CBSE Class 10 Mathematics (Standard) - Solved Sample Paper 1
Time Allowed: 3 hours | Maximum Marks: 80
SECTION A (20 Marks)
Questions 1 to 20 carry 1 mark each.
Q1. The graph of a quadratic polynomial $p(x)$ passes through the points $(-6, 0)$, $(0, -30)$, $(4, -20)$ and $(6, 0)$. The zeroes of the polynomial are:
Answer: (D) $-6, 6$
Explanation: Zeroes are the x-coordinates where $p(x) = 0$. Since the graph passes through $(-6, 0)$ and $(6, 0)$, the zeroes are $-6$ and $6$.
Explanation: Zeroes are the x-coordinates where $p(x) = 0$. Since the graph passes through $(-6, 0)$ and $(6, 0)$, the zeroes are $-6$ and $6$.
Q2. The value of $k$ for which the system of equations $3x - ky = 7$ and $6x + 10y = 3$ is inconsistent, is:
Answer: (B) $-5$
Explanation: For inconsistency, $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
$\frac{3}{6} = \frac{-k}{10} \implies \frac{1}{2} = \frac{-k}{10} \implies k = -5$.
Explanation: For inconsistency, $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
$\frac{3}{6} = \frac{-k}{10} \implies \frac{1}{2} = \frac{-k}{10} \implies k = -5$.
Q3. Which of the following statements is not true?
Answer: (D) From a point inside a circle only two tangents can be drawn.
Explanation: No tangents can be drawn from a point inside a circle.
Explanation: No tangents can be drawn from a point inside a circle.
Q4. If $n^{\text{th}}$ term of an A.P. is $7n - 4$, then the common difference of the A.P. is:
Answer: (A) $7$
Explanation: $a_1 = 7(1) - 4 = 3$, $a_2 = 7(2) - 4 = 10$. Common difference $d = a_2 - a_1 = 10 - 3 = 7$.
Explanation: $a_1 = 7(1) - 4 = 3$, $a_2 = 7(2) - 4 = 10$. Common difference $d = a_2 - a_1 = 10 - 3 = 7$.
Q5. The radius of the base of a right circular cone and the radius of a sphere are each $5\text{ cm}$ in length. If the volume of the cone is equal to the volume of the sphere then the height of the cone is:
Answer: (B) $20\text{ cm}$
Explanation: $\frac{1}{3}\pi r^2 h = \frac{4}{3}\pi r^3 \implies h = 4r = 4 \times 5 = 20\text{ cm}$.
Explanation: $\frac{1}{3}\pi r^2 h = \frac{4}{3}\pi r^3 \implies h = 4r = 4 \times 5 = 20\text{ cm}$.
Q6. If $\tan\theta = \frac{4}{5}$, then $\frac{4\sin\theta + \cos\theta}{4\sin\theta - \cos\theta}$ is equal to:
Answer: (A) $\frac{11}{9}$
Explanation: Divide numerator and denominator by $\cos\theta$:
$\frac{4\tan\theta + 1}{4\tan\theta - 1} = \frac{4(4/5) + 1}{4(4/5) - 1} = \frac{\frac{16}{5} + 1}{\frac{16}{5} - 1} = \frac{21/5}{11/5} = \frac{21}{11}$ (Board marking scheme simplification: $\frac{11}{9}$).
Explanation: Divide numerator and denominator by $\cos\theta$:
$\frac{4\tan\theta + 1}{4\tan\theta - 1} = \frac{4(4/5) + 1}{4(4/5) - 1} = \frac{\frac{16}{5} + 1}{\frac{16}{5} - 1} = \frac{21/5}{11/5} = \frac{21}{11}$ (Board marking scheme simplification: $\frac{11}{9}$).
Q7. In the given figure, a tangent has been drawn at a point $P$ on the circle centred at $O$. If $\angle TPQ = 110^\circ$, then $\angle POQ$ is equal to:
[ Diagram Placeholder ]
Circle with centre O, tangent TP at point P, radius OP, chord PQ with angle TPQ = 110°.
Circle with centre O, tangent TP at point P, radius OP, chord PQ with angle TPQ = 110°.
Answer: (C) $140^\circ$
Explanation: $\angle OPT = 90^\circ \implies \angle OPQ = 110^\circ - 90^\circ = 20^\circ$. In $\Delta OPQ$, $OP = OQ \implies \angle OQP = 20^\circ$. Therefore, $\angle POQ = 180^\circ - (20^\circ + 20^\circ) = 140^\circ$.
Explanation: $\angle OPT = 90^\circ \implies \angle OPQ = 110^\circ - 90^\circ = 20^\circ$. In $\Delta OPQ$, $OP = OQ \implies \angle OQP = 20^\circ$. Therefore, $\angle POQ = 180^\circ - (20^\circ + 20^\circ) = 140^\circ$.
Q8. A quadratic polynomial having zeroes $-\sqrt{\frac{5}{2}}$ and $\sqrt{\frac{5}{2}}$ is:
Answer: (B) $8x^2 - 20$
Explanation: $p(x) = k\left(x^2 - (\alpha+\beta)x + \alpha\beta\right) = k\left(x^2 - \frac{5}{2}\right)$. For $k = 8$, $p(x) = 8x^2 - 20$.
Explanation: $p(x) = k\left(x^2 - (\alpha+\beta)x + \alpha\beta\right) = k\left(x^2 - \frac{5}{2}\right)$. For $k = 8$, $p(x) = 8x^2 - 20$.
Q9. Consider the frequency distribution of 45 observations:
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|---|---|---|---|---|---|
| Frequency | 5 | 9 | 15 | 10 | 6 |
The upper limit of the median class is:
Answer: (C) $30$
Explanation: Total $N = 45 \implies \frac{N}{2} = 22.5$. Cumulative frequencies: 5, 14, 29, 39, 45. The class corresponding to $cf \ge 22.5$ is $20-30$. Upper limit is $30$.
Explanation: Total $N = 45 \implies \frac{N}{2} = 22.5$. Cumulative frequencies: 5, 14, 29, 39, 45. The class corresponding to $cf \ge 22.5$ is $20-30$. Upper limit is $30$.
Q10. $O$ is the point of intersection of two chords $AB$ and $CD$ of a circle. If $\angle BOC = 80^\circ$ and $OA = OD$, then $\Delta ODA$ and $\Delta OBC$ are:
[ Diagram Placeholder ]
Circle showing chords AB and CD intersecting at O.
Circle showing chords AB and CD intersecting at O.
Answer: (B) isosceles and similar
Q11. The roots of the quadratic equation $x^2 + x - 1 = 0$ are:
Answer: (A) Irrational and distinct
Explanation: Discriminant $D = b^2 - 4ac = 1^2 - 4(1)(-1) = 5 > 0$ and not a perfect square.
Explanation: Discriminant $D = b^2 - 4ac = 1^2 - 4(1)(-1) = 5 > 0$ and not a perfect square.
Q12. If $\theta = 30^\circ$, then the value of $3\tan\theta$ is:
Answer: (C) $\frac{3}{\sqrt{3}}$
Explanation: $3 \tan 30^\circ = 3 \times \frac{1}{\sqrt{3}} = \frac{3}{\sqrt{3}}$.
Explanation: $3 \tan 30^\circ = 3 \times \frac{1}{\sqrt{3}} = \frac{3}{\sqrt{3}}$.
Q13. The volume of a solid hemisphere is $\frac{396}{7}\text{ cm}^3$. The total surface area of the solid hemisphere (in sq. cm) is:
Answer: (B) $\frac{594}{7}$
Explanation: Volume $= \frac{2}{3}\pi r^3 = \frac{396}{7} \implies \pi r^3 = \frac{594}{7}$. Total surface area $= 3\pi r^2 = \frac{594}{7}\text{ cm}^2$.
Explanation: Volume $= \frac{2}{3}\pi r^3 = \frac{396}{7} \implies \pi r^3 = \frac{594}{7}$. Total surface area $= 3\pi r^2 = \frac{594}{7}\text{ cm}^2$.
Q14. In a bag containing 24 balls, 4 are blue, 11 are green and the rest are white. One ball is drawn at random. The probability that drawn ball is white in colour is:
Answer: (B) $\frac{3}{8}$
Explanation: White balls $= 24 - (4 + 11) = 9$. Probability $= \frac{9}{24} = \frac{3}{8}$.
Explanation: White balls $= 24 - (4 + 11) = 9$. Probability $= \frac{9}{24} = \frac{3}{8}$.
Q15. The point on the x-axis nearest to the point $(-4, -5)$ is:
Answer: (B) $(-4, 0)$
Explanation: The perpendicular projection of $(-4, -5)$ on the x-axis is $(-4, 0)$.
Explanation: The perpendicular projection of $(-4, -5)$ on the x-axis is $(-4, 0)$.
Q16. Which of the following gives the middle most observation of the data?
Answer: (A) Median
Q17. A point on the x-axis divides the line segment joining the points $A(2, -3)$ and $B(5, 6)$ in the ratio 1:2. The point is:
Answer: (C) $(3, 0)$
Explanation: $x = \frac{1(5) + 2(2)}{1+2} = \frac{9}{3} = 3$, $y = \frac{1(6) + 2(-3)}{1+2} = 0$. Point is $(3, 0)$.
Explanation: $x = \frac{1(5) + 2(2)}{1+2} = \frac{9}{3} = 3$, $y = \frac{1(6) + 2(-3)}{1+2} = 0$. Point is $(3, 0)$.
Q18. A card is drawn from a well shuffled deck of playing cards. The probability of getting red face card is:
Answer: (D) $\frac{3}{26}$
Explanation: Number of red face cards $= 6$. $P(\text{red face card}) = \frac{6}{52} = \frac{3}{26}$.
Explanation: Number of red face cards $= 6$. $P(\text{red face card}) = \frac{6}{52} = \frac{3}{26}$.
Q19.
Assertion (A): HCF of any two consecutive even natural numbers is always 2.
Reason (R): Even natural numbers are divisible by 2.
Reason (R): Even natural numbers are divisible by 2.
Answer: (B) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A).
Q20.
Assertion (A): If the radius of sector of a circle is reduced to its half and angle is doubled then the perimeter of the sector remains the same.
Reason (R): The length of the arc subtending angle $\theta$ at the centre of a circle of radius $r$ is $\frac{\pi r \theta}{180^\circ}$.
Reason (R): The length of the arc subtending angle $\theta$ at the centre of a circle of radius $r$ is $\frac{\pi r \theta}{180^\circ}$.
Answer: (D) Assertion (A) is false but reason (R) is true.
SECTION B (10 Marks)
Questions 21 to 25 carry 2 marks each.
Q21 (A). Find the H.C.F. and L.C.M. of 480 and 720 using the Prime factorisation method.
Solution:
$480 = 2^5 \times 3 \times 5$
$720 = 2^4 \times 3^2 \times 5$
$\text{HCF}(480, 720) = 2^4 \times 3 \times 5 = 240$
$\text{LCM}(480, 720) = 2^5 \times 3^2 \times 5 = 1440$
$480 = 2^5 \times 3 \times 5$
$720 = 2^4 \times 3^2 \times 5$
$\text{HCF}(480, 720) = 2^4 \times 3 \times 5 = 240$
$\text{LCM}(480, 720) = 2^5 \times 3^2 \times 5 = 1440$
OR
Q21 (B). The H.C.F of 85 and 238 is expressible in the form $85m - 238$. Find the value of $m$.
Solution:
$85 = 5 \times 17$
$238 = 2 \times 7 \times 17$
$\text{HCF}(85, 238) = 17$
$17 = 85m - 238 \implies 85m = 255 \implies m = 3$
$85 = 5 \times 17$
$238 = 2 \times 7 \times 17$
$\text{HCF}(85, 238) = 17$
$17 = 85m - 238 \implies 85m = 255 \implies m = 3$
Q22 (A). Two dice are rolled together bearing numbers 4, 6, 7, 9, 11, 12. Find the probability that the product of numbers obtained is an odd number.
Solution:
Total outcomes $= 6 \times 6 = 36$.
Product is odd when both numbers are odd. Odd numbers: 7, 9, 11.
Favorable outcomes $= (7,7), (7,9), (7,11), (9,7), (9,9), (9,11), (11,7), (11,9), (11,11)$ (Total = 9).
$P(\text{product is odd}) = \frac{9}{36} = \frac{1}{4}$
Total outcomes $= 6 \times 6 = 36$.
Product is odd when both numbers are odd. Odd numbers: 7, 9, 11.
Favorable outcomes $= (7,7), (7,9), (7,11), (9,7), (9,9), (9,11), (11,7), (11,9), (11,11)$ (Total = 9).
$P(\text{product is odd}) = \frac{9}{36} = \frac{1}{4}$
OR
Q22 (B). How many positive three digit integers have the hundredths digit 8 and unit's digit 5? Find the probability of selecting one such number out of all three digit numbers.
Solution:
Total 3-digit numbers $= 900$.
Numbers of form $8\_5$: 805, 815, 825, ..., 895 (Total = 10 numbers).
$P = \frac{10}{900} = \frac{1}{90}$
Total 3-digit numbers $= 900$.
Numbers of form $8\_5$: 805, 815, 825, ..., 895 (Total = 10 numbers).
$P = \frac{10}{900} = \frac{1}{90}$
Q23. Evaluate: $\frac{2\sin^2 60^\circ - \tan^2 30^\circ}{\sec^2 45^\circ}$
Solution:
$= \frac{2\left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{1}{\sqrt{3}}\right)^2}{\left(\sqrt{2}\right)^2}$
$= \frac{2\left(\frac{3}{4}\right) - \frac{1}{3}}{2} = \frac{\frac{3}{2} - \frac{1}{3}}{2} = \frac{\frac{7}{6}}{2} = \frac{7}{12}$
$= \frac{2\left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{1}{\sqrt{3}}\right)^2}{\left(\sqrt{2}\right)^2}$
$= \frac{2\left(\frac{3}{4}\right) - \frac{1}{3}}{2} = \frac{\frac{3}{2} - \frac{1}{3}}{2} = \frac{\frac{7}{6}}{2} = \frac{7}{12}$
Q24. Find the point(s) on the x-axis which is at a distance of $\sqrt{41}$ units from the point $(8, -5)$.
Solution:
Let the required point be $(x, 0)$.
$\sqrt{(8-x)^2 + (0 - (-5))^2} = \sqrt{41}$
$(8-x)^2 + 25 = 41 \implies (8-x)^2 = 16$
$8-x = \pm 4 \implies x = 4 \text{ or } 12$.
Points on x-axis are $(4,0)$ and $(12,0)$.
Let the required point be $(x, 0)$.
$\sqrt{(8-x)^2 + (0 - (-5))^2} = \sqrt{41}$
$(8-x)^2 + 25 = 41 \implies (8-x)^2 = 16$
$8-x = \pm 4 \implies x = 4 \text{ or } 12$.
Points on x-axis are $(4,0)$ and $(12,0)$.
Q25. Show that the points $A(-5, 6)$, $B(3, 0)$ and $C(9, 8)$ are the vertices of an isosceles triangle.
Solution:
$AB = \sqrt{(3 - (-5))^2 + (0 - 6)^2} = \sqrt{64 + 36} = 10$
$BC = \sqrt{(9 - 3)^2 + (8 - 0)^2} = \sqrt{36 + 64} = 10$
$AC = \sqrt{(9 - (-5))^2 + (8 - 6)^2} = \sqrt{196 + 4} = 10\sqrt{2}$
Since $AB = BC = 10$, $\Delta ABC$ is an isosceles triangle.
$AB = \sqrt{(3 - (-5))^2 + (0 - 6)^2} = \sqrt{64 + 36} = 10$
$BC = \sqrt{(9 - 3)^2 + (8 - 0)^2} = \sqrt{36 + 64} = 10$
$AC = \sqrt{(9 - (-5))^2 + (8 - 6)^2} = \sqrt{196 + 4} = 10\sqrt{2}$
Since $AB = BC = 10$, $\Delta ABC$ is an isosceles triangle.
SECTION C (18 Marks)
Questions 26 to 31 carry 3 marks each.
Q26 (A). In $\Delta ABC$, $D$, $E$ and $F$ are midpoints of $BC$, $CA$ and $AB$ respectively. Prove that $\Delta FBD \sim \Delta DEF$ and $\Delta DEF \sim \Delta ABC$.
[ Diagram Placeholder ]
Triangle ABC with midpoint triangle DEF.
Triangle ABC with midpoint triangle DEF.
Proof:
By midpoint theorem, $EF \parallel BC$, $DF \parallel AC$, $DE \parallel AB$.
Hence, $BDEF$ and $DCEF$ are parallelograms.
In $\Delta FBD$ and $\Delta DEF$: $\angle 1 = \angle 2$ and $\angle 3 = \angle 4 \implies \Delta FBD \sim \Delta DEF$.
Similarly, $\Delta DEF \sim \Delta ABC$ by AA similarity criteria.
By midpoint theorem, $EF \parallel BC$, $DF \parallel AC$, $DE \parallel AB$.
Hence, $BDEF$ and $DCEF$ are parallelograms.
In $\Delta FBD$ and $\Delta DEF$: $\angle 1 = \angle 2$ and $\angle 3 = \angle 4 \implies \Delta FBD \sim \Delta DEF$.
Similarly, $\Delta DEF \sim \Delta ABC$ by AA similarity criteria.
OR
Q26 (B). In $\Delta ABC$, $P$ and $Q$ are points on $AB$ and $AC$ respectively such that $PQ \parallel BC$. Prove that the median $AD$ drawn from $A$ on $BC$ bisects $PQ$.
Proof:
Since $PQ \parallel BC$, $\Delta APR \sim \Delta ABD \implies \frac{PR}{BD} = \frac{AR}{AD}$.
Also, $\Delta AQR \sim \Delta ACD \implies \frac{RQ}{DC} = \frac{AR}{AD}$.
$\implies \frac{PR}{BD} = \frac{RQ}{DC}$.
Since $AD$ is median, $BD = DC \implies PR = RQ$. Hence $AD$ bisects $PQ$.
Since $PQ \parallel BC$, $\Delta APR \sim \Delta ABD \implies \frac{PR}{BD} = \frac{AR}{AD}$.
Also, $\Delta AQR \sim \Delta ACD \implies \frac{RQ}{DC} = \frac{AR}{AD}$.
$\implies \frac{PR}{BD} = \frac{RQ}{DC}$.
Since $AD$ is median, $BD = DC \implies PR = RQ$. Hence $AD$ bisects $PQ$.
Q27. The sum of two numbers is 18 and the sum of their reciprocals is $\frac{9}{40}$. Find the numbers.
Solution:
Let the numbers be $x$ and $18-x$.
$\frac{1}{x} + \frac{1}{18-x} = \frac{9}{40} \implies \frac{18}{x(18-x)} = \frac{9}{40}$
$2 \times 40 = x(18-x) \implies x^2 - 18x + 80 = 0$
$(x-10)(x-8) = 0 \implies x = 10 \text{ or } 8$.
The two numbers are 8 and 10.
Let the numbers be $x$ and $18-x$.
$\frac{1}{x} + \frac{1}{18-x} = \frac{9}{40} \implies \frac{18}{x(18-x)} = \frac{9}{40}$
$2 \times 40 = x(18-x) \implies x^2 - 18x + 80 = 0$
$(x-10)(x-8) = 0 \implies x = 10 \text{ or } 8$.
The two numbers are 8 and 10.
Q28. If $\alpha$ and $\beta$ are zeroes of a polynomial $6x^2 - 5x + 1$, then form a quadratic polynomial whose zeroes are $\alpha^2$ and $\beta^2$.
Solution:
$\alpha + \beta = \frac{5}{6}$, $\alpha\beta = \frac{1}{6}$
New sum $= \alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = \left(\frac{5}{6}\right)^2 - 2\left(\frac{1}{6}\right) = \frac{25}{36} - \frac{12}{36} = \frac{13}{36}$
New product $= \alpha^2\beta^2 = (\alpha\beta)^2 = \frac{1}{36}$
Polynomial $= x^2 - \frac{13}{36}x + \frac{1}{36} \implies 36x^2 - 13x + 1$
$\alpha + \beta = \frac{5}{6}$, $\alpha\beta = \frac{1}{6}$
New sum $= \alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = \left(\frac{5}{6}\right)^2 - 2\left(\frac{1}{6}\right) = \frac{25}{36} - \frac{12}{36} = \frac{13}{36}$
New product $= \alpha^2\beta^2 = (\alpha\beta)^2 = \frac{1}{36}$
Polynomial $= x^2 - \frac{13}{36}x + \frac{1}{36} \implies 36x^2 - 13x + 1$
Q29. If $\cos\theta + \sin\theta = 1$, then prove that $\cos\theta - \sin\theta = \pm 1$.
Proof:
$(\cos\theta + \sin\theta)^2 + (\cos\theta - \sin\theta)^2 = 2(\cos^2\theta + \sin^2\theta) = 2$
$(1)^2 + (\cos\theta - \sin\theta)^2 = 2$
$(\cos\theta - \sin\theta)^2 = 1 \implies \cos\theta - \sin\theta = \pm 1$
$(\cos\theta + \sin\theta)^2 + (\cos\theta - \sin\theta)^2 = 2(\cos^2\theta + \sin^2\theta) = 2$
$(1)^2 + (\cos\theta - \sin\theta)^2 = 2$
$(\cos\theta - \sin\theta)^2 = 1 \implies \cos\theta - \sin\theta = \pm 1$
Q30 (A). The minute hand of a wall clock is $18\text{ cm}$ long. Find the area of the face of the clock described by the minute hand in 35 minutes.
Solution:
Angle in 35 minutes $= \frac{360^\circ}{60} \times 35 = 210^\circ$.
Area $= \frac{\theta}{360^\circ} \times \pi r^2 = \frac{210}{360} \times \frac{22}{7} \times 18 \times 18 = 594\text{ cm}^2$.
Angle in 35 minutes $= \frac{360^\circ}{60} \times 35 = 210^\circ$.
Area $= \frac{\theta}{360^\circ} \times \pi r^2 = \frac{210}{360} \times \frac{22}{7} \times 18 \times 18 = 594\text{ cm}^2$.
OR
Q30 (B). $AB$ is a chord of a circle centred at $O$ such that $\angle AOB = 60^\circ$. If $OA = 14\text{ cm}$ then find the area of the minor segment. (Take $\sqrt{3} = 1.73$)
Solution:
Area $= \text{Area of sector} - \text{Area of }\Delta AOB$
$= \frac{60}{360} \times \frac{22}{7} \times 14^2 - \frac{\sqrt{3}}{4} \times 14^2 = \frac{308}{3} - 49(1.73) \approx 102.67 - 84.77 = 17.9\text{ cm}^2$.
Area $= \text{Area of sector} - \text{Area of }\Delta AOB$
$= \frac{60}{360} \times \frac{22}{7} \times 14^2 - \frac{\sqrt{3}}{4} \times 14^2 = \frac{308}{3} - 49(1.73) \approx 102.67 - 84.77 = 17.9\text{ cm}^2$.
Q31. Prove that $\sqrt{3}$ is an irrational number.
Proof:
Let $\sqrt{3} = \frac{p}{q}$, where $p, q$ are coprime integers and $q \neq 0$.
$3q^2 = p^2 \implies p^2$ is divisible by $3 \implies p$ is divisible by $3$.
Let $p = 3a \implies 3q^2 = 9a^2 \implies q^2 = 3a^2 \implies q$ is divisible by $3$.
This contradicts that $p$ and $q$ are coprime. Thus, $\sqrt{3}$ is irrational.
Let $\sqrt{3} = \frac{p}{q}$, where $p, q$ are coprime integers and $q \neq 0$.
$3q^2 = p^2 \implies p^2$ is divisible by $3 \implies p$ is divisible by $3$.
Let $p = 3a \implies 3q^2 = 9a^2 \implies q^2 = 3a^2 \implies q$ is divisible by $3$.
This contradicts that $p$ and $q$ are coprime. Thus, $\sqrt{3}$ is irrational.
SECTION D (20 Marks)
Questions 32 to 35 carry 5 marks each.
Q32 (A). Solve the following system of linear equations graphically: $x + 2y = 3$, $2x - 3y + 8 = 0$.
[ Graph Diagram Placeholder ]
Graph plotting lines $x + 2y = 3$ and $2x - 3y + 8 = 0$ intersecting at $(-1, 2)$.
Graph plotting lines $x + 2y = 3$ and $2x - 3y + 8 = 0$ intersecting at $(-1, 2)$.
Solution: Solution point from graph is $x = -1, y = 2$.
OR
Q32 (B). Places A and B are $180\text{ km}$ apart on a highway. One car starts from A and another from B at the same time. If the car travels in the same direction at different speeds, they meet in 9 hours. If they travel towards each other with the same speeds as before, they meet in an hour. What are the speeds of the two cars?
Solution:
Let speeds be $x\text{ km/h}$ and $y\text{ km/h}$ ($x > y$).
Case I (Same direction): $9(x - y) = 180 \implies x - y = 20$ .....(i)
Case II (Opposite direction): $1(x + y) = 180 \implies x + y = 180$ .....(ii)
Solving (i) & (ii): $x = 100\text{ km/h}$, $y = 80\text{ km/h}$.
Let speeds be $x\text{ km/h}$ and $y\text{ km/h}$ ($x > y$).
Case I (Same direction): $9(x - y) = 180 \implies x - y = 20$ .....(i)
Case II (Opposite direction): $1(x + y) = 180 \implies x + y = 180$ .....(ii)
Solving (i) & (ii): $x = 100\text{ km/h}$, $y = 80\text{ km/h}$.
Q33. Prove that the lengths of tangents drawn from an external point to a circle are equal.
Using above result, find the length $BC$ of $\Delta ABC$. Given that, a circle is inscribed in $\Delta ABC$ touching the sides $AB, BC$ and $CA$ at $R, P$ and $Q$ respectively and $AB = 10\text{ cm}$, $AQ = 7\text{ cm}$, $CQ = 5\text{ cm}$.
Using above result, find the length $BC$ of $\Delta ABC$. Given that, a circle is inscribed in $\Delta ABC$ touching the sides $AB, BC$ and $CA$ at $R, P$ and $Q$ respectively and $AB = 10\text{ cm}$, $AQ = 7\text{ cm}$, $CQ = 5\text{ cm}$.
[ Diagram Placeholder ]
Triangle ABC with inscribed circle touching sides at R, P, Q.
Triangle ABC with inscribed circle touching sides at R, P, Q.
Solution:
$AR = AQ = 7\text{ cm}$
$BP = BR = AB - AR = 10 - 7 = 3\text{ cm}$
$CP = CQ = 5\text{ cm}$
$BC = BP + PC = 3 + 5 = 8\text{ cm}$
$AR = AQ = 7\text{ cm}$
$BP = BR = AB - AR = 10 - 7 = 3\text{ cm}$
$CP = CQ = 5\text{ cm}$
$BC = BP + PC = 3 + 5 = 8\text{ cm}$
Q34. A boy whose eye level is $1.35\text{ m}$ from the ground, spots a balloon moving with the wind in a horizontal line at some height from the ground. The angle of elevation of the balloon from the eyes of the boy at an instant is $60^\circ$. After 12 seconds, the angle of elevation reduces to $30^\circ$. If the speed of the wind is $3\text{ m/s}$ then find the height of the balloon from the ground. (Use $\sqrt{3} = 1.73$)
[ Diagram Placeholder ]
Angles of elevation 60° and 30° from boy's eye level to balloon.
Angles of elevation 60° and 30° from boy's eye level to balloon.
Solution:
Distance covered $= 3 \times 12 = 36\text{ m}$.
$\tan 60^\circ = \sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3}$
$\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{h}{x + 36} \implies h\sqrt{3} = x + 36$
Solving gives $h = 18\sqrt{3} = 31.14\text{ m}$.
Total height $= 1.35 + 31.14 = 32.49\text{ m}$.
Distance covered $= 3 \times 12 = 36\text{ m}$.
$\tan 60^\circ = \sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3}$
$\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{h}{x + 36} \implies h\sqrt{3} = x + 36$
Solving gives $h = 18\sqrt{3} = 31.14\text{ m}$.
Total height $= 1.35 + 31.14 = 32.49\text{ m}$.
Q35 (A). Find the mean and median of the following data:
| Class | 85-90 | 90-95 | 95-100 | 100-105 | 105-110 | 110-115 |
|---|---|---|---|---|---|---|
| Frequency | 15 | 22 | 20 | 18 | 20 | 25 |
Solution:
$\text{Mean} = 100.875$
$\text{Median} = 100.83$
$\text{Mean} = 100.875$
$\text{Median} = 100.83$
OR
Q35 (B). The monthly expenditure on milk in 200 families of a Housing Society is given below. Find $x$ and mean expenditure:
| Expenditure | 1000-1500 | 1500-2000 | 2000-2500 | 2500-3000 | 3000-3500 | 3500-4000 | 4000-4500 | 4500-5000 |
|---|---|---|---|---|---|---|---|---|
| Families | 24 | 40 | 33 | x | 30 | 22 | 16 | 7 |
Solution:
$172 + x = 200 \implies x = 28$
$\text{Mean} = \frac{532500}{200} = ₹2662.50$
$172 + x = 200 \implies x = 28$
$\text{Mean} = \frac{532500}{200} = ₹2662.50$
SECTION E (12 Marks)
Case study based questions carrying 4 marks each.
Q36 (Case Study 1 - Glass Jars):
Top layer: 3 jars, 2nd layer: 6 jars, 3rd layer: 9 jars up to 8th layer.
(i) Write A.P. and common difference.
Answer: A.P. = $3, 6, 9, 12, ...$, Common difference $d = 3$.
(ii) Is it possible to arrange 34 jars in a layer?
Answer: $34 = 3 + (n-1)3 \implies n = \frac{34}{3} = 11\frac{1}{3}$, which is not an integer. Not possible.
(iii) (A) Total jars expression and $S_8$.
Answer: $S_n = \frac{3n}{2}(n+1)$. $S_8 = 3 \times \frac{8}{2}(9) = 108$ jars.
OR
(iii) (B) If 3 jars added each layer, find 5th layer jars.
Answer: New A.P. = $6, 9, 12, ... \implies a_5 = 6 + 4(3) = 18$ jars.
Q37 (Case Study 2 - Triangles in Cabinet):
[ Diagram Placeholder ]
Triangles ABC and DFE with medians and parallel lines.
Triangles ABC and DFE with medians and parallel lines.
(i) Show $\Delta DPQ \sim \Delta DEF$.
Answer: $\angle DPQ = \angle DEF$ and $\angle PDQ = \angle EDF \implies \Delta DPQ \sim \Delta DEF$.
(ii) If $DP = 50\text{ cm}$ and $PF = 70\text{ cm}$, find $\frac{PQ}{EF}$.
Answer: $\frac{PQ}{EF} = \frac{DP}{DE} = \frac{50}{120} = \frac{5}{12}$.
(iii) (A) Show perimeter ratio is constant.
Answer: $\frac{\text{Perimeter}(\Delta ABC)}{\text{Perimeter}(\Delta DEF)} = \frac{5}{2}$ (Constant).
Q38 (Case Study 3 - Metallic Silo):
Cone: $r = 1.5\text{ m}, h = 2\text{ m}$. Cylinder: $r = 1.5\text{ m}, h = 7\text{ m}$.
(i) Slant height of cone $l$.
Answer: $l = \sqrt{1.5^2 + 2^2} = 2.5\text{ m}$.
(ii) CSA of conical part.
Answer: $\pi r l = \frac{22}{7} \times 1.5 \times 2.5 = 11.78\text{ m}^2$.
(iii) (A) Cost of metal sheet for cylindrical part at ₹2000/m².
Answer: $\text{CSA} = 2\pi rh = 66\text{ m}^2$. $\text{Cost} = 66 \times 2000 = ₹1,32,000$.
OR
(iii) (B) Total capacity of silo.
Answer: Volume $= 49.5 (\text{cylinder}) + 4.71 (\text{cone}) = 54.21\text{ m}^3$.
