l+b=12
l=12-b     (1)
Now, the area of rectangle is 35 cm2
l×b=35     (2)
Substituting (1) in (2)
(12-b)b=35
12b-b2=35
b2-12b+35=0
b2-12b=-35
Adding (frac {b^2}{4}) on both sides, where b=-12
b2 – 12b + (frac {-12^2}{4}=frac {-12^2}{4}) – 35
b2 – 12b + (frac {144}{4}=frac {144}{4}) – 35
b2 – 12b + (frac {144}{4}=frac {4}{4})
((b-frac {12}{2}))2=((frac {2}{2}))2
b-(frac {12}{2})=±(frac {2}{2})
b = (frac {2}{2}+frac {12}{2}=frac {14}{2}) = 7 and b = (frac {-2}{2}+frac {12}{2}=frac {10}{2}) = 5
The length of the rectangle is l=12-b=12-7=5 and l=12-5=7
The length and breadth of the rectangle are 7 and 5 or 5 and 7.

9. The sum of ages of Eera and her sister is 46 and the product of their ages is 465. What are their ages?
a) Eera 31, Her sister 15
b) Eera 5, Her sister 10
c) Eera 12, Her sister 37
d) Eera 57, Her sister 12
Answer: a
Clarification: Let the age of Eera be x years. The age of her sister will be 46-x years.
The product of their ages is 465.
x(46-x)=465
46x-x2=465
x2-46x+465=0
x2-46x=-465
Adding (frac {b^2}{4}) on both sides, where b=-46
x2 – 46x + (frac {-46^2}{4}=frac {-46^2}{4}) – 465
x2 – 46x + (frac {2116}{4}=frac {2116}{4}) – 465
x2 – 46x + (frac {2116}{4}=frac {256}{4})
((x-frac {46}{2}))2=((frac {16}{2}))2
x-(frac {46}{2})=±(frac {16}{2})
x = (frac {16}{2}+frac {46}{2}=frac {62}{3}) = 31 and x = (frac {-16}{2}+frac {46}{2}=frac {30}{2}) = 15
The ages of Eera and her sister are 31 and 15 respectively or 15 and 31 years.

10. The sum of a number and its square root is 110. What is the number?
a) 100
b) 102
c) 99
d) 98
Answer: a
Clarification: Let the number be x. It square root will be √x
Sum of the number and its square root is 110.
x+√x=110
Let √x=y, x=y2
The equation becomes,
y2+y=110
Adding (frac {b^2}{4}) on both sides, where b=1
y2 + y + (frac {1^2}{4}=frac {1^2}{4}) + 110
y2 + y + (frac {1}{4}=frac {1}{4}) + 110
y2 + y + (frac {1}{4}=frac {441}{4})
((y+frac {1}{2}))2 = ((frac {21}{2}))2
y + (frac {1}{2}) = ±(frac {21}{2})
y = (frac {21}{2}-frac {1}{2}=frac {20}{2}) = 10 and y = (frac {-20}{2}-frac {1}{2}=frac {-21}{2}) = -10.5
x = y2 = 102 = 100 and x = y2 = -10.52 = 110.25
The numbers are 100 and 110.25.

[CLASS 10] Mathematics MCQs on Trigonometric Ratios of Specific Angles

Mathematics Written Test Questions and Answers for Class 10 on “Trigonometric Ratios of Specific Angles – 2”.

1. What is the value of tan θ when θ is 30°?
a) ( frac {1}{sqrt 3})
b) 1
c) 0
d) 2
Answer: a
Clarification: Tan θ = ( frac {Sin , theta}{Cos , theta} = frac {Sin , 30^{circ }}{Cos , 30^{circ }} )
= ( frac {1/2}{sqrt 3/2})
= ( frac {1}{2} times frac {2}{sqrt 3})
= ( frac {1}{sqrt 3})

2. What is the value of sec 0° + cosec 90°?
a) 0
b) 2
c) 1
d) ∞
Answer: b
Clarification: The value of sec 0° is 1 and the value of cosec 90° is 1.
sec 0° + cosec 90° = 1 + 1
= 2

3. Evaluate sin2 30.
a) 2
b) 0
c) ( frac {1}{4})
d) ∞
Answer: c
Clarification: Sin2 30° = (( frac {1}{2}))2
= ( frac {1}{4})

4. What is the value of cos 45° + sin 45°?
a) 1
b) ( frac {1}{4})
c) ( frac {3}{4})
d) √2
Answer: d
Clarification: cos 45° + sin 45° = ( frac {1}{sqrt 2} + frac {1}{sqrt 2})
= 2 × ( frac {1}{sqrt 2}) ( Since, 2 = √2 × √2 )
= √2

5. 135° lies in the third quadrant.
a) True
b) False
Answer: b
Clarification: A plane is divided into four quadrants. The second quadrant ranges from 90° to 180°. So, 135° lies between 90° to 180° which is in the second quadrant of a plane not in the third quadrant.

6. What is the value of sec 60° – ( frac {1}{cosec , 90^{circ }})?
a) Cot 30°
b) Cot 45°
c) Cot 60°
d) Tan 60°
Answer: b
Clarification: sec 60° – ( frac {1}{cosec , 90^{circ }}) = 2 – ( frac {1}{1})
= 1
= cot 45°

7. Tan 90° is _____
a) Not defined
b) 1
c) 0
d) 2
Answer: a
Clarification: Tan 90° = (frac {Sin , 90^{circ }}{Cos , 90^{circ }} = frac {1}{0})
Any number divided with 0 is always not defined.

8. The value of sin2 90° + √2 cos 45° + √3 cot 30° is_____
a) 2
b) 0
c) 4
d) 5
Answer: d
Clarification: sin2 90° + √2 cos 45° + √3 cot 30° = 12 + √2 (( frac {1}{sqrt 2})) + √3 (√3)
= 1 + 1 + 3
= 5

9. The product of sec 30° and cos 60° is _____
a) 0
b) 2
c) 1
d) ( frac {1}{sqrt 3})
Answer: d
Clarification: (sec 30°) (cos 60°) = (( frac {2}{sqrt 3}) ( frac {1}{2} ))
= ( frac {1}{sqrt 3})

10. ( frac {1 – cot , 45^{circ}}{1+ cot , 45^{circ}}) is _____
a) 0
b) 2
c) 3
d) 1
Answer: a
Clarification: ( frac {1 – cot , 45^{circ}}{1 + cot , 45^{circ}} = frac {1-1}{1+1})
= ( frac {0}{1})
= 0

To practice Mathematics Written Test Questions and Answers for Class 10,