Solved 2027 Specimen Paper ICSE Class 10 Mathematics

A customer pays a total GST of ₹ 600 on an article in an intra-state sale, and the tax rate is 12%. The price of the article is :

1. ₹2,000

2. ₹3,000

3. ₹5,000

4. ₹7,200

Answer

Given,

Total GST paid = ₹ 600

Rate of GST = 12%

Let the price of the article be ₹ $x$.

By formula,

GST = Rate of GST × Price of the article

$\Rightarrow 600 = \dfrac{12}{100} \times x \\[1em] \Rightarrow x = \dfrac{600 \times 100}{12} \\[1em] \Rightarrow x = \dfrac{60000}{12} \\[1em] \Rightarrow x = 5000.$

Thus, the price of the article = ₹ 5,000.

Hence, option 3 is the correct option.

Assertion (A): In a recurring deposit account, a man deposits ₹ 1000 per month for 1 year. On maturity he receives ₹ 13500. The amount of interest he earns is ₹ 1500.

Reason (R): Interest earned = Maturity value − Total amount deposited.

1. (A) is true and (R) is false.

2. (A) is false and (R) is true.

3. Both (A) and (R) are true, and (R) is the correct explanation for (A).

4. Both (A) and (R) are true, but (R) is not the correct explanation for (A).

Answer

Given,

Deposit per month (P) = ₹ 1000

Time (n) = 1 year = 12 months

Maturity value = ₹ 13,500

Total amount deposited = P × n

= 1000 × 12

= ₹ 12,000

By formula,

Interest earned = Maturity value − Total amount deposited

= 13500 − 12000

= ₹ 1,500

So assertion (A) is true.

We know that,

Maturity value = Total amount deposited + Interest earned

⇒ Interest earned = Maturity value − Total amount deposited

So reason (R) is true and it correctly explains the assertion (A).

Thus, Both (A) and (R) are true, and (R) is the correct explanation for (A).

Hence, option 3 is the correct option.

A man buys 50, ₹ 100 shares of a company at ₹ 125 paying 6% dividend per annum. At the end of the year, the company will pay a total dividend on :

1. ₹100

2. ₹125

3. ₹5000

4. ₹6250

Answer

Given,

Number of shares bought = 50

Nominal value (N.V.) of each share = ₹ 100

Market value (M.V.) of each share = ₹ 125

Rate of dividend = 6%

We know that,

Dividend is always paid by the company on the nominal value (face value) of the shares and not on their market value.

Total nominal value of the shares = No. of shares × N.V. of 1 share

= 50 × 100

= ₹ 5,000

Thus, the company will pay the total dividend on ₹ 5,000.

Hence, option 3 is the correct option.

If the roots of the quadratic equation $x^2 − 8x + p^2$ = 0 are equal, then the positive value of $p$ is :

1. 16

2. 8

3. 4

4. 2

Answer

Given,

The roots of the quadratic equation $x^2 - 8x + p^2$ = 0 are equal.

Comparing the given equation with $ax^2 + bx + c$ = 0, we get :

$a = 1, b = -8 \text{ and } c = p^2$

We know that,

For equal roots, discriminant (D) = 0.

$\Rightarrow b^2 - 4ac = 0 \\[1em] \Rightarrow (-8)^2 - 4 \times 1 \times p^2 = 0 \\[1em] \Rightarrow 64 - 4p^2 = 0 \\[1em] \Rightarrow 4p^2 = 64 \\[1em] \Rightarrow p^2 = 16 \\[1em] \Rightarrow p = \pm 4$

Since, the positive value of $p$ is required.

∴ $p = 4$.

Hence, option 3 is the correct option.

12, 18 and $x$ are in continued proportion. The value of $x$ is :

1. 24

2. 27

3. 30

4. 36

Answer

Given,

12, 18 and $x$ are in continued proportion.

We know that,

If $a, b$ and $c$ are in continued proportion, then $b^2 = ac$.

$\Rightarrow 18^2 = 12 \times x \\[1em] \Rightarrow 324 = 12x \\[1em] \Rightarrow x = \dfrac{324}{12} \\[1em] \Rightarrow x = 27.$

Hence, option 2 is the correct option.

$(x + 2)$ and $(x + 3)$ are the factors of $x^3 + 4x^2 + x - 6$. The third factor of the given polynomial is :

1. $(x - 1)$

2. $(x - 4)$

3. $(x + 1)$

4. $(x + 4)$

Answer

Given,

$(x + 2)$ and $(x + 3)$ are the factors of $x^3 + 4x^2 + x - 6$.

⇒ $(x + 2)(x + 3) = x^2 + 5x + 6$

Thus, $x^2 + 5x + 6$ is also the factor of $x^3 + 4x^2 + x - 6$.

$\begin{array}{l} \phantom{x^2 + 5x + 6\big)\quad }\phantom{x^3 +} x - 1 \\ x^2 + 5x + 6\overline{\smash{\big)}\quad x^3 + 4x^2 + x - 6} \\ \phantom{+++++()}\underline{\underset{-}{+}x^3 \underset{-}{+}5x^2 \underset{-}{+}6x} \\ \phantom{+++++())}\phantom{x^3 +} -x^2 - 5x - 6 \\ \phantom{+++++())}\phantom{x^3 +} \underline{\underset{+}{-}x^2 \underset{+}{-}5x \underset{+}{-}6} \\ \phantom{x^2 + 5x + 6\big)\quad }\phantom{x^3 + 4x^2 +} \times \end{array}$

Thus,

$x^3 + 4x^2 + x - 6 = (x^2 + 5x + 6)(x - 1) \\[1em] = (x + 2)(x + 3)(x - 1).$

Thus, the third factor of the given polynomial is $(x - 1)$.

Hence, option 1 is the correct option.

Given two matrices A and B, where $A = \begin{bmatrix} -1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 3 \\ 4 \end{bmatrix}$, then product AB is :

1. $\begin{bmatrix} -3 \\ 8 \end{bmatrix}$

2. 5

3. $\begin{bmatrix} 5 \end{bmatrix}$

4. $\begin{bmatrix} -3 & 8 \end{bmatrix}$

Answer

Given,

Matrix A is of order 1 × 2 and matrix B is of order 2 × 1.

We know that,

If matrix A is of order $m \times n$ and matrix B is of order $n \times p$, then the order of the product AB is $m \times p$.

∴ Order of AB = 1 × 1

Calculating,

$\Rightarrow AB = \begin{bmatrix*}[r] -1 & 2 \end{bmatrix*}\begin{bmatrix*}[r] 3 \\ 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -1 \times 3 + 2 \times 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3 + 8 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 \end{bmatrix*}.$

Hence, option 3 is the correct option.

If the $n^{th}$ term of a Geometric Progression (G.P.) is given by $5 \times 2^n$ then the common ratio is :

1. 2

2. 5

3. 10

4. 20

Answer

Given,

$n^{th} \text{ term of the G.P.} = 5 \times 2^n \\[1em] \therefore t_n = 5 \times 2^n$

Replacing $n$ by $(n + 1)$, we get :

⇒ $t_{n + 1} = 5 \times 2^{n + 1}$

By formula,

$\text{Common ratio } (r) = \dfrac{t_{n + 1}}{t_n} \\[1em] = \dfrac{5 \times 2^{n + 1}}{5 \times 2^n} \\[1em] = 2^{n + 1 - n} \\[1em] = 2.$

Hence, option 1 is the correct option.

The line $y = 2x$ passes through which pair of coordinates?

1. (0, 0) and (2, 2)

2. (0, 2) and (2, 4)

3. (1, 2) and (−1, −2)

4. (2, 2) and (−2, −2)

Answer

Given,

Equation of the line is $y = 2x$.

A point lies on the line only if its coordinates satisfy the equation of the line.

Substituting the coordinates (1, 2) in $y = 2x$, we get :

⇒ 2 = 2 × 1

⇒ 2 = 2

Since, L.H.S. = R.H.S., the point (1, 2) lies on the line.

Substituting the coordinates (−1, −2) in $y = 2x$, we get :

⇒ −2 = 2 × (−1)

⇒ −2 = −2

Since, L.H.S. = R.H.S., the point (−1, −2) lies on the line.

Thus, the line $y = 2x$ passes through the points (1, 2) and (−1, −2).

Hence, option 3 is the correct option.

On a map of scale 1 : 50,000, a rectangular plot has sides 4 cm by 3 cm. The actual area of the plot on ground is :

1. 3 km²

2. 9 km²

3. 12 km²

4. 16 km²

Answer

Given,

Scale of the map = 1 : 50,000

Length of the plot on the map = 4 cm

Breadth of the plot on the map = 3 cm

Actual length of the plot = 4 × 50000 = 2,00,000 cm

= $\dfrac{200000}{100000}$ = 2 km [∵ 1 km = 1,00,000 cm]

Actual breadth of the plot = 3 × 50000 = 1,50,000 cm

= $\dfrac{150000}{100000}$ = 1.5 km

Actual area of the plot = Actual length × Actual breadth

= 2 × 1.5

= 3 km².

Hence, option 1 is the correct option.

In the given diagram, the radius of the circle with centre O is 3 cm. PA and PB are the tangents to the circle which are at right angle to each other. The length of OP is:

1. $\dfrac{3}{\sqrt{2}}$ cm

2. 3 cm

3. $3\sqrt{2}$ cm

4. $6\sqrt{2}$ cm

Answer

Join OA and OP.

Given,

Radius of circle (OB) = 3 cm

PB and AP are at right angles.

We know that,

Radius and tangent at point of contact are perpendicular to each other.

OA ⊥ AP

From figure,

⇒ OA = OB = 3 cm [Radius of circle]

⇒ OB is perpendicular to PB

⇒ ∠OBP = 90°

This shows APBO is square.

OA = PB = AP = OB = 3 cm.

In right angled triangle OBP,

⇒ OP² = OB² + PB²

⇒ OP² = 3² + 3²

⇒ OP² = 9 + 9

⇒ OP² = 18

⇒ OP = $\sqrt{18}$

⇒ OP = $3\sqrt{2}$ cm.

Hence, option 3 is the correct option.

A conical tent with a capacity of 616 m³ has a base area 154 m². The height of the tent is :

1. 3 m

2. 4 m

3. 7 m

4. 12 m

Answer

Given,

Capacity (volume) of the conical tent = 616 m³

Base area of the conical tent = 154 m²

By formula,

Volume of cone = $\dfrac{1}{3} \times \text{Base area} \times \text{Height}$

Substituting values, we get :

$\Rightarrow 616 = \dfrac{1}{3} \times 154 \times h \\[1em] \Rightarrow 154h = 616 \times 3 \\[1em] \Rightarrow h = \dfrac{1848}{154} \\[1em] \Rightarrow h = 12 \text{ m}.$

Thus, the height of the tent = 12 m.

Hence, option 4 is the correct option.

Assertion (A): If $\sec\theta + \tan\theta = a$ and $\sec\theta - \tan\theta = b$ then $ab = 1$.

Reason (R): $\sec^2\theta - \tan^2\theta = 1$

1. (A) is true and (R) is false.

2. (A) is false and (R) is true.

3. Both (A) and (R) are true and (R) is the correct explanation of (A).

4. Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Answer

Given,

$\Rightarrow \sec\theta + \tan\theta = a \\[1em] \Rightarrow \sec\theta - \tan\theta = b \\[1em] \Rightarrow ab = (\sec\theta + \tan\theta)(\sec\theta - \tan\theta) \\[1em] \Rightarrow ab = \sec^2\theta - \tan^2\theta \\[1em] \Rightarrow ab = 1$

So assertion (A) is true.

We know that,

⇒ $\sec^2\theta - \tan^2\theta = 1$

This is a fundamental trigonometric identity.

Thus, Both (A) and (R) are true and (R) is the correct explanation of (A).

Hence, option 3 is the correct option.

Consider the following distribution 30, 34, 35, 36, 37, 38, 39, 40. If the value 35 is removed from the given data, then the median changes by :

1. 0.5

2. 1.0

3. 1.5

4. 2.0

Answer

Given,

The distribution is 30, 34, 35, 36, 37, 38, 39, 40.

The data is already arranged in ascending order.

Here, $n = 8$, which is even.

By formula,

$\text{Median} = \dfrac{1}{2}\Big[\Big(\dfrac{n}{2}\Big)^{th} \text{term} + \Big(\dfrac{n}{2} + 1\Big)^{th} \text{term}\Big] \\[1em] = \dfrac{1}{2}[4^{th} \text{ term} + 5^{th} \text{ term}] \\[1em] = \dfrac{36 + 37}{2} \\[1em] = \dfrac{73}{2} \\[1em] = 36.5$

On removing the value 35, the distribution becomes 30, 34, 36, 37, 38, 39, 40.

Here, $n = 7$, which is odd.

By formula,

$\text{Median} = \Big(\dfrac{n + 1}{2}\Big)^{th} \text{ term} \\[1em] = 4^{th} \text{ term} \\[1em] = 37$

Change in median = 37 − 36.5 = 0.5.

Hence, option 1 is the correct option.

A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the numbers 1, 2, 3……12. The probability that it will point to an odd number is :

1. $\dfrac{1}{6}$

2. $\dfrac{1}{12}$

3. $\dfrac{1}{2}$

4. $\dfrac{5}{12}$

Answer

Given,

The arrow can come to rest pointing to one of the numbers 1, 2, 3, ......, 12.

Total number of possible outcomes = 12

Favourable outcomes = Arrow pointing to an odd number

{1, 3, 5, 7, 9, 11}

Number of favourable outcomes = 6

By formula,

Probability = $\dfrac{\text{No. of favourable outcomes}}{\text{Total number of outcomes}}$

= $\dfrac{6}{12}$

= $\dfrac{1}{2}$.

Hence, option 3 is the correct option.

Three students A, B and C of Class 10 are each given an equal amount of Play-Dough to form different shaped solids for a mathematics project. A forms a cone of radius 7 cm and height 28 cm. B forms a sphere of radius $r$ and C forms a cylinder of radius 14 cm. Find the :

(a) radius $r$ of the sphere so formed.

(b) height of the cylinder so formed.

(c) Volume of the dough given to each of them.

Give your answers to (b) and (c) correct to the nearest whole number. $\left(\pi = \dfrac{22}{7}\right)$

Answer

Given,

For the cone formed by A,

Radius (R) = 7 cm

Height (H) = 28 cm

Since, all the three students are given an equal amount of Play-Dough, the volumes of the cone, the sphere and the cylinder are equal.

(a) Volume of sphere = Volume of cone

$\Rightarrow \dfrac{4}{3}\pi r^3 = \dfrac{1}{3}\pi R^2 H \\[1em] \Rightarrow \dfrac{4}{3}\pi r^3 = \dfrac{1}{3} \times \pi \times 7^2 \times 28 \\[1em] \Rightarrow 4r^3 = 7^2 \times 28 \\[1em] \Rightarrow 4r^3 = 49 \times 28 \\[1em] \Rightarrow r^3 = \dfrac{1372}{4} \\[1em] \Rightarrow r^3 = 343 \\[1em] \Rightarrow r^3 = 7^3 \\[1em] \Rightarrow r = 7 \text{ cm}.$

Hence, radius of the sphere = 7 cm.

(b) Given,

Radius of the cylinder (R') = 14 cm

Let height of cylinder be H'.

Volume of cylinder = Volume of sphere

$\Rightarrow \pi R'^2 H' = \dfrac{4}{3}\pi r^3 \\[1em] \Rightarrow 14^2 \times H' = \dfrac{4}{3} \times 7^3 \\[1em] \Rightarrow 196 \times H' = \dfrac{4 \times 343}{3} \\[1em] \Rightarrow H' = \dfrac{1372}{3 \times 196} \\[1em] \Rightarrow H' = \dfrac{7}{3} \\[1em] \Rightarrow H' = 2.33 \text{ cm} \approx 2 \text{ cm}.$

Hence, height of the cylinder = 2 cm.

(c) By formula,

Volume of the dough given to each of them = Volume of cone

$= \dfrac{1}{3}\pi R^2 H \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 7^2 \times 28 \\[1em] = \dfrac{1}{3} \times 22 \times 7 \times 28 \\[1em] = \dfrac{4312}{3}\\[1em] = 1437.33 \text{ cm}^3 ≈ 1437 \text{ cm}^3.$

Hence, volume of the dough given to each of them = 1437 cm³.

Solve the quadratic equation $(x - 2)^2 + 11x - 7$ = 0 and give your answer correct to 3 significant figures.

Answer

Given,

$\Rightarrow (x - 2)^2 + 11x - 7 = 0 \\[1em] \Rightarrow x^2 - 4x + 4 + 11x - 7 = 0 \\[1em] \Rightarrow x^2 + 7x - 3 = 0$

Comparing the above equation with $ax^2 + bx + c$ = 0, we get :

$a = 1, b = 7 \text{ and } c = -3$

By quadratic formula,

$\Rightarrow x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \dfrac{-7 \pm \sqrt{7^2 - 4 \times 1 \times (-3)}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{-7 \pm \sqrt{49 + 12}}{2} \\[1em] \Rightarrow x = \dfrac{-7 \pm \sqrt{61}}{2} \\[1em] \Rightarrow x = \dfrac{-7 \pm 7.8102}{2}$

Solving further,

$\Rightarrow x = \dfrac{-7 + 7.8102}{2} \text{ or } x = \dfrac{-7 - 7.8102}{2} \\[1em] \Rightarrow x = \dfrac{0.8102}{2} \text{ or } x = \dfrac{-14.8102}{2} \\[1em] \Rightarrow x = 0.4051 \text{ or } x = -7.4051$

Correct to 3 significant figures,

$x = 0.405 \text{ or } x = -7.41$.

Hence, $\bm{x}$ = 0.405 or $\bm{x}$ = −7.41.

A line is formed joining the two points P(−4, 1) and Q(11, −2). Straight line PQ intersects the x-axis at R. Find the :

(a) ratio PR : RQ.

(b) coordinates of the point R.

(c) equation of the line perpendicular to PQ at R.

Answer

(a) Since, R lies on the x-axis, its ordinate is zero.

Let the point R$(x, 0)$ divide the line segment joining P(−4, 1) and Q(11, −2) in the ratio $m : n$.

By section formula,

$y = \dfrac{my_2 + ny_1}{m + n}$

Substituting values, we get :

$\Rightarrow 0 = \dfrac{m(-2) + n(1)}{m + n} \\[1em] \Rightarrow -2m + n = 0 \\[1em] \Rightarrow n = 2m \\[1em] \Rightarrow \dfrac{m}{n} = \dfrac{1}{2}.$

Hence, PR : RQ = 1 : 2.

(b) From part (a),

P(−4, 1), Q(11, −2) and $m : n$ = 1 : 2

By section formula,

$x = \dfrac{mx_2 + nx_1}{m + n}$

Substituting values, we get :

$\Rightarrow x = \dfrac{1(11) + 2(-4)}{1 + 2} \\[1em] \Rightarrow x = \dfrac{11 - 8}{3} \\[1em] \Rightarrow x = \dfrac{3}{3} \\[1em] \Rightarrow x = 1.$

R = $(x, 0)$ = (1, 0).

Hence, the coordinates of the point R are (1, 0).

(c) By formula,

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

Substituting values, we get :

$\Rightarrow m_{PQ} = \dfrac{-2 - 1}{11 - (-4)} \\[1em] \Rightarrow m_{PQ} = \dfrac{-3}{15} \\[1em] \Rightarrow m_{PQ} = -\dfrac{1}{5}$

We know that,

The product of the slopes of two perpendicular lines is −1.

∴ Slope of the line perpendicular to PQ = $\dfrac{-1}{-\dfrac{1}{5}}$ = 5

By point-slope form,

$y - y_1 = m(x - x_1)$

Equation of the line perpendicular to PQ and passing through R(1, 0) is :

$\Rightarrow y - 0 = 5(x - 1) \\[1em] \Rightarrow y = 5x - 5 \\[1em] \Rightarrow 5x - y = 5.$

Hence, the equation of the line perpendicular to PQ at R is $\bm{5x − y = 5}$.

A shopkeeper marked an article at ₹ 2000. The rate of GST on the article is 18%. The customer has only ₹ 1888 with him and he requests the shopkeeper to reduce the price so that he can buy the article at ₹ 1888 including GST. What percent discount must the shopkeeper give to do so?

Answer

Given,

Marked price of the article = ₹ 2000

Rate of GST = 18%

Amount available with the customer = ₹ 1888 (including GST)

Let the reduced price of the article be ₹ $x$.

⇒ $x + 18$

$\Rightarrow x + \dfrac{18}{100} \times x = 1888 \\[1em] \Rightarrow x\Big(1 + \dfrac{18}{100}\Big) = 1888 \\[1em] \Rightarrow x \times \dfrac{118}{100} = 1888 \\[1em] \Rightarrow x = \dfrac{1888 \times 100}{118} \\[1em] \Rightarrow x = 1600.$

Reduced price of the article = ₹ 1,600

Reduction in price = ₹ 2000 − ₹ 1600 = ₹ 400

By formula,

Discount % = $\dfrac{\text{Reduction in price}}{\text{Marked price}} \times 100$

= $\dfrac{400}{2000} \times 100$

= 20%.

Hence, the shopkeeper must give a discount of 20%.

An Arithmetic Progression (A.P.) consists of all whole numbers which are divisible by 3 and 5.

(a) Write its first term and the common difference.

(b) Find the 10th term of the A.P.

(c) Find the sum of the first 10 terms of the A.P.

Answer

(a) Given,

The A.P. consists of all whole numbers which are divisible by 3 and 5.

A number which is divisible by both 3 and 5 is divisible by their L.C.M.

L.C.M. of 3 and 5 = 15

Thus, the A.P. is 15, 30, 45, 60, ......

First term (a) = 15

Common difference (d) = 30 − 15 = 15.

Hence, first term = 15 and common difference = 15.

(b) By formula,

t_n = a + (n - 1)d

Substituting values, we get :

⇒ t₁₀ = 15 + (10 − 1) × 15

⇒ t₁₀ = 15 + 9 × 15

⇒ t₁₀ = 15 + 135

⇒ t₁₀ = 150.

Hence, the 10^th term of the A.P. = 150.

(c) By formula,

$S_n = \dfrac{n}{2}[2a + (n - 1)d]$

Substituting values, we get :

⇒ S₁₀ = $\dfrac{10}{2}[2 \times 15 + (10 - 1) \times 15]$

⇒ S₁₀ = 5[30 + 9 × 15]

⇒ S₁₀ = 5[30 + 135]

⇒ S₁₀ = 5 × 165

⇒ S₁₀ = 825.

Hence, the sum of the first 10 terms of the A.P. = 825.

Study the graph and answer the given questions :

(a) Write down the coordinates of A, B, C, D and E.

(b) Write down the image of A under reflection on the y-axis.

(c) Name the point, with its coordinates, which is the image of C under reflection through the origin.

(d) What is the image of the polygon ABCDEFGHI when reflected on the y-axis?

(e) Write down the equation of the line joining the points N and D.

Answer

(a) From the graph,

The coordinates of the points are A(0, 3), B(1, 2), C(2, 2), D(2, 1) and E(3, 0).

(b) We know that,

The image of a point $(x, y)$ under reflection on the y-axis is $(-x, y)$.

∴ Image of A(0, 3) on reflection in y-axis is (0, 3).

Since, A lies on the y-axis, it remains invariant under reflection on the y-axis.

Hence, the image of A under reflection on the y-axis is A(0, 3).

(c) We know that,

The image of a point $(x, y)$ under reflection through the origin is $(-x, -y)$.

∴ Image of C(2, 2) on reflection in origin is (−2, −2).

From the graph, the point (−2, −2) is the point K.

Hence, the image of C under reflection through the origin is K(−2, −2).

(d) We know that,

The image of a point $(x, y)$ under reflection on the y-axis is $(-x, y)$.

So, on reflecting the polygon ABCDEFGHI on the y-axis, each vertex is mapped to the point lying at the same distance from the y-axis but on the opposite side of it.

Since, the vertices A and I lie on the y-axis, they remain invariant under reflection on the y-axis.

From the graph, the images of the vertices of the polygon under reflection in y-axis are :

A → A, B → P, C → O, D → N, E → M, F → L, G → K, H → J and I → I

Thus, on joining these image points in order, we get the polygon APONMLKJI, which is the mirror image of the polygon ABCDEFGHI in the y-axis.

Hence, the image of the polygon ABCDEFGHI under reflection on the y-axis is the polygon APONMLKJI.

(e) From the graph,

N = (−2, 1) and D = (2, 1)

Since, both the points have the same ordinate (y-coordinate) equal to 1, the line joining them is parallel to the x-axis.

Hence, the equation of the line joining the points N and D is $\bm{y}$ = 1.

A man opened a recurring deposit account in a bank. He deposits ₹ 1000 per month for 2 years. The matured value is equal to 26 times the amount he pays per month. Find the :

(a) interest paid by the bank.

(b) rate of interest per annum that the bank was paying for the recurring deposit account.

Answer

Given,

Deposit per month (P) = ₹ 1000

Time (n) = 2 years = 24 months

Maturity value = 26 × ₹ 1000 = ₹ 26,000

(a) Total amount deposited = P × n

= 1000 × 24

= ₹ 24,000

By formula,

Interest = Maturity value − Total amount deposited

= ₹ 26,000 − ₹ 24,000

= ₹ 2,000.

Hence, interest paid by the bank = ₹ 2,000.

(b) Let the rate of interest be r% per annum.

By formula,

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

Substituting values, we get :

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

Hence, rate of interest (r%) = 8% per annum.

Given, matrix $A = \begin{bmatrix} x & 1 \\ 2 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} x \\ x - 2 \end{bmatrix}$, such that AB is a null matrix. Find the :

(a) order of the null matrix.

(b) value of $x$, given that $x \gt 0$.

Answer

(a) Given,

Matrix A is of order 2 × 2 and matrix B is of order 2 × 1.

We know that,

If matrix A is of order $m \times n$ and matrix B is of order $n \times p$, then the order of the product AB is $m \times p$.

∴ Order of AB = 2 × 1

Since, AB is a null matrix, the order of the null matrix is the same as the order of AB.

Hence, the order of the null matrix is 2 × 1.

(b) Calculating AB,

$\Rightarrow AB = \begin{bmatrix*}[r] x & 1 \\ 2 & 2 \end{bmatrix*}\begin{bmatrix*}[r] x \\ x - 2 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] x \times x + 1 \times (x - 2) \\ 2 \times x + 2 \times (x - 2) \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] x^2 + x - 2 \\ 2x + 2x - 4 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] x^2 + x - 2 \\ 4x - 4 \end{bmatrix*}.$

Since, AB is a null matrix,

$\Rightarrow \begin{bmatrix*}[r] x^2 + x - 2 \\ 4x - 4 \end{bmatrix*} = \begin{bmatrix*}[r] 0 \\ 0 \end{bmatrix*}$

Comparing the corresponding elements, we get :

$\Rightarrow 4x - 4 = 0 \\[1em] \Rightarrow 4x = 4 \\[1em] \Rightarrow x = 1$

Also,

$\Rightarrow x^2 + x - 2 = 0 \\[1em] \Rightarrow x^2 + 2x - x - 2 = 0 \\[1em] \Rightarrow x(x + 2) - 1(x + 2) = 0 \\[1em] \Rightarrow (x + 2)(x - 1) = 0 \\[1em] \Rightarrow (x + 2) = 0 \text{ or } (x - 1) = 0 \qquad [\text{Using zero product rule}] \\[1em] \Rightarrow x = -2 \text{ or } x = 1$

The value of $x$ which satisfies both the equations and is greater than 0 is 1.

Hence, $\bm{x}$ = 1.

Prove the following trigonometric identity :

$\dfrac{\cos^2\theta}{1 - \tan\theta} + \dfrac{\sin^2\theta}{1 - \cot\theta} = 1 + \cos\theta,\sin\theta$

Answer

Solving L.H.S.,

$\Rightarrow \dfrac{\cos^2\theta}{1 - \tan\theta} + \dfrac{\sin^2\theta}{1 - \cot\theta} \\[1em] \Rightarrow \dfrac{\cos^2\theta}{1 - \dfrac{\sin\theta}{\cos\theta}} + \dfrac{\sin^2\theta}{1 - \dfrac{\cos\theta}{\sin\theta}} \\[1em] \Rightarrow \dfrac{\cos^2\theta}{\dfrac{\cos\theta - \sin\theta}{\cos\theta}} + \dfrac{\sin^2\theta}{\dfrac{\sin\theta - \cos\theta}{\sin\theta}} \\[1em] \Rightarrow \dfrac{\cos^3\theta}{\cos\theta - \sin\theta} + \dfrac{\sin^3\theta}{\sin\theta - \cos\theta} \\[1em] \Rightarrow \dfrac{\cos^3\theta}{\cos\theta - \sin\theta} - \dfrac{\sin^3\theta}{\cos\theta - \sin\theta} \\[1em] \Rightarrow \dfrac{\cos^3\theta - \sin^3\theta}{\cos\theta - \sin\theta} \\[1em] \Rightarrow \dfrac{(\cos\theta - \sin\theta)(\cos^2\theta + \cos\theta,\sin\theta + \sin^2\theta)}{\cos\theta - \sin\theta} \\[1em] \Rightarrow \cos^2\theta + \sin^2\theta + \cos\theta,\sin\theta \\[1em] \Rightarrow 1 + \cos\theta,\sin\theta \qquad [\because \sin^2\theta + \cos^2\theta = 1]$

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

Hence, proved that $\mathbf{\dfrac{\cos^2\theta}{1 - \tan\theta} + \dfrac{\sin^2\theta}{1 - \cot\theta} = 1 + \cos\theta,\sin\theta}$.

Ms. Kaur invested ₹ 8,000 in buying ₹100 shares of a company paying 6% dividend at ₹ 80. After a year, she sold these shares at ₹75 each and invested the proceeds including the dividend received during the first year in buying ₹ 20 shares, paying 15% dividend at ₹ 27 each. Find the :

(a) dividend received by her during the first year.

(b) number of shares she purchased using the total proceeds including the dividend.

Answer

Given,

For initial investment,

Investment = ₹ 8,000

Face Value = ₹ 100

Market Value = ₹ 80

Dividend Rate = 6%

By formula,

Number of shares = $\dfrac{\text{Investment}}{\text{Market value of each share}}$

= $\dfrac{8000}{80}$

= 100

By formula,

Dividend for the first year = No. of shares × Rate of div. × N.V. of 1 share

= 100 × $\dfrac{6}{100}$ × 100

= ₹ 600

Hence, dividend for first year = ₹ 600.

(b) Given,

Number of shares sold = 100

Selling price per share = ₹ 75

Proceeds from sale = Number of shares × selling price

= 100 × 75

= ₹ 7,500

Total proceeds = Proceeds from sale + Dividend received = 7500 + 600 = ₹ 8,100

Total investment = ₹ 8,100

Market value per share = ₹ 27

Number of new shares = $\dfrac{\text{Investment}}{\text{Market value of each share}}$

= $\dfrac{8100}{27}$

= 300.

Hence, number of shares purchased by Ms. Kaur = 300.

Solve the following inequation, write the solution set and represent it on the real number line.

$-6 + \dfrac{12x}{5} \leq \dfrac{7x}{5} - 3 \lt \dfrac{2}{5} + 2x, ; x \in R$

Answer

Solving L.H.S. of the inequation, we get :

$\Rightarrow -6 + \dfrac{12x}{5} \le \dfrac{7x}{5} - 3 \\[1em] \Rightarrow \dfrac{12x}{5} - \dfrac{7x}{5} \le 6 - 3 \\[1em] \Rightarrow \dfrac{12x - 7x}{5} \le 3 \\[1em] \Rightarrow \dfrac{5x}{5} \le 3 \\[1em] \Rightarrow x \le 3 \text{ ......(1)}$

Solving R.H.S. of the inequation, we get :

$\Rightarrow \dfrac{7x}{5} - 3 \lt \dfrac{2}{5} + 2x \\[1em] \Rightarrow \dfrac{7x}{5} - 2x \lt \dfrac{2}{5} + 3 \\[1em] \Rightarrow \dfrac{7x - 10x}{5} \lt \dfrac{2 + 15}{5} \\[1em] \Rightarrow \dfrac{-3x}{5} \lt \dfrac{17}{5} \\[1em] \Rightarrow -3x \lt 17 \\[1em] \Rightarrow x \gt \dfrac{-17}{3} \\[1em] \Rightarrow x \gt -5\dfrac{2}{3} \text{ ......(2)}$

From equation (1) and (2),

Solution set = ${x : -5\dfrac{2}{3} \lt x \le 3, x \in R}$

Hence, solution set = {$\bm{x} : \mathbf{-5\dfrac{2}{3}} \lt \bm{x} \le 3, \bm{x}$ ∈ R}.

In the given figure, O is the centre of the circle and AD = AB. If ∠ABC = 32°, find the angles $x, y, z, v$ and $w$.

Answer

Given,

O is the centre of the circle, AD = AB and ∠ABC = 32°.

From the figure, BC passes through the centre O and hence BC is a diameter of the circle.

We know that,

Angle in a semi-circle is a right angle.

∴ ∠BAC = 90°

In ∆ABC,

⇒ ∠ACB + ∠BAC + ∠ABC = 180° [Angle sum property of triangle]

$\Rightarrow x + 90\degree + 32\degree = 180\degree \\[1em] \Rightarrow x = 180\degree - 122\degree \\[1em] \Rightarrow x = 58\degree.$

We know that,

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

$\Rightarrow y = \angle ACB \\[1em] \Rightarrow y = x \\[1em] \Rightarrow y = 58\degree.$

We know that,

Angles in the same segment of a circle are equal.

Here, ∠ADB and ∠ACB are the angles in the same segment subtended by the chord AB.

$\Rightarrow z = \angle ADB = \angle ACB \\[1em] \Rightarrow z = x \\[1em] \Rightarrow z = 58°.$

In ∆ADB,

⇒ AD = AB [Given]

⇒ ∠ABD = ∠ADB [Angles opposite to equal sides are equal]

⇒ $\angle ABD = z = 58°$

By angle sum property of triangle,

$\Rightarrow v + \angle ADB + \angle ABD = 180° \\[1em] \Rightarrow v + 58° + 58° = 180° \\[1em] \Rightarrow v = 180° - 116° \\[1em] \Rightarrow v = 64°.$

In cyclic quadrilateral ADEB,

We know that,

The sum of the opposite angles of a cyclic quadrilateral is 180°.

$\Rightarrow v + w = 180° \\[1em] \Rightarrow 64° + w = 180° \\[1em] \Rightarrow w = 180° - 64° \\[1em] \Rightarrow w = 116°.$

Hence, $\bm{x = 58°}, \bm{y = 58°}, \bm{z = 58°}, \bm{v = 64°}$ and $\bm{w} = 116°$.

Using Remainder and Factor Theorem, factorise the following polynomial :

$2x^3 + 5x^2 - 28x - 15$

Answer

Given,

$f(x) = 2x^3 + 5x^2 - 28x - 15$

Substituting $x = 3$ in $f(x)$, we get :

$\Rightarrow f(3) = 2(3)^3 + 5(3)^2 - 28(3) - 15 \\[1em] \Rightarrow f(3) = 2 \times 27 + 5 \times 9 - 84 - 15 \\[1em] \Rightarrow f(3) = 54 + 45 - 84 - 15 \\[1em] \Rightarrow f(3) = 99 - 99 \\[1em] \Rightarrow f(3) = 0$

Since, the remainder is zero.

∴ $(x - 3)$ is a factor of $f(x)$.

Dividing $2x^3 + 5x^2 - 28x - 15$ by $(x - 3)$, we get :

$\begin{array}{l} \phantom{x - 3\big)\quad } 2x^2 + 11x + 5 \\ x - 3\overline{\smash{\big)}\quad 2x^3 + 5x^2 - 28x - 15} \\ \phantom{+++(}\underline{\underset{-}{+}2x^3 \underset{+}{-}6x^2} \\ \phantom{x - 3\big)\quad }\phantom{2x^3 +} 11x^2 - 28x \\ \phantom{+++( }\phantom{2x^3 +} \underline{\underset{-}{+}11x^2 \underset{+}{-}33x} \\ \phantom{++++( }\phantom{2x^3 + 5x^2 -} 5x - 15 \\ \phantom{++++) }\phantom{2x^3 + 5x^2 -} \underline{\underset{-}{+}5x \underset{+}{-}15} \\ \phantom{x - 3\big)\quad }\phantom{2x^3 + 5x^2 - 28x} \times \end{array}$

∴ $2x^3 + 5x^2 - 28x - 15 = (x - 3)(2x^2 + 11x + 5)$

Factorising $2x^2 + 11x + 5$, we get :

$= (x - 3)[2x^2 + 10x + x + 5] \\[1em] = (x - 3)[2x(x + 5) + 1(x + 5)] \\[1em] = (x - 3)(x + 5)(2x + 1).$

Hence, $\bm{{2x^3} + 5x^2 − 28x − 15 = (x − 3)(x + 5)(2x + 1)}$.

In the given diagram, AD ∥ GE ∥ BC, DE = 18 cm, EC = 3 cm, AD = 35 cm. Find :

(a) AF : FC and AG : GB

(b) length of EF

(c) area (trapezium ADEF) : area (∆ EFC)

Answer

Given,

AD ∥ GE ∥ BC, DE = 18 cm, EC = 3 cm and AD = 35 cm

(a) In ∆ADC,

Since, GE ∥ AD, we have FE ∥ AD.

By Basic Proportionality Theorem (B.P.T.),

$\Rightarrow \dfrac{AF}{FC} = \dfrac{DE}{EC} \\[1em] \Rightarrow \dfrac{AF}{FC} = \dfrac{18}{3} \\[1em] \Rightarrow \dfrac{AF}{FC} = \dfrac{6}{1}.$

∴ AF : FC = 6 : 1

In ∆ABC,

Since, GE ∥ BC, we have GF ∥ BC.

By Basic Proportionality Theorem (B.P.T.),

$\dfrac{AG}{GB} = \dfrac{AF}{FC} = \dfrac{6}{1}$

∴ AG : GB = 6 : 1.

Hence, AF : FC = 6 : 1 and AG : GB = 6 : 1.

(b) In ∆ADC and ∆FEC,

⇒ ∠ADC = ∠FEC [Corresponding angles as AD ∥ FE]

⇒ ∠ACD = ∠FCE [Common angle]

∴ ∆ADC ~ ∆FEC (By A.A. axiom)

We know that,

Corresponding sides of similar triangles are proportional.

$\therefore \dfrac{EF}{AD} = \dfrac{EC}{DC} \\[1em] \Rightarrow \dfrac{EF}{35} = \dfrac{3}{18 + 3} \\[1em] \Rightarrow \dfrac{EF}{35} = \dfrac{3}{21} \\[1em] \Rightarrow EF = \dfrac{35 \times 3}{21} \\[1em] \Rightarrow EF = 5 \text{ cm}.$

Hence, length of EF = 5 cm.

(c) We know that,

The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

⇒ $\dfrac{\text{Area of ∆ADC}}{\text{Area of ∆EFC}} = \dfrac{AD^2}{EF^2}$

From the figure,

Area of trapezium ADEF = Area of ∆ADC − Area of ∆EFC

Dividing each side by Area of ∆EFC,

$\therefore \dfrac{\text{Area of trapezium ADEF}}{\text{Area of ∆EFC}} = \dfrac{\text{Area of ∆ADC} - \text{Area of ∆EFC}}{\text{Area of ∆EFC}} \\[1em] = \dfrac{\text{Area of ∆ADC}}{\text{Area of ∆EFC}} - \dfrac{\text{Area of ∆EFC}}{\text{Area of ∆EFC}}\\[1em] = \dfrac{AD^2}{EF^2} - 1 \\[1em] = \dfrac{35^2}{5^2} - 1 \\[1em] = \dfrac{1225}{25} - 1 \\[1em] = \dfrac{1225 - 25}{25} \\[1em] = \dfrac{1200}{25} \\[1em] = \dfrac{48}{1}.$

Hence, area (trapezium ADEF) : area (∆ EFC) = 48 : 1.

Use ruler and compass for this question :

(a) Draw line segment AB of length 6 cm.

(b) Construct the locus of points which are equidistant from A and B. Mark a point on the locus which is 4.5 cm from A. Name this point as O.

(c) Construct the locus of all points which are 4.5 cm from O.

(d) Mark the point of intersection of loci (b) and (c) as P, which is on the same side of AB as O.

(e) Construct the locus of points which are equidistant from AP and AB.

Answer

Steps of construction :

1. Draw a line segment AB of length 6 cm.

2. Draw XY, the perpendicular bisector of AB. XY is the locus of points which are equidistant from A and B.

3. With A as centre and radius 4.5 cm, draw an arc cutting XY at the point O. Thus, O is the point on the locus which is 4.5 cm from A.

4. With O as centre and radius 4.5 cm, draw a circle. This circle is the locus of all points which are 4.5 cm from O.

5. Mark P, the point of intersection of the perpendicular bisector XY and the circle, which lies on the same side of AB as O.

6. Join AP.

7. Draw AZ, the bisector of ∠PAB. AZ is the locus of points which are equidistant from AP and AB.

A bag contains identical cards numbered 1 to 50. A card is selected randomly. Find the probability that the card selected, is :

(a) divisible by 6.

(b) not divisible by 2 and 3.

Answer

Given,

The bag contains identical cards numbered 1 to 50.

Total number of possible outcomes = 50

(a) Favourable outcomes = Getting a card with a number divisible by 6.

Numbers divisible by 6 = {6, 12, 18, 24, 30, 36, 42, 48}

Number of favourable outcomes = 8

By formula,

Probability = $\dfrac{\text{No. of favourable outcomes}}{\text{Total number of outcomes}}$

P(card divisible by 6) = $\dfrac{8}{50} = \dfrac{4}{25}$.

Hence, probability that the card selected is divisible by 6 = $\mathbf{\dfrac{4}{25}}$.

(b) We know that,

A number which is divisible by both 2 and 3 is divisible by 6.

∴ P(card divisible by 2 and 3) = P(card divisible by 6) = $\dfrac{4}{25}$

By formula,

P(card not divisible by 2 and 3) = 1 − P(card divisible by 2 and 3)

$= 1 - \dfrac{4}{25} \\[1em] = \dfrac{25 - 4}{25}\\[1em] = \dfrac{21}{25}.$

Hence, probability that the card selected is not divisible by 2 and 3 = $\mathbf{\dfrac{21}{25}}$.

In the given figure, the parallelogram ABCD circumscribes a circle, touching the circle at P, Q, R and S. Prove that :

(a) AB = BC

(b) diagonals AC and BD bisect each other at right angles.

Answer

Join AC and BD.

(a) We know that,

If two tangents are drawn to a circle from an exterior point, the tangents are equal in length.

∴ AP = AS, BP = BQ, CR = CQ and DR = DS

Adding the equations, we get :

⇒ AP + BP + CR + DR = AS + BQ + CQ + DS

⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

⇒ AB + CD = AD + BC ......(1)

We know that,

Opposite sides of a parallelogram are equal.

∴ AB = CD and AD = BC

Substituting the above values in equation (1), we get :

⇒ AB + AB = BC + BC

⇒ 2AB = 2BC

⇒ AB = BC.

Hence, proved that AB = BC.

(b) From part (a),

⇒ AB = BC

Also, AB = CD and AD = BC [Opposite sides of a parallelogram are equal]

∴ AB = BC = CD = AD

Since, all the four sides of the parallelogram ABCD are equal, ABCD is a rhombus.

We know that,

The diagonals of a rhombus bisect each other at right angles.

Hence, proved that the diagonals AC and BD bisect each other at right angles.

Given, $\dfrac{a + 2b + x}{a - 2b + x} = \dfrac{(a + b)^2}{(a - b)^2}, a \neq 0$

Using properties of proportion, find the value of $x$.

Answer

Given,

$\dfrac{a + 2b + x}{a - 2b + x} = \dfrac{(a + b)^2}{(a - b)^2}$

Applying componendo and dividendo, we get :

$\Rightarrow \dfrac{(a + 2b + x) + (a - 2b + x)}{(a + 2b + x) - (a - 2b + x)} = \dfrac{(a + b)^2 + (a - b)^2}{(a + b)^2 - (a - b)^2} \\[1em] \Rightarrow \dfrac{2a + 2x}{4b} = \dfrac{2(a^2 + b^2)}{4ab} \\[1em] \Rightarrow \dfrac{2(a + x)}{4b} = \dfrac{2(a^2 + b^2)}{4ab} \\[1em] \Rightarrow a + x = \dfrac{a^2 + b^2}{a} \\[1em] \Rightarrow x = \dfrac{a^2 + b^2}{a} - a \\[1em] \Rightarrow x = \dfrac{a^2 + b^2 - a^2}{a} \\[1em] \Rightarrow x = \dfrac{b^2}{a}.$

Hence, $\bm{x} = \mathbf{\dfrac{b^2}{a}}$.

The mean of the following distribution is 27 and the sum of all the frequencies is 50. Find the missing frequencies x and y.

Answer

Given,

Mean = 27 and sum of all the frequencies = 50

Given,

Sum of all the frequencies = 50

⇒ 33 + x + y = 50

⇒ x + y = 17 ......(1)

By formula,

Mean = $\dfrac{∑f_ix_i}{∑f_i}$

Substituting values, we get :

⇒ 27 = $\dfrac{(795 + 15x + 45y)}{50}$

⇒ 795 + 15x + 45y = 1350

⇒ 15x + 45y = 555

⇒ x + 3y = 37 ......(2)

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

⇒ (x + 3y) − (x + y) = 37 − 17

⇒ 2y = 20

⇒ y = 10

Substituting y = 10 in equation (1), we get :

⇒ x + 10 = 17

⇒ x = 7.

Hence, the missing frequencies are x = 7 and y = 10.

In the given diagram, the area of the shaded region is 57 m². Form an equation in x and solve it to find the possible value of x.

Answer

Given,

Area of the shaded region = 57 m²

From the figure,

Length of the outer rectangle = 12 m

Breadth of the outer rectangle = 10 m

Area of the outer rectangle = 12 × 10 = 120 m²

Length of the inner (unshaded) rectangle = (12 − x) m

Breadth of the inner (unshaded) rectangle = (10 − x) m

Area of the inner rectangle = (12 − x)(10 − x) m²

From the figure,

Area of shaded region = Area of outer rectangle − Area of inner rectangle

⇒ 12 × 10 − (12 − x)(10 − x) = 57

⇒ 120 − (120 − 12x − 10x + x²) = 57

⇒ 120 − 120 + 22x − x² = 57

⇒ 22x − x² = 57

⇒ x² − 22x + 57 = 0

⇒ x² − 19x − 3x + 57 = 0

⇒ x(x − 19) − 3(x − 19) = 0

⇒ (x − 19)(x − 3) = 0

[Using zero product rule]

⇒ (x − 19) = 0 or (x − 3) = 0

⇒ x = 19 or x = 3

Since, the breadth of the rectangle is 10 m, the width x cannot be 19 m.

∴ x = 3 m.

Hence, the possible value of x = 3 m.

Two beakers (drinking glasses) are cylindrical in shape having diameter and height 7 cm and 10 cm respectively. Glass A has hemispherical raised bottom, and glass B has a conical raised bottom of height 3.5 cm.

Determine which glass has a greater capacity and calculate the difference in their capacities. Give your answer correct to 2 decimal places. $\left(\pi = \dfrac{22}{7}\right)$

Answer

Given,

Diameter of each cylindrical glass = 7 cm

Radius $(r) = \dfrac{7}{2}$ = 3.5 cm

Height of each glass ($h$) = 10 cm

By formula,

$\text{Volume of cylinder} = \pi r^2h \\[1em] = \dfrac{22}{7} \times 3.5 \times 3.5 \times 10 \\[1em] = 385 \text{ cm}^3$

For glass A, the raised bottom is a hemisphere of radius 3.5 cm.

$\text{Volume of hemisphere} = \dfrac{2}{3}\pi r^3 \\[1em] = \dfrac{2}{3} \times \dfrac{22}{7} \times 3.5 \times 3.5 \times 3.5 \\[1em] = 89.83 \text{ cm}^3$

Capacity of glass A = Volume of cylinder − Volume of hemisphere

= 385 − 89.83

= 295.17 cm³

For glass B, the raised bottom is a cone of radius 3.5 cm and height ($h'$) = 3.5 cm.

$\text{Volume of cone} = \dfrac{1}{3}\pi r^2 h' \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 3.5 \times 3.5 \times 3.5 \\[1em] = 44.92 \text{ cm}^3$

Capacity of glass B = Volume of cylinder − Volume of cone

= 385 − 44.92

= 340.08 cm³

Since, 340.08 cm³ > 295.17 cm³, glass B has the greater capacity.

Difference in their capacities = Capacity of glass B − Capacity of glass A

$= \Big(\pi r^2h - \dfrac{1}{3}\pi r^2h'\Big) - \Big(\pi r^2h - \dfrac{2}{3}\pi r^3\Big) \\[1em] = \dfrac{2}{3}\pi r^3 - \dfrac{1}{3}\pi r^2h' \\[1em] = \dfrac{2}{3} \times \dfrac{22}{7} \times 3.5^3 - \dfrac{1}{3} \times \dfrac{22}{7} \times 3.5^2 \times 3.5 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 3.5^3 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 42.875 \\[1em] = 44.9166.... \\[1em] = 44.92 \text{ cm}^3.$

Hence, glass B has the greater capacity and the difference in their capacities = 44.92 cm³.

The distribution given below represents marks obtained by 200 students in a Mathematics Examination with maximum marks 80.

Using graph paper draw an ogive for the given distribution. Along one axis take 2 cm equal to 10 marks and along the other axis take 1 cm equal to 10 students.

Use your graph to find the following :

(a) Median

(b) Number of students who scored above 75 marks

(c) If the pass percentage is 35%, find the number of students who scored 35% and below.

Answer

Steps :

1. Take 2 cm = 10 marks on the x-axis.

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

3. Plot the points (0, 0), (10, 8), (20, 20), (30, 36), (40, 60), (50, 89), (60, 127), (70, 169) and (80, 200).

4. Join the points by a free hand curve.

(a) Here, $n$ (no. of students) = 200, which is even.

Median = $\dfrac{n}{2} = \dfrac{200}{2}$ = 100^th term.

Through point A = 100 on the y-axis draw a horizontal line parallel to the x-axis touching the graph at point B, through B draw a vertical line parallel to the y-axis touching the x-axis at point C = 53.

Hence, median = 53.

(b) Through point D = 75 on the x-axis draw a vertical line parallel to the y-axis touching the graph at point E, through E draw a horizontal line parallel to the x-axis touching the y-axis at point F = 186.

∴ 186 students scored 75 marks and below.

∴ Number of students who scored above 75 marks = 200 − 186 = 14.

Hence, 14 students scored above 75 marks.

(c) Maximum marks = 80

Pass marks = 35% of 80 = $\dfrac{35}{100} \times 80$ = 28 marks

Through point G = 28 on the x-axis draw a vertical line parallel to the y-axis touching the graph at point H, through H draw a horizontal line parallel to the x-axis touching the y-axis at point I = 33.

Hence, 33 students scored 35% and below.

Draw the necessary diagram for this question.

A man on the top of a lighthouse observes the angle of depression of two ships on the opposite sides of the lighthouse as 32° and 58° respectively. If the height of the lighthouse is 120 m, find the distance between the two ships. Give your answer correct to the nearest meter.

Answer

Let AD be the lighthouse and B and C be the two ships on the opposite sides of the lighthouse.

Given,

Height of the lighthouse (AD) = 120 m

The angles of depression of the two ships are 32° and 58°.

From figure,

⇒ ∠FAB = ∠ABD = 32° (alternate angles are equal)

⇒ ∠EAC = ∠ACD = 58° (alternate angles are equal)

In right angled triangle ABD,

$\Rightarrow \tan 32° = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \tan 32° = \dfrac{AD}{BD} \\[1em] \Rightarrow BD = \dfrac{AD}{\tan 32°} \\[1em] \Rightarrow BD = \dfrac{120}{0.6249} \\[1em] \Rightarrow BD = 192.03 \text{ m}.$

In right angled triangle ACD,

$\Rightarrow \tan 58° = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \tan 58° = \dfrac{AD}{CD} \\[1em] \Rightarrow CD = \dfrac{AD}{\tan 58°} \\[1em] \Rightarrow CD = \dfrac{120}{1.6003} \\[1em] \Rightarrow CD = 74.99 \text{ m}.$

From the figure,

Distance between the two ships (BC) = BD + CD

= 192.03 + 74.99

= 267.02 m ≈ 267 m.

Hence, the distance between the two ships = 267 m.

A finite Geometric Proportion (G.P.) series contains 10 terms. The sum of the first three terms is 7 and that of the last three terms is 896. Find the first term and the common ratio.

Answer

Let the first term of the G.P. be $a$ and the common ratio be $r$.

Since, the G.P. contains 10 terms, the terms are $a, ar, ar^2, ......, ar^9$.

Given,

Sum of the first three terms = 7

$\Rightarrow a + ar + ar^2 = 7 \\[1em] \Rightarrow a(1 + r + r^2) = 7 \text{ ......(1)}$

Sum of the last three terms = 896

$\Rightarrow ar^7 + ar^8 + ar^9 = 896 \\[1em] \Rightarrow ar^7(1 + r + r^2) = 896 \text{ ......(2)}$

Dividing equation (2) by equation (1), we get :

$\Rightarrow \dfrac{ar^7(1 + r + r^2)}{a(1 + r + r^2)} = \dfrac{896}{7} \\[1em] \Rightarrow r^7 = 128 \\[1em] \Rightarrow r^7 = 2^7 \\[1em] \Rightarrow r = 2.$

Substituting $r = 2$ in equation (1), we get :

$\Rightarrow a(1 + 2 + 2^2) = 7 \\[1em] \Rightarrow a(1 + 2 + 4) = 7 \\[1em] \Rightarrow 7a = 7 \\[1em] \Rightarrow a = 1.$

Hence, first term = 1 and common ratio = 2.

In a circle with centre O, PA and PB are tangents drawn from an external point P. If PA = 9 cm and ∠APB = 60°, find the :

(a) length of PB.

(b) radius of the circle.

Answer

(a) We know that,

The lengths of two tangents drawn from an external point to a circle are equal.

∴ PB = PA

⇒ PB = 9 cm.

Hence, length of PB = 9 cm.

(b) Join OA, OB and OP.

We know that,

The tangents drawn from an external point are equally inclined to the line joining the centre of the circle and that point.

∴ OP bisects ∠APB.

⇒ ∠OPA = $\dfrac{1}{2} \times ∠APB = \dfrac{1}{2} \times 60° = 30°$

We know that,

Radius and tangent at the point of contact are perpendicular to each other.

∴ ∠OAP = 90°

In right angled triangle OAP,

$\Rightarrow \tan 30° = \dfrac{OA}{PA} \\[1em] \Rightarrow OA = PA \times \tan 30° \\[1em] \Rightarrow OA = 9 \times \dfrac{1}{\sqrt{3}} \\[1em] \Rightarrow OA = \dfrac{9 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\[1em] \Rightarrow OA = \dfrac{9\sqrt{3}}{3} \\[1em] \Rightarrow OA = 3\sqrt{3} \\[1em] \Rightarrow OA = 3 \times 1.7321 \\[1em] \Rightarrow OA = 5.1963 \\[1em] \Rightarrow OA = 5.2 \text{ cm}.$

Hence, radius of the circle = 5.2 cm.

The table given below shows a record of the weight in kilogram of 100 students of a school.

Draw a histogram and find the modal weight.

[Take 2 cm = 5 kg along one axis and 2 cm = 5 students along the other axis]

Answer

Steps :

1. Take 2 cm = 5 kg along the x-axis.

2. Take 2 cm = 5 students along the y-axis.

3. Draw the rectangles (bars) for the given class intervals, taking the corresponding frequencies as their heights.

4. The class interval 50 - 55 has the highest frequency 25, so 50 - 55 is the modal class.

5. In the highest rectangle, draw two lines AD and BC from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let P be the point of intersection of AD and BC.

6. From the point of intersection P, draw a perpendicular to the x-axis. The point R = 51.5, where this perpendicular meets the x-axis gives the mode.

From the histogram,

Mode = 51.5 kg.

Hence, modal weight = 51.5 kg.