Class 10 Maths Sample Paper(Standard) for 2021 CBSE Board Exam -Term 1 with Solutions
Class 10 Maths Sample Paper(Standard) for 2021 CBSE Board Exam -Term 1 with Solutions created here for helping the students to boost their preparation for the exam of term 1 Class 10 CBSE board which is going to be held on november -december 2021.Class 10 Maths Sample Paper(Standard) for 2021 CBSE Board Exam -Term 1 with Solutions is the reference from CBSE Board circular no-Acad 75/2021 dated 02-09-2021 regarding sample paper for CBSE board Examination.
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Class 10 Maths Sample Paper(Standard) for 2021 CBSE Board Exam -Term 1 with Solutions
Q1. The ratio of LCM and HCF of the least composite and the least prime numbers is
(a) 1 : 2 (b) 2 : 1 (c) 1 : 1 (d) 1 : 3
Ans. (b) 2 : 1
Explanation. The least composite number is 4 and the least prime number is 2
HCF(4,2) = 2 and LCM(4,2) = 4
Ratio of LCM(4,2) : Ratio of HCF(4,2) = 4 : 2 = 2 : 1
Q2.The value of k for which the lines 5x + 7y =3 and 15x +21 y = k coincide is
(a) 9 (b) 5 (c) 7 (d) 18
Ans.(a) 9
Explanation. The lines a_{1}x + b_{1}y + c_{1} =0 and a_{1}x + b_{1}y + c_{1} =0 coincide to each other when
The given equations are 5x + 7y =3 and 15x +21 y = k
Arranging the equations into the standard equation
5x + 7y – 3 =0 and 15x +21 y – k =0
5k = 15 ×3 =45
k = 45/5 = 9
See the video -Complete solutions of Pre-board class 10 Maths Exam for Term 1 2021
Q3. A girl walks 200 m towards east and then 150m towards north .The distance of the girl from the starting point is
(a) 350 m (b) 250m (c) 300m (d) 225m
Ans. (b) 250 m
Let initial point of the girl is A ,she walks 200m towards the east upto B point and then turn 150 m towards the north upto C
Therefore the distance of the girl from the initial point is = AC
The triangle ABC so formed is a right triangle,so applying pythogorus theorem
AC² =AB² + BC² = 200² +150² = 40000 + 22500
AC² = 62500
AC =√62500
AC = 250
Hence the distance of the girl from the starting point is 250 m
Q4. Length of the diagonals of a rhombus are 24 cm and 32 cm,then the length of the altitude of the rhombus is
(a) 12 cm (b) 12.8 cm (c) 19 cm (d) 19.2 cm
Ans.
Since the diagonal of the rhombus bisect each other at 90°, therefore ΔAOB is a right triangle
So, AB² =OA² + OB² = 16² +12²=256 +144
AB² = 400
AB =√400 = 20
Area of the rhombus = Base × Atitude =AB ×h = 20h
The area of the rhombus = (1/2) (Product of the diagonals)= (32×24)/2=384 cm²
20h =384
h = 384/20 = 19.2
Hence altitude of the rhombus is 19.2 cm
Class 10 Maths Sample Paper(Standard) for 2021 CBSE Board Exam -Term 1 with Solutions
Q5.Two fair coins are tossed .What is the probability of getting at the most one head ?
(a) 3/4 (b) 1/4 (c) 1/2 (d) 3/8
Ans. (a) 3/4
When two fair coins are tossed,then the total outcomes are (HH,TT,HT,TH)
The favourable outcomes are = At the most one head(i.e one head and less than 1 head)
So,the favourable outcomes are (TT,HT,TH)
∴ P(at the most one head) = 3/4
Q6.ΔABC ∼ΔPQR,If AM and PN are altitudes of ΔABC and ΔPQR respectively and AB² : PQ²=4: 9,then AM : PN =
Ans. We know the corresponding sides and medians of two similar triangles are in proportion
AB/PQ = BC/QR =AC/PR = AM/PN….(i)
We are given AB² : PQ²=4: 9
AB² / PQ²=4/ 9
AB/PQ = 2 / 3
Therefore from equation (i) AM /PN = 2 /3
Q7. If 2 sin²β -cos²β =2,then β is
(a) 0° (b) 90° (c) 45° (d) 30°
Ans. We are given
2 sin²β -cos²β =2
2(1-cos²β) -cos²β =2 (since sin²θ = 1- cos²θ)
2 -2cos²β -cos²β = 2
-3cos²β = 0
cosβ =0 =cos 90°
β =90°
Q8.Prime factors of the denominator of a rational number with the decimal expansion 44.123 are
(a) 2 ,3 (b) 2,3,5 (c) 2,5 (d) 3,5
Ans. (c) 2,5
The decimal expansion of the given rational number is 44.123
44.123 = 44123/1000 =44123 /10³= 44123/(2×5)³
Hence the prime factors of the denominator are 2,5
Q9. The lines x = a and y =b, are
(a) Intersecting (b) Parallel (c) Overlapping (d) Non-of these
Ans (a) Intersecting
The given lines x =a and y =b are drawn graphically as below.
It is clear from the figure that both lines intersect at P
Q10.The distance of the point A(-5,6) from the origin is
(a) 11 units (b) 61 units (c) √11 units (d) √61 units
Ans. (d) √61 units
The coordinates of the origin are (0,0) and the given point is (-5,6)
We know the distance (d) between (x_{1},y_{1}) and (x_{2},y_{2}) is
d = √[(x_{2}-x_{1})²+(y_{2}-y_{1})²] =√[(-5-0)²+(6-0)²] =√[(-5)²+6²] =√(25 +36) =√61
Q11.If a² = 23/25,then a is
(a) rational (b) irrational (c) whole number (d) integer
Ans. (b) irrational
We are given
a² = 23/25
a =√(23/25) = √23/5
√23 is irrational number and 5 is rational
An irrational number divided by a rational number results an irrational number
See the video for Answers of The questions from Q1-Q11
Q12. If LCM(x,18) =36 and HCF(x,18) =2,then x is
(a) 2 (b) 3 (c) 4 (d) 5
Ans.(c) 4
We know
LCM × HCF = Product of the numbers
The numbers given to us are x and 18,and their LCM =36,HCF =2
So,
36×2 = 18x
72 = 18x
x = 72/18 =4
Class 10 Maths Sample Paper(Standard) for 2021 CBSE Board Exam -Term 1 with Solutions
Q13. In ΔABC right angled at B,if tan A = √3,then cos A cos C – sin A sin C =
(a) -1 (b) 0 (c) 1 (d) √3/2
Ans.(b) 0
We are given
tan A = √3⇒ P/B =BC/AB =√3/1
H = √(P² + B²) =AC=√(BC²+AB²)=√[(√3)² +1²) =√(3+1) =√4 =2
cos A =AB/AC = 1/2, cos C =BC/AC =√3/2, sin A =BC/AC =√3/2, sin C = AB/AC= 1/2
Putting the values in
cos A cos C – sin A sinC= (1/2)(√3/2) – (√3/2)(1/2) =√3/4 – √3/4 =0
Q14. the angle sum property of the triangle)
4x =180°If the angles of a ΔABC are in the ratio 1 : 1 : 2, respectively( the largest angle being angle C) ,then the value of (sec A/cosec B) – (tan A/cot B) is
(a) 0 (b) 1/2 (c) 1 (d) √3/2
Ans. (a) 0
Let the angles of the given triangle are x,x and 2x
x + x +2x =180° (As per
x = 180°/4 = 45
∴ ∠A = 45°, ∠B =45° and ∠C = 2x =2×45=90°
Putting the values of all angles in the expression (sec A/cosec B) – (tan A/cot B) is
(sec 45°/cosec 45°) – (tan 45°/cot 45)
=(√2/√2 ) -(1/1)
=1 -1 = 0
Q15. The number of revolutions made by a circular wheel of the radius 0.7 m in rolling a distance of 176 m is
(a) 22 (b) 24 (c) 75 (d) 40
Ans. (d) 40
The distance covered by the wheel in one revolution = Circumference of the wheel
Circumference of the wheel = 2πr = 2×(22/7)×0.7 =4.4 m
Number of revolutions = Distance/Distance covered in one revolution =176/4.4 =40
Q16.ΔABC is such that AB = 3cm, BC = 2 cm, CA = 2.5 cm. If ΔABC ∼ΔDEF and EF = 4 cm,then perimeter of the ΔDEF is
(a) 7.5 cm (b) 15 cm (c) 22.5 cm (d) 30 cm
Ans. (b) 15 cm
We are given ΔABC ∼ΔDEF
According to the rule of similarity of the triangles
AB/DE = BC/EF = CA/DF
3/DE = 2/4 = 2.5/DF
Taking
3/DE = 2/4
2DE =12
DE = 12/2 = 6 cm
Taking
2/4 = 2.5/DF
2DF = 2.5×4 =10
DF =10/2 = 5 cm
Now, we have EF = 4cm, DE = 6 cm and DF = 5 cm
Perimeter of ΔDEF = 4 + 6 + 5 = 15 cm
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Q17.In the figure if DE ∥ BC,AD = 3 cm,BD = 4 cm and BC = 14 cm, then DE =
(a) 7 cm (b) 6 cm (c) 4 cm (d) 3 cm
Ans. (b) 6 cm
It is given to us DE ∥ BC
∠A = ∠A (common)
∠ADE =∠B (corresponding angle)
ΔABC ∼ΔADE (AA rule)
Applying the rule of similarity
AB/AD = BC/DE = AC/AE
(3+4)/3 = 14/DE
14/DE = 7/3
7DE = 14×3 =42
DE = 42/7 =6 cm
Class 10 Maths Sample Paper(Standard) for 2021 CBSE Board Exam -Term 1 with Solutions
Class 10 Maths MCQ -Trigonometry for Term 1 CBSE 2021
Q18.If 4 tan β =3, then
(a) 0 (b) 1/3 (c) 2/3 (d) 3/4
Ans. (a) 0
We are given that
4 tan β =3
tan β =3/4 =P/B
∴ H= √(P² +B²) = √(3² +4²)=√(9 +16)=√25 = 5
sin β = P/H = 3/5, cos β = B/H = 4/5
Q19.One equation of a pair of dependent linear equations is -5x +7y =2. The second equation can be
(a) 10x +14y +4 =0 (b) -10x -14y +4 =0 (c) -10x +14y +4 =0 (d) 10x -14y =-4
Ans. (d) 10x -14y =-4
Two equations are called dependent on each other if their coeffiecients are multiple of each other.
The given equation is –5x +7y =2, observing all equations, we get equation (d) 10x -14y =-4 is a dependent equation of the given equation because if we divide equation (d) by -2,we get -5x +7y =2
Q20. A letter of english alphabets is chosen at random .What is probabilty that it is a letter of the word ‘Mathematics’
(a) 4/13 (b) 9/26 (c) 5/13 (d) 11/26
Ans. (a) 4/13
Total letters in the english alphabets are = 26
The letters in the word ‘Mathematics’ are (m,a,t,h,e,i,c,s)
P(letter in the word ‘Mathematics’) = 8/26 = 4/13
Class 10 Maths Sample Paper(Standard) for 2021 CBSE Board Exam -Term 1 with Solutions
Section-B
Q21. If the sum of two numbers is 1215 and their HCF is 81, then the possible number of pairs of such numbers are
(a) 2 (b) 3 (c) 4 (5) 5
Ans. (c) 4
The given HCF of both numbers is=81
Since the numbers are multiple of their HCF, so let the numbers are 81x and 81y where x and y are coprime numbers
The sum of the number is given =1215
81x + 81y = 1215
x + y = 1215/81 =15
If x =1 then y =14 if x is 2 then y =13, if x =4 then y =11,if x =7 then y =8
Q22.Given below is the graph representing two linear equations by lines AB and CD respectively .What is the area of the triangle formed by these two lines and the line x =0
(a) 3 sq. units (b) 4 sq.units (c) 6 sq.units (d) 8 sq.units
Ans. The line shown in the graph are AB and DC, which are intersecting at C(2,2) and the line x =0(i.2 Y axis) at A(0,4) and D(0.-2)
The triangle formed by the lines and x =0 is ΔADC
The altitude of the ΔADC is the distance of C from the base AD = 2 units and the length of the base is 4 -(-2) = 4+2 = 6 units
Area of the ΔADC = (1/2) ×6×2 = 12/2 = 6 sq.units
Q23. If tan α + cot α =2, then tan^{20}α + cot^{20}α =
(a) 0 (b) 2 (c) 20 (d) 2^{20}
Ans. (b) 2
We are given
tan α + cot α =2
Simplifying it
tan α + 1/tan α =2
(1 +tan ²α)/tan α =2
tan²α +1 =2 tan α
tan ²α -2 tanα +1 = 0
(tan α -1)² =0
tan α -1 =0
tan α = 1
Therefore tan^{20}α + cot^{20}α =
1^{20} + 1^{20} = 1 +1 =2
Q24. If 217x + 131y = 913,131x + 217y =827,then x +y is
(a) 5 (b) 6 (c) 7 (d) 8
Ans.(a) 5
The given equations are 217x + 131y = 913……(i) 131x + 217y =827….(ii)
Adding both of the equations (i) and (ii)
348x + 348y = 1740
348(x +y) =1740
x +y = 5
Class 10 Maths Sample Paper(Standard) for 2021 CBSE Board Exam -Term 1 with Solutions
Q25. The LCM of two prime numbers is p and q (p > q) is 221 . Find the value of 3p – q
(a) 4 (b) 28 (c) 38 (d) 48
Ans. (c) 38
Prime factors of 221 = 13×17
Since it is given to us that p > q, so p =17 and q=13
3p -q =3×17 – 13 = 51 -13 = 38
Q27. A card is drawn from a well-suffled deck of cards . What is the probability that the card drawn is neither a king nor a queen ?
(a) 11/13 (b) 12/13 (c) 11/26 (d) 11/52
Ans.(a) 11/13
The total cards are 52,so total outcomes are =52
The number of queens are 4 and number of kings are 4, favourable outcomes are 52 -8 =44
P(neither a king nor a queen) = 44/52 = 11/13
Q27. Two fair dice are rolled simultaneously .The probability that 5 will come up at least once is
(a) 5/36 (b) 11/36 (c) 12/36 (d) 23/36
Ans.(b) 11/36
When two fair dice are rolled then total possible outcomes are 36
Favourable outcomes are ⇒ 5 will come up at least once
∴Favourable outcomes of this event are (1,5),(2,5),(3,5),(4,5),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,5),hence favourable outcomes are 11
P( 5 comes up at least once) = 11/36
Q28. If 1 + sin²α = 3 sin α cos α,then values of cot α are
(a) -1,1 (b) 0,1 (c) 1,2 (d) -1,-1
Ans. We are given that
1 + sin²α = 3 sin α cos α
sin²α + cos²α + sin²α = 3 sin α cos α
2 sin²α -3 sin α cos α + cos²α = 0
2 sin²α -2sin α cos α- sin α cos α+ cos²α = 0
2 sin α( sin α -cos α) -cos α( sin α – cos α) = 0
( sin α -cos α) ( 2 sin α -cos α) = 0
sin α -cos α= 0
sin α = cos α
cos α /sin α =1
cot α = 1
Taking
2 sin α -cos α= 0
2 sin α = cos α
cos α/sin α =2
cot α =2
Hence the values of cot α are 1,2
Q29. The vertices of a parallelogram in order are A(1,2),B(4,y),C(x,6) and D(3,5). Then (x,y) is
(a) (6,3) (b) (3,6) (c) (5,6) (d) (1,4)
Ans.(a) (6,3)
The vetrtices of the parallogram are given in order A(1,2),B(4,y),C(x,6) and D(3,5)
Therefore parallogram is ABCD, where AC and BD are the diagonals
AC and BD bisect each other,therefore their mid points coincide to each other
The coordinates of the mid point of AC = Coordinates of the mid point of BD
(1 +x)/2, (2 +6)/2 = (4 +3)/2, (y +5)/2
(1 +x)/2 , 4 = 7/2, (y+5)/2
(1 +x)/2 =7/2
x +1 =7
x = 7-1 =6
(y+5)/2 =4
y +5 = 8
y =8-5 =3
Hence the value of x is 6 and of y is 3,so the (x,y) are (6,3)
You can also see Utube video-Class 10 Maths Preboard-2 Question Paper(Standard) for 2021-22 CBSE Board Exam -Term 1 with Solutions
Q30. In the given figure ∠ACB =∠CDA ,AC = 8 cm,AD =3 cm,then BD is
(a) 22/3 (b) 26/3 (c) 55/3 (d) 64/3
Ans.(c) 55/3
From the figure
In ΔABC and ΔADC
∠ACB =∠CDA (given)
∠A = ∠A (common)
ΔABC ∼ ΔACD (AA rule)
According to the rule of similarity
AB/AC = BC /CD = AC /AD
AB/8 = 8/3
3AB = 8×8 =64
AB = 64/3
BD = AB -AD = 64/3 -3 = (64 -9)/3 = 55/3
Q31. The equation of the perpendicular bisector of line segment joining points A(4,5) and B(-2,3) is
(a) 2x -y +7 = 0 (b) 3x + 2y -7 =0 (c) 3x – y -7 =0 (d) 3x + y-7 =0
Ans.(d) 3x + y -7 =0
Let the mid point of the line segment joining A(4,5) and B(-2,3)
Perpendicular bisector is equidistant from the point A and point B
√[(x -4)² + (y -5)²] = √[(x +2)² + (y -3)²]
Squaring both sides
(x -4)² + (y -5)² = (x +2)² + (y -3)²
x² + 16 -8x + y² + 25 -10y = x² + 4 + 4x + y²+9 -6y
-8x -10y +41 = 4x -6y +13
-8x -4x -10y +6y =13 -41 = -28
-12x -4y =-28
3x + y =7
3x + y -7 =0
Q32. In the given figure ,D is the mid point of BC , then the value of cot y°/ cot x° is
(a) 2 (b) 1/2 (c) 1/3 (d) 1/4
Ans. (b) 1/2
AC ⊥BC and AC ⊥ DC,so ΔACD and ΔACB are the right triangles
cot y° = B/P = AC/BC….(i)
cot x° = B/P = AC/DC = AC/(BC/2) = 2AC/BC….(ii)
From equations (i) and (ii)
cot y°/cot x° = (AC/BC)/(2AC/BC) = (AC/2AC).BC/BC = 1/2
Utube Video-Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22
Q33. The smallest number by which 1/13 should be multiplied so that its decimal expansion terminates after two decimal places is
(a) 13/100 (b) 13/10 (c) 10/13 (d) 100/13
Ans. (a) 13/100
The number given to us is 1/13
For terminating decimal, the factor of the denominator should be in the form of 2^{n} × 5^{n}
For, this first of all eliminate the denominator by multiplying 13, and dividing it by 100 so that number terminates up to 2 places
1/13 ×13/100= 1/100 = 0.01
Q34.Sides AB and BE of a right triangle, right-angled at B are of lengths 16 cm and 8 cm respectively. The length of the side of the largest square FDGB that can be inscribed in the triangle ABE is
(a) 32/3 cm (b) 16/3 cm (c) 8/3 cm (d) 4/3 cm
Ans.(b) 16/3
The given triangle is ABE ,in which largest square FDGB is inscribed
Let the length of the square is x
So,BG = FD =x
∠ADF =∠DEG (corresponding angle)
∠AFD =∠DGE =90°(FD⊥AF and DG ⊥GE)
ΔAFD ∼ΔDGE(AA rule)
Applying the rule of similarity
AF/DG =FD/GE
(16 -x)/x =x/(8-x)
(16 -x)(8-x) = x²
128 -16x -8x +x²=x²
24x =128
x = 128/24 =16/3
Class 10 Maths Sample Paper(Standard) for 2021 CBSE Board Exam -Term 1 with Solutions
Q35. Point P divides the line segment joining R(-1,3) and S(9,8) in ratio k : 1. If P lies on the line x -y +2 =0,then value of k is
(a) 2/3 (b) 1/2 (c) 1/3 (d) 1/4
Ans. (a) 2/3
It is given to us taht point P divides the line segment joining two points R(-1,3) and S(9,8) in ratio k : 1
Applying the section formula
x =(mx_{2}+nx_{1})(m+n), y=(my_{2}+ny_{1})(m+n)
Where (x,y) is the point P, m : n = k : 1, (x_{1},y_{1}) =(-1,3) and (x_{2},y_{2}) =(9,8)
x =(9k-1)/k+1), y =(8k+3)/k+1)
Therefore coordinates of the point P are [(9k-1)/k+1), (8k+3)/k+1)]
Since point P lies on the line x -y +2 =0,so the coordinates of point P should satisfy the equation
(9k-1)/k+1) -(8k+3)/k+1) +2 =0
9k -1 -8k -3 +2k +2 =0
3k -2 =0
3k =2
k =2/3
Q36. In the figure given below ,ABCD is a square of side 14 cm with E ,F,G and H as the mid points of AB,BC,CD and DA respectively. The area of the shaded portion is
(a) 44 cm² (b) 49 cm² (c) 98 cm² (d) 49π/2 cm²
Ans.(c) 98 cm²
Area of shaded portion =Area of semicircle HEF + Area of HFG
Area of shaded portion HFG =Area of the rectangle HFCD -Area of two-quadrant HGD and FGC =7×14 -2×(1/4)πr² (r=7) = 98 -(22/7)(1/2)×7×7 =98 -77 =21 cm²
Area of the shaded semicircle HEF is =( 1/2)π×r² (where r=7, is half of the side of the square)
=(22/14)×7×7 =77 cm²
Area of shaded portion = 21 + 77 =98 cm²
Q37. Given below is the picture of Olympic rings made by taking five congruent circles of radius 1 cm each , intersecting in such a way that the chord formed by joining the point of intersection of two circles is also of length 1 cm. The total area of all the dotted regions assuming the thickness of the rings to be negligible is
(a) 4(π/12 -√3/4) cm² (b) (π/6 -√3/4) cm² (c) 4(π/6 -√3/4) cm² (d) 8(π/6-√3/4) cm²
Ans.(d) 8(π/6-√3/4)
The length of the chord is =1 cm, length of the radii of the circles is =1 cm
Joining A and B to the centre of circle,we get an equilateral triangle ABO of the side 1 cm
Area of the minor segment = Area of minor sector AOB-Area of ΔAOB
Area of ΔAOB = side²(√3/4) ( side =1 cm) = 1²×√3/4 =√3/4
Area of minor sector = (60°/360°)×πr² = (1/6)π×1²=π/6 cm²
Area of the minor segment = (π/6- √3/4) cm²
Since chord drawn in the circle divides the dotted portion in two equal parts
Therefore area of the dotted portion =2 (π/6-√3/4) cm²
The area of 4 dotted portion =4×2(π/6-√3/4) =8(π/6-√3/4)
Q38. If 2 and 1/2 are the zeroes of px² + 5x + r,then
(a) p =r =2 (b) p =r =-2 (c) p = 2 ,r =-2 (d) p =-2, r =2
Ans.(b) p = r =-2
In a quadratic equation ax² +bx + c=0,the relationship between coefficients and the zeroes (α and β) are as follows
α + β = -b/a and α β = c/a
The given equation is px² + 5x + r
2 +1/2 = -5/p
5/2 = -5/p
p = -5×2/5 = -2
2(1/2) = r/-2
r =-2
Hence p=r =-2
Q39.The circumference of a circle is 100 cm. The side of a square inscribed in the circle is
(a) 50√2 cm (b) 100/π cm (c) 50√2/π cm (d) 100√2/π cm
Ans.(c) 50√2/π cm
The circumference of the circle is =2πr =100 ⇒r = 100/2π
Since the square is inscribed in the circle
Therefore the radius of the circle = half of the diagonal of the square =100/2π
Diagonals of the square bisect each other at 90°
∴ Side of the square =√[(100/2π)² +(100/2π)²]=√2(100/2π)²)=√2(10000/4π²) =50√2/π cm
Class 10 Maths Sample Paper(Standard) for 2021 CBSE Board Exam -Term 1 with Solutions
Q40.The number of solutions of 3^{x+y}=243 and 243^{x+y }=3 is
(a) 0 (b) 1 (c) 2 (d) infinite
Ans.(a) 0
We are given the equations 3^{x+y}=243 and 243^{x+y }=3
Factorizing 243
243 = 3×3×3×3×3 = 3^{5}
Substituting 243 =3^{5 }in both of the equations
3^{x+y}=3^{5} and( 3^{5})^{x+y }=3
x + y =5 …(i) and 5x +5y =1…(ii)
From (i) and (ii), we have
1/5 =1/5 ≠-5/-1
Therefore the given equations have infinite solutions
Class 10 Maths Sample Paper(Standard) for 2021 CBSE Board Exam -Term 1 with Solutions
Section-C
Q41 -Q45 are based on Case Study I
The figure given alongside shows the path of a diver ,when she takes a jump from the diving board .Clearly it is a parabola.
Anni was standing on a diving board ,48 feet above the water level. She took a dive into the pool. Her height (in feet) above the water level at any time ‘t’ in seconds is given by the polynomial h(t) such that
h(t) =-16 t²+8t + k
Q41. What is the value of k?
(a) 0 (b) -48 (c) 48 (d) 48/-16
Ans.(c) 48
Inially Anni was standing on a diving board which is 48 above the water level
So,at time ,t =0 the height is 48 feet
h(0) = 16 ×0² +8×0 +k
48 =k
Q42. At what time will she touch the water in the pool ?
(a) 30 seconds (b) 2 seconds (c) 1.5 seconds (d) 0.5 seconds
Ans.(b) 2 seconds
Let she touch the water in the pool in t time
Since at t=0,her height above the water level was 48 feet,so after t time ,her height above the water level will be 0, means h(t) =0
So,
-16 t²+8t + k =0
16t² -8t -48 =0
2t² -t -6 =0
2t² -4t +3t -6 =0
2t(t -2) +3(t -2) =0
(2t +3)(t-2) =0
Taking 2t +3=0⇒t =-3/2 and t-2 =0⇒t =2
Neglecting negative root because time can’t be negative,therefore time taken by Annie to reach the water level is 2 seconds
Q43. Rita’s height (in feet) above the water level is given by another polynomial p(t) with zeroes -1 and 2. Then p(t) is given by-
(a) t² +t -2 (b) t² +2t -1 (c)24t² -24t +48 (d) -24t² +24t +48
Ans.(c)24t² -24t +48
The quadratic(i.e of power 2) polynomial p(x) with zeroes α and β is given as
p(x)=x² – (α + β)x + αβ
α =-1 and β=2
∴ p(t) = t² -(-1 +2)t + 2×-1
p(t) = t² -t – 2
If we multiply the polynomail by 24(constant) then polynomial is not effected
24p(t) = 24t² -24t – 48
p(t) = 24t² -24t – 48
Hence the required polynomial is 24t² -24t – 48
Q44. A polynomial q(t) with sum of zeroes as 1 and the product as -6 is modelling Anu’s height in feet above the water at any time t(in seconds). Then q(t) is given by
(a) t² +t +6 (b) t² +t -6 (c)-8t² +8t +48 (d) 8t² -8t +48
Ans.(c) -8t² +24t + 48
The quadratic(i.e of power 2) polynomial p(x) with zeroes α and β is given as
p(x)=x² – (α + β)x + αβ
The sum of the zeroes and the product of zeroes given to us ,
α + β =1 and αβ =-6
q(t)=t² – t + (-6)
q(t) =t² – t -6
If we multiply the polynomail by -8(constant) then polynomial is not effected
-8p(t) = -8t² +24t + 48
Hence the required polynomial is -8t² +24t + 48
Q45. The zeroes of the polynomial r(t) =-12t² +(k-3)t +48 are negative of each other.Then k is
(a) 3 (b) 0 (c) -1.5 (d) -3
Ans.(a) 3
Let one of the zeroes of the given polynomial -12t² +(k-3)t +48 is α then other is -α
The relationship between the coefficients and the zeroes of the polynomial is given by
α +β = -b/a and αβ = c/a
α -α =-(k-3)/-12
0 =(k-3)/12
k-3 =0
k =3
Hence the value of k is 3
Case Study-2
A hockey field is the playing surface for the game of hockey. Historically, the game was played on natural turf (grass) but nowadays it is predominantly played on an artificial turf.
It is rectangular in shape – 100 yards by 60 yards. Goals consist of two upright posts placed equidistant from the centre of the backline, joined at the top by a horizontal crossbar. The inner edges of the posts must be 3.66 metres (4 yards) apart, and the lower edge of the crossbar must be 2.14 metres (7 feet) above the ground.
Each team plays with 11 players on the field during the game icluding the goalie.
Positions you might play include-
- Forward: As shown by players A,B,C and D
- Midfielders: As shown by players E,F and G
- Fullbacks: As shown by players H,I and J
- Goalie: As shown by player K
Using the picture of a hocky field below ,answer the questions that follow.
Q46.The coordinates of the centroid of ΔEHJ are
(a) (-2/3,1) (b) (1,-2/3) (c) (2/3,1) (d) -2/3,-1)
Ans.(a) (-2/3,1)
The vertices of ΔEHJ are E(2,1),H(-2,4) and J(-2,-2)
The centroid of the triangle is =[(2-2-2)/3,(1+4-2)/3]=(-2/3,1)
Q47.If a player needs to be at equal distances from A and G,such that A,P and G are in straight line ,then position of P will be given by
(a) (-3/2,2) (b) (2,-3/2) (c) (2,3/2) (d) (-2,-3)
Ans.(c) (2,3/2)
The position of player A(3,6) and of player G(1,-3)
A,P and G are in straight line and P is equidistant from A and G,means P is the mid point od A and G
Coordinates of P = [(3+1)/2,(6-3)/2]=(4/2,3/2) =(2,3/2)
Hence position of P is (2,3/2)
Q48.The point on x axis equidistant from I and E is
(a) (1/2,0) (b) (0,-1/2) (c) (-1/2,0) (d) 0,1/2)
Ans.(a) (1/2,0)
Let the point on axis is (x,0) equidistant from I and E
The coordinates of E are (2,1) and of I are (-1,1)
√[(x-2)² +(0-1)²]=√[(x+1)² +(0-1)²]
(x-2)² +(0-1)²=(x+1)² +(0-1)²
x² +4-4x +1 =x² +1 +2x +1
-4x -2x +5 -2 =0
-6x +3 =0
-6x =-3
x = -3/-6 = 1/2
Hence the required point on x axis is (1/2,0)
Q49.What are the coordinates of the position of a player Q such that his distance from k is twice his distance from E and K,Q and E are collinear?
(a) (1,0) (b) (0,1) (c) (-2,1) (d) (-1,0)
Ans.(b) (0,1)
It is given that K,Q and E are collinear
Let the coordinates of Q are (x,y)
Distance of player Q from K =2×Distance of player Q from E
KQ = 2QE
KQ/QE =2/1⇒ KQ : QE =2: 1
The points are K(-4,1),E(2,1)
Applying section formula
x = (2×2-4)(2+1) and y =(2×1+1×1)(2+1)
x =(4-4)/3 and y =(2+1)/3
x =0 and y =1
Hence the position of the player Q is (0,1)
Q50.The point on Y axis equidistant from B and C is
(-1,0) (b) (0,-1) (c) (1,0) (d) (0,1)
Ans.(d) (0,1)
Let point on y-axis is (0,y)
Sice point (0, y) is eqidistant from points B(4, 3) and C(4, -1)
Squaring both sides
The distance of the point (0,y) and (4,3) =The distance of the point (0,y) and C(4,-1)
√[(4-0)²+(y-3)²]=√[(4-0)²+(y+1)²]
4² + (y – 3)² = 4² + (y + 1)²
(y – 3)² = (y + 1)²
y² + 9 – 6y = y² + 1 + 2y
9 – 6y = 1 + 2y
9 – 1 = 2y + 6y
8 = 8y
y = 8/8
y = 1
∴ Required coordinates on Y axis = (0, y) = (0, 1)
So, the correct answer is (d)
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