Oswal 61 Sample Question Papers ICSE Class 10 Maths Solutions

Section-A

Answer 1.

(i) (b) ₹ 600

Explanation :

$$\text{MV = P × n + P ×}\\\frac{n(n+1)}{2×12}×\frac{r}{100}\\\Rarr\space\\\text{7,668 = P}\bigg(12 + \frac{12×(12+1)}{2×12}×\frac{12}{100}\bigg)$$

[∵ n = 1year = 12 months]

$$\Rarr\space 7,668 = \text{P}\bigg(12 + \frac{78}{100}\bigg)\\\Rarr\space\text{7,668 = P}\bigg(\frac{1,278}{100}\bigg)\\\Rarr\space\text{P = 600}$$

$$\textbf{(ii)\space}\textbf{(b)}\begin{vmatrix}\textbf{7} &\textbf{11}\\\textbf{7} &\textbf{3}\end{vmatrix}$$

Explanation :

Given,

$$\text{A} =\begin{bmatrix}\normalsize-2 &3\\4 &1\end{bmatrix}\space\text{and B =}\begin{bmatrix}1 &2\\3 &5\end{bmatrix}\\\text{AB} = \begin{bmatrix}\normalsize-2 &3\\4 &1\end{bmatrix}\begin{bmatrix}1 &2\\3 &5\end{bmatrix}\\=\\\begin{bmatrix}-2×1 +3×3 &-2×2+3×5\\4×1 +1×3 &4×2+1×5\end{bmatrix}\\=\begin{bmatrix}7 &11\\7 &13\end{bmatrix}$$

(iii) (c) 10 : 21

Explanation :

Compounded ratio of 2 : 3 and 5 : 7

$$=\frac{2}{3}×\frac{5}{7} =\frac{10}{21}$$

(iv) (c) 6

Explanation :

Given : (x – 1) is a factor of (x³ – kx² + 11x – 6)

Since, (x – 1) is a factor, so, using Remainder’s theorem,

Putting the value 1 in place of x, the remainder equal to zero.

1 – k + 11 – 6 = 0

⇒ k = 6

(v) (b) –2

Explanation :

$$\text{Given: x =}\frac{1}{2}\space\text{as root of} \\\text{the equation x}^{2} -mx-\frac{5}{4} = 0.\\\therefore\space\bigg(\frac{1}{2}\bigg)^{2}-m\bigg(\frac{1}{2}\bigg)-\frac{5}{4}=0\\\Rarr\space\frac{1}{4}-\frac{m}{2}-\frac{5}{4}= 0\\\Rarr\space\text{m = –2}$$

(vi) (b) {x : x ∈ R, –2 < x < 3}

(vii) (c) 4.5 cm

Explanation :

In similar triangles, the corresponding sides are in proportion.

Given: ΔABC ~ ΔDEF

$$\therefore\space\frac{\text{AB}}{\text{DE}} =\frac{\text{BC}}{\text{EF}} =\frac{\text{AC}}{\text{DF}}\\\Rarr\space\frac{\text{AB}}{\text{DE}} = \frac{\text{AC}}{\text{DF}}\\\Rarr\space\frac{x -0.5}{9} =\frac{1.5x}{3x}\\\Rarr\space x -0.5 = 4.5\\\Rarr\space x = 5\space\text{cm}$$

So, length of AB = (x – 0.5)

= 5 – 0.5

= 4.5 cm

(viii) (d) 990 cm²

Explanation :

$$\text{Radius (r) =}\frac{1}{2}×\text{diameter}\\=\frac{21}{2}\space\text{cm}$$

Height (h) = 15 cm

∴ Curved surface area = 2πrh

$$= 2×\frac{22}{7}×\frac{21}{2}×15$$

= 990 cm².

(ix) (c) (1, – 1)

Explanation :

If A′(x, y) be the reflection of point A(5, – 3) in the point P(3, – 2), then P will be mid-point of AA′.

$$x =\frac{x_1 + x_2}{2}\space\text{and y =}\frac{y_{1} + y_{2}}{2}\\3 =\frac{x+5}{2}\space\text{and}-2 =\frac{y-3}{2}$$

⇒ x = 1 and y = – 1

∴ Reflection point will be (1, – 1).

(x) (b) 120°

Explanation :

reflex ∠AOC = 360° – 120° = 240°

∵ since, the angle subtended at the centre by an arc of a circle is double the angle which this arc subtends at any remaining part of the circumference.

∴ reflex ∠AOC = 2 × ∠ABC

$$\Rarr\space\angle\text{ABC} =\frac{1}{2}\text{reflex}\space\angle\text{AOC}\\=\frac{1}{2}×240\degree = 120\degree$$

(xi) (b) 1

Explanation :

Given, m = sec A + tan A

and, n = sec A – tan A

By multiplying to each other, we get

mn = (sec A + tan A) (sec A – tan A)

mn = sec²A – tan²A = 1 (using identity)

(xii) (b) 20-30

Explanation :

Class	Frequency	Cumulative frequency (less than)
0-10	9	9
10-20	3	12
20-30	12	24
30-40	4	28
40-50	12	40
Total (N)	40

N = 40

$$\therefore\space\frac{\text{N}}{2} =\frac{\text{40}}{2} = 20 $$

∵ 20 belongs to the class 20-30.

Hence, the median class is 20-30.

$$\text{(xiii) (d)\space}\frac{35}{36}$$

Explanation :

The possible outcomes are {(1, 1), (1, 2), (1, 3), … (6, 6)}

Total No. of possible outcomes = 36

∵ The maximum number as sum less than 12 is 6 + 5 = 11

∴ No. of favourable outcomes = 35

$$\text{Hence, P(getting sum less than 12)}\\=\frac{35}{36}.$$

(xiv) (d) 3 : 4

Explanation :

Let the X-axis divides the given line in the ratio m : 1. Then the co-ordinates of the point of division are:

$$x =\frac{2m-3}{m+1}\space\text{and y} =\frac{-8m +6}{m+1}$$

Since, this is a point on the X-axis. So,

$$\frac{-8m +6}{m+1} = 0\\\Rarr\space\text{– 8m + 6 = 0 or m =}\frac{3}{4}$$

Hence, the X-axis divides the line internally in the ratio 3 : 4.

(xv) (a) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Explanation :

We have, t₂ + t₇ = 30

$$\Rarr\space\text{(a + d) + (a + 6d) = 30}\\\Rarr\space\text{2a + 7d = 30}\qquad\text{…(i)}$$

Also, t₁₅ = 2t₈ – 1

$$\Rarr\space a+14d = 2(a + 7d)-1\\\Rarr\space a=1$$

Putting a = 1 in eq. (i), we get

d = 4

∴ The A.P. is 1, 1 + 4, 1 + 2(4), … i.e., 1, 5, 9, …

Answer 2.

(i) (a) Since, number of months (n) = 24 and rate of interest (r) = 6%

$$\text{I = P}×\frac{n(n+1)}{2}×\frac{r}{100}\\\Rarr\space 1200 =\text{P}×\frac{24(24+1)}{2×12}×\frac{6}{100}\\\Rarr\space\text{P =}\frac{1,200×24×100}{6×24×25}$$

= ₹ 800

∴ Monthly instalment = ₹ 800 Ans.

(b) Sum deposited = ₹ 800 × 24

= ₹ 19200

Amount on maturity = ₹ 19,200 + ₹ 1,200

= ₹20,400 Ans.

(ii) Given, a, b, c, d are in continued proportion.

$$\therefore\space\frac{a}{b} =\frac{b}{c}= \frac{c}{d}=k\space\text{(say)}\\\Rarr\space\text{c = kd, b = kc = k}^{2}d,\\\text{a =kb = k}^{3}\text{d}\\\text{Now,\space L.H.S = a:d}=\frac{a}{d} =\frac{k^{3}d}{d} = k^{3} $$

R.H.S. = Triplicate ratio of (a – b) : (b – c)

$$=\frac{(a-b)^{3}}{(b-c)^{3}} =\bigg(\frac{k^{3}d -k^{2}d}{k^{2}d - kd}\bigg)^{3}\\=\bigg(\frac{k^{2}d(k-1)}{kd(k-1)}\bigg)^{3} = k^{3}$$

Thus, L.H.S. = R.H.S. Hence Proved.

(iii) (i) L.H.S.

$$=\sqrt{\frac{\text{1 - \text{cos}}\space\theta}{1 + \text{cos}\space\theta}×\frac{\text{1 - \text{cos}}\space\theta}{\text{1 - cos}\theta}}\\=\sqrt{\frac{\text{(1 - cos}\space \theta)^{2}}{\text{1 - cos}^{2}\theta}}\\=\frac{\text{1 - cos}\theta}{\sqrt{1 - \text{cos}^{2}\theta}}\\=\frac{1 -\text{cos}\space\theta}{\sqrt{\text{sin}^{2}\theta}}$$

$$=\frac{\text{1-cos}\space\theta}{\text{sin}\space\theta}\\=\frac{1}{\text{sin}\space\theta} -\frac{\text{cos}\space\theta}{\text{sin}\space\theta}$$

= cosec θ – cot θ

= R.H.S. Hence Proved.

$$\text{(ii)\space L.H.S =}\\\sqrt{\frac{\text{1 + sin}\space\theta}{\text{1 - sin}\space\theta}×\frac{\text{1 + sin}\space\theta}{\text{1 + sin}\space\theta}}\\=\sqrt{\frac{(1 + sin\space\theta)^{2}}{\text{1 - sin}^{2}\theta}}\\=\sqrt{\frac{(1 + sin\space\theta)^{2}}{\text{cos}^{2}\theta}}\\=\frac{\text{1 + sin}\space\theta}{\text{cos}\space\theta}\\=\frac{1}{\text{cos}\space\theta} + \frac{\text{sin\space}\theta}{\text{cos}\space\theta}$$

= sec θ + tan θ = R.H.S. Hence Proved.

Answer 3.

(i) Given, Surface area of sphere = 616 cm²

$$\Rarr\space\text{4}\pi r^{2} = 616\\\Rarr\space 4×\frac{22}{7}×r^{2} = 616\\\Rarr\space r^{2} =\frac{616×7}{4×22} = 49\\\Rarr\space r = 7\space\text{cm.}\\\therefore\space\text{Volume(V}_{1}) =\frac{4}{3}\pi r^{3}\\=\frac{4}{3}×\frac{22}{7}×7^{3}\\=\frac{4}{3}×\frac{22}{7}×343\space\text{cm}^{3}$$

Diameter of small sphere = 3.5 cm

$$\therefore\space\text{Its radius (r}_1) =\frac{3.5}{2}\text{cm}\\\therefore\space\text{Its volume (V}_2) =\frac{4}{3}\pi r^{3}_{1}\\=\frac{4}{3}×\frac{22}{7}×\bigg(\frac{3.5}{2}\bigg)^{3}\\=\frac{4}{3}×\frac{22}{7}×\frac{343}{8×8}\text{cm}^{3}\\\therefore\space\text{No. of smaller spheres recast}\\ =\frac{\text{V}_{1}}{\text{V}_{2}}\\=\frac{\frac{4}{3}×\frac{22}{7}×343}{\frac{4}{3}×\frac{22}{7}×\frac{343}{8×8}} = 64.\space\textbf{Ans.}$$

(ii) (i) Co-ordinates of the points A, B and C are (1, 3), (– 3, – 2) and (3, 0) respectively.

$$\text{(ii) Slope of AB =}\frac{\text{-2-3}}{-3-1}\\=\frac{5}{4}.$$

(iii) Line through C (3, 0) and parallel to AB

$$\therefore\space\text{Slope =}\frac{5}{4}.$$

∴ Equation to the line is

y – y₁ = m(x – x₁)

$$\Rarr\space\text{y - 0} =\frac{5}{4}(x-3)\\\Rarr\space\text{4y = 5x – 15}$$

This line intersects Y-axis at D.

∴ On solving

4y = 5x – 15

and x = 0 (Equation to Y-axis)

we get, 4y = – 15

$$\Rarr\space y =-\frac{15}{4}\\\therefore\space\text{Co-ordinates of point D are}\\\bigg(0,\frac{\normalsize-15}{4}\bigg).\space\textbf{Ans.}$$

(iii) (a) Coordinates of P′ = (3, – 4)

Coordinates of O′ = (6, 0)

(b) PP′ = 8 units, OO′ = 6 units

$$\textbf{(c)\space}\text{OP}= \sqrt{3^{2} + 4^{2}}\\=\sqrt{25} = \text{5 units}$$

∴ Perimeter of POP′O′ = 4 × 5 = 20 units

(d) Special name of POP′O′ is rhombus.

Scale : At X-axis : 1 cm = 1 unit

At Y axis : 1 cm = 1unit

Section-B

Answer 4.

(i) According to the question,

GST on ticket of ₹ 90 = 18% of 90

$$=\frac{18}{100}×90$$

= ₹ 16.20

GST on ticket of ₹ 140 = 28% of 140

$$=\frac{28}{100}×140$$

= ₹ 39.20

Difference between both GST = 39.20 – 16.20

= ₹23.00 Ans.

(ii) Given equation is, x² – 3 (x + 3) = 0

⇒ x² – 3x – 9 = 0

On comparing the equation with ax² + bx + c = 0, we get

∴ a = 1, b = – 3, c = – 9

b² – 4ac = (– 3)² – 4 (1) (– 9)

= 9 + 36

= 45

$$x = \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\\=\frac{-(-3)\pm\sqrt{45}}{2×1}\\=\frac{3\pm3\sqrt{5}}{2}\\x =\frac{3 + 3×2.236}{2}\space\text{and}\space\\x =\frac{3-3×2.236}{2}\\x =\frac{3 + 6.708}{2}\space\text{and}\\ x =\frac{3 -6.708}{2}$$

$$x =\frac{9.708}{2}\space\text{and}\space x =-\frac{3.708}{2}$$

x = 4·854 and x = – 1·854

∴ x = 4·9 and x = – 1·9 Ans.

(iii)

Number of match sticks (x_i)	Number of boxes (f_i)	f_i x_i
35	6	210
36	10	360
37	18	666
38	25	950
39	21	819
40	12	480
41	8	328
	Σf_i = 100	Σf_i x_i = 3813

$$\textbf{(i)}\space\text{Mean = }\frac{Sf_ix_i}{Sf_i} =\frac{3813}{100}$$

= 38·13 ≈ 38·1 Ans.

(ii) Now, the number of extra sticks to be added.

= 39 × 100 – 38·13 × 100

= 3900 – 3813 = 87 Ans.

Answer 5.

$$\text{(i)\space Given,\space}\begin{bmatrix}4 &2\\\normalsize-1 &1\end{bmatrix}\text{M = 6}\text{I}\\\Rarr\space\begin{bmatrix}4 &2\\\normalsize-1 &1\end{bmatrix}\text{M = 6}\begin{bmatrix}1 &0\\0 &1\end{bmatrix}\\\Rarr\space\begin{bmatrix}4 &2\\\normalsize-1 &1\end{bmatrix}\text{M} = \begin{bmatrix}6 &0\\0 &6\end{bmatrix}\\\text{...(i)}$$

(i) (2 × 2) (m × n) = (2 × 2) → Order of matrix, M = 2 × 2. Ans.

$$\textbf{(ii)\space}\text{Let,\space M =}\begin{bmatrix}a &b\\c &d\end{bmatrix}\\\therefore\space\begin{bmatrix}4 &2\\\normalsize-1 &1\end{bmatrix}\begin{bmatrix}a &b\\c &d\end{bmatrix} =\begin{bmatrix}6 &0\\0 &6\end{bmatrix}\\\text{[using (i)]}\\\Rarr\space\begin{bmatrix}4a +2c &4b +2d\\-a+c &-b+d\end{bmatrix} =\begin{bmatrix}6 &0\\0 &6\end{bmatrix}$$

∴ 4a + 2c = 6 ...(ii)

– a + c = 0 ...(iii) × 4

Solving equations (ii) and (iii),

4a + 2c = 6

– 4a + 4c = 0

6c = 6

$$\Rarr\space\text{c = 1}$$

From equation (iii),

– a + 1 = 0

⇒ a = 1

and 4b + 2d = 0 ...(iv)

⇒ – b + d = 6 ...(v) × 4

Solving equations (iv) and (v),

4b + 2d = 0

– 4b + 4d = 24

6d = 24

⇒ d = 4

From equation (iv),

– b + 4 = 6

⇒ – b = 2

⇒ b = – 2

$$\therefore\space\text{M =}\begin{bmatrix}1 &\normalsize-2\\1 &4\end{bmatrix}\space\textbf{Ans.}$$

(ii) Let, f (x) = x³ + (kx + 8) x + k.

By Remainder theorem,

when, f (x) is divided by (x + 1).

Remainder

f (– 1) = (– 1)³ + {k (– 1) + 8} (– 1) + k

= – 1 + (– k + 8) (– 1) + k

= – 1 + k – 8 + k

= 2k – 9

when, f (x) is divided by (x – 2).

Remainder,

f (2) = (2)³ + (k.2 + 8) 2 + k

= 8 + 4k + 16 + k

= 5k + 24

Also, sum of remainders = 1 (Given)

f (– 1) + f (2) = 1

$$\Rarr\space 2k-9+5k+24 = 1\\\Rarr\space 7k+15=1\\\Rarr\space 7k =1-15\\\Rarr\space k =\frac{-14}{7} =-2$$

Ans.

(iii) Given, AC is diameter, BC||AE, and ∠BAC = 50°

(i) ∠ABC = 90° (Q Angle at circumference of a semicircle)

In ΔABC,

∴ ∠ACB + ∠BAC + ∠ABC = 180° (Angles sum property)

⇒ ∠ACB + 50° + 90° = 180°

⇒ ∠ACB = 180° – 140°

∠ACB = 40° Ans.

(ii) ∠CAE = ∠ACB (Alternate angles as BC || AE)

= 40°

∴ ∠EDC + ∠CAE = 180° (Sum of opposite angles of a cyclic quadrilateral is 180°)

⇒ ∠EDC + 40° = 180°

⇒ ∠EDC = 180° – 40°

∠EDC = 140° Ans.

(iii) ∠BEC = ∠BAC (Angles on same segment are equal)

= 50° Ans.

Now, ∠BAE = ∠BAC + ∠CAE

= 50° + 40°

= 90°

We know that, if an angle of a triangle in a circle is 90°. Then, the hypotenuse must be the diameter of the circle.

Hence, BE is a diameter (∵ ∠BAE = 90°)

Hence Proved.

Answer 6.

(i) (i) y = x + 1

⇒ m₁ = tan θ₁ = 1 = tan 45°

⇒ θ₁ = 45°.

$$\text{(ii)\space y =}\sqrt{3}x-1\\\Rarr\space m_{2} =\text{tan}\space\theta_{2} =\sqrt{3} =\text{tan 60\degree}\\\Rarr\space\theta_{2}= 60\degree.$$

(iii) ∵ Exterior angle = Sum of interior opposite angles

∴ θ = θ₂ – θ₁

= 60° – 45°

= 15°. Ans.

(iv) Put y = 0 in y = x + 1, we get

0 = x + 1

⇒ x = – 1

∴ The required point is (– 1, 0). Ans.

$$\text{(v)\space}\text{Put x = 0 in}\space y =\sqrt{3}x-1,\\\text{we get y =-1}$$

∴ The required point is (0, – 1). Ans

(ii) We have,

$$\frac{\text{tan A}}{\text{1 - cot A}} +\frac{\text{cot A}}{\text{1 - tan A}}\\=\text{sec A cosec A +1}\\\text{L.H.S =}\frac{\text{tan A}}{\text{1 - cot A}} + \frac{\text{cot A}}{\text{1 - tan A}}\\=\frac{\frac{\text{sin A}}{\text{cos A}}}{\frac{\text{1 - cos A}}{sin A}} + \frac{\frac{\text{cos A}}{\text{sin A}}}{\frac{\text{1 - sin A}}{\text{cos A}}}\\=\frac{\text{sin A}}{\text{cos A}}×\frac{\text{sin A}}{\text{sin A - cos A}} +\\\frac{\text{cos A}}{\text{sin A}}×\frac{\text{cos A}}{\text{cos A - sin A}}\\=\frac{\text{sin}^{2}\text{A}}{\text{cos A}(\text{sin A - cos A})} -\\\frac{\text{cos}^{2}A}{\text{sin A(sin A - cos A)}} $$

$$=\frac{\text{sin}^{3}\text{A} -\text{cos}^{3}\text{A}}{\text{sin A cos A}(\text{sin A - cos A})}\\=\\\frac{(\text{sin A - cos A})(\text{sin}^{2}\text{A} + \text{sin A cos A + cos}^{2}\text{A})}{\text{sin A cos A(sin A - cos A)}}\\\lbrack\because\space a^{3}- b^{3} =(a-b)(a^{2} + ab+ b^{2})\rbrack\\=\frac{\text{1 + sin A × cos A}}{\text{sin A cos A}}\\=\frac{1}{\text{cos A sin A}} + \frac{\text{sin A cos A}}{\text{sin A cos A}}\\\lbrack\because\space\text{cos}^{2}A + \text{sin}^{2}A =1\rbrack$$

= sec A cosec A + 1 = R.H.S.

Hence Proved.

(iii) Let a and d be the first term and common difference of the given A.P. respectively

Then, a₄ = 8 and a₆ = 14

⇒ a + 3d = 8 ...(i)

and a + 5d = 14 ...(ii)

On subtracting equation (i) from (ii), we get

2d = 6

⇒ d = 3

On putting d = 3 in equation (i), we get

a + 3 × 3 = 8

⇒ a = 8 – 9 = – 1

(a) First term (a) = – 1. Ans.

(b) Common difference (d) = 3. Ans.

$$\because\space\text{S}_{n} =\frac{n}{2}\lbrack2a + (n-1)d\rbrack\\\text{S}_{20} =\frac{20}{2}\lbrack2×(\normalsize-1) + (20-1)×3\rbrack$$

= 10 (– 2 + 57)

= 550 Ans.

Answer 7.

(i) Here, C.P. of x articles = ₹1200

$$\therefore\space\text{C.P. of 1 article =}\space ₹\frac{1200}{x}\\\therefore\space\text{S.P. of 1 article} =\\ ₹\bigg(\frac{1200}{x} + 2\bigg)$$

Since, 10 articles were damaged, the number of articles left = x – 10.

$$\therefore\space\text{Total S.P. =}\\\text{ ₹ (x – 10)}\bigg(\frac{1200}{x} + 2\bigg)$$

∵ Profit = ₹60

$$\therefore\space(x-10)\bigg(\frac{1200}{x} + 2\bigg)\\-1200 = 60\\\Rarr\space(x-10)\bigg(\frac{1200 + 2x}{x}\bigg)\\ = 1260$$

⇒ (x – 10)(1200 + 2x) = 1260x

⇒ 1200x + 2x² – 12000 – 20x – 1260x = 0

⇒ 2x² – 80x – 12000 = 0

⇒ x² – 40x – 6000 = 0

⇒ x² – 100x + 60x – 6000 = 0

⇒ x(x – 100) + 60(x – 100) = 0

⇒ (x – 100)(x + 60) = 0

⇒ x – 100 = 0 or x + 60 = 0

⇒ x = 100 or x = – 60

∴ Number of articles are 100

(∵ x cannot be negative). Ans.

(ii)

Wages (in ₹)	No. of workers	Cumulative Frequency
400 – 450	2	2
450 – 500	6	8
500 – 550	12	20
550 – 600	18	38
600 – 650	24	62
650 – 700	13	75
700 – 750	5	80
	n = 80

$$\text{(a)\space Median wage} =\frac{n}{2}\text{th value}\\=\frac{80}{2}\text{th value}$$

= 40th value

= ₹ 600. Ans.

$$\text{(b) Lower quartile =}\frac{n}{4}\\\text{th value =}\\\text{20th value = ₹ 550}\space\textbf{Ans.}$$

Answer 8.

$$\text{(i)\space}\frac{-x}{3}-4\leq\frac{x}{2}-\frac{7}{3}\lt-\frac{7}{6},\\\text{x}\epsilon \text{R}\\\frac{\normalsize-x}{3}-4\leq\frac{x}{2}-\frac{7}{3}\\\frac{x}{2} - \frac{7}{3}\lt\frac{\normalsize-7}{6}\\\frac{\normalsize-x}{3}-\frac{x}{2}\leq\frac{\normalsize-7}{3} + 4\\\frac{x}{2}\lt\frac{\normalsize-7}{6} + \frac{7}{3}\\=\frac{-2x-3x}{6}\leq\frac{-7+12}{3}\\\frac{x}{2}\lt\frac{-7+14}{6}$$

$$\frac{-5x}{6}\leq\frac{5}{3}\space\\\frac{x}{2}\lt\frac{7}{6}\\x\geq-\frac{5×6}{5×3}\qquad x\lt\frac{7×2}{6}\\x\geq-2\qquad x\lt\frac{7}{3} $$

x ≥ –2 x < 2.3

– 2 ≤ x < 2.3

(ii)

Sol.

Distance in m	Frequency (f)	c.f.
12 – 13	3	3
13 – 14	9	12
14 – 15	12	24
15 – 16	9	33
16 – 17	4	37
17 – 18	2	39
18 – 19	1	40

Note : Instead of 2 cm = 1 m and 2 cm = 5 students, we have used 1 cm = 1 m and 1 cm = 5 students on X and Y axes, respectively.

$$\text{(i) Median =}\space\bigg(\frac{N}{2}\bigg)^{\text{th}}\space\text{term}\\=\bigg(\frac{40}{2}\bigg)^{\text{th}}\space\text{term}$$

= 20th term

On the graph, through a point 20 on y-axis, draw a horizontal line which meets the ogive at point A.
Through A, draw a vertical line which meets the x-axis at 14.7.

∴ Median = 14.7 Ans.

(ii) Upper quartile (Q₃) =

$$\bigg(\frac{3N}{4}\bigg)^{\text{th}}\space\text{term}\\=\bigg(\frac{3×40}{4}\bigg)^{th}\space\text{term}$$

= 30th term

= 15.7 Ans.

(iii) Number of students who cover more than

$$16\frac{1}{2}m = 40-35 = 5 $$

Ans.

(iii) (a) In ΔADF and ΔCFE

∠DAF = ∠FCE [Alternate angles]

∠AFD = ∠CFE [Vertically opposite angles]

∠ADF = ∠CEF [Alternate angles]

∴ ΔADF ~ ΔCEF [By A.A. axiom]

Hence Proved.

(b) ΔADF ~ ΔCEF

$$\therefore\space\frac{\text{AD}}{\text{CE}} =\frac{\text{AF}}{\text{FC}}\qquad\text{...(i)}$$

We have, AF : AC = 5 : 8, CE = 6 cm

Let AF = 5x, AC = 8x

∴ FC = AC – AF

= 8x – 5x = 3x

$$\text{From (i),\space}\frac{\text{AD}}{6} = \frac{\text{5x}}{3x}$$

⇒ AD = 10 cm. Ans.

Answer 9.

(i) Given, Total number of outcomes i.e., n(S) = 25

(i) Let A be the event of getting an odd number.

∴ A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25}

∴ n(A) = 13

$$\therefore\space\text{P(A) =}\frac{\text{n(A)}}{\text{n(S)}} =\frac{13}{25}$$

Ans.

(ii) Let B be the event of getting a number divisible by 2 and 3 both.

∴ B = {6, 12, 18, 24}

∴ n(B) = 4

$$\therefore\space\text{P(B) =}\frac{\text{n(B)}}{\text{n(S)}} =\frac{4}{25}$$

Ans.

(iii) Let C be the event of getting a number less than 16.

∴ C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}

∴ n(C) = 15

$$\therefore\space\text{P(C) =}\frac{\text{n(C)}}{\text{n(S)}} =\frac{15}{25} =\frac{3}{5}$$

Ans.

(ii) Time of ascent = 1 minute 28 seconds

= 88 seconds

Distance moved by the bucket = Speed × Time

= [1.1 × 88] metre

= 96.8 metre

Circumference of wheel = 2πr

$$= 2×\frac{22}{7}×\frac{77}{2}\text{cm}\\\begin{bmatrix}\because\space r =\frac{d}{2}=\frac{77}{2}\text{cm}\end{bmatrix}$$

= 242 cm = 2.42 metre

∴ Number of complete revolutions the wheel makes in raising the bucket

$$= \frac{\text{Distance}}{\text{Circumference}}\\=\bigg(\frac{96.8}{2.42}\bigg) = 40.$$

Ans.

(iii) (a) In the figure, ∠DAB + ∠BCD = 180°

[Opposite angles of a cyclic quadrilateral]

∴ ∠DAB + 130° = 180°

⇒ ∠DAB = 180° – 130°

⇒ ∠DAB = 50°. Ans.

(b) Now, ∠ADB = 90° [Angle in semi-circle]

In DADB,

∠DAB + ∠ADB + ∠DBA = 180° [Angle sum property]

⇒ 50° + 90° + ∠DBA = 180°

⇒ ∠DBA = 180° – 140°

⇒ ∠DBA = 40°. Ans.

Answer 10.

(i) We have, (2x – 5) : (3x + 1) is duplicate ratio of 2 : 3.

∴ (2x – 5) : (3x + 1) = 2² : 3²

$$\Rarr\space\frac{\text{2x - 5}}{\text{3x+1}} =\frac{4}{9}$$

⇒ 18x – 45 = 12x + 4

⇒ 18x – 12x = 4 + 45

⇒ 6x = 49

$$\Rarr\space x =\frac{49}{6}= 8\frac{1}{6}\space\textbf{Ans.}$$

(ii) Given radius = 3.5 cm

Steps of construction :

1. Draw PO = 6 cm. Draw a circle of radius 3.5 cm with centre O.

2. Draw perpendicular bisector of OP which intersects OP at M.

3. Taking MO radius and M as centre, draw arcs which cut the circle at A and B.

4. Join AP and BP which are the required tangents.

5. Measure the length of tangents PA = PB = 4.5 cm

(iii) From right angle Δ ADC,

$$\frac{\text{AD}}{\text{CD}} =\text{tan 36}\degree\\\Rarr\space\frac{100}{y} = \text{tan 36\degree}\\\Rarr\space y =\frac{100}{\text{tan 36}\degree}\\=\frac{100}{0·7265}\\\Rarr\space y = 137.646\space m$$

From right angle Δ ADB,

$$\frac{100}{x} =\text{tan 48}\degree\\\Rarr\space x =\frac{100}{1.1106} =\text{90.04 m.} $$

∴ Distance between the ships

= x + y

= 137·638 + 90·04

= 227·678 m

= 228 m. (approx.).

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