JKBOSE MATHEMATICS: Solved Model Question Paper for Class 10TH MATHEMATICS NCERT (2023)
Question Number 1 to 15 are MCQs of 1 marks each. Select the correct option:
Q1. The HCF of 2 and 11 is
a) 2 b) 11 c) 22 d) 1
Answer: d) 1
Q2. A polynomial of degree 2′ is called
a) Quadratic poly b) Zero Poly C) Quartic Poly D) None of these
Answer: A) Quadratic Poly
Q3. A Quadratic Equation aX+ bX+ C=0, a≠0 has two equal roots if
a) D>o b) D=0 C) D <0 D) N.O.T
Answer: B) D=0
Q4. The Common difference of the AP.6, 9,12,15 —is:
a) 6 b) -3 c) 9 d 3
Answer: d) 3
Q5. A line which touches a circle at one point is called
a) Secant b) Chord c) tangent d N.O.T
Answer: c) tangent
Q6. The area of the circle is given by
a) Ar b) 2Ar c}R d) N.O.T
Answer: b) 2Ar
Q7. Which of the following cannot be the probability of an event
A) 2/3 b) -1.5 c) 15% d) 0.7
Answer: -1.5
Q8. The value of Sin 18/Cos 72′ is:
a) -1 b) o c) 1 d) v3
Answer: c) 1
Q.Nos: Fill in the blanks. Each question is of 1 marks.
Q9. LCM (a,b) x HCF (a,b) = —————-(axb/a+b).
Answer: Yes, this is correct.
Q10. x=1, y=2 is the solution of the pair of linear equation
Answer: The solution of the pair of the linear equation x + y = 3 and x – y = -1 is x = 1, y = 2.
Q11. x+2y=3 and x+y=3 Yes/No)
Answer: Yes
Q12. qn=q+(n+1)d is the general term of an AP————- (True/False)
Answer: True
Q13.The sum of the first p term of an A.P is given by
Answer: The sum of the first p terms of an Arithmetic Progression (AP) is given by, S_p = (p/2) [2a + (p-1)d] Where p = number of terms a = first term d = common difference
Q14. Sp= p/2 (2a+(p-1)D L (True/False)
Answer: True
Q15.All—————– triangles are similár (lsoscèles/Equilateral).
Answer: Triangles can be either isosceles or equilateral.
Short Answer type questions of 1 marks each
Q16. Define Collinear Points
Answer: Collinear points are points that lie in a straight line on the same plane. They can be three or more points, and they all share the same slope.
Q17. Write a formula for finding the area of a triangle ABC with Coordinates of the Vertices as A (xLyi), B {x2, Y2), C (x3.y).
Answer: Area = 1/2 * | x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) |
Q18. Write One application of Trigonometry
Answer: Trigonometry is used to measure the distance, direction, and angle of various objects in relation to a ship or aircraft, allowing for more precise navigation.
Q19. State Pythagoras theorem?
Answer: Pythagoras’ theorem states that in a right triangle, the sum of the squares of the two shorter sides of the triangle is equal to the square of the hypotenuse.
Mathematically, it can be written as a2 + b2 = c2, where a and b are the lengths of the two shorter sides and c is the length of the hypotenuse.
Q21. Given r=1 unit, find the vol. of Sphere.
Answer: The volume of a sphere with a radius of 1 unit is 4.1887902047863905.
Section-B (Carry 2 marks each)
Q22: 2 Cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting Cuboid.
Solution
The volume of cube 64cm3
Side of cube 3√64=4cm
Length of resulting cuboid 4+4=8
Surface area =2 (lb+hl+bh)
=2(4(4) +4(8)+8(4))=2(16+32+32)=2(80)=160cm2
Q23. Given that HCF (306,657)=9, find LCM(306,657)
We know that
HCF (a, b)×LCM (a, b) = a × b.
i.e., 9×LCM=306×657
⇒LCM=306 × 6579
=2010429
=22338
∴ LCM of 306 and 657 is 22338.
Q24. Check for Consistency 5x-4y=8=0, 10x-8y+16=0
Solution
For the given pair of equations 5x+4y=20 and 10x+8y=16;
a2a1=105=21
c2b1=84=21
c2c1=−16−20=45
As the given pair of linear equations satisfies –
a2a1=b2b1=c2c1
The graph will be a pair of parallel lines and so the pair of equations will have no solution.
Q25. One A die is thrown Once. Find the probabilitý of getting an odd number
Solution
The total outcomes that can occur are 1,2,3,4,5,6
Number of possible outcomes of a dice=6
Numbers which are odd=1,3,5
Total numbers which are odd=3
Probability of getting an odd number
=(Number of outcomes where there is an odd)÷(Total number of outcomes)
=6/3=2/1
Section-C
Q26. Find the zero’s of the quadratic polynomial and verity. the relationship between the zero’s and the coefficients 4S cm2-4+1
Solution
Step 1: Form the equation
The given polynomial is 4s2 – 4s + 1
We are aware that a polynomial’s zeroes are evaluated by equating them to zero.
p(s) = 0
Hence the above equation becomes 4s2 – 4s + 1 = 0.
Step 2: Solve to find the zeroes
4s2 – 4s + 1 = 0
The above equation can be written as:
2s)2 – 2 (2s) + 12 = 0
(2s + 1)2 = 0 [As (a – b)2 = a2 – 2ab + b2]
s = ½, ½
Step 3: Verification
We know that for a given polynomial as2 + bs + c
Sum of the zeroes = -b/a and
product of the roots = c/a
Substituting the values in the above formula we get:
Sum of the zeroes = ½ + ½ = 1
Again, -b/a = – (-4)/4 = 1
Product of the zeroes = ½ x ½ = ¼
Again, c/a = ¼
Thus, the relationship between the zeroes and the coefficients of the polynomial is verified.
Hence, the zeroes of this polynomial are ½ and ½.
Question: Divide the polynomial p(x)=x3−3×2+5x−3 by the polynomial g(x)=x2−2. Find the quotient and remainder.
Solution
Given:-
p(x)=x3−3×2+5x−3
g(x)=x2−2
p(x)÷g(x)
(x3−3×2+5x−3)÷(x2−2)
x2−2×3−3×2+5x−3=(x−3)+(x2−2)(7x−9)
Quotient =(x−3)
Remainder =(7x−9)
Q28. Solve the pair of linear equations by substitution method.
X + Y = 14, X – Y = 4
Solution
The correct option is D)
x+y=14 ….(i)
x−y=4 …..(ii)
From (ii) y=x−4 …(iii)
Substituting y from (iii) in (i), we get
x+x−4=14
⇒2x=18
⇒x=9
Substituting x=9 in (iii), we get
y=9−4=5,
i.e, y=5
x=9,y=5
Q29. Which term of an AP:3,8,13,18. Is———— 78?
Solution
The given sequence is 3, 8, 13, 18, ….78.
‘a’ = 3, ‘d’ = 5, last term = 78, ‘n’ = ?
tn=a+(n−1) d
78=3+(n−1)5
78=3+5n−5
78= 5n−2
5n=80 or n=16.
The 16th term is 78.
Q30. Find the sum of the first 15 multiples of 8
Solution
The multiples of 8 are 8,16,24,32…
These are in an A.P., having the first term as 8 and the common difference as 8.
Therefore, a=8,d=8, S15=?
Sn=2/n[2a+(n−1)d]
S15=15/2[2(8)+(15−1)8]
=15/2[16+(14)(8)]
=15×128/2
=960
Q31. Prove that the tangents drawn at the ends of a diameter of a circle are parallel
To prove: PQ∣∣ RS
Given: A circle with centre O and diameter AB. Let PQ be tangent at point A & Rs be point B.
Solution
Proof: Since PQ is a tangent at point A.
OA⊥ PQ(Tangent at any point of the circle is perpendicular to the radius through the point of contact).
∠OQP=90o …………(1)
OB⊥ RS
∠OBS=90o ……………(2)
From (1) & (2)
∠OAP=∠OBS
i.e., ∠BAP=∠ABS
for lines PQ & RS and transversal AB
∠BAP=∠ABS i.e., both alternate angles are equal.
So, lines are parallel.
$$\therefore PQ||RS.
Q32. Prove that the |lgm circumscribing a circle is a rhombus
Solution
Since ABCD is a parallelogram circumscribed in a circle
AB=CD……..(1)
BC=AD……..(2)
DR=DS (Tangents on the circle from same point D)
CR=CQ(Tangent on the circle from same point C)
BP=BQ (Tangent on the circle from same point B )
AP=AS (Tangents on the circle from same point A)
Adding all these equations we get
DR+CR+BP+AP=DS+CQ+BQ+AS
(DR+CR)+(BP+AP)=(CQ+BQ)+(DS+AS)
CD+AB=AD+BC
Putting the value of equations 1 and 2 in the above equation we get
2AB=2BC
AB=BC………..(3)
From equations (1), (2) and (3) we get
AB=BC=CD=DA
∴ ABCD is a Rhombus
Q33. Find the area of a sector of a circle with a radius of 6cm if the angle of the sector is 60 degrees
Solution:
We use the formula for the area of the sector of a circle.
The formula for the area of the sector of a circle with radius ‘r’ and angle θ = (θ/360°) × πr2
Given, θ = 60°, Radius = 6 cm
Area of the sector = (θ/360°) × πr2
= 60°/360° × 22/7 × 6 × 6
= 132/7 cm2
34. A drinking glass is in the shape of a frustum with a core of height 14cm. the diameter of its two circular ends is 4cm and 2cm. find the capacity of the glass.
Solution
Upper base diameter, D =4 cm
Lower base diameter, d =2cm
Height, h =14 cm
So, R (upper base) =2 cm,
r(lower =1cm
The capacity of glass = Volume of a frustum of a cone
=πh/3[R2+Rr+r2]
=π×14/3[22+2×1+12]
=22×14[4+1+2]/3×7
=3308/3=10 2/ 3 cm 3
So, the capacity of the glass is 10 3/2 cm3
Section-D
Q35. Find two numbers whose sum is 27 and whose product is 182.
Solution:
Let us consider one of the numbers to be x. Then the other number will be 27 – x.
The product of the two numbers is given as 182
This can be written in the form of the following quadratic equation
x(27 – x) = 182
x(27 – x) = 182
27x – x2 = 182
27x – x2 – 182 = 0
x2 – 27 x + 182 = 0 [Rearranging the terms and multiplying both sides by a negative sign]
x2 – 14x – 13x + 182 = 0
x (x – 14) – 13 (x – 14) = 0
(x -13) (x – 14) = 0
x – 13 = 0 and x – 14 = 0
x = 13 and x = 14
Therefore, the required numbers are 13, and 14.
Q36.The of angle elevation of the top of a tower from a point on the ground which is 30m away from the foot of the tower is 30 degrees. Find the height of the tower.
Solution
We have to find the height of the tower.
Let us consider the height of the tower as AB and the distance between the foot of the tower to the point on the ground as BC.
In ΔABC, the trigonometric ratio involving AB, BC and ∠C is tan θ.
tan C = AB/BC
tan 30° = AB/30
1/√3 = AB/30
AB = 30/√3
= (30 × √3) / (√3 × √3)
= (30√3)/3
= 10√3
Height of tower AB = 10√3 m.
Q 37. Find the points on the x-axis which is equidistant from (2,-5) and (-2, 9)
Solution:
The distance between any two points can be measured using the Distance Formula which is given by: Distance Formula = √ [(x₂ – x₁)2 + (y₂ – y₁)2]
Let’s assume a point P on the x-axis which is of the form P(x, 0).
We have to find a point on the x-axis which is equidistant from A (2, – 5) and B (- 2, 9).
To find the distance between P and A, substitute the values of P (x, 0) and A (2, – 5) in the distance formula.
PA = √(x – 2)² + (0 – (- 5))²
= √(x – 2)² + (5)² ——— (1)
To find the distance between P and B, substitute the values of P (x, 0) and B (- 2, 9) in the distance formula.
PB = √(x – (- 2))² + (0 – 9)²
= √(x + 2)² + (- 9)² ———- (2)
By the given condition, these distances are equal in measure.
Hence, PA = PB
√(x – 2)² + (5)² = √(x + 2)² + (- 9)² [From equations (1) and (2)]
Squaring on both sides, we get
(x – 2)2 + 25 = (x + 2)2 + 81
x2 + 4 – 4x + 25 = x2 + 4 + 4x + 81
8x = 25 – 81
8x = – 56
x = – 7
Therefore, the point equidistant from the given points on the x-axis is (- 7, 0).
Q 38. Find the ratio in which the line segment joining the points (-3,10) and (6,-8) is divided by (-1,6)
Solution:
The coordinates of the point P(x, y) which divides the line segment joining the points A(x₁, y₁) and B(x₂, y₂), internally, in the ratio m₁: m₂ is given by the Section Formula: P(x, y) = [(mx₂ + nx₁) / m + n, (my₂ + ny₁) / m + n]
Let the ratio in which the line segment joining A(- 3, 10) and B(6, – 8) be divided by point C(- 1, 6) be k : 1.
By Section formula, C(x, y) = [(mx₂ + nx₁) / m + n, (my₂ + ny₁) / m + n]
m = k, n = 1
Therefore,
– 1 = (6k – 3) / (k + 1)
– k – 1 = 6k – 3
7k = 2
k = 2 / 7
Hence, point C divides the line segment AB in the ratio 2: 7.
Q40. Question: ABC is an issosless right angled at C. prove that AB2= 2AC2?
Solution:
We know that, in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
In ΔABC, ∠ACB = 90° and AC = BC [Since ABC is an isosceles triangle right angled at C]
Using Pythagoras’ theorem,
⇒ AB2 = AC2 + BC2
⇒ AB2 = AC2 + AC2 [Since AC = BC]
Therefore, AB2 = 2 AC2
Q41. Question: construct a triangle with sides 5cm, 6cm, & 7cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle.
Solution:
Draw the line segment of the largest length of 7 cm. Measure 5 cm and 6 cm separately and cut arcs from both ends of the line segment such that they cross each other at one point. Connect this point from both the ends
Then draw another line that makes an acute angle with the given line (7 cm). Divide the line into m + n parts where m and n are the ratios given.
Two triangles are said to be similar if their corresponding angles are equal and said to satisfy the AA criteria.
The basic proportionality theorem states that “If a straight line is drawn parallel to a side of a triangle, then it divides the other two sides proportionally”.
Steps of construction:
Draw BC = 7cm with B and C as centres and radii 5 cm and 6 cm respectively. Draw arcs to intersect at A. ΔABC is obtained.
Draw ray BX making ∠CBX an acute angle.
Mark 7 points (greater of 7 and 5 in 7/5 ) B₁, B₂,………B₇ on BX such that BB₁ = B₁B₂ =…………… = B₆B₇
Join B₅ (smaller of 7 and 5 in 7/5 which is the 5th point) to C and draw B₇C’ parallel to B₅C intersecting the extension of BC at C’.
Through C’ draw C’A’ parallel to CA to meet the extension of BA at A’. Now, ΔA’ B’C’ is the required triangle similar to ΔABC where BA’/BA = C’A’/CA = BC’/BC = 7/5
Proof:
In ΔBB₇C’, B₃C is parallel to B₇C’
Hence by the Basic proportionality theorem,
B₅B₇/BB₅ = CC’/BC = 2/5
Adding 1 to both the sides of CC’/BC = 2/5,
CC’/BC + 1 = 2/5 + 1
(BC + CC’)/BC = 7/5
BC’/BC = 7/5
Consider ΔBAC and ΔBA’C’
∠ABC = ∠A’BC’ (Common)
∠BCA = ∠BC’A’ (Corresponding angles ∵ CA || C’A’)
By AA criteria, ΔBAC ∼ ΔBA’C’
∴ The corresponding sides are proportional
Hence,
BA’/BA = C’A’/A = BC’/BC = 7/5
Q42. The distribution below gives the weight of 3 students in a class. Find the median weight of the students
Solution:
We know that,
Median = l + [(n/2 – cf)/f] × h
Class size, h
Number of observations, n
The lower limit of median class, l
Frequency of median class, f
Cumulative frequency of class preceding median class, cf
n = 30 ⇒ n/2 = 15
From the table, it can be observed that cumulative frequency (cf) just greater than 15 is 19, belonging to class 55 – 60.
Therefore, median class = 55 – 60
Class size, h = 5
The lower limit of median class, l = 55
Frequency of median class, f = 6
Cumulative frequency of class preceding median class, cf = 13
Median = l +[ (n/2 – cf)/f] × h
= 55 + [(15 – 13)/6] × 5
= 55 + (2/6) × 5
= 55 + 5/3
= 55 + 1.67
= 56.67
Therefore, the median weight is 56.67 kg.