The Scottish Church Collegiate School | School Test Question Paper | Target Book

(b) 50%

Explanation:

27\over 8P = P(1 + r\over 100)³

or, (3\over 2)³ = (1 + r\over 100)³

or, r\over 100 = 3\over 2 – 1

or, r\over 100 = 1\over 2

or, r = 1\over 2 × 100 % = 50%

(b) -2

(c) abc = constant

(b) 7 cm

(d) 1

Explanation:

1\over \text{cos α} = 1\over \text{sin β}

or, cos α = sin β

or, cos α = cos (90º – β)

or, α = 90º – β

or, α + β = 90º

sin (α + β) = sin 90º = 1

(b) 27

Explanation:

y = 2x – 5

or, y = 2(16) – 5 = 27

(i) ₹0.01

(ii) 8

(iii) 3

(iv) 70°

(v) π/6

(vi) 1

(vii) ax̄

(i) False; At the same rate of interest, the compound interest for 2 years is more than the simple interest on the same principal due to interest being calculated on the accumulated amount.

(ii) False; The number of real roots of the equation (x – 1)² + (x – 2)² + (x – 3)² = 0 is zero because the sum of squared terms cannot be zero unless all terms are zero, which is not possible here.

(iii) True

(iv) False; The value of sinθ lies between −1 and 1, so it can never be greater than 1 1.

(v) True

(vi) False; In a frequency distribution, the mode is the term with the highest frequency, not the median.

Ratio of capitals of A, B and C = 1\over4 : 1\over3 : 1\over 8 = 6 : 8 : 3

Profit of A = 6x

Profit of B = 8x

Profit of C = 3x

ATP:

Profit of B – Profit of C = 500

or, 8x – 3x = 500

or, 5x = 500

or, x = 100

Total profit = 6x + 8x + 3x = 17x

= 17 × 100 = ₹ 1700

Let principal (P) = x

Amount (A) = 2x

2x = x(1+r)ⁿ

or, 2 = (1+r)ⁿ

Squaring Both Side

or, 4 = (1+r)²ⁿ

It will take 2n years for the sum of money to become 4 times.

x = ± 2

For x = 2

(5 + 2√6)² + (5 – 2√6)²

= 25 + 20√6 + 24 + 25 – 20√6 + 24

= 98

(1 + m²)x² + 2mcx + (c² – p²) = 0.

For real and equal roots, the discriminant Δ must be zero.

Discriminant formula:

Δ = b² – 4ac,

where a = 1 + m², b = 2mc, and c = c² – p².

Substitute:

Δ = (2mc)² – 4(1 + m²)(c² – p²).

Δ = 4m²c² – 4[(1 + m²)(c² – p²)].

Δ = 4[m²c² – c² + p² + m²p²].

Set Δ = 0:

m²c² – c² + p² + m²p² = 0.

c² = p² + m²p².

c² = p²(1 + m²).

Thus, c² = p²(1 + m²).

(sin²α)³ + (cos²α)³ + 3 sin²α cos²α

= (sin²α + cos²α)(sin⁴α -sin²α cos²α + cos⁴α) + 3 sin²α cos²α

= (sin²α)² – sin²α cos²α + (cos²α) + 3 sin²α cos²α

= (sin²α)² + (cos²α)² + 2 sin²α cos²α

= (sin²α + cos²α)² – 2 sin²α cos²α + 2 sin²α cos²α

= 1 + 0 = 1

Using l² = h² + r²

or, l² – h² = (d/2)²

or, \text{l}² - \text{h}² = \text{d}² \over 4

or, \text{l}² - \text{h}² \over \text{d}² = 1 \over 4

Volume of a sphere = 4\over3πr³

Volume of Cylinder = πr²h = 2πr³

Ratio = Volume of a sphere : Volume of Cylinder

= 4\over3πr³ : 2πr³

= 2\over3 = 2 : 3

The ratio of the volume of the sphere to the volume of the cylinder is 2:3.

Volume of cube side 3 cm = 3³ = 27 cm³

Volume of cube side 4 cm = 4³ = 64 cm³

Volume of cube side 5 cm = 5³ = 125 cm³

Total volume = 27 cm³ + 64 cm³ + 125 cm³ = 216 cm³

Side of cube = \sqrt[3]{216} = 6 cm

r cos θ = 2√3

or, (r cos θ)² = (2√3)²

or, r² cos² θ = 12 — (a)

r sin θ = 2

or, r² sin² θ = 2²

or, r² sin² θ = 4 — (b)

Now adding (a) and (b)

r² sin² θ + r² cos² θ = 4 + 12

or, r² (sin² θ + cos² θ) = 16

or, r² = 16

or, r = 4

Now, r sin θ = 2

sin θ = 2/4 = 1/2

or, sin θ = sin 30º

or, θ = 30º

∴ θ = 30º and r = 4

Mean = Σfx / Σf.

Given:

Mean = 24, Σfx = 1700 + 35p, and Σf = 80 + p.

24 = (1700 + 35p) / (80 + p)

or, 24(80 + p) = 1700 + 35p

or, 1920 + 24p = 1700 + 35p

or, 1920 – 1700 = 35p – 24p

or, 220 = 11p

or, p = 220 / 11 = 20

Present Population (V) = ₹ 33,750

rate (r) = 25\over4%

time (n) = 3 years

Initial Population (V_o) = ?

Present Population (V) = V_o(1 – 25\over4×100)³

or, 33750 = V_o (1 – 1\over 16)³

or, 33750 = V_o (15\over 16)³

or, V_o = 33750 × 16 × 16 × 16\over 15 × 15 × 15 = 40,960.

The number of smokers three years ago was approximately 40,960.

Investment of A = ₹ 60,000 × 12 = ₹ 7,20,000

Investment of B = ₹ 75,000 × 12 = ₹ 9,00,000

Investment of C = ₹ 40,500 × 8 = ₹ 3,24,000

Ratio of capital = 7,20,000 : 9,00,000 : 3,24,000

= 40 : 50 : 18

Sum of ratio = 40 + 50 + 18 = 108

A’s profit share = 40\over 108× 40 = ₹7,250.

B’s profit share = 40\over 108 × 50 = ₹9,062.50.

C’s profit share = 40\over 108 × 18 = ₹3,262.50.

x+3\over x-3 + x-3\over x+3 = 5\over 2

or, (x+3)² + (x-3)²\over (x-3)(x + 3) = 5\over 2

or, x² + 6x + 9 + x² - 6x + 9\over x² - 9 = 5\over 2

or, 2x² + 9\over x² - 9 = 5\over 2

or, 4x² + 36 = 5x² − 45.

or,  -x² = – 45 – 36

or, – x² = – 81

or, x² = 81

or, x = ± 9

Sun of roots (α + β) = – \text {b} \over \text {a}

Product of roots (α × β) = \text {c} \over \text {a}

Now, 1\over \text{(bα + c)} + 1\over \text{(bβ + c)}

= \text{(bβ+c)+(bα+c)}​ \over \text{(bα+c)(bβ+c)}

= \text{b(α+β)+2c.}​ \over \text{b²αβ+bc(α+β)+c²}

= \text{- b(b/a)+2c.}​ \over \text{b²(c/a)-bc(b/a)+c²}

= \text{−b²+2ac}\over \text{ac²}

beyond syllabus

beyond syllabus

beyond syllabus

beyond syllabus

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.

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  (∠s opp to equal side are equal)

Again, 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)

Given: 

In right angled ΔABC, A = 90º and AD is perpendicular on hypotenuse BC.

To Prove: \text{ΔABC}\over \text{ΔACD} = \text{BC²}\over \text{AC²}

Proof: Δ ACD ∼ Δ ABC

\text{AC}\over \text{BC} = \text{AD}\over \text{AB} = \text{CD}\over \text{AC}

or, \text{AC}\over \text{BC} = \text{CD}\over \text{AC}

∴ AC² = BC. CD

Now, \text{ΔABC}\over \text{ΔACD} = ½ \text{BC × AD}\over ½ \text{CD × AD} = \text{BC}\over \text{CD}

= \text{BC²}\over \text{CD.BC}

= \text{BC²}\over \text{CD.BC}

Substitute AC² = BC. CD

= \text{BC²}\over \text{AC^²}

Given: In cyclic quadrilateral ABCD, the extended two sides of AB and DC meet at the point P.

To prove: PA · PB = PC · PD

Proof:

ABCD is a cyclic quadrilateral.

∴ ∠DAB + ∠DCB = 180°

Again, ∠DCB + ∠BCP = 180°

∴ ∠DAB + ∠DCB = ∠DCB + ∠BCP

∴ ∠DAB = ∠BCP … (i)

In ΔAPD and ΔCPB, ∠APD = ∠CPB [Same angle]
and ∠DAP = ∠BCP [from (i)]

∴ ΔAPD and ΔCPB are similar.

∴ PA / PC = PD / PB

∴ PA · PB = PC · PD (Proved)

Not in syllabus

LHS : \text{1/cos α + sin β}\over \text{sin α}

= {\text{1 + sin β cos α} \over \text{cos α}} \over \text{sin α}

= \text{1 + cos α × sin β} \over \text{sin α cos α}

= 1\over \text{sin α cos α} + \text{cos α × cos α} \over \text{sin α cos α}

= 1\over \text{sin α cos α} + \text{cos α × cos α} \over \text{sin α cos α}

= \text{sin² α + cos² α}\over \text{sin α cos α} + cot α

= \text{sin² α} \over \text{sin α cos α} + \text{cos² α} \over \text{sin α cos α} + cot α

= tan α + cot α + cot α

= tan α + 2 cot α

= tan α + 2 cot β = RHS (Proved)

LHS: \text{tan α + sec α - 1}\over \text{tan α - sec α + 1}

= \text{tan α + sec α - (sec² α - tan² α)}\over \text{tan α - sec α + 1}

= \text{(sec α + tan α) - (sec α + tan α)(sec α - tan α)}\over \text{tan α - sec α + 1}

= \text{(sec α + tan α)[1 - (sec α - tan α)]}\over \text{tan α - sec α + 1}

= \text{(sec α + tan α)(tan α - sec α + 1)}\over \text{tan α - sec α + 1}

= sec α + tan α

= \text{1}\over \text{cos α} + \text{sin α}\over \text{cos α}

= \text{1 + sin α}\over \text{cos α} = RHS (Proved)

81^sin²θ + 81^cos²θ = 30

or, 81^sin²θ + 81^{1 – sin²θ}  = 30

or, 81^sin²θ +81\over 81^ {sin²θ}  = 30

Let 81^sin²θ = x, we get

x + 81\over x  = 30

or, x² + 81 = 30x

or, x² – 30x + 81 = 0

or, (x – 27)(x – 3) = 0

∴ x = 27 or 3

∴ θ = 30º and 60º

The tank has dimensions:

• Length = 21 decimeters,
• Breadth = 11 decimeters,
• Depth = 6 decimeters.

Volume = Length×Breadth×Depth.

= 21 × 11 × 6 = 1386 dcm³.

Initial water volume = 1\over2 × 1386 = 693dcm³.

The diameter of each iron sphere is 21 cm = 2.1 dcm

radius = 1.05 dcm

Volume of a sphere = 4\over3πr³

= 4\over3π(1.05)³ cm³

Total volume of 100 spheres = 4\over3π(1.05)³ × 100 cm³

Base area of the tank = Length × Breadth × h

= 21 × 11 × h = 231h dcm³ .

ATP: 231h = 4\over3π(1.05)³

or, h = 4\over3π(1.05)³ ÷ 231 = 2.1 decimeters

Initial height = 2r

Final height = 6r

The difference in volumes = 539 cubic cm.

πr²(6r) – πr²(2r) = 539

or, 6πr³ – 2πr³ = 539

or, 4πr³ = 539

or, 4 × 22\over 7r³ = 539

or, r³ = 539 × 7\over 22 × 4 = 42.875

or, r = 3.5 cm

or, h = 6r = 6 × 3.5 = 21 cm

The height of the cylinder is 21 cm.

Cone 1: Base radius = 3 cm, Height = 4 cm.

Volume = 1\over3πr²h

= 1\over3 × π × 3² × 4 = 12π cm³

Cone 2: Base radius = 4 cm, Height = 3 cm.

Volume = 1\over3πr²h

= 1\over3 × π × 4² × 3 = 16π cm³

The total volume of the two cones = 12π cm³ + 16π cm³

= 28π cm³

Median Class = 30-40

Median = L + {N\over 2 - CF}\over f × h

or, 32 = 30 + {100\over 2 - (41 + x)}\over 25 × 10

or, {50 - (41 + x)}\over 25 × 10 = 32 – 30

or, {9 - x}\over 25 × 10 = 2

or, (9 – x) × 10 = 2 × 25

or, (9 – x) × 10 = 2 × 25

or, 9 – x = 5

or, x = 4

Now, 76 + x + y = 100

or, 76 + 4 + y = 100

or, y = 100 – 80 = 20

∴ x = 4 and y = 20

Mode = L + f_m - f_{m-1}\over 2f_m - f_{m-1} - f_{m+1} × h

The mode is given as 24, which lies in the class 20-30. Thus:

• L = 20,
• f_m = 21,
• f_m-1 = 27,
• f_m+1 = x,
• h = 10.

or, 24 = 20 + 21 - 27\over 2 × 21 - 27 - x × 10

or, 24 = 20 + -6\over 15 - x × 10

or, 24 = 20 – 60\over 15 - x

or, 24 – 20  = – 60\over 15 - x

or, 4  = – 60\over 15 - x

or, 4(15 – x) = -60.

or, 60 – 4x = -60.

or, 4x = 120.

or, x = 30.

∴ Final Answer: x = 30