Tamluk Hamilton High School | School Test Question Paper | Target Book

(b) 12 %

Explanation:

Amount (A) = 28x

Principal (P) = 25x

Time (T) = 1 year

or, 28 = 25 × (1 + R \over 100)

or, 28 \over 25 = (1 + R \over 100)

or, 28 \over 25 – 1 = R \over 100

or, 3 \over 25 = R \over 100

or, R = 3 \over 25 × 100 = 12 %

(c) c/a < 0

(d) 4

(c) 1\over √2

Explanation:

sin A + sin B + sin C = 3

sin A + B + C\over 6 = sin 90º + 90º + 90º\over 6

= sin 270º\over 6

= sin 45º = 1\over √2

(b) 3 : 1 : 2

Explanation:

V_cylinder = πr²h = 2πr³

V_cone = 1\over3πr²h = 2\over3πr³

V_sphere = 4\over3πr³

V_cylinder : V_cone : V_sphere = 2πr³ : 2\over3πr³ : 4\over3πr³

= 2 : 2\over3 : 4\over3

= 1 : 1\over3 : 2\over3

= 3 : 1 : 2

The ratio of the volumes is 3 : 1 : 2.

(d) 5

Explanation:

The given data is: 2, 8, 2, 3, 8, 5, 9, 5, 6.

Arranging in ascending order: 2, 2, 3, 5, 5, 6, 8, 8, 9.

Median position = \text{n + 1}\over2 = \text{9 + 1}\over2 = 5

The 5th value in the ordered data is 5.

(i) General partnership and limited partnership

(ii) z

(iii) Rectangle

(iv) 0

(v) 4 units

(vi) Median

(i) True

(ii) False

(iii) False

(iv) True

(v) True

(vi) True

Principal = P

Rate (r) = 4 % – 3 ¾ = 1\over 4 %

Interest (I) = ₹ 60

Time (t) = 1 year

P = \text{I} × 100\over \text{r × t} = 60 × 100\over {1\over 4} × 1 = ₹ 24,000

Ratio of capitals of A, B, and C = 1\over6 : 1\over5 : 1\over4

To simplify, take the LCM of the denominators (6, 5, and 4), which is 60. Multiply each term in the ratio by 60

= 1\over6 × 60 : 1\over5 × 60 : 1\over4 × 60

= 10 : 12 : 15

Sum of Ratio = 10 + 12 + 15 = 37

The profit of the business is ₹3700.

Profit of C = ₹ 15\over37 × 3700 = ₹ 1500

The profit share of C is ₹1500.

Sum of the roots (α + β) = –1\over6

Product of roots (αβ) = k\over6

Now, α² + β² = (α + β)² – 2αβ

= (-1\over6)² – 2 × k\over6

= 1\over36 – k\over3

or, 25\over26 = 1 - 12k\over36

or, 1 – 12k = 25\over26 × 36

or, 12 k = 1 – 900\over26

or, k = 26 - 900\over26 × 12

or, k = – 874\over312 = – 437\over156

Given: \sqrt{4+ \sqrt{15} } -{1\over2}\sqrt{10}

= \sqrt{4 \times 4+ 4 \times\sqrt{15} \over4} – {1\over2}\sqrt{10}

= {1\over2}\sqrt{16+ 4\sqrt{15}} – {1\over2}\sqrt{10}

= {1\over2}\sqrt{(\sqrt{10})^2 + (\sqrt{6})^2+ 2\sqrt{10}\sqrt{6}} – {1\over2}\sqrt{10}

= {1\over2} \sqrt{(\sqrt{10} + \sqrt{6})^2} – {1\over2}\sqrt{10}

= {1\over 2} (\sqrt{10} + \sqrt{6}) – {1\over2}\sqrt{10}

= {\sqrt{10}\over2} + {\sqrt{6}\over2} – {1\over2}\sqrt{10}

= {\sqrt{6}\over2}

PC = 2 cm

OP = r – 2

OB = r

PB = 1\over2AB cm = 3 cm

Applying pythagoras theorem:

OB² = PB² + OP²

or, r² = 3² + (r – 2)²

or, r² = 3² + r² – 4r + 4

or, 4r = 13

or, r = 3.25 cm

In the right-angled triangle ABC, with ∠ABC = 90°, the altitude BD is drawn from the right angle B to the hypotenuse AC.

This divides AC into two segments AD and DC.

BD² = AD × DC

Given: BD = 8 cm and AD = 4 cm.

Substitute these values into the relation: (8)² = 4 × DC

64 = 4 × DC

DC = 64 ÷ 4

DC = 16 cm

Thus, the length of CD is 16 cm.

Given that ∠AOD = 140°

∴ reflex ∠AOD = 360° – 140° = 220°

Again, the central angle produced by the arc ÂBD is reflex ∠AOD and angle in circle = ∠ACD

∴ reflex ∠AOD = 2∠ACD

or, 220° = 2∠ACD

∴ ∠ACD = 220°\over2 = 110°

Given that ∠CAB = 50°

Now, in ΔACE,

∠AEC + ∠EAC + ∠ACE = 180°

or, ∠AEC = 180° – 50° – 110° = 20°

Hence ∠BED = 20°

Let PB = x. Given that PB = AQ, we also have AQ = x.

From the given data:

• AP = 9 units
• QC = 4 units

Since P and Q lie on AB and AC respectively:

• AB = AP + PB = 9 + x
• AC = AQ + QC = x + 4

Because PQ is drawn parallel to BC, the triangles APQ and ABC are similar.

This gives us the proportion: AP / AB = AQ / AC

9 / (9 + x) = x / (x + 4)

9(x + 4) = x(9 + x) 9x + 36 = 9x + x²

9x + 36 = 9x + x²

36 = x²

Therefore: x = √36 = 6

Since PB = x, PB = 6 units.

cos θ + sec θ = 2

or, cos θ + 1\over \text{cos θ} = 2

or, cos² θ + 1 = 2cos θ

or, cos² θ – 2cos θ + 1 = 0

or, (cos θ – 1)² = 0

or, cos θ – 1 = 0

or, cos θ = 1 and sec θ = 1

Now, cos²⁰²⁴θ + sec²⁰²⁴θ = (1) ²⁰²⁴ + (1) ²⁰²⁴

= 1 + 1 = 2

The two given walls that meet at a corner (12 m and 8 m) represent the length and width of the rectangular floor.

Length = 12 m and Width = 8 m

Area = 12 m × 8 m = 96 m²

The height of the room (4 m) is not needed to find the area of the floor. The height would be relevant if we were asked for the volume of the room, but for the floor area alone, only the length and width matter.

Therefore, the area of the floor is 96 m².

Given a solid sphere with radius r, we know:

Surface area S = 4πr²

Volume V = (4/3)πr³

Now, S^3\over V^2.

= (4πr²)^3\over ({({4\over3})}πr³)^2

= 64π³r⁶\over {({16\over9})}π²r⁶

= 64π\over {({16\over9})}

= 64π × 9\over 16

= 36 π

Given that the mean of x1, x2, x3, …, x50 is x̄, we know:

x̄ = (x1 + x2 + x3 + … + x50) / 50

Consider the sum: (x1 – x̄) + (x2 – x̄) + … + (x50 – x̄)

Distribute the summation: = (x1 + x2 + … + x50) – 50x̄

Since x̄ is the mean: x1 + x2 + … + x50 = 50x̄

Substitute this: = 50x̄ – 50x̄ = 0

Therefore, the value of (x1 – x̄) + (x2 – x̄) + … + (x50 – x̄) is 0.

Let after x years he will repayment the amount.

Interest of ₹ 240000 at 12% for x years

= \frac{240000 × 12 × x}{100} = 28800 x

Amount = Principal + Interest

= ₹ (240000+28800x)

Now, house rent for 1 year 12m = ₹ 5200 × 12

∴ House rent for (x – 1) yrs = ₹ 5200 × 12 × (x – 1)

= 62400 (x – 1)

According to the problem,

62400 (x – 1) = 240000 + 28800 x

or, 62400 x – 62400 = 240000 + 28800 x

or, 62400 x – 28800 x = 240000 + 62400

or, 33600 x = 302400

∴ x =\frac{302400}{33600} = 9

∴ After 9 years he will repay his loan with interest.

Investment of Srikanta = ₹30,000 × 6 + ₹34,000 × 6

= ₹ 384,000

Investment of Soifuddin = ₹50,000 × 6 + ₹49,000 × 6

= ₹ 594,000

Ratio of investments = ₹ 384,000 : ₹ 594,000 = 64 : 99

Sum of ratio parts = 64 + 99 = 163

Srikanta’s share = 64\over163 × 19,000 = ₹ 7,460

Soifuddin’s share = 99\over163 × 19,000 = ₹ 11,540

x - 2\over x + 2 + 6 x - 2\over x - 6 = 1

or, (x - 2)(x - 6) + 6(x - 2)(x + 2)\over (x + 2)(x - 6) = 1

or, (x² – 8x + 12) + 6(x² – 4) = x² – 4x – 12

or, 7x² – 8x – 12 = x² – 4x – 12

or, 7x² – 8x – 12 – x² + 4x + 12 = 0

or, 6x² – 4x = 0

or, 2x(3x – 2) = 0

or, x = 0 or x = 2/3

Let the roots be α and 2α

Sum of roots = α + 2α = –5\over a

or, 3α = –5\over a

or, α = –5\over 3a — (1)

Product of root = α × 2α = c\over a

or, 2α² = c\over a

or, α² = c\over 2a — (2)

Put α = –5\over 3a

(-5\over 3a)² = c\over 2a

or, 25\over 9a² = c\over 2a

or, 25\over 9a = c\over 2

or, 50 = 9 ac

or, ac = 50\over9

Now, b^2\over ac = 5^2\over {50\over9}

= 9\over 2

x = √7 + √3\over√7 - √3 × √7 + √3\over√7 + √3

= (√7 + √3)²\over (√7)² - (√3)²

= 7 + 2√21 + 3\over 7 - 3

= 10 + 2√21\over 4

= 5 + √21\over 2

and y = 1\over x = 5 - √21\over 2

Finding (x + y) :

x + y = 5 + √21\over 2 + 5 - √21\over 2

= 5

Finding x² + y²: 

x² + y² = (x + y)² – 2xy

= (5)² – 2(1)

= 23

Now, x² + xy + y²\over x² - xy + y² = x² + y² + xy\over x² + y² - xy

= 23 + 1\over 23 - 1

= 24\over 22

= 12\over 11

(3x – 4y) ∝ √xy

or, (3x – 4y)² ∝ (√xy)²

or, 9x² – 24xy + 16y² ∝ xy

or, 9x² – 25xy + 16y² ∝ 0

or, 9x² – 16xy – 9xy + 16y² ∝ 0

or, x (9x – 16y) – y (9x – 16y) ∝ 0

or, (9x – 16y)(x – y) ∝ 0

or, (x – y) ∝ 0 —- (1)

Multiplying both sides by x + y

(x – y)(x + y) ∝ 0

or, (x + y) ∝ 0

We know, x³ + y³ = (x + y)(x² – xy + y²)

or, (x + y) = x³ + y³\over x² - xy + y²

or, x³ + y³\over x² - xy + y² ∝ 0

or, x³ + y³ ∝ 0

or, (x + y)(x² – xy + y²) ∝ 0

or, x² – xy + y² ∝ 0

or, x² + y² ∝ xy

ay - bx\over c = cx - az\over b = bz - cy\over a = k

or, ay - bx\over c = k

or, ay – bx = kc

or, ay - bx\over ab = kc\over ab

or, {y\over b} - {x\over a} = kc\over ab — (1)

cx - az\over b = k

or, cx – az = kb

or, cx - az\over ac = kb\over ac

or, {x\over a} - {z\over c} = kb\over ac — (2)

bz - cy\over a = k

or, bz – cy = ka

or, bz - cy\over bc = ka\over bc

or, {z\over c} - {y\over b} = ka\over bc — (3)

Adding (1), (2) and (3)

{y\over b} - {x\over a} + {x\over a} - {z\over c} + {z\over c} - {y\over b} = kc\over ab + kb\over ac + ka\over bc

or, kc\over ab + kb\over ac + ka\over bc = =

or, (a² + b² + c²)k\over abc = 0

or, k = 0 × abc\over (a² + b² + c²) = 0

From eq (1)

cx – az = kb

or, cx – az = 0 × b

or, cx = az

or, x\over a = z\over c —- (4)

From eq (2)

bz – cy = ka

or, bz – cy = 0 × a

or, bz – cy = 0

or, bz = cy

or, \text{b}\over \text{y} = \text{z}\over \text{c} —- (5)

From (4) and  (5)

\text{x}\over \text{a} = \text{b}\over \text{ y} = \text{z}\over \text{c}

\text{a}\over \text{b} = \text{b}\over \text{c} = \text{c}\over \text{d} = k

c = dk

b = ck = dk²

a = bk = dk³

LHS: (dk² – dk)² + (dk – dk³)² + (dk² – d)²

= d²k²(k – 1)² + d²k²(1 – k²)² + d²(k² – 1)²

= d²{k²(k – 1)² + k²(1 – k²)² + (k² – 1)²}

= d²{k²(k² – 2k + 1) + k²(k⁴- 2k² + 1) + (k⁴- 2k² + 1)}

= d²(k⁴- 2k³ + k² + k⁶ – 2k⁴ + k² + k⁴- 2k² + 1)

= d²(k⁶ – 2k³ + 1)

= d²k⁶ – 2d²k³ + d²

= (dk³)² – 2(dk²)(d) + d²

= (dk³ – d)² = (a – d)² RHS (Proved)

Given: ∠AOB is the angle at the center of the circle with center O, and ∠ACB is the angle at any point on the circle formed by the circular arc APB.

To Prove: ∠AOB = 2∠ACB

Construction: C, O are joined, and CO is extended up to the point D.

Proof:

In △AOC,

OA = OC in each case (radii of the same circle).

∴ ∠OCA = ∠OAC

Again, in each case, side CO of Δ AOC is extended up to point D.

Exterior ∠AOD = ∠OAC + ∠OCA = 2∠OCA — (I)

In ΔBOC

OB = OC in each case (radii of the same circle).

Again, since side CO of ΔBOC is extended up to point D:

∴ Exterior ∠BOD = ∠OCB + ∠OBC = 2 ∠OCB — (II)

∠AOD + ∠BOD = 2∠OCA + 2∠OCB

∴ ∠AOB = 2(∠OCA + ∠OCB) = 2∠ACB — (III)

∴ ∠AOB = 2∠ACB (Proved)

Pythagoras Theorem: In any right-angled triangle, the area of the square drawn on the hypotenuse is equal to the sum of the areas of the squares drawn on the other two sides.

In △ABC, AB = AC,

∴ ∠ABC = ∠ACB ………..(1)

Now, ABCD is a cyclic quadrilateral,

∴ ∠ABC + ∠ADC = 180°

or, ∠ACB + ∠ADC = 180° ……..(2) [from (1)]

Again, ∠ACB + ∠ACE = 180° ……..(3)

From (2) and (3), we get:

∠ACB + ∠ADC = ∠ACB + ∠ACE

or, ∠ADC = ∠ACE

or ∠DCE + ∠DEC = ∠ACD + ∠DCE

or ∠DEC = ∠ACD

∴ ∠ACD = ∠DEC

Hence, ∠ACD = ∠AEC.

Contruction: Extent AG to Point D.

G is the centroid of Δ ABC. AD is a median

Proof: AG : GD = 2 : 1

\text{AG}\over \text{GD} = 2\over 1

or, \text{AO + OG}\over \text{OD - OG} = 2\over 1

or, 2 OD – 2OG = OA + OG

or, 2 OD – OA =  3OG —– (i)

In Δ ABD, OF || BD

\text{AF}\over \text{BF} = \text{AO}\over \text{OD}

or, \text{AF}\over \text{AF} = \text{AO}\over \text{OD}

or, 1 = \text{AO}\over \text{OD}

or, OD = OA

Put OD = OA

2 OA – OA =  3OG

or, OA = 3OG (Proved)

A + B = 135º —- (1)

A – B = π/12 = 15º—- (2)

Adding (1) and (2)

(A + B) + (A – B) = 150º

or, 2A = 150º

or, A = 75º

Put A in equation (1)

A + B = 135º

or, 75º + B = 135º

or, B = 135º – 75º = 60º

In Degree, A = 75º and B = 60º

In Circular values, A = 5π/12 and B = π/3

\text{sinθ}\over \text{x} = \text{cosθ}\over \text{y} = k

∴ sinθ = kx — (1)  and cosθ = ky — (2)

Squaring and adding (1) and (2)

(kx)² + (ky)² = sin²θ + cos²θ

or, k²(x² + y²) = 1

or, k = 1\over \sqrt{\text{x}^2+\text{y}^2}

sinθ = x\over \sqrt{\text{x}^2+\text{y}^2} and cosθ = y\over \sqrt{\text{x}^2+\text{y}^2}

∴ sinθ – cosθ = \text{x - y}\over \sqrt{\text{x}^2+\text{y}^2}

x 1\over √3 + y 1\over √3 = 0

or, x + y = 0 — (1)

2x – y tan 45° = 1

or, 2x – y = 1 — (2)

Adding (1) and (2) we get

3x = 1 ⇒ x = 1\over 3

Substitutuing x in equation (1)

x + y = 0

or, 1\over 3+ y = 0

or, y = –1\over 3

∴ x = 1\over 3 and y = –1\over 3

Height of House (AB) = 15 m

Total height of chimney (CE) = CD + DE

In Δ ABC,

tan 30° = AB/BC

or, 1/√3 = 15/BC

or, BC = 15√3 m

In fig, BC = AD = 15√3 m

In ΔADE,

tan 60° = DE/AD

or, √3 = DE/15√3

or, DE = 15√3 × √3 = 45 m

∴ Total height of the chimney = CD + DE

= 15 + 45  = 60 m (Ans)

In ABC,

tan 60° = 11\over \text{y}

or, √3 = 11\over \text{y}

or, y = 11\over √3

In ADE,

tan 30° = \text{DE}\over \text{AD}

or, tan 30° = \text{11 - x}\over \text{y}

or, tan 30° =\text{11 - x}\over {11\over √3}

or, 1\over √3 = (11 - x)√3\over 11

or, 11 = (11 – x)√3 × √3

or, 33 – 3x = 11

or, 3x = 22

or, x = 22/3 = 7.33 m

Radius = 3.2 dcm = 32 cm

Length of rod = x cm (Let)

∴ the volume of the rod = π × (32)² × x

The radius of each sphere = 8 cm

∴ The volume of each sphere = 4\over 3 × π × 8³

So, the volume of 21 spheres = 21 × 4\over 3 × π × 8³

= 28 × 512π

ATP

π × (32)² × x = 28 × 512π

or, x = 28 × 512\over 32 × 32 = 14 cm

Slant height = 5 m

Radius of buoy = r  m

∴ TSA of a buoy = 22\over 7r(r+5)

or, 22\over 7r(r+5) = 75 3\over7

or, 22\over 7r(r+5) = 528\over7

or, r(r+5) = 528 × 7\over7 × 22

or, r² + 5r = 24

or, r² + 5r – 24 = 0

or, (r + 8)(r – 3) = 0

∴ r = 3 or -8

Height of the buoy (h) = \sqrt{5^2-3^2} = √16 = 4 cm

(a) Volume of buoy = {1\over 3}×{22\over 7} × 3 × 3 × 4

= 37 {5\over 7} cm³

(b) Cost of colouring = 2.80 × 528\over7 = ₹ 211.20

Mean = ∑fx\over ∑f = 6060\over 80 = 75.75

The mean of the given data is 75.75.

Median Class = N\over 2 = 70\over 2 = 35th Observation

∴ Median Class = 40 – 50

Median = 40 + {70\over 2}- 29\over 13 × 10

= 40 + 35- 29\over 13 × 10

= 40 + 4.615 = 44.615

The modal class is 20 − 25 with f_m = 22.

Given:

• L = 20 (lower boundary of modal class)
• f_m = 22 (frequency of modal class)
• f₁ = $16$ (frequency of previous class)
• f₂ = $11$ (frequency of next class)
• h = 5 (class width)

Mode = 20 + {22-16 \over 2 (22)-16-11} \times 5

= 20 + 1.7645 = 21.76

The mode of the given data is approximately 21.76.