CBSE Class 10 Mathematics Basic Question Paper 2024 Available- Download Solution PDF with Answer Key (Set 1- 430/1/1)

For a quadratic equation ax² + bx + c, the product of its zeroes is given by c/a.
For the given polynomial kx² - 4x - 7, Product of zeroes = -7/k.
We are given that the product of the zeroes is -2, so -7/k = -2 ⇒ k = -7/-2 = 7/2.

The nth term of an A.P. is given by the formula: a_n = a + (n - 1) · d.
For a₁₀ = -19, we have: -19 = 8 + (10 - 1) · d ⇒ -19 = 8 + 9d.
Solving for d: -19 - 8 = 9d ⇒ -27 = 9d ⇒ d = -27/9 = -3.

The formula for the midpoint of a line segment joining two points (x₁, y₁) and (x₂, y₂) is:
(midpoint) = ((x₁ + x₂)/2, (y₁ + y₂)/2).
Substituting the given points (-1, 3) and (8, 3/2), we get:
Midpoint = ((-1 + 8)/2, (3 + 3/2)/2) = (7/2, (6/2 + 3/2)/2) = (7/2, 9/4).

We know that sec θ = 1/cos θ.
From the identity sin² θ + cos² θ = 1, we can find cos θ.
sin θ = 1/3 → sin² θ = 1/9.
Thus, cos² θ = 1 - 1/9 = 8/9 → cos θ = √8/3 = 2√2/3.
Therefore, sec θ = 1/cos θ = 3/(2√2).

To find the Highest Common Factor (HCF) of 132 and 77, we use the Euclidean algorithm.
First, divide 132 by 77:
132 = 1 × 77 + 55
Now, divide 77 by 55:
77 = 1 × 55 + 22
Next, divide 55 by 22:
55 = 2 × 22 + 11
Finally, divide 22 by 11:
22 = 2 × 11 + 0
Since the remainder is now 0, the HCF is the last non-zero remainder, which is 11.

For a quadratic equation ax² + bx + c = 0, the condition for real and equal roots is:
Δ = b² - 4ac = 0
For the given quadratic equation 4x² - 5x + k = 0, we have:
- a = 4
- b = -5
- c = k
Substitute the values into the discriminant formula:
Δ = (-5)² - 4(4)(k) = 0
25 - 16k = 0
Solve for k:
16k = 25 ⟹ k = 25/16

The total probability of all events must sum to 1. If the probability of winning is p, then the probability of losing is:
1 - p
Thus, the probability of losing the game is 1 - p.

The distance between two points (x₁, y₁) and (x₂, y₂) is given by the distance formula:
d = √((x₂ - x₁)² + (y₂ - y₁)²)
Substituting the given points (x₁, y₁) = (2, -3) and (x₂, y₂) = (-2, 3):
d = √((-2 - 2)² + (3 - (-3))²)
d = √((-4)² + (6)²) = √(16 + 36) = √52 = 2√13
Thus, the distance is 2√13 units.

We know that:
sin²θ + cos²θ = 1
The given expression is:
sin²θ + sinθ + cos²θ = 1 + sinθ
For this to equal 2:
1 + sinθ = 2 ⟹ sinθ = 1
The value of θ for which sinθ = 1 is 90°.

There are 2 red queens in a deck of 52 cards (one from hearts and one from diamonds). The probability of drawing a red queen is:
2/52 = 1/26
Thus, the probability is 1/26.

The value of x that divides the statistical data into two equal parts is called the median. It is the middle value in an ordered data set.

The volume V of a sphere is given by the formula:
V = (4/3)πr³
Substitute the radius r = 7/2 cm into the formula:
V = (4/3)π(7/2)³ = (4/3)π × 343/8
Simplifying:
V = (4 × 343)/(3 × 8)π = 1372/24π = 343/6π
Approximating π ≈ 3.14:
V ≈ (343/6) × 3.14 = 179.39 cu cm
The exact answer is 539/3 cu cm, which matches option (c).

In statistics, the relationship between the mean, median, and mode is given by:
Mode = 3 × Median - 2 × Mean
Substitute the given values:
Mode = 3 × 23 - 2 × 21 = 69 - 42 = 27
Thus, the mode of the data is 27.

The slant height l of a right circular cone can be calculated using the Pythagorean theorem, as it forms a right triangle with the height and radius:
l = √(r² + h²)
Substitute the given values r = 7 cm and h = 24 cm:
l = √(7² + 24²) = √(49 + 576) = √625 = 25 cm
Thus, the slant height is 25 cm.

Let the quadratic polynomial be f(x) = (α - 1)x² + αx + 1.
Given that one of the zeroes is -3, we can substitute x = -3 into the equation:
f(-3) = (α - 1)(-3)² + α(-3) + 1 = 0
Simplifying:
(α - 1)(9) - 3α + 1 = 0
9α - 9 - 3α + 1 = 0
6α - 8 = 0
Solve for α:
6α = 8 ⟹ α = 8/6 = 4/3
Thus, α = 4/3.

We are given that the length of the diameter of the circle is 6 cm, and one end of the diameter is at (-4, 0), and the other end lies on the x-axis.
Since the center of the circle lies at the midpoint of the diameter, the midpoint can be found by averaging the x-coordinates of the two points:
Let the other end of the diameter be (x₂, 0), and the midpoint is the average of the coordinates:
(((-4 + x₂)/2), (0 + 0)/2).
This midpoint lies on the x-axis, meaning that its y-coordinate is 0. Therefore, we can equate:
(-4 + x₂)/2 = 0.
Solving for x₂:
-4 + x₂ = 0 ⟹ x₂ = 4.
Thus, the other end of the diameter is at (2, 0).

For a pair of linear equations to have no solution, the condition is that the determinant of the coefficient matrix should be zero. The general form of two linear equations is:
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
For the given equations:
1) 5x + 2y - 7 = 0
2) 2x + ky + 1 = 0
The coefficient matrix is:
(5, 2)
(2, k)
The determinant of the coefficient matrix is:
Determinant = (5)(k) - (2)(2) = 5k - 4.
For no solution, the determinant must be zero:
5k - 4 = 0.
Solving for k:
5k = 4 ⟹ k = 4/5.

A doublet refers to a situation where both dice show the same number. When two dice are rolled, there are a total of 6 × 6 = 36 possible outcomes. The doublets are:
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).
Thus, there are 6 favorable outcomes for getting a doublet. Therefore, the probability is:
6/36 = 1/6.
So, the correct answer is 1/6.

- Assertion (A): The quadrilateral OAPB is indeed cyclic. This is a property of tangents drawn from an external point to a circle. The angle between the tangents is supplementary to the angle subtended by the chord at the center, hence forming a cyclic quadrilateral.
- Reason (R): The reason states that in a cyclic quadrilateral, opposite angles are equal. However, this is not necessarily true for the given scenario. The angles in the cyclic quadrilateral formed by tangents from an external point may not be equal, as the quadrilateral is not necessarily inscribed in a circle in the general sense. Therefore, this statement is false.

- The given polynomial is p(x) = x² - 2x - 3. To find the zeroes, we solve x² - 2x - 3 = 0 by factoring:
x² - 2x - 3 = (x - 3)(x + 1) = 0.
Thus, the zeroes of the polynomial are x = 3 and x = -1.
- The graph of a quadratic polynomial intersects the x-axis at its zeroes. Therefore, the points where the graph intersects the x-axis are (-1, 0) and (3, 0), as given in the reason.
Since both the assertion and the reason are true, and the reason explains the assertion, the correct answer is (a).

We are given that ∠ADC = ∠BAC, which means triangle ADC is similar to triangle ABC by the AA (Angle-Angle) similarity criterion.
We can apply the following proportionality rule:
AC / BC = DC / AC
Now, cross-multiply to get:
AC² = BC × DC.
This is the required result.

From the first equation:
x + 2y = 9 ⟹ x = 9 - 2y
Substitute this expression for x into the second equation:
y - 2(9 - 2y) = 2
Simplifying:
y - 18 + 4y = 2 ⟹ 5y = 20 ⟹ y = 4
Now substitute y = 4 into x = 9 - 2y:
x = 9 - 2(4) = 9 - 8 = 1
Thus, the solution is x = 1 and y = 4.

For the first equation x + y + 1 = 0:
Substitute x = -4 and y = 3:
(-4) + 3 + 1 = 0
This is true, so the point lies on the first line.
For the second equation x - y = 1:
Substitute x = -4 and y = 3:
(-4) - 3 = -7
This is false, so the point does not lie on the second line.
Thus, the point (-4, 3) lies on the first line but not on the second line.

Let x = 6 - 4√5.
Assume for contradiction that x is rational, which means 6 - 4√5 is a rational number.
Rearranging:
4√5 = 6 - x
√5 = (6 - x) / 4
Since x is assumed to be rational, the right-hand side is rational, which implies that √5 must be rational.
But √5 is irrational, which is a contradiction. Therefore, 6 - 4√5 must be irrational.

Factor out the common term 11 from the expression:
11 × 19 × 23 + 3 × 11 = 11(19 × 23 + 3)
Now simplify the expression inside the parentheses:
19 × 23 = 437
437 + 3 = 440
Thus, the expression becomes:
11 × 440 = 4840
Since 4840 is divisible by 11, it is not a prime number.

Using the formula for sin(A + B):
sin(A + B) = sin A cos B + cos A sin B
Substituting the values A = 30° and B = 45°:
sin 30° cos 45° + cos 30° sin 45°
Now, calculate the trigonometric values:
sin 30° = 1/2, cos 30° = √3/2, sin 45° = 1/√2, cos 45° = 1/√2
Substitute these values into the expression:
(1/2) × (1/√2) + (√3/2) × (1/√2)
Simplify:
= (1 + √3) / 2√2.
Thus, the final answer is (√3 + 1) / 2√2.

Let the number of yellow balls be x.
The total number of balls in the bag is 4 + 5 + x = 9 + x.
The probability of drawing a red ball is given by:
4 / (9 + x) = 1 / 5.
Now, solve for x:
4 × 5 = 1 × (9 + x) ⟹ 20 = 9 + x ⟹ x = 11.
Thus, the total number of balls is 9 + 11 = 20.
The probability of drawing a yellow ball is: 11 / 20.

The alarm clocks will beep together again at the Least Common Multiple (LCM) of 20 and 25 minutes.
LCM of 20 and 25 = 100 minutes.
Thus, the next time they beep together will be 100 minutes after 12:00 PM, which is 1 hour and 40 minutes later, or 1:40 PM.

Let the smaller angle be x. Then, the larger angle is x + 18°.
Since the angles are supplementary, their sum is 180°:
x + (x + 18°) = 180°
Simplifying:
2x + 18° = 180° ⟹ 2x = 162° ⟹ x = 81°
Thus, the smaller angle is 81°, and the larger angle is:
81° + 18° = 99°.

The points divide the line segment into three equal parts, so the ratio of division is 1:2 for the first point and 2:1 for the second point.
For the first point, using the section formula, the coordinates dividing the segment (-2, 2) and (7, -4) in the ratio 1:2 are:
x = (1 × 7 + 2 × (-2)) / (1 + 2) = (7 - 4) / 3 = 1, y = (1 × (-4) + 2 × 2) / (1 + 2) = (-4 + 4) / 3 = 0.
For the second point, dividing the segment in the ratio 2:1, we get:
x = (2 × 7 + 1 × (-2)) / (2 + 1) = (14 - 2) / 3 = 4, y = (2 × (-4) + 1 × 2) / (2 + 1) = (-8 + 2) / 3 = -2.
So, the coordinates of the trisection points are (1, 0) and (4, -2).

In this problem, we are given two concentric circles. The radius of the smaller circle is OQ = 6 cm, and the radius of the larger circle is OA = r cm. The chord CD of the larger circle is tangent to the smaller circle at point Q, and we are asked to find the length of CD.
We are given the following information:
- PA = 16 cm, where PA is the tangent to the larger circle at point A.
- OP = 20 cm, where O is the center of both circles, and P is a point outside the larger circle.
Now, let’s use the property that the tangent to a circle from an external point is perpendicular to the radius at the point of tangency. Therefore, the length of PA is perpendicular to the radius OA.
We can now apply the Pythagorean theorem to the right triangle OPA, where OP = 20 cm and PA = 16 cm:
OP² = OA² + PA²
Substituting the given values:
20² = r² + 16²
400 = r² + 256
r² = 144
r = 12 cm.
Now, we know the radius OA = 12 cm.
Next, let’s use the property of tangents and the fact that the chord CD of the larger circle is tangent to the smaller circle at Q. From geometry, the length of the chord CD can be found using the following formula:
CD = 2√(OP² - OQ²)
Substituting the known values OP = 20 cm and OQ = 6 cm:
CD = 2√(20² - 6²)
CD = 2√(400 - 36)
CD = 2√364
CD = 12√3 cm.

- Let PT and QT be two tangents drawn from the external point T to the circle with center O.
- Since PT and QT are tangents, the angle between a tangent and the radius is always 90°, so:
∠OTP = ∠OTQ = 90°
- Also, we know that ∠PTQ = ∠OTP + ∠OTQ, which means:
∠PTQ = 90° + 90° = 180°
Therefore, ∠PTQ = 2∠OPQ.

- The solid consists of a cylindrical part and two hemispherical ends. The total height of the solid is the sum of the height of the cylinder and the height of the two hemispheres.
- Let the radius of the cylinder be r = 14 / 2 = 7 cm.
- The height of the cylinder is h = 20 - 2r = 20 - 2(7) = 6 cm.
- Surface area of the solid is the sum of the curved surface area of the cylinder and the surface area of the two hemispheres:
Surface Area = 2πr² + 2πrh + 2πr²
Substituting values:
Surface Area = 2π(7)² + 2π(7)(6) + 2π(7)² = 2π(49) + 2π(42) + 2π(49)
Surface Area = 2π(49 + 42 + 49) = 2π(140) = 280π cm².
Therefore, the surface area is:
Surface Area ≈ 880 cm².

Radius of the glass r = 10 / 2 = 5 cm
Capacity of glass = volume of cylinder - volume of hemisphere
Volume of cylinder = πr²h
Volume of hemisphere = (2/3)πr³
Capacity of glass = π × 5² × 14 - (2/3)π × 5³
= 3.14 × 5 × 5 × 14 - (2/3) × 3.14 × 5 × 5 × 5
= 2512 cm³ or 837.33 cm³ (approx)

LHS = (cos θ / sin θ - 1 / sin θ)²
= (1 / sin² θ) (cos θ - 1)²
= (cos θ - 1)² / (sin θ)²
= (cos θ - 1)² / (1 - cos θ)(1 + cos θ)
= (1 - cos θ) / (1 + cos θ) (RHS)
Hence proved.

Given: In △ABC, DE ∥ BC
To Prove: (AD / DB) = (AE / EC)
Construction: Join BE, DC. Draw DM ⊥ AC and EN ⊥ AB.
Proof:
(ar(△ADE) / ar(△DBE)) = (1 / 2) × AD × EN / (1 / 2) × DB × EN = AD / DB (i)
(ar(△ADE) / ar(△DCE)) = (1 / 2) × AE × DM / (1 / 2) × EC × DM = AE / EC (ii)
Conclusion: △BDE and △CDE are on the same base DE and between the same parallels DE and BC.
Thus, ar(△BDE) = ar(△CDE) (iii)
From (i), (ii), and (iii), we get: (AD / DB) = (AE / EC).
Hence proved.

△ADB ≅ △EDC ⟹ AB = CE, similarly PQ = RN
Given: (AB / PQ) = (AC / PR) = (AD / PM)
⟹ (CE / RN) = (AC / PR) = (2 / PN) ⟹ △AEC ~ △PNR
⟹ ∠1 = ∠2, similarly ∠3 = ∠4
Therefore, ∠1 + ∠3 = ∠2 + ∠4 ⟹ ∠BAC = ∠QPR
Also, (AB / PQ) = (AC / PR) (given)
Therefore, △ABC ~ △PQR.

The given arithmetic progression is 27, 24, 21, ..., with the first term a = 27 and the common difference d = -3.
The sum of the first n terms of an A.P. is given by:
Sn = (n / 2) × [2a + (n - 1)d]
Substituting the known values:
105 = (n / 2) × [2(27) + (n - 1)(-3)]
105 = (n / 2) × [54 - 3n + 3]
105 = (n / 2) × (57 - 3n)
Multiplying both sides by 2:
210 = n(57 - 3n)
Solving the quadratic equation:
210 = 57n - 3n²
3n² - 57n + 210 = 0
n² - 19n + 70 = 0
Solving for n:
n = 7 or n = 10
Therefore, n = 7 gives the sum as 105.
To find the term that is zero, we use the formula for the n-th term:
an = a + (n - 1)d = 27 + (n - 1)(-3) = 0
Solving:
27 + (n - 1)(-3) = 0
27 - 3n + 3 = 0
30 = 3n
n = 10.
So, the term is zero at n = 10.

Let h be the height of the tower and x be the length of the original shadow.
From the first situation (altitude 30°):
tan 30° = h / (x + 40)
1 / √3 = h / (x + 40)
h = (x + 40) / √3
From the second situation (altitude 60°):
tan 60° = h / x
√3 = h / x
h = √3x
Equating the two expressions for h:
(x + 40) / √3 = √3x
x + 40 = 3x
40 = 2x
x = 20
Thus, the height of the tower is:
h = √3 × 20 = 20√3 ≈ 34.64 m.
The length of the original shadow is x = 20 m.

In △AEC, tan 30° = (h - 8) / x
h - 8 = x / √3 (i)
In △ABD, tan 45° = h / x
h = x (ii)
From (i) and (ii):
h = x = 12 + 4√3 = 18.92 m.

Area of minor segment:
Area of minor segment = (1/4) × π × 14 × 14 - (1/2) × 14 × 14
= 154 - 98 = 56 sq. cm.
Area of major segment:
Area of major segment = π × 14 × 14 - 56
= 560 - 56 = 504 sq. cm.

(i) Quadratic equation involving R and r :
We are given that the sum of the areas of the two circles is \( 130\pi \). The area of a circle is given by \( \pi r^2 \). Thus, for the two circles, we have:
πR² + πr² = 130π
Dividing through by π, we get:
R² + r² = 130 (Equation 1)
(ii) Quadratic equation in terms of \( r^2 \):
Next, we are given that the distance between the centers of the circles is 14 m. Using the distance formula for the two radii of the circles, we know:
R + r = 14
Squaring both sides:
(R + r)² = 14²
Expanding:
R² + 2Rr + r² = 196 (Equation 2)
Now subtract Equation 1 from Equation 2:
(R² + 2Rr + r²) - (R² + r²) = 196 - 130
Simplifying:
2Rr = 66
Rr = 33 (Equation 3)
(iii) (a) Find \( r \):
From Equation 3, we have Rr = 33. Substitute R = 33 / r into Equation 1:
(33 / r)² + r² = 130
Simplifying:
1089 / r² + r² = 130
Multiply through by r² to clear the denominator:
1089 + r⁴ = 130r²
Rearranging:
r⁴ - 130r² + 1089 = 0
This is a quadratic equation in terms of r². Let x = r², then:
x² - 130x + 1089 = 0
Solve for x using the quadratic formula:
x = (130 ± √(130² - 4 × 1 × 1089)) / (2 × 1)
x = (130 ± √12544) / 2
x = (130 ± 112) / 2
Thus, the two solutions for x are:
x = 121 or x = 9
Therefore, r² = 9, so r = 3 m.
Now substitute r = 3 into Equation 3 to find R:
R × 3 = 33 ⟹ R = 11 m.
The area irrigated by the smaller circle is:
Area = πr² = 9π m².
OR
(iii) (b) Find R:
Substitute r = 3 into Equation 3:
R × 3 = 33 ⟹ R = 11 m.
The area irrigated by the larger circle is:
Area = πR² = 121π m².

(i) Median Class:
The cumulative frequency is calculated as follows:

(i) Find the length of PQ:
We are given that AP = AQ = 30 cm, and the angle between the tangents, ∠PAQ = 60°. To find PQ, we can use the law of cosines in triangle PAQ, where:
PQ² = AP² + AQ² - 2 × AP × AQ × cos(60°).
Substitute the values:
PQ² = 30² + 30² - 2 × 30 × 30 × 0.5.
PQ² = 900 + 900 - 900 = 900.
PQ = √900 = 30 cm.
Thus, the length of PQ is 30 cm. (ii) Find ∠POQ:
Since AP and AQ are tangents to the circle from the point A, the angle between the tangents at P and Q is equal to the angle at the center of the circle subtended by the chord PQ. Therefore:
∠POQ = 2 × ∠PAQ = 2 × 60° = 120°. (iii) (a) Find the length of OA:
To find OA, use the fact that ∠OAP = 90° (the angle between the radius and the tangent). Thus, triangle OAP is a right triangle, and we can use the Pythagorean theorem:
OA² = OP² + AP².
Since OP = r (the radius), and AP = 30 cm, we have:
OA² = r² + 30².
OA = 2r.
Therefore, 2r² = r² + 900, leading to r = 30 cm.