Detailed Solutions

Detailed Solutions: CBSE Class 10 Maths Sample Paper 2027 – Set 5

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

Answer.

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

Question 22 (2 marks)

Find a quadratic polynomial whose zeroes are $3 + \sqrt{2}$ and $3 - \sqrt{2}$.

Answer.

1. Sum of the zeroes $= (3 + \sqrt{2}) + (3 - \sqrt{2}) = 6$; product $= (3 + \sqrt{2})(3 - \sqrt{2}) = 9 - 2 = 7$ — 1 mark
2. With the zeroes as $\alpha$ and $\beta$, the polynomial is $x^2 - (\alpha + \beta)x + \alpha\beta = x^2 - 6x + 7$ (or any non-zero constant multiple of it) — 1 mark

Question 23 (2 marks)

In the figure, $S$ and $T$ are points on the sides $PQ$ and $PR$ of $\triangle PQR$ such that $PS = 4$ cm, $PQ = 10$ cm, $PT = 6$ cm and $TR = 9$ cm. Show that $ST \parallel QR$.

Answer.

1. $SQ = 10 - 4 = 6$ cm; $\frac{PS}{SQ} = \frac{4}{6} = \frac{2}{3}$ and $\frac{PT}{TR} = \frac{6}{9} = \frac{2}{3}$ — 1 mark
2. Since $\frac{PS}{SQ} = \frac{PT}{TR}$, $ST \parallel QR$ by the converse of the Basic Proportionality Theorem — 1 mark

OR

$ABCD$ is a parallelogram and $E$ is a point on the side $BC$. The line $DE$, when produced, meets $AB$ produced at $F$. Prove that $\triangle DCE \sim \triangle FBE$.

Answer.

1. In $\triangle DCE$ and $\triangle FBE$: $\angle DEC = \angle FEB$ (vertically opposite angles) — 1 mark
2. $DC \parallel AF$ and $BC$ is a transversal, so $\angle DCE = \angle FBE$ (alternate angles); hence $\triangle DCE \sim \triangle FBE$ (AA similarity) — 1 mark

Question 24 (2 marks)

Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.

Answer.

1. Let $AB$ be a chord of a circle with centre $O$. $OA = OB$ (radii), so $\angle OAB = \angle OBA$ — 1 mark
2. The tangents at $A$ and $B$ are perpendicular to $OA$ and $OB$, so they make angles $90^\circ - \angle OAB$ and $90^\circ - \angle OBA$ with the chord; these are equal — 1 mark

OR

$PA$ is a tangent at $A$ to a circle with centre $O$ and radius $r$, and $OP = 2r$. The segment $OP$ meets the circle at $B$. Find $\angle APO$ and show that $\triangle OAB$ is equilateral.

Answer.

1. $OA \perp PA$, so in right $\triangle OAP$, $\sin \angle APO = \frac{OA}{OP} = \frac{r}{2r} = \frac{1}{2}$; hence $\angle APO = 30^\circ$ — 1 mark
2. $\angle AOB = 90^\circ - 30^\circ = 60^\circ$ and $OA = OB = r$, so $\angle OAB = \angle OBA = 60^\circ$ and $\triangle OAB$ is equilateral — 1 mark

Question 25 (2 marks)

If $\sqrt{3}\tan 2\theta = 3$, where $2\theta$ is an acute angle, find $\theta$. Hence find the value of $\sin^2 3\theta + \cos^2 \theta$.

Answer.

1. $\tan 2\theta = \frac{3}{\sqrt{3}} = \sqrt{3} = \tan 60^\circ$, so $2\theta = 60^\circ$ and $\theta = 30^\circ$ — 1 mark
2. $\sin^2 90^\circ + \cos^2 30^\circ = 1 + \frac{3}{4} = \frac{7}{4}$ — 1 mark

Section C (18 marks)

Question 26 (3 marks)

Anjali and Farhan start running together from the same point on a circular track, in the same direction. Anjali takes 4 minutes 12 seconds to complete one round and Farhan takes 4 minutes 40 seconds. After how many minutes will they next meet at the starting point? How many rounds will each of them have completed by then?

Answer.

1. In seconds: $252 = 2^2 \times 3^2 \times 7$ and $280 = 2^3 \times 5 \times 7$ — 1 mark
2. They next meet at the start after $\text{LCM}(252, 280) = 2^3 \times 3^2 \times 5 \times 7 = 2520$ seconds $= 42$ minutes — 1 mark
3. Rounds: Anjali $\frac{2520}{252} = 10$, Farhan $\frac{2520}{280} = 9$ — 1 mark

Question 27 (3 marks)

In the figure, $AB$ is a diameter of a circle with centre $O$, and the tangent at a point $C$ of the circle meets $AB$ produced at $P$. If $\angle PCA = 115^\circ$, find (i) $\angle CAB$, (ii) $\angle CBA$, (iii) $\angle CPA$.

Answer.

1. (i) $OC \perp CP$, so $\angle OCA = 115^\circ - 90^\circ = 25^\circ$; $OA = OC$, so $\angle CAB = \angle OCA = 25^\circ$ — 1 mark
2. (ii) $\angle ACB = 90^\circ$ (angle in a semicircle), so $\angle CBA = 180^\circ - 90^\circ - 25^\circ = 65^\circ$ — 1 mark
3. (iii) In $\triangle PCA$, $\angle CPA = 180^\circ - 115^\circ - 25^\circ = 40^\circ$ — 1 mark

Question 28 (3 marks)

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

Answer.

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

OR

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

Answer.

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

Question 29 (3 marks)

A circular paper disc of radius 42 cm is cut along a chord $AB$ which subtends an angle of $60^\circ$ at the centre $O$. Find the area of the smaller piece (the minor segment). (Use $\pi = \frac{22}{7}$ and $\sqrt{3} = 1.73$)

Answer.

1. Area of the sector $= \frac{60}{360} \times \frac{22}{7} \times 42 \times 42 = 924 \text{ cm}^2$ — 1 mark
2. $OA = OB$ and $\angle AOB = 60^\circ$, so $\triangle OAB$ is equilateral; its area $= \frac{\sqrt{3}}{4} \times 42^2 = 441\sqrt{3} = 762.93 \text{ cm}^2$ — 1 mark
3. Area of minor segment $= 924 - 762.93 = 161.07 \text{ cm}^2$ — 1 mark

Question 30 (3 marks)

A wooden block for a children's play corner is a cube of edge 14 cm, with a solid hemisphere of diameter 7 cm fixed on the middle of its top face. The whole outer surface of the block is to be painted at ₹ 10 per 100 $\text{cm}^2$. Find the total surface area of the block and the cost of painting it. (Use $\pi = \frac{22}{7}$)

Answer.

1. Surface area of the cube $= 6 \times 14 \times 14 = 1176 \text{ cm}^2$ — 1 mark
2. The hemisphere ($r = 3.5$ cm) adds its curved surface $2\pi r^2 = 77 \text{ cm}^2$ but covers a circle of area $\pi r^2 = 38.5 \text{ cm}^2$ on the top face; total $= 1176 + 77 - 38.5 = 1214.5 \text{ cm}^2$ — 1 mark
3. Cost $= \frac{1214.5 \times 10}{100} =$ ₹ 121.45 — 1 mark

OR

A wooden pen holder is a cuboid with a square base of side 10 cm and height 12 cm. A cylindrical hole of radius 3.5 cm and depth 10 cm is scooped out of it from the top to hold pens. Find the volume of wood left in the pen holder. If 1 $\text{cm}^3$ of the wood has a mass of 0.8 g, find the mass of the pen holder. (Use $\pi = \frac{22}{7}$)

Answer.

1. Volume of the cuboid $= 10 \times 10 \times 12 = 1200 \text{ cm}^3$ — 1 mark
2. Volume of the hole $= \pi r^2 h = \frac{22}{7} \times 3.5 \times 3.5 \times 10 = 385 \text{ cm}^3$; wood left $= 1200 - 385 = 815 \text{ cm}^3$ — 1 mark
3. Mass $= 815 \times 0.8 = 652$ g — 1 mark

Question 31 (3 marks)

A die is thrown twice. Find the probability that (i) the sum of the two numbers obtained is a prime number, (ii) the product of the two numbers is odd, (iii) the two numbers differ by 2.

Answer.

1. There are 36 equally likely outcomes. (i) The prime sums 2, 3, 5, 7, 11 occur in $1 + 2 + 4 + 6 + 2 = 15$ outcomes; $P = \frac{15}{36} = \frac{5}{12}$ — 1 mark
2. (ii) The product is odd only when both numbers are odd: $3 \times 3 = 9$ outcomes; $P = \frac{9}{36} = \frac{1}{4}$ — 1 mark
3. (iii) Difference 2: (1, 3), (3, 1), (2, 4), (4, 2), (3, 5), (5, 3), (4, 6), (6, 4), i.e. 8 outcomes; $P = \frac{8}{36} = \frac{2}{9}$ — 1 mark

Section D (20 marks)

Question 32 (5 marks)

A swimming pool charges a fixed monthly fee and, in addition, a fixed amount for every visit. In June, Ritu visited the pool 12 times and paid ₹ 1300 in all, while Kabir visited it 20 times and paid ₹ 1700 in all. Find the fixed monthly fee and the charge per visit. Meera has ₹ 2000 to spend at the pool in July. At most how many times can she visit the pool that month?

Answer.

1. Let the fixed fee be ₹ $x$ and the charge per visit ₹ $y$: $x + 12y = 1300$ and $x + 20y = 1700$ — 2 marks
2. Subtracting, $8y = 400$, so $y = 50$ — 1 mark
3. $x = 1300 - 12 \times 50 = 700$ — 1 mark
4. Meera: $700 + 50n = 2000$ gives $n = 26$ visits — 1 mark

OR

Sunita invested a part of her savings in a scheme paying 6% simple interest per annum and the rest in a scheme paying 8% simple interest per annum. Her total interest for one year was ₹ 3600. Had she invested the two parts the other way round, her interest for the year would have been ₹ 200 less. How much did she invest in each scheme? What would her interest for one year have been if she had invested all her savings at 7% per annum?

Answer.

1. Let ₹ $x$ be at 6% and ₹ $y$ at 8%: $6x + 8y = 360000$ and $8x + 6y = 340000$, i.e. $3x + 4y = 180000$ and $4x + 3y = 170000$ — 2 marks
2. Adding, $7(x + y) = 350000$, so $x + y = 50000$; subtracting, $y - x = 10000$ — 1 mark
3. $y = 30000$ and $x = 20000$: ₹ 20000 at 6% and ₹ 30000 at 8% — 1 mark
4. At 7% on ₹ 50000 the interest is $\frac{7 \times 50000}{100} =$ ₹ 3500 — 1 mark

Question 33 (5 marks)

Two taps together can fill an overhead water tank in $3\frac{3}{4}$ hours. The larger tap alone takes 4 hours less than the smaller tap alone to fill the tank. Find the time each tap takes to fill the tank alone. One morning, with the tank empty, both taps were opened together for 2 hours and then the larger tap was closed. How much more time did the smaller tap take to fill the tank?

Answer.

1. Let the larger tap take $x$ hours; the smaller takes $(x + 4)$ hours. In 1 hour: $\frac{1}{x} + \frac{1}{x + 4} = \frac{4}{15}$ — 1 mark
2. $15(2x + 4) = 4x(x + 4)$, i.e. $2x^2 - 7x - 30 = 0$ — 1 mark
3. $(x - 6)(2x + 5) = 0$, so $x = 6$ ($x = -\frac{5}{2}$ rejected); larger tap 6 hours, smaller tap 10 hours — 1 mark
4. In 2 hours both taps fill $2\left(\frac{1}{6} + \frac{1}{10}\right) = \frac{8}{15}$ of the tank, so $\frac{7}{15}$ is left — 1 mark
5. The smaller tap needs $\frac{7}{15} \times 10 = \frac{14}{3}$ hours $= 4$ hours 40 minutes — 1 mark

Question 34 (5 marks)

In the figure, $AD$ and $BE$ are altitudes of the acute-angled triangle $ABC$, with $D$ on $BC$ and $E$ on $AC$.

(i) Prove that $\triangle ADC \sim \triangle BEC$.

(ii) Hence show that $CA \times CE = CB \times CD$.

(iii) If $AC = 10$ cm, $BC = 15$ cm and $CD = 6$ cm, find $CE$ and $BE$.

Answer.

1. In $\triangle ADC$ and $\triangle BEC$: $\angle ADC = \angle BEC = 90^\circ$ and $\angle ACD = \angle BCE$ (common angle) — 1 mark
2. So $\triangle ADC \sim \triangle BEC$ (AA similarity), with $A \leftrightarrow B$, $D \leftrightarrow E$, $C \leftrightarrow C$ — 1 mark
3. Hence $\frac{AD}{BE} = \frac{DC}{EC} = \frac{AC}{BC}$; from $\frac{DC}{EC} = \frac{AC}{BC}$, $CA \times CE = CB \times CD$ — 1 mark
4. $10 \times CE = 15 \times 6$, so $CE = 9$ cm — 1 mark
5. In right $\triangle ADC$, $AD = \sqrt{10^2 - 6^2} = 8$ cm; $\frac{BE}{AD} = \frac{BC}{AC}$ gives $BE = \frac{8 \times 15}{10} = 12$ cm — 1 mark

Question 35 (5 marks)

The daily water consumption of 50 households in a housing society is given below.

Find the mean daily water consumption using the step-deviation method. Also find the median daily water consumption.

Answer.

1. Class marks 125, 175, 225, 275, 325, 375; take $a = 225$ and $h = 50$, so $u_i = -2, -1, 0, 1, 2, 3$ — 1 mark
2. $\sum f_i u_i = -12 - 7 + 0 + 10 + 14 + 15 = 20$ — 1 mark
3. Mean $= a + h \times \frac{\sum f_i u_i}{\sum f_i} = 225 + 50 \times \frac{20}{50} = 245$ litres — 1 mark
4. Cumulative frequencies 6, 13, 28, 38, 45, 50; $\frac{n}{2} = 25$, so the median class is 200-250 with $l = 200$, $cf = 13$, $f = 15$, $h = 50$ — 1 mark
5. Median $= 200 + \frac{25 - 13}{15} \times 50 = 200 + 40 = 240$ litres — 1 mark

OR

The table shows the mobile data (in GB) used in a month by 60 families of a colony.

Find the median and the mode of the data.

Answer.

1. Cumulative frequencies 4, 9, 21, 39, 53, 60; $\frac{n}{2} = 30$, so the median class is 40-50 — 1 mark
2. $l = 40$, $cf = 21$, $f = 18$, $h = 10$: Median $= 40 + \frac{30 - 21}{18} \times 10 = 45$ GB — 1.5 marks
3. Modal class 40-50; $l = 40$, $f_1 = 18$, $f_0 = 12$, $f_2 = 14$, $h = 10$ — 1 mark
4. Mode $= 40 + \frac{18 - 12}{36 - 12 - 14} \times 10 = 40 + 6 = 46$ GB — 1.5 marks

Section E (12 marks)

Question 36 (4 marks)

Kavya is training for a half marathon. In the first week of her training plan she runs a total of 12 km, and in every week after that she runs 1.5 km more than in the week before.

(i) How many kilometres does she run in the 8th week? (1 mark)

Answer.

1. $a_8 = 12 + 7 \times 1.5 = 22.5$ km — 1 mark

(ii) In which week does she run 30 km for the first time? (1 mark)

Answer.

1. $12 + (n - 1) \times 1.5 = 30$ gives $n - 1 = 12$, $n = 13$; the 13th week — 1 mark

(iii) Find the total distance she runs in the first 12 weeks. (2 marks)

Answer.

1. $S_{12} = \frac{12}{2}[2 \times 12 + 11 \times 1.5]$ — 1 mark
2. $= 6 \times 40.5 = 243$ km — 1 mark

OR

(iii) Find the total distance she runs from the 5th week to the 12th week, both weeks included. (2 marks)

Answer.

1. $a_5 = 12 + 4 \times 1.5 = 18$ km and $a_{12} = 12 + 11 \times 1.5 = 28.5$ km — 1 mark
2. These are 8 weeks: $\frac{8}{2}(18 + 28.5) = 4 \times 46.5 = 186$ km — 1 mark

Question 37 (4 marks)

At a winter fair, a hot-air balloon is rising vertically from a point $Q$ on level ground. Anil is standing at the edge $A$ of the flat roof of a building 50 m high, on the side facing the balloon. At a certain instant, the angle of depression of $Q$ from $A$ is $45^\circ$ and the angle of elevation of the balloon $B$ from $A$ is $60^\circ$. (Take $A$ as the point of observation, and use $\sqrt{2} = 1.41$ and $\sqrt{3} = 1.73$)

(i) How far is $Q$ from the foot of the building? (1 mark)

Answer.

1. $\tan 45^\circ = \frac{50}{d}$, so $d = 50$ m — 1 mark

(ii) Find the distance $AQ$. (1 mark)

Answer.

1. $\cos 45^\circ = \frac{50}{AQ}$, so $AQ = 50\sqrt{2} = 70.5$ m — 1 mark

(iii) Find the height of the balloon above the ground at that instant. (2 marks)

Answer.

1. Let $M$ be the point on $QB$ at the level of the roof; $AM = 50$ m and $\tan 60^\circ = \frac{BM}{50}$, so $BM = 50\sqrt{3} = 86.5$ m — 1 mark
2. Height $= QM + MB = 50 + 86.5 = 136.5$ m — 1 mark

OR

(iii) Find the distance $AB$ between Anil and the balloon at that instant. (2 marks)

Answer.

1. Let $M$ be the point on $QB$ at the level of the roof; $AM = 50$ m and $\angle BAM = 60^\circ$ — 1 mark
2. $\cos 60^\circ = \frac{AM}{AB}$, so $AB = \frac{50}{1/2} = 100$ m — 1 mark

Question 38 (4 marks)

The map of a town is drawn on a coordinate plane in which 1 unit = 1 km (see the figure). A delivery company's warehouse is at $W(2, 1)$, one of its regular customers lives at $C(8, 9)$, and its second store is at $S(12, 7)$. The x-axis runs along the town's ring road.

(i) Find the distance between the warehouse and the customer's house. (1 mark)

Answer.

1. $WC = \sqrt{(8 - 2)^2 + (9 - 1)^2} = \sqrt{36 + 64} = 10$ km — 1 mark

(ii) A delivery rider stops at the point $P$ on the straight path from $W$ to $C$ for which $WP : PC = 3 : 2$. Find the coordinates of $P$. (1 mark)

Answer.

1. $P = \left(\frac{3 \times 8 + 2 \times 2}{5}, \frac{3 \times 9 + 2 \times 1}{5}\right) = (5.6, 5.8)$ — 1 mark

(iii) The company wants to open a depot $D$ such that $W$, $C$, $S$ and $D$, taken in order, are the vertices of a parallelogram. Find the coordinates of $D$. (2 marks)

Answer.

1. The diagonals $WS$ and $CD$ bisect each other; the mid-point of $WS$ is $\left(\frac{2 + 12}{2}, \frac{1 + 7}{2}\right) = (7, 4)$ — 1 mark
2. If $D = (x, y)$, then $\frac{8 + x}{2} = 7$ and $\frac{9 + y}{2} = 4$, so $D = (6, -1)$ — 1 mark

OR

(iii) A fuel station $F$ is to be built on the ring road so that it is equidistant from $C$ and $S$. Find the coordinates of $F$. (2 marks)

Answer.

1. Let $F = (x, 0)$; $FC = FS$ gives $(x - 8)^2 + 81 = (x - 12)^2 + 49$ — 1 mark
2. $-16x + 145 = -24x + 193$, so $8x = 48$ and $x = 6$; $F = (6, 0)$ — 1 mark