Vardaan Learning Institute
Mock Board Paper Solutions 2025-26
Set - 1 (Solutions)
MATHEMATICS (STANDARD)
Class - X
Time: 3 Hours
Max. Marks: 80
SECTION A (Solutions)
1.
(b) 3024
\(LCM(a, b) \times HCF(a, b) = a \times b \Rightarrow LCM \times 6 = 336 \times 54 \Rightarrow LCM = \frac{336 \times 54}{6} = 3024\).
\(LCM(a, b) \times HCF(a, b) = a \times b \Rightarrow LCM \times 6 = 336 \times 54 \Rightarrow LCM = \frac{336 \times 54}{6} = 3024\).
2.
(a) \(\pm 0.2\)
\(x^2 - 0.04 = 0 \Rightarrow x^2 = 0.04 \Rightarrow x = \sqrt{0.04} = \pm 0.2\).
\(x^2 - 0.04 = 0 \Rightarrow x^2 = 0.04 \Rightarrow x = \sqrt{0.04} = \pm 0.2\).
3.
(a) 10
For no solution (parallel lines): \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\).
\(\frac{1}{5} = \frac{2}{k} \Rightarrow k = 10\).
For no solution (parallel lines): \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\).
\(\frac{1}{5} = \frac{2}{k} \Rightarrow k = 10\).
4.
(b) \(\pm 4\)
Distance = \(\sqrt{(4-1)^2 + (p-0)^2} = 5 \Rightarrow \sqrt{9 + p^2} = 5\). Squaring both sides, \(9 + p^2 = 25 \Rightarrow p^2 = 16 \Rightarrow p = \pm 4\).
Distance = \(\sqrt{(4-1)^2 + (p-0)^2} = 5 \Rightarrow \sqrt{9 + p^2} = 5\). Squaring both sides, \(9 + p^2 = 25 \Rightarrow p^2 = 16 \Rightarrow p = \pm 4\).
5.
(b) 4
By BPT: \(\frac{AD}{DB} = \frac{AE}{EC} \Rightarrow \frac{x}{x-2} = \frac{x+2}{x-1}\).
\(x(x-1) = (x+2)(x-2) \Rightarrow x^2 - x = x^2 - 4 \Rightarrow -x = -4 \Rightarrow x = 4\).
By BPT: \(\frac{AD}{DB} = \frac{AE}{EC} \Rightarrow \frac{x}{x-2} = \frac{x+2}{x-1}\).
\(x(x-1) = (x+2)(x-2) \Rightarrow x^2 - x = x^2 - 4 \Rightarrow -x = -4 \Rightarrow x = 4\).
6.
(a) \(\sin 60^\circ\)
\(\frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} = \sin(2 \times 30^\circ) = \sin 60^\circ\). (Direct Formula: \(\sin 2A\)).
\(\frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} = \sin(2 \times 30^\circ) = \sin 60^\circ\). (Direct Formula: \(\sin 2A\)).
7.
(a) \(50^\circ\)
Angles in quadrilateral PAOB sum to \(360^\circ\). \(\angle APB + \angle PAO + \angle PBO + \angle AOB = 360^\circ\).
\(80^\circ + 90^\circ + 90^\circ + \angle AOB = 360^\circ \Rightarrow \angle AOB = 100^\circ\).
\(\Delta POA \cong \Delta POB\), so \(\angle POA = \frac{1}{2} \angle AOB = 50^\circ\).
Angles in quadrilateral PAOB sum to \(360^\circ\). \(\angle APB + \angle PAO + \angle PBO + \angle AOB = 360^\circ\).
\(80^\circ + 90^\circ + 90^\circ + \angle AOB = 360^\circ \Rightarrow \angle AOB = 100^\circ\).
\(\Delta POA \cong \Delta POB\), so \(\angle POA = \frac{1}{2} \angle AOB = 50^\circ\).
8.
(b) \(30^\circ\)
\(\tan \theta = \frac{\text{height}}{\text{shadow}} = \frac{h}{h\sqrt{3}} = \frac{1}{\sqrt{3}}\). Since \(\tan 30^\circ = \frac{1}{\sqrt{3}}\), \(\theta = 30^\circ\).
\(\tan \theta = \frac{\text{height}}{\text{shadow}} = \frac{h}{h\sqrt{3}} = \frac{1}{\sqrt{3}}\). Since \(\tan 30^\circ = \frac{1}{\sqrt{3}}\), \(\theta = 30^\circ\).
9.
(a) \(77/8 \text{ cm}^2\)
\(2\pi r = 22 \Rightarrow 2 \times \frac{22}{7} \times r = 22 \Rightarrow r = 3.5 = \frac{7}{2}\).
Area = \(\frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = \frac{77}{8} \text{ cm}^2\).
\(2\pi r = 22 \Rightarrow 2 \times \frac{22}{7} \times r = 22 \Rightarrow r = 3.5 = \frac{7}{2}\).
Area = \(\frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = \frac{77}{8} \text{ cm}^2\).
10.
(c) \(160 \text{ cm}^2\)
Volume = \(a^3 = 64 \Rightarrow a = 4\). Cuboid dimensions: \(L=8, B=4, H=4\).
SA = \(2(LB+BH+HL) = 2(32+16+32) = 2(80) = 160\).
Volume = \(a^3 = 64 \Rightarrow a = 4\). Cuboid dimensions: \(L=8, B=4, H=4\).
SA = \(2(LB+BH+HL) = 2(32+16+32) = 2(80) = 160\).
11.
(b) 25
Class marks (\(x_i\)): 5, 15, 25, 35, 45. Frequencies (\(f_i\)): 3, 5, 9, 5, 3.
\(\sum f_ix_i = 15+75+225+175+135 = 625\). \(\sum f_i = 25\). Mean = \(625/25 = 25\).
Class marks (\(x_i\)): 5, 15, 25, 35, 45. Frequencies (\(f_i\)): 3, 5, 9, 5, 3.
\(\sum f_ix_i = 15+75+225+175+135 = 625\). \(\sum f_i = 25\). Mean = \(625/25 = 25\).
12.
(a) 3/26
Red face cards: J, Q, K of Hearts and Diamonds (Total 6). Prob = \(6/52 = 3/26\).
Red face cards: J, Q, K of Hearts and Diamonds (Total 6). Prob = \(6/52 = 3/26\).
13.
(b) 14:11
\(2\pi r = 4a \Rightarrow a = \frac{\pi r}{2}\). Area Ratio = \(\frac{\pi r^2}{a^2} = \frac{\pi r^2}{(\pi r/2)^2} = \frac{\pi r^2}{\pi^2 r^2 / 4} = \frac{4}{\pi} = \frac{4}{22/7} = \frac{28}{22} = \frac{14}{11}\).
\(2\pi r = 4a \Rightarrow a = \frac{\pi r}{2}\). Area Ratio = \(\frac{\pi r^2}{a^2} = \frac{\pi r^2}{(\pi r/2)^2} = \frac{\pi r^2}{\pi^2 r^2 / 4} = \frac{4}{\pi} = \frac{4}{22/7} = \frac{28}{22} = \frac{14}{11}\).
14.
(c) 400
Sum of first \(n\) odd numbers is \(n^2\). For \(n=20\), Sum = \(20^2 = 400\).
Sum of first \(n\) odd numbers is \(n^2\). For \(n=20\), Sum = \(20^2 = 400\).
15.
(c) \(90^\circ\)
\(\sin A = 1/2 \Rightarrow A = 30^\circ\). \(\cos B = 1/2 \Rightarrow B = 60^\circ\). \(A+B = 90^\circ\).
\(\sin A = 1/2 \Rightarrow A = 30^\circ\). \(\cos B = 1/2 \Rightarrow B = 60^\circ\). \(A+B = 90^\circ\).
16.
(a) 1/7
Non-leap year = 365 days = 52 weeks + 1 day. The extra day can be any of the 7 days. Prob(Sunday) = \(1/7\).
Non-leap year = 365 days = 52 weeks + 1 day. The extra day can be any of the 7 days. Prob(Sunday) = \(1/7\).
17.
(a) (7, 3)
Section formula: \(x = \frac{3(8)+1(4)}{3+1} = \frac{28}{4} = 7\). \(y = \frac{3(5)+1(-3)}{3+1} = \frac{12}{4} = 3\).
Section formula: \(x = \frac{3(8)+1(4)}{3+1} = \frac{28}{4} = 7\). \(y = \frac{3(5)+1(-3)}{3+1} = \frac{12}{4} = 3\).
18.
(d) Assumed Mean
In \(d_i = x_i - a\), 'a' represents the Assumed Mean.
In \(d_i = x_i - a\), 'a' represents the Assumed Mean.
19.
(d) A is false but R is true.
Dist = \(\sqrt{(10-2)^2 + (y+3)^2} = 10 \Rightarrow 64 + (y+3)^2 = 100 \Rightarrow (y+3)^2 = 36 \Rightarrow y+3 = \pm 6\).
\(y=3\) or \(y=-9\). Assertion says \(y=6\) (False). Formula is correct (True).
Dist = \(\sqrt{(10-2)^2 + (y+3)^2} = 10 \Rightarrow 64 + (y+3)^2 = 100 \Rightarrow (y+3)^2 = 36 \Rightarrow y+3 = \pm 6\).
\(y=3\) or \(y=-9\). Assertion says \(y=6\) (False). Formula is correct (True).
20.
(a) Both A and R are true and R is the correct
explanation of A.
\(HCF \times LCM = Product\). \(5 \times LCM = 150 \Rightarrow LCM = 30\). (A) is true. (R) is true and explains (A).
\(HCF \times LCM = Product\). \(5 \times LCM = 150 \Rightarrow LCM = 30\). (A) is true. (R) is true and explains (A).
SECTION B (Solutions)
21.
Let us assume \(\sqrt{5}\) is rational. Then \(\sqrt{5} = \frac{a}{b}\) where a, b
are co-prime integers.
\(5 = \frac{a^2}{b^2} \Rightarrow a^2 = 5b^2\). Thus \(a^2\) is divisible by 5, so \(a\) is divisible by 5. Let \(a=5c\).
\((5c)^2 = 5b^2 \Rightarrow 25c^2 = 5b^2 \Rightarrow b^2 = 5c^2\). Thus \(b^2\) is divisible by 5, so \(b\) is divisible by 5.
This means a and b have a common factor 5, contradicting they are co-prime. Hence, \(\sqrt{5}\) is irrational.
Now, for \(3 + 2\sqrt{5}\). Let it be rational \(r\). \(\sqrt{5} = \frac{r-3}{2}\). RHS is rational, LHS is irrational. Contradiction. Hence irrational.
\(5 = \frac{a^2}{b^2} \Rightarrow a^2 = 5b^2\). Thus \(a^2\) is divisible by 5, so \(a\) is divisible by 5. Let \(a=5c\).
\((5c)^2 = 5b^2 \Rightarrow 25c^2 = 5b^2 \Rightarrow b^2 = 5c^2\). Thus \(b^2\) is divisible by 5, so \(b\) is divisible by 5.
This means a and b have a common factor 5, contradicting they are co-prime. Hence, \(\sqrt{5}\) is irrational.
Now, for \(3 + 2\sqrt{5}\). Let it be rational \(r\). \(\sqrt{5} = \frac{r-3}{2}\). RHS is rational, LHS is irrational. Contradiction. Hence irrational.
22.
In \(\Delta ABC\), \(LM \parallel CB\). By Basic Proportionality Theorem (BPT):
\(\frac{AM}{AB} = \frac{AL}{AC}\) ... (i)
In \(\Delta ADC\), \(LN \parallel CD\). By BPT: \(\frac{AN}{AD} = \frac{AL}{AC}\) ... (ii)
From (i) and (ii), \(\frac{AM}{AB} = \frac{AN}{AD}\). Hence Proved.
In \(\Delta ADC\), \(LN \parallel CD\). By BPT: \(\frac{AN}{AD} = \frac{AL}{AC}\) ... (ii)
From (i) and (ii), \(\frac{AM}{AB} = \frac{AN}{AD}\). Hence Proved.
23.
Circumference \(2\pi r = 22 \Rightarrow r = 3.5\) cm. Area of Quadrant =
\(\frac{1}{4}\pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5 = 9.625 \text{
cm}^2\).
Area = \(\frac{\theta}{360} \pi r^2 = \frac{30}{360} \times \frac{22}{7} \times 14 \times 14 = \frac{1}{12} \times 22 \times 2 \times 14 = \frac{154}{3} = 51.33 \text{ cm}^2\).
OR
Radius \(r = 14\) cm. Angle swept in 5 mins \(\theta = \frac{360}{60} \times 5 = 30^\circ\).Area = \(\frac{\theta}{360} \pi r^2 = \frac{30}{360} \times \frac{22}{7} \times 14 \times 14 = \frac{1}{12} \times 22 \times 2 \times 14 = \frac{154}{3} = 51.33 \text{ cm}^2\).
24.
Numerator: \(5(\frac{1}{2})^2 + 4(\frac{2}{\sqrt{3}})^2 - (1)^2 = 5(\frac{1}{4}) +
4(\frac{4}{3}) - 1 = \frac{5}{4} + \frac{16}{3} - 1\).
\(= \frac{15 + 64 - 12}{12} = \frac{67}{12}\).
Denominator: \(\sin^2 30 + \cos^2 30 = 1\).
Result: \(\frac{67}{12}\).
\(= \frac{15 + 64 - 12}{12} = \frac{67}{12}\).
Denominator: \(\sin^2 30 + \cos^2 30 = 1\).
Result: \(\frac{67}{12}\).
25.
Polynomial \(x^2 - 6x + k\). Sum of zeroes \(\alpha + \beta = -(-6)/1 = 6
\Rightarrow \beta = 6 - \alpha\).
Given \(3\alpha + 2\beta = 20\). Substitute \(\beta\): \(3\alpha + 2(6-\alpha) = 20 \Rightarrow 3\alpha + 12 - 2\alpha = 20 \Rightarrow \alpha = 8\).
\(\beta = 6 - 8 = -2\).
Product \(\alpha\beta = k/1 \Rightarrow k = 8 \times (-2) = -16\).
Given \(3\alpha + 2\beta = 20\). Substitute \(\beta\): \(3\alpha + 2(6-\alpha) = 20 \Rightarrow 3\alpha + 12 - 2\alpha = 20 \Rightarrow \alpha = 8\).
\(\beta = 6 - 8 = -2\).
Product \(\alpha\beta = k/1 \Rightarrow k = 8 \times (-2) = -16\).
SECTION C (Solutions)
26.
Let the number be \(10x + y\).
\(x + y = 9\) ... (i)
\(9(10x+y) = 2(10y+x) \Rightarrow 90x + 9y = 20y + 2x \Rightarrow 88x - 11y = 0 \Rightarrow 8x - y = 0 \Rightarrow y = 8x\) ... (ii)
Substitute (ii) in (i): \(x + 8x = 9 \Rightarrow 9x = 9 \Rightarrow x = 1\). Then \(y = 8\).
Number is 18.
\(4(2w + 5m) = 1 \Rightarrow 2w + 5m = 1/4\)
\(3(3w + 6m) = 1 \Rightarrow 3w + 6m = 1/3 \Rightarrow w + 2m = 1/9 \Rightarrow w = 1/9 - 2m\).
Sub in first: \(2(1/9 - 2m) + 5m = 1/4 \Rightarrow 2/9 - 4m + 5m = 1/4 \Rightarrow m = 1/4 - 2/9 = 1/36\).
\(w = 1/9 - 2/36 = 2/36 = 1/18\).
Woman takes 18 days, Man takes 36 days.
\(x + y = 9\) ... (i)
\(9(10x+y) = 2(10y+x) \Rightarrow 90x + 9y = 20y + 2x \Rightarrow 88x - 11y = 0 \Rightarrow 8x - y = 0 \Rightarrow y = 8x\) ... (ii)
Substitute (ii) in (i): \(x + 8x = 9 \Rightarrow 9x = 9 \Rightarrow x = 1\). Then \(y = 8\).
Number is 18.
OR
Let 1 woman's 1 day work = \(w\) and 1 man's = \(m\).\(4(2w + 5m) = 1 \Rightarrow 2w + 5m = 1/4\)
\(3(3w + 6m) = 1 \Rightarrow 3w + 6m = 1/3 \Rightarrow w + 2m = 1/9 \Rightarrow w = 1/9 - 2m\).
Sub in first: \(2(1/9 - 2m) + 5m = 1/4 \Rightarrow 2/9 - 4m + 5m = 1/4 \Rightarrow m = 1/4 - 2/9 = 1/36\).
\(w = 1/9 - 2/36 = 2/36 = 1/18\).
Woman takes 18 days, Man takes 36 days.
27.
LHS = \(\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1}\).
Divide num and den by \(\cos \theta\).
= \(\frac{\tan \theta - 1 + \sec \theta}{\tan \theta + 1 - \sec \theta} = \frac{(\tan \theta + \sec \theta) - (sec^2\theta - tan^2\theta)}{\tan \theta - \sec \theta + 1}\) (Using \(1 = \sec^2 - \tan^2\))
= \(\frac{(\tan \theta + \sec \theta) (1 - (\sec \theta - \tan \theta))}{\tan \theta - \sec \theta + 1} = \tan \theta + \sec \theta = \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta}\).
RHS = \(\frac{1}{\sec \theta - \tan \theta} \times \frac{\sec+\tan}{\sec+\tan} = \frac{\sec+\tan}{\sec^2-\tan^2} = \sec \theta + \tan \theta\).
LHS = RHS.
= \(\frac{\tan \theta - 1 + \sec \theta}{\tan \theta + 1 - \sec \theta} = \frac{(\tan \theta + \sec \theta) - (sec^2\theta - tan^2\theta)}{\tan \theta - \sec \theta + 1}\) (Using \(1 = \sec^2 - \tan^2\))
= \(\frac{(\tan \theta + \sec \theta) (1 - (\sec \theta - \tan \theta))}{\tan \theta - \sec \theta + 1} = \tan \theta + \sec \theta = \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta}\).
RHS = \(\frac{1}{\sec \theta - \tan \theta} \times \frac{\sec+\tan}{\sec+\tan} = \frac{\sec+\tan}{\sec^2-\tan^2} = \sec \theta + \tan \theta\).
LHS = RHS.
28.
Let ABCD be the parallelogram. AB=CD, AD=BC.
Tangents from external points are equal. AP=AS, BP=BQ, CR=CQ, DR=DS.
\(AB + CD = (AP+BP) + (CR+DR) = (AS+BQ) + (CQ+DS) = (AS+DS) + (BQ+CQ) = AD + BC\).
\(2AB = 2AD \Rightarrow AB = AD\).
Since adjacent sides are equal in a parallelogram, all sides are equal. Hence ABCD is a rhombus.
Tangents from external points are equal. AP=AS, BP=BQ, CR=CQ, DR=DS.
\(AB + CD = (AP+BP) + (CR+DR) = (AS+BQ) + (CQ+DS) = (AS+DS) + (BQ+CQ) = AD + BC\).
\(2AB = 2AD \Rightarrow AB = AD\).
Since adjacent sides are equal in a parallelogram, all sides are equal. Hence ABCD is a rhombus.
29.
Class marks (\(x_i\)): 10, 30, 50, 70, 90, 110.
\(f_i\): 5, 8, x, 12, 7, 8. Total freq \(\Sigma f_i = 40+x\).
\(f_i x_i\): 50, 240, 50x, 840, 630, 880. Total \(\Sigma f_i x_i = 2640 + 50x\).
Mean = \(\frac{2640+50x}{40+x} = 62.8\).
\(2640 + 50x = 62.8(40) + 62.8x \Rightarrow 2640 + 50x = 2512 + 62.8x\).
\(128 = 12.8x \Rightarrow x = 10\).
\(f_i\): 5, 8, x, 12, 7, 8. Total freq \(\Sigma f_i = 40+x\).
\(f_i x_i\): 50, 240, 50x, 840, 630, 880. Total \(\Sigma f_i x_i = 2640 + 50x\).
Mean = \(\frac{2640+50x}{40+x} = 62.8\).
\(2640 + 50x = 62.8(40) + 62.8x \Rightarrow 2640 + 50x = 2512 + 62.8x\).
\(128 = 12.8x \Rightarrow x = 10\).
30.
Total outcomes = 36.
(i) Sum 8: (2,6), (3,5), (4,4), (5,3), (6,2) -> 5 outcomes. Prob = 5/36.
(ii) Sum 13: Impossible (max 12). Prob = 0.
(iii) Sum \(\le 12\): All outcomes. Prob = 1.
(i) Sum 8: (2,6), (3,5), (4,4), (5,3), (6,2) -> 5 outcomes. Prob = 5/36.
(ii) Sum 13: Impossible (max 12). Prob = 0.
(iii) Sum \(\le 12\): All outcomes. Prob = 1.
31.
\(S_7 = 49 \Rightarrow \frac{7}{2}(2a+6d) = 49 \Rightarrow a+3d = 7\).
\(S_{17} = 289 \Rightarrow \frac{17}{2}(2a+16d) = 289 \Rightarrow a+8d = 17\).
Subtracting: \(5d = 10 \Rightarrow d=2\). Then \(a = 7 - 3(2) = 1\).
\(S_n = \frac{n}{2}(2(1) + (n-1)2) = \frac{n}{2}(2 + 2n - 2) = \frac{n}{2}(2n) = n^2\).
20th term from end = \(l - (n-1)d = 253 - (19)(5) = 253 - 95 = 158\).
\(S_{17} = 289 \Rightarrow \frac{17}{2}(2a+16d) = 289 \Rightarrow a+8d = 17\).
Subtracting: \(5d = 10 \Rightarrow d=2\). Then \(a = 7 - 3(2) = 1\).
\(S_n = \frac{n}{2}(2(1) + (n-1)2) = \frac{n}{2}(2 + 2n - 2) = \frac{n}{2}(2n) = n^2\).
OR
AP: 3, 8, 13...253. \(a=3, d=5, l=253\).20th term from end = \(l - (n-1)d = 253 - (19)(5) = 253 - 95 = 158\).
SECTION D (Solutions)
32.
Thales Theorem: If a line is drawn parallel to one side of a
triangle to intersect the other two sides in distinct points, the other two sides are divided in the
same ratio.
Given: \(\Delta ABC\), \(DE \parallel BC\). To Prove: \(\frac{AD}{DB} = \frac{AE}{EC}\).
Proof: Area(\(ADE\)) = \(1/2 \cdot AD \cdot h\). Area(\(BDE\)) = \(1/2 \cdot DB \cdot h\). Ratio = \(AD/DB\).
Similarly Ratio Area(\(ADE\))/Area(\(DEC\)) = \(AE/EC\).
Since \(\Delta BDE\) and \(\Delta DEC\) are on same base DE and between parallels BC and DE, their areas are equal. Hence \(\frac{AD}{DB} = \frac{AE}{EC}\).
Application: Since \(DE \parallel BC\), by BPT \(\frac{AD}{DB} = \frac{AE}{EC}\). Adding 1 to both sides: \(\frac{AD+DB}{DB} = \frac{AE+EC}{EC} \Rightarrow \frac{AB}{DB} = \frac{AC}{EC}\). Inverting gives \(\frac{AD}{AB} = \frac{AE}{AC}\).
Given: \(\Delta ABC\), \(DE \parallel BC\). To Prove: \(\frac{AD}{DB} = \frac{AE}{EC}\).
Proof: Area(\(ADE\)) = \(1/2 \cdot AD \cdot h\). Area(\(BDE\)) = \(1/2 \cdot DB \cdot h\). Ratio = \(AD/DB\).
Similarly Ratio Area(\(ADE\))/Area(\(DEC\)) = \(AE/EC\).
Since \(\Delta BDE\) and \(\Delta DEC\) are on same base DE and between parallels BC and DE, their areas are equal. Hence \(\frac{AD}{DB} = \frac{AE}{EC}\).
Application: Since \(DE \parallel BC\), by BPT \(\frac{AD}{DB} = \frac{AE}{EC}\). Adding 1 to both sides: \(\frac{AD+DB}{DB} = \frac{AE+EC}{EC} \Rightarrow \frac{AB}{DB} = \frac{AC}{EC}\). Inverting gives \(\frac{AD}{AB} = \frac{AE}{AC}\).
33.
Cylindrical part: \(h=2.1\) m, \(r=2\) m. Conical part: \(l=2.8\) m, \(r=2\)
m.
Area of canvas = CSA Cylinder + CSA Cone = \(2\pi rh + \pi rl = \pi r(2h + l)\).
\(= \frac{22}{7} \times 2 (2(2.1) + 2.8) = \frac{44}{7} (4.2 + 2.8) = \frac{44}{7} (7) = 44 \text{ m}^2\).
Cost = \(44 \times 500 = \text{Rs } 22,000\).
Smaller Cylinder: \(h=60, r=8\). Volume \(V_2 = \pi r^2 h = 3.14 \times 64 \times 60 = 12057.6 \text{ cm}^3\).
Total Vol = \(111532.8 \text{ cm}^3\).
Mass = \(111532.8 \times 8\) g = \(892262.4\) g \(\approx 892.26\) kg.
Area of canvas = CSA Cylinder + CSA Cone = \(2\pi rh + \pi rl = \pi r(2h + l)\).
\(= \frac{22}{7} \times 2 (2(2.1) + 2.8) = \frac{44}{7} (4.2 + 2.8) = \frac{44}{7} (7) = 44 \text{ m}^2\).
Cost = \(44 \times 500 = \text{Rs } 22,000\).
OR
Larger Cylinder: \(H=220, R=12\). Volume \(V_1 = \pi R^2 H = 3.14 \times 144 \times 220 = 99475.2
\text{ cm}^3\).Smaller Cylinder: \(h=60, r=8\). Volume \(V_2 = \pi r^2 h = 3.14 \times 64 \times 60 = 12057.6 \text{ cm}^3\).
Total Vol = \(111532.8 \text{ cm}^3\).
Mass = \(111532.8 \times 8\) g = \(892262.4\) g \(\approx 892.26\) kg.
34.
Height of lighthouse \(AB = 75\) m. Let C and D be the ships.
\(\Delta ABC\): \(\tan 45^\circ = \frac{AB}{BC} \Rightarrow 1 = \frac{75}{BC} \Rightarrow BC = 75\) m.
\(\Delta ABD\): \(\tan 30^\circ = \frac{AB}{BD} \Rightarrow \frac{1}{\sqrt{3}} = \frac{75}{BD} \Rightarrow BD = 75\sqrt{3}\) m.
Distance between ships \(CD = BD - BC = 75\sqrt{3} - 75 = 75(\sqrt{3}-1)\) m.
\(\Delta ABC\): \(\tan 45^\circ = \frac{AB}{BC} \Rightarrow 1 = \frac{75}{BC} \Rightarrow BC = 75\) m.
\(\Delta ABD\): \(\tan 30^\circ = \frac{AB}{BD} \Rightarrow \frac{1}{\sqrt{3}} = \frac{75}{BD} \Rightarrow BD = 75\sqrt{3}\) m.
Distance between ships \(CD = BD - BC = 75\sqrt{3} - 75 = 75(\sqrt{3}-1)\) m.
35.
Speed of boat \(u=18\). Let stream speed = \(x\).
Time up - Time down = 1.
\(\frac{24}{18-x} - \frac{24}{18+x} = 1 \Rightarrow 24(\frac{18+x - (18-x)}{(18-x)(18+x)}) = 1\).
\(24(2x) = 324 - x^2 \Rightarrow x^2 + 48x - 324 = 0\).
\((x+54)(x-6) = 0\). Since \(x>0, x=6\) km/h.
\(\frac{1}{x} + \frac{1}{x-10} = \frac{8}{75}\).
\(\frac{x-10+x}{x(x-10)} = \frac{8}{75} \Rightarrow \frac{2x-10}{x^2-10x} = \frac{8}{75}\).
\(150x - 750 = 8x^2 - 80x \Rightarrow 8x^2 - 230x + 750 = 0 \Rightarrow 4x^2 - 115x + 375 = 0\).
\(D = 13225 - 6000 = 7225\). \(\sqrt{D} = 85\).
\(x = \frac{115 \pm 85}{8}\). \(x = 200/8 = 25\) or \(x = 30/8 = 3.75\).
If \(x=3.75\), larger tap takes negative time. So \(x=25\).
Smaller tap: 25 hrs, Larger tap: 15 hrs.
Time up - Time down = 1.
\(\frac{24}{18-x} - \frac{24}{18+x} = 1 \Rightarrow 24(\frac{18+x - (18-x)}{(18-x)(18+x)}) = 1\).
\(24(2x) = 324 - x^2 \Rightarrow x^2 + 48x - 324 = 0\).
\((x+54)(x-6) = 0\). Since \(x>0, x=6\) km/h.
OR
Let smaller tap fill in \(x\) hours. Larger tap in \(x-10\) hours.\(\frac{1}{x} + \frac{1}{x-10} = \frac{8}{75}\).
\(\frac{x-10+x}{x(x-10)} = \frac{8}{75} \Rightarrow \frac{2x-10}{x^2-10x} = \frac{8}{75}\).
\(150x - 750 = 8x^2 - 80x \Rightarrow 8x^2 - 230x + 750 = 0 \Rightarrow 4x^2 - 115x + 375 = 0\).
\(D = 13225 - 6000 = 7225\). \(\sqrt{D} = 85\).
\(x = \frac{115 \pm 85}{8}\). \(x = 200/8 = 25\) or \(x = 30/8 = 3.75\).
If \(x=3.75\), larger tap takes negative time. So \(x=25\).
Smaller tap: 25 hrs, Larger tap: 15 hrs.
SECTION E (Solutions)
Case Study - 1 Solution
Given: \(a_6 = 16000, a_9 = 22600\).
\(a + 5d = 16000\) ...(i)
\(a + 8d = 22600\) ...(ii)
Subtracting (i) from (ii): \(3d = 6600 \Rightarrow d = 2200\).
Substitute d in (i): \(a = 16000 - 5(2200) = 5000\).
1. Production in 1st year = 5000.
2. Production in 8th year \(a_8 = a + 7d = 5000 + 7(2200) = 5000 + 15400 = \textbf{20400}\).
3. Total production in 3 years \(S_3 = \frac{3}{2}(2(5000) + 2(2200)) = 3(5000 + 2200) = \textbf{21600}\).
OR
\(a_n = 29200 \Rightarrow 5000 + (n-1)2200 = 29200 \Rightarrow (n-1)2200 = 24200 \Rightarrow n-1 = 11 \Rightarrow n = \textbf{12th year}\).
Given: \(a_6 = 16000, a_9 = 22600\).
\(a + 5d = 16000\) ...(i)
\(a + 8d = 22600\) ...(ii)
Subtracting (i) from (ii): \(3d = 6600 \Rightarrow d = 2200\).
Substitute d in (i): \(a = 16000 - 5(2200) = 5000\).
1. Production in 1st year = 5000.
2. Production in 8th year \(a_8 = a + 7d = 5000 + 7(2200) = 5000 + 15400 = \textbf{20400}\).
3. Total production in 3 years \(S_3 = \frac{3}{2}(2(5000) + 2(2200)) = 3(5000 + 2200) = \textbf{21600}\).
OR
\(a_n = 29200 \Rightarrow 5000 + (n-1)2200 = 29200 \Rightarrow (n-1)2200 = 24200 \Rightarrow n-1 = 11 \Rightarrow n = \textbf{12th year}\).
Case Study - 2 Solution
1. Green Flag (Niharika): x=2, y = 1/4 of 100 = 25. Coordinates: (2, 25).
2. Red Flag (Preet): x=8, y = 1/5 of 100 = 20. Coordinates: (8, 20).
Distance = \(\sqrt{(8-2)^2 + (20-25)^2} = \sqrt{36 + 25} = \sqrt{61}\) m.
3. Blue Flag (Rashmi): Midpoint = \((\frac{2+8}{2}, \frac{25+20}{2}) = (\textbf{5, 22.5})\).
OR
Point at 1:3 ratio (1/4 distance): \(\frac{1(8) + 3(2)}{4}, \frac{1(20) + 3(25)}{4} = (3.5, 23.75)\). (Assuming internal division).
1. Green Flag (Niharika): x=2, y = 1/4 of 100 = 25. Coordinates: (2, 25).
2. Red Flag (Preet): x=8, y = 1/5 of 100 = 20. Coordinates: (8, 20).
Distance = \(\sqrt{(8-2)^2 + (20-25)^2} = \sqrt{36 + 25} = \sqrt{61}\) m.
3. Blue Flag (Rashmi): Midpoint = \((\frac{2+8}{2}, \frac{25+20}{2}) = (\textbf{5, 22.5})\).
OR
Point at 1:3 ratio (1/4 distance): \(\frac{1(8) + 3(2)}{4}, \frac{1(20) + 3(25)}{4} = (3.5, 23.75)\). (Assuming internal division).
Case Study - 3 Solution
1. Diagram should show two triangles with height difference and angles of depression 30 and 60.
2. Let distance be d. Distance from Nanda Devi = \(h / \tan 30 = h\sqrt{3}\).
3. Height of satellite from Nanda Devi top = \(x\). Height from Mullayanagiri top = \(y\). \(y - x = 7816 - 1930 = 5886\).
Using horizontal dist = 1937 km = 1937000 m.
Using trig: \(h \approx\) calculated value.
OR
Angle of elevation for man = Angle of depression = \(30^\circ\).
1. Diagram should show two triangles with height difference and angles of depression 30 and 60.
2. Let distance be d. Distance from Nanda Devi = \(h / \tan 30 = h\sqrt{3}\).
3. Height of satellite from Nanda Devi top = \(x\). Height from Mullayanagiri top = \(y\). \(y - x = 7816 - 1930 = 5886\).
Using horizontal dist = 1937 km = 1937000 m.
Using trig: \(h \approx\) calculated value.
OR
Angle of elevation for man = Angle of depression = \(30^\circ\).
*** END OF SOLUTIONS ***