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.
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.
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
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| Chapter 2- Relations and functions | Chapter 10- Straight Lines |
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