# 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.

**Class 10 Maths Sample Paper (Basic) with Solutions for Term 1 CBSE Board Exam 2021-22**

**MCQ’s on Real Numbers for Term 1 CBSE with Solutions**

**MCQ’s on Class 10 Maths Co-ordinate Geometry for Term 1 CBSE**

**Class 10 Maths MCQ’s on Trigonometry for Term 1 CBSE with Solutions**

**Class 10 MCQ’s questions with solutions-Polynomial**

## 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 5 ^{13} and 2^{6 }will be**

**(a) 0 (b) 1 (c) 13 (d) 26**

Ans.(b) 1

The numbers given 5^{13} and 2^{6}are co-prime numbers(1 is a common factor between them)

- Q7. Equations a
_{1}x + b_{1}y + c_{1}= 0 and a_{2}x + b_{2}y + c_{2}= 0 has infinite many solutions if :

(a) a_{1}/ a_{2}≠b_{1}/ b_{2 }(b) a_{1}/ a_{2}=b_{1}/ b_{2 }(c) a_{1}/ a_{2}= b_{1}/ b_{2}=c_{1}/ c_{2 }(d) a_{1}/ a_{2}= b_{1}/ b_{2}≠ c_{1}/ c_{2}

_{Ans. }(c) a_{1}/ a_{2}= b_{1}/ b_{2}=c_{1}/ c_{2 }

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 r_{1 }and r_{2}

πr_{1}²/πr_{2}² = 4/9

r_{1}²/r_{2}² = 4/9

r_{1}/r_{2} = 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 (x_{1},y_{1 }) and (x_{2},y_{2}) is

d= √[(x_{2}-x_{1})²+ (y_{2}– y_{1})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 (x_{1},y_{1 }) and (x_{2},y_{2}) is = (x_{1}+x_{2 })/2, (y_{1}+y_{2})/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 = (mx_{2}+nx_{1})(m+n), y = (my_{2}+ny_{1})(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 a_{1}=6,b_{1}=-7, c_{1}=-1,a_{2}=3,b_{2}=-4,c_{2}= -5

a_{1}/ a_{2}= 6/3 = 2, b_{1}/ b_{2}=(-7)/(-4) = 7/4, c_{1}/c_{2}= (-1)/(-5) = 1/5

a_{1}/ a_{2}≠ b_{1}/ b_{2}

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 (x_{1},y_{1 }) and (x_{2},y_{2}) is

d= [(x_{2}-x_{1})²+ √(y_{2}– y_{1}])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=√ [(x_{2}-x_{1})²+ √(y_{2}– y_{1}])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 = (mx_{2}+nx_{1})(m+n), y = (my_{2}+ny_{1})(m+n)

Putting the value , x = K , y = 4, x_{1}

=2,x_{2 }= 5, y_{1}= 6,y_{2}=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 r_{1 }and r_{2}

The ratio of their circumference is given as 2 : 3

2πr_{1}/2πr_{2}= 2/3

r_{1}/r_{2}= 2/3

Ratio between their areas

= πr_{1}²/πr_{2}² = (r_{1}/r_{2})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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