Oswal 61 Sample Question Papers ICSE Class 10 Maths Solutions
Section-A
Answer 1.
(i) (c) ₹ 1,050
Let the price of the article be ₹ x.
So, x + GST = 1,239
x + 18% of x = 1,239
x+18x/100=1,239
118x/100=1,239
⇒ x = ₹ 1,050
√2/3 is a solution of the equation
m =-2√6
(iii) (c) 1/3
⇒ 3a – 1 = 0
⇒ a = 1/3
$$\text{Given, \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space}\text{A =}\begin{bmatrix}5&3\\-1&2\\\end{bmatrix}\\\text{}\\\text{We Know, \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space}\text{I = }\begin{bmatrix}1&0\\0&1\\\end{bmatrix}\\\text{}\\\therefore\text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space}\text{Required matrix = A – 2I}\\\text{}\\\text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space}\text{=}\begin{bmatrix}5&3\\-1&2\end{bmatrix}-2\begin{bmatrix}1&0\\0&1\end{bmatrix}\\\text{}\\\text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space}\text{=}\begin{bmatrix}5&3\\-1&2\end{bmatrix}-\begin{bmatrix}2&0\\0&2\end{bmatrix}\\\text{}\\\text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space}\text{=}\begin{bmatrix}5-2&3-0\\-1-0&2-2\end{bmatrix}\\\text{}\\\text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space}\text{A – 2I =}\begin{bmatrix}3&3\\-1&0\end{bmatrix}$$
(v) (c) 945
List of first 14 natural numbers, where each number is divisible by 9 is :
9, 18, 27, .................. .
Clearly, this list forms an A. P.
First term of A.P. (a) = 9
Common difference (d) = 9
We know,
= 7[18 + 117]
= 945
Required sum = 945
(vii) (b) 6 cm
In ΔADE and DABC
∠ADE = ∠ABC [Corresponding angles]
∠A = ∠A [Common angle]
∠ADE ~ ∠ABC [AA similarity rule]
AD/AB=AE/AC
⇒ 2/6=AE/9
⇒ AE = 3 cm
CE = AC – AE
= 9 – 3 = 6 cm.
(viii) (b) 2.5
Inequality 7 – 3x ≥ (-1/2)
⇒ – 3x ≥ (-1/2) - 7
⇒ – 3x ≥ (-15/2)
⇒ x ≤ 15/6
⇒ x ≤ 2.5
Greatest value of x = 2.5
(ix) (d) 8
No. of possible outcomes n(S) = n(HHH, HHT, HTH, HTT, THH, THT, TTH, TTT)
= 8.
(x) (b) 48π cm2
(xi) (d) 2
$$\begin{bmatrix}x-2y & 5\\ 3 & y\end{bmatrix} = \begin{bmatrix}6 & 5\\3 & -2\end{bmatrix}\\\text{}\\\text{Comparing, we get}\\\text{}\\\text{y = -2}$$
and x – 2y = 6
⇒ x – 2(–2) = 6
⇒ x + 4 = 6
⇒ x = 6 – 4 = 2
(xii) (b) y = 2x
Let the equation be y = mx + c
As the slope is 2, y = 2x + c
as it is passing through (1,2), substituting this in the given equation
2 = 2 × 1 + c
⇒ c = 0 Hence, the equation is y = 2x
(xiii) (b) ∠A = ∠B
The angles at the circumference subtended by the same arc are equal.
(xiv) (c) n(n+1)/√2
Given: series is, √2+√8+√18+√32+.........
⇒ √2+2√2+3√2+4√2+......+n√2
⇒√2(1+2+3+4+.....+n)
We know that,
$$\sum^n_{i=1}k = \text{1+2+3+…..+n.}\\\text{}\\=\frac{\text{n(n+1)}}{\text{2}}\\\text{}\\=\sqrt{2}\bigg[{\frac{n(n+1)}{2}}\bigg]\text{ We know that,}\\\text{}\\ \therefore \text{}=\sqrt{2}\text{(1+2+3+…..+n)}\\ \text{}\\ = \frac{\text{n(n+1)}}{\sqrt{2}}$$
(xv) (c) Mode
$$\text{(i) Given:P=₹ 200, M.V. = ₹ 5150, r = 7}\%\\\text{}\\\text{Let the number of instalments be n.}\\\text{}\\\therefore\text{ }\text{ I = p}×\frac{\text{n(n+1)}}{2×12}×\frac{r}{100}-200×\frac{\text{n}^2+\text{n}}{24}×\frac{7}{100}\\\text{}\\=\frac{7}{12}(\text{n}^2+\text{n})\\\text{}\\ \therefore\text{ }\text{ M.V. = }\text{P}_\text{n}\text{+I=}\text{200}_\text{n}+\frac{7}{12}(\text{n}^2+\text{n)} = \frac{\text{2400n}+\text{7n}^{2}+\text{7n}}{12}\\\text{}\text{ }\\=\frac{ \text{7n}^2+2407n}{12}\\ \text{According to the question.}\\\text{}\\ \therefore \text{}\frac{\text{7n}^\text{2}+\text{2407n}\\\text{}}{12}=\text{5150}\\\text{}\\ \Rightarrow \text{ }\text{7n}^2+2407\text{n}=61800\\\text{}\\ \Rightarrow \text{ }\text{7n}^{2}+2575\text{n}-168\text{n}-61800 = 0 \\\text{}\\ \Rightarrow \text{}\text{n}(\text{7n}+2575)-24(\text{7n}+2575) = 0\\\text{}\\ \Rightarrow \text{}(7\text{n}+2575)(\text{n}-24) = 0\\\text{}\\ \therefore \text{n}=24 \text{ }[\because \text{n cannot be negative}]\\ \text{}\\ \therefore \text{The required number of instalments = 24.}\text{ Ans.}$$
(iii) (cosec θ – sin θ)(sec θ – cos θ) (tan θ + cot θ) = 1
$$L.H.S. = (cosec\space \theta -sin\space \theta)(sec\space \theta -cos\space \theta) (tan\space \theta+cot\space \theta) \\ =\left(\frac{1}{sin\space \theta}-sin\space \theta)\right)\left(\frac{1}{cos\space \theta}-cos\space \theta)\right)\left(\frac{sin \space \theta}{cos\space \theta}+\frac{cos\space \theta}{sin\space \theta}\right) \\ = \left(\frac{1-sin^2 \space \theta}{sin\space \theta}\right)\left(\frac{1-cos^2 \space \theta}{cos\space \theta}\right)\left(\frac{sin^2\space \theta+cos^2 \space \theta}{sin\space \theta \space cos\space \theta}\right) \\ = \frac{cos^2\space \theta}{sin\space \theta}×\frac{sin^2\space \theta}{cos\space \theta}× \frac{1}{sin\space \theta cos\space \theta}\space\space [\because sin^2 \theta + cos^2\theta = 1] \\ = \frac{sin^2 \space \theta\space cos^2 \space \theta}{sin^2 \space \theta\space cos^2 \space \theta} \\ = 1= R.H.S. \space\space Hence. Proved.$$
Answer 3.
(i) For cylinder 1 :
Diameter = 24 cm
Radius, R = 24/2 = 12 cm
Height, H = 220 cm
For cylinder 2 :
Radius, r = 8 cm
Height, h = 60 cm
= π(144 × 220 + 64 × 60) cm3
= 3.14 × 35520 = 111532.8 cm3
Total mass of iron = 111532.8 × 8 g
=111532.8 ×8/1000 kg
892.26 kg (approximately). Ans.
(ii) For Point P :
$$\therefore \text{Using section formula}\\\text{}\\\text{P =}\bigg(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}\bigg)\\\text{}\\=\bigg(\frac{2×3+3×(-2)}{2+3},\frac{2×(-4)+3×6}{2+3}\bigg)\\\text{}\\ = \bigg(\frac{6-6}{5},\frac{-8+18}{5}\bigg)\\\text{}\\ = \bigg(0,\frac{10}{5}\bigg)=(0,2)\\\text{}\\ \text{For the required line :}\\\text{}\\m=\frac{3}{2}\text{and}(x_1,y_1)=\text{P}(0,2)\\\text{}\\\text{Equation of line is given as,}\\\text{}\\ y-y_1 = m(x – x_1)\\\text{}\\ \Rightarrow y-2=\frac{3}{2}(x-0)\\\text{}\\ \Rightarrow 2y-4=3x \space i.e.\space \space 3x-2y+4 = 0\text{Ans.}$$
(iii) (i) A (3, 5), B (– 2, – 4) are plotted on graph.
$$(ii)\space A \space(3,5) \xrightarrow{R_x}A'(3,-5)$$
$$(iii) B(-2,-4)\xrightarrow{R_y}(2,-4)\xrightarrow{R_O}B'(-2,4)$$
(iv) AA′BB′ is an isosceles trapezium.
(v) Two invariant points under reflection in X-axis are C (3, 0) and D (– 2, 0).
Section-B
Answer 4.
(i) When the product is sold from Agra to Kanpur (intra - state transaction) :
For dealer in Agra :
S.P = ₹ 20,000
CGST = 9% of ₹ 20,000 = ₹ 1,800
SGST = 9% of ₹ 20,000 = ₹1,800
When product is sold from Kanpur to Jaipur (inter- state transaction):
For the dealer in Kanpur :
Input tax credit = ₹1,800 + ₹ 1,800 = ₹ 3,600
C.P = ₹ 20,000 and Profit = ₹ 5,000
S.P = ₹ 20,000 + ₹ 5,000 = ₹ 25,000
IGST = 18% of 25,000 = ₹ 4,500
Net GST paid by the dealer at Kanpur
= Output GST – Input GST
= 4,500 – 3,600 = ₹ 900
The cost of goods/services at Jaipur
= S.P in Kanpur + IGST
= 25,000 + 18% of 25,000
= 25,000 + 4,500 = ₹ 29,500
IGST = 18% of 25,000 = ₹ 4,500
= Output GST – Input GST
= 4,500 – 3,600 = ₹ 900
The cost of goods/services at Jaipur
= S.P in Kanpur + IGST
= 25,000 + 18% of 25,000
= 25,000 + 4,500 = ₹ 29,500
(ii) Given equation is,
Here, a = (k + 1), b = – 4k, c = 9
For real and equal roots,
16k2 – 36k – 36 = 0
Dividing both sides by 4, we get
⇒ 4k(k – 3) + 3(k – 3) = 0
⇒ (k – 3)(4k + 3) = 0
k – 3 = 0 4k + 3 = 0
k = 3 k = − 3/4
For k = 3, the equation will be
⇒ 2x(2x – 3) – 3(2x – 3) = 0
⇒ x = 3/2, 3/2
For k = – 3/4 , the equation will be
⇒ x(x + 6) + 6(x + 6) = 0
⇒ x = – 6, – 6.
Hence, x = 3 /2 , − 6.
(iii)
Class Interval | Frequency |
0 – 5 | 2 |
5 – 10 | 7 |
10 – 15 | 18 |
20 – 25 | 8 |
25 – 30 | 5 |
15 – 20 | 10 |
∴ Mode = 13 (using histogram).
Answer 4.
$$\text{(i) We have}\\\text{}\\ \text{x+y}=\begin{bmatrix}5&2\\0&9\end{bmatrix}\\ \text{}\\ \text{and }\text{X-Y=}\begin{bmatrix}3&6\\0&-1\end{bmatrix}\\ \text{}\\ \text{Now }\text{(X+Y)+(X-Y)=}\begin{bmatrix}5&2\\0&9\end{bmatrix}+\begin{bmatrix}3&6\\0&-1\end{bmatrix}\\ \text{}\\ \Rightarrow 2X = \begin{bmatrix}8&8\\0&8\end{bmatrix}\\ \\ \text{}\\ \Rightarrow X = \frac{1}{2} \begin{bmatrix}8&8\\0&8\end{bmatrix}= \begin{bmatrix}4&4\\0&4\end{bmatrix}\\\text{}\\\text{}\text{Also }\text{(X+Y)-(X-Y)=}\begin{bmatrix}5&2\\0&9\end{bmatrix}-\begin{bmatrix}3&6\\0&-1\end{bmatrix}\\\text{}\\ \Rightarrow \text{ }\text{2Y}=\begin{bmatrix}2&-4\\0&10\end{bmatrix}\\ \text{}\\ \Rightarrow \text{ }\text{Y=}\frac{1}{2}\begin{bmatrix}2&-4\\0&10\end{bmatrix}=\begin{bmatrix}1&-2\\0&5\end{bmatrix}\\\text{}\\ \text{Thus, } X = \begin{bmatrix}4&4\\0&4\end{bmatrix}\text{and Y =}\begin{bmatrix}1&-2\\0&5\end{bmatrix}\text{ Ans.}$$
(ii) Given, ∠SRT = 65° and SP is tangent
∠TSR = 90° [Angle between the radius and tangent]
∴ ∠SRT + ∠STR = ∠TSR = 90°
⇒ ∠STR = ∠TSR – ∠SRT
∠STR = x° = 90° – 65° = 25°
∠y° = 2∠x°
[Angle subtended at the centre is double that of the angle subtended by the arc at same circle]
∴ ∠y° = 2 × 25° = 50°
and ∠SPO = ∠OSP – ∠SOP
∴ ∠SPO = ∠z = 90° – 50° = 40°
Hence, ∠x = 25°, ∠y = 50° and ∠z = 40°. Ans.
(a) Put x – 2 = 0 ⇒ x = 2 in expression, we get
⇒ 16 – 4 – 2p – 2 = 0
⇒ 10 – 2p = 0
⇒ p = 5 Ans.
Dividing the expression by (x – 2),
$$\space\space\space\space\space\space\space\space\space\space\space(2x^2+3x+1)\\x-2)\overline{2x^3-x^2-5x-2}\\\space\space\space\space\space\space\space\space\space\space\space 2x^3-4x^2\\ \space\space\space\space\space\space\space\space\space\space -\space\space\space\space\space+\\\space\space\space\space\space\space\space\space\space\overline{\space\space\space\space\space\space\space\space\space\space\space3x^2-5x}\\\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space3x^2-6x\\\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space-\space\space\space\space\space+\\\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\overline{\space\space\space\space\space\space\space\space\space\space\space\space\space\space x-2}\\\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space x-2\\\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space-\space\space\space+\\\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\overline{\space\space\space\space\space ×\space\space\space\space\space\space}$$
= (x – 2) {2x2 + (2 + 1) x + 1}
= (x – 2) {2x (x + 1) +1 (x + 1)}
= (x – 2) (2x + 1) (x + 1) Ans.
Answer 6.
(i) Given, A(7, – 3) and B(1, 9).
(a) Slope of AB = y2-y1/x2-x1
= 9-(-3) /1-7
=12/-6
=-2
(b) Slope of perpendicular bisector of AB (m) = -1/(-2)=1/2
Mid-point of AB =((x1+x2)/2,(y1+y2)/2)
=((7+1)2,(-3+9)/2)=(4,3)
Let the equation of required line be
y – y1 = m(x – x1)
⇒ y – 3 = 1/2 (x – 4)
⇒ 2y – 6 = x – 4
⇒ x – 2y – 4 + 6 = 0
⇒ x – 2y + 2 = 0, which is the required line. Ans.
(c) Since (– 2, p) lies on x – 2y + 2 = 0
Putting x = 2, y = p, we get
– 2 – 2p + 2 = 0
⇒ 2p = 0
⇒ p = 0 Ans.
$$\text{(ii) To prove},\frac{1}{\text{cos }\theta+\text{sin} \theta}+\frac{1}{\text{sin }\theta-\text{cos} \theta}=\frac{2\text{sin } \theta}{\text{1-2cos}^2 \theta}\\ \text{}\\ \text{L.H.S = }\frac{1}{\text{cos }\theta+\text{sin }\theta}+\frac{1}{\text{sin}\theta-\text{cos}\theta}\\\text{}\\=\frac{\text{sin }\theta-\text{cos }\theta+\text{cos }\theta+\text{sin }\theta}{(\text{cos }\theta+\text{sin }\theta )(\text{sin }\theta-\text{cos }\theta )}\\\text{}\\=\frac{2\text{sin} \theta}{\text{sin}^2 \theta-\text{cos}^2\theta}\\\text{}\\=\frac{2\text{sin}\theta}{1-\text{cos}^2\theta-\text{cos}^2 \theta}\text{ }[\because \text{sin}^2 \theta -1-\text{cos}^2 \theta]\\ \text{}\\=\frac{2\text{sin} \theta}{1-2\text{cos}^2\theta}\\\text{}\\=\text{R.H.S }\text{Hence Proved}$$
(iii) Let the first term of A.P. be a and the common difference be d.
Given,
$$\text{Given}, \text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space}\text{T}_{24}-\text{2T}_{10}\\ \Rightarrow \text{\space\space\space\space\space\space\space\space} \{\text{a+(24-1)d}\}=2\{\text{a+(10-1)d}\}\\ \Rightarrow \text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space} \text{a+23d}=2\text{a+18d}\\ \Rightarrow \text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space} \text{a}=5\text{d}\text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space …(i)}\\ \text{Now},\text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space} \text{T}_{15}=\text{\{a+(15-1)d\}}\\ \Rightarrow \text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space} \text{T}_{15}=\text{a+14d}\\ \\ \Rightarrow \text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space} \text{T}_{15}=\text{5d+14d}\text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space}[\because \text{a=5d}]\\ \\ \Rightarrow \text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space} \text{T}_{15}=\text{19d}\\ \text{and}\text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space} \text{T}_{72}=\text{\{a+(72-1)d\}}\\ \Rightarrow \text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space} \text{T}_{72}=\text{a+71d}\\ \Rightarrow \text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space} \text{T}_{72}=\text{5d+71d}\\ \Rightarrow \text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space} \text{T}_{72}=\text{76d-4(19d)}\\ \Rightarrow \text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space} \text{T}_{72}=\text{4T}_{15.}\text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space}\text{Hence Proved}$$
Answer 7.
(i) After removing king, queen and jack of clubs, 49 cards are left in the deck.
∴ Total number of possible outcomes = 49
(a) There are 3 kings left in the deck.
∴ Number of favourable outcomes = 3
∴ P (a king) = 3/49. Ans.
(b) After removing king, queen and jack of clubs, number of clubs cards left = 13 – 3 = 10.
∴ No. of favourable outcomes = 10
∴ P (a club) = 10/49. Ans.
(c) There is only one ‘10’ of hearts
∴ No. of favourable outcomes = 1
∴ P (a ‘10’ of hearts) = 1/49. Ans.
(ii) According to the question,
2/3 (Volume of hemisphere) = Volume of cone
h =4 × 3.5 × 3.5 × 3.5 × 3/3.5 × 3.5 × 9
42.0/9
14m/3 = 4.67 m
35m/6
Now, Surface area of buoy = Surface area of right cone + Surface area of hemisphere
= pr(l + 2r)
22/7 × 3.5 ((35/6)+2 × 3.5)
= 11 × (5.83 + 7)
= 11 × 12.83
= 141.13 m2 Ans.
(iii) Given, AB is a diameter.
and ∠BAC = 40°.
In ΔABC
∴ ∠ABC + ∠ACB + ∠BAC = 180° [Sum of angles in a D is 180°]
⇒ ∠ABC + 90° + 40° = 180°
∠ABC = 180° – 130° = 50°
Now, ∠ACT = ∠ABC [Angles in alternate segment]
⇒ ∠ACT = 50°. Ans.
Answer 8.
$$\text{(i) We have-}\frac{x}{3}\le\frac{x}{2}-1\frac{1}{3}<\frac{1}{6},\text{x}\isin\text{R}\\\text{}\\\text{Now}, \text{\space\space\space\space\space\space} -\frac{x}{3}\le\frac{x}{2}-1\frac{1}{3} \text{\space\space\space\space\space\space}\\-\frac{x}{3}\le\frac{x}{2}-\frac{4}{3}\\\text{}\\\text{\space\space\space\space\space\space} \frac{4}{3}\le\frac{x}{2}+\frac{x}{3}\\ \text{}\\\text{\space\space\space\space\space\space\space}\frac{4}{3}\le\frac{5x}{6}\\\text{}\\\text{\space\space\space\space\space\space}\frac{6}{5}×\frac{4}{3}\le\text{x}\\\text{}\\\text{\space\space\space\space\space\space}\frac{8}{5}\le\text{x}\text{ \space\space\space\space\space\space…(i)}\\\text{}\\\text{Thus}, \space\space\space\space\space\space\space\frac{8}{5}\le\text{x}<3\\\text{}\\\text{or}\space\space\space\space\space\space\text{1.6}\le\text{x<3}\\\text{}\\ \therefore\text{ Solution set = \{1 x : 1.6} \le \text{x <3, x } \epsilon \text{ R\}}.$$
$$\frac{x}{2}-1\frac{1}{3}<\frac{1}{6}\\\text{}\\\text{\space\space\space\space\space\space\space\space\space\space}\frac{x}{2}<\frac{1}{6}+\frac{4}{3}\\\text{}\\\text{\space\space\space\space\space\space\space\space\space\space}\frac{x}{2}<\frac{1+8}{6}\\\text{}\\\text{\space\space\space\space\space\space\space\space\space\space}\text{x}<\frac{9×2}{6}\\\text{}\\\text{\space\space\space\space\space\space\space\space\space\space}\text{x}<\text{3}\text{\space\space\space\space\space\space\space}\text{…(ii)}$$
Marks : Less than | Classes | Mid Values (x) | No. of students (c.f.) | Frequency (f) | fx |
10 | 0 – 10 | 5 | 2 | 2 | 10 |
20 | 10 – 20 | 15 | 7 | 5 | 75 |
30 | 20 – 30 | 25 | 15 | 8 | 200 |
40 | 30 – 40 | 35 | 18 | 3 | 105 |
50 | 40 – 50 | 45 | 20 | 2 | 90 |
Σf = 20 | Σfx = 480 |
∴ Arithmetic mean = Σfx/Σf=480/20=24
(iii) Given, ∠ABC = ∠DAC = x (say)
AB = 8 cm, AC = 4 cm, AD = 5 cm.
(a) In ΔACD andΔBCA
∠ABC = ∠DAC [Given]
∠ACD = ∠BCA [Common]
∴ ΔACD ~ ΔBCA [AA axiom]
Hence ΔACD is similar to ΔBCA. Hence Proved.
(b) ΔACD ~ ΔBCA
∴ AC/BC=CD/CA=AD/BA [Corresponding parts of similar triangles]
⇒ 4/BC=CD/4=5/8
⇒ 4/BC = 5/8
⇒ BC = 8 x 4/5=32/5=6.4 CM
and CD/4 = 5/8
⇒ CD = 5 x 4 /8
⇒ CD = 2.5 cm. Ans.
Answer 9.
(i) The original cost of each book = ₹ x.
Total cost = ₹ 960
No. of books purchased = 960/x
If the price was ₹ 8 less, the cost of each book = ₹ (x – 8)water.
No. of books purchased = 960/x -8
According to question,
(960/x-8)-(960/x) = 4
960(1/x-8)-(1/x) = 4
960 x (x-x+8)/x(x-8)=4
⇒ 960 × 8 = 4x (x – 8)
⇒ x (x – 48) + 40 (x – 48) = 0
⇒ (x – 48) (x + 40) = 0
⇒ x – 48 = 0 or x + 40 = 0
⇒ x = 48 or x = – 40 (not possible)
x = 48
The original cost of each book = ₹ 48.
(ii)
Marks | No. of Students | c.f. |
0 – 10 | 5 | 5 |
10 – 20 | 11 | 16 |
20 – 30 | 10 | 26 |
30 – 40 | 20 | 46 |
50 – 60 | 37 | 111 |
60 – 70 | 40 | 151 |
70 – 80 | 29 | 180 |
80 – 90 | 14 | 194 |
90 – 100 | 6 | 200 |
∴ N = 200 [even]
(a) Median = Nth/2 value
= 100th value
= 57 marks (from ogive) Ans.
(b) No. of students scoring less than 40 marks = 45.
Answer 10.
(i) We have,
⇒ x = ± 2. Ans.
(ii) Steps of construction :
(a) Draw DABC with each side 5 cm.
(b) Bisect ∠ABC and ∠ACB.
(c) Let the bisectors meet at O.
(d) Draw a perpendicular OD from O to BC.
(e) Taking OD as radius, draw the required incircle.
(iii) Let the aeroplane be at position A and BC be the river. Drop a perpendicular from A on BC. Let it intersect BC at D.
In DADB, tan 60° = AD/BD
$$\sqrt{3} = \frac{250}{x}\\\text{}\\\text{x = }\frac{250}{\sqrt{3}}\text{m}\\\text{}\\\text{In }\Delta\text{ADC},\text{ tan 45\textdegree} = \frac{\text{AD}}{\text{DC}}\\\text{}\\\text{1}=\frac{\text{250}}{\text{y}}\\\text{}\\\text{y }=\text{ 250m}\\\text{}\\ Thus,\space width\space of\space the\space river \space = 250+\frac{250}{\sqrt{3}} = 250\left(1+\frac{1}{\sqrt{3}}\right) = 250 \left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)\\ \text{= 250}×\frac{2.732}{1.732}\\\text{}\\\text{= 394.34 = 394m}\text{ \space\space\space\space Ans.}$$
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