Mock Test Paper for Class 10 Board Exam (2025-26)
📝 Exam Instructions:
- This is the Final Practice Set. Solve it like a real exam.
- Questions are based on the latest CBSE pattern (Competency-based).
- Do not use calculators.
- Use the "View Answer" button to check your solution after attempting.
Section A (MCQs - 20 Marks)
If HCF(26, 169) = 13, then LCM(26, 169) is:
(A) 26
(B) 52
(C) 338
(D) 13
View Answer
Option (C)
Explanation: $LCM = \frac{a \times b}{HCF} = \frac{26 \times 169}{13} = 26 \times 13 = 338$.
Explanation: $LCM = \frac{a \times b}{HCF} = \frac{26 \times 169}{13} = 26 \times 13 = 338$.
If $\alpha$ and $\beta$ are zeroes of the polynomial $x^2 - p(x+1) - c$, then $(\alpha+1)(\beta+1)$ is equal to:
(A) c
(B) c - 1
(C) 1 - c
(D) 1 + c
View Answer
Option (C)
Explanation: Poly is $x^2 - px - (p+c)$. $\alpha+\beta = p$, $\alpha\beta = -(p+c)$.
$(\alpha+1)(\beta+1) = \alpha\beta + (\alpha+\beta) + 1 = -(p+c) + p + 1 = 1-c$.
Explanation: Poly is $x^2 - px - (p+c)$. $\alpha+\beta = p$, $\alpha\beta = -(p+c)$.
$(\alpha+1)(\beta+1) = \alpha\beta + (\alpha+\beta) + 1 = -(p+c) + p + 1 = 1-c$.
For what value of $k$ do the equations $3x - y + 8 = 0$ and $6x - ky = -16$ represent coincident lines?
(A) 1/2
(B) -1/2
(C) 2
(D) -2
View Answer
Option (C)
Explanation: $\frac{3}{6} = \frac{-1}{-k} = \frac{8}{16}$.
$1/2 = 1/k \Rightarrow k=2$.
Explanation: $\frac{3}{6} = \frac{-1}{-k} = \frac{8}{16}$.
$1/2 = 1/k \Rightarrow k=2$.
The nature of roots of the quadratic equation $2x^2 - 4x + 3 = 0$ is:
(A) Real and Equal
(B) Real and Distinct
(C) No Real Roots
(D) None of these
View Answer
Option (C)
Explanation: $D = b^2 - 4ac = (-4)^2 - 4(2)(3) = 16 - 24 = -8$. Since $D < 0$, no real roots.
Explanation: $D = b^2 - 4ac = (-4)^2 - 4(2)(3) = 16 - 24 = -8$. Since $D < 0$, no real roots.
If the common difference of an AP is 5, then the value of $a_{18} - a_{13}$ is:
(A) 5
(B) 20
(C) 25
(D) 30
View Answer
Option (C)
Explanation: $a_{18} - a_{13} = (a+17d) - (a+12d) = 5d = 5(5) = 25$.
Explanation: $a_{18} - a_{13} = (a+17d) - (a+12d) = 5d = 5(5) = 25$.
It is given that $\Delta ABC \sim \Delta PQR$ with $\frac{BC}{QR} = \frac{1}{3}$. Then $\frac{ar(\Delta PQR)}{ar(\Delta ABC)}$ is:
(A) 9
(B) 3
(C) 1/3
(D) 1/9
View Answer
Option (A)
Explanation: Ratio of areas = square of ratio of corresponding sides.
$\frac{Area(PQR)}{Area(ABC)} = (\frac{QR}{BC})^2 = (\frac{3}{1})^2 = 9$.
Explanation: Ratio of areas = square of ratio of corresponding sides.
$\frac{Area(PQR)}{Area(ABC)} = (\frac{QR}{BC})^2 = (\frac{3}{1})^2 = 9$.
The point on x-axis which is equidistant from points A(-1, 0) and B(5, 0) is:
(A) (0, 2)
(B) (2, 0)
(C) (3, 0)
(D) (0, 3)
View Answer
Option (B)
Explanation: Midpoint of AB = $(\frac{-1+5}{2}, \frac{0+0}{2}) = (2, 0)$.
Explanation: Midpoint of AB = $(\frac{-1+5}{2}, \frac{0+0}{2}) = (2, 0)$.
If $\tan \theta = \frac{a}{b}$, then $\frac{a \sin \theta - b \cos \theta}{a \sin \theta + b \cos \theta}$ is:
(A) $\frac{a^2+b^2}{a^2-b^2}$
(B) $\frac{a^2-b^2}{a^2+b^2}$
(C) $\frac{a^2}{a^2+b^2}$
(D) $\frac{b^2}{a^2+b^2}$
View Answer
Option (B)
Explanation: Divide num and den by $\cos \theta$.
Expression becomes $\frac{a \tan \theta - b}{a \tan \theta + b}$. Substitute $\tan \theta = a/b$.
Result = $\frac{a(a/b)-b}{a(a/b)+b} = \frac{a^2-b^2}{a^2+b^2}$.
Explanation: Divide num and den by $\cos \theta$.
Expression becomes $\frac{a \tan \theta - b}{a \tan \theta + b}$. Substitute $\tan \theta = a/b$.
Result = $\frac{a(a/b)-b}{a(a/b)+b} = \frac{a^2-b^2}{a^2+b^2}$.
The value of $9 \sec^2 A - 9 \tan^2 A$ is:
(A) 1
(B) 9
(C) 8
(D) 0
View Answer
Option (B)
Explanation: $9(\sec^2 A - \tan^2 A) = 9(1) = 9$.
Explanation: $9(\sec^2 A - \tan^2 A) = 9(1) = 9$.
If angle between two radii of a circle is $130^{\circ}$, the angle between the tangents at the ends of the radii is:
(A) $90^{\circ}$
(B) $50^{\circ}$
(C) $70^{\circ}$
(D) $40^{\circ}$
View Answer
Option (B)
Explanation: Angles are supplementary. $180^{\circ} - 130^{\circ} = 50^{\circ}$.
Explanation: Angles are supplementary. $180^{\circ} - 130^{\circ} = 50^{\circ}$.
The length of the minute hand of a clock is 14 cm. The area swept by the minute hand in 5 minutes is:
(A) 154 cm²
(B) 44 cm²
(C) 51.33 cm²
(D) 54 cm²
View Answer
Option (C)
Explanation: Angle in 5 mins = $30^{\circ}$. Area = $\frac{30}{360} \times \frac{22}{7} \times 14 \times 14 = \frac{154}{3} = 51.33$.
Explanation: Angle in 5 mins = $30^{\circ}$. Area = $\frac{30}{360} \times \frac{22}{7} \times 14 \times 14 = \frac{154}{3} = 51.33$.
Two cubes each of volume 64 cm³ are joined end to end. The surface area of the resulting cuboid is:
(A) 128 cm²
(B) 160 cm²
(C) 192 cm²
(D) 320 cm²
View Answer
Option (B)
Explanation: Side $a=4$. Cuboid dimensions: $L=8, B=4, H=4$.
SA = $2(LB+BH+HL) = 2(32+16+32) = 160$.
Explanation: Side $a=4$. Cuboid dimensions: $L=8, B=4, H=4$.
SA = $2(LB+BH+HL) = 2(32+16+32) = 160$.
The class mark of the class interval 10 - 25 is:
(A) 17
(B) 17.5
(C) 18
(D) 15
View Answer
Option (B)
Explanation: $\frac{10+25}{2} = \frac{35}{2} = 17.5$.
Explanation: $\frac{10+25}{2} = \frac{35}{2} = 17.5$.
A die is thrown once. The probability of getting a prime number is:
(A) 2/3
(B) 1/3
(C) 1/2
(D) 1/6
View Answer
Option (C)
Explanation: Primes on die: 2, 3, 5. Total 3 outcomes. $P=3/6=1/2$.
Explanation: Primes on die: 2, 3, 5. Total 3 outcomes. $P=3/6=1/2$.
The distance of the point P(2, 3) from the x-axis is:
(A) 2
(B) 3
(C) 1
(D) 5
View Answer
Option (B)
Explanation: Distance from x-axis is the y-coordinate.
Explanation: Distance from x-axis is the y-coordinate.
If $x=a, y=b$ is the solution of equations $x-y=2$ and $x+y=4$, then values of a and b are:
(A) 3 and 5
(B) 5 and 3
(C) 3 and 1
(D) -1 and -3
View Answer
Option (C)
Explanation: Add equations: $2x=6 \Rightarrow x=3$. Put in $x+y=4 \Rightarrow 3+y=4 \Rightarrow y=1$.
Explanation: Add equations: $2x=6 \Rightarrow x=3$. Put in $x+y=4 \Rightarrow 3+y=4 \Rightarrow y=1$.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is:
(A) 12 cm
(B) 13 cm
(C) 8.5 cm
(D) $\sqrt{119}$ cm
View Answer
Option (D)
Explanation: $PQ^2 = OQ^2 - OP^2 = 12^2 - 5^2 = 144 - 25 = 119$. $PQ = \sqrt{119}$.
Explanation: $PQ^2 = OQ^2 - OP^2 = 12^2 - 5^2 = 144 - 25 = 119$. $PQ = \sqrt{119}$.
If 1/2 is a root of the equation $x^2 + kx - 5/4 = 0$, then the value of k is:
(A) 2
(B) -2
(C) 1/4
(D) 1/2
View Answer
Option (A)
Explanation: Put $x=1/2$. $(1/4) + k(1/2) - 5/4 = 0 \Rightarrow k/2 = 1 \Rightarrow k=2$.
Explanation: Put $x=1/2$. $(1/4) + k(1/2) - 5/4 = 0 \Rightarrow k/2 = 1 \Rightarrow k=2$.
Assertion (A): The point (0, 4) lies on y-axis.
Reason (R): The x-coordinate on the point on y-axis is zero.
Reason (R): The x-coordinate on the point on y-axis is zero.
(A) Both true, R explains A
(B) Both true, R doesn't explain A
(C) A is true, R is false
(D) A is false, R is true
View Answer
Option (A)
Explanation: Both statements are correct and Reason explains why (0,4) is on y-axis.
Explanation: Both statements are correct and Reason explains why (0,4) is on y-axis.
Assertion (A): If the circumference of two circles are in the ratio 2:3, then their areas are in the ratio 4:9.
Reason (R): Area of circle = $2\pi r$.
Reason (R): Area of circle = $2\pi r$.
(A) Both true, R explains A
(B) Both true, R doesn't explain A
(C) A is true, R is false
(D) A is false, R is true
View Answer
Option (C)
Explanation: A is True (ratio squared). R is False because Area = $\pi r^2$.
Explanation: A is True (ratio squared). R is False because Area = $\pi r^2$.
Section B (2 Marks Each)
Prove that $3\sqrt{2}$ is irrational, assuming $\sqrt{2}$ is irrational.
Show Solution
Let $3\sqrt{2} = a/b$ (rational).
$\sqrt{2} = a/3b$.
Since $a, b, 3$ are integers, $a/3b$ is rational.
But $\sqrt{2}$ is irrational. This is a contradiction.
Hence, $3\sqrt{2}$ is irrational.
$\sqrt{2} = a/3b$.
Since $a, b, 3$ are integers, $a/3b$ is rational.
But $\sqrt{2}$ is irrational. This is a contradiction.
Hence, $3\sqrt{2}$ is irrational.
In a trapezium ABCD, $AB \parallel DC$ and its diagonals intersect at O. Show that $\frac{AO}{BO} = \frac{CO}{DO}$.
Show Solution
In $\Delta AOB$ and $\Delta COD$:
1. $\angle OAB = \angle OCD$ (Alt. interior)
2. $\angle OBA = \angle ODC$ (Alt. interior)
By AA Similarity, $\Delta AOB \sim \Delta COD$.
$\therefore \frac{AO}{CO} = \frac{BO}{DO} \Rightarrow \frac{AO}{BO} = \frac{CO}{DO}$.
1. $\angle OAB = \angle OCD$ (Alt. interior)
2. $\angle OBA = \angle ODC$ (Alt. interior)
By AA Similarity, $\Delta AOB \sim \Delta COD$.
$\therefore \frac{AO}{CO} = \frac{BO}{DO} \Rightarrow \frac{AO}{BO} = \frac{CO}{DO}$.
Evaluate: $\frac{2 \tan 30^{\circ}}{1 + \tan^2 30^{\circ}}$.
Show Solution
$\tan 30 = 1/\sqrt{3}$.
Num: $2(1/\sqrt{3}) = 2/\sqrt{3}$.
Den: $1 + (1/\sqrt{3})^2 = 1 + 1/3 = 4/3$.
Result: $\frac{2/\sqrt{3}}{4/3} = \frac{2}{\sqrt{3}} \times \frac{3}{4} = \frac{\sqrt{3}}{2}$ (which is $\sin 60$).
Num: $2(1/\sqrt{3}) = 2/\sqrt{3}$.
Den: $1 + (1/\sqrt{3})^2 = 1 + 1/3 = 4/3$.
Result: $\frac{2/\sqrt{3}}{4/3} = \frac{2}{\sqrt{3}} \times \frac{3}{4} = \frac{\sqrt{3}}{2}$ (which is $\sin 60$).
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Show Solution
Let $R=5, r=3$. The chord is tangent to smaller circle.
Half chord length = $\sqrt{R^2 - r^2} = \sqrt{25 - 9} = \sqrt{16} = 4$ cm.
Total chord length = $2 \times 4 = 8$ cm.
Half chord length = $\sqrt{R^2 - r^2} = \sqrt{25 - 9} = \sqrt{16} = 4$ cm.
Total chord length = $2 \times 4 = 8$ cm.
Find the area of a sector of a circle of radius 21 cm and central angle $120^{\circ}$.
Show Solution
Area = $\frac{\theta}{360} \pi r^2$
$= \frac{120}{360} \times \frac{22}{7} \times 21 \times 21$
$= \frac{1}{3} \times 22 \times 3 \times 21$
$= 22 \times 21 = 462$ cm².
$= \frac{120}{360} \times \frac{22}{7} \times 21 \times 21$
$= \frac{1}{3} \times 22 \times 3 \times 21$
$= 22 \times 21 = 462$ cm².
Section C (3 Marks Each)
Find the zeroes of the polynomial $f(x) = \sqrt{3}x^2 - 8x + 4\sqrt{3}$.
Show Solution
Factorize: $\sqrt{3}x^2 - 6x - 2x + 4\sqrt{3}$
$\sqrt{3}x(x - 2\sqrt{3}) - 2(x - 2\sqrt{3})$
$(\sqrt{3}x - 2)(x - 2\sqrt{3}) = 0$
Zeroes: $2/\sqrt{3}$ and $2\sqrt{3}$.
$\sqrt{3}x(x - 2\sqrt{3}) - 2(x - 2\sqrt{3})$
$(\sqrt{3}x - 2)(x - 2\sqrt{3}) = 0$
Zeroes: $2/\sqrt{3}$ and $2\sqrt{3}$.
Taxi charges consist of a fixed charge and charge per km. For 10km, the charge is ₹105. For 15km, the charge is ₹155. Find the charge for 25km.
Show Solution
Let fixed = x, per km = y.
1) $x + 10y = 105$
2) $x + 15y = 155$
Subtract (1) from (2): $5y = 50 \Rightarrow y=10$.
Put in (1): $x + 100 = 105 \Rightarrow x=5$.
For 25km: $x + 25y = 5 + 25(10) = 255$.
Ans: ₹255.
1) $x + 10y = 105$
2) $x + 15y = 155$
Subtract (1) from (2): $5y = 50 \Rightarrow y=10$.
Put in (1): $x + 100 = 105 \Rightarrow x=5$.
For 25km: $x + 25y = 5 + 25(10) = 255$.
Ans: ₹255.
Find the point on the x-axis which is equidistant from $(2, -5)$ and $(-2, 9)$.
Show Solution
Let point P(x, 0).
$(x-2)^2 + (0+5)^2 = (x+2)^2 + (0-9)^2$
$x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81$
$-4x + 29 = 4x + 85$
$-8x = 56 \Rightarrow x = -7$.
Point is $(-7, 0)$.
$(x-2)^2 + (0+5)^2 = (x+2)^2 + (0-9)^2$
$x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81$
$-4x + 29 = 4x + 85$
$-8x = 56 \Rightarrow x = -7$.
Point is $(-7, 0)$.
Prove that: $\frac{\cos A}{1 - \tan A} + \frac{\sin A}{1 - \cot A} = \sin A + \cos A$.
Show Solution
Change to sin/cos.
LHS = $\frac{\cos A}{1 - \frac{\sin A}{\cos A}} + \frac{\sin A}{1 - \frac{\cos A}{\sin A}}$
$= \frac{\cos^2 A}{\cos A - \sin A} + \frac{\sin^2 A}{\sin A - \cos A}$
$= \frac{\cos^2 A - \sin^2 A}{\cos A - \sin A} = \frac{(\cos A - \sin A)(\cos A + \sin A)}{\cos A - \sin A}$
$= \cos A + \sin A$. (Proved)
LHS = $\frac{\cos A}{1 - \frac{\sin A}{\cos A}} + \frac{\sin A}{1 - \frac{\cos A}{\sin A}}$
$= \frac{\cos^2 A}{\cos A - \sin A} + \frac{\sin^2 A}{\sin A - \cos A}$
$= \frac{\cos^2 A - \sin^2 A}{\cos A - \sin A} = \frac{(\cos A - \sin A)(\cos A + \sin A)}{\cos A - \sin A}$
$= \cos A + \sin A$. (Proved)
A quadrilateral ABCD is drawn to circumscribe a circle. Prove that $AB + CD = AD + BC$.
Show Solution
Let points of contact be P, Q, R, S.
AP=AS, BP=BQ, CR=CQ, DR=DS (Tangents from external point).
Add all: $(AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ)$.
$\Rightarrow AB + CD = AD + BC$.
AP=AS, BP=BQ, CR=CQ, DR=DS (Tangents from external point).
Add all: $(AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ)$.
$\Rightarrow AB + CD = AD + BC$.
All the face cards (Kings, Queens, Jacks) are removed from a deck of 52 playing cards. A card is drawn at random. Find probability of getting: (i) A red card (ii) An ace.
Show Solution
Removed = $4 \times 3 = 12$. Remaining = 40 cards.
(i) Red cards = 26 (total) - 6 (face) = 20. Prob = $20/40 = 1/2$.
(ii) Aces are not removed. Total 4. Prob = $4/40 = 1/10$.
(i) Red cards = 26 (total) - 6 (face) = 20. Prob = $20/40 = 1/2$.
(ii) Aces are not removed. Total 4. Prob = $4/40 = 1/10$.
Section D (5 Marks Each)
A motorboat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
Show Solution
Let stream speed = x.
Upstream = $18-x$, Downstream = $18+x$.
Time diff: $\frac{24}{18-x} - \frac{24}{18+x} = 1$
$24[\frac{18+x - (18-x)}{324-x^2}] = 1$
$24(2x) = 324 - x^2$
$x^2 + 48x - 324 = 0$
$(x+54)(x-6)=0$. Speed cannot be negative.
Stream speed = 6 km/h.
Upstream = $18-x$, Downstream = $18+x$.
Time diff: $\frac{24}{18-x} - \frac{24}{18+x} = 1$
$24[\frac{18+x - (18-x)}{324-x^2}] = 1$
$24(2x) = 324 - x^2$
$x^2 + 48x - 324 = 0$
$(x+54)(x-6)=0$. Speed cannot be negative.
Stream speed = 6 km/h.
A vertical pole of length 6m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28m long. Find the height of the tower.
Show Solution
Let height of tower be h.
At the same time, sun's elevation is same.
$\Delta ABC \sim \Delta PQR$.
$\frac{\text{Height of Pole}}{\text{Shadow of Pole}} = \frac{\text{Height of Tower}}{\text{Shadow of Tower}}$
$\frac{6}{4} = \frac{h}{28}$
$h = \frac{6 \times 28}{4} = 6 \times 7 = 42$ m.
At the same time, sun's elevation is same.
$\Delta ABC \sim \Delta PQR$.
$\frac{\text{Height of Pole}}{\text{Shadow of Pole}} = \frac{\text{Height of Tower}}{\text{Shadow of Tower}}$
$\frac{6}{4} = \frac{h}{28}$
$h = \frac{6 \times 28}{4} = 6 \times 7 = 42$ m.
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used. Also find the cost at ₹500 per m².
Show Solution
$r=2$. $h_{cyl} = 2.1$, $l=2.8$.
Area = CSA Cylinder + CSA Cone
$= 2\pi rh + \pi rl = \pi r(2h+l)$
$= \frac{22}{7} \times 2 (2(2.1) + 2.8)$
$= \frac{44}{7} (4.2+2.8) = \frac{44}{7}(7) = 44$ m².
Cost = $44 \times 500 = ₹22,000$.
Area = CSA Cylinder + CSA Cone
$= 2\pi rh + \pi rl = \pi r(2h+l)$
$= \frac{22}{7} \times 2 (2(2.1) + 2.8)$
$= \frac{44}{7} (4.2+2.8) = \frac{44}{7}(7) = 44$ m².
Cost = $44 \times 500 = ₹22,000$.
The mean of the following distribution is 18. Find the missing frequency f.
Class: 11-13 (7), 13-15 (6), 15-17 (9), 17-19 (13), 19-21 (f), 21-23 (5), 23-25 (4).
Class: 11-13 (7), 13-15 (6), 15-17 (9), 17-19 (13), 19-21 (f), 21-23 (5), 23-25 (4).
Show Solution
1. Class marks ($x_i$): 12, 14, 16, 18, 20, 22, 24.
2. $\Sigma f_i = 44 + f$.
3. $\Sigma f_i x_i = 752 + 20f$.
4. Mean formula: $18 = \frac{752+20f}{44+f}$
$792 + 18f = 752 + 20f$
$40 = 2f \Rightarrow f = 20$.
2. $\Sigma f_i = 44 + f$.
3. $\Sigma f_i x_i = 752 + 20f$.
4. Mean formula: $18 = \frac{752+20f}{44+f}$
$792 + 18f = 752 + 20f$
$40 = 2f \Rightarrow f = 20$.
Section E (Case Study - 12 Marks)
School Auditorium: In an auditorium, seats are arranged in rows. The first row has 20 seats, the second row has 22 seats, the third has 24, and so on.
(i) How many seats are there in the 15th row? (1 M)
(ii) If there are 30 rows, find total seats. (2 M)
(iii) Which row has 60 seats? (1 M)
(i) How many seats are there in the 15th row? (1 M)
(ii) If there are 30 rows, find total seats. (2 M)
(iii) Which row has 60 seats? (1 M)
Show Answer
AP: 20, 22, 24... ($a=20, d=2$)
(i) $a_{15} = 20 + 14(2) = 48$.
(ii) $S_{30} = \frac{30}{2}[2(20) + 29(2)] = 15(40+58) = 1470$.
(iii) $60 = 20 + (n-1)2 \Rightarrow 40 = 2(n-1) \Rightarrow n=21$.
(i) $a_{15} = 20 + 14(2) = 48$.
(ii) $S_{30} = \frac{30}{2}[2(20) + 29(2)] = 15(40+58) = 1470$.
(iii) $60 = 20 + (n-1)2 \Rightarrow 40 = 2(n-1) \Rightarrow n=21$.
GPS Mapping: Three villages A, B, and C are located at coordinates A(1, 2), B(4, 6) and C(5, 7) on a map.
(i) Find the distance between village A and B. (1 M)
(ii) Find the midpoint of the path connecting B and C. (1 M)
(iii) A well is to be dug at a point P on AB such that AP:PB = 1:2. Find coordinates of P. (2 M)
(i) Find the distance between village A and B. (1 M)
(ii) Find the midpoint of the path connecting B and C. (1 M)
(iii) A well is to be dug at a point P on AB such that AP:PB = 1:2. Find coordinates of P. (2 M)
Show Answer
(i) $\sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9+16} = 5$.
(ii) $(\frac{4+5}{2}, \frac{6+7}{2}) = (4.5, 6.5)$.
(iii) Section formula: $(\frac{1(4)+2(1)}{3}, \frac{1(6)+2(2)}{3}) = (2, 10/3)$.
(ii) $(\frac{4+5}{2}, \frac{6+7}{2}) = (4.5, 6.5)$.
(iii) Section formula: $(\frac{1(4)+2(1)}{3}, \frac{1(6)+2(2)}{3}) = (2, 10/3)$.
Hot Air Balloon: A girl 1.2 m tall spots a balloon moving with the wind in a horizontal line at a height of 88.2 m. The angle of elevation is $60^{\circ}$. After some time, angle reduces to $30^{\circ}$.
(i) What is the effective height for calculation? (1 M)
(ii) Find the distance traveled by the balloon during the interval. (2 M)
(iii) Find distance of balloon from girl in first case. (1 M)
(i) What is the effective height for calculation? (1 M)
(ii) Find the distance traveled by the balloon during the interval. (2 M)
(iii) Find distance of balloon from girl in first case. (1 M)
Show Answer
(i) $88.2 - 1.2 = 87$ m.
(ii) Dist 1: $x = 87/\tan 60 = 87/\sqrt{3} = 29\sqrt{3}$.
Dist 2: $y = 87/\tan 30 = 87\sqrt{3}$.
Distance travelled = $87\sqrt{3} - 29\sqrt{3} = 58\sqrt{3}$ m.
(iii) $\sin 60 = 87/H \Rightarrow H = 174/\sqrt{3} = 58\sqrt{3}$.
(ii) Dist 1: $x = 87/\tan 60 = 87/\sqrt{3} = 29\sqrt{3}$.
Dist 2: $y = 87/\tan 30 = 87\sqrt{3}$.
Distance travelled = $87\sqrt{3} - 29\sqrt{3} = 58\sqrt{3}$ m.
(iii) $\sin 60 = 87/H \Rightarrow H = 174/\sqrt{3} = 58\sqrt{3}$.
--- End of Set 2 ---