These Sample papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 12. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.
CBSE Sample Papers for Class 10 Maths Paper 12
Board | CBSE |
Class | X |
Subject | Maths |
Sample Paper Set | Paper 12 |
Category | CBSE Sample Papers |
Students who are going to appear for CBSE Class 10 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 12 of Solved CBSE Sample Paper for Class 10 Maths is given below with free PDF download solutions.
Time Allowed: 3 hours
Maximum Marks: 80
General Instructions:
- All questions are compulsory.
- The question paper consists of 30 questions divided into four sections A, B, C and D.
- Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.
- There is no overall choice. However, an internal choice has been provided in four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
- Use of calculators is not permitted.
SECTION-A
Question 1.
Complete the missing entries in the following factor tree:
Question 2.
If (x+a) is a factor of 2x2 +2ax+5x+10,find a.
Question 3.
The first term of an AP is p and its common difference is q. Find its 10th term.
Question 4.
A die is thrown once. Find the probability of getting a number less than 3.
Question 5.
Is the following pair of linear equations consistent? Justify your answer.
2ax + by -a, 4 ax + 2by – 2a = 0; a, b ≠ 0
Question 6.
From an external point P, tangents PA and PB are drawn to a circle with centre O. If ∠PAB = 50°, then find ∠AOB.
SECTION-B
Question 7.
Find all the zeros of the polynomial x4 + x3 – 34x2 – 4x +120, if two of its zeros are 2 and -2.
Question 8.
If sec 4A – cosec (A – 20°), where 4A is an acute angle, find the value of A
Question 9.
In a ΔABC, right-angled at C, if tan A = \(\cfrac { 1 }{ \sqrt { 3 } } \) , find the value of sin A cos B + cos A sin
Question 10.
Using Euclid’s division algorithm, prove that 847, 2160 are coprime.
Question 11.
If -5 is a root of the quadratic equation 2x2 + px – 15 = 0 and the quadratic equation p(x2 +x) + k=0 has equal roots ,then find the value of k.
Question 12.
Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
SECTION—C
Question 13.
Use Euclid’s Division Lemma to show that the square of any positive integer is either of the form 3m or (3m + 1) for some integer m.
Question 14.
Represent the following pair of equations graphically and write the coordinates of points where the lines intersect y-axis:
x + 3y = 6
2x-3y = 12
Question 15.
For what value of n are the nth terms of two APs 63, 65, 67,… and 3,10,17,… equal?
OR
If m times the mth terms of an AP is equal to n times its nth term, find the (m + n )th term of the AP.
Question 16.
If P divides the join of A(-2, -2) and B(2, -4) such that, find the coordinates of \(\cfrac { AP }{ AB } \) = \(\cfrac { 3 }{ 7 } \)
Question 17.
The mid-points of the sides of a triangle are (3,4), (4,6) and (5,7). Find the coordinates of the vertices of the triangle.
OR
Solve the following pair of equation for x and y:
\(\cfrac { { a }^{ 2 } }{ x } -\cfrac { b^{ 2 } }{ y } =0;\cfrac { { a }^{ 2 }b }{ x } +\cfrac { { b }^{ 2 }a }{ y } =a+b,x\neq 0,y\neq 0 \)
Question 18.
In Figure AD⊥ BC. Prove that
AB2 + CD2 = BD2 + AC2
Prove that the equilateral triangles described on the two sides of a right angled triangle are together equal to the equilateral triangle described on the hypotenuse in term of their areas.
Question 19.
What must be added to f(x) = 4x4 + 2x3 – 2x2 + x – 1 so that the resulting polynomial is divisible by
g(x) -x2 + 2x-3?
OR
If one zero of polynomial (a2 + 9) x2 + 13x + 6a is reciprocal of the other, find the value of a.
Question 20.
A circle is touching the side BC of ΔABC at P and touching AB and AC produced at Q and R respectively. Prove that AQ = \(\cfrac { 1 }{ 2 } \) (Perimeter of ΔABC).
Question 21.
A die is thrown once. Find the probability of getting:
(1) a prime number.
(2) a number lying between 2 and 6.
(3) an odd number.
Question 22.
One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card will:
(1) be an ace.
(2) not be an ace.
SECTION—D
Question 23.
A peacock is sitting on the top of a pillar, which is 9 m high. From a point 27 m away from the bottom of the pillar, a snake is coming to its hole at the base of the pillar. Seeing the snake the peacock pounces on it. If their speeds are equal, at what distance from the hole is the snake caught?
OR
The difference of two numbers is 4. If the difference of their reciprocals is \(\cfrac { 4 }{ 21 } \) , find the two numbers.
Question 24.
The angle of elevation of an aeroplane from a point A on the ground is 60°. After a flight of 30 seconds, the angle of elevation changes to 30°. If the plane is flying at a constant height of 3600 \( \sqrt { 3 } \)m, find the speed, in km/hour, of the plane.
Question 25.
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio.
Using the above, prove the following:
In Figure AB||DE and BC || EF. Prove that AC || DE
Question 26.
If the radii of the circular ends of a conical bucket, which is 16 cm high, are 20 cm and 8 cm, find the capacity and total surface area of the bucket. [Use π=\(\cfrac { 22 }{ 7 } \) ]
Question 27.
Draw a circle of radius 4 cm. Construct a pair of tangents to it, the angle between which is 60°. Also justify the construction. Measure the distance between the centre of the circle and the point of intersection of tangents.
Question 28.
If cos θ = \(\cfrac { 7 }{ 8 } \);evaluate :
(1) \(\cfrac { (1+sin\theta )(1-sin\theta ) }{ (1+cos\theta )(1-cos\theta ) } \)
(2) Cot2 θ
OR
If 3 Cot A = 4,check whether \(\cfrac { 1-{ tan }^{ 2 }A }{ 1+{ tan }^{ 2 }A } \) = cos2 A -sin2 A or not
Question 29.
Figure depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:
(1) the distance around the track along its inner edge.
(2) the area of the track.
Water flows through a cylindrical pipe, whose inner diameter is 7 cm, at the rate of 6 km/h in an empty cylindrical tank, the radius of whose base is 40 cm and height is 4.9 m. How long will it take to fill the whole tank?
Question 30.
Some people of a society decorated their area with flags and tricolour ribbons on Republic Day. The following data shows the number of persons in different age group who participated in the decoration:
Find the mode of the above data.
What values do these persons possess?
SECTION—A
Answer 1.
Answer 2.
Since (x + a) is a factor of P(x) = 2x2 + 2ax + 5x + 10
⇒ P (-a) = 0
⇒ 2a + 2a(-a) + 5(-a) + 10 = 0
⇒ 2a2 – 2a2 – 5a + 10 = 0
⇒ -5a + 10 = 0
⇒ a = 2
Answer 3.
We have
an – a + (n -1). d
Here, a = p,n = 10 and d = q
∴ a10 = p + (10 -1).q -p + 9q
Answer 4.
When a die is thrown then S = {1,2,3,4,5, 6}
⇒ n (S) = 6
If E be an event getting a number less than 3, then E = {1,2}
⇒ n (E) = 2
∴ p(E) = \(\cfrac { n(E) }{ n(S) } \) = \(\cfrac { 1 }{ 2 } \) = \(\cfrac { 1 }{ 3 } \)
Answer 5.
Answer 6.
∠APB = 180° – 50° – 50° = 80°
∴ ∠AOB=180°-80°=100°
SECTION—B
Answer 7.
Since 2 and -2 are zeros of P(x) = x4 + x3 – 34x2 -4x + 120
⇒ (x – 2) (x + 2) are factors of P(x)
⇒ (x2 – 4) is a factor, of P(x)
Obviously, other two zeros of P(x) will be the zeros of quotient obtained by dividing P(x) by x2 – 4.
Now,
Answer 8.
Given
sec 4A = cosec (A – 20°)
⇒ cosec (90° – 4A) = cosec (A – 20°) [∵ sec θ = cosec (90° – θ)]
⇒ 90° -4A- A- 20° [Comparing coefficients both the sides]
⇒ 5A = 110° ⇒ A = \(\cfrac { { 110 }^{ o } }{ 5 } \) ⇒ A = 22°
Answer 9.
Answer 10
Let us find HCF of 847,2160 by Euclid’s division algorithm
2160 = 847 x 2 + 466
847 = 466 x 1 + 381
466 = 381 x 1 + 85
381 = 85 x 4 + 41
85 = 41 x 2 + 3
41 = 3 x 13 + 2
3 = 2x 1 + 1
2 = 1 x 2 + 0
⇒ HCF of 2160 and 847 is 1
⇒ 2160, 847 are coprime.
Answer 11.
Since – 5 is a root of the equation 2x2 + px-15 = 0
∴ 2(-5)2 + p(-5) -15 = 0
50 – 5p – 15 = 0 or 5p = 35 or p = 7
⇒ Again p(x2 + x) + k = 0 or 7x2 + 7x + k = 0 has equal roots
∴ D = 0
i.e., b2 – 4ac = 0 or 49 – 4 x 7k = 0
k = \(\cfrac { 49 }{ 28 } \) = \(\cfrac { 7 }{ 4 } \)
Answer 12.
SECTION-C
Answer 13.
Let a be an arbitrary positive integer.
Then by Euclid’s division algorithm, corresponding to the positive integers a and 3 there exist non-negative integers q and r such that
Answer 14.
Given pair of linear equations is
x + 3y – 6 …(i)
2x-3y -12 …..(ii)
Lines (i) and (ii) intersect y-axis at (0,2) and (0, -4)
Answer 15.
nth term of AP 63, 65, 67,… = 63 + 2 (n -1)
nth term of AP 3,10,17,… = 3 + 7 (n – 1)
From question
∴ 63 + 2n – 2 = 3 + 7n – 7
⇒ 5n = 65
⇒ n = 13
Hence 13th term of both AP are equal.
OR
Let first term = a and common difference = d
∴ (mth term) = am = a + (m -1) d
nth term = an = a + (n- 1 )d
According to question,
∴ m [a + (m -1 )d] = n [a + (n -1 )d]
⇒ (m – n) [a + (m + n -1 )d] = 0
⇒ a + (m + n-1 )d = 0 [∵ m ≠ n⇒ m-n ≠ 0]
or ∴ (m + n)th term = am + n = 0
Answer 16.
Here, P(x, y) divides AB such that
\(\cfrac { AP }{ AB } \) = \(\cfrac { 3 }{ 7 } \)
Answer 17.
Let ABC be triangle and D (3,4), E (4,6) and F (5,7) are mid points of AB, BC and CD respectively.
Let the co-ordinates of A, B, C are (xv y,), (x2, y2) and (x3, y3) respectively
As D is mid point of AB
Answer 18.
Given in ΔABC, AD ⊥ BC
We have to prove AB2 + CD2 = BD2 + AC2
Now, AB2 + CD2 = AD2 + BD2 + CD
= BD2 + (AD2 + CD2)
= BD2 + AC2
OR
Given: A right angled AABC with right angle at B. Equilateral As PAB, QBC and RAC are described on the sides AB, BC and CA respectively.
To prove: ar (Δ PAB) + ar (Δ QBC) = ar (Δ RAC).
Proof: Since Δs PAB, QBC and RAC are equilateral therefore they are equiangular and hence similar.
Answer 19.
By division algorithm, we have
f(x) = g(x) x q (x) + r (x)
⇒ f(a) – r (x) =g(x) x q(x)
⇒ f(x) + {-r (x)} = g (x) x q(x)
Clearly, RHS is divisible by g (x). Therefore, LHS is also divisible by g (x). Thus, if we add -r (x) to f(x), then the resulting polynomial is divisible by g (x). Let us now find the remainder when g (x) is is divided by g(x)
Answer 20.
Since tangents from an exterior point to a circle are equal in length.
BP = BQ [Tangents from B] …(i)
CP = CR [Tangents from C] … (ii)
and AQ = AR [Tangents from A] … (iii)
From (iii), we have
AQ = AR ⇒ AB +BQ=AC+CR
⇒ AB+BP=AC+CP [using (i) and (ii)] ………..(iv)
Now, perimeter of Δ ABC – AB + BC + AC
AB + (BP + PC) + AC = (AB + BP) + (AC + PC)
=2(AB+BP) [using (iv)]
= 2(AB+BQ) = 2AQ [using (i)]
∴ AQ = \(\cfrac { 1 }{ 2 } \) ( Δ perimeter of ABC)
Answer 21.
We have, the total number of possible outcomes associated with the random experiment of throwing a die is 6 (i.e., 1,2,3,4,5, 6).
(1) Let E denotes the event of getting a prime number.
So, favourable number of outcomes = 3 (i.e, 2, 3,5)
∴ P(E) = \(\cfrac { 3 }{ 6 } \) = \(\cfrac { 1 }{ 2 } \)
(2) Let E be the event of getting a number lying between 2 and 6.
∴ Favourable number of elementary events (outcomes) = 3 (i.e., 3,4,5)
∴ P(E) = \(\cfrac { 3 }{ 6 } \) = \(\cfrac { 1 }{ 2 } \)
(3) Let E be the event of getting an odd number.
∴ Favourable number of elementary events = 3 (i.e., 1, 3,5)
∴ P(E) = \(\cfrac { 3 }{ 6 } \) = \(\cfrac { 1 }{ 2 } \)
Answer 22.
Well-shuffling ensures equally likely outcomes.
(1) There are 4 aces in a deck. Let E be the event ‘the card is an ace’.
The number of outcomes favourable to E= 4.
The number of possible outcomes = 52
Therefore, P(E) = \(\cfrac { 4 }{ 52 } \) = \(\cfrac { 1 }{ 13 } \)
(2) Let \(\bar { E } \) be the event ‘card drawn is not an ace’.
The number of outcomes favourable to the event \(\bar { E } \) = 52 – 4 = 48 .
The number of possible outcomes = 52
Therefore, P(\(\bar { E } \))=\(\cfrac { 48 }{ 52 } \) = \(\cfrac { 12 }{ 13 } \)
SECTION-D
Answer 23.
Let BC be the pillar and C be the position of peacock and A the initial position of snake.
Let D be the position at which snake is caught by peacock.
As the speeds of peacock and snake are equal
⇒ CD = AD = x (Say)
⇒ BD – 27 – x
From right triangle DBC, 92 + (27 – x)2 = x2
⇒ 54x = 810 or x = 15
∴ CD = 15m and BD = 12m
∴ Snake is caught at a distance of 12m from its hole
OR
Answer 24.
Let B be the initial position of plane and D be its position after 30 sec. A be the position of observer from where angle of elevations of both positions are observed.
Answer 25.
Given: A triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively.
To Prove: \(\cfrac { AD }{ DB } \) = \(\cfrac { AE }{ CE } \)
Construction: Join BE and CD and then draw DM ⊥ AC and EN⊥ AB
Answer 26.
Let ABCD be the bucket in the form of frustum in which
Answer 27.
Steps of Construction
Step I : Draw the circle with centre o and radius 4 cm
Step II : Draw any diameter AOB.
Step III : Make ∠AOP – 60°. OP is radius which intersect the circle at P.
Step IV : Draw PQ⊥ LOP and BE ⊥ OB. PQ and BE intersect at R.
Step V: Hence, RB and RP are the required tangents.
Step VI :Measure of OR = 8 cm.
Justification:
∴ ∠OPQ = 90 ⇒ PR is a tangent to the circle.
Also ∠OBR = 90 ⇒ BR is a tangent to the circle.
Now ∠POB = 180 – 60 = 120°
In □ BOPR,
∠BRP = 360° – (120° + 90° + 90°) = 60°.
Answer 28.
Answer 29.
Here, we have
OE = 0′ G = 30 m
AE = CG = 10 m
0A = 0’C = (30 + 10) m
= 40 m AC = EG = FH = BD = 106 m
OR
Answer 30.
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