Detailed Solutions

Detailed Solutions: CBSE Class 10 Maths Sample Paper 2027 – Set 3

Every written question from the paper (2, 3, 4 and 5 marks) is solved here in order, with the marks given for each step, so you can check your answer sheet the way an examiner would. Where a question has an internal choice (OR), both options are solved. The 1-mark questions are answered, with explanations, in the quiz on the Question Paper page.

Section B (10 marks)

Question 21 (2 marks)

The floor of a school's computer lab is a rectangle 5.4 m long and 4.2 m wide. It is to be covered with identical square tiles without cutting any tile. Find the side of the largest square tile that can be used and the number of such tiles needed.

Answer.

1. In cm: $540 = 2^2 \times 3^3 \times 5$ and $420 = 2^2 \times 3 \times 5 \times 7$; largest side = HCF $= 2^2 \times 3 \times 5 = 60$ cm — 1 mark
2. Number of tiles $= \frac{540}{60} \times \frac{420}{60} = 9 \times 7 = 63$ — 1 mark

Question 22 (2 marks)

One zero of the quadratic polynomial $x^2 - 9x + k$ is twice the other. Find the zeroes and the value of $k$.

Answer.

1. Let the zeroes be $\alpha$ and $2\alpha$; sum $= 3\alpha = 9$, so $\alpha = 3$ and the zeroes are $3$ and $6$ — 1 mark
2. $k =$ product of the zeroes $= 3 \times 6 = 18$ — 1 mark

OR

Find a quadratic polynomial the sum and product of whose zeroes are $-\frac{1}{4}$ and $-\frac{3}{4}$ respectively. Also find its zeroes.

Answer.

1. Polynomial $= k\left(x^2 + \frac{1}{4}x - \frac{3}{4}\right)$; taking $k = 4$, it is $4x^2 + x - 3$ — 1 mark
2. $4x^2 + x - 3 = 4x^2 + 4x - 3x - 3 = (4x - 3)(x + 1)$, so the zeroes are $\frac{3}{4}$ and $-1$ — 1 mark

Question 23 (2 marks)

In the figure, $D$ and $E$ are points on the sides $AB$ and $AC$ of $\triangle ABC$ such that $DE \parallel BC$, and $F$ is a point on $AD$ such that $EF \parallel CD$. If $AF = 4$ cm and $FD = 6$ cm, find $DB$.

Answer.

1. In $\triangle ADC$, $FE \parallel DC$, so $\frac{AF}{FD} = \frac{AE}{EC}$ (BPT); in $\triangle ABC$, $DE \parallel BC$, so $\frac{AD}{DB} = \frac{AE}{EC}$ — 1 mark
2. Hence $\frac{AD}{DB} = \frac{AF}{FD}$, i.e. $\frac{10}{DB} = \frac{4}{6}$, so $DB = 15$ cm — 1 mark

Question 24 (2 marks)

If $\sec A = \frac{29}{21}$, where $A$ is an acute angle, find the value of $\frac{1 + \sin A}{\cos A}$.

Answer.

1. $\cos A = \frac{21}{29}$ and $\sin A = \sqrt{1 - \frac{441}{841}} = \frac{20}{29}$ — 1 mark
2. $\frac{1 + \sin A}{\cos A} = \frac{1 + \frac{20}{29}}{\frac{21}{29}} = \frac{49}{21} = \frac{7}{3}$ — 1 mark

Question 25 (2 marks)

In the figure, a circle with centre $O$ is inscribed in $\triangle ABC$, which is right-angled at $B$. The circle touches $AB$, $BC$ and $CA$ at $P$, $Q$ and $R$ respectively. If $AB = 8$ cm and $BC = 15$ cm, find the radius of the circle.

Answer.

1. $AC = \sqrt{8^2 + 15^2} = 17$ cm; $OP \perp AB$, $OQ \perp BC$ and $OP = OQ$, so $OPBQ$ is a square and $BP = BQ = r$ — 1 mark
2. Tangents from $A$ and $C$: $AR = AP = 8 - r$ and $CR = CQ = 15 - r$; $(8 - r) + (15 - r) = 17$, so $r = 3$ cm — 1 mark

OR

From an external point $P$, two tangents $PA$ and $PB$ are drawn to a circle with centre $O$. If $PA = 12$ cm and $OP = 15$ cm, find the perimeter of the quadrilateral $OAPB$.

Answer.

1. $OA \perp PA$, so $OA = \sqrt{15^2 - 12^2} = 9$ cm; also $OB = 9$ cm and $PB = PA = 12$ cm — 1 mark
2. Perimeter $= 9 + 12 + 12 + 9 = 42$ cm — 1 mark

Section C (18 marks)

Question 26 (3 marks)

Prove that $\sqrt{5}$ is an irrational number.

Answer.

1. Assume $\sqrt{5}$ is rational: $\sqrt{5} = \frac{a}{b}$, where $a$ and $b$ are co-prime integers and $b \neq 0$; squaring, $5b^2 = a^2$ — 1 mark
2. So 5 divides $a^2$, and hence 5 divides $a$. Write $a = 5c$: then $25c^2 = 5b^2$, i.e. $b^2 = 5c^2$, so 5 divides $b^2$ and hence 5 divides $b$ — 1 mark
3. 5 is a common factor of $a$ and $b$, which contradicts that they are co-prime; hence $\sqrt{5}$ is irrational — 1 mark

Question 27 (3 marks)

Find the coordinates of the points of trisection of the line segment joining $A(-2, -3)$ and $B(7, 9)$. Also verify that the distance from $A$ of the point of trisection nearer to $A$ is one-third of $AB$.

Answer.

1. The points divide $AB$ in the ratios $1 : 2$ and $2 : 1$; $P = \left(\frac{1(7) + 2(-2)}{3}, \frac{1(9) + 2(-3)}{3}\right) = (1, 1)$ — 1 mark
2. $Q = \left(\frac{2(7) + 1(-2)}{3}, \frac{2(9) + 1(-3)}{3}\right) = (4, 5)$ — 1 mark
3. $AP = \sqrt{3^2 + 4^2} = 5$ and $AB = \sqrt{9^2 + 12^2} = 15$, so $AP = \frac{1}{3}AB$ — 1 mark

Question 28 (3 marks)

Prove that $\sec^4 A - \sec^2 A = \tan^4 A + \tan^2 A$.

Answer.

1. LHS $= \sec^2 A(\sec^2 A - 1)$ — 1 mark
2. $= (1 + \tan^2 A)\tan^2 A$, using $\sec^2 A = 1 + \tan^2 A$ — 1 mark
3. $= \tan^2 A + \tan^4 A =$ RHS — 1 mark

OR

Prove that $(\sec A - \cos A)(\cot A + \tan A) = \tan A \sec A$.

Answer.

1. $\sec A - \cos A = \frac{1 - \cos^2 A}{\cos A} = \frac{\sin^2 A}{\cos A}$ — 1 mark
2. $\cot A + \tan A = \frac{\cos^2 A + \sin^2 A}{\sin A \cos A} = \frac{1}{\sin A \cos A}$ — 1 mark
3. LHS $= \frac{\sin^2 A}{\cos A} \times \frac{1}{\sin A \cos A} = \frac{\sin A}{\cos A} \times \frac{1}{\cos A} = \tan A \sec A =$ RHS — 1 mark

Question 29 (3 marks)

Prove that the lengths of the tangents drawn from an external point to a circle are equal. Using this result, find the perimeter of $\triangle PXY$ in the figure, where $PA$ and $PB$ are tangents from $P$ to a circle with centre $O$, and the tangent at the point $C$ of the circle meets $PA$ at $X$ and $PB$ at $Y$. It is given that $PA = 10$ cm.

Answer.

1. Figure, given and to prove; join $OA$, $OB$ and $OP$; $\angle OAP = \angle OBP = 90^\circ$ (radius is perpendicular to the tangent) — 1 mark
2. In $\triangle OAP$ and $\triangle OBP$: $OA = OB$ (radii) and $OP$ is common, so $\triangle OAP \cong \triangle OBP$ (RHS) and $PA = PB$ — 1 mark
3. $XA = XC$ and $YB = YC$; perimeter $= PX + XC + CY + YP = PX + XA + YB + PY = PA + PB = 20$ cm — 1 mark

Question 30 (3 marks)

In the figure, $PQ$ is a chord of a circle with centre $O$ and radius 6 cm, and $\angle POQ = 120^\circ$. Find the area of the shaded minor segment. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)

Answer.

1. Area of the sector $= \frac{120}{360} \times 3.14 \times 6 \times 6 = 37.68 \text{ cm}^2$ — 1 mark
2. Draw $OM \perp PQ$: $\angle POM = 60^\circ$, $OM = 6\cos 60^\circ = 3$ cm, $PM = 6\sin 60^\circ = 3\sqrt{3}$ cm; area of the triangle $= \frac{1}{2} \times 6\sqrt{3} \times 3 = 9\sqrt{3} = 15.57 \text{ cm}^2$ — 1 mark
3. Area of minor segment $= 37.68 - 15.57 = 22.11 \text{ cm}^2$ — 1 mark

OR

An arc of a circle of radius 14 cm is 22 cm long. Find (i) the angle subtended by the arc at the centre, (ii) the area of the minor sector formed by the arc, (iii) the area of the corresponding major sector. (Use $\pi = \frac{22}{7}$)

Answer.

1. $\frac{\theta}{360} \times 2 \times \frac{22}{7} \times 14 = 22$, i.e. $\frac{\theta}{360} \times 88 = 22$, so $\theta = 90^\circ$ — 1 mark
2. Minor sector $= \frac{90}{360} \times \frac{22}{7} \times 14 \times 14 = 154 \text{ cm}^2$ — 1 mark
3. Area of the circle $= 616 \text{ cm}^2$; major sector $= 616 - 154 = 462 \text{ cm}^2$ — 1 mark

Question 31 (3 marks)

Two dice, one red and one blue, are thrown together. Find the probability that (i) the sum of the numbers on the two dice is 9, (ii) the product of the numbers on the two dice is a perfect square, (iii) the number on the red die is less than the number on the blue die.

Answer.

1. Total outcomes $= 36$. (i) Sum 9: $(3, 6), (4, 5), (5, 4), (6, 3)$; $P = \frac{4}{36} = \frac{1}{9}$ — 1 mark
2. (ii) Product a perfect square: $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (1, 4), (4, 1)$; $P = \frac{8}{36} = \frac{2}{9}$ — 1 mark
3. (iii) Red less than blue: $5 + 4 + 3 + 2 + 1 = 15$ outcomes; $P = \frac{15}{36} = \frac{5}{12}$ — 1 mark

Section D (20 marks)

Question 32 (5 marks)

The length of a rectangular garden in a housing society is 16 m more than its breadth, and its area is 1700 $\text{m}^2$. A gravel path of uniform width is laid inside the garden along all four sides, leaving a rectangular lawn of area 1092 $\text{m}^2$ in the middle. Find the length and breadth of the garden and the width of the path.

Answer.

1. Let the breadth be $x$ m: $x(x + 16) = 1700$, i.e. $x^2 + 16x - 1700 = 0$ — 1 mark
2. $(x + 50)(x - 34) = 0$, so $x = 34$ ($x = -50$ rejected); the garden is 50 m long and 34 m wide — 1 mark
3. Let the width of the path be $w$ m: $(50 - 2w)(34 - 2w) = 1092$ — 1 mark
4. $4w^2 - 168w + 608 = 0$, i.e. $w^2 - 42w + 152 = 0$ — 1 mark
5. $(w - 4)(w - 38) = 0$; $w = 38$ is rejected since $2 \times 38 > 34$, so the path is 4 m wide — 1 mark

Question 33 (5 marks)

Solve the following pair of linear equations graphically: $2x + y = 8$ and $x - y + 2 = 0$. Also find the coordinates of the vertices of the triangle formed by these lines and the y-axis, and find its area.

Answer.

1. Table of values for $2x + y = 8$: $(0, 8), (4, 0), (2, 4)$ — 1 mark
2. Table of values for $x - y + 2 = 0$: $(0, 2), (-2, 0), (2, 4)$ — 1 mark
3. Both lines correctly drawn on the same graph — 1 mark
4. Solution $x = 2$, $y = 4$; vertices $(2, 4)$, $(0, 8)$ and $(0, 2)$ — 1 mark
5. Area $= \frac{1}{2} \times (8 - 2) \times 2 = 6$ square units — 1 mark

OR

At a school fete, a food stall sold 45 veg rolls and 60 glasses of lemonade on the first day and collected ₹ 2250. On the second day it sold 60 veg rolls and 40 glasses of lemonade and collected ₹ 2400. Find the price of one veg roll and of one glass of lemonade. How much would a customer pay for 4 veg rolls and 3 glasses of lemonade?

Answer.

1. Let a roll cost ₹ $x$ and a glass cost ₹ $y$: $45x + 60y = 2250$ and $60x + 40y = 2400$, i.e. $3x + 4y = 150$ and $3x + 2y = 120$ — 2 marks
2. Subtracting, $2y = 30$, so $y = 15$ — 1 mark
3. $3x = 150 - 60 = 90$, so $x = 30$ — 1 mark
4. Cost $= 4 \times 30 + 3 \times 15 =$ ₹ 165 — 1 mark

Question 34 (5 marks)

A devotee whose eyes are 1.5 m above the ground sees the top of a temple gopuram at an angle of elevation of $30^\circ$. After walking 50 m straight towards the gopuram on level ground, he finds that the angle of elevation of its top is $60^\circ$. Find the height of the gopuram and his distance from the gopuram at the second position. (Use $\sqrt{3} = 1.73$)

Answer.

1. Correct figure: top $A$ of the gopuram, eye-level line meeting the gopuram at $E$, eye positions $C$ and $D$ with $CD = 50$ m; let $AE = h$ m and $DE = x$ m — 1 mark
2. In right $\triangle AED$: $\tan 60^\circ = \frac{h}{x}$, so $h = \sqrt{3}x$ — 1 mark
3. In right $\triangle AEC$: $\tan 30^\circ = \frac{h}{x + 50}$, so $\sqrt{3}h = x + 50$ — 1 mark
4. $3x = x + 50$, so $x = 25$ m and $h = 25\sqrt{3} = 43.25$ m — 1 mark
5. Height of the gopuram $= 43.25 + 1.5 = 44.75$ m; distance at the second position $= 25$ m — 1 mark

Question 35 (5 marks)

A wooden top (lattu) is in the shape of a cone mounted on a hemisphere of the same radius. The diameter of the hemisphere is 4.2 cm and the total height of the top is 4.1 cm. Find the total surface area of the top and the volume of wood in it. (Use $\pi = \frac{22}{7}$)

Answer.

1. $r = 2.1$ cm; height of the cone $= 4.1 - 2.1 = 2$ cm; slant height $l = \sqrt{2.1^2 + 2^2} = \sqrt{8.41} = 2.9$ cm — 1 mark
2. CSA of hemisphere $= 2\pi r^2 = 2 \times \frac{22}{7} \times 2.1 \times 2.1 = 27.72 \text{ cm}^2$; CSA of cone $= \pi r l = \frac{22}{7} \times 2.1 \times 2.9 = 19.14 \text{ cm}^2$ — 1 mark
3. Total surface area $= 27.72 + 19.14 = 46.86 \text{ cm}^2$ — 1 mark
4. Volume of hemisphere $= \frac{2}{3}\pi r^3 = 19.404 \text{ cm}^3$; volume of cone $= \frac{1}{3}\pi r^2 h = \frac{1}{3} \times 13.86 \times 2 = 9.24 \text{ cm}^3$ — 1 mark
5. Volume of wood $= 19.404 + 9.24 = 28.644 \text{ cm}^3$ — 1 mark

OR

A memento is made of a solid wooden cube of edge 10 cm with a solid hemisphere of radius 4.2 cm fixed on the middle of its top face. The memento stands on its bottom face, which is not polished; every other outer surface, including the curved surface of the hemisphere, is to be polished. Find the area to be polished and the volume of the memento. (Use $\pi = \frac{22}{7}$)

Answer.

1. Faces of the cube other than the bottom: $5 \times 10 \times 10 = 500 \text{ cm}^2$; the circle of radius 4.2 cm under the hemisphere is hidden — 1 mark
2. $\pi r^2 = \frac{22}{7} \times 4.2 \times 4.2 = 55.44 \text{ cm}^2$; CSA of hemisphere $= 2\pi r^2 = 110.88 \text{ cm}^2$ — 1 mark
3. Area to be polished $= 500 - 55.44 + 110.88 = 555.44 \text{ cm}^2$ — 1 mark
4. Volume of cube $= 10^3 = 1000 \text{ cm}^3$; volume of hemisphere $= \frac{2}{3} \times \frac{22}{7} \times 4.2^3 = 155.232 \text{ cm}^3$ — 1 mark
5. Volume of the memento $= 1000 + 155.232 = 1155.232 \text{ cm}^3$ — 1 mark

Section E (12 marks)

Question 36 (4 marks)

At a brick kiln, bricks are stacked in rows, one row on top of another. The bottom row has 42 bricks, the row above it has 39, the next one has 36, and so on: every row has 3 bricks fewer than the row just below it. The top row has 6 bricks.

(i) How many bricks are there in the 6th row from the bottom? (1 mark)

Answer.

1. $a = 42$, $d = -3$; $a_6 = 42 + 5(-3) = 27$ — 1 mark

(ii) How many rows are there in the stack? (1 mark)

Answer.

1. $42 + (n - 1)(-3) = 6$ gives $n - 1 = 12$, so $n = 13$ rows — 1 mark

(iii) Find the total number of bricks in the stack. (2 marks)

Answer.

1. $S_{13} = \frac{13}{2}(42 + 6)$ — 1 mark
2. $= 13 \times 24 = 312$ bricks — 1 mark

OR

(iii) Find the total number of bricks in all the rows that have more than 30 bricks each. (2 marks)

Answer.

1. Rows with more than 30 bricks: 42, 39, 36, 33 (the 5th row has exactly 30), i.e. 4 rows — 1 mark
2. Sum $= \frac{4}{2}(42 + 33) = 150$ bricks — 1 mark

Question 37 (4 marks)

For his science project, Kabir made a pinhole camera: a closed box with a tiny hole $O$ in the front face and a screen on the back face, 20 cm behind the hole. He points it at a tree $AB$ which is 12 m tall and stands 40 m away from the hole. Light from the top $A$ of the tree passes through $O$ and reaches the screen at $A'$, so an inverted image $A'B'$ of the tree is formed. Take the foot $B$ of the tree, the hole $O$ and the point $B'$ to lie on one horizontal line, with the tree and the screen both vertical.

(i) By which criterion is $\triangle OAB \sim \triangle OA'B'$? (1 mark)

Answer.

1. $\angle ABO = \angle A'B'O = 90^\circ$ and $\angle AOB = \angle A'OB'$ (vertically opposite angles), so the triangles are similar by the AA criterion — 1 mark

(ii) Find the height of the image $A'B'$. (1 mark)

Answer.

1. $\frac{A'B'}{AB} = \frac{OB'}{OB}$, so $A'B' = \frac{1200 \times 20}{4000} = 6$ cm — 1 mark

(iii) Kabir moves the camera 10 m closer to the tree. Find the new height of the image. (2 marks)

Answer.

1. Now $OB = 30$ m $= 3000$ cm, and $\frac{A'B'}{1200} = \frac{20}{3000}$ — 1 mark
2. $A'B' = \frac{1200 \times 20}{3000} = 8$ cm — 1 mark

OR

(iii) Kabir's friend, who is 1.5 m tall, stands in front of the camera, and the image of the friend on the screen is 5 cm tall. How far from the hole is the friend standing? (2 marks)

Answer.

1. Heights in cm: $\frac{5}{150} = \frac{20}{OB}$ — 1 mark
2. $OB = \frac{150 \times 20}{5} = 600$ cm $= 6$ m — 1 mark

Question 38 (4 marks)

An electricity company surveyed 60 households in a colony. The table shows the number of units of electricity they used in a month.

(i) How many households used 200 units or more in the month? (1 mark)

Answer.

1. 15 + 12 + 6 = 33 households — 1 mark

(ii) Write the modal class and the median class. (1 mark)

Answer.

1. Modal class 150-200 (highest frequency 16); cumulative frequencies 4, 11, 27, 42, … and $\frac{n}{2} = 30$, so the median class is 200-250 — 1 mark

(iii) Find the median of the data. (2 marks)

Answer.

1. $\frac{n}{2} = 30$; median class 200-250: $l = 200$, $cf = 27$, $f = 15$, $h = 50$ — 1 mark
2. Median $= 200 + \frac{30 - 27}{15} \times 50 = 200 + 10 = 210$ units — 1 mark

OR

(iii) Find the mode of the data. (2 marks)

Answer.

1. Modal class 150-200: $l = 150$, $f_1 = 16$, $f_0 = 7$, $f_2 = 15$, $h = 50$ — 1 mark
2. Mode $= 150 + \frac{16 - 7}{32 - 7 - 15} \times 50 = 150 + 45 = 195$ units — 1 mark