Oswal Practice Papers CBSE Class 10 Mathematics Solutions (Practice Paper - 10)
Section-A
1. (b) 4
Explanation :
Prime factors of 196 = 22 × 72
So, sum of exponents = 2 + 2 = 4.
2. (c) Irrational number
Explanation :
Product or sum of rational and irrational number will always be a irrational number.
$$\text{e.g, 2 +}\sqrt{3}\space\\\text{is irrational number}\\5\sqrt{7}\\\space\text{is also a irrational number. }$$
3. (b) 4 : 7
Explanation :
Let, K : 1 be the ratio in which X-axis divides the line joining (6, 4) and (1, – 7).
Let, P be the point on X-axis that divides the line. Then
$$\text{P(x,y)} =\bigg[\frac{\text{K+6}}{\text{K+1}},\frac{\text{-7K + 4}}{\text{K + 1}}\bigg]\\\text{As P (x, y) lies on X-axis}\\\text{P(x,0)} =\bigg(\frac{\text{K + 6}}{\text{K + 1}},\frac{-7K + 4}{\text{K + 1}}\bigg)\\\lbrack\because\space \text{x-coordinate will be x, 0}\rbrack\\\Rarr\space \frac{-7K + 4}{K+1} = 0$$
⇒ – 7K + 4 = 0
⇒ 7K = 4
$$\Rarr\space\text{K =}\space \frac{4}{7}$$
So, the required ratio be
$$K : 1 =\frac{4}{7}:1 = 4:7$$
4. (b) x2 – 3x – 2
Explanation :
Given, sum of zeroes, S = 3
Product of zeroes, P = – 2
Required polynomial is given by
f(x) = x2 – (Sum of zeroes)x + Product of zeroes
⇒ f(x) = x2 – 3x – 2
5. (a) – 2
Explanation :
As, a, a –2, 3a are AP.
Then, a – 2 – a = 3a – (a – 2)
⇒ – 2 = 2a + 2
⇒ 2a = – 2 – 2
⇒ 2a = – 4
⇒ a = – 2
6. (c) 6
Explanation :
Given system of equation is
2x + 3y – 5 = 0
4x + ky – 10 = 0
Here, a1 = 2, b1 = 3, c1 = – 5
a2 = 4, b2 = k, c2 = – 10
For infinite solutions,
$$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\\\Rarr\space\frac{2}{4}=\frac{3}{k}=\frac{5}{10}\\\Rarr\space\frac{2}{4} =\frac{3}{k}$$
2k = 3 × 4
k = 6
Hence, for k = 6 given system of equation has infinitely many solutions.
7. (d) 12
Explanation :
Let x be the natural number than according to question,
x2 – 84 = 3 (x + 8)
⇒ x2 – 84 = 3x + 24
⇒ x2 – 3x – 108 = 0
⇒ x2 – 12x + 9x – 108 = 0
⇒ x(x – 12) + 9(x – 12) = 0
⇒ (x – 12) (x + 9) = 0
⇒ x – 12 = 0 or x + 9 = 0
⇒ x = 12 or x = – 9
Since – 9 is not a natural number. So, the required number is 12.
$$\textbf{8.\space}(a)\space\frac{1}{7}$$
Explanation :
A non-leap year has 365 days and so it has 52 weeks and 1 day.
This 1 day can be Sunday or Monday or Tuesday or Wednesday or Thursday or Fridays or Saturday. Thus, total outcomes = 7
Favourable outcome = 1
$$\text{Probability =}\\\frac{\text{Number of favourable outcomes}}{\text{Number of total outcomes}}=\frac{1}{7}$$
9. (b) 3
Explanation :
Given: Σfi= 15, Σfixi = 3p + 36 and Mean = 3
$$\text{We know,}\space\text{Mean =}\frac{\Sigma f_{i}x_{i}}{\Sigma f_{i}}\\\Rarr\space 3 =\frac{3p + 36}{15}$$
⇒ 45 = 3p + 36
⇒ 3p = 45 – 36
⇒ 3p = 9
∴ p = 3.
10. (b) 0.95
Explanation :
P(E) = 0.05
P(not E) = 1 – 0.05 = 0.95
The probability of ‘not E’ is 0.95.
$$\textbf{11.}\space\text{(c)}\space\frac{1}{6}$$
Explanation :
Total number of possible outcomes =
{(1, 1), (1, 2), (1, 3), (1,4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} = 36
Thus, the favourable outcomes are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).
Hence, the total number of favourable outcomes = 6
$$\Rarr\space\text{P(E)} =\frac{6}{36}=\frac{1}{6}$$
12. (a) 1 : 1
Explanation :
Given Let K : 1 be the ratio in which X-axis divides the line joining A (1, – 5) and
B (– 4, 5).
Let P be the point on X-axis that divides the line AB, then
$$\text{P(x,y)} =\bigg(\frac{-4K +1}{\text{K+1}},\frac{5\text{K-5}}{\text{K+1}}\bigg)$$
As P(x, y) lies on X-axis
$$\text{P(x, 0)} =\bigg(\frac{-4\text{K +1}}{\text{K +1}},\frac{5K - 5}{\text{K + 1}}\bigg)\\\Rarr\space\frac{\text{5K - 5}}{\text{K +1}} = 0$$
5 = 5K
K = 1
∴ Point (x, 0) on X-axis divides the line segment in 1 : 1 ratio.
13. (c) co-primes
Explanation :
Two numbers are said to be co-primes if they have only one common factor, namely 1.
14. F(c) – 61
Explanation :
Given A.P. is 19, 14, 9, .......
Here First term, a = 19
Common difference, d = 14 – 19 = – 5
We know that, nth term of an A.P. is given as,
an = a + (n – 1)d
∴ a17 = a + (17 – 1)d
= a + 16d
= 19 + 16(– 5) = 19 – 80
⇒ a17 = – 61.
15. (a) 2
Explanation :
Given quadratic equation is,
6x2 – x – k = 0
$$\text{sin},\frac{2}{3}\space\text{is one of the root}\\\text{of given quadratic equation}\\\therefore\space 6\bigg(\frac{2}{3}\bigg)^{2}-\frac{2}{3}-k = 0\\\Rarr\space 6×\frac{4}{9}-\frac{2}{3}-k =0\\\Rarr\space \frac{8}{3}-\frac{2}{3}-k=0\\\Rarr\space\frac{6}{3}-k =0$$
⇒ 2 – k = 0
⇒ k = 2
16. (a) 3
Explanation :
Given integers are 81 and 237
Prime factors of 81 = 3 × 3 × 3 × 3
Prime factors of 237 = 3 × 79
Hence, HCF (81, 237) = 3.
17. (b) 8.25 cm
Explanation :
Since, PS is the internal bisector of ∠P and it meets QR at S.
$$\therefore\space\frac{\text{PQ}}{\text{QS}}=\frac{\text{PR}}{\text{SR}}\\\Rarr\space \frac{15}{3 +x} =\frac{7}{x-3}$$
⇒ 7(3 + x) = 15(x – 3)
⇒ 21 + 7x = 15x – 45
⇒ 15x – 7x = 45 + 21
⇒ 8 = 66
⇒ 4x = 33
⇒ x = 8.25 cm
18. (d) 90°
Explanation :
Let PO meet AB at O such that PO is a tangent.
Thus, AO = PO = OB and ∠AOP = ∠BOP = 90°.
As DAOP and DBOP are right-angled isosceles triangles.
So, ∠OAP = ∠OPA = ∠OBP = ∠OPB = 45°
Thus, ∠APB = ∠APO + ∠OPB
= 45° + 45° = 90°
19. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
Explanation :
Let’s construct a circle whose radius is 3 cm and distance of a point from the centre of a circle is 5 cm
As per the above diagram, by joining the distant point A, we get DAOB and AB is the tangent to the given circle.
∵ tangent is perpendicular to the radius.
∴ using the Pythagoras theorem,
(Hypotenuse)2 = (Base)2 + (Perpendicular height)2
⇒ (OA)2 = (AB)2 + (OB)2
⇒ (AB)2 = (OA)2 – (OB)2
= (5)2 – (3)2
= 25 – 9 = 16
$$\Rarr\space\text{AB} =\sqrt{16}\space = 4\space\text{cm}$$
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. 17 × 5 × 11 × 3 × 2 + 2 × 11 = 17 × 5 × 3 × 22 + 22
= 22 (17 × 5 × 3 + 1)
= 22 (255 + 1)
= 2 × 11 × 256 ...(i)
Now equation (i) is divisible by 2, 11 and 256, which means it has more than 2 prime factors.
∴ (17 × 5 × 11 × 3 × 2 + 2 × 11) is a composite number. Ans.
22. Given quadratic equation is,
$$x^{2} + 2\sqrt{2x}-6 =0\\\Rarr\space x^{2} + 3\sqrt{2}x -\sqrt{2}x =0\\\Rarr\space x(x + 3\sqrt{2})-\sqrt{2}(x + 3\sqrt{2}) = 0\\\Rarr\space(x + 3\sqrt{2})(x -\sqrt{2}) = 0\\\Rarr\space x + 3\sqrt{2} =0\\\text{or} \space x-\sqrt{2} = 0\\\Rarr\space x=-3\sqrt{2}\space\text{or}\space x =\sqrt{2}\space\textbf{Ans.}$$
OR
Given quadratic equation is,
2x2 + ax – a2 = 0
⇒ 2x2 + 2ax – ax – a2 = 0
⇒ 2x (x + a) – a (x + a) = 0
⇒ (2x – a) (x + a) = 0
$$\Rarr\space x =\frac{a}{2}, x =-a\space\textbf{Ans.}$$
23. Given, the distance between the points (3, 0) and (0, y) is 5 units.
$$\text{Distance =}\sqrt{(x_{2} -x_{1})^{2} +(y_{2} -y_{1})^{2}}\\\text{Thus,}\\5=\sqrt{(3-0)^{2} + (0-y)^{2}}\\=\sqrt{9 + y^{2}}$$
⇒ 9 + y2 = 25 ⇒ y2 = 16
⇒ y2 = 42 ⇒ y = ± 4
But as y is positive so the value of y = 4. Ans.
24. Let the coordinates of point A be (x, y) and point O (2, – 3)be the centre, then
By mid-point formula,
$$\frac{x +1}{2} = 2\text{and}\space\frac{y +4}{2} = -3$$
or x = 4 – 1 and y = – 6 – 4
⇒ x = 3 and y = – 10
∴ The coordinates of point A are (3, – 10) Ans.
25. Savita may have any one of the 365 days of the year as her birthday. Similarly, Hamida may have any one of the 365 days of the year as her birthday.
Total number of ways in which Savita and Hamida may have their birthday = 365 × 365
(i) Savita and Hamida may have same birthday on any one of 365 days of the year.
∴ Number of ways in which Savita and Hamida will have birthday on same day = 365
Probability that Savita and Hamida will have birthday on same day =
$$\frac{365}{365×365} =\frac{1}{365}.\space\textbf{Ans.}$$
(ii) Probability that Savita and Hamida will have different birthday
= 1 – Probability that Savita and Hamida will have birthday on same day
$$= 1 -\frac{1}{365} =\frac{364}{365}$$
OR
We have,
Sample space = {11, 12, 13, ......., 123}
⇒ n(S) = 113
(i) Let E1 be the event that the card drawn is a square number.
So, E1 = { 16, 25, 36, ...., 121}
⇒ n(E1) = 8
$$\therefore\space\text{P(E}_{1}) =\frac{8}{113}\space\textbf{Ans.}$$
(ii) Let E2 be the event that card drawn is a multiple of 8.
So, E2 = {16, 24, 32, ......, 120}
⇒ n(E2) = 14
$$\therefore\space\text{P(E}_{2}) =\frac{14}{113}\space\textbf{Ans.}$$
Section-C
26. Let ACB be the given arc subtending an angle of 60° at the centre.
Here, r = 14 cm and θ = 60°.
Area of the minor segment ACBA = (Area of the sector OACBO) – (Area of ∆OAB)
$$=\frac{\pi r^{2}\theta}{360\degree}-\frac{1}{2}r^{2}\text{sin}\space\theta\\=\frac{22}{7}×14×14×\frac{60\degree}{360\degree}-\\\frac{1}{2}×14×14×\text{sin\space 60}\degree\\=\frac{308}{3}-7×14×\frac{\sqrt{3}}{2}-\\49\sqrt{3} = 17.89\space\text{cm}^{2}$$
Area of the major segment BDAB = Area of circle – Area of minor segment ACBA
= πr2 – 17.89
$$=\frac{22}{7}×14×14-17.89$$
= 616 – 17.89 = 598.11 ≈ 598 cm2. Ans.
$$\textbf{27.\space}(i)\space\frac{\text{AO}}{\text{DO}}=\frac{16}{9}\text{and}\\\frac{\text{BO}}{\text{CO}} =\frac{9}{5}\\\because\space\frac{\text{AO}}{\text{DO}}\neq\frac{\text{BO}}{\text{CO}}$$
∴ Given triangles are not similar. Ans.
(ii) In ∆PQR, ∠P + ∠Q + ∠R = 180° [Angle-sum property of a triangle]
⇒ 45° + 78° + ∠R = 180°
⇒ ∠R = 180° – 45° – 78° = 57°
In ∆LMN, ∠L + ∠M + ∠N = 180° [Angle-sum property of a triangle]
⇒ 57° + 45° + ∠N = 180°
⇒ ∠N = 180° – 57° – 45°
= 78°
Now, in ∆PQR and ∆LMN ∠P = ∠M [Each 45°]
∠Q = ∠N [Each 78°]
∠R = ∠L [Each 57°]
By AAA similarity criterion ∆PQR ~ ∆MNL Ans.
OR
(i) In ∆AGF and ∆DBG, we have
∠GAF = ∠BDG = 90°
and ∠AGF = ∠GBD [Corresponding angles between parallel lines GF and BC with AB being the transversal as GDEF is a square]
Thus, by AA similarity criterion,
∆AGF ~ ∆DBG. Hence proved.
(ii) In ∆AGF and ∆EFC, we have
∠GAF = ∠FEC = 90°
and ∠AFG = ∠FCE [Corresponding angles between parallel lines GF and BC with AC being the transversal]
Thus, ∆AGF ~ ∆EFC Hence proved.
(iii) From (i) and (ii), we have ∆AGF ~ ∆DBG
and ∆AGF ~ ∆EFC
Thus ∆DBG ~ ∆EFC Hence proved.
(iv) From (iii), we have
∆DBG ~ ∆EFC
$$\text{Thus,}\\\frac{\text{BD}}{\text{FE}} =\frac{\text{DG}}{\text{EC}}\\\text{[C.P.C.T.]}\\\Rarr\space\frac{\text{BD}}{\text{DE}} = \frac{\text{DE}}{\text{EC}}$$
[DE = DG = FE = GF as they are the sides of a square]
⇒ DE2 = BD × EC. Hence proved.
28. Let ABCD be a parallelogram such that its sides touch a circle with centre O.
Now, we know that tangents to a circle from an exterior point are equal in length.
∴ AP = AS …(i)
BP = BQ …(ii)
CR = CQ …(iii)
DR = DS …(iv)
Adding equations (i), (ii), (iii) and (iv), we get
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
⇒ AB + AB = BC + BC
(Since ABCD is a parallelogram ∴ AB = CD, BC = DA)
⇒ 2AB = 2BC
⇒ AB = BC
Thus, AB = BC = CD = AD
Hence, ABCD is a rhombus. Hence Proved.
29. L.H.S. = 2 sec2 θ – sec4 θ – 2 cosec2 θ + cosec4 θ
= 2 sec2 θ – (sec2 θ)2 – 2 cosec2 θ + (cosec2 θ)2
= 2(1 + tan2 θ) – (1 + tan2 θ)2 – 2(1 + cot2 θ) + (1 + cot2 θ)2
= 2 + 2 tan2 θ – 1 – tan4 θ – 2 tan2 θ – 2 – 2 cot2 θ + 1 + cot4 θ + 2 cot2 θ
= cot4 θ – tan4 θ
= R.H.S. Hence Proved.
OR
$$\text{L.H.S}=\\\frac{\text{cosec A}}{\text{cosec A}-1} + \frac{\text{cosec A}}{\text{cosec A + 1}}\\=\\\frac{\text{cosec A(\text{cosec A + 1}) + \text{cosec A}(\text{cosec A -1})}}{(\text{cosec A -1})(\text{cosec A + 1})}\\=\\\frac{\text{cosec A(\text{cosec A + 1 + cosec A - 1})}}{\text{cosec}^{2}A -1}\\=\\\frac{\text{2 cosec}^{2}A}{\text{1 + cot}^{2}A -1}\\=\frac{\text{2 cosec}^{2}A}{\text{cot}^{2}A}$$
[∵ cosec2 θ – cot2 θ = 1]
= 2 cosec2 A tan2 A
= 2(1 + cot2 A) tan2 A
= 2 tan2 A + 2 tan2 A cot2A
= 2 + 2 tan2 A = 2(1 + tan2 A)
= 2 sec2 A
= R.H.S. Hence Proved.
30. Let the ten’s digit be x and the unit’s digit be y.
Thus, the given number is (10x + y) and its reverse is (10y + x).
So, xy = 20
$$\Rarr\space x =\frac{20}{y}\space\text{...(i)}$$
and 10x + y + 9 = 10y + x
⇒ 9x – 9y = – 9
⇒ x – y = – 1
$$\Rarr\space\frac{20}{y}-y =-1$$
⇒ 20 – y2 = – y
⇒ y2 – y – 20 = 0
⇒ y2 – 5y + 4y – 20 = 0
⇒ y(y – 5) + 4(y – 5) = 0
⇒ (y – 5)(y + 4) = 0
⇒ y = 5, – 4
$$\text{So,\space}x =\frac{20}{5},-\frac{20}{4}$$
⇒ x = 4, – 5
Thus, the number is 45 or – 54. Ans.
31. Length of each side of the cube = 7 cm
∴ Largest possible diameter of the sphere = 7 cm
∴ Radius of the sphere = 3.5 cm
Thus, Volume of cube = (7)3 cm3 = 343 cm3
and Volume of sphere =
$$\frac{4}{3}×\frac{22}{7}(3.5)^{3}\space\text{cm}^{3}\\=\frac{539}{3}\text{cm}^{3}$$
$$=\frac{539}{3}\text{cm}^{3}\\\therefore\space\text{Remaining volume of wood = }\\\bigg(343 -\frac{539}{3}\bigg)\text{cm}^{3}\\=\frac{\text{1029 - 539}}{3}\text{cm}^{3}$$
= 163.33 cm3. Ans.
Section-D
32. Given : a (first term) = a,
d = b – a, an = c
By the formula of nth term, a + (n – 1)d = c
(n – 1)(b – a) = c – a
$$(n-1) =\frac{\text{c-a}}{\text{b-a}}\\n =\frac{\text{(c-a)}}{\text{(b-a)}} +1\\=\frac{\text{c - a + b - a}}{\text{b - a}}\\n = \frac{\text{b +c-2a}}{b - a}\space\text{...(i)}$$
Now, find the sum of n terms.
$$\text{S}_{n}=\frac{n}{2}[2a + (n-1)d]\\=\frac{(b + c - 2a)}{2(b-a)}\\\bigg[2a +\begin{Bmatrix}\frac{b +c - 2a}{b-a}-1\end{Bmatrix}(b-a)\bigg]\\=\frac{(b + c - 2a)}{2(b-a)}\\\bigg[2a +\begin{Bmatrix}\frac{\text{b + c - 2a - b +a}}{\text{b-a}}\end{Bmatrix}(b-a)\bigg]\\=\frac{(b + c - 2a)}{2(b-a)}\\\lbrack2a + c - a\rbrack\\=\frac{(b + c -2a)}{2(b-a)}(a + c)$$
Hence Proved.
OR
Let a be the first term and d be the common difference of an AP.
a11 : a18 = 2 : 3
$$\Rarr\space\frac{\text{a + 10d}}{\text{a + 17d}}=\frac{2}{3}$$
⇒ 3a + 30d = 2a + 34d
⇒ 3a – 2a = 34d – 30d
⇒ a = 4d
To find
$$\frac{a_{5}}{a_{21}}=\frac{a + 4d}{a +20 d}\\=\frac{4d +4d}{4d + 20d}\\=\frac{8d}{24 d}=\frac{1}{3}$$
Now, the ratio of
a5 : a21 = 1 : 3
Now, the ratio of sum
$$\frac{\text{S}_{5}}{\text{S}_{21}}=\frac{\frac{5}{2}[2a +(5-1)d]}{\frac{21}{2}[2a + (21-1)d]}\\\Rarr\space\frac{\text{S}_{5}}{\text{S}_{21}}=\frac{5}{21}\frac{[2a + 4d]}{[2a + 20d]}\\=\frac{5}{21}\frac{[2(4d) + 4d]}{[2(4d) + 20d]}\\\text{[from (i)]}\\=\frac{5}{21}\frac{[\text{8d + 4d}]}{[8d + 20 d]}\\=\frac{5×12 d}{21×28 d}\\=\frac{5}{7×7}=\frac{5}{49}=5:49$$
Hence, S5 : S21 = 5 : 49. Ans.
33. Given, a circle with centre O touches the opposite sides of a quadrilateral at points P, Q, R and S.
In order to prove the above, the following need to be satisfied :
∠AOB + ∠COD = 180° and ∠AOD + ∠BOC = 180°
Join OS, OP, OQ and OR.
In DOQC and DOCR, QC = CR [Tangents from external point C]
OC = OC [Common]
OQ = OR [Radii of circle]
Thus DOQC @ DORC [by S.S.S. congruency]
Hence ∠QOC = ∠COR [Corresponding angles of congruent triangles]
Similarly, ∠ROB = ∠SOB, ∠SOA = ∠POA and ∠POD = ∠QOD
and ∠POD = ∠QOD
Now ∠QOC + ∠COR + ∠ROB + ∠SOB + ∠SOA + ∠POA + ∠POD + ∠QOD = 360° (∵ O is point)
⇒ 2(∠QOC + ∠ROB + ∠SOA + ∠POD) = 360°
and 2(∠COR + ∠SOB + ∠POA + ∠QOD) = 360°
⇒ ∠QOC + ∠ROB + ∠SOA + ∠POD = 180°
and ∠COR + ∠SOB + ∠POA + ∠QOD = 180°
⇒ ∠QOC + ∠SOB + ∠SOA + ∠QOD = 180°
(∵ ∠ROB = ∠SOB, ∠POD = ∠QOD)
and ∠COR + ∠ROB + ∠POA + ∠POD = 180°
⇒ ∠QOC + ∠QOD + ∠SOA + ∠SOB = 180°
and ∠COR + ∠ROB + ∠POA + ∠POD = 180°
⇒ ∠AOB + ∠COD = 180°
and ∠AOD + ∠BOC = 180° Hence Proved.
34. Given : sec θ + tan θ = p
Now,
$$\text{L.H.S} =\frac{\text{p}^{2}-1}{\text{p}^{2}+1}\\=\frac{(\text{sec}\space\theta +\text{tan}\space\theta)^{2}-1}{(\text{sec}\space\theta + \text{tan}\theta)^{2} +1}\\=\\\frac{\text{sec}^{2}\theta +\text{tan}^{2}\theta +\text{2 sec}\theta\space\text{tan}\theta -1}{\text{sec}^{2}\theta +\text{tan}^{2}\theta +2\text{sec}\theta\space\text{tan}\space\theta +1}\\=\\\frac{\text{(sec}^{2}\theta-1) +\text{tan}^{2}\theta + 2\text{sec}\theta\text{tan}\theta}{\text{sec}^{2}\theta +(\text{tan}^{2}\theta +1) + 2\text{sec}\theta\text{tan}\theta}\\=\frac{\text{tan}^{2}\theta + \text{tan}^{2}\theta + \text{2 sec}\theta\text{tan}\theta}{\text{sec}^{2}\theta + \text{sec}^{2}\theta + \text{2 sec}\theta\text{tan}\theta}$$
(∵ 1 + tan2 θ = sec2 θ)
$$=\frac{\text{2 tan}^{2}\theta + 2\text{sec}\theta\text{tan}\theta}{\text{2 sec}^{2}\theta + \text{2 sec}\theta\text{tan}\theta}\\=\frac{2\text{tan}\theta(\text{tan}\space\theta + \text{sec}\space\theta)}{2\text{sec}\theta(\text{tan}\space\theta + \text{sec}\theta)}\\=\frac{\text{tan}\space\theta}{\text{sec}\space\theta}\\=\frac{\text{sin}\theta}{\text{cos}\theta}×\frac{\text{cos}\theta}{1} =\text{sin}\space\theta$$
= R.H.S. Hence Proved.
35.
| Class Interval | Frequency (fi) | Cumulative Frequency (c.f.) |
Class Marks (xi) | (fi × xi) |
| 0 – 10 | 5 | 5 | 5 | 25 |
| 10 – 20 | 10 | 15 | 15 | 150 |
| 20 – 30 | 18 | 33 | 25 | 450 |
| 30 – 40 | 30 | 63 | 35 | 1050 |
| 40 – 50 | 20 | 83 | 45 | 900 |
| 50 – 60 | 12 | 95 | 55 | 660 |
| 60 – 70 | 5 | 100 | 65 | 325 |
| N = Σfi = 100 | Σ(fi × xi) = 3560 |
$$\text{Mean} =\frac{\Sigma(f_{i}×x_{i})}{\Sigma f_{i}}\\=\frac{3560}{100}=35.6\\\textbf{Ans.}$$
Now, N = Σfi = 100
$$\text{So,\space}\frac{\text{N}}{2} = 50$$
The cumulative frequency just above 50 is 63.
Hence, the median class is 30 – 40.
So, l = 30, h = 10, f = 30,
$$\text{F = 33}\space\text{and}\space\frac{\text{N}}{2}= 50\\\text{Median = }l +\frac{\frac{\text{N}}{2}-\text{F}}{f}×h\\= 30 +\frac{50-33}{30}×10$$
= 30 + 5.67
= 35.67 Ans.
We know, Mode = 3 Median – 2 Mean
⇒ Mode = 3(35.67) – 2(35.6)
= 107.01 – 71.2
= 35.81. Ans.
OR
As this series is an inclusive one, we can make it exclusive by adding 0.5 to the upper limit and subtracting 0.5 from the lower limit of each class interval. Thus, we have
| Class Interval | Frequency (fi) | Mid-value (xi) | $$u_{i}=\frac{x_{i}- A}{h}$$ h = upper limit – lower limit |
fi × ui |
| 24.5-29.5 | 14 | 27 | – 3 | – 42 |
| 29.5-34.5 | 22 | 32 | – 2 | – 44 |
| 34.5-39.5 | 16 | 37 | – 1 | – 16 |
| 39.5-44.5 | 6 | 42 = A | 0 | 0 |
| 44.5-49.5 | 5 | 47 | 1 | 5 |
| 49.5-54.5 | 3 | 52 | 2 | 6 |
| 54.5-59.5 | 4 | 57 | 3 | 12 |
| Σfi = 70 | Σ(fi × ui) = – 79 |
Thus, A = 42, h = 5, Σfi= 70, Σ(fi × ui) = – 79
$$\text{Mean} = \text{A} + \begin{Bmatrix}h×\frac{\Sigma(f_{i}×u_{i})}{\Sigma f_{i}}\end{Bmatrix}\\= 42 + \begin{Bmatrix}5×\frac{(-79)}{70}\end{Bmatrix}$$
= 42 – 5.64
= 36.36. Ans.
Section-E
36. (i) Given, height of building =
$$\sqrt{3}\space\text{times}×\text{Length of shadow (l)}\\\text{or}\space h =\sqrt{3}\space l\\\text{so,\space tan}\space\theta=\frac{\sqrt{3}l}{l}\\\text{or}\space\text{tan}\space\theta =\sqrt{3}\\\therefore\space\theta = 60\degree\space\\(\because \text{tan 60\degree}=\sqrt{3})\space\textbf{Ans.}\\\text{(ii) Here, tan 60\degree}=\\\frac{\text{Height of building (h)}}{\text{Length of the shadow}}\\\sqrt{3}=\frac{h}{160}\\\text{So,\space} h = 160\sqrt{3}\space m\space\textbf{Ans.}$$
$$\text{(iii)\space Here}\\\text{tan}\space\theta =\frac{\text{Height of building}}{\text{Length of shadow}}\\\Rarr\space\text{tan}\space\theta =\frac{80\space m}{120\space m}\\\Rarr\space\text{tan}\space\theta=\frac{2}{3}\\\text{or}\space\theta =\text{tan}^{\normalsize-1}\bigg(\frac{2}{3}\bigg).\\\textbf{Ans.}$$
OR
Here
$$\text{tan}\space\theta=\frac{\text{Height of building}}{\text{Length of shadow}}\\\Rarr\space \text{tan}\space\theta=\frac{160\sqrt{3}m}{80m}\\\because\space\theta = \text{tan}^{\normalsize-1}(2\sqrt{3})$$
Ans.
37. (i) Length of outer frame = (11 + 2x) cm
Breadth of outer frame = (6 + 2x) cm Ans.
(ii) Here, Area of frame = Area of outer rectangle – Area of Inner rectangle
= (6 + 2x) (11 + 2x) – 6 × 11
= 66 + 22x + 12x + 4x2 – 66
= 34x + 4x2
⇒ 34x + 4x2 = 28
⇒ 2x2 + 17x – 14 = 0 Ans.
(iii) Here, 9x2 + 6kx + 4 = 0
For equal roots, D = b2 – 4ac = 0
= (6k)2 – 4 × 9 × 4 = 0
⇒ 36k2 – 144 = 0
⇒ k2 = 4
⇒ k = + 2 Ans.
OR
On the basis of equation obtained in part (ii),
we get
2x2 + 17x – 14 = 0
$$\therefore\space x =\frac{-b\pm\sqrt{b^{2}-4ac}}{\text{2a}}\\=\frac{-17\pm\sqrt{289 + 112}}{2×2}\\=\frac{-17\pm\sqrt{401}}{4}\\=\frac{-17\pm20.02}{4}$$
= 0.8 or – 9.3
≈ 1 cm Ans.
38. (i) In resulting cuboid, length gets doubled, breadth and height remains same
Then l = 8 + 8 = 16 cm
b = 8 cm, h = 8 cm Ans.
(ii) Total surface area of cuboid = 2 (lb + bh + hl)
= 2 (16 × 8 + 8 × 8 + 16 × 8)
= 2 (128 + 64 + 128)
= 2 × 320 = 640 cm2 Ans.
(iii) Dimension of cuboid = 16 cm, 8 cm, 8 cm
Volume of cuboid = lbh
= 16 × 8 × 8 = 1024 cm3 Ans.
OR
T.S.A. of paper used = 640 cm2
Rate of paper = ₹0.05/cm2
∴ Cost of paper = ₹640 × 0.05
$$= \frac{640×5}{100} = ₹32\space\textbf{Ans.} $$
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