Ramakrishna Mission Vidyalaya, Narendrapur | School Test Question Paper | Target Book

(a) P (1 + r/25)ⁿ

(d) ± 5

Explanation:

Discriminant (Δ) = b² – 4ac

From the equation 5x² – 2kx + 5 = 0

a = 5, b = -2k, c = 5

Substituting into the discriminant formula:

Δ = (-2k)² – 4(5)(5)

Δ = 4k² – 100

Setting the discriminant to zero:

⇒ 4k² – 100 = 0

⇒ 4k² = 100

⇒ k² = 25

⇒ k = ±5

(d) √5 : 1

Explanation:

Given:

• Radius (r) = 2 × Height (h)

Formula for slant height (l):

The slant height of a cone is given by the Pythagoras theorem: l = √(r² + h²)

Substituting r = 2h:

l = √((2h)² + h²)

l = √(4h² + h²)

l = √(5h²)

l = h√5

Ratio of slant height to perpendicular height:

Ratio = l : h

Ratio = h√5 : h = √5 : 1

(b) 90°

Explanation:

If ABCD is both cyclic and a parallelogram, then:

∠A = ∠C

Since ∠A + ∠C = 180° (from the cyclic property),

2∠A = 180°

∠A = 90°

Thus, ∠C = 90°.

(c) π/2n

Explanation:

cot A = tan(90° – A)

Thus, comparing the given equation:

A = (π/2) – (n – 1)A

A + (n – 1)A = π/2

nA = π/2

A = π / 2n

(a) 68

(i) 50x/(yt)

(ii) p

(iii) parallel to the third side

(iv) 15 cm

(v) 61π/360

(vi) 2.1 cm

(i) True

(ii) True

(iii) False; (The correct volume is 16π/3 cubic meters.)

(iv) False; (The number of common tangents when two circles touch externally is 3.)

(v) True

(vi) False (Mode is the value of data having maximum frequency.)

Simple Interest (SI) = (Principal × Rate × Time) / 100

Given:

• SI = 832
• Rate = 8%
• Time = 2 years

Substitute the values into the formula:

832 = (Principal × 8 × 2) / 100

or, 832 = (Principal × 16) / 100

or, 832 × 100 = Principal × 16

or, 83200 = Principal × 16

or, Principal = 83200 / 16

or, Principal = 5200

Ratio of capitals of A, B, and C = 1 : 1/2 : 3/4,

= 4 : 2 : 3.

Sum of Ratio  = 4 + 2 + 3 = 9.

Since B’s share is ₹240 and corresponds to 2 parts, the value of one part = ₹240 / 2 = ₹120.

Total profit = 9 × ₹120 = ₹1080.

Given y ∝ x/2,

⇒ y = k(x/2), where k is the proportionality constant.

When x = 2 and y = 2, substitute into y = k(x/2):

⇒ 2 = k(2/2)

⇒ k = 2.

Thus, y = kx/2.

⇒ y = 2(x/2) = x

Now, x + y = 3

⇒ x + x = 3

⇒ 2x = 3

⇒ x = 3/2

∴ x² + y² = (3/2)² + (3/2)² = 9/2

Given (2x – y) : (x + 2y) = 4 : 7

⇒ (2x – y) / (x + 2y) = 4 / 7

⇒ 7(2x – y) = 4(x + 2y)

⇒ 14x – 7y = 4x + 8y

⇒ 14x – 4x = 8y + 7y

⇒ 10x = 15y.

⇒ x / y = 15 / 10 = 3 / 2

For a cuboid:

• Number of surfaces (S) = 6
• Number of edges (E) = 12
• Number of vertices (V) = 8
• Total number of diagonals (D) = 16 (12 face diagonals + 4 space diagonals).

Substitute these values into the expression

S – E + V – D = 6 – 12 + 8 – 16.

= -14.

The base area of a cone is πr², and the base area of a hemisphere is also πr². Since the base areas are the same, the radius of the cone and the radius of the hemisphere are equal, so r₁ = r₂ = r.

The volume of a cone = (1/3)πr²h, and

The volume of a hemisphere = (2/3)πr³.

Since the volumes are the same:

⇒ (1/3)πr²h = (2/3)πr³.

⇒ (1/3)h = (2/3)r.

⇒ h = 2r.

The ratio of height to radius of the cone is 2 : 1.

In Δ ABC, D and E are points on AB and AC such that DE is parallel to BC.

By the Basic Proportionality Theorem,

AD / AB = AE / AC.

Substitute the given values:

• AD = x
• AB = 2x + 2,
• AE = x – 2,
• AC = AE + EC = (x – 2) + (x – 1) = 2x – 3.

The equation becomes:

x / (2x + 2) = (x – 2) / (2x – 3)

⇒ x(2x – 3) = (x – 2)(2x + 2)

⇒ 2x² – 3x = 2x² + 2x – 4x – 4

⇒ 2x² – 3x = 2x² – 2x – 4.

⇒ -3x = -2x – 4

⇒ -x = -4

⇒ x = 4.

ABCD is a cyclic quadrilateral.

∠ BDC = 30°

∠ ABC = 100°

To find: ∠ ACB.

In a cyclic quadrilateral, angles subtended by the same arc are equal.

∠ ABC and ∠ ADC subtend the same arc AC

Therefore, ∠ ABC = ∠ ADC = 100°

In ∠ BDC, the sum of the angles is 180°.

⇒ ∠ BDC + ∠ DBC + ∠ CDB = 180°.

Substitute ∠ BDC = 30°:

⇒ 30° + ∠ DBC + ∠ CDB = 180°.

⇒ ∠ DBC + ∠ CDB = 150°.

Angle ACB is subtended by the arc AB in triangle ABC.

Using the cyclic property and triangle ABC, angle ACB = 50°.

Radius of the smaller circle (r) = 6 cm

Radius of the bigger circle (R) = 8 cm

The length of the tangent to the smaller circle = 10 cm.

The distance ‘d’ from the external point to the center of the circles satisfies the Pythagorean theorem:

For the smaller circle:

d² = R² + r²

d² = 10² + 6²

d² = 136

For the bigger circle:

The length of the tangent l₂ satisfies:

d² = R² + l₂²

Where R = 8 cm (radius of the bigger circle).

Substitute d² = 136 and R = 8:

136 = R² + l₂²

136 = 64 + l₂²

l₂² = 136 – 64

l₂² = 72.

l₂ = √72 = 6√2 cm.

sin θ + cosec θ = 2.

⇒ sin θ + (1 / sin θ) = 2.    [∵ cosec θ = 1 / sin θ]

⇒ sin² θ + 1 = 2sin θ.

⇒ sin² θ – 2sin θ + 1 = 0.

⇒ (sin θ – 1)² = 0

⇒ sin θ – 1 = 0

⇒ sin θ = 1 and cosec θ = 1

Now, sin² θ + cosec² θ = 1 + 1 = 2.

In Δ ABC:

• ∠ ABC = 90°
• BC = √3 × AB.

We need to find the value of sin C.

Use the Pythagorean theorem to find AC:

In a right triangle,

AC² = AB² + BC².

Substitute BC = √3 × AB:

AC² = AB² + (√3 × AB)².

⇒ AC² = AB² + 3AB².

⇒ AC² = 4AB².

⇒ AC = 2AB.

In Δ ABC,

sin C = opposite side / hypotenuse.

So, sin C = BC / AC.

Substitute BC = √3 × AB and AC = 2AB:

sin C = (√3 × AB) / (2AB)

sin C = √3 / 2

\sum_{i=2}^{10}{20i} = 20 \sum_{i=2}^{10}{i}

= 20 (2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)

= 1080

Given

• CI – SI = ₹228.75
• Rate of interest (r) = 5% per annum.
• Time = 3 years.

We need to find the principal (P).

SI = P × r × t/100 = 15P/100

CI = P { (1+r/100)ⁿ – 1}

= P {(5/100)³– 1}

= P {(105/100)³– 1}

= P {(105/100)³– 1}

= P {(105/100)³– 1}

= 157625P/1000000

Now, CI – SI = ₹228.75

⇒ 157625P/1000000 – 15P/100 = ₹228.75

⇒ 7625P / 1000000 = 228.75

⇒ 7625P = 228750000

⇒ P = 228750000 / 7625

⇒ P = 30000.

Final Answer: The principal (sum) is ₹ 30000.

Total cost of the land = ₹ 15,000

A’s investment = ₹ 6,000

B’s profit = ₹ 2,775

C’s profit = ₹ 2,175

Ratio of Profit B and C = 275 : 175

= 11 : 7

Ratio of insvestment of B and C = 11 : 7

Sum of ratio = 11 + 7 = 18

Total Investment by B and C = ₹ 15,000 – ₹ 6,000

= ₹ 9,000

Investment by B = (11/18) × 9,000 = ₹ 5500

Investment by C = (7/18) × 9,000 = ₹ 3500

Given the equation 3x^2 + 6x + 2 = 0, the roots are p and q.

By the properties of quadratic equations:

• Sum of roots (p + q) = -b/a = -6/3 = -2.
• Product of roots (pq) = c/a = 2/3.

We need to find the equation whose roots are -p²/q and -q²/p.

Step 1: Sum of the new roots:

Let the new roots be α = -p²/q and β = -q²/p.

The sum of the new roots is:

α + β = -p^2/q – q^2/p.

Take the common denominator pq:

α + β = -(p³ + q³) / pq.

Using the identity p³ + q³ = (p + q)(p² – pq + q²):

p³ + q³ = (p + q)((p + q)² – 3pq)

Substitute p + q = -2 and pq = 2/3:

p³ + q³ = (-2)((-2)² – 3(2/3))

p³ + q³ = -2(4 – 2)

p³ + q³ = -4

Substitute into α + β:

α + β = -(-4) / (2/3)

α + β = 4 × 3 / 2

α + β = 6

Step 2: Product of the new roots:

The product of the new roots is:

α × β = (-p²/q) × (-q²/p)

α × β = p² × q² / pq

α × β = (pq)² / pq

α × β = pq = 2/3

Step 3: Form the new quadratic equation:

The quadratic equation with roots alpha and beta is:

x² – (sum of roots)x + (product of roots) = 0

x² – 6x + 2/3 = 0

3x² – 18x + 2 = 0

The difference between the fraction and its reciprocal is 9/20:

1/x – x = 9/20

1 – x² = (9x) / 20

20(1 – x²) = 9x

20 – 20x² = 9x

20x² + 9x – 20 = 0

20x² + 25x – 16x – 20 = 0

5x(4x + 5) – 4(4x + 5) = 0

(5x – 4)(4x + 5) = 0

x = 4/5

∴ The proper fraction is 4/5

x = 4√15 \over √5 + √3

⇒ x = √240 \over √5 + √3

⇒ x = √12 × √20 \over √5 + √3

⇒ x\over √20 = √12 \over √5 + √3

⇒ x + √20\over x - √20 = 2√3 + √5 + √3\over 2√3 - √5 - √3

⇒ x + √20\over x - √20 = 3√3 + √5\over √3 - √5 —- (i)

Again, x\over √12 = √20 \over √5 + √3

⇒ x + √12\over x - √12 = 2√5 + √5 + √3\over 2√5 - √5 - √3

⇒ x + √12\over x - √12 = 3√5 + √3\over √5 - √3 —- (ii)

Adding (i) and  (ii)

⇒ x + √20\over x - √20 + x + √12\over x - √12 = 3√3 + √5\over √3 - √5 – 3√5 + √3\over √3 - √5

⇒ x + √20\over x - √20 + x + √12\over x - √12 = 3√5 + √3 - 3√5 - √3\over √3 - √5

⇒ x + √20\over x - √20 + x + √12\over x - √12 = 2(√3 - √5)\over √3 - √5 = 2

(a + b) ∝ √ab

⇒ (a + b) = k√ab

⇒ {a\over \sqrt{ab}}+{b\over \sqrt{ab}} = k

⇒ {\sqrt{a\over b}}+{\sqrt{b\over a}} = k

Let x = {\sqrt{a\over b}}, then {\sqrt{b\over a}} = {1\over x}

Now considered (√a + √b) and (√a – √b)

(√a + √b) = √b ({\sqrt{a\over b}}+1)

(√a – √b) = √b ({\sqrt{a\over b}}-1)

The ratio is: {√a + √b\over√a - √b} = {{\sqrt{a\over b}}+1\over {\sqrt{a\over b}}-1}

⇒ {√a + √b\over√a - √b} = {x+1\over x-1}

⇒ {√a + √b\over√a - √b} = {x+1\over x-1}

⇒ √a + √b = {x+1\over x-1} × (√a – √b)

⇒ (√a + √b) ∝ (√a – √b)

If a, b, c, d are in continued proportion, then:

b\over a = c\over b = d\over c = k

where k is the common ratio. From this, we can express b, c and d in terms of $a$:

b = ka, c = k²a, d = k³a

Simplify the left-hand side:

{ab + bc + c² \over bc + cd + d²} = a^2(k+k^3+k^4)\over a^2(k^3+k^5+k^6)

⇒ {ab + bc + c² \over bc + cd + d²} = k+k^3+k^4\over k^3+k^5+k^6

⇒ {ab + bc + c² \over bc + cd + d²} = k(1+k^2+k^3)\over k^3(1+k^2+k^3)

⇒ {ab + bc + c² \over bc + cd + d²} = 1\over k^2

Simplify the right-hand side:

{a² + b² + c² \over b² + c² + d²} = a^2(1+k^2+k^4)\over a^2(k^2+k^4+k^6)

⇒ {a² + b² + c² \over b² + c² + d²} = 1+k^2+k^4\over k^2(1 +k^2+k^4)

⇒ {a² + b² + c² \over b² + c² + d²} = 1\over k^2

Hence LHS = RHS (Proved)

a \over b + c = b \over c + a = c \over a + b = k

a \over b + c = k ⇒ a = k(b + c)

b \over c + a = k ⇒ b = k(c + a)

c \over a + b = k ⇒ c = k(a + b)

Add all three equations:

a + b + c = k(b + c) + k(c + a) + k(a + b)

⇒ a + b + c = k(2a + 2b + 2c)

⇒ a + b + c = 2k(a + b + c)

⇒ 2k = 1

⇒ k = 1\over 2

 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

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)

Given:

• Two circles intersect at points P and Q.
• PA and PB are the diameters of the two circles, respectively.

To Prove: A, Q, and B are collinear.

Proof:

The centers of the two circles are O₁ and O₂.

• PA is the diameter of the first circle, so point A lies on the circumference of the first circle.
• PB is the diameter of the second circle, so point B lies on the circumference of the second circle.

In any circle, the angle subtended by a diameter at any point on the circle is 90º.

• For the first circle, ∠ PQA = 90º.
• For the second circle, ∠ PQB = 90º.

Adding the two angles, we have:

• ∠ PQA + ∠ PQB = 90º + 90º = 180º.

Since the sum of the angles is 180º, points A, Q, and B lie on the same straight line.

Given: In triangle ABC, AD is perpendicular to BC, and AD² = BD × CD.

To Prove: Angle BAC = 90°.

Proof: In the right-angled triangle ADB,

∠ ADB = 90°.

Using the Pythagoras theorem:

AB² = AD² + BD² …….. (i)

Again, in the right-angled triangle ADC,

∠ ADC = 90°.

Using the Pythagoras theorem:

AC² = AD² + CD² …….. (ii)

By adding (i) and (ii), we get:

AB² + AC² = BD² + CD² + 2AD².

= BD² + CD² + 2BD × CD [since AD² = BD × CD].

= (BD + CD)² = BC².

Therefore, AB² + AC² = BC².

From the converse of the Pythagoras theorem, we get that triangle ABC is a right-angled triangle, where angle BAC = 90°.

First angle = 105º = 105 × π/180 = 7π/12

Second angle = π/12

Third angle = π – 7π/12 – π/12 = π/3

The circular measure of the third angle is π/3 radians.

LHS: \sqrt{\text{{sec} θ }-1\over {\text{sec θ }}+1}+\sqrt{\text{{sec} θ }+1\over {\text{sec θ }}-1}

= \sqrt{{\text{{sec} θ }-1\over {\text{sec θ }}+1} \times {\text{{sec} θ }-1\over {\text{sec θ }}-1}} + \sqrt{{\text{{sec} θ }+1\over {\text{sec θ }}-1} \times {\text{{sec} θ }+1\over {\text{sec θ }}+1}}

= \sqrt{{\text{({sec} θ }-1)^2\over {(\text{sec θ }})^2-1}}+\sqrt{{\text{({sec} θ }+1)^2\over {(\text{sec θ }})^2-1}}

= \sqrt{{\text{({sec} θ }-1)^2\over {(\text{tan θ }})^2}}+\sqrt{{\text{({sec} θ }+1)^2\over {(\text{tan θ }})^2}}

= {\text{{sec} θ }-1\over {\text{tan θ }}}+{\text{{sec} θ }+1\over {\text{tan θ }}}

= {\text{{sec} θ }-1 + \text{{sec} θ }+1\over {\text{tan θ }}}

= {\text{{sec} θ }-1 + \text{{sec} θ }+1\over {\text{tan θ }}}

= {\text{2\ {sec} θ }\over {\text{tan θ }}}

= {{2\over \text{{cos} θ }}\over {\text {sin θ}\over \text{{cos} θ }}}

= {2\over \text{{sin} θ }}

= 2 cosec θ = RHS

tan (π/16) = cot (7π/16)

tan (3π/16) = cot (5π/16)

tan (4π/16) = tan (π/4) = 1

Now, tan (π/16) × tan (3π/16) × tan (5π/16) × tan (7π/16) × tan (4π/16)

= cot (7π/16) × cot (5π/16) × tan (5π/16) × tan (7π/16) × 1

= 1

Tower Height (AB) = 12 m

Let CE = x

Chimney Height (CD) = 12 + x

In Δ BCE,

tan 30° = CE/BE

⇒ 1/√3 = CE/BE

⇒ BE = √3CE = √3x

In Δ ADC,

tan 60° = CD/AD

⇒ √3 = (12 + x)/√3x

⇒ 3x = 12 + x

⇒ x = 12/2 = 6

∴ Chimney Height (CD) = 12 + x

= 12 + 6 = 18 m

Radius of sphere = 3 cm

External radius of Cylinder (R) = 5 cm

Internal radius of Cylinder = r cm

Volume of sphere = 4\over 3π × 3³

Volume of Cylinder = π(R² – r²) × h

According to problem,

π(R² – r²) × h= 4\over 3π × 3³

⇒ (R² – r²) × h = 4 × 3²

⇒ (5² – r²) × 4 = 36

⇒ (5² – r²) = 9

⇒ 25 – 9 = r²

⇒  r = √16 = 4 cm

∴ The thickness = 4 – 3 = 1 cm.

Height of cone (h₁) = 20 cm

Slant height of cone (l) = 25 cm

Height of cylinder (h₂) = 15 cm

Volume of cone = Volume of cylinder

Let the radius of the cone be r₁ ​and the radius of the cylinder be r₂.

From the Pythagorean theorem:

r₁² = 25² – 20² = 625−400 = 225

⇒ r₁ = √225 = 25 cm

Volume of cone = 1\over 3π × r₁² × h₁

= 1\over 3π × 15² × 20

= 1500 π cm³

Volume of cylinder = π × r₂² × h₂

⇒ 1500 π = π × r₂² × 15

⇒ r₂² = 100 cm

⇒ r₂ = √100 = 10 cm

The diameter of the cylinder = 2 × 10 = 20 cm

Dimensions of the water tank:

Length = 21 dm = 210 cm,

Breadth = 11 dm = 110 cm,

Height = 6 dm = 60 cm

Volume of the tank = Length × Breadth × Height

= 210 × 110 × 60 cm ³

The tank is half-filled with water, so the initial water volume is:

Initial water volume = 210 × 110 × 60 ÷ 2

= 693000 cm³

Number of iron spheres = 50,

Diameter of each sphere = 21 cm,

Radius of each sphere (r) = 21/2 = 10.5 cm

• L: Lower boundary of the median class = 75
• N: Total frequency = 90
• CF: Cumulative frequency of the class before the median class = 40
• f: Frequency of the median class = 32
• h: Class width = 10

Median = 75 + {90\over 2}- 40\over 32 × 10

= 75 + 45- 40\over 32 × 10

= 75 + 5\over 32 × 10

= 75+1.5625 = 76.5625

• L: Lower boundary of the median class = 60
• N: Total frequency = 80
• CF: Cumulative frequency of the class before the median class = 27
• f: Frequency of the median class = 20
• h: Class width = 30

Median = 60 + 40- 27\over 20 × 30

= 60 + 13\over 20 × 30

= 60 + 19.5

Median = 79.5

The median of the given frequency distribution is 79.5 years.