Vardaan Learning Institute
Mock Board Paper Solutions 2025-26
Set - 2 (Solutions)
MATHEMATICS (STANDARD)
Class - X
Time: 3 Hours
Max. Marks: 80
SECTION A (Solutions)
1.
(a) 84
Product of numbers = HCF \(\times\) LCM. \(504 = 6 \times \text{LCM} \Rightarrow \text{LCM} = 504/6 = 84\).
Product of numbers = HCF \(\times\) LCM. \(504 = 6 \times \text{LCM} \Rightarrow \text{LCM} = 504/6 = 84\).
2.
(b) -10
Let \(p(x) = x^2 + 3x + k\). Since 2 is a zero, \(p(2) = 0\).
\(2^2 + 3(2) + k = 0 \Rightarrow 4 + 6 + k = 0 \Rightarrow 10 + k = 0 \Rightarrow k = -10\).
Let \(p(x) = x^2 + 3x + k\). Since 2 is a zero, \(p(2) = 0\).
\(2^2 + 3(2) + k = 0 \Rightarrow 4 + 6 + k = 0 \Rightarrow 10 + k = 0 \Rightarrow k = -10\).
3.
(d) Intersecting at (a, b)
\(x=a\) is a vertical line, \(y=b\) is a horizontal line. They intersect at the point \((a, b)\).
\(x=a\) is a vertical line, \(y=b\) is a horizontal line. They intersect at the point \((a, b)\).
4.
(a) No real roots
\(D = b^2 - 4ac = (-\sqrt{5})^2 - 4(2)(1) = 5 - 8 = -3\). Since \(D < 0\), no real roots.
\(D = b^2 - 4ac = (-\sqrt{5})^2 - 4(2)(1) = 5 - 8 = -3\). Since \(D < 0\), no real roots.
5.
(a) 32
\(a=5, d=3\). \(a_{10} = a + 9d = 5 + 9(3) = 5 + 27 = 32\).
\(a=5, d=3\). \(a_{10} = a + 9d = 5 + 9(3) = 5 + 27 = 32\).
6.
(b) 4.5 cm
\(\frac{\text{area}(\Delta ABC)}{\text{area}(\Delta PQR)} = (\frac{BC}{QR})^2 \Rightarrow \frac{81}{144} = (\frac{BC}{6})^2 \Rightarrow \frac{9}{12} = \frac{BC}{6} \Rightarrow BC = \frac{54}{12} = 4.5\).
\(\frac{\text{area}(\Delta ABC)}{\text{area}(\Delta PQR)} = (\frac{BC}{QR})^2 \Rightarrow \frac{81}{144} = (\frac{BC}{6})^2 \Rightarrow \frac{9}{12} = \frac{BC}{6} \Rightarrow BC = \frac{54}{12} = 4.5\).
7.
(b) 3
The distance of a point from the x-axis is the absolute value of its y-coordinate. \(|3| = 3\).
The distance of a point from the x-axis is the absolute value of its y-coordinate. \(|3| = 3\).
8.
(b) 3/4
\(\cos A = 4/5\). By Pythagorean triplet (3, 4, 5), Perpendicular = 3. \(\tan A = P/B = 3/4\).
\(\cos A = 4/5\). By Pythagorean triplet (3, 4, 5), Perpendicular = 3. \(\tan A = P/B = 3/4\).
9.
(b) 2
\((\frac{\sqrt{3}}{2})^2 + 2(1) - (\frac{\sqrt{3}}{2})^2 = \frac{3}{4} + 2 - \frac{3}{4} = 2\).
\((\frac{\sqrt{3}}{2})^2 + 2(1) - (\frac{\sqrt{3}}{2})^2 = \frac{3}{4} + 2 - \frac{3}{4} = 2\).
10.
(a) \(70^\circ\)
Angle between tangents + Angle at center = \(180^\circ\). \(110^\circ + \theta = 180^\circ \Rightarrow \theta = 70^\circ\).
Angle between tangents + Angle at center = \(180^\circ\). \(110^\circ + \theta = 180^\circ \Rightarrow \theta = 70^\circ\).
11.
(b) Area of the circle > Area of the
square
Let Circumference = Perimeter = C. Circle area = \(C^2/4\pi\). Square area = \(C^2/16\). Since \(4\pi \approx 12.56 < 16\), Circle Area is larger.
Let Circumference = Perimeter = C. Circle area = \(C^2/4\pi\). Square area = \(C^2/16\). Since \(4\pi \approx 12.56 < 16\), Circle Area is larger.
12.
(d) 16:9
\(\frac{V_1}{V_2} = \frac{r_1^3}{r_2^3} = \frac{64}{27} \Rightarrow \frac{r_1}{r_2} = \frac{4}{3}\). Ratio of Areas = \((\frac{r_1}{r_2})^2 = (\frac{4}{3})^2 = \frac{16}{9}\).
\(\frac{V_1}{V_2} = \frac{r_1^3}{r_2^3} = \frac{64}{27} \Rightarrow \frac{r_1}{r_2} = \frac{4}{3}\). Ratio of Areas = \((\frac{r_1}{r_2})^2 = (\frac{4}{3})^2 = \frac{16}{9}\).
13.
(c) 24
Mode = 3 Median - 2 Mean. \(16 = 3M - 2(28) \Rightarrow 16 = 3M - 56 \Rightarrow 3M = 72 \Rightarrow M = 24\).
Mode = 3 Median - 2 Mean. \(16 = 3M - 2(28) \Rightarrow 16 = 3M - 56 \Rightarrow 3M = 72 \Rightarrow M = 24\).
14.
(d) 0
Probability of an impossible event is 0.
Probability of an impossible event is 0.
15.
(c) 12/13
Probability(Ace) = 4/52 = 1/13. Probability(Not Ace) = \(1 - 1/13 = 12/13\).
Probability(Ace) = 4/52 = 1/13. Probability(Not Ace) = \(1 - 1/13 = 12/13\).
16.
(a) -8
Midpoint of y-coordinates: \(\frac{(b-2) + 4}{2} = -3 \Rightarrow b+2 = -6 \Rightarrow b = -8\).
Midpoint of y-coordinates: \(\frac{(b-2) + 4}{2} = -3 \Rightarrow b+2 = -6 \Rightarrow b = -8\).
17.
(b) Two decimal places
Denominator is \(2^2 \times 5^1\). The highest power of 2 or 5 is 2. So, it terminates after 2 decimal places.
Denominator is \(2^2 \times 5^1\). The highest power of 2 or 5 is 2. So, it terminates after 2 decimal places.
18.
(c) \(15/2\) m
\(\cos 60^\circ = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{\text{Height}}{15}\). \(1/2 = h/15 \Rightarrow h = 15/2\). Note: Ladder angle with WALL is 60, so angle with ground is 30. Wait, question says "angle... with the wall". So Base is horizontal, Height is vertical. \(\cos 60 = \text{Height} / 15 \Rightarrow h = 15/2\). Correct.
\(\cos 60^\circ = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{\text{Height}}{15}\). \(1/2 = h/15 \Rightarrow h = 15/2\). Note: Ladder angle with WALL is 60, so angle with ground is 30. Wait, question says "angle... with the wall". So Base is horizontal, Height is vertical. \(\cos 60 = \text{Height} / 15 \Rightarrow h = 15/2\). Correct.
19.
(a) Both A and R are true and R is the correct
explanation of A.
For a number to end in 0, prime factors must include 2 and 5. \(3^n\) only has 3.
For a number to end in 0, prime factors must include 2 and 5. \(3^n\) only has 3.
20.
(a) Both A and R are true and R is the correct
explanation of A.
x-coordinate is 0, so it lies on the y-axis.
x-coordinate is 0, so it lies on the y-axis.
SECTION B (Solutions)
21.
For infinitely many solutions: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} =
\frac{c_1}{c_2}\).
\(\frac{2}{k-1} = \frac{3}{k+2} = \frac{7}{3k}\).
From first two: \(2(k+2) = 3(k-1) \Rightarrow 2k + 4 = 3k - 3 \Rightarrow k = 7\).
Check with third: \(\frac{3}{7+2} = \frac{3}{9} = \frac{1}{3}\) and \(\frac{7}{3(7)} = \frac{1}{3}\). Correct.
Value of k is 7.
\(\frac{2}{k-1} = \frac{3}{k+2} = \frac{7}{3k}\).
From first two: \(2(k+2) = 3(k-1) \Rightarrow 2k + 4 = 3k - 3 \Rightarrow k = 7\).
Check with third: \(\frac{3}{7+2} = \frac{3}{9} = \frac{1}{3}\) and \(\frac{7}{3(7)} = \frac{1}{3}\). Correct.
Value of k is 7.
22.
In \(\Delta ABC\), since \(DE \parallel AC\), by Basic Proportionality
Theorem (BPT): \(\frac{BD}{DA} = \frac{BE}{EC}\) ... (i)
In \(\Delta ABE\), since \(DF \parallel AE\), by BPT: \(\frac{BD}{DA} = \frac{BF}{FE}\) ... (ii)
From (i) and (ii), \(\frac{BF}{FE} = \frac{BE}{EC}\). Hence Proved.
In \(\Delta ABE\), since \(DF \parallel AE\), by BPT: \(\frac{BD}{DA} = \frac{BF}{FE}\) ... (ii)
From (i) and (ii), \(\frac{BF}{FE} = \frac{BE}{EC}\). Hence Proved.
23.
Numerator: \(\sin 30^\circ + \tan 45^\circ - \text{cosec } 60^\circ =
\frac{1}{2} + 1 - \frac{2}{\sqrt{3}} = \frac{3}{2} - \frac{2}{\sqrt{3}}\).
Denominator: \(\sec 30^\circ + \cos 60^\circ + \cot 45^\circ = \frac{2}{\sqrt{3}} + \frac{1}{2} + 1 = \frac{2}{\sqrt{3}} + \frac{3}{2}\).
\(\frac{3\sqrt{3}-4}{2\sqrt{3}} \div \frac{4+3\sqrt{3}}{2\sqrt{3}} = \frac{3\sqrt{3}-4}{3\sqrt{3}+4}\).
Rationalizing: \(\frac{(3\sqrt{3}-4)^2}{(3\sqrt{3})^2 - 4^2} = \frac{27 + 16 - 24\sqrt{3}}{27 - 16} = \frac{43 - 24\sqrt{3}}{11}\).
Expression: \(\frac{4\sin \theta - \cos \theta + 1}{4\sin \theta + \cos \theta - 1}\). Divide num. and den. by \(\cos \theta\):
\(= \frac{4\tan \theta - 1 + \sec \theta}{4\tan \theta + 1 - \sec \theta}\). Given \(\tan \theta = 3/4\), so \(\sec \theta = 5/4\).
\(= \frac{4(3/4) - 1 + 5/4}{4(3/4) + 1 - 5/4} = \frac{3 - 1 + 1.25}{3 + 1 - 1.25} = \frac{3.25}{2.75} = \frac{325}{275} = \frac{13}{11}\).
Denominator: \(\sec 30^\circ + \cos 60^\circ + \cot 45^\circ = \frac{2}{\sqrt{3}} + \frac{1}{2} + 1 = \frac{2}{\sqrt{3}} + \frac{3}{2}\).
\(\frac{3\sqrt{3}-4}{2\sqrt{3}} \div \frac{4+3\sqrt{3}}{2\sqrt{3}} = \frac{3\sqrt{3}-4}{3\sqrt{3}+4}\).
Rationalizing: \(\frac{(3\sqrt{3}-4)^2}{(3\sqrt{3})^2 - 4^2} = \frac{27 + 16 - 24\sqrt{3}}{27 - 16} = \frac{43 - 24\sqrt{3}}{11}\).
OR
\(4 \tan \theta = 3 \Rightarrow \tan \theta = 3/4\).Expression: \(\frac{4\sin \theta - \cos \theta + 1}{4\sin \theta + \cos \theta - 1}\). Divide num. and den. by \(\cos \theta\):
\(= \frac{4\tan \theta - 1 + \sec \theta}{4\tan \theta + 1 - \sec \theta}\). Given \(\tan \theta = 3/4\), so \(\sec \theta = 5/4\).
\(= \frac{4(3/4) - 1 + 5/4}{4(3/4) + 1 - 5/4} = \frac{3 - 1 + 1.25}{3 + 1 - 1.25} = \frac{3.25}{2.75} = \frac{325}{275} = \frac{13}{11}\).
24.
The chord of the larger circle touches the smaller circle, so the radius
of the smaller circle is perpendicular to the chord and bisects it.
Let length of chord be \(2x\). Radius of larger \(R=5\), smaller \(r=3\).
In right triangle: \(r^2 + x^2 = R^2 \Rightarrow 3^2 + x^2 = 5^2 \Rightarrow 9 + x^2 = 25 \Rightarrow x^2 = 16 \Rightarrow x = 4\).
Length of chord = \(2x = 8\) cm.
Let length of chord be \(2x\). Radius of larger \(R=5\), smaller \(r=3\).
In right triangle: \(r^2 + x^2 = R^2 \Rightarrow 3^2 + x^2 = 5^2 \Rightarrow 9 + x^2 = 25 \Rightarrow x^2 = 16 \Rightarrow x = 4\).
Length of chord = \(2x = 8\) cm.
25.
Let number of blue balls be \(x\). Total balls = \(5 + x\).
P(Red) = \(\frac{5}{5+x}\). P(Blue) = \(\frac{x}{5+x}\).
Given P(Blue) = 2 \(\times\) P(Red).
\(\frac{x}{5+x} = 2 (\frac{5}{5+x}) \Rightarrow x = 10\).
Number of blue balls = 10.
P(Red) = \(\frac{5}{5+x}\). P(Blue) = \(\frac{x}{5+x}\).
Given P(Blue) = 2 \(\times\) P(Red).
\(\frac{x}{5+x} = 2 (\frac{5}{5+x}) \Rightarrow x = 10\).
Number of blue balls = 10.
SECTION C (Solutions)
26.
Let us assume \(5 - \sqrt{3}\) is rational, say \(r\).
\(5 - \sqrt{3} = r \Rightarrow \sqrt{3} = 5 - r\).
Since 5 is rational and \(r\) is rational, \(5 - r\) is rational. This implies \(\sqrt{3}\) is rational.
But this contradicts the given fact that \(\sqrt{3}\) is irrational. Therefore, our assumption is wrong.
\(5 - \sqrt{3}\) is irrational.
HCF = Product of smallest power of common primes = \(2^2 = 4\).
LCM = Product of highest power of all primes = \(2^5 \times 3 \times 101 = 32 \times 3 \times 101 = 9696\).
Verification: HCF \(\times\) LCM = \(4 \times 9696 = 38784\).
Product of numbers = \(404 \times 96 = 38784\). Hence verified.
\(5 - \sqrt{3} = r \Rightarrow \sqrt{3} = 5 - r\).
Since 5 is rational and \(r\) is rational, \(5 - r\) is rational. This implies \(\sqrt{3}\) is rational.
But this contradicts the given fact that \(\sqrt{3}\) is irrational. Therefore, our assumption is wrong.
\(5 - \sqrt{3}\) is irrational.
OR
Prime factorization: \(404 = 2^2 \times 101\), \(96 = 2^5 \times 3\).HCF = Product of smallest power of common primes = \(2^2 = 4\).
LCM = Product of highest power of all primes = \(2^5 \times 3 \times 101 = 32 \times 3 \times 101 = 9696\).
Verification: HCF \(\times\) LCM = \(4 \times 9696 = 38784\).
Product of numbers = \(404 \times 96 = 38784\). Hence verified.
27.
Polynomial: \(6x^2 - 7x - 3\).
Splitting middle term: \(6x^2 - 9x + 2x - 3 = 3x(2x-3) + 1(2x-3) = (3x+1)(2x-3)\).
Zeroes are \(x = -1/3\) and \(x = 3/2\).
Sum of zeroes: \(-1/3 + 3/2 = (-2+9)/6 = 7/6\). From coeff: \(-b/a = -(-7)/6 = 7/6\). Verified.
Product of zeroes: \((-1/3)(3/2) = -1/2\). From coeff: \(c/a = -3/6 = -1/2\). Verified.
Splitting middle term: \(6x^2 - 9x + 2x - 3 = 3x(2x-3) + 1(2x-3) = (3x+1)(2x-3)\).
Zeroes are \(x = -1/3\) and \(x = 3/2\).
Sum of zeroes: \(-1/3 + 3/2 = (-2+9)/6 = 7/6\). From coeff: \(-b/a = -(-7)/6 = 7/6\). Verified.
Product of zeroes: \((-1/3)(3/2) = -1/2\). From coeff: \(c/a = -3/6 = -1/2\). Verified.
28.
Let fraction be \(x/y\).
Case 1: \(\frac{x-1}{y} = \frac{1}{3} \Rightarrow 3x - 3 = y \Rightarrow 3x - y = 3\) ... (i)
Case 2: \(\frac{x}{y+8} = \frac{1}{4} \Rightarrow 4x = y + 8 \Rightarrow 4x - y = 8\) ... (ii)
Subtract (i) from (ii): \((4x - y) - (3x - y) = 8 - 3 \Rightarrow x = 5\).
Substitute x in (i): \(3(5) - y = 3 \Rightarrow 15 - 3 = y \Rightarrow y = 12\).
Fraction is \(5/12\).
Case 1: \(\frac{x-1}{y} = \frac{1}{3} \Rightarrow 3x - 3 = y \Rightarrow 3x - y = 3\) ... (i)
Case 2: \(\frac{x}{y+8} = \frac{1}{4} \Rightarrow 4x = y + 8 \Rightarrow 4x - y = 8\) ... (ii)
Subtract (i) from (ii): \((4x - y) - (3x - y) = 8 - 3 \Rightarrow x = 5\).
Substitute x in (i): \(3(5) - y = 3 \Rightarrow 15 - 3 = y \Rightarrow y = 12\).
Fraction is \(5/12\).
29.
LHS = \(\frac{\sin \theta / \cos \theta}{1 - \cos \theta / \sin \theta} +
\frac{\cos \theta / \sin \theta}{1 - \sin \theta / \cos \theta}\)
= \(\frac{\sin^2 \theta}{\cos \theta(\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta(\cos \theta - \sin \theta)}\)
= \(\frac{\sin^2 \theta}{\cos \theta(\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta(\sin \theta - \cos \theta)}\)
= \(\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}\)
Use \(a^3 - b^3 = (a-b)(a^2+b^2+ab)\):
= \(\frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}\)
= \(\frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} + 1 = \sec \theta \text{ cosec } \theta + 1\). LHS = RHS.
= \(\frac{\sin^2 \theta}{\cos \theta(\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta(\cos \theta - \sin \theta)}\)
= \(\frac{\sin^2 \theta}{\cos \theta(\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta(\sin \theta - \cos \theta)}\)
= \(\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}\)
Use \(a^3 - b^3 = (a-b)(a^2+b^2+ab)\):
= \(\frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}\)
= \(\frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} + 1 = \sec \theta \text{ cosec } \theta + 1\). LHS = RHS.
30.
Given: Circle with centre O, point P outside. Two tangents PQ and
PR.
To Prove: PQ = PR.
Construction: Join OP, OQ, OR.
Proof: In \(\Delta OQP\) and \(\Delta ORP\):
1. \(\angle OQP = \angle ORP = 90^\circ\) (Tangent perpendicular to radius)
2. OQ = OR (Radii)
3. OP = OP (Common hypotenuse)
By RHS congruence, \(\Delta OQP \cong \Delta ORP\).
By CPCT, PQ = PR. Hence Proved.
To Prove: PQ = PR.
Construction: Join OP, OQ, OR.
Proof: In \(\Delta OQP\) and \(\Delta ORP\):
1. \(\angle OQP = \angle ORP = 90^\circ\) (Tangent perpendicular to radius)
2. OQ = OR (Radii)
3. OP = OP (Common hypotenuse)
By RHS congruence, \(\Delta OQP \cong \Delta ORP\).
By CPCT, PQ = PR. Hence Proved.
31.
| Class | Freq (f) | Cum Freq (cf) |
|---|---|---|
| 40-45 | 2 | 2 |
| 45-50 | 3 | 5 |
| 50-55 | 8 | 13 |
| 55-60 | 6 | 19 |
| 60-65 | 6 | 25 |
| 65-70 | 3 | 28 |
| 70-75 | 2 | 30 |
\(l = 55, cf = 13, f = 6, h = 5\).
Median = \(l + \frac{N/2 - cf}{f} \times h = 55 + \frac{15 - 13}{6} \times 5 = 55 + \frac{10}{6} = 55 + 1.67 = 56.67\) kg.
SECTION D (Solutions)
32.
Let h be height of tower AB. C is first position (angle 30), D is second
(angle 60).
In \(\Delta ABD\) (right angled at B): \(\tan 60^\circ = AB/BD \Rightarrow \sqrt{3} = h/BD \Rightarrow BD = h/\sqrt{3}\).
In \(\Delta ABC\): \(\tan 30^\circ = AB/BC \Rightarrow 1/\sqrt{3} = h/BC \Rightarrow BC = h\sqrt{3}\).
Distance travelled CD = \(BC - BD = h\sqrt{3} - h/\sqrt{3} = \frac{3h-h}{\sqrt{3}} = \frac{2h}{\sqrt{3}}\).
Time taken for CD = 6 sec. Speed = Distance/Time = \(\frac{2h/\sqrt{3}}{6} = \frac{h}{3\sqrt{3}}\) m/s.
Time to cover remaining distance BD: \(Time = \frac{Distance}{Speed} = \frac{h/\sqrt{3}}{h/3\sqrt{3}} = 3\) seconds.
Total time from C to foot = 6 + 3 = 9 seconds.
In \(\Delta ABD\) (right angled at B): \(\tan 60^\circ = AB/BD \Rightarrow \sqrt{3} = h/BD \Rightarrow BD = h/\sqrt{3}\).
In \(\Delta ABC\): \(\tan 30^\circ = AB/BC \Rightarrow 1/\sqrt{3} = h/BC \Rightarrow BC = h\sqrt{3}\).
Distance travelled CD = \(BC - BD = h\sqrt{3} - h/\sqrt{3} = \frac{3h-h}{\sqrt{3}} = \frac{2h}{\sqrt{3}}\).
Time taken for CD = 6 sec. Speed = Distance/Time = \(\frac{2h/\sqrt{3}}{6} = \frac{h}{3\sqrt{3}}\) m/s.
Time to cover remaining distance BD: \(Time = \frac{Distance}{Speed} = \frac{h/\sqrt{3}}{h/3\sqrt{3}} = 3\) seconds.
Total time from C to foot = 6 + 3 = 9 seconds.
33.
Let speed be \(x\) km/h. Time = \(360/x\).
New speed \(x+5\). New Time = \(360/(x+5)\).
Difference is 1 hour: \(\frac{360}{x} - \frac{360}{x+5} = 1\).
\(360 (\frac{x+5-x}{x(x+5)}) = 1 \Rightarrow 360(5) = x^2 + 5x\).
\(x^2 + 5x - 1800 = 0\).
Factoring: \((x+45)(x-40) = 0\). Since speed cannot be negative, \(x = 40\) km/h.
\(\frac{x - (a+b+x)}{x(a+b+x)} = \frac{a+b}{ab} \Rightarrow \frac{-(a+b)}{x^2+ax+bx} = \frac{a+b}{ab}\).
\(-ab = x^2 + ax + bx \Rightarrow x^2 + ax + bx + ab = 0\).
\(x(x+a) + b(x+a) = 0 \Rightarrow (x+a)(x+b) = 0\).
\(x = -a\) or \(x = -b\).
New speed \(x+5\). New Time = \(360/(x+5)\).
Difference is 1 hour: \(\frac{360}{x} - \frac{360}{x+5} = 1\).
\(360 (\frac{x+5-x}{x(x+5)}) = 1 \Rightarrow 360(5) = x^2 + 5x\).
\(x^2 + 5x - 1800 = 0\).
Factoring: \((x+45)(x-40) = 0\). Since speed cannot be negative, \(x = 40\) km/h.
OR
\(\frac{1}{a+b+x} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b}\).\(\frac{x - (a+b+x)}{x(a+b+x)} = \frac{a+b}{ab} \Rightarrow \frac{-(a+b)}{x^2+ax+bx} = \frac{a+b}{ab}\).
\(-ab = x^2 + ax + bx \Rightarrow x^2 + ax + bx + ab = 0\).
\(x(x+a) + b(x+a) = 0 \Rightarrow (x+a)(x+b) = 0\).
\(x = -a\) or \(x = -b\).
34.
Radius \(r = 2.5\) mm. Total Length = 14 mm. Length of cylinder \(h = 14 -
2.5 - 2.5 = 9\) mm.
Surface Area = CSA Cylinder + 2 \(\times\) CSA Hemisphere.
\(= 2\pi rh + 2(2\pi r^2) = 2\pi r (h + 2r)\).
\(= 2 \times \frac{22}{7} \times 2.5 \times (9 + 5) = \frac{110}{7} \times 14 = 220 \text{ mm}^2\).
Total Surface Area = CSA Cylinder + Base Area Cylinder + CSA Hemisphere.
\(= 2\pi rh + \pi r^2 + 2\pi r^2 = 2\pi rh + 3\pi r^2 = \pi r (2h + 3r)\).
\(= \frac{22}{7} \times 3.5 \times (2(10) + 3(3.5)) = 11 \times (20 + 10.5) = 11 \times 30.5 = 335.5 \text{ cm}^2\).
Surface Area = CSA Cylinder + 2 \(\times\) CSA Hemisphere.
\(= 2\pi rh + 2(2\pi r^2) = 2\pi r (h + 2r)\).
\(= 2 \times \frac{22}{7} \times 2.5 \times (9 + 5) = \frac{110}{7} \times 14 = 220 \text{ mm}^2\).
OR
Height \(h=10\) cm, Radius \(r=3.5\) cm.Total Surface Area = CSA Cylinder + Base Area Cylinder + CSA Hemisphere.
\(= 2\pi rh + \pi r^2 + 2\pi r^2 = 2\pi rh + 3\pi r^2 = \pi r (2h + 3r)\).
\(= \frac{22}{7} \times 3.5 \times (2(10) + 3(3.5)) = 11 \times (20 + 10.5) = 11 \times 30.5 = 335.5 \text{ cm}^2\).
35.
Total Frequency \(N=60\). \(5 + x + 20 + 15 + y + 5 = 60 \Rightarrow x + y
= 15\) ... (i)
Median is 28.5, so median class is 20-30.
\(l=20, h=10, f=20, cf=5+x\).
\(Median = l + \frac{N/2 - cf}{f} \times h\).
\(28.5 = 20 + \frac{30 - (5+x)}{20} \times 10\).
\(8.5 = \frac{25-x}{2} \Rightarrow 17 = 25 - x \Rightarrow x = 8\).
From (i), \(8 + y = 15 \Rightarrow y = 7\).
Median is 28.5, so median class is 20-30.
\(l=20, h=10, f=20, cf=5+x\).
\(Median = l + \frac{N/2 - cf}{f} \times h\).
\(28.5 = 20 + \frac{30 - (5+x)}{20} \times 10\).
\(8.5 = \frac{25-x}{2} \Rightarrow 17 = 25 - x \Rightarrow x = 8\).
From (i), \(8 + y = 15 \Rightarrow y = 7\).
SECTION E (Solutions)
Case Study - 1 Solution
1. Initial time \(a = 51\). Reduction \(d = -2\).
Terms in AP: 51, 49, 47, ...
2. Target time \(a_n = 31\).
\(a + (n-1)d = 31 \Rightarrow 51 + (n-1)(-2) = 31\).
\(-2(n-1) = -20 \Rightarrow n-1 = 10 \Rightarrow n = 11\).
Minimum number of days practice = 11.
3. \(a_n = 2n + 3\).
\(a_1 = 2(1)+3 = 5\). \(a_2 = 2(2)+3 = 7\).
Common difference \(d = a_2 - a_1 = 7 - 5 = 2\).
1. Initial time \(a = 51\). Reduction \(d = -2\).
Terms in AP: 51, 49, 47, ...
2. Target time \(a_n = 31\).
\(a + (n-1)d = 31 \Rightarrow 51 + (n-1)(-2) = 31\).
\(-2(n-1) = -20 \Rightarrow n-1 = 10 \Rightarrow n = 11\).
Minimum number of days practice = 11.
3. \(a_n = 2n + 3\).
\(a_1 = 2(1)+3 = 5\). \(a_2 = 2(2)+3 = 7\).
Common difference \(d = a_2 - a_1 = 7 - 5 = 2\).
Case Study - 2 Solution
Coordinates: A(3, 4), B(6, 7), C(9, 4), D(6, 1).
1. Distance AB = \(\sqrt{(6-3)^2 + (7-4)^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}\) units.
2. Distance AC = \(\sqrt{(9-3)^2 + (4-4)^2} = \sqrt{36} = 6\) units.
3. Sides: AB=\(3\sqrt{2}\), BC=\(\sqrt{3^2+(-3)^2}=3\sqrt{2}\), CD=\(3\sqrt{2}\), DA=\(3\sqrt{2}\). All sides equal.
Diagonals: AC=6, BD=\(\sqrt{(6-6)^2+(7-1)^2}=6\). Diagonals are equal.
Since sides are equal and diagonals are equal, ABCD is a Square.
OR
Midpoint of AC (Ratio 1:1): \((\frac{3+9}{2}, \frac{4+4}{2}) = (\frac{12}{2}, \frac{8}{2}) = (6, 4)\).
Coordinates: A(3, 4), B(6, 7), C(9, 4), D(6, 1).
1. Distance AB = \(\sqrt{(6-3)^2 + (7-4)^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}\) units.
2. Distance AC = \(\sqrt{(9-3)^2 + (4-4)^2} = \sqrt{36} = 6\) units.
3. Sides: AB=\(3\sqrt{2}\), BC=\(\sqrt{3^2+(-3)^2}=3\sqrt{2}\), CD=\(3\sqrt{2}\), DA=\(3\sqrt{2}\). All sides equal.
Diagonals: AC=6, BD=\(\sqrt{(6-6)^2+(7-1)^2}=6\). Diagonals are equal.
Since sides are equal and diagonals are equal, ABCD is a Square.
OR
Midpoint of AC (Ratio 1:1): \((\frac{3+9}{2}, \frac{4+4}{2}) = (\frac{12}{2}, \frac{8}{2}) = (6, 4)\).
Case Study - 3 Solution
Height of India Gate = 42 m.
1. Angle of elevation \(\theta\). Distance = 42m.
\(\tan \theta = \frac{Height}{Distance} = \frac{42}{42} = 1 \Rightarrow \theta = 45^\circ\).
2. Angle = \(60^\circ\). Distance = \(x\).
\(\tan 60^\circ = \frac{42}{x} \Rightarrow \sqrt{3} = \frac{42}{x} \Rightarrow x = \frac{42}{\sqrt{3}} = 14\sqrt{3}\) m.
3. Angle \(60^\circ\), Shadow length 20m. Height \(h\).
\(\tan 60^\circ = h/20 \Rightarrow h = 20\sqrt{3}\) m.
OR
Ratio Rod:Shadow = \(1:\sqrt{3}\).
\(\tan \theta = \frac{1}{\sqrt{3}} \Rightarrow \theta = 30^\circ\).
Height of India Gate = 42 m.
1. Angle of elevation \(\theta\). Distance = 42m.
\(\tan \theta = \frac{Height}{Distance} = \frac{42}{42} = 1 \Rightarrow \theta = 45^\circ\).
2. Angle = \(60^\circ\). Distance = \(x\).
\(\tan 60^\circ = \frac{42}{x} \Rightarrow \sqrt{3} = \frac{42}{x} \Rightarrow x = \frac{42}{\sqrt{3}} = 14\sqrt{3}\) m.
3. Angle \(60^\circ\), Shadow length 20m. Height \(h\).
\(\tan 60^\circ = h/20 \Rightarrow h = 20\sqrt{3}\) m.
OR
Ratio Rod:Shadow = \(1:\sqrt{3}\).
\(\tan \theta = \frac{1}{\sqrt{3}} \Rightarrow \theta = 30^\circ\).