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Solutions of 10 class maths questions frequently asked in  CBSE board exams

Frequently asked maths questions 10 board exam

Solutions of 10 class maths questions frequently asked in  CBSE board exams

Here the solutions are given of the questions of maths that are asked frequently in CBSE X board exams .These questions have been examined by a team of experts by the study of last year’s maths question papers of CBSE board exam, the expert team of future study point found that these questions are asked frequently in the last year’s 10 class CBSE board  exams in maths question papers. The solutions to these questions will give an idea to the students about the type of questions asked in the exam and the way of solving them in the exam. The frequently asked question in the exam will help the students in gaining the  idea that how often the maths questions are repeated and why is it important to study last year’s question papers . The study of this set of questions and answers will provide you a confidence that is necessary for achieving excellent marks in the exam.

Frequently asked maths questions 10 board exam

 

Solutions of 10 class maths questions frequently asked in  CBSE board exams

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   Section-A (1 marks)

Q1.If the quadratic equation   has two equal roots,  then find the value of p.

Solution.

When two roots are equal in a quadratic equation we have

The given equation is

 

b = –2√5, c = 15, a = P

20 – 60p = 0

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Q2. In fig, a tower AB is 20 m high and BC, its shadow on the ground, is 20√3 m long . Find the Sun’s altitude.

Solutions of 10 class maths questions frequently asked in  CBSE board exams Q2

 

Solution. Let altitude of the Sun is at θ

From the fig. we have

tanθ = tan30°

θ = 30°

Therefore the altitude of the sun is at 30°.

Q3. If the area of the circle is equal to the sum of areas of the two circles of diameter 10 cm and 24 cm, then the diameter of the larger circle (in cm) is :

(A) 34

(B) 26

(C) 17

(D) 14

Solution.

Area of the circle is  = πr²

radii of two circle = 10/2 = 5 cm and 24/2 = 12 cm

Let the radius of the larger circle is = R

The  sum of the areas of both circle = Area of the larger circle

π × 5² +  π × 12 ² = πR²

R² = 25 + 144

R² = 169

R = 13

Hence the diameter of the larger circle = 2R = 2× 13 = 26 cm

Q4. Two dice are thrown together . The probability of getting the same number on both dice is :

Solution.

Total outcome when die is thrown alone = 6

When two dice are thrown together then total possible outcome = 6² = 36

The favorable outcome ( the number is same on both dice) → (1,1), (2,2), (3,3), (4,4),(5,5), (6,6) which are 6 in number

Q5. In fig. the sides AB, BC, and CA of a triangle ABC touches a circle at P, Q and R respectively. If PA = 4 cm ,BP = 3 cm and AC =11 cm,then the length of BC(in cm) is:

Solution.

AP = AR ( tangents drown from the same external points)

Q6. If co-ordinates of the one end of the diameter of a circle are (2,3) and the coordinates of its centers are (–2,5) , then co-ordinates of other end of the diameter are :

(A) (–6,7)

(B) (6,–7)

(C) (6,7)

(D) (–6,–7)

Solution.

Let co-ordinates of another end of the diameter are = (x,y)

As the center of the circle is the midpoint of the diameter so applying the section formula

The coordinates of the midpoint of the diameter are = (–2,5)

The coordinates of the endpoint of the diameter are (2,3) and (x,y)

x = –6, y = 7

Therefore co-ordinates of another end are = A(–6,7)

Q7. The sum of the first 20 odd natural numbers is:

(A) 100

(B) 210

(C) 400

(D) 420

Solution.

The series formed by first odd natural number is 1,3, 5,7,9………upto 20 th term

d = 2 , n = 20

Applying the formula used for sum of the AP

= 10 [2 +38]

= 400

Hence the sum of the first 20 odd natural number is = C(400)

Q8. If 1 is a root of the equations ay² + ay + 3 = 0  and y² + y + b = 0 then ab equals :

(A) 3

(B) 

(C) 6

(D) –3

Solution.

1 is the root of the equation ay² + ay + 3 = 0

Therefore placing the value y = 1

a +a + 3 = 0

2a = –3 ⇒ a = –3/2

1 is also the root of the equation y²+ y + b = 0, putting  y = 1

1² + 1 +b = 0 ⇒ b = –2

ab = –3/2×–2 = –3

The rquired answer is (D) –3

Section B(2 marks)

 

Q9. If a point A (0,2) is equidistant from the points B(3,p) and C(p,5) then find the value of p.

Solution.

The distance between A(0,2) and B(3,p)

The distance between A(0,2) and C(p,5)

AB = AC

9 + 4 +P² –4P = P² + (–3)²

–4P = 9 –13

–4P = –4

P = 1

Therefore value of p = 1

Q10. The volume of a hemisphere is   cm³ . Find its curved surface area.

Solution.

The volume of hemisphere =

Volume given  cm³

The curved surface area of hemisphere = 2πr²

Hence the curved surface area of the hemisphere is 693 cm²

 

Q11. The tangents PA and PB are drawn from an external point P to two concentric circles with center O and radii 8 cm and 5cm respectively, as shown in fig.if AP = 15 cm , then find the length of  BP.

Solution.

Drawing the line by joining O to P

In ΔAOP, AP = 15 cm, OA = 8 cm, OB = 5 cm

∠OAP = 90°

Applying the Pythagoras theorem

OP² = OA² + AP²

OP² = 15² + 8²

OP² = 225 + 64

OP = √(289)

OP = 17

In ΔOPB, ∠OPB = 90°

Applying the pythogorus theorem

BP² = OP² – OB² = 17² – 5²

BP² = 289 – 25 = 264

BP = 2√66

Therefore the length of the tangent BP is 2√66 cm

Q12.How many four digits numbers are divisible by 7.

Solutions.

The smallest four-digit number is = 1000

Dividing it by 7 we get reminder 6

The first 4 digit number divisible by 7 = 1000 + (7–6) = 1001

The largest four-digit number = 9999

Dividing it by 7 we get  reminder 3

Subtracting it from 9999 we get 9996 the largest four-digit number divisible by 7

Therefore series of the 4 digit number divisible by 7

1001, 1008,1015, 1022……….9996

Applying the formula for n th term of an AP

9996 = 1001 + (n – 1)×7

7n – 7 = 9996 –1001

7n –7= 8995

7n = 9002

Therefore the number of 4 digit numbers divisible by 7 are 1286.

Q13. Solve the following quadratic equation for x: 4√3 x² + 5x – 2√3 = 0.

Solution.

4√3 x² + 5x – 2√3 = 0

The product of 4√3 × 2√3 = 8 × 3 = 24

Splitting up 5 into 8 – 3

4√3 x² + (8–3) x – 2√3 = 0

4√3 x² + 8x – 3x – 2√3

4x (√3x + 2) – √3( √3x + 2)

(√3x + 2)(4x –√3)

Q15. A card is drawn at random from a well-suffled pack of 52 playing cards. Find the probability that the drawn card is neither a king nor a queen.

Solution.

Total number of possible outcome = 52

Number of queens  and kings = 4 + 4 = 8

Number of the cards in which neither a king nor a queen = 52 – 8 = 44

 

So, the required probabilty is .

                                 Section C(3 marks)

Q16. A  vessel is in the form of a hemispherical bowl surmounted by a hollow cylinder of the same diameter of the hemispherical bowl is 14 cm and the total height of the vessel is 13 cm. Find the total surface area of the vessel.

Solution.

The radius of cylinder (r) = 14/2 = 7 cm

The height of the cylinder is (h) = 13 – 7 = 6 cm

The total surface area of the vessel = inner surface area + outer surface area

∵ inner surface area = outer surface area ( here thickness of the vessel is not given so inner  and outer radii are supposed the same)

The total surface area of the vessel = 2(curved surface area of cylinder + curved surface area of the hemisphere) (∵ the vessel is hollow)

The surface area of the vessel = 2(2πrh + 2πr²) = 4πr(h + r)

The surface area of the vessel 

The surface area of vessel = 88 × 13 = 1154

The surface area of the vessel is = 1154 sq.cm

Q17.Find the ratio in which the y-axis divides the line segments joining the points (–4,–6) and (10,12) also, find the coordinates of the point of division.

Solution.

The abscissa of the co-ordinates of the point of intersection of the y-axis and the line segment is 0, so let the co-ordinates  in which the y-axis divides the line segment are (0,y)

Applying the section formula

 

Applying the formula for the value of x because abscissa of intersecting coordinates is known i.e o and substituting the following values.

Substituting the in the formula for the value of y

The required co-ordinates are and the ratio in which these coordinates divides the line segments is

Q19.For what values of k,  the roots of the quadratic equation (k+4)x² + (k+1)x +1 = 0 are equal.

Solution.

(k+4)x² + (k+1)x +1 = 0

When roots of the quadratic equation ax² + bx + c = 0 are equal then we have b² – 4ac = 0

Substituting the value of a,b and c from the given quadratic equation

(k+4)x² + (k+1)x +1 = 0

(k+1)² – 4(k+4) = 0

k² + 1 + 2k – 4k –16 = 0

k² – 2k – 15 = 0

k² –5k + 3k – 15 = 0

k(k –5) + 3(k –5) = 0

(k –5)(k +3) = 0

k = 5, = –3

Q18. In fig. a circle is inscribed in triangle ABC touches its sides AB,BC and AC at points D,E and F respectively . If AB = 12 cm, BC = 8 cm and AC =10 cm,then find the length of AD, BE and CF.

Solution.AB = 12 cm, BC = 8 cm, AC = 10 cm

Let the AD = x

Applying the theorem that the tangents drawn from an external point to the circle are equal

BD = 12 – x, BD = BE = 12 – x , CE = BC – BE = 8 – (12 – x) = 8 – 12 + x = x – 4

CE = CF = x – 4, AF = AC – CF = 10 – (x – 4) = 10 – x + 4 = 14 – x

AF = AD

14  – x = x

2x = 14⇒ x = 7

AD =x = 7 cm, BE = 12 – x = 12 – 7 = 5 cm. CF = x – 4 = 7 – 4 = 3 cm

Therefore length of AD, BE and CF are  7 cm, 5 cm and 3 cm respectively.

Section C (3 marks)

Q19. In a circle of radius 21 cm, an arc subtends an angle of 60° at the center. Find (i) the length of the arc (ii) area of the sector formed by the arc.

Solution. Length of an arc (l) is given by the following formula

r = 21 cm, θ = 60°

l= 22

The length of arc = 22 cm

Area of the sector formed by the arc =

= 231

Therefore the length of arc is 22 cm and the area of the sector is 231 square cm.

Q20. The first and last terms of an AP are 5 and 45 respectively . If the sum of all its terms is 400 , find its common difference.

Solution.

Applying the sum of  n term formula of an AP

  where l (last term) = a + (n – 1)d

In the question, we are given a = 5, =45 and

25n = 400

n = 16

For calculating the value of common difference(d), applying the n th term formula of an AP.

45 = 5 + (16 – 1)d

15d = 40

Thus the common difference of given AP is .

Q21. Prove that the line segment joining the points of contact of two parallel tangents of a circle passes through its center.

Solution.

Given- A circle with center O in which two parallel tangents exist XY∥PQ

Construction- Drawing a radius OC∥XY∥PQ

To prove – AB passes through O

Proof – ∵XA ⊥ OA ( the angle between tangent and radius =90°)

∴ ∠XAO = 90°

∠AOC = 180°–  90° = 90°(OC∥ XA)……….(i)(sum of co-interior angles)

∵ OB⊥ PB ( the angle between tangent and radius =90°)

∴ ∠PBO = 90°

∠BOC = 180°–  90° = 90°(OC∥ PB)………(ii)(sum of co-interior angles)

From (i) and (ii) we have

∠AOC + ∠BOC = 90°  + 90°  =180°

∴ AOB is a straight line which passes through O.

Q22. Rahim tosses two coins simultaneously. Find the probability of getting.

(i) At least 1 tail

(ii) At most 2 head

(iii) Exactly two head

Solution. The total possible outcomes after tossing two coins =(H,H),(T,T),(H,T),(T,H)

(i) The outcomes of at least 1 tail =(T,H),(H,T),(T,T)

Hint: at least means 1 or more than 1 tail

(ii) The outcomes of almost two heads = (H,H),(T,T),(T,H),(H,T)

At most two heads mean two or less than two heads

(iii) The outcomes of exactly two head =(H,H)

Q23. Solve the following equation.

Solution.

(4–3x)(2x + 3) = 5x

8x + 12 –6x ²– 9x = 5x

–6x² –6x + 12 = 0

6x² + 6x – 12 = 0

x² + x – 2 = 0

x² + 2x – x – 2 = 0

x(x + 2) –1(x +2) =0

(x + 2)(x –1) = 0

x = –2, 1

Therefore the solution of the given equation is –2 and 1

Q24. If the coordinates of the points A and B are (–2,–2) and (2,–4) respectively, find the co-ordinates of P such that where P lies on the line segment AB.

Solution. We are given that

The P point lies in the line segment AB ,so getting its posion we have to determine AP : BP

∵ AP +BP = AB⇒AB–AP = BP

P divides the line segment joining the points A(–2,–2)  and B(2,–4) in ratio of 3: 4, let co-ordinates of P are(x,y)

Hence co-ordinates of P are

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Section D (4 marks)

Q25. The probability of selecting a red ball at random from a jar that contains only red, blue and the orange ball is . The probability of selecting a blue ball at random from the same jar . If the jar contains 10 orange balls, find the total number of the balls in the jar.

Solution.

Let the  number of blue balls in the jar = x

The number of orange balls in the jar = 10

The number of  red balls = y

Total number of balls in the jar = x +y+ 10

According to question

3x = x +y +10

2x –y = 10………….(i)

according to question

4y = x +y +10

3y –x = 10…………(ii)

Solving equation number (ii) for y

Substituting the value of y in equation number (i)

6x – 10 –x = 30

5x = 40

x = 8

Putting the value x in eq.(i),we get y= 6.

Therefore the number of blue balls are 8 and the number of red balls are 6 and total balls in the jar =x+y+10=8+6+10=24

Q26.A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72 cylindrical bottles of diameter 6 cm. Find the height of each bottle, if 10% liquid is wasted in this transfer.

Solution. Let the height of cylindrical bottles is h cm

The volume of each bottle =πr²h

r = 6 /2 = 3cm

The volume of each bottle =π ×3²h =9πh

The volume of 72 bottles = 72×9πh=648πh

Volume of spherical bowl

10% of the volume of a spherical bowl = 388.8π cubic cm

The water-filled in the bottles = Volume of bowl – Water wasted

= 3888π – 388.8π =3499.2π cubic cm

The volume of 72 bottles =648πh = 3499.2π

h = 5.4

Hence height of bottle = 5.4 cm

Q27.Find the area of the minor segment of a circle of radius 14 cm, when its central angle is 60°. Also, find the area of the corresponding major segment(use).

Solution.

Area of minor segment = Area of minor sector – Area of ΔAOB

In ΔAOB , ∠AOB = 60°

Since OA = OB (radii of the same circle)

∠BAO = ∠ABO=x (OA = OB)

x + x + 60 = 180

x =60°, therefore ΔAOB is an equilateral triangle

OA = 14 cm

The area of major segment is

Q28.If the point P(x,y) is equidistant from the points A(a+b,b–a) and B(a –b,a +b), prove that bx = ay.

Solution.As the point P(x,y) is equidistant from the points A(a+b,b–a) and B(a –b,a +b)

∴PA = PB

[x –(a +b)]² + [y –(b–a)]² =[x –(a –b)]² + [y –(a +b)]²

x² + (a +b)² –2x(a +b) + y² + (b–a)²– 2y(b– a) =x² + (a –b)² –2x(a– b) + y² + (a +b)²– 2y(a +b)

–2x(a +b) + (b–a)²–2y(b–a) = (a –b)²–2x(a– b)– 2y(a +b)

–2xa –2xb + b² + a² –2ab –2yb + 2ya = a² + b²–2ab –2xa +2xb –2ya –2yb

–2xb–2xb = –2ya–2ya

–4xb = –4ya

bx = ay, Hence proved.                          

Q29. A motorboat whose speed is 24 km/hr in still water takes 1 hr more to go 32 km upstream than to return downstream to the same spot. Find the speed of the stream.

Solution.

Let the speed of the stream = x km/hr

Distance covered  by the boat in upstream or downstream =32 km

 The speed of the river in still water = 24 km/hr

The speed of the boat in upstream = (x – 24) km/h

The time is taken by the boat to go up to 32 km in upstream =

The time is taken by the boat to go down to 32 km in downstream =

According to question

64x = 576 –x²

x²  + 64x – 576=0

x² + 72x – 8x – 576 =0

x(x + 72) –8(x +72) =0

(x + 72)(x –8) =0

x = –72, 8

Neglecting the negative sign, therefore the speed of the stream =8 km/hr

Q30.In fig.the vertices of ΔABC are A(4,6), B(1,5) and C(7,2).A line segment DE is drawn to intersect the sides AB and AC at D and E respectively such that  .Calculate the area of ΔADE  and compare it with the area of ΔABC.

Solution.

We are given that

The line DE is intersecting the sides AB and AC of the Δ ABC in the same ratio of 1 : 3

∴ DE ∦BC (opposite of BPT theorem)

∠ADE = ∠ABC (corresponding angle)

∠AED = ∠ACB (corresponding angle)

∠A = ∠A (common)

ΔABC ∼ ΔADE (AA rule)

According to the theorem that the ratio of areas of two similar triangles is in the ratio of the square of their corresponding sides.

…………….(i)

The coordinates of the vertices of ΔABC are A(4,6), B(1,5) and C(7,2).

                         

                       

arΔABC = 9arΔADE

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 maths

Chapter 1- Number System Chapter 9-Areas of parallelogram and triangles
Chapter 2-Polynomial Chapter 10-Circles
Chapter 3- Coordinate Geometry Chapter 11-Construction
Chapter 4- Linear equations in two variables Chapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s Geometry Chapter 13-Surface Areas and Volumes
Chapter 6-Lines and Angles Chapter 14-Statistics
Chapter 7-Triangles Chapter 15-Probability
Chapter 8- Quadrilateral

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Chapter 1-Matter in our surroundings Chapter 9- Force and laws of motion
Chapter 2-Is matter around us pure? Chapter 10- Gravitation
Chapter3- Atoms and Molecules Chapter 11- Work and Energy
Chapter 4-Structure of the Atom Chapter 12- Sound
Chapter 5-Fundamental unit of life Chapter 13-Why do we fall ill ?
Chapter 6- Tissues Chapter 14- Natural Resources
Chapter 7- Diversity in living organism Chapter 15-Improvement in food resources
Chapter 8- Motion Last years question papers & sample papers

NCERT Solutions for class 10 maths

Chapter 1-Real number Chapter 9-Some application of Trigonometry
Chapter 2-Polynomial Chapter 10-Circles
Chapter 3-Linear equations Chapter 11- Construction
Chapter 4- Quadratic equations Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression Chapter 13-Surface areas and Volume
Chapter 6-Triangle Chapter 14-Statistics
Chapter 7- Co-ordinate geometry Chapter 15-Probability
Chapter 8-Trigonometry

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Chapter 1- Chemical reactions and equations Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds Chapter 12- Electricity
Chapter 5-Periodic classification of elements Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process Chapter 14-Sources of Energy
Chapter 7-Control and Coordination Chapter 15-Environment
Chapter 8- How do organisms reproduce? Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-Sets Chapter 9-Sequences and Series
Chapter 2- Relations and functions Chapter 10- Straight Lines
Chapter 3- Trigonometry Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations Chapter 15- Statistics
Chapter 8- Binomial Theorem  Chapter 16- Probability

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NCERT Solutions for Class 11 Physics

Chapter 1- Physical World

chapter 3-Motion in a Straight Line

NCERT Solutions for Class 11 Chemistry

Chapter 1-Some basic concepts of chemistry

Chapter 2- Structure of Atom

NCERT Solutions for Class 11 Biology

Chapter 1 -Living World

NCERT solutions for class 12 maths

Chapter 1-Relations and Functions Chapter 9-Differential Equations
Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra
Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry
Chapter 4-Determinants Chapter 12-Linear Programming
Chapter 5- Continuity and Differentiability Chapter 13-Probability
Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals

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