Oswal Practice Papers CBSE Class 10 Mathematics Solutions (Practice Paper - 9)
Section-A
1. (a) – 10, – 4
Explanation :
x2 + 14x + 40 = 0
⇒ x2 + 10x + 4x + 40 = 0 (on splitting the middle term)
⇒ x(x + 10) + 4(x + 10) = 0
⇒ (x + 10)(x + 4) = 0
⇒ x + 10 = 0 or x + 4 = 0
⇒ x = – 10 or x = – 4
Hence, roots of equation are – 10, – 4.
2. (b) 40°
Explanation :
Since, ∆ABC ~ ∆PQR
∠A = ∠P = 80° [Given]
⇒ ∠B = ∠Q = 60° [Given]
∠C = ∠R
By angle sum property, we have
∠A + ∠B + ∠C = 180°
⇒ 80° + 60° + ∠C = 180°
⇒ 140° + ∠C = 180°
⇒ ∠C = 180º – 140°
⇒ ∠C = 40°
3. (c) 55
Explanation :
$$\text{Here, a = -5},\space d =-\frac{5}{2}-(\normalsize-2)\\=\frac{\normalsize-5}{2} +5 =\frac{5}{2}$$
We know that, nth term of A.P. is given as,
an = a + (n – 1)d
$$\therefore\space a_{25} =-5 +(25-1)\frac{5}{2}\\=-5 + 24×\frac{5}{2} \\= -5+60$$
= 55
4. (b) 12
Explanation :
Since tangent to a circle is perpendicular to radius.
So, ΔPOQ is right angled at P.
So, ΔPOQ is right angled at P.
By Pythagoras theorem OQ2 = PQ2 + PO2
⇒ (13)2 = (PQ)2 + (5)2
⇒ 169 – 25 = (PQ)2
⇒ 144 = PQ2
$$\Rarr\space\text{PQ} =\sqrt{144}$$
= 12 cm
5. (c) 7.2 cm
Explanation :
Given : ∆PQR ~ ∆PST
Using SS similarity criterion,
$$\frac{\text{PQ}}{\text{PS}} =\frac{\text{QR}}{\text{ST}}\\\text{From the figure,}\\\frac{10}{4}=\frac{18}{\text{ST}}$$
10 × ST = 4 × 18
$$\text{ST}=\frac{72}{10}$$
= 7.2 cm
Hence, the length of ST is 7.2 cm.
6. (a) – 1
Explanation :
$$\text{Given, cot}^{2}\theta-\frac{1}{\text{sin}^{2}\theta}\\\Rarr\space\text{cot}^{2}\theta-\text{cosec}^{2}\theta\\\bigg(\because\space\text{cosec}\theta =\frac{1}{\text{sin}\theta}\bigg)\\\Rarr\space-(\text{cosec}^{2}\theta - cot^{2}\theta)\\\Rarr\space -1\\(\because\space \text{cosec}^{2}\theta - \text{cot}^{2}\theta = 1)$$
7. (d) 94.64 m
Explanation :
Given, in ∆ABC
$$\text{tan 45}\degree =\frac{\text{BC}}{\text{AC}}\\1 =\frac{60}{\text{AC}}$$
AC = 60 m ...(i)
In ∆BCD,
$$\text{tan\space 60\degree}=\frac{\text{BC}}{\text{CD}}\\=\sqrt{3}=\frac{60}{\text{CD}}\\\text{CD}=\frac{60}{\sqrt{3}}\\=\frac{60\sqrt{3}}{3}\\= 20\sqrt{3}$$
= 20 × 1.732
= 34.64 m
∴ Distance between two points AD = AC + CD
= 60 + 34.64
= 94.64 m
$$\textbf{8.\space}(c)\space\frac{12}{13}$$
Explanation :
Total number of possible outcomes = 52
Number of aces in the pack = 4
Thus, the probability of not drawing an ace
$$\therefore\space \text{P(E)}=\frac{52-4}{52}\\=\frac{48}{52}=\frac{12}{13}$$
$$\textbf{9.\space}(c)\space\frac{1}{8}\pi d^{2}$$
Explanation :
As we know that,
$$\text{Area of semi-circle =}\frac{\pi r^{2}}{2}$$
and
$$\text{Radius (r) = }\frac{\text{Diameter (d)}}{2}\\\therefore\space r=\frac{d}{2}$$
Thus,
$$\text{Area of semi-circle =}\frac{\pi}{2}\bigg(\frac{d}{2}\bigg)^{2}\\=\frac{\pi rl}{2×4}=\frac{\pi d^{2}}{8}$$
10. (d) 1000 cm2
Explanation :
Length of resulting cuboid = 20 cm
Breadth of resulting cuboid = 10 cm
Height of resulting cuboid = 10 cm
Surface area of cuboid = 2(lb + bh + hl)
= 2(20 × 10 + 10 × 10 + 20 × 10)
= 2[200 + 100 + 200]
= 2 × 500
= 1000 cm2
11. (b) a ≠ 0
Explanation :
Equation ax2 + bx + c = 0 is said to be quadratic when a ≠ 0.
12. (c) 1
Explanation :
$$\text{Maximum value of}\frac{1}{\text{cosec}\space\theta}\\=\text{sin}\space\theta = 1,\text{when}\space\theta=90\degree$$
$$\textbf{13.}\space\text{(b)\space}\frac{2}{3}$$
Explanation :
$$\text{cos}^{4}\theta - \text{sin}^{4}\theta=\frac{2}{3}\\\Rarr\space(\text{cos}^{2}\theta)^{2}-(\text{sin}^{2}\theta)^{2}=\frac{2}{3}\\\Rarr\space(\text{cos}^{2}\theta + \text{sin}^{2}\theta)(\text{cos}^{2}\theta -\text{sin}^{2}\theta)=\frac{2}{3}\\\Rarr\space\text{cos}^{2}\theta -\text{sin}^{2}\theta =\frac{2}{3}\\(\because\space\text{sin}^{2}\theta + \text{cos}^{2}\theta =1)\\\Rarr\space 1 -\text{sin}^{2}\theta-\text{sin}^{2}\theta=\frac{2}{3}\\(\because\space\text{cos}^{2}\theta = 1 -\text{sin}^{2}\theta)\\\text{1 - 2\text{sin}}^{2}\theta=\frac{2}{3}.$$
14. (d) 2
Explanation :
Given polynomials is x4 – 1
⇒ x4 – 1 = (x2 – 1)(x2 + 1)
[∵ a2 – b2 = (a – b)(a + b)]
⇒ x4 – 1 = (x – 1)(x + 1)(x2 + 1)
So, x4 – 1 has two real zeroes.
15. (b) 54
Explanation :
a × b = H.C.F. × L.C.M.
= 2 × 27
= 54.
16. (b) 30
Explanation :
tn = n(n + 3)
∴ t5 = 5(5 + 3) = 40
∴ t2 = 2(2 + 3) = 10
∴ t5 – t2 = 40 – 10 = 30
17. (a) 11
Explanation :
Given, AB = 29 cm, AD = 23 cm, DS = 5 cm and ∠B = 90°
Now, DS = DR = 5 cm [tangents from point, D]
So, AR = AD – DR = (23 – 5) cm = 18 cm
Again, AR = AQ = 18 cm [tangents from points, A]
Now, QB = AB – AQ = (29 – 18) cm = 11 cm
∠OPB = ∠OQB = 90°
[∵ OQ and OP are radius of circle and PB and BQ are tangents]
Thus, in the quadrilateral OQBP,
∠POQ = 360° – (90° + 90° + 90°) = 90°
Hence, it is a right-angled quadrilateral with adjacent opposite sides are equal.
∴ OQ = QB = 11 cm = r
18. (b) 2
Explanation :
QR is parallel to AB.
According to basic proportionality theorem,
$$\frac{\text{PQ}}{\text{PA}}=\frac{\text{PR}}{\text{PB}}\\\text{...(1)}$$
ΔPQR ~ ΔPAB
DR is also parallel to QB. [Given]
According to basic proportionality theorem,
$$\frac{\text{PD}}{\text{PQ}} = \frac{\text{PR}}{\text{PB}}\space\text{...(2)}$$
ΔPDR ~ ΔPQB
Therefore, there are two distinct pair of similar triangles in the given figure.
19. (c) Assertion (A) is true, but reason (R) is false.
Explanation :
For assertion, the area enclosed by a sector is proportional to the arc length of the sector.
$$\text{So, A =}\frac{\text{RL}}{2},$$
A = area, R= radius and L = arc length
Hence, the resulting enclosed area is a major segment if it is a major arc.
So assertion is true.
For reason,
We basically denote an arc as a major arc when it is greater than the semicircle and if we divide a circle into 3 arc each of it will be less than a semicircle so it is a minor arc.
20. (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A).
Explanation :
As per the two-tangent theorem, if two tangent segments are drawn to one circle from the same external point, then they are congruent. So, the assertion is proved to be correct. For the reason, as per the tangent to a circle theorem “A line is tangent to a circle if and only if the line is perpendicular to the radius drawn to the point of tangency.” So, the statement in the reason is also true. But they don’t support or define each other.
Section-B
21. We have, 198 = 2 × 32 × 11
and 144 = 24 × 32
So, LCM(198, 144) = 24 × 32 × 11 = 1584
and HCF (198, 144) = 2 × 32 = 18
Now, LCM(198, 144) × HCF (198, 144) = 1584 × 18 = 28512
Also, Product of 198 and 144 = 28512
Thus, LCM (198, 144) × HCF (198, 144) =
Product of 198 and 144. Hence Proved.
22. We have,
$$\frac{3}{2}x + \frac{5}{3}y -7 = 0 $$
and 9x – 10y – 14 = 0
Here,
$$a_{1}=\frac{3}{2},b_{1}=\frac{5}{3},c_{1}=-7$$
a2 = 9, b2 = – 10, c2 = – 14
Thus,
$$\frac{a_{1}}{a_{2}} =\frac{3}{2×9}=\frac{1}{6},\\\frac{b_{1}}{b_{2}}=\frac{5}{3(\normalsize-10)}=-\frac{1}{6}\\\text{Since\space}\frac{a_{1}}{a_{2}}\neq\frac{b_{1}}{b_{2}}$$
So, given system of equations has unique solution and it is consistent. Ans.
23. (i) Let E be the event of getting a number greater than 4.
So, favourable outcomes of event E = {5, 6}
∴ n(E) = 2
⇒ Total number of outcomes, n(S) = 6
$$\text{So,\space}\text{P(E)}=\frac{n(E)}{n(S)}\\=\frac{2}{6}=\frac{1}{3}$$Ans.
(ii) Let F be the event of getting a number less than or equal to 4.
Then clearly, P(E) + P(F) = 1
⇒ P(F) = 1 – P(E)
$$= 1-\frac{1}{3}=\frac{2}{3}\space\textbf{Ans.}$$
OR
Number of possible outcomes = 100
(i) Let E1 be the event of getting a number divisible by 9 and is a perfect square.
∴ Favourable outcomes = {9, 36, 81}
⇒ n(E1) = 3
$$\therefore\space\text{P(E}_{2})=\frac{3}{100}\space\textbf{Ans.}$$
24. Given A.P. is 3, 8, 13, …, 253.
Here, a = 3, d = 8 – 3 = 5, l = 253
We know that, nth term from last = l – (n – 1)d
∴ 15th term from last = 253 – (15 – 1)5
= 253 – 70
⇒ = 183 Ans.
OR
Multiples of 7 lying between 500 and 900 are 504, 511, 518, 525,..., 896
We observe that the above series is an A.P.
∴ a = 504, d = 7, an = 896
Now, we know that, an = a + (n – 1)d
⇒ 896 = 504 + (n – 1) 7
⇒ 392 = 7(n – 1)
⇒ n – 1 = 56
⇒ n = 57
$$\text{Thus,}\\\text{S}_{57}=\frac{57}{2}[504 + 896]\\\bigg[\because\space\text{S}_{n}=\frac{n}{2}(a + l)\bigg]\\=\frac{57}{2}×1400$$
= 57 × 700 = 39900 Ans.
$$\textbf{25.}\text{Given : A(– 2, 2), B(2, – 4)}\\ \text{and AP =}\frac{3}{7}\text{AB}$$
⇒ 7AP = 3AB = 3AP + 3BP
4AP = 3BP
$$\Rarr\space\frac{\text{AP}}{\text{BP}}=\frac{3}{4}$$
Thus, AP : BP = 3 : 4
Let, the coordinates of P be (x, y)
$$\text{So,\space}x =\frac{3(2) +4(\normalsize-2)}{3+4}\\\Rarr\space x =\frac{6-8}{7}\\\Rarr\space x=\frac{\normalsize-2}{7}\\ y =\frac{3(\normalsize-4) + 4(2)}{3 + 4}\\\Rarr\space y =\frac{-12 + 8}{7}\\\Rarr\space y =\frac{\normalsize-4}{7}$$
Hence, the coordinates of P are
$$\bigg(\frac{\normalsize-2}{7},\frac{\normalsize-4}{7}\bigg).$$
Ans.
Section-C
26. Total number of cards = 52
(i) Let, E be the event that chosen card is a non-ace.
Number of non-ace cards = 52 – 4 = 48
Thus,
$$\text{P(E)}=\frac{\text{Number of non-ace cards}}{\text{Total number of cards}}\\=\frac{48}{52}=\frac{12}{13}\space\textbf{Ans.}$$
(ii) Let, E be the event that chosen card is neither a king nor a jack.
Number of cards in which there is neither a king nor a jack = 52 – 8 = 44
Thus,
$$\text{P(E)}=\frac{\text{Number of cards neither a king nor a jack}}{\text{Total number of cards}}\\=\frac{44}{52}=\frac{11}{13}\space\textbf{Ans.}$$
27. Let us assume, to the contrary that \sqrt{7}\space\text{is rational number.}
Then there exists co-prime positive integers p and q such that
$$\sqrt{7}=\frac{p}{q},q\neq0\\\Rarr p =\sqrt{7}q$$
On squaring both sides,
p2 = 7q2 …(i)
⇒ 7 divides p2
⇒ 7 divides p
So, we have p = 7r, (where r is any integer)
Putting value of p = 7r in equation (i), we have
49r2 = 7q2
⇒ 7r2 = q2
⇒ 7 divides q2 so, 7 divides q.
So, 7 is a common factor of both p and q, which is a contradiction to our assumption.
So our assumption that
$$\sqrt{7}\space\text{is rational is wrong.}\\\text{Hence,}\space\sqrt{7}\space\text{is an irrational number.}$$Hence Proved.
OR
Let HCF be H, and the other number be x
LCM = 4H
Now, the sum of LCM and HCF is 45,
4H + H = 45
5H = 45
H = 9
LCM = 4H = 4 × 9 = 36
We know that,
Product of LCM and HCF = Product of two numbers
9 × 36 = 9 × x
x = 36
Now, the sum of the squares of those numbers is
(9)2 + (36)2 = 81 + 1296
= 1377
28. Let P(x) = 2x2 + 3x + λ
$$\text{Its one zero is}\frac{1}{2}\\\text{So,\space}\text{P}\bigg(\frac{1}{2}\bigg) = 0\\\Rarr\space 2\bigg(\frac{1}{2}\bigg)^{2} + 3\bigg(\frac{1}{2}\bigg) +\lambda = 0\\\Rarr\space\frac{1}{2} + \frac{3}{2} +\lambda = 0\\\frac{4}{2} +\lambda = 0$$
⇒ 2 + λ = 0
⇒ λ = – 2
Let other zero be α. Then,
$$\alpha + \frac{1}{2} =\frac{\normalsize-3}{2}\\\Rarr\space \alpha =\frac{\normalsize-3}{2}-\frac{1}{2}=\normalsize-2$$
Hence, λ = – 2 and the other zero is – 2. Ans.
29. The first number greater than 10 which when divided by 4 leaves a remainder 3 is 11.
So, the next number will be 11 + 4 = 15
The other numbers will be 15 + 4 = 19; 19 + 4 = 23 ......
Thus, A.P. = 11, 15, 19, 23......
The last term of this A.P. will be 299.
We have to find n i.e., number of terms in the A.P.
We know, a + (n – 1)d = an
⇒ 11 + (n – 1)4 = 299
⇒ 7 + 4n = 299
⇒ 4n = 299 – 7
⇒ 4n = 292
$$\Rarr\space n =\frac{292}{4}$$
⇒ n = 73
∴ 73 numbers lie between 10 and 300 which when divided by 4 leaves a remainder 3. Ans.
30. As XP and XQ are two tangents from X, so
XP = XQ ...(i)
Also, AP = AR [Tangents from the external point A] ...(ii)
and BR = BQ [Tangents from the external point B] ...(iii)
Now, XP = XQ [from (i)]
⇒ XA + AP = XB + BQ
⇒ XA + AR = XB + BR [from (ii) and (iii)]
Hence Proved.
31. Given :
$$\text{L.H.S}=\frac{\text{sin}\space\theta}{\text{1 - tan}\space\theta}-\frac{\text{cos}\space\theta}{\text{1 - cot}\space\theta}\\=\frac{\text{sin}}{1 -\frac{\text{sin}\space\theta}{\text{cos}\space\theta}}-\frac{\text{cos}\space\theta}{1 -\frac{\text{cos}\theta}{\text{sin}\space\theta}}\\=\frac{\text{sin}\space\theta}{\frac{\text{cos}\theta -\text{sin}\theta}{\text{cos}\space\theta}}-\frac{\text{cos}\space\theta}{\frac{\text{sin}\theta - cos\theta}{\text{sin}\space\theta}}\\=\frac{\text{sin}\space\theta\text{cos}\space\theta}{\text{cos}\theta -\text{sin}\space\theta}-\\\frac{\text{sin}\theta \text{cos}\theta}{\text{sin}\theta -\text{cos}\space\theta}\\=\frac{\text{sin}\theta\text{cos}\theta}{\text{cos}\theta -\text{sin}\space\theta} +\\\frac{\text{sin}\theta\text{cos}\theta}{\text{cos}\theta-\text{sin}\space\theta}$$
$$=\frac{\text{2 sin}\space\theta\text{cos}\theta}{\text{cos}\space\theta - \text{sin}\space\theta}$$
= R.H.S. Hence Proved.
OR
Let sec θ + tan θ = λ …(i)
We know that sec2 θ – tan2 θ = 1
⇒ (sec θ + tan θ) (sec θ – tan θ) = 1
⇒ λ(sec θ – tan θ) = 1 [from (i)]
$$\Rarr\space\text{sec}\space\theta -\text{tan}\space\theta=\frac{1}{\lambda}\\\text{...(ii)}$$
On adding equations (i) and (ii), we get
$$\text{2 sec}\space\theta =\lambda +\frac{1}{\lambda}\\\Rarr\space 2\bigg(x +\frac{1}{4x}\bigg) =\lambda +\frac{1}{\lambda}\\\bigg[\because\space\text{sec}\space\theta= x +\frac{1}{4x};\text{Given}\bigg]\\\Rarr\space 2x +\frac{1}{2x} =\lambda +\frac{1}{\lambda}\\\frac{4x^{2} + 1}{2x} =\frac{\lambda^{2} + 1}{\lambda}$$
4x2λ + λ = 2xλ2 + 2x
4x2λ – 2xλ2 + λ – 2x = 0
2xλ (2x – λ) –1(2x – λ) = 0
(2x – λ) (2xλ – 1) = 0
$$\lambda = 2x\text{or}\space\lambda=\frac{1}{2x}\\\Rarr\space\text{sec}\space\theta + \text{tan}\space\theta=2x\\\text{or}\frac{1}{2x}\space\textbf{Hence Proved.}$$
Section-D
32. Given, the height of the lighthouse AC is 200 m. Let the distance of the first ship B from the light house be x m and the distance of the second ship D from the lighthouse be y m.
Since, the angle of depression is 60° for the first ship.
In ∆ABC, we have
$$\text{tan\space 60}\degree =\frac{200}{x}\\\Rarr\space \sqrt{3}=\frac{200}{x}\\\Rarr\space x =\frac{200}{\sqrt{3}}\space\text{…(i)}$$
Also, as the angle of depression is 45° with the second ship.
In ∆ACD, we have
$$\text{tan 45}\degree=\frac{200}{y}\\\Rarr\space 1 =\frac{200}{y}\\\Rarr\space\text{y = 200 m}\\\text{...(ii)}$$
Thus, from equations (i) and (ii), the distance between the two ships is
$$\text{x + y}=\frac{200}{\sqrt{3}} + 200\\= 200\bigg(1 + \frac{1}{\sqrt{3}}\bigg)\\= 200\bigg(\frac{\sqrt{3}+1}{\sqrt{3}}\bigg)\\= 200\bigg(\frac{1.73 + 1}{1.73}\bigg)\\=200\bigg(\frac{2.73}{1.73}\bigg)$$
= 200(1.578)
= 315.6 m. Ans.
33. Let XOX’ and YOY’ be the X-axis and Y-axis respectively.
Now, 2x + 3y = 2 ⇒ 2x = 2 – 3y
$$\Rarr\space x=\frac{2-3y}{2}$$
If y = 0, x = 1
If y = 2, x = – 2
If y = – 2, x = 4
Therefore,
| x | -2 | 1 | 4 |
| y | 2 | 0 | -2 |
Thus, plotting the points P(– 2, 2), Q(1, 0) and R(4, – 2) on the graph paper, we get the graph of 2x + 3y = 2, which is represented by PR.
Now, x – 2y = 8 ⇒ 2y = x – 8
$$\Rarr\space y=\frac{x-8}{2}$$
If x = 4, y = – 2
If x = 6, y = – 1
If x = 8, y = 0
Therefore,
| x | 4 | 6 | 8 |
| y | -2 | -1 | 0 |
Thus, plotting the points R(4, – 2), S(6, – 1) and T(8, 0) on the graph paper, we get the graph of x – 2y = 8, which is represented by RT.
Hence, the solution is (4, – 2) as this is the point of intersection of both the lines. Ans.
OR
First line
x + 3y = 6
⇒ x = 6 – 3y
⇒
| x | 6 | 3 | 0 |
| y | 0 | 1 | 2 |
(6, 0), (3, 1), (0, 2)
| x | 6 | 3 | 0 |
| y | 0 | -2 | -4 |
(6, 0), (3, – 2), (0, – 4),
$$\text{Area of triangle =}\\\frac{1}{2}×\text{Base × Corresponding altitude}\\\therefore\space\frac{\text{Area of AOB}}{\text{Area of AOC}}\\=\frac{1/2×\text{OA}×\text{OB}}{1/2×\text{OA}×\text{OC}}\\\Rarr\space\frac{\text{OB}}{\text{OC}}=\frac{2}{4}=\frac{1}{2}$$
∴ Required ratio = 1 : 2 Ans.
34. Let EO || DC meet AD at E.
So, in ∆ADC, EO || DC
$$\text{So,}\space\frac{\text{AO}}{\text{OC}}=\frac{\text{AE}}{\text{ED}}\\\text{[Thales’ theorem] …(i)}\\\text{But\space}\frac{\text{AO}}{\text{OC}}=\frac{\text{BO}}{\text{OD}}$$
[Given] …(ii)
By (i) and (ii),
$$\text{So,\space}\frac{\text{BO}}{\text{OD}}=\frac{\text{AE}}{\text{ED}}$$
$$\text{Hence,}\space\frac{\text{BO}}{\text{OD}} =\frac{\text{AE}}{\text{ED}}\space\text{and BO, OD,}$$
AE and ED are line segments of ∆DAB.
So, EO || AB [By converse of Thales’ theorem]
But EO || DC
Thus, AB || DC
As there are only two parallel sides in this quadrilateral, it is a trapezium.
Hence Proved.
OR
In ∆BMC and ∆EMD, we have
[Vertically opposite angles]
MC = MD [M being the mid-point of CD]
∠BCM = ∠EDM [Alternate angles]
Thus, ∆BMC ≅ ∆EMD [by ASA]
⇒ BC = DE (cpct)
Again, BC = AD [Opposite sides of the parallelogram ABCD]
∴ BC = AD = DE
So, AE = AD + DE = 2BC …(i)
Again, in ∆AEL and ∆CBL, ∠5 = ∠6
[Vertically opposite angles]
∠3 = ∠4 [Alternate angles]
So, ∆AEL ~ ∆CBL
$$\therefore\space\frac{\text{EL}}{\text{BL}}=\frac{\text{AE}}{\text{BC}}=\frac{\text{2BC}}{\text{BC}}=2$$
[From (i)]
Thus, EL = 2BL. Hence Proved.
35.
| Marks | No. of Students (fi) | Cumulative Frequency |
| 0 – 5 | 4 | 4 |
| 5 – 10 | 6 | 10 |
| 10 – 15 | 10 | 20 |
| 15 – 20 | 10 | 30 |
| 20 – 25 | 25 | 55 |
| 25 – 30 | 22 | 77 |
| 30 – 35 | 18 | 95 |
| 35 – 40 | 5 | 100 |
| N = ∑fi = 100 |
Now, N = ∑fi= 100
$$\text{So,\space}\frac{\text{N}}{2} = 50$$
The cumulative frequency just above 50 is 55.
Hence, the median class is 20–25.
$$\text{So, l = 20, h= 5, c = 30, f = 25}\\\text{and}\space \frac{\text{N}}{2} = 50\\\text{Median =}\space l + \frac{\frac{\text{N}}{2}-c}{f}×h\\ = 20 +\frac{50-30}{25}×5\\= 20 +\frac{100}{25} $$
= 20 + 4 = 24 Ans.
Section-E
36. (i) Distance between C(2, – 2) and D(– 2, – 3) is
$$\text{CD =}\sqrt{(-2-2)^{2} + (-3+2)^{2}}\\\text{CD} =\sqrt{16 + 1}\\\text{CD} =\sqrt{17}\space \textbf{Ans.}$$
(ii) DA + AH = DH Ans.
OR
Given that P divides the line segment joining B(1, 4) and C(2, – 2) in the ratio k : 1 so the
coordinates of P are
$$\bigg(\frac{2k+1}{k+1},\frac{-2k+4}{k+1}\bigg)\\\text{P}\bigg(\frac{2k+1}{k+1},\frac{-2k+4}{k+1}\bigg)\\\text{lies on the line segment x – y = – 8}\\\Rarr\space\bigg(\frac{2k+1}{k+1}\bigg)-\bigg(\frac{-2k+4}{k+1}\bigg) =-8$$
⇒ 2k + 1 + 2k – 4 = – 8k – 8
⇒ 12k = – 5
$$\Rarr\space k =\frac{\normalsize-5}{12}\space\textbf{Ans.}$$
(iii) Given point F(– 4, 3) and E(12, 3).
Coordinates of mid-point are
$$\bigg(\frac{-4 + 12}{2},\frac{3+3}{2}\bigg)$$
$$\Rarr\space (4,3)\space\textbf{Ans.}$$
37. (i) Since, AB and AC are tangents to a given circle. Then, lengths of tangents from an external point to a circle are equal in length
∴ AB = AC = x Ans.
(ii) Since, AB ⊥ OB
Then, by Pythagoras theorem in ∆AOB,
AB2 = AO2 + OB2
= 242 + 72
= 576 + 49
AB2 = 625
AB = 25 m Ans.
OR
Total distance available = Length of 2 tangents
= AB + AC
= 2AB (∵ AB = AC)
$$= 2\sqrt{\text{OA}^{2} + \text{OB}^{2}}\\= 2\sqrt{24^{2} + 7^{2}}$$
= 2 × 25 = 50 m Ans.
(iii) Total length of line segments= 50 m = 50 × 100 cm
∴ Number of trees can be planted =
$$\frac{50×100}{20} = 250\space\text{trees}\\\textbf{Ans.}$$
38. (i) Curved surface area of hemisphere = 2πr2
$$= 2×\frac{22}{7}×\bigg(\frac{5.6}{2}\bigg)^{2}\\= 2×\frac{22}{7}×2.8×2.8$$
= 49.28 cm2
= 50 cm2 (Approx) Ans.
(ii) C.S.A. of cone = πrl
$$=\pi rl\sqrt{h^{2} + r^{2}}$$
Here, r = 2.8 cm, h = 7 – 2.8 = 4.2 m
$$\therefore\space\text{C.S.A}=\\\frac{22}{7}×2.8\sqrt{(4.2)^{2} + (2.8)^{2}}\\= 22×0.4×\sqrt{17.64 + 7.84}\\= 8.8×\sqrt{25.48}$$
= 44.42 cm2
= 45 cm2 (Approx) Ans.
(iii) Height of cone = 7 – 2.8 = 4.2 cm
Height of hemisphere = 2.8 cm (radius of cone)
$$\therefore\space\text{Ratio =}\frac{4.2}{2.8}=\frac{6}{4} =\frac{3}{2}$$ Ans.
OR
Total area to be painted = 45 + 50 = 95 cm2
Cost = 95 × 0.05
= ₹ 4.75 Ans.
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