Detailed Solutions

Detailed Solutions: CBSE Class 10 Maths Sample Paper 2027 – Set 4

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 $\sqrt{2}$ is irrational, prove that $7 - 3\sqrt{2}$ is irrational.

Answer.

1. Assume $7 - 3\sqrt{2} = r$, a rational number; then $\sqrt{2} = \frac{7 - r}{3}$ — 1 mark
2. $\frac{7 - r}{3}$ is rational because $r$ is rational, but $\sqrt{2}$ is irrational - a contradiction; hence $7 - 3\sqrt{2}$ is irrational — 1 mark

Question 22 (2 marks)

A circle has its centre at $C(2, -3)$ and passes through the point $P(-4, 5)$. Find the radius of the circle. Does the point $Q(10, 3)$ lie on this circle? Give a reason.

Answer.

1. Radius $= CP = \sqrt{(2 + 4)^2 + (-3 - 5)^2} = \sqrt{36 + 64} = 10$ units — 1 mark
2. $CQ = \sqrt{(10 - 2)^2 + (3 + 3)^2} = \sqrt{64 + 36} = 10$ units, equal to the radius, so $Q$ lies on the circle — 1 mark

Question 23 (2 marks)

Two vertical poles $AB$ and $CD$, of heights 6 m and 9 m, stand on level ground at $B$ and $D$. Wires are stretched from $A$ to $D$ and from $C$ to $B$, and they cross at $P$, as shown in the figure. Find the height $PM$ of the crossing point above the ground.

Answer.

1. Let $PM = h$, with $M$ on $BD$. $\triangle BMP \sim \triangle BDC$ and $\triangle DMP \sim \triangle DBA$ (AA), so $\frac{h}{9} = \frac{BM}{BD}$ and $\frac{h}{6} = \frac{MD}{BD}$ — 1 mark
2. Adding, $\frac{h}{9} + \frac{h}{6} = \frac{BM + MD}{BD} = 1$, so $\frac{5h}{18} = 1$ and $h = 3.6$ m — 1 mark

OR

In $\triangle ABC$ and $\triangle PQR$, $\angle A = \angle P$ and $\angle B = \angle Q$. If $AB = 6$ cm, $BC = (2x + 1)$ cm, $PQ = 4$ cm and $QR = (x + 2)$ cm, find $x$ and the lengths of $BC$ and $QR$.

Answer.

1. $\triangle ABC \sim \triangle PQR$ (AA), so $\frac{AB}{PQ} = \frac{BC}{QR}$, i.e. $\frac{6}{4} = \frac{2x + 1}{x + 2}$ — 1 mark
2. $3(x + 2) = 2(2x + 1)$ gives $x = 4$; $BC = 9$ cm and $QR = 6$ cm — 1 mark

Question 24 (2 marks)

In the figure, a circle touches the side $BC$ of $\triangle ABC$ at $P$, and touches $AB$ produced and $AC$ produced at $Q$ and $R$ respectively. If $AB = 7$ cm, $BC = 6$ cm and $CA = 5$ cm, find the length of $AQ$.

Answer.

1. Tangents from an external point are equal: $AQ = AR$, $BQ = BP$ and $CR = CP$ — 1 mark
2. $AQ + AR = (AB + BQ) + (AC + CR) = AB + AC + (BP + CP) = 7 + 5 + 6 = 18$, so $2AQ = 18$ and $AQ = 9$ cm — 1 mark

Question 25 (2 marks)

A solid model of a rocket is made of a right circular cylinder of radius 5 cm and height 20 cm, with a right circular cone of the same radius and height 12 cm fixed on its top, as shown in the figure. Find the total surface area of the model in terms of $\pi$.

Answer.

1. Slant height of the cone $l = \sqrt{5^2 + 12^2} = 13$ cm; the surface is the cone's curved surface, the cylinder's curved surface and the circular base — 1 mark
2. Total surface area $= \pi r l + 2\pi r h + \pi r^2 = 65\pi + 200\pi + 25\pi = 290\pi \text{ cm}^2$ — 1 mark

OR

A measuring vessel is a hollow cylinder of internal radius 3 cm and height 10 cm, closed at the bottom by a hemispherical bowl of the same radius. How much liquid can the vessel hold? Give your answer in terms of $\pi$.

Answer.

1. Cylinder: $\pi r^2 h = \pi \times 9 \times 10 = 90\pi \text{ cm}^3$; hemisphere: $\frac{2}{3}\pi r^3 = \frac{2}{3}\pi \times 27 = 18\pi \text{ cm}^3$ — 1 mark
2. Capacity $= 90\pi + 18\pi = 108\pi \text{ cm}^3$ — 1 mark

Section C (18 marks)

Question 26 (3 marks)

A school librarian has 336 Hindi books, 240 English books and 216 Science books. She wants to arrange them in stacks so that every stack has the same number of books and each stack has books of only one subject. What is the greatest number of books she can put in each stack? How many stacks will she make in all?

Answer.

1. $336 = 2^4 \times 3 \times 7$, $240 = 2^4 \times 3 \times 5$, $216 = 2^3 \times 3^3$ — 1 mark
2. Greatest number of books in a stack = HCF $= 2^3 \times 3 = 24$ — 1 mark
3. Number of stacks $= \frac{336}{24} + \frac{240}{24} + \frac{216}{24} = 14 + 10 + 9 = 33$ — 1 mark

Question 27 (3 marks)

If $\alpha$ and $\beta$ are the zeroes of the polynomial $x^2 - 3x - 10$, find a quadratic polynomial whose zeroes are $\alpha^2$ and $\beta^2$.

Answer.

1. $\alpha + \beta = 3$ and $\alpha\beta = -10$ — 1 mark
2. $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 9 + 20 = 29$ and $\alpha^2\beta^2 = (\alpha\beta)^2 = 100$ — 1 mark
3. The required polynomial is $x^2 - 29x + 100$ (or any non-zero constant multiple of it); check: the zeroes of $x^2 - 3x - 10$ are $5$ and $-2$, and $25 + 4 = 29$, $25 \times 4 = 100$ — 1 mark

Question 28 (3 marks)

Meera was asked to find the sum of all the multiples of 6 between 100 and 200. Her working is shown below.

First term $a = 102$, last term $l = 198$, common difference $d = 6$.

Number of terms $n = \frac{198 - 102}{6} = 16$.

Sum $= \frac{16}{2}(102 + 198) = 2400$.

Is her answer correct? If not, point out the mistake and find the correct sum.

Answer.

1. The mistake is in the number of terms: $198 = 102 + (n - 1) \times 6$ gives $n - 1 = 16$, so $n = 17$, not 16 — 1.5 marks
2. Correct sum $= \frac{17}{2}(102 + 198) = 17 \times 150 = 2550$ — 1.5 marks

OR

The 5th term of an AP is 17 and the sum of its first 10 terms is 185. Find the AP.

Answer.

1. $a_5 = a + 4d = 17$ and $S_{10} = \frac{10}{2}(2a + 9d) = 185$, i.e. $2a + 9d = 37$ — 1 mark
2. Doubling the first, $2a + 8d = 34$; subtracting, $d = 3$, and then $a = 17 - 12 = 5$ — 1 mark
3. The AP is $5, 8, 11, 14, \ldots$ — 1 mark

Question 29 (3 marks)

The point $P(x, 6)$ lies on the line segment joining $A(-6, 9)$ and $B(10, -3)$. Find the ratio in which $P$ divides $AB$ and the value of $x$. Verify the ratio by finding $AP$ and $PB$ with the distance formula.

Answer.

1. Let the ratio be $k : 1$. From the y-coordinate, $\frac{-3k + 9}{k + 1} = 6$, so $-3k + 9 = 6k + 6$ and $k = \frac{1}{3}$; the ratio is $1 : 3$ — 1 mark
2. $x = \frac{1 \times 10 + 3 \times (-6)}{4} = \frac{-8}{4} = -2$, so $P$ is $(-2, 6)$ — 1 mark
3. $AP = \sqrt{4^2 + 3^2} = 5$ and $PB = \sqrt{12^2 + 9^2} = 15$, so $AP : PB = 1 : 3$; verified — 1 mark

Question 30 (3 marks)

Prove that $\frac{\cos^2 A}{1 - \sin A} + \frac{\sin^2 A}{1 + \cos A} = 2 + \sin A - \cos A$.

Answer.

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

OR

Prove that $\frac{\tan^2 A}{\sec A - 1} + \frac{\cot^2 A}{\text{cosec}\,A + 1} = \sec A + \text{cosec}\,A$.

Answer.

1. $\frac{\tan^2 A}{\sec A - 1} = \frac{\sec^2 A - 1}{\sec A - 1} = \frac{(\sec A - 1)(\sec A + 1)}{\sec A - 1} = \sec A + 1$ — 1 mark
2. $\frac{\cot^2 A}{\text{cosec}\,A + 1} = \frac{\text{cosec}^2 A - 1}{\text{cosec}\,A + 1} = \frac{(\text{cosec}\,A - 1)(\text{cosec}\,A + 1)}{\text{cosec}\,A + 1} = \text{cosec}\,A - 1$ — 1 mark
3. LHS $= (\sec A + 1) + (\text{cosec}\,A - 1) = \sec A + \text{cosec}\,A =$ RHS — 1 mark

Question 31 (3 marks)

A card is drawn at random from a well-shuffled deck of 52 playing cards. Find the probability that the card drawn is (i) a red king, (ii) neither a heart nor a king, (iii) a black card that is not a face card.

Answer.

1. (i) There are 2 red kings; $P = \frac{2}{52} = \frac{1}{26}$ — 1 mark
2. (ii) Hearts or kings: $13 + 3 = 16$ cards (the king of hearts is counted once), so $52 - 16 = 36$ cards are neither; $P = \frac{36}{52} = \frac{9}{13}$ — 1 mark
3. (iii) There are 26 black cards, of which 6 are face cards, leaving 20; $P = \frac{20}{52} = \frac{5}{13}$ — 1 mark

Section D (20 marks)

Question 32 (5 marks)

The roll numbers of three students sitting next to each other in a row are three consecutive natural numbers. The product of the smallest and the largest roll number, increased by 5 times the middle roll number, is 233. Find the three roll numbers.

Answer.

1. Let the roll numbers be $n$, $n + 1$ and $n + 2$ — 1 mark
2. $n(n + 2) + 5(n + 1) = 233$ — 1 mark
3. $n^2 + 7n + 5 = 233$, i.e. $n^2 + 7n - 228 = 0$ — 1 mark
4. $(n - 12)(n + 19) = 0$, so $n = 12$ ($n = -19$ is rejected, as a roll number is a natural number) — 1 mark
5. The roll numbers are 12, 13 and 14 (check: $12 \times 14 + 5 \times 13 = 168 + 65 = 233$) — 1 mark

OR

A mother is 26 years older than her daughter. Five years from now, the product of their ages (in years) will be 615. Find their present ages.

Answer.

1. Let the daughter's present age be $x$ years; then the mother's present age is $(x + 26)$ years — 1 mark
2. After 5 years: $(x + 5)(x + 31) = 615$ — 1 mark
3. $x^2 + 36x + 155 = 615$, i.e. $x^2 + 36x - 460 = 0$ — 1 mark
4. $(x - 10)(x + 46) = 0$, so $x = 10$ ($x = -46$ is rejected, as an age cannot be negative) — 1 mark
5. The daughter is 10 years old and the mother is 36 years old — 1 mark

Question 33 (5 marks)

State and prove the Basic Proportionality Theorem (take $\triangle PQR$ with $ST \parallel QR$). Using it, solve the following: in the figure, $ABCD$ is a trapezium with $AB \parallel DC$, and $E$ and $F$ are points on $AD$ and $BC$ such that $EF \parallel AB$. If $AE = (x - 1)$ cm, $ED = (x + 2)$ cm, $BF = x$ cm and $FC = 2x$ cm, find $x$.

Answer.

1. Correct statement of the theorem — 1 mark
2. Given: in $\triangle PQR$, $ST \parallel QR$ with $S$ on $PQ$ and $T$ on $PR$; to prove: $\frac{PS}{SQ} = \frac{PT}{TR}$; construction: join $QT$ and $RS$, draw $TN \perp PQ$ and $SM \perp PR$; figure — 1 mark
3. $\frac{ar(PST)}{ar(QST)} = \frac{\frac{1}{2} \times PS \times TN}{\frac{1}{2} \times SQ \times TN} = \frac{PS}{SQ}$ (same height $TN$) and $\frac{ar(PST)}{ar(RST)} = \frac{\frac{1}{2} \times PT \times SM}{\frac{1}{2} \times TR \times SM} = \frac{PT}{TR}$ (same height $SM$); $ar(QST) = ar(RST)$ (same base $ST$, between the same parallels $ST$ and $QR$) - hence $\frac{PS}{SQ} = \frac{PT}{TR}$ — 1 mark
4. Join $AC$, meeting $EF$ at $G$. In $\triangle ADC$, $EG \parallel DC$ gives $\frac{AE}{ED} = \frac{AG}{GC}$; in $\triangle CAB$, $GF \parallel AB$ gives $\frac{AG}{GC} = \frac{BF}{FC}$; so $\frac{AE}{ED} = \frac{BF}{FC}$ — 1 mark
5. $\frac{x - 1}{x + 2} = \frac{x}{2x} = \frac{1}{2}$, so $2x - 2 = x + 2$ and $x = 4$ (then $AE = 3$ cm, $ED = 6$ cm, $BF = 4$ cm, $FC = 8$ cm) — 1 mark

Question 34 (5 marks)

An aeroplane is flying horizontally at a constant height of 2500 m above the ground, moving away from an observer standing on the ground. At one instant the angle of elevation of the aeroplane from the observer is $45^\circ$, and 10 seconds later it is $30^\circ$. Find the speed of the aeroplane in km/h. Also find the distance of the aeroplane from the observer at the second instant. (Use $\sqrt{3} = 1.73$)

Answer.

1. Correct figure: observer $O$ on the ground; aeroplane at $P$ and then at $Q$, 2500 m above the ground points $M$ and $N$ — 1 mark
2. In $\triangle OMP$: $\tan 45^\circ = \frac{2500}{OM}$, so $OM = 2500$ m — 1 mark
3. In $\triangle ONQ$: $\tan 30^\circ = \frac{2500}{ON}$, so $ON = 2500\sqrt{3}$ m — 1 mark
4. Distance flown $= MN = 2500(\sqrt{3} - 1) = 2500 \times 0.73 = 1825$ m in 10 s; speed $= 182.5$ m/s $= 182.5 \times \frac{18}{5} = 657$ km/h — 1 mark
5. At the second instant, $\sin 30^\circ = \frac{2500}{OQ}$, so $OQ = 5000$ m — 1 mark

Question 35 (5 marks)

The weights (in kg) of 60 students of Class X of a school are given below.

Find the median and the mode of the weights.

Answer.

1. Cumulative frequencies: 5, 12, 22, 38, 52, 60; $\frac{n}{2} = 30$, so the median class is 50-55 — 1 mark
2. $l = 50$, $cf = 22$, $f = 16$, $h = 5$ — 1 mark
3. Median $= 50 + \frac{30 - 22}{16} \times 5 = 50 + 2.5 = 52.5$ kg — 1 mark
4. Modal class 50-55; $l = 50$, $f_1 = 16$, $f_0 = 10$, $f_2 = 14$, $h = 5$ — 1 mark
5. Mode $= 50 + \frac{16 - 10}{32 - 10 - 14} \times 5 = 50 + 3.75 = 53.75$ kg — 1 mark

OR

The table shows the quantity of milk (in litres) supplied in a day by the farmers of a village to a dairy cooperative. The median quantity is 24 litres. Find the missing frequency $p$, and then find the mode of the data.

Answer.

1. $n = 42 + p$; cumulative frequencies 8, 20, 35, $35 + p$, $42 + p$; the median 24 lies in the class 20-30 — 1 mark
2. $24 = 20 + \frac{\frac{42 + p}{2} - 20}{15} \times 10$ — 1 mark
3. $\frac{42 + p}{2} - 20 = 6$, so $42 + p = 52$ and $p = 10$ — 1 mark
4. Modal class 20-30; $l = 20$, $f_1 = 15$, $f_0 = 12$, $f_2 = 10$, $h = 10$ — 1 mark
5. Mode $= 20 + \frac{15 - 12}{30 - 12 - 10} \times 10 = 20 + 3.75 = 23.75$ litres — 1 mark

Section E (12 marks)

Question 36 (4 marks)

A city zoo sells entry tickets at one price for an adult and a lower price for a child. The Sharma family bought tickets for 2 adults and 3 children and paid ₹ 330. The Iyer family bought tickets for 3 adults and 4 children and paid ₹ 470. Let the price of an adult ticket be ₹ $x$ and that of a child ticket be ₹ $y$.

(i) Write the pair of linear equations that describes this situation. (1 mark)

Answer.

1. $2x + 3y = 330$ and $3x + 4y = 470$ — 1 mark

(ii) Will the lines representing these equations intersect, be parallel or coincide? Give a reason. (1 mark)

Answer.

1. $\frac{a_1}{a_2} = \frac{2}{3}$ and $\frac{b_1}{b_2} = \frac{3}{4}$ are unequal, so the lines intersect at exactly one point — 1 mark

(iii) Find the price of an adult ticket and the price of a child ticket. (2 marks)

Answer.

1. Multiplying by 3 and 2: $6x + 9y = 990$ and $6x + 8y = 940$; subtracting, $y = 50$ — 1 mark
2. $2x = 330 - 150 = 180$, so $x = 90$; adult ticket ₹ 90, child ticket ₹ 50 — 1 mark

OR

(iii) Inside the zoo, a toy-train ride costs ₹ $a$ for an adult and ₹ $b$ for a child. The fare for 4 adults and 2 children is ₹ 200, and the fare for 2 adults and 5 children is ₹ 220. Find $a$ and $b$. (2 marks)

Answer.

1. $4a + 2b = 200$, i.e. $2a + b = 100$, and $2a + 5b = 220$; subtracting, $4b = 120$, so $b = 30$ — 1 mark
2. $2a = 100 - 30 = 70$, so $a = 35$; adult ₹ 35, child ₹ 30 — 1 mark

Question 37 (4 marks)

A park has a circular fountain with centre $O$ and radius 7 m. A gate $P$ of the park is 25 m from $O$. Two straight paths $PA$ and $PB$ go from the gate and just touch the edge of the fountain at $A$ and $B$, as shown in the figure.

(i) Find the length of the path $PA$. (1 mark)

Answer.

1. $OA \perp PA$ (the radius is perpendicular to the tangent), so $PA = \sqrt{25^2 - 7^2} = \sqrt{576} = 24$ m — 1 mark

(ii) Find the area of the quadrilateral $OAPB$. (1 mark)

Answer.

1. $PB = PA = 24$ m (tangents from $P$), so $\triangle OAP \cong \triangle OBP$ and the area $= 2 \times \frac{1}{2} \times 7 \times 24 = 168 \text{ m}^2$ — 1 mark

(iii) A straight water pipe is laid from $A$ to $B$. It meets $OP$ at $M$. Find the length of the pipe $AB$. (2 marks)

Answer.

1. $PA = PB$ and $OA = OB$, so $OP$ is the perpendicular bisector of $AB$: $AM \perp OP$ and $AB = 2AM$ — 1 mark
2. $\frac{1}{2} \times OP \times AM = \frac{1}{2} \times OA \times PA$ (both are the area of $\triangle OAP$), so $AM = \frac{7 \times 24}{25} = 6.72$ m and $AB = 13.44$ m — 1 mark

OR

(iii) Another gate $G$ is placed so that the two straight paths from $G$ that just touch the fountain make an angle of $60^\circ$ with each other. How far is $G$ from the centre $O$, and how long is each of these paths? (Use $\sqrt{3} = 1.73$) (2 marks)

Answer.

1. Let one path touch the fountain at $T$. $OG$ bisects the angle between the paths, so $\angle OGT = 30^\circ$, and $\angle OTG = 90^\circ$; $\sin 30^\circ = \frac{7}{OG}$ gives $OG = 14$ m — 1 mark
2. $GT = \sqrt{14^2 - 7^2} = \sqrt{147} = 7\sqrt{3} = 12.11$ m — 1 mark

Question 38 (4 marks)

A rotating sprinkler is fixed at a point $O$ on the lawn of a park. It throws water up to a distance of 21 m and turns back and forth through an angle of $120^\circ$, so that it waters the sector $OAB$ of the lawn shown in the figure. (Use $\pi = \frac{22}{7}$)

(i) Find the length of the arc $AB$. (1 mark)

Answer.

1. Arc $= \frac{120}{360} \times 2 \times \frac{22}{7} \times 21 = 44$ m — 1 mark

(ii) Find the area of the lawn watered by the sprinkler. (1 mark)

Answer.

1. Area $= \frac{120}{360} \times \frac{22}{7} \times 21 \times 21 = \frac{1}{3} \times 1386 = 462 \text{ m}^2$ — 1 mark

(iii) To save water, the gardener changes the setting so that the sprinkler turns through only $90^\circ$, with the same range. By how much does the watered area decrease? In the new setting, find the area of the watered part that lies between the arc and the chord joining its two ends. (2 marks)

Answer.

1. New area $= \frac{1}{4} \times \frac{22}{7} \times 21 \times 21 = 346.5 \text{ m}^2$; decrease $= 462 - 346.5 = 115.5 \text{ m}^2$ — 1 mark
2. Segment $= 346.5 - \frac{1}{2} \times 21 \times 21 = 346.5 - 220.5 = 126 \text{ m}^2$ — 1 mark

OR

(iii) A second sprinkler on another part of the lawn has a range of 28 m and turns through $90^\circ$. Find the area watered by it and the total length of the boundary of the region it waters. (2 marks)

Answer.

1. Area $= \frac{90}{360} \times \frac{22}{7} \times 28 \times 28 = 616 \text{ m}^2$ — 1 mark
2. Arc $= \frac{90}{360} \times 2 \times \frac{22}{7} \times 28 = 44$ m; boundary $= 28 + 28 + 44 = 100$ m — 1 mark