Solved 2023 Question Paper ICSE Class 10 Mathematics

If $\begin{bmatrix*}[r] 2 & 0 \\ 0 & 4 \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 2 \\ -8 \end{bmatrix*}$, the values of x and y respectively are :

1. 1, -2

2. -2, 1

3. 1, 2

4. -2, -1

Answer

Solving,

$\Rightarrow \begin{bmatrix*}[r] 2 & 0 \\ 0 & 4 \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 2 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2.x + 0.y \\ 0.x + 4.y \end{bmatrix*} = \begin{bmatrix*}[r] 2 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x + 0 \\ 0 + 4y \end{bmatrix*} = \begin{bmatrix*}[r] 2 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x \\ 4y \end{bmatrix*} = \begin{bmatrix*}[r] 2 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow 2x = 2 \text{ and } 4y = -8 \\[1em] \Rightarrow x = \dfrac{2}{2} \text{ and } y = -\dfrac{8}{4} \\[1em] \Rightarrow x = 1 \text{ and } y = -2.$

Hence, Option 1 is the correct option.

If x - 2 is a factor of x³ - kx - 12, then the value of k is :

1. 3

2. 2

3. -2

4. -3

Answer

By factor theorem,

If x - a is a factor of f(x), then f(a) = 0.

Given,

x - 2 is a factor of x³ - kx - 12

⇒ x - 2 = 0

⇒ x = 2.

∴ 2³ - 2k - 12 = 0

⇒ 8 - 2k - 12 = 0

⇒ -2k - 4 = 0

⇒ 2k = -4

⇒ k = $-\dfrac{4}{2}$ = -2.

Hence, Option 3 is the correct option.

In the given diagram, RT is a tangent touching the circle at S. If ∠PST = 30° and ∠SPQ = 60° then ∠PSQ is equal to :

1. 40°

2. 30°

3. 60°

4. 90°

Answer

We know that,

Angle between tangent and the chord at the point of contact is equal to angle of the alternate segment.

∴ ∠PQS = ∠PST = 30°

In △ PQS,

By angle sum property of triangle,

⇒ ∠PQS + ∠QPS + ∠PSQ = 180°

⇒ 30° + 60° + ∠PSQ = 180°

⇒ ∠PSQ + 90° = 180°

⇒ ∠PSQ = 180° - 90° = 90°.

Hence, Option 4 is the correct option.

A letter is chosen at random from all the letters of the English alphabets. The probability that the letter chosen is a vowel, is :

1. $\dfrac{4}{26}$

2. $\dfrac{5}{26}$

3. $\dfrac{21}{26}$

4. $\dfrac{5}{24}$

Answer

All the letters of English alphabets have 5 vowels (a, e, i, o and u) and total number of English alphabets = 26.

Probability that letter chosen is a vowel = $\dfrac{\text{No. of vowels}}{\text{No. of alphabets}} = \dfrac{5}{26}$.

Hence, Option 2 is the correct option.

If 3 is a root of the quadratic equation x² - px + 3 = 0 then p is equal to :

1. 4

2. 3

3. 5

4. 2

Answer

Given,

3 is a root of the quadratic equation x² - px + 3 = 0

∴ 3² - 3p + 3 = 0

⇒ 9 - 3p + 3 = 0

⇒ 12 - 3p = 0

⇒ 3p = 12

⇒ p = $\dfrac{12}{3}$ = 4.

Hence, Option 1 is the correct option.

In the given figure ∠BAP = ∠DCP = 70°, PC = 6 cm and CA = 4 cm, then PD : DB is :

1. 5 : 3

2. 3 : 5

3. 3 : 2

4. 2 : 3

Answer

Given,

∠BAP = ∠DCP = 70°

From figure,

∠BAP and ∠DCP are corresponding angles and since they are equal.

∴ AB // CD.

From figure,

PA = PC + AC = 6 + 4 = 10 cm.

In △ PAB and △ PCD,

⇒ ∠PAB = ∠PCD (Both equal to 70°)

⇒ ∠BPA = ∠DPC (Both are equal)

∴ △ PAB ~ △ PCD (By A.A. axiom)

We know that,

Corresponding sides of similar triangle are proportional to each other.

$\Rightarrow \dfrac{PB}{PD} = \dfrac{PA}{PC} \\[1em] \Rightarrow \dfrac{PB}{PD} = \dfrac{10}{6} \\[1em] \Rightarrow \dfrac{PB}{PD} = \dfrac{5}{3}.$

Let PB = 5x and PD = 3x.

From figure,

DB = PB - PD = 5x - 3x = 2x.

$\Rightarrow \dfrac{PD}{DB} = \dfrac{3x}{2x} = \dfrac{3}{2}$

⇒ PD : DB = 3 : 2.

Hence, Option 3 is the correct option.

The printed price of an article is ₹ 3080. If the rate of GST is 10% then the GST charged is :

1. ₹ 154

2. ₹ 308

3. ₹ 30.80

4. ₹ 15.40

Answer

GST charged = GST (%) × M.P.

= $\dfrac{10}{100} \times 3080$

= ₹ 308.

Hence, Option 2 is the correct option.

(1 + sin A)(1 - sin A) is equal to :

1. cosec² A

2. sin² A

3. sec² A

4. cos² A

Answer

Solving,

⇒ (1 + sin A)(1 - sin A)

⇒ 1 - sin A + sin A - sin² A

⇒ 1 - sin² A

⇒ cos² A.

Hence, Option 4 is the correct option.

The coordinates of the vertices of △ABC are respectively (-4, -2), (6, 2) and (4, 6). The centroid G of △ABC is :

1. (2, 2)

2. (2, 3)

3. (3, 3)

4. (0, -1)

Answer

By formula,

Centroid of triangle = $\Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big)$

Centroid of triangle ABC (G)

$= \Big(\dfrac{-4 + 6 + 4}{3}, \dfrac{-2 + 2 + 6}{3}\Big) \\[1em] = \Big(\dfrac{6}{3}, \dfrac{6}{3}\Big) \\[1em] = (2, 2).$

Hence, Option 1 is the correct option.

The n^th term of an Arithmetic Progression (A.P.) is 2n + 5. The 10^th term is :

1. 7

2. 15

3. 25

4. 45

Answer

Given,

⇒ a_n = 2n + 5

⇒ a₁₀ = 2(10) + 5 = 20 + 5 = 25.

Hence, Option 3 is the correct option.

The mean proportional between 4 and 9 is :

1. 4

2. 6

3. 9

4. 36

Answer

Let x be the mean proportional between 4 and 9 then,

$\Rightarrow \dfrac{4}{x} = \dfrac{x}{9} \\[1em] \Rightarrow x^2 = 4 \times 9 \\[1em] \Rightarrow x^2 = 36 \\[1em] \Rightarrow x = \sqrt{36} = 6.$

Hence, Option 2 is the correct option.

Which of the following cannot be determined graphically for a grouped frequency distribution ?

1. Median

2. Mode

3. Quartiles

4. Mean

Answer

We know that,

To determine mean we do not need any graphical method.

Hence, Option 4 is the correct option.

Volume of a cylinder of height 3 cm is 48π. Radius of the cylinder is :

1. 48 cm

2. 16 cm

3. 4 cm

4. 24 cm

Answer

Let radius of cylinder be r cm.

Given,

Height of cylinder (h) = 3 cm

By formula,

⇒ Volume of cylinder = πr²h

⇒ 3πr² = 48π

⇒ 3r² = 48

⇒ r² = $\dfrac{48}{3}$

⇒ r² = $\dfrac{48}{3}$

⇒ r² = 16

⇒ r = $\sqrt{16}$ = 4 cm.

Hence, Option 3 is the correct option.

Naveen deposits ₹ 800 every month in a recurring deposit account for 6 months. If he receives ₹ 4884 at the time of maturity, then the interest he earns is :

1. ₹ 84

2. ₹ 42

3. ₹ 24

4. ₹ 284

Answer

By formula,

Interest earned = Maturity value - Sum deposited

= ₹ 4884 - ₹ (800 × 6)

= ₹ 4884 - ₹ 4800

= ₹ 84.

Hence, Option 1 is the correct option.

The solution set for the inequation 2x + 4 ≤ 14, x ∈ W is :

1. {1, 2, 3, 4, 5}

2. {0, 1, 2, 3, 4, 5}

3. {1, 2, 3, 4}

4. {0, 1, 2, 3, 4}

Answer

Solving,

⇒ 2x + 4 ≤ 14

⇒ 2x ≤ 14 - 4

⇒ 2x ≤ 10

⇒ x ≤ $\dfrac{10}{2}$

⇒ x ≤ 5.

Since, x ∈ W,

∴ x = {0, 1, 2, 3, 4, 5}.

Hence, Option 2 is the correct option.

Find the value of 'a' if x - a is a factor of the polynomial 3x³ + x² - ax - 81.

Answer

By factor theorem,

If x - a is a factor of f(x), then f(a) = 0.

Given,

x - a is a factor of the polynomial 3x³ + x² - ax - 81.

⇒ x - a = 0

⇒ x = a.

∴ 3a³ + a² - a.a - 81 = 0

⇒ 3a³ + a² - a² - 81 = 0

⇒ 3a³ - 81 = 0

⇒ 3a³ = 81

⇒ a³ = $\dfrac{81}{3}$

⇒ a³ = 27

⇒ a³ = 3³

⇒ a = 3.

Hence, value of a = 3.

Salman deposits ₹ 1,000 every month in a recurring deposit account for 2 years. If he receives ₹ 26,000 on maturity, find :

(a) total interest Salman earns

(b) the rate of interest.

Answer

(a) Given,

Salman deposits ₹ 1000 every month in a recurring deposit account for 2 years.

Total deposit = ₹ 1,000 × 2 × 12 = ₹ 24,000.

By formula,

Total interest earned = Maturity value - Total deposit

= ₹ 26,000 - ₹ 24,000

= ₹ 2,000.

Hence, interest earned = ₹ 2,000.

(b) Let rate of interest be r%. Time (n) = 24 months

By formula,

Interest = $\dfrac{P \times n \times (n + 1)}{2 \times 12} \times \dfrac{r}{100}$

Substituting values we get :

$\Rightarrow 2000 = 1000 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{r}{100} \\[1em] \Rightarrow 2000 = 1000 \times 25 \times \dfrac{r}{100} \\[1em] \Rightarrow 2000 = 250r \\[1em] \Rightarrow r = \dfrac{2000}{250} \\[1em] \Rightarrow r = 8$

Hence, rate of interest earned = 8%.

In the given figure O, is the center of the circle. CE is a tangent to the circle at A. If ∠ABD = 26°, then find :

(a) ∠BDA

(b) ∠BAD

(c) ∠CAD

(d) ∠ODB

Answer

(a) We know that,

Angle in a semi-circle is a right angle.

∴ ∠BDA = 90°.

Hence, ∠BDA = 90°.

(b) In △ BAD,

By angle sum property of triangle,

⇒ ∠BDA + ∠BAD + ∠ABD = 180°

⇒ 90° + ∠BAD + 26° = 180°

⇒ ∠BAD + 116° = 180°

⇒ ∠BAD = 180° - 116° = 64°.

Hence, ∠BAD = 64°.

(c) From figure,

CE is tangent to the circle.

Angle between tangent and radius of the circle is 90°.

From figure,

⇒ ∠CAD + ∠BAD = 90°

⇒ ∠CAD + 64° = 90°

⇒ ∠CAD = 90° - 64° = 26°.

Hence, ∠CAD = 26°.

(d) Join OD.

In △ OBD,

⇒ OD = OB (Radius of same circle)

We know that,

Angles opposite to equal sides are equal.

⇒ ∠ODB = ∠OBD = 26°.

Hence, ∠ODB = 26°.

Solve the following quadratic equation :

x² + 4x - 8 = 0.

Give your answer correct to one decimal place.

Answer

Comparing x² + 4x - 8 = 0 with ax² + bx + c, we get :

a = 1, b = 4 and c = -8.

By formula,

$\Rightarrow x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[1em] = \dfrac{-4 \pm \sqrt{4^2 - 4 \times 1 \times (-8)}}{2 \times 1} \\[1em] = \dfrac{-4 \pm \sqrt{16 + 32}}{2} \\[1em] = \dfrac{-4 \pm \sqrt{48}}{2} \\[1em] = \dfrac{-4 \pm 4\sqrt{3}}{2} \\[1em] = -2 \pm 2\sqrt{3} \\[1em] = -2 \pm 2 \times 1.732 \\[1em] = -2 \pm 3.464 \\[1em] = -2 + 3.464 \text{ or } -2 - 3.464 \\[1em] = 1.464 \text{ or } -5.464 \\[1em] = 1.5 \text{ or } -5.5$

Hence, x = 1.5 or 5.5

Prove the following identity :

(sin² θ - 1)(tan² θ + 1) + 1 = 0

Answer

Solving L.H.S. of the given equation :

⇒ (sin² θ - 1)(tan² θ + 1) + 1

⇒ (1 - cos² θ - 1).sec² θ + 1

⇒ -cos² θ. sec² θ + 1

⇒ -cos² θ $\times \dfrac{1}{\text{cos}^2 \text{θ}}$ + 1

⇒ -1 + 1

⇒ 0.

Since, L.H.S. = R.H.S.

Hence, proved that (sin² θ - 1)(tan² θ + 1) + 1 = 0.

Use graph sheet to answer this question. Take 2 cm = 1 unit along both the axes.

(a) Plot A, B, C where A(0, 4), B(1, 1) and C(4, 0).

(b) Reflect A and B on the x-axis and name them as E and D respectively.

(c) Reflect B through the origin and name it F. Write down the coordinates of F.

(d) Reflect B and C on the y-axis and name them as H and G respectively.

(e) Join points A, B, C, D, E, F, G, H and A in order and name the closed figure formed.

Answer

The points are shown in the graph below:

(c) From graph,

F = (-1, -1)

Hence, co-ordinates of F = (-1, -1).

(e) From graph,

The closed figure formed is a star.

If $A = \begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}, B = \begin{bmatrix*}[r] 1 & 2 \\ 2 & 4 \end{bmatrix*}, C = \begin{bmatrix*}[r] 4 & 1 \\ 1 & 5 \end{bmatrix*} \text{ and } I = \begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}$. Find A(B + C) - 14I.

Answer

B + C = $\begin{bmatrix*}[r] 1 & 2 \\ 2 & 4 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & 1 \\ 1 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 3 \\ 3 & 9 \end{bmatrix*}$

Substituting values we get :

$A(B + C) - 14I = \begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 5 & 3 \\ 3 & 9 \end{bmatrix*} - 14\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 5 + 3 \times 3 & 1 \times 3 + 3 \times 9 \\ 2 \times 5 + 4 \times 3 & 2 \times 3 + 4 \times 9 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 + 9 & 3 + 27 \\ 10 + 12 & 6 + 36 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 14 & 30 \\ 22 & 42 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 14 - 14 & 30 - 0 \\ 22 - 0 & 42 - 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & 30 \\ 22 & 28 \end{bmatrix*}.$

Hence, A(B + C) - 14I = $\begin{bmatrix*}[r] 0 & 30 \\ 22 & 28 \end{bmatrix*}.$

ABC is a triangle whose vertices are A(1, -1), B(0, 4) and C(-6, 4), D is the mid-point of BC. Find the :

(a) coordinates of D.

(b) equation of the median AD.

Answer

(a) By formula,

Mid-point (M) = $\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$

Given,

D is the mid-point of BC.

$\therefore \text{Co-ordinates of D} = \Big(\dfrac{0 + (-6)}{2}, \dfrac{4 + 4}{2}\Big) \\[1em] = \Big(\dfrac{-6}{2}, \dfrac{8}{2}\Big) \\[1em] = (-3, 4).$

Hence, co-ordinates of D = (-3, 4).

(b) By two-point form :

Equation of a line :

y - y₁ = $\dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

Substituting values we get :

Equation of AD :

⇒ y - (-1) = $\dfrac{4 - (-1)}{(-3) - 1}(x - 1)$

⇒ y + 1 = $\dfrac{5}{-4}(x - 1)$

⇒ -4(y + 1) = 5(x - 1)

⇒ -4y - 4 = 5x - 5

⇒ 5x + 4y = -4 + 5

⇒ 5x + 4y = 1.

Hence, equation of median AD is 5x + 4y = 1.

In the given figure, O is the center of the circle. PQ is a tangent to the circle at T. Chord AB produced meets the tangent at P. AB = 9 cm, BP = 16 cm, ∠PTB = 50°, ∠OBA = 45°. Find :

(a) length of PT

(b) ∠BAT

(c) ∠BOT

(d) ∠ABT

Answer

(a) We know that,

If a chord and a tangent intersect externally, then the product of the lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

Since, PT is tangent from point P and PAB is secant.

∴ PT² = PA × PB

⇒ PT² = (9 + 16) × 16

⇒ PT² = 25 × 16

⇒ PT² = 400

⇒ PT = $\sqrt{400}$ = 20 cm.

Hence, PT = 20 cm.

(b) Since, angle between tangent and chord at point of contact is equal to angle of alternative segment.

∴ ∠BAT = ∠PTB = 50°.

Hence, ∠BAT = 50°.

(c) We know that,

Angle subtended by an arc at the centre of the circle is twice the angle subtended by the same arc at some point on the remaining circumference.

∴ ∠BOT = 2 × ∠BAT = 2 × 50° = 100°.

Hence, ∠BOT = 100°.

(d) From figure,

⇒ OB = OT (Radii of same circle)

⇒ ∠OBT = ∠OTB = x (let) (Angles opposite to equal sides are equal)

In △ BOT,

By angle sum property of triangle,

⇒ ∠OBT + ∠OTB + ∠BOT = 180°

⇒ x + x + 100° = 180°

⇒ 2x = 180° - 100°

⇒ 2x = 80°

⇒ x = $\dfrac{80°}{2}$ = 40°.

∴ ∠OBT = 40°

From figure,

⇒ ∠ABT = ∠ABO + ∠OBT = 45° + 40° = 85°.

Hence, ∠ABT = 85°.

Mrs. Arora bought the following articles from a departmental store :

Find the :

(a) Total GST paid.

(b) Total bill amount including GST.

Answer

(a) By formula,

GST on an article = (Price of item - Discount) × Rate of GST

GST on hair oil = (1200 - 100) × 18%

= 1100 × $\dfrac{18}{100}$

= 11 × 18

= ₹ 198.

GST on cashew nuts = (600 - 0) × 12%

= 600 × $\dfrac{12}{100}$

= 6 × 12

= ₹ 72.

Total GST paid = ₹ 198 + ₹ 72 = ₹ 270.

Hence, total GST paid = ₹ 270.

(b) Total bill amount including GST = Total cost price + Total GST paid

= [(₹ 1200 - ₹ 100) + ₹ 600] + ₹ 270

= ₹ 1100 + ₹ 600 + ₹ 270

= ₹ 1970.

Hence, total bill amount = ₹ 1970.

Solve the following inequation. Write down the solution set and represent it on the real number line.

-5(x - 9) ≥ 17 - 9x > x + 2, x ∈ R.

Answer

Given, equation :

-5(x - 9) ≥ 17 - 9x > x + 2

Solving L.H.S. of the given equation :

⇒ -5(x - 9) ≥ 17 - 9x

⇒ -5x + 45 ≥ 17 - 9x

⇒ -5x + 9x ≥ 17 - 45

⇒ 4x ≥ -28

⇒ x ≥ $-\dfrac{28}{4}$

⇒ x ≥ -7 ............(1)

Solving L.H.S. of the given equation :

⇒ 17 - 9x > x + 2

⇒ x + 9x < 17 - 2

⇒ 10x < 15

⇒ x < $\dfrac{15}{10}$

⇒ x < $\dfrac{3}{2}$

⇒ x < 1.5 .............(2)

From equation (1) and (2), we get :

Solution set : {x : -7 ≤ x < 1.5}

Solution set on the number line :

In the given figure, AC // DE // BF. If AC = 24 cm, EG = 8 cm, GB = 16 cm, BF = 30 cm.

(a) Prove △ GED ~ △ GBF

(b) Find DE

(c) Find DB : AB.

Answer

(a) Given,

DE // BF

∴ ∠DEG = ∠GBF (Alternate angles are equal)

∠EDG = ∠GFB (Alternate angles are equal)

⇒ △GED ~ △ GBF (By A.A. axiom)

Hence, proved that △GED ~ △ GBF.

(b) We know that,

Corresponding sides of similar triangle are proportional.

Since, △ GED ~ △ GBF

$\Rightarrow \dfrac{DE}{BF} = \dfrac{EG}{BG} \\[1em] \Rightarrow \dfrac{DE}{30} = \dfrac{8}{16} \\[1em] \Rightarrow DE = 30 \times \dfrac{8}{16} \\[1em] \Rightarrow DE = 15 \text{ cm}.$

Hence, DE = 15 cm.

(c) In △ BAC and △ BDE,

⇒ ∠BAC = ∠BDE (Corresponding angles are equal)

⇒ ∠ABC = ∠DBE (Common angle)

⇒ △ BAC ~ △ BDE (By A.A. axiom)

We know that,

Corresponding sides of similar triangle are proportional.

Since, △ BAC ~ △ BDE

$\Rightarrow \dfrac{AB}{BD} = \dfrac{AC}{DE} \\[1em] \Rightarrow \dfrac{BD}{AB} = \dfrac{DE}{AC} \\[1em] \Rightarrow \dfrac{BD}{AB} = \dfrac{15}{24} \\[1em] \Rightarrow \dfrac{BD}{AB} = \dfrac{5}{8}$

⇒ BD : AB = 5 : 8.

Hence, BD : AB = 5 : 8.

The following distribution gives the daily wages of 60 workers of a factory.

Use graph paper to answer this question.

Take 2 cm = ₹ 100 along one axis and 2 cm = 2 workers along the other axis. Draw a histogram and hence find the mode of the give distribution.

Answer

Steps of construction :

1. Draw a histogram of the given distribution.

2. Inside the highest rectangle, which represents the maximum frequency (or modal class), draw two lines AC and BD diagonally from the upper corners C and D of adjacent rectangles.

3. Through the point K (the point of intersection of diagonals AC and BD), draw KL perpendicular to the horizontal axis.

4. The value of point L on the horizontal axis represents the value of mode.

From graph,

L = ₹ 530

Hence, required mode = ₹ 530.

The 5^th and 9^th term of an Arithmetic Progression are 4 and -12 respectively. Find :

(a) the first term

(b) common difference

(c) sum of 16 terms of the A.P.

Answer

If first term is a and common difference is d of the A.P.

By formula,

n^th term = a_n = a + (n - 1)d

Given,

⇒ 5^th term = 4

⇒ a₅ = 4

⇒ a + (5 - 1)d = 4

⇒ a + 4d = 4 .........(1)

⇒ 9^th term = -12

⇒ a₉ = -12

⇒ a + (9 - 1)d = -12

⇒ a + 8d = -12 .........(2)

Subtracting equation (1) from (2), we get :

⇒ (a + 8d) - (a + 4d) = -12 - 4

⇒ a - a + 8d - 4d = -16

⇒ 4d = -16

⇒ d = $-\dfrac{16}{4}$

⇒ d = -4.

Substituting value of d in equation (1), we get :

⇒ a + 4d = 4

⇒ a + 4(-4) = 4

⇒ a - 16 = 4

⇒ a = 4 + 16 = 20.

(a) Hence, first term of A.P. = 20.

(b) Common difference of A.P. = -4.

(c) By formula,

Sum of n terms of A.P. = $\dfrac{n}{2}(a + l)$

Sum of 16 terms of A.P. = $\dfrac{n}{2}(a + a_{16})$

$= \dfrac{16}{2}[a + a + (16 - 1)d] \\[1em] = 8(2a + 15d) \\[1em] = 8 \times (2 \times 20 + 15 \times -4) \\[1em] = 8 \times (40 - 60) \\[1em] = 8 \times -20 \\[1em] = -160.$

Hence, sum of 16 terms of A.P. = -160.

A and B are two points on the x-axis and y-axis respectively.

(a) Write down the co-ordinates of A and B.

(b) P is a point on AB such that AP : PB = 3 : 1. Using section formula find the coordinates of point P.

(c) Find the equation of a line passing through P and perpendicular to AB.

Answer

(a) From figure,

A = (4, 0) and B = (0, 4).

(b) Let coordinates of P be (x, y).

By section formula,

(x, y) = $\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Substituting values we get :

$\Rightarrow (x, y) = \Big(\dfrac{3 \times 0 + 1 \times 4}{3 + 1}, \dfrac{3 \times 4 + 1 \times 0}{3 + 1}\Big) \\[1em] = \Big(\dfrac{0 + 4}{4}, \dfrac{12 + 0}{4}\Big) \\[1em] = \Big(\dfrac{4}{4}, \dfrac{12}{4}\Big) \\[1em] = (1, 3).$

Hence, coordinates of P = (1, 3).

(c) By formula,

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get :

Slope of AB = $\dfrac{4 - 0}{0 - 4} = \dfrac{4}{-4}$ = -1.

We know that,

Product of slope of perpendicular lines = -1.

∴ Slope of AB × Slope of line perpendicular to AB = -1

⇒ -1 × Slope of line perpendicular to AB = -1

⇒ Slope of line perpendicular to AB = $\dfrac{-1}{-1}$ = 1.

Line passing through P and perpendicular to AB :

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = 1(x - 1)

⇒ y - 3 = x - 1

⇒ y = x - 1 + 3

⇒ y = x + 2.

Hence, required equation is y = x + 2.

A bag contains 25 cards, numbered through 1 to 25. A card is drawn at random. What is the probability that the number on the card drawn is :

(a) a multiple of 5

(b) a perfect square

(c) a prime number ?

Answer

Total number of outcomes = 25

(a) Number of multiples of 5 between 1 to 25 = 5 (5, 10, 15, 20, 25).

P(that number on card is a multiples of 5)

= $\dfrac{\text{No. of multiples of 5}}{\text{Total number of outcomes}} = \dfrac{5}{25} = \dfrac{1}{5}$.

Hence, the required probability is $\dfrac{1}{5}$.

(b) Number of perfect squares between 1 to 25 = 5 (1, 4, 9, 16, 25).

P(that number on card is a perfect square)

= $\dfrac{\text{No. of perfect squares}}{\text{Total number of outcomes}} = \dfrac{5}{25} = \dfrac{1}{5}$.

Hence, the required probability is $\dfrac{1}{5}$.

(c) Number of prime numbers between 1 to 25 = 9 (2, 3, 5, 7, 11, 13, 17, 19, 23).

P(that number is a prime number)

= $\dfrac{\text{No. of prime numbers}}{\text{Total number of outcomes}} = \dfrac{9}{25}$.

Hence, the required probability is $\dfrac{9}{25}$.

A man covers a distance of 100 km, travelling with a uniform speed of x km/hr. Had the speed been 5 km/hr more it would have taken 1 hour less. Find x the original speed.

Answer

In first case :

Distance = 100 km

Speed = x km/hr

Time = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{100}{x}$

In second case :

Distance = 100 km

Speed = (x + 5) km/hr

Time = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{100}{x + 5}$

According to question,

The difference between the time taken between first and second case is 1 hour.

$\therefore \dfrac{100}{x} - \dfrac{100}{x + 5} = 1 \\[1em] \Rightarrow \dfrac{100(x + 5) - 100x}{x(x + 5)} = 1 \\[1em] \Rightarrow \dfrac{100x + 500 - 100x}{x^2 + 5x} = 1 \\[1em] \Rightarrow \dfrac{500}{x^2 + 5x} = 1 \\[1em] \Rightarrow 500 = x^2 + 5x \\[1em] \Rightarrow x^2 + 5x - 500 = 0 \\[1em] \Rightarrow x^2 + 25x - 20x - 500 = 0 \\[1em] \Rightarrow x(x + 25) - 20(x + 25) = 0 \\[1em] \Rightarrow (x - 20)(x + 25) = 0 \\[1em] \Rightarrow x - 20 = 0 \text{ or } x + 25 = 0 \\[1em] \Rightarrow x = 20 \text{ or } x = -25.$

Since, speed cannot be negative.

∴ x = 20 km/hr.

Hence, original speed = 20 km/hr.

A solid is in the shape of a hemisphere of radius 7 cm, surmounted by a cone of height 4 cm. The solid is immersed completely in a cylindrical container filled with water to a certain height so that the solid is completely submerged in water. If the radius of the cylinder is 14 cm, find the rise in the water level.

Answer

Let the rise in water level be x cm.

∴ Volume of water that rises by x cm in the cylindrical container = Volume of hemisphere submerged + Volume of cone submerged

⇒ π × (14)² × x = $\dfrac{2}{3} \times π \times 7^3 + \dfrac{1}{3} \times π \times 7^2 \times 4$

⇒ 196πx = $\dfrac{2}{3} \times π \times 343 + \dfrac{1}{3} \times π \times 196$

⇒ 196x = $\dfrac{2}{3} \times 343 + \dfrac{1}{3} \times 196$

⇒ 196x = $\dfrac{686 + 196}{3}$

⇒ 196x = $\dfrac{882}{3}$

⇒ x = $\dfrac{882}{3 \times 196} = \dfrac{882}{588}$ = 1.5 cm.

Hence, rise in water level = 1.5 cm

The following table gives the marks scored by a set of students in an examination. Calculate the mean of the distribution by using the short cut method.

Answer

By formula,

Mean = A + $\dfrac{Σfd}{Σf} = 25 + \dfrac{90}{40}$

= 25 + 2.25

= 27.25

Hence, required mean = 27.25

What number must be added to each of the number 4, 6, 8, 11 in order to get the four numbers in proportion ?

Answer

Let x be added to each number.

So, 4 + x, 6 + x, 8 + x and 11 + x must be in proportion.

$\therefore \dfrac{4 + x}{6 + x} = \dfrac{8 + x}{11 + x} \\[1em] \Rightarrow (4 + x) \times (11 + x) = (8 + x) \times (6 + x) \\[1em] \Rightarrow 44 + 4x + 11x + x^2 = 48 + 8x + 6x + x^2\\[1em] \Rightarrow 44 + 15x + x^2 = 48 + 14x + x^2 \\[1em] \Rightarrow 15x - 14x = 48 - 44 \\[1em] \Rightarrow x = 4.$

Hence, required number is 4.

Using ruler and compass construct a triangle ABC in which AB = 6 cm. ∠BAC = 120° and AC = 5 cm. Construct a circle passing through A, B and C. Measure and write down the radius of the circle.

Answer

Steps of construction :

1. Draw AB = 6 cm.

2. Draw AP so that angle BAP = 120°.

3. From AP cut AC = 5 cm ⇒ ∠BAC = 120°.

4. Join B and C to get triangle ABC.

5. Draw perpendicular bisectors of AB and AC which cut each other at point O.

6. With O as center and OA (or OB or OC) as radius draw circle which passes through points A, B and C.

Hence, radius of circle obtained = 5 cm.

Using componendo and dividendo solve for x : $\dfrac{\sqrt{2x + 2} + \sqrt{2x - 1}}{\sqrt{2x + 2} - \sqrt{2x - 1}} = 3$.

Answer

Given : $\dfrac{\sqrt{2x + 2} + \sqrt{2x - 1}}{\sqrt{2x + 2} - \sqrt{2x - 1}} = 3$

Applying componendo and dividendo, we get :

$\Rightarrow \dfrac{\sqrt{2x + 2} + \sqrt{2x - 1} + \sqrt{2x + 2} - \sqrt{2x - 1}}{\sqrt{2x + 2} + \sqrt{2x - 1} - (\sqrt{2x + 2} - \sqrt{2x - 1})} = \dfrac{3 + 1}{3 - 1} \\[1em] \Rightarrow \dfrac{2\sqrt{2x + 2}}{2\sqrt{2x - 1}} = \dfrac{4}{2} \\[1em] \Rightarrow \dfrac{\sqrt{2x + 2}}{\sqrt{2x - 1}} = 2 \\[1em] \Rightarrow \sqrt{2x + 2} = 2\sqrt{2x - 1}$

Squaring both sides we get :

⇒ 2x + 2 = 4(2x - 1)

⇒ 2x + 2 = 8x - 4

⇒ 8x - 2x = 2 + 4

⇒ 6x = 6

⇒ x = $\dfrac{6}{6}$ = 1.

Hence, x = 1.

Which term of the Arithmetic Progression (A.P.) 15, 30, 45, 60 ..... is 300 ? Hence find the sum of all terms of the Arithmetic Progression (A.P.).

Answer

Given,

A.P. : 15, 30, 45, 60 .....

First term (a) = 15

Common difference (d) = 30 - 15 = 15

Let n^th term be 300.

⇒ a_n = 300

⇒ a + (n - 1)d = 300

⇒ 15 + 15(n - 1) = 300

⇒ 15 + 15n - 15 = 300

⇒ 15n = 300

⇒ n = $\dfrac{300}{15}$ = 20.

Sum of the terms of the given A.P. :

⇒ S_n = $\dfrac{n}{2}(a + a_n)$

⇒ S₂₀ = $\dfrac{20}{2}(15 + 300)$

= 10 × 315

= 3150.

Hence, sum of all terms of the given A.P. = 3150.

From the top of a tower 100 m high, a man observes the angle of depression of two ships A and B, on opposite sides of the tower as 45° and 38° respectively. If the foot of tower and the ships are in the same horizontal line, find the distance between two ships A and B to the nearest metre.

Answer

Let CD be the tower.

From figure,

⇒ ∠A = ∠EDA = 45° (Alternate angles are equal)

⇒ ∠B = ∠FDB = 38° (Alternate angles are equal)

In △ ACD,

⇒ tan A = $\dfrac{CD}{AC}$

⇒ tan 45° = $\dfrac{100}{AC}$

⇒ AC = $\dfrac{100}{\text{tan 45°}} = \dfrac{100}{1}$ = 100 m.

In △ BCD,

⇒ tan B = $\dfrac{CD}{BC}$

⇒ tan 38° = $\dfrac{100}{BC}$

⇒ BC = $\dfrac{100}{\text{tan 38°}} = \dfrac{100}{0.7813}$ = 127.99 m.

From figure,

The distance between ships A and B = AC + BC

= 100 + 127.99

= 227.99 m ≈ 228 m.

Hence, the distance between the two ships A and B = 228 m.

Factorize completely using factor theorem : 2x³ - x² - 13x - 6.

Answer

Substituting x = -2 in 2x³ - x² - 13x - 6, we get :

⇒ 2(-2)³ - (-2)² - 13(-2) - 6

⇒ 2(-8) - 4 + 26 - 6

⇒ -16 - 4 + 20

⇒ -20 + 20

⇒ 0.

∴ x + 2 is a factor of the polynomial 2x³ - x² - 13x - 6.

Dividing, 2x³ - x² - 13x - 6 by x + 2, we get :

$\begin{array}{l} \phantom{x + 2)}{\quad 2x^2 -5x - 3} \\ x + 2\overline{\smash{\big)}\quad 2x^3 - x^2 - 13x - 6} \\ \phantom{x + 2)}\phantom{)}\underline{\underset{-}{+}2x^3 \underset{-}{+}4x^2} \\ \phantom{{x + 2}x^3-2}-5x^2 - 13x \\ \phantom{{x + 2)}x^3-2}\underline{\underset{+}{-}5x^2 \underset{+}{-} 10x} \\ \phantom{{x + 2)}{x^3-2x^{2}(3)}}-3x - 6 \\ \phantom{{x + 2)}{x^3-2x^{2}(31)}}\underline{\underset{+}{-}3x \underset{+}{-} 6} \\ \phantom{{x + 2)}{x^3-2x^{2}(31)}{-2x}}\times \end{array}$

∴ 2x³ - x² - 13x - 6 = (x + 2)(2x² - 5x - 3)

= (x + 2)(2x² - 6x + x - 3)

= (x + 2)[2x(x - 3) + 1(x - 3)]

= (x + 2)(2x + 1)(x - 3).

Hence, 2x³ - x² - 13x - 6 = (x + 2)(2x + 1)(x - 3).

Use graph paper to answer this question.

During a medical checkup of 60 students in a school, weights were recorded as follows :

Taking 2 cm = 2 kg along one axis and 2 cm = 10 students along the other axis draw an ogive. Use your graph to find the :

(a) median

(b) upper quartile

(c) number of students whose weight is above 37 kg.

Answer

(a)

Here, n = 60, which is even.

Median = $\dfrac{n}{2}$ th term = $\dfrac{60}{2}$ = 30th term.

Steps of construction :

1. Take 1 cm = 2 kg on x-axis.

2. Take 1 cm = 10 students on y-axis.

3. Since, x axis starts at 28 hence, a kink is drawn at the starting of x-axis. Plot the point (28, 0) as ogive starts on x-axis representing lower limit of first class.

4. Plot the points (30, 2), (32, 6), (34, 16), (36, 29), (38, 44), (40, 53), (42, 58) and (44, 60).

5. Join the points by a free-hand curve.

6. Draw a line parallel to x-axis from point A (no. of students) = 30, touching the graph at point B. From point B draw a line parallel to y-axis touching x-axis at point C.

From graph, C = 36.2

Hence, median = 36.2 kg

(ii) Here, n = 60, which is even.

By formula,

Upper quartile = $\dfrac{3n}{4} = \dfrac{3 \times 60}{4}$ = 45th term.

Draw a line parallel to x-axis from point D (no. of students) = 45, touching the graph at point E. From point E draw a line parallel to y-axis touching x-axis at point F.

From graph, F = 38.2 kg

Hence, upper quartile = 38.2 kg.

(iii) Draw a line parallel to y-axis from point G (weight) = 37 kg, touching the graph at point H. From point H draw a line parallel to x-axis touching y-axis at point I.

From graph, I = 38.

∴ 38 students have weight less than or equal to 37 kg.

No. of students whose weight is more than 37 kg = 60 - 38 = 22.

Hence, no. of students whose weight is more than 37 kg = 22.