Vardaan Learning Institute

Solution:
p(x) = x² + 7x + 10
To find zeroes, let p(x) = 0.
x² + 5x + 2x + 10 = 0
x(x + 5) + 2(x + 5) = 0
(x + 5)(x + 2) = 0
x = -5 or x = -2.

Solution:
Equation: 2x² - 4x + 3 = 0. Here a=2, b=-4, c=3.
Discriminant D = b² - 4ac
D = (-4)² - 4(2)(3) = 16 - 24 = -8.

Solution:
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Area ratio = (4/9)² = 16/81.

Solution:
tan 45° = 1.
(1 - 1²) / (1 + 1²) = (1 - 1) / 2 = 0/2 = 0.

Solution:
Let O be the centre. OQ = 25 cm, Length of tangent PQ = 24 cm.
Radius OP ⊥ PQ (Tangent is perpendicular to radius at point of contact).
In right ΔOPQ: OP² + PQ² = OQ²
OP² + 24² = 25²
OP² = 625 - 576 = 49
OP = 7 cm.

Solution:
Volume of Cylinder = πr²h
Volume of Cone = (1/3)πr²h
Ratio = πr²h : (1/3)πr²h = 1 : 1/3 = 3 : 1.

Solution:
P(not E) = 1 - P(E) = 1 - 0.05 = 0.95.

Solution:
Sum of zeroes = -3 + 4 = 1.
Product of zeroes = (-3)(4) = -12.
Quadratic Polynomial: k[x² - (sum)x + product]
x² - (1)x + (-12) = x² - x - 12.

Solution:
Equation: x² + 4x + 5 = 0.
D = b² - 4ac = 4² - 4(1)(5) = 16 - 20 = -4.
Since D < 0, roots are not real.

Solution:
In ΔABC right angled at B, AB=24, BC=7.
By Pythagoras theorem, AC² = 24² + 7² = 576 + 49 = 625 ⇒ AC = 25.
sin A = Opposite/Hypotenuse = BC/AC = 7/25.

Solution:
By Basic Proportionality Theorem (BPT): AD/DB = AE/EC.
1.5 / 3 = 1 / EC
1/2 = 1 / EC ⇒ EC = 2 cm.

Solution:
Tangents drawn at the endpoints of a diameter are always parallel to each other.

Solution:
Total Surface Area of solid hemisphere = Curved Surface Area + Area of Base
= 2πr² + πr² = 3πr².

Solution:
Outcomes on a die: {1, 2, 3, 4, 5, 6}.
Prime numbers: {2, 3, 5} (Total 3).
Probability = 3/6 = 1/2.

Solution:
p(x) = (x + 1)(x² - x + 1) = x³ + 1.
The highest power of x is 3. So, degree is 3.

Solution:
For real equal roots, D = 0.
b² - 4ac = 0 ⇒ k² - 4(2)(3) = 0
k² - 24 = 0 ⇒ k² = 24
k = ±√24 = ±2√6.

Solution:
This is the expansion of sin(60° + 30°) = sin 90° = 1.
Alternatively: (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1.

Solution:
Perimeter = Area ⇒ 2πr = πr²
Dividing by πr: 2 = r.
Radius = 2 units.

Solution:
Probability of an event cannot be negative. It must lie between 0 and 1.

Solution:
In right ΔOPQ (right angled at P): OP² + PQ² = OQ²
5² + PQ² = 12²
PQ² = 144 - 25 = 119
PQ = √119 cm.

Solution:
x² - 2x - 8 = x² - 4x + 2x - 8
= x(x - 4) + 2(x - 4) = (x - 4)(x + 2).
Zeroes are 4 and -2.
Sum of zeroes = 4 + (-2) = 2. Also -b/a = -(-2)/1 = 2. Verified.
Product of zeroes = 4 × (-2) = -8. Also c/a = -8/1 = -8. Verified.

Solution:
2x² - x + 1/8 = 0
Multiply by 8: 16x² - 8x + 1 = 0
(4x)² - 2(4x)(1) + 1² = 0 ⇒ (4x - 1)² = 0.
4x - 1 = 0 ⇒ x = 1/4.
Roots are 1/4, 1/4.

Solution:
Given AB || DC. In ΔAOB and ΔCOD:
∠OAB = ∠OCD (Alternate interior angles)
∠OBA = ∠ODC (Alternate interior angles)
∴ ΔAOB ~ ΔCOD (AA Similarity)
Hence, corresponding sides are proportional: AO/CO = BO/DO.
Rearranging: AO/BO = CO/DO.

Solution:
tan 2A = cot (A - 18°)
Since tan θ = cot (90° - θ), we can write:
cot (90° - 2A) = cot (A - 18°)
Comparing angles: 90° - 2A = A - 18°
90° + 18° = A + 2A
108° = 3A
A = 108° / 3 = 36°.

Solution:
Total cards = 52.
(i) Number of aces = 4. P(Ace) = 4/52 = 1/13.
(ii) Number of non-aces = 52 - 4 = 48. P(Not Ace) = 48/52 = 12/13.

Solution:
Let consecutive odd positive integers be x and x + 2.
According to question: x² + (x + 2)² = 290.
x² + x² + 4x + 4 = 290
2x² + 4x - 286 = 0
x² + 2x - 143 = 0
x² + 13x - 11x - 143 = 0
x(x + 13) - 11(x + 13) = 0 ⇒ (x + 13)(x - 11) = 0.
Since x is positive integer, x = 11. (Reject -13)
Numbers are 11 and 13.

Solution:
Let ABCD be a parallelogram circumscribing a circle.
Property of tangents: Tangents from an external point are equal in length.
Sum of opposite sides of quadrilateral circumscribing a circle are equal:
AB + CD = AD + BC.
Since ABCD is a parallelogram, AB = CD and AD = BC.
2AB = 2AD ⇒ AB = AD.
A parallelogram with adjacent sides equal is a rhombus.
Hence proved.

Solution:
LHS = √[(1 + sin A)/(1 - sin A)]
Multiply numerator and denominator inside root by (1 + sin A):
= √[(1 + sin A)² / (1 - sin² A)]
= √[(1 + sin A)² / cos² A] (Since 1 - sin² A = cos² A)
= (1 + sin A) / cos A
= (1/cos A) + (sin A/cos A)
= sec A + tan A = RHS.

Solution:
Volume of new sphere = Sum of volumes of three spheres.
(4/3)πR³ = (4/3)π(r₁³ + r₂³ + r₃³)
R³ = 6³ + 8³ + 10³
R³ = 216 + 512 + 1000 = 1728
R = ∛1728 = 12 cm.

Solution:
In right ΔABC (at A): BC² = AB² + AC².
In right ΔABL: BL² = AB² + AL². Since L is midpoint, AL = AC/2.
BL² = AB² + AC²/4 ⇒ 4BL² = 4AB² + AC².
In right ΔAMC: CM² = AC² + AM². Since M is midpoint, AM = AB/2.
CM² = AC² + AB²/4 ⇒ 4CM² = 4AC² + AB².
Adding both equations:
4(BL² + CM²) = 5AB² + 5AC² = 5(AB² + AC²) = 5BC².

Solution:
Total outcomes = 90.
(i) Two-digit numbers (10 to 90): Total = 90 - 9 = 81.
P(Two-digit) = 81/90 = 9/10.
(ii) Perfect square numbers: 1, 4, 9, 16, 25, 36, 49, 64, 81. Total = 9.
P(Perfect Square) = 9/90 = 1/10.

Solution:
Let uniform speed = x km/h. Distance = 480 km.
Time taken = 480/x.
New speed = (x - 8) km/h. New time = 480/(x - 8).
According to condition: 480/(x - 8) - 480/x = 3
480 [ (x - (x - 8)) / x(x - 8) ] = 3
480(8) / (x² - 8x) = 3
160(8) = x² - 8x
x² - 8x - 1280 = 0
Factorizing: x² - 40x + 32x - 1280 = 0
(x - 40)(x + 32) = 0.
Since speed cannot be negative, x = 40.
Speed of train = 40 km/h.

Solution:
Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Proof: Given ΔABC with line DE || BC intersecting AB at D and AC at E.
Construction: Join BE and CD. Draw DM ⊥ AC and EN ⊥ AB.
Area(ΔADE) = (1/2) × AD × EN.
Area(ΔBDE) = (1/2) × DB × EN.
Ratio 1: Area(ADE)/Area(BDE) = AD/DB.
Similarly, Area(ADE) = (1/2) × AE × DM and Area(DEC) = (1/2) × EC × DM.
Ratio 2: Area(ADE)/Area(DEC) = AE/EC.
Since ΔBDE and ΔDEC are on the same base DE and between parallel lines BC and DE, Area(BDE) = Area(DEC).
Therefore, AD/DB = AE/EC.

Solution:
Cylinder: h = 2.1 m, d = 4 m ⇒ r = 2 m.
Cone: l = 2.8 m, r = 2 m.
Area of canvas = CSA of Cylinder + CSA of Cone
= 2πrh + πrl = πr(2h + l)
= (22/7) × 2 × (2(2.1) + 2.8)
= (44/7) × (4.2 + 2.8) = (44/7) × 7 = 44 m².
Cost = Area × Rate = 44 × 500 = ₹ 22,000.

Solution:
LHS = (sin A + cosec A)² + (cos A + sec A)²
= sin²A + cosec²A + 2sinA.cosecA + cos²A + sec²A + 2cosA.secA
= (sin²A + cos²A) + cosec²A + sec²A + 2(1) + 2(1)
= 1 + (1 + cot²A) + (1 + tan²A) + 4
= 1 + 1 + cot²A + 1 + tan²A + 4
= 7 + tan² A + cot² A = RHS.

Solution:
(i) The graph of a quadratic polynomial is a Parabola.
(ii) p(x) = x² - 2x - 8. Factorizing: (x - 4)(x + 2) = 0.
Zeroes are 4 and -2.
(iii) α = 4, β = -2.
α + β + αβ = (4 + (-2)) + (4 × -2)
= 2 + (-8) = -6.

Solution:
(i) Yes, the triangles are similar by AA Similarity criterion (Angle of elevation of sun is same at same time, and both stand vertical at 90°).
(ii) Ratio of corresponding sides: Height1/Height2 = Shadow1/Shadow2.
(iii) Let height of tower be h.
6 / h = 4 / 28
h = (6 × 28) / 4 = 6 × 7 = 42 m.

Solution:
Diameter = 2.8 cm ⇒ Radius r = 1.4 cm.
Total length = 5 cm. Length of cylinder part (h) = 5 - (1.4 + 1.4) = 2.2 cm.
(i) Volume of Cylinder = πr²h = (22/7) × (1.4)² × 2.2 = 13.55 cm³.
(ii) Volume of 2 hemispheres = 2 × (2/3)πr³ = (4/3) × (22/7) × (1.4)³ = 11.50 cm³.
(iii) Total volume of 1 Gulab Jamun = 13.55 + 11.50 = 25.05 cm³.
Volume of 45 Gulab Jamuns = 45 × 25.05 = 1127.25 cm³.
Volume of Syrup = 30% of 1127.25 = 0.30 × 1127.25 ≈ 338 cm³.