Class 10 Maths(Basic) Preboard Exam (First) 2021-22 CBSE with Solutions
10 Maths(Basic) Preboard Exam (First) 2021-22 CBSE with Solutions is created for the class 10 Maths students who has chosen Maths -Basic, question paper of maths basic consists of 50 multiple questions. The students have to solve 40 questions. Each question carries 1 mark. The basic maths question paper with solutions is helpful for the students in their Term -1 exam CBSE Board 2021.The importance of studying these solutions of basic maths question paper preboard exam is to get the idea about the type of the questions in Term 1 CBSE Board exam 2021.
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Class 10 Maths(Basic) Preboard Exam (First) 2021-22 CBSE with Solutions
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Q1. After how many places of decimal will the decimal expansion of the number 27/(2²×5³) terminate ?
(a) 3 (b) 4 (c) 4 (d) 1
Ans. (a) 3
(27)/(2²×5³)
Multiplying the denominator and numerator by 2
54/(2×5)³= 0.054
Q2. A die is thrown once. The probability of getting an even number is :
(a) 1/3 (b) 1/6 (c) 1/4 (d) 1/2
Ans.(d) 1/2
When a die is thrown once,the total possible outcomes are = 6(1,2,3,4,5,6)
Favourable outcomes = 2,4 and 6
P(E) = favourable outcomes/total possible outcomes
P(even number) = 3/6 = 1/2
Q3. Sum of two numbers is 35 and their difference is 13,then the numbers are
(a) 24,13 (b) 24,11 (c) 12,11 (d) 12,23
Ans (b) 24,11
(Let the numbers are x and y
x + y = 35…….(i) and x – y = 13……(ii)
Adding both equation
2x = 48 ⇒x = 24
Putting the value in equation (i)
24 + y = 35
y = 11
Q4. The value of sin 60°/cos 30° is
(a) √3/2 (b) 1/2 (c) 1 (d) 2
Ans. (c) 1
Q5. If a bag contains 3 red and 7 black balls, then what will be the probability of getting a black ball ?
(a) 3/10 (b) 4/10 (c) 7/10 (d) 5/10
Ans. (c) 7/10
Total number of balls are = 3 red balls + 7 black ball = 10 balls
The number of black balls = 7
P(black ball) = 7/10
Q6.The HCF of 513 and 26 will be
(a) 0 (b) 1 (c) 13 (d) 26
Ans.(b) 1
The numbers given 513 and 26are co-prime numbers(1 is a common factor between them)
- Q7. Equations a1x + b1y + c1= 0 and a2x + b2y + c2= 0 has infinite many solutions if :
(a) a1/ a2≠b1/ b2 (b) a1/ a2=b1/ b2 (c) a1/ a2= b1/ b2=c1/ c2 (d) a1/ a2= b1/ b2≠ c1/ c2
Ans. (c) a1/ a2= b1/ b2=c1/ c2
Class 10 Maths(Basic) Preboard Exam (First) 2021-22 CBSE with Solutions
Q8. sin 2A = 2 sin A is true when A is equal to
(a) 0° (b) 30° (c) 45° (d) 60°
Ans. (a) 0°
The value of LHS sin (2×0°) =sin 0° = 0
The value of RHS 2 sin A = 2× sin 0° = 2×0 = 0
Hence,at A = 0,sin 2A = 2sin A
Q9. If the ratio of areas of two circle are 4 : 9, then the ratio of their radii will be:
(a) 4 : 9 (b) 2 : 3 (c) 8 : 27 (d) 3 : 2
Ans. (b) 2 : 3
Area of the circle is = πr²
Let the radii of two circles are r1 and r2
πr1²/πr2² = 4/9
r1²/r2² = 4/9
r1/r2 = 2/3
Q10. Distance between the points (-1,3) and (2,-1) is
(a) 1 unit (b) 6 units (c) 5 units (d) 7 units
Ans. (c) 5 units
The distance(d) between two points (x1,y1 ) and (x2,y2) is
d= √[(x2-x1)²+ (y2– y1)²]
The distance between (-1,3) and (2,-1) is
d = √[(2+1)²+ (-1- 3)²] =√[3² +(-4)²] =√(9 +16) =√25 = 5 units
Q11. How many rational numbers are there between any two rational numbers ?
(a) 1 (b) 2 (c) 3 (d) infiitely many
Ans. (d) infinitely many
There are numberless numbers and any sets of numbers are infinite, hence between two rational numbers, there are infinite numbers.
Q12. If coordinates of the diameter of a circle are (-6,3) and (6,9), then the coordinates of the centre is:
(a) (8,-8) (b) (12,6) (c) (0,6) (d) (0,3)
Ans. (c) (0,6)
The mid point between two points (x1,y1 ) and (x2,y2) is = (x1+x2 )/2, (y1+y2)/2
The midpoints of the diameter is the centre of the circle
The endpoints of the diameter are (-6,3) and (6,9)
The coordinates of the centre of the circle are = (-6+6)/2,( 3+9)/2 = (0,6)
Class 10 Maths(Basic) Preboard Exam (First) 2021-22 CBSE with Solutions
Q13.If quadratic polynomial p(x) = x² – x + 4 has roots α and β, then the value of α + β is :
(a) -1 (b) 1 (c) 4 (d) 0
Ans. (b) 1
In the standard quadratic equation ax² + bx + c, if the roots are α and β
α + β = -b/a
The given quadratic polynomial is
p(x) = x² – x + 4 ,where a = 1 and b = -1
α + β = -(-1)/1 = 1
Q14. is a /an
(a) Integer (b) Rational Number (c) Irrational Number (d) Natural Number
Ans.(b) Rational Number
= 2.353535……, which is a recurrent (repeating) infinite decimal numbers,it can be written in the form of P/Q,where P and Q are co-prime numbers
x = 2.353535…..(i)
Multiplying it by 100,we get equation (ii)
100 x = 235.3535….(ii)
Subtracting equation (i) from (ii)
99x = 233
x = 233/99
Q15. If tan (3x -15°) = 1,then the value of x is:
Ans.
tan (3x -15°) = 1
tan (3x -15°) = tan 45°
3x – 15° = 45°
3x = 60°
x = 20°
Q16.From a well shuffled cards,the probability of getting a red face card is:
(a) 3/26 (b) 6/13 (c) 3/52 (d) 1/13
Ans. Total number of cards are = 52
The number of red face cards are = 26
P( getting a red face card) = (number of red face cards)/(total number of cards) = 26/52 =1/2
Q17. In what ratio the point P(0,0) divides the line segment joining the points A(3,3) and B(-3,-3)?
(a) 1 : 2 (b) 1 : 3 (c) 1 : 4 (d) 1 : 1
Ans. (d) 1 : 1
Applying the section formula
x = (mx2+nx1)(m+n), y = (my2+ny1)(m+n)
We are given that the point P(0,0) divides the line segment A(3,3) and B(-3,-3)
0 = (m×-3 +n×3)/(m+n)
-3m + 3n = 0
-3m = -3n
m : n = 1 : 1
Q18. The pair of linear equation 6x – 7y = 1 and 3x – 4y = 5 has solution:
(a) Two solutions (b) Infinitely many solutions (c) Unique solution (d) No solution
Ans. (c) Uniques solution
From the given equation 6x -7y =1 and 3x -4y =5,we have the coefficients
are a1=6,b1=-7, c1=-1,a2=3,b2=-4,c2= -5
a1/ a2= 6/3 = 2, b1/ b2=(-7)/(-4) = 7/4, c1/c2= (-1)/(-5) = 1/5
a1/ a2≠ b1/ b2
Therefore both equations has a uniques solution
Class 10 Maths(Basic) Preboard Exam (First) 2021-22 CBSE with Solutions
Q19.If the distance between the points A (4,P) and B(1,0) is 5 units, then the value of P are :
(a) 4,0 (b) -4,0 (c) 4, -4 (d) 0, 1
Ans. (c) 4, -4
The distance(d) between two points (x1,y1 ) and (x2,y2) is
d= [(x2-x1)²+ √(y2– y1])²
The distance between the points A (4,P) and B(1,0) is 5 units
5 = [(1-4)²+ √(0- P])²
25 = (-3)² + (-P)²
P² = 25 -9 = 16
P = ± 4
Q20. 225 can be expressed as :
(a) 5² × 3² (b) 5²× 3 (c) 5²× 3² (d) 5³× 3
Ans. 5²×3²
225 = 3×3×5×5 = 5²× 3²
Q21. In an isosceles triangle ABC, if AC = BC and AB² = 2AC², then value of C is:
(a) 30° (b) 45° (c) 60° (d) 90°
Ans.
The given triangle is isosceles triangle in which AC = BC and we are given AB² = 2AC²
AB² = 2AC² = AC² + AC²
Putting AC = BC
AB² = BC² + AC²
AB =Hyptenuse ,BC and AC are perpendicular to each other
∴ ∠C = 90°
Q22. In a triangle ABC ,DE parallel to BC, then the value of EC is
(a) 5 cm (b) 10 cm (c) 8 cm (d) 10 cm
Ans.(c) 8 cm
In ΔABC and ΔADE
DE parallel to BC and AB is the transversal
∠ADE = ∠ABC (corresponding angle)
∠AED = ∠ACB (corresponding angle)
∴ΔABC ∼ΔADE (AA rule criteria)
Applying the rule of similar triangles
AD/AB = AE/AC = DE/BC
3/(3 +4) = 6/AC
3/7 = 6/AC
3AC = 42
AC = 14
EC = AC – AE = 14 – 6 = 8 cm
Q23. The parimeter of semicircular protector whose radius ‘r ’ is
(a) π + r (b) πr (c) π + 2r (d) πr + 2r
Ans. The perimeter of semicircular protector = Length of the semicircular arc + Diameter
Length of semicircular arc = (2πr)/2 =πr
Diameter = 2r
Hence perimeter of semicircular protector = πr + 2r
Q24. What is the area of a sector of a circle with radius 14 cm and central angle 45°?
(a)76 cm² (b) 77 cm² (c)66 cm² (d) 55 cm²
Ans. (b) 77 cm²
Area of sector of a circle is given as
Where θ = 45°, r = 14 cm
Class 10 Maths(Basic) Preboard Exam (First) 2021-22 CBSE with Solutions
Q25. What will be radius of the circle whose circumference is 44 m ?
(a) 14 m (b) 7 m (c) 5 m (d) 44 m
Ans.(b) 7 m
The circumerence of the circle =2πr
The circumference of the given circle is = 44 m
2πr = 44
2 (22/7) r = 44
r = (44×7)/(44) = 7 m
Q26. (1 +tan²A)/cosec²A will be equal to:
(a) sec²A (b) cosec²A (c) cot²A (d) tan²A
Ans. (d) tan²A
The given expression is
(1 +tan²A)/cosec²A
Converting tan A and cosec A into sin A and cos A
(1 + sin²A/cos²A)
Q27. If p and q are co-prime numbers, then the HCF of p³q² and p²q will be:
(a) p³q² (b) p²q (c) p²q² (d) pq
Ans. (b) p²q
Since p and q are co-prime numbers,therefore in p and q there is no common factor except to 1.
The HCF of p³q² and p²q is = the lower power of p in both terms × lower power of q in both terms =p²q
Q28. In equation x + 2y =9 ,if x = 5,then the value of y will be:
(a) 1 (b) -2 (c) 2 (d) 4
Ans. (c) 2
Putting x = 5 in given equation x + 2y =9
5 + 2y =9
2y = 9 -5 = 4
y = 4/2 = 2
Q29. If (2K -1,K) is the solution of 10 x -9y =12, then the value of K will be:
(a) 1 (b) 2 (c) 3 (d) 4
Ans. (b) 2
The solution of the given equation is 10x – 9y =12 is =(2K-1,K)
Therefore putting x = 2K-1 and y = k in the given equation
10(2K-1) – 9K =12
20 K -10 -9K = 12
11K = 10 + 12 = 22
K = 2
Q30. If x = a cosθ and y = b sinθ, then the value of b²x² +a²y² is :
(a) a² +b² (b) a²/b² (c) a²b² (d) 1/a² + 1/b²
Ans. (c) a²b²
Putting the value of x = a cos θ and y = b sin θ in the expression b²x² + a²y²
b² a² cos²θ + a²b² sin²θ
a²b²( cos²θ + sin²θ) = a²b²
Q31. Distance of the point Q(3,-4) from the origin is :
(a) 3 units (b) 4 units (c) 5 units (d) 7 units
Ans. (c) 5 units
The coordinates of the origins are (0,0) and the point given to us is Q(3, -4),distance(d) between two two points is given as
d=√ [(x2-x1)²+ √(y2– y1])²
= √[(3 -0)² + (-4 -0)²] = √(3² + 4²) = √(9 + 16) = √25 = 5 units
Q32. If the point C(K,4) divides the line segement joining the two points A(2,6) and B(5,1) in the ratio 2 : 3, then the value of K is:
(a) 4/5 (b) 16/5 (c) 11/5 (d) 19/5
Ans.(b) 16/5
Applying the section formula
x = (mx2+nx1)(m+n), y = (my2+ny1)(m+n)
Putting the value , x = K , y = 4, x1
=2,x2 = 5, y1= 6,y2=1, m : n = 2: 3
K=(2×5 + 3×2)/(2+3) =(10 +6)/5 = 16 /5
Q33. An event A is very unlikely to happen ,then its probabilty will be closest to :
(a) 0.0001 (b) 0.001 (c) 0.01 (d) 0.1
Ans.
The probabilty of an event is very likely to happen = P(A) = 1
The sum of probabilty of an event and not happening of the probabilty is equal to 1
Q34.In a linear equation ax +by +c = 0,a,b and c are:
(a) Natural number (b) Whole number (c) Integers (d) Real Number
Ans. (d) Real Number
As an example √2 x + 3y -4 = 0 ,3x + 3y – 5 = 0 ,both are linear equations,no matters if a,b and c are natural numbers,whole numbers,integers ,rational or irrational numbers ,so a,b and c should be real numbers.
Q35. If a pair of dice is thrown once,then probability of getting a sum of 8 will be :
(a) 1/6 (b) 5/36 (c) 1/9 (d) 1/12
Ans. (b) 5/36
If a pair of dice is thrown once then the total outcomes are = 6² = 36
The outcomes in which sum of the digits in both dice is 8 are = (2,6),(3,5),(4,4),(5,3),(6,2) [i.e 5)
P(getting the sum of 8) = 5/36
Class 10 Maths(Basic) Preboard Exam (First) 2021-22 CBSE with Solutions
Q36.The perimeter of rectangle is 44 cm.Its length exceeds twice its breadth by 4 cm. Find the area of rectangle:
(a) 46 cm² (b) 49 cm² (c) 96 cm² (d) 69 cm²
Ans. (c) 96 cm²
Let the breadth of rectangle = x cm then length = (2x + 4 ) cm
Perimeter of the rectangle given to us = 44 cm
Perimeter of the rectangle is = 2 ( L+ B) = 2( 2x + 4+x) = 2(3x + 4)
According to question
2(3x + 4) = 44
3x + 4 = 22
3x = 22 – 4 = 18
x = 6
Therefore breadth =6 cm and the length of rectangle = 2x + 4 = 2×6 + 4 =16 cm
Area of the rectangle = L × B = 16 ×6 = 96 cm²
Q37. If sin θ = cos θ, then the value of tan ²θ + cot²θ is:
(a) 2 (b) 4 (c) 1 (d) 10/3
Ans. We are given
sin θ = cos θ
sinθ/cos θ = 1
tan θ = 1 ⇒ cot θ = 1/tan θ = 1
The value of
tan²θ + cot²θ = 1² + 1² = 1 + 1 = 2
Te questions 38 -40 consists of Assertion (A) and Reason (R) . Answer these questions selecting the appropriate option given below:
(a) Both A and R are true and R is the correct exolanation of A.
(b) Both A and R are true but R is not the correct exolanation of A.
(c) A is true,but R is false
(d) A is false, but R is true
Q38.Assertion(A): The point (0,4) lies on the y-axis.
Reason(R): The x coordinate of the point on the y-axis is zero.
Ans.(a) Both A and R are true and R is the correct explanation of A.
The point (0,4) lies on the y -axis because x = 0
Class 10 Maths(Basic) Preboard Exam (First) 2021-22 CBSE with Solutions
Q39. Assertion(A): If the circumference of two circles are in the ratio 2 : 3, then the ratio of their areas is 4 : 9.
Reason(R): The circumference of a circle of radius is 2πr and its area is πr²
Ans.(a) Both A and R are true and R is the correct explanation of A.
Let the radii of two circles are r1 and r2
The ratio of their circumference is given as 2 : 3
2πr1/2πr2= 2/3
r1/r2= 2/3
Ratio between their areas
= πr1²/πr2² = (r1/r2)² =(2/3)² = 4/9
Q40. Assertion(A): For any two positive integers a and b, LCM (a,b) ×HCF(a,b) = a× b.
Reason (R) : Product of two numbers is 150 and HCF is 5,then their LCM is 40.
Ans.(a) A is true but R is false
The relationship between HCF(a,b) and LCM(a,b) and the product of numbers a and b is given as
LCM (a,b) ×HCF(a,b) = ab
LCM×5 = 150
LCM = 150/5 =30
SECTION – C
(Case Study Based Questions)
Section ‘C’ consists of 10 questions. Do any 8 questions.
Points A and B are on the opposite edges of a pond a shown in below figure. To find distance between the two points, Ramesh makes a right angled triangle using the rope. Distance between point B and C is 12 m, distance between C and D is 40 m and A and D is 30 m such that ∠ADC = 90°.
Q41. Which property of geometry is used to find the distance in the figure ?
(a) Similarity of triangles (b) Thales theorem
(c) Pythogorus theorem (d) Converse of thales
Ans.Pythogorus theorem
Q42. What is total length of pond AB and rope BC in the figure ?
(a) 50 m (b) 70 m (c) 100 m (d) 38 m
Ans. (a) 50 m
∠ADC = 90°
AD and DC are perpendicular to each other
AC² = AD² + DC²
AC² = 30² + 40² = 900 + 1600 = 2500
AC = √(2500) = 50
AB = AC – BC = 50 – 12 = 38
Q43. Length of the total rope used is:
(a) 82 m (b) 70 m (c) 120 m (d) 100 m
Ans.(a) 82 m
The lenghth of total rope = BC +DC + AD =12 + 40 + 30 = 82 m
Q44. Which of the following doesn’t form Pythagoras triplet ?
(a) (20,21,28) (b) (8,15,17) (c) (5,12,13) (d) (7,24,25)
Ans.(a) (20,21,28)
(a) 28² = 21² +20²
784 ≠ 441 + 400=841
(b) 17² = 15² + 8²
289 = 225 + 64 =289
(c) 13² = 12² + 5²
169 = 144 + 25 =169
(d) 25² = 24² + 7²
625 = 576 +49 = 625
(a) doesn’t show Pythogoras triplet because 28² ≠ 21²+20²
Q45. Length of the Pond AB is:
(a) 50 m (b) 62 m (c) 70 m (d) 38 m
(d) 38 m
Case Study -II
The path moved by a group of ants has been traced on graph. Aditi is a student of class 10. She compared the ant’s path with the diagram she learned in maths class then she drew the sketch on the graph.
Q46. The shape of the path formed by ants is:
(a) Spiral (b) Oval (c) Parabola (d) Linear
Ans. (c) Parabola
Q47. How many zeroes are possible for this shape ?
(a) 2 (b) 3 (c) 4 (d) 0
Ans. (a) 2
A parabola can intersects any of the axises in two points,so maximum possible zeroes of the parabolic polynomial is 2.
Q48. According to graph ,the zeroes of the polynomial are:
(a) -3,-1 (b) 3,-1 (c) -3, 1 (d) 0,0
Ans.(c) -3,1
The parabola is intersecting the x-axis at x = -3 and at x = 1,therefore its zeroes are -3,1
Q49. What will be the expression of the polynomial shown in figure?
(a) x² -2x +3 (b) x² -3x +2 (c) x² +3x +2 (d) x² +2x – 3
Ans. (d) x² +2x -3
The algebraic expression of a quadratic polynomial with zeroes α and β is given as
x² – (α +β) x + αβ
Here α = -3 and β = 1
x² -(-3 +1)x +(-3) 1
x² +2x -3
Q50. What will be the value of the polynomial if x = -1 ?
(a) -5 (b) 6 (c) -4 (d) 0
Ans.(c) -4
The polynomial,P(x) = x² +2x -3
P(-1) = (-1)² + 2(-1) -3 = 1 -2 -3 = 1-5 = -4
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