Detailed Solutions

Detailed Solutions: CBSE Class 10 Maths Sample Paper 2027 – Set 2

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)

Given that $\sqrt{5}$ is irrational, prove that $3 + 2\sqrt{5}$ is irrational.

Answer.

1. Assume $3 + 2\sqrt{5} = r$, a rational number; then $\sqrt{5} = \frac{r - 3}{2}$ — 1 mark
2. RHS is rational but $\sqrt{5}$ is irrational - a contradiction; hence $3 + 2\sqrt{5}$ is irrational — 1 mark

Question 22 (2 marks)

In the figure, $AB \parallel DE$ and the line segments $AE$ and $BD$ intersect at $C$. If $AB = 7.5$ cm, $AC = 6$ cm, $CE = 4$ cm and $CD = 3.2$ cm, find $DE$ and $BC$.

Answer.

1. $\angle BAC = \angle DEC$ (alternate angles) and $\angle ACB = \angle ECD$ (vertically opposite), so $\triangle ACB \sim \triangle ECD$ (AA) — 1 mark
2. $\frac{AB}{ED} = \frac{AC}{EC} = \frac{BC}{DC}$: $DE = \frac{7.5 \times 4}{6} = 5$ cm and $BC = \frac{3.2 \times 6}{4} = 4.8$ cm — 1 mark

OR

$\triangle ABC \sim \triangle DEF$. If $AB = 4$ cm, $BC = 6$ cm, $CA = 5$ cm, $DE = 6$ cm and $EF = (x + 3)$ cm, find $x$ and the perimeter of $\triangle DEF$.

Answer.

1. $\frac{DE}{AB} = \frac{EF}{BC} = \frac{FD}{CA} = \frac{6}{4} = \frac{3}{2}$, so $EF = 9$ cm and $x + 3 = 9$, $x = 6$ — 1 mark
2. $FD = \frac{3}{2} \times 5 = 7.5$ cm; perimeter $= 6 + 9 + 7.5 = 22.5$ cm — 1 mark

Question 23 (2 marks)

Find the zeroes of the quadratic polynomial $3x^2 + 4x - 4$ and verify the relationship between the zeroes and the coefficients.

Answer.

1. $3x^2 + 6x - 2x - 4 = (3x - 2)(x + 2)$, so the zeroes are $\frac{2}{3}$ and $-2$ — 1 mark
2. Sum $= \frac{2}{3} - 2 = -\frac{4}{3} = -\frac{b}{a}$; product $= \frac{2}{3} \times (-2) = -\frac{4}{3} = \frac{c}{a}$ — 1 mark

Question 24 (2 marks)

Priya evaluated $\frac{2\tan 30^\circ}{1 - \tan^2 30^\circ}$ as follows.

Step 1: $\frac{2\tan 30^\circ}{1 - \tan^2 30^\circ} = \frac{2 \times \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}}$

Step 2: $= \frac{\frac{2}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{2}{\sqrt{3} - 1}$

In which step did she make a mistake? Find the correct value of the expression.

Answer.

1. The mistake is in Step 1: $\tan^2 30^\circ = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}$, not $\frac{1}{\sqrt{3}}$ — 1 mark
2. Correct value $= \frac{\frac{2}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{2}{\sqrt{3}} \times \frac{3}{2} = \sqrt{3}$ — 1 mark

OR

Evaluate: $\frac{4\sin^2 60^\circ + 3\tan^2 30^\circ + 8\sin 45^\circ \cos 45^\circ}{\text{cosec}^2 30^\circ - \cot^2 45^\circ}$

Answer.

1. Numerator $= 4 \times \frac{3}{4} + 3 \times \frac{1}{3} + 8 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 3 + 1 + 4 = 8$ — 1 mark
2. Denominator $= 4 - 1 = 3$; value $= \frac{8}{3}$ — 1 mark

Question 25 (2 marks)

In the figure, a circle with centre $O$ is inscribed in $\triangle ABC$ and touches the sides $AB$, $BC$ and $CA$ at $P$, $Q$ and $R$ respectively. If $AB = 9$ cm, $CA = 11$ cm, $AP = 5$ cm and $BC = (3x + 1)$ cm, find the value of $x$.

Answer.

1. Tangents from an external point are equal: $AR = AP = 5$ cm, so $BQ = BP = 9 - 5 = 4$ cm and $CQ = CR = 11 - 5 = 6$ cm — 1 mark
2. $BC = 4 + 6 = 10$ cm, so $3x + 1 = 10$ and $x = 3$ — 1 mark

Section C (18 marks)

Question 26 (3 marks)

Three bells in a temple ring at intervals of 18 minutes, 24 minutes and 32 minutes respectively. They ring together at 6:00 a.m. At what time will they next ring together? How many times in all will they ring together from 6:00 a.m. to 6:00 p.m. on the same day, counting the ring at 6:00 a.m.?

Answer.

1. $18 = 2 \times 3^2$, $24 = 2^3 \times 3$, $32 = 2^5$ — 1 mark
2. LCM $= 2^5 \times 3^2 = 288$ minutes $=$ 4 h 48 min; next together at 10:48 a.m. — 1 mark
3. In 12 hours (720 minutes) they ring together at 6:00 a.m., 10:48 a.m. and 3:36 p.m. (the next would be 8:24 p.m.), so 3 times — 1 mark

Question 27 (3 marks)

Prove that $\frac{1 - \sin\theta}{1 + \sin\theta} = (\sec\theta - \tan\theta)^2$.

Answer.

1. RHS $= \left(\frac{1}{\cos\theta} - \frac{\sin\theta}{\cos\theta}\right)^2 = \frac{(1 - \sin\theta)^2}{\cos^2\theta}$ — 1 mark
2. $= \frac{(1 - \sin\theta)^2}{1 - \sin^2\theta} = \frac{(1 - \sin\theta)^2}{(1 - \sin\theta)(1 + \sin\theta)}$ — 1 mark
3. $= \frac{1 - \sin\theta}{1 + \sin\theta} =$ LHS — 1 mark

Question 28 (3 marks)

$PA$ and $PB$ are tangents drawn from an external point $P$ to a circle with centre $O$. Prove that $PO$ bisects $\angle APB$.

Answer.

1. Figure; join $OA$ and $OB$. $\angle OAP = \angle OBP = 90^\circ$ (the radius is perpendicular to the tangent at the point of contact) — 1 mark
2. In $\triangle OAP$ and $\triangle OBP$: $OA = OB$ (radii), $OP$ common and the right angles are equal; so $\triangle OAP \cong \triangle OBP$ (RHS) — 1.5 marks
3. Hence $\angle APO = \angle BPO$ (CPCT), i.e. $PO$ bisects $\angle APB$ — 0.5 marks

OR

$PQ$ is a chord of length 24 cm of a circle of radius 20 cm. The tangents at $P$ and $Q$ intersect at a point $T$. Find the length of $TP$.

Answer.

1. $TP = TQ$ and $OP = OQ$, so $OT$ is the perpendicular bisector of $PQ$; let it meet $PQ$ at $R$. $PR = 12$ cm, $OR = \sqrt{20^2 - 12^2} = 16$ cm — 1 mark
2. Let $TR = y$. In right $\triangle OPT$, $OT^2 = TP^2 + OP^2$: $(y + 16)^2 = (y^2 + 144) + 400$, so $32y = 288$, $y = 9$ — 1 mark
3. $TP = \sqrt{9^2 + 12^2} = 15$ cm — 1 mark

Question 29 (3 marks)

A grain storage bin on a farm is in the shape of a right circular cylinder surmounted by a hemispherical dome of the same radius. The diameter of the bin is 4.2 m and the height of the cylindrical part is 4.9 m. The outer curved surface of the bin (the cylindrical wall and the dome, not the base) is to be painted at the rate of ₹ 20 per $\text{m}^2$. Find the cost of painting. (Use $\pi = \frac{22}{7}$)

Answer.

1. $r = 2.1$ m; CSA of cylinder $= 2\pi rh = 2 \times \frac{22}{7} \times 2.1 \times 4.9 = 64.68 \text{ m}^2$ — 1 mark
2. CSA of hemisphere $= 2\pi r^2 = 2 \times \frac{22}{7} \times 2.1 \times 2.1 = 27.72 \text{ m}^2$ — 1 mark
3. Area to be painted $= 64.68 + 27.72 = 92.4 \text{ m}^2$; cost $= 92.4 \times 20 =$ ₹ 1848 — 1 mark

OR

From a solid cylinder of radius 7 cm and height 30 cm, a conical cavity of the same radius and of height 24 cm is hollowed out from one end. Find the total surface area of the remaining solid in terms of $\pi$.

Answer.

1. Slant height of the cone $= \sqrt{7^2 + 24^2} = 25$ cm — 1 mark
2. Surfaces: CSA of cylinder $2\pi \times 7 \times 30 = 420\pi$, the other flat end $\pi \times 7^2 = 49\pi$, CSA of the cavity $\pi \times 7 \times 25 = 175\pi$ — 1 mark
3. Total surface area $= 420\pi + 49\pi + 175\pi = 644\pi \text{ cm}^2$ — 1 mark

Question 30 (3 marks)

All the black face cards are removed from a well-shuffled pack of 52 playing cards. A card is then drawn at random from the remaining cards. Find the probability that the card drawn is (i) a red card, (ii) a face card, (iii) a spade or an ace.

Answer.

1. Black face cards $= 6$, so 46 cards remain. (i) Red cards $= 26$: $P = \frac{26}{46} = \frac{13}{23}$ — 1 mark
2. (ii) Only the 6 red face cards remain: $P = \frac{6}{46} = \frac{3}{23}$ — 1 mark
3. (iii) Spades left $= 13 - 3 = 10$ (including the ace of spades), other aces $= 3$: $P = \frac{13}{46}$ — 1 mark

Question 31 (3 marks)

In a T20 match, a batter hit 16 boundaries, each of them a four or a six, and scored 76 runs from these boundaries. How many fours and how many sixes did the batter hit? How many runs came from sixes?

Answer.

1. Let the fours be $x$ and the sixes be $y$: $x + y = 16$ and $4x + 6y = 76$ — 1 mark
2. Substituting $x = 16 - y$: $64 - 4y + 6y = 76$, so $y = 6$ and $x = 10$ — 1.5 marks
3. Runs from sixes $= 6 \times 6 = 36$ — 0.5 marks

Section D (20 marks)

Question 32 (5 marks)

A State Transport bus covers a distance of 240 km between two towns at a uniform speed. If its speed had been 10 km/h more, it would have taken 2 hours less for the journey. Find the usual speed of the bus and the time it takes for the journey.

Answer.

1. Let the usual speed be $x$ km/h; time taken $= \frac{240}{x}$ hours — 1 mark
2. $\frac{240}{x} - \frac{240}{x + 10} = 2$ — 1 mark
3. $240 \times 10 = 2x(x + 10)$, i.e. $x^2 + 10x - 1200 = 0$ — 1 mark
4. $(x + 40)(x - 30) = 0$, so $x = 30$ ($x = -40$ rejected, speed cannot be negative) — 1 mark
5. Usual speed 30 km/h; time $= \frac{240}{30} = 8$ hours — 1 mark

Question 33 (5 marks)

In the figure, $ABCD$ is a trapezium in which $AB \parallel DC$, and its diagonals $AC$ and $BD$ intersect at $O$.

(i) Prove that $\triangle AOB \sim \triangle COD$.

(ii) Hence show that $OA \times OD = OB \times OC$.

(iii) If $OA = (2x + 1)$ cm, $OC = (x + 2)$ cm, $OB = (3x - 6)$ cm and $OD = x$ cm, find $x$ and the length of $BD$.

Answer.

1. (i) $AB \parallel DC$: $\angle OAB = \angle OCD$ and $\angle OBA = \angle ODC$ (alternate angles) — 1.5 marks
2. So $\triangle AOB \sim \triangle COD$ (AA similarity) — 0.5 marks
3. (ii) $\frac{OA}{OC} = \frac{OB}{OD}$, so $OA \times OD = OB \times OC$ — 1 mark
4. (iii) $\frac{2x + 1}{x + 2} = \frac{3x - 6}{x}$, so $2x^2 + x = 3x^2 - 12$, i.e. $x^2 - x - 12 = 0$ — 1 mark
5. $(x - 4)(x + 3) = 0$, so $x = 4$ ($x = -3$ rejected, as $OD = x$ is a length); $BD = OB + OD = 6 + 4 = 10$ cm — 1 mark

OR

In the figure, $AB$ and $CD$ are two vertical poles standing on level ground $BD$. The wires $AD$ and $CB$ cross each other at $P$, and $PQ \perp BD$.

(i) Prove that $\triangle DPQ \sim \triangle DAB$ and $\triangle BPQ \sim \triangle BCD$.

(ii) Hence show that $\frac{1}{AB} + \frac{1}{CD} = \frac{1}{PQ}$.

(iii) If $AB = 12$ m, $CD = 6$ m and $BD = 15$ m, find $PQ$ and $BQ$.

Answer.

1. (i) $\angle PQD = \angle ABD = 90^\circ$ and $\angle D$ is common, so $\triangle DPQ \sim \triangle DAB$ (AA) — 1 mark
2. $\angle PQB = \angle CDB = 90^\circ$ and $\angle B$ is common, so $\triangle BPQ \sim \triangle BCD$ (AA) — 1 mark
3. (ii) $\frac{PQ}{AB} = \frac{DQ}{DB}$ and $\frac{PQ}{CD} = \frac{BQ}{BD}$; adding, $\frac{PQ}{AB} + \frac{PQ}{CD} = \frac{DQ + BQ}{BD} = 1$, so $\frac{1}{AB} + \frac{1}{CD} = \frac{1}{PQ}$ — 1 mark
4. (iii) $\frac{1}{PQ} = \frac{1}{12} + \frac{1}{6} = \frac{1}{4}$, so $PQ = 4$ m — 1 mark
5. $\frac{BQ}{BD} = \frac{PQ}{CD}$: $BQ = \frac{4 \times 15}{6} = 10$ m — 1 mark

Question 34 (5 marks)

From the top $A$ of a vertical cliff $AB$ 90 m high, the angles of depression of two boats $C$ and $D$ on the sea, due east of the cliff and in line with its foot $B$, are $60^\circ$ and $30^\circ$ respectively. (i) Find the distance between the two boats. (ii) The boat at $D$ sails towards the cliff at a uniform speed and reaches the position of the other boat in 3 minutes. Find its speed in metres per minute. (Use $\sqrt{3} = 1.73$)

Answer.

1. Figure: with $AX$ horizontal, $\angle XAC = 60^\circ$ and $\angle XAD = 30^\circ$, so $\angle ACB = 60^\circ$ and $\angle ADB = 30^\circ$ (alternate angles); $AB = 90$ m — 1 mark
2. In $\triangle ABC$: $\tan 60^\circ = \frac{90}{BC}$, so $BC = \frac{90}{\sqrt{3}} = 30\sqrt{3}$ m — 1 mark
3. In $\triangle ABD$: $\tan 30^\circ = \frac{90}{BD}$, so $BD = 90\sqrt{3}$ m — 1 mark
4. $CD = 90\sqrt{3} - 30\sqrt{3} = 60\sqrt{3} = 60 \times 1.73 = 103.8$ m — 1 mark
5. Speed $= \frac{60\sqrt{3}}{3} = 20\sqrt{3} = 34.6$ m per minute — 1 mark

Question 35 (5 marks)

The table shows the distance (in km) that 50 employees of a company travel daily to reach their office. The mean distance is 23.4 km.

Find the missing frequencies $p$ and $q$.

Answer.

1. $8 + p + 12 + q + 6 = 50$, so $p + q = 24$ — 1 mark
2. Class marks 5, 15, 25, 35, 45; $\sum f_i x_i = 40 + 15p + 300 + 35q + 270 = 610 + 15p + 35q$ — 1.5 marks
3. $\frac{610 + 15p + 35q}{50} = 23.4$ gives $15p + 35q = 560$, i.e. $3p + 7q = 112$ — 1 mark
4. Using $p = 24 - q$: $72 + 4q = 112$, so $q = 10$ and $p = 14$ — 1.5 marks

OR

The mode of the following distribution of the time (in seconds) taken by some students to solve a puzzle is 54 seconds. Find the missing frequency $x$.

Answer.

1. Mode 54 lies in 50-60, so the modal class is 50-60: $l = 50$, $f_1 = x$, $f_0 = 16$, $f_2 = 12$, $h = 10$ — 1.5 marks
2. $54 = 50 + \frac{x - 16}{2x - 16 - 12} \times 10$ — 1 mark
3. $4(2x - 28) = 10(x - 16)$, i.e. $8x - 112 = 10x - 160$ — 1 mark
4. $2x = 48$, so $x = 24$ — 1.5 marks

Section E (12 marks)

Question 36 (4 marks)

Ramesh runs a tea stall. He joins a monthly savings scheme at the post office and deposits ₹ 500 in the first month. Every month after that he deposits ₹ 100 more than in the previous month.

(i) How much does he deposit in the 10th month? (1 mark)

Answer.

1. $a_{10} = 500 + 9 \times 100 =$ ₹ 1400 — 1 mark

(ii) In which month does he deposit ₹ 2500? (1 mark)

Answer.

1. $500 + (n - 1)100 = 2500$ gives $n = 21$; the 21st month — 1 mark

(iii) Find the total amount he deposits in the first 2 years. (2 marks)

Answer.

1. $S_{24} = \frac{24}{2}[2 \times 500 + 23 \times 100]$ — 1 mark
2. $= 12 \times 3300 =$ ₹ 39600 — 1 mark

OR

(iii) Find the total amount he deposits in the second year, that is, from the 13th month to the 24th month. (2 marks)

Answer.

1. $a_{13} = 500 + 12 \times 100 = 1700$, $a_{24} = 500 + 23 \times 100 = 2800$ — 1 mark
2. Sum $= \frac{12}{2}(1700 + 2800) =$ ₹ 27000 — 1 mark

Question 37 (4 marks)

The map of a village is drawn on a coordinate grid, with the Panchayat Bhawan at the origin $O$ and the main road along the x-axis. One unit on the grid stands for 100 m. The school is at $S(-4, 5)$, the health centre at $H(5, -7)$, the market at $M(-5, -2)$ and the temple at $T(7, 4)$.

(i) How far is the health centre from the school, in metres? (1 mark)

Answer.

1. $SH = \sqrt{(5 + 4)^2 + (-7 - 5)^2} = \sqrt{81 + 144} = 15$ units $= 1500$ m — 1 mark

(ii) A bus stop is to be built exactly midway between the market and the temple. Find its coordinates. (1 mark)

Answer.

1. Mid-point of $MT$ $= \left(\frac{-5 + 7}{2}, \frac{-2 + 4}{2}\right) = (1, 1)$ — 1 mark

(iii) A straight road joins the market and the temple. In what ratio does the main road (the x-axis) divide the line segment $MT$? At which point do the two roads cross? (2 marks)

Answer.

1. Let the ratio be $k : 1$; y-coordinate $\frac{4k - 2}{k + 1} = 0$ gives $k = \frac{1}{2}$, i.e. $1 : 2$ — 1 mark
2. $x = \frac{1 \times 7 + 2 \times (-5)}{3} = -1$; the roads cross at $(-1, 0)$ — 1 mark

OR

(iii) A water tank $W$ is to be built beside the lane that runs along the y-axis, at the same distance from the market and the temple. Find the coordinates of $W$. (2 marks)

Answer.

1. Let $W(0, y)$. $WM^2 = WT^2$: $(0 + 5)^2 + (y + 2)^2 = (0 - 7)^2 + (y - 4)^2$ — 1 mark
2. $25 + 4y + 4 = 49 - 8y + 16$, so $12y = 36$, $y = 3$; $W(0, 3)$ — 1 mark

Question 38 (4 marks)

The windscreen wiper of a car has an arm $OB$ of length 42 cm, pivoted at $O$. The rubber blade $AB$ is fixed on the outer half of the arm, so $OA = AB = 21$ cm. In one sweep the arm turns through an angle of $120^\circ$ and the blade wipes the region between the two arcs (shaded in the figure). (Use $\pi = \frac{22}{7}$)

(i) Find the length of the arc traced by the tip $B$ in one sweep. (1 mark)

Answer.

1. $\frac{120}{360} \times 2 \times \frac{22}{7} \times 42 = 88$ cm — 1 mark

(ii) Find the area of the sector swept by the whole arm $OB$ in one sweep. (1 mark)

Answer.

1. $\frac{120}{360} \times \frac{22}{7} \times 42 \times 42 = 1848 \text{ cm}^2$ — 1 mark

(iii) Find the area wiped by the blade in one sweep. The car has a second, identical wiper, and the regions wiped by the two blades are separate. What total area do the two blades wipe in one sweep? (2 marks)

Answer.

1. Sector of radius 21 cm $= \frac{120}{360} \times \frac{22}{7} \times 21 \times 21 = 462 \text{ cm}^2$; area wiped by one blade $= 1848 - 462 = 1386 \text{ cm}^2$ — 1 mark
2. Both blades: $2 \times 1386 = 2772 \text{ cm}^2$ — 1 mark

OR

(iii) Find the perimeter of the region wiped by one blade in one sweep. (2 marks)

Answer.

1. Outer arc $= 88$ cm; inner arc $= \frac{120}{360} \times 2 \times \frac{22}{7} \times 21 = 44$ cm — 1 mark
2. Perimeter $= 88 + 44 + 21 + 21 = 174$ cm — 1 mark