CBSE Class 10 Maths Standard 2023 Question Paper (Set 30/5/1) with Solutions

1. The polynomial $k(x - 3)(x - 5)$ has the zeroes $3$ and $5$ for every non-zero constant $k$.
2. There are infinitely many choices of $k$, so there are infinitely many such polynomials.

Answer: (B) infinite

1. Parallel lines need $\dfrac{a}{3} = \dfrac{2}{b} \neq \dfrac{9}{18}$.
2. $\dfrac{a}{3} = \dfrac{2}{b}$ gives $ab = 6$. (The condition $\dfrac{2}{b} \neq \dfrac{1}{2}$ just rules out $b = 4$.)

Answer: (D) ab = 6

1. $a_{n+1} - a_n = (3(n + 1) + 7) - (3n + 7) = 3$.

Answer: (B) 3

1. $DE \parallel BC$, so $\triangle ADE \sim \triangle ABC$ and $\dfrac{DE}{BC} = \dfrac{AD}{AB}$.
2. $AB = 2 + 3 = 5$ cm. $\dfrac{4}{x} = \dfrac{2}{5}$, so $x = 10$.

Answer: (D) 10

1. Sum of roots $= (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$. Product $= 4 - 3 = 1$.
2. The equation is $x^2 - (\text{sum})x + \text{product} = x^2 - 4x + 1 = 0$.

Answer: (A) $x^2 - 4x + 1 = 0$

1. Divide the numerator and the denominator by $\cos\theta$: $\dfrac{\tan\theta + 1}{\tan\theta - 1}$.
2. $= \dfrac{\frac{5}{12} + 1}{\frac{5}{12} - 1} = \dfrac{17/12}{-7/12} = -\dfrac{17}{7}$.

Answer: (A) $-\dfrac{17}{7}$

1. The $y$-coordinates are equal, so the distance is the difference of the $x$-coordinates: $\left|-\dfrac{2}{3} + \dfrac{11}{3}\right| = \dfrac{9}{3} = 3$ units.

Answer: (D) 3 units

1. Tangents from a point to a circle are equal: $AP = AR$ and $CQ = CR$. Also $AB = BC$ makes the figure symmetric, so $AR = CR = \dfrac{7}{2} = 3.5$ cm.
2. $AP = 3.5$ cm, so $BP = AB - AP = 10 - 3.5 = 6.5$ cm.

Answer: (C) 6.5 cm

1. Speed $= 2$ km/h $= 2000$ m per $60$ min. In $2$ minutes the water moves $\dfrac{2000}{60} \times 2 = \dfrac{200}{3}$ m.
2. Volume $= 3 \times 40 \times \dfrac{200}{3} = 8000$ m³.

Answer: (C) $8000$ m³

1. $\text{Mode} = 3\,\text{Median} - 2\,\text{Mean} = 3 \times 15 - 2 \times 12 = 45 - 24 = 21$.

Answer: (B) 21

1. $OB \perp AB$ (radius and tangent), so $\triangle OBA$ is right-angled at $B$.
2. $\sin 30^\circ = \dfrac{OB}{OA}$, so $OB = 6 \times \dfrac{1}{2} = 3$ cm.

Answer: (A) 3 cm

1. $\tan 30^\circ = \dfrac{1}{\sqrt{3}}$, so $\dfrac{2/\sqrt{3}}{1 + 1/3} = \dfrac{2/\sqrt{3}}{4/3} = \dfrac{\sqrt{3}}{2}$.
2. $\dfrac{\sqrt{3}}{2} = \sin 60^\circ$. (This is $\sin 2A = \dfrac{2\tan A}{1 + \tan^2 A}$ with $A = 30^\circ$.)

Answer: (A) $\sin 60^\circ$

1. The ratio $\dfrac{AB}{DE} = \dfrac{BC}{FD}$ pairs $AB$ with $DE$ and $BC$ with $FD$. The angle between $AB$ and $BC$ is $\angle B$. The angle between $DE$ and $FD$ is $\angle D$.
2. For SAS similarity the included angles must be equal: $\angle B = \angle D$.

Answer: (B) $\angle B = \angle D$

1. Reverse the A.P.: first term $l = -62$ and common difference $d = +3$. The $11$th term from the end is $-62 + 10 \times 3 = -32$.
2. (Check: the A.P. has $n$ terms with $10 - 3(n - 1) = -62$, so $n = 25$, and the $15$th term is $10 - 42 = -32$.)

Answer: (C) −32

1. Outcomes $HH, HT, TH, TT$. At least one tail: $HT, TH, TT$. $P = \dfrac{3}{4}$.

Answer: (C) $\dfrac{3}{4}$

1. $A$, $O$, $D$ are in a straight line, so $\angle AOC = 180^\circ - 120^\circ = 60^\circ$.
2. $OC \perp AC$ (radius and tangent). In $\triangle ACO$: $\angle OAC = 90^\circ - 60^\circ = 30^\circ$. $AO$ bisects the angle between the tangents, so $\angle BAO = \angle OAC = 30^\circ$.

Answer: (A) $30^\circ$

1. A probability lies between $0$ and $1$ (both included).
2. $\dfrac{7}{0.01} = 700 > 1$, so it cannot be a probability. The others, $0$, $0.07$ and $\dfrac{0.07}{3} \approx 0.023$, lie in $[0, 1]$.

Answer: (B) $\dfrac{7}{0.01}$

1. If every one of $n$ values drops by $2$, the sum drops by $2n$ and the mean $\dfrac{\text{sum}}{n}$ drops by $2$.

Answer: (A) decreases by 2

1. (A) $OA = \sqrt{(4 - 2)^2 + 0} = 2$, so the radius is $2$. $OB = 2$: $(x - 2)^2 + (5 - 3)^2 = 4$, so $(x - 2)^2 = 0$ and $x = 2$. True.
2. (R) The centre is the mid-point only of a diameter, not of every chord. False.

Answer: (C) A is true, R is false

1. (A) A number ends in $0$ only if it is divisible by both $2$ and $5$. $5^n$ has no factor $2$, so it never ends in $0$. True.
2. (R) The prime factorisation of $5^n$ is $5 \times 5 \times \dots \times 5$ ($n$ factors). It is not "only two factors, $1$ and $5$" (that is true only for $n = 1$). So R is false.

Answer: (C) A is true, R is false

1. (a) $AP : AB = 2 : 5$ means $AP : PB = 2 : 3$. Section formula: $x = \dfrac{2 \times 4 + 3 \times 4}{5} = 4$ and $y = \dfrac{2 \times 5 + 3 \times (-5)}{5} = \dfrac{10 - 15}{5} = -1$. $P = (4, -1)$.
2. (b) $PA = PB$, so $PA^2 = PB^2$: $(x - 5)^2 + (y - 1)^2 = (x - 1)^2 + (y - 5)^2$.
3. $-10x - 2y + 26 = -2x - 10y + 26$, so $8y = 8x$ and $x = y$.

Answer: (a) $P(4, -1)$   (b) proved

1. $OC \perp AB$, and the perpendicular from the centre bisects the chord, so $AC = BC$.
2. $PA = PC - AC$ and $PB = PC + BC = PC + AC$ (as $BC = AC$).
3. $PA \cdot PB = (PC - AC)(PC + AC) = PC^2 - AC^2$.

Answer: Proved

1. $96 = 2^5 \times 3$ and $120 = 2^3 \times 3 \times 5$.
2. HCF $=$ smallest powers $= 2^3 \times 3 = 24$. LCM $=$ greatest powers $= 2^5 \times 3 \times 5 = 480$.
3. (Check: $24 \times 480 = 11520 = 96 \times 120$ ✓.)

Answer: HCF = 24 and LCM = 480

1. Let the $y$-axis ($x = 0$) divide the segment in the ratio $k : 1$: $\dfrac{k(-1) + 1(5)}{k + 1} = 0$.
2. $-k + 5 = 0$, so $k = 5$. The ratio is $5 : 1$.

Answer: $5 : 1$

1. (a) $m^2 + n^2 = (a\cos\theta + b\sin\theta)^2 + (a\sin\theta - b\cos\theta)^2 = a^2(\cos^2\theta + \sin^2\theta) + b^2(\sin^2\theta + \cos^2\theta) + 2ab\cos\theta\sin\theta - 2ab\sin\theta\cos\theta$.
2. The cross terms cancel and each bracket is $1$: $m^2 + n^2 = a^2 + b^2$.
3. (b) Put over a common root: $\dfrac{(\sec A - 1) + (\sec A + 1)}{\sqrt{(\sec A + 1)(\sec A - 1)}} = \dfrac{2\sec A}{\sqrt{\sec^2 A - 1}} = \dfrac{2\sec A}{\tan A}$.
4. $= \dfrac{2}{\cos A} \times \dfrac{\cos A}{\sin A} = \dfrac{2}{\sin A} = 2\,\mathrm{cosec}\,A$.

Answer: Both proved

1. (a) Suppose $\sqrt{3} = \dfrac{p}{q}$ with $p, q$ co-prime and $q \neq 0$. Then $3q^2 = p^2$, so $3$ divides $p$. Write $p = 3a$.
2. $9a^2 = 3q^2$, so $q^2 = 3a^2$ and $3$ divides $q$. Now $3$ divides both $p$ and $q$, contradicting that they are co-prime. So $\sqrt{3}$ is irrational.
3. (b) They change together again after the LCM of $48, 72, 108$. $48 = 2^4 \times 3$, $72 = 2^3 \times 3^2$, $108 = 2^2 \times 3^3$. LCM $= 2^4 \times 3^3 = 432$ seconds.
4. $432$ seconds $= 7$ minutes $12$ seconds. They next change together at $7{:}07{:}12$ a.m.

Answer: (a) proved   (b) $7{:}07{:}12$ a.m.

1. $a_p = a + (p - 1)d = q$. … (i) $a_q = a + (q - 1)d = p$. … (ii)
2. Subtract (ii) from (i): $(p - q)d = q - p$, so $d = -1$ (as $p \neq q$).
3. Put in (i): $a - (p - 1) = q$, so $a = p + q - 1$.
4. $a_n = a + (n - 1)d = (p + q - 1) - (n - 1) = p + q - n$.

Answer: Proved: aₙ = p + q − n

1. (a) $EF \perp CD$ and $AB \perp CD$, so $EF \parallel AD$. In $\triangle EFG$ and $\triangle ADG$: $\triangle EFG \sim \triangle ADG$ (AA), so $\dfrac{EF}{AD} = \dfrac{FG}{DG}$. … (i)
2. $EF \parallel BD$ too, so $\triangle CEF \sim \triangle CBD$ and $\dfrac{EF}{BD} = \dfrac{CF}{CD}$. Since $BD = AD$ ($D$ is the mid-point of $AB$): $\dfrac{EF}{AD} = \dfrac{CF}{CD}$. … (ii)
3. From (i) and (ii): $\dfrac{CF}{CD} = \dfrac{FG}{DG}$.
4. (b) $AE \parallel BC$, so $\triangle ALE \sim \triangle CLB$ and $\dfrac{EL}{BL} = \dfrac{AL}{CL}$. … (i)
5. $AB \parallel CM$ (as $AB \parallel CD$), so $\triangle CLM \sim \triangle ALB$ and $\dfrac{AL}{CL} = \dfrac{AB}{CM} = \dfrac{CD}{CM} = 2$ (since $CD = 2CM$). … (ii)
6. From (i) and (ii): $\dfrac{EL}{BL} = 2$, so $EL = 2BL$.

Answer: (a) $\dfrac{CF}{CD} = \dfrac{FG}{DG}$   (b) $EL = 2BL$

1. Let the speeds be $x$ km/h and $y$ km/h with $x > y$.
2. Towards each other: they cover the $16$ km together in $2$ h: $2x + 2y = 16$, so $x + y = 8$. … (i)
3. In the same direction: the faster one gains $16$ km in $8$ h: $8x - 8y = 16$, so $x - y = 2$. … (ii)
4. Add: $2x = 10$, so $x = 5$, and $y = 3$.

Answer: $5$ km/h and $3$ km/h

1. $\dfrac{\tan\theta}{1 - \cot\theta} = \dfrac{\sin^2\theta}{\cos\theta(\sin\theta - \cos\theta)}$ and $\dfrac{\cot\theta}{1 - \tan\theta} = \dfrac{\cos^2\theta}{\sin\theta(\cos\theta - \sin\theta)} = -\dfrac{\cos^2\theta}{\sin\theta(\sin\theta - \cos\theta)}$.
2. Sum $= \dfrac{\sin^3\theta - \cos^3\theta}{\sin\theta\cos\theta(\sin\theta - \cos\theta)} = \dfrac{\sin^2\theta + \cos^2\theta + \sin\theta\cos\theta}{\sin\theta\cos\theta}$.
3. $= \dfrac{1}{\sin\theta\cos\theta} + 1 = 1 + \sec\theta\,\mathrm{cosec}\,\theta$ = RHS.

Answer: Proved

1. Class marks $x_i$: $27.5, 32.5, 37.5, 42.5, 47.5, 52.5, 57.5$. Take $a = 42.5$ and $h = 5$: $u_i = \dfrac{x_i - 42.5}{5} = -3, -2, -1, 0, 1, 2, 3$.
2. $f_i u_i$: $-42, -44, -16, 0, 5, 6, 12$. $\sum f_i = 70$ and $\sum f_i u_i = -79$.
3. Mean $= a + \dfrac{\sum f_i u_i}{\sum f_i} \times h = 42.5 - \dfrac{79}{70} \times 5 = 42.5 - 5.64 = 36.86$.

Answer: Mean $\approx 36.86$

1. Take both observers on the same side of the basket. Let the first (nearer) observer be at horizontal distance $x$ from the point below the basket, and the height be $h$.
2. $\tan 60^\circ = \dfrac{h}{x}$, so $h = \sqrt{3}\,x$. … (i) $\tan 30^\circ = \dfrac{h}{x + 100}$, so $x + 100 = \sqrt{3}\,h$. … (ii)
3. Put (i) in (ii): $x + 100 = \sqrt{3} \times \sqrt{3}\,x = 3x$, so $x = 50$ m and $h = 50\sqrt{3}$ m.
4. (a) Height of the basket $= 50\sqrt{3}$ m $\approx 86.6$ m.
5. (b) $\sin 60^\circ = \dfrac{h}{y}$, so $y = \dfrac{50\sqrt{3}}{\sqrt{3}/2} = 100$ m.
6. (c) Horizontal distance of the second observer $= x + 100 = 150$ m.

Answer: (a) $50\sqrt{3}$ m   (b) $100$ m   (c) $150$ m

Common mistake: CBSE also accepts the case where the observers are on opposite sides of the basket ($h = 25\sqrt{3}$ m, $75$ m from the second observer). The nearer-observer figure above is the standard one.

1. (a) Let the circle touch $AB$ at $F$ and $AC$ at $E$. Tangents from a point are equal: $BF = BD = 10$, $CE = CD = 8$, and $AF = AE = x$.
2. The sides are $BC = 18$, $AB = 10 + x$, $AC = 8 + x$. The area is the sum of the areas of $\triangle OBC$, $\triangle OCA$, $\triangle OAB$ (each with height $4$): $90 = \dfrac{1}{2} \times 4 \times (BC + CA + AB) = 2(18 + 8 + x + 10 + x)$.
3. $45 = 36 + 2x$, so $x = 4.5$. $AB = 14.5$ cm and $AC = 12.5$ cm.
4. (b) $OP \perp O'P$ (tangents), so in the right triangle $OPO'$: $OO' = \sqrt{6^2 + 8^2} = 10$ cm.
5. The common chord $PQ$ is perpendicular to $OO'$ at $A$ and is bisected there. Let $OA = x$, so $O'A = 10 - x$. $AP^2 = 36 - x^2$ and $AP^2 = 64 - (10 - x)^2$.
6. $36 - x^2 = 64 - 100 + 20x - x^2$, so $20x = 72$ and $x = 3.6$. $AP^2 = 36 - 12.96 = 23.04$, so $AP = 4.8$ cm and $PQ = 2 \times 4.8 = 9.6$ cm.

Answer: (a) $AB = 14.5$ cm, $AC = 12.5$ cm   (b) $PQ = 9.6$ cm

1. (a) Let the first speed be $x$ km/h. $\dfrac{54}{x} + \dfrac{63}{x + 6} = 3$.
2. $54(x + 6) + 63x = 3x(x + 6)$, so $117x + 324 = 3x^2 + 18x$ and $x^2 - 33x - 108 = 0$.
3. $(x - 36)(x + 3) = 0$. Speed is positive, so $x = 36$ km/h.
4. (b) Let the smaller pipe take $x$ hours and the larger $(x - 2)$ hours. $\dfrac{1}{x - 2} + \dfrac{1}{x} = \dfrac{8}{15}$.
5. $15(2x - 2) = 8x(x - 2)$, so $8x^2 - 46x + 30 = 0$, that is $4x^2 - 23x + 15 = 0$ and $(4x - 3)(x - 5) = 0$.
6. $x = \dfrac{3}{4}$ is rejected because $x - 2$ would be negative. So $x = 5$: the smaller pipe takes $5$ hours and the larger takes $3$ hours.

Answer: (a) $36$ km/h   (b) $5$ hours and $3$ hours

1. At a corner the horse can graze a quarter circle, since the field covers a $90^\circ$ angle there.
2. With the $5$ m rope: $\dfrac{1}{4} \times 3.14 \times 5^2 = 19.625$ m².
3. With the $10$ m rope (still less than the side $15$ m): $\dfrac{1}{4} \times 3.14 \times 10^2 = 78.5$ m².
4. Increase $= 78.5 - 19.625 = 58.875$ m².

Answer: $19.625$ m²; the increase is $58.875$ m²

1. All lengths in mm: ball radius $R = 21$ mm and dimple radius $r = 2$ mm.
2. (i) Curved surface of one dimple $= 2\pi r^2 = 2 \times \dfrac{22}{7} \times 4 = \dfrac{176}{7} \approx 25.1$ mm².
3. (ii) Volume dug out $= \dfrac{2}{3}\pi r^3 = \dfrac{2}{3} \times \dfrac{22}{7} \times 8 = \dfrac{352}{21} \approx 16.76$ mm³.
4. (iii)(a) Each dimple takes away a flat circle $\pi r^2$ of the sphere and adds a curved surface $2\pi r^2$. Exposed area $= 4\pi R^2 - 315\pi r^2 + 315 \times 2\pi r^2 = 4\pi R^2 + 315\pi r^2 = 5544 + 3960 = 9504$ mm².
5. (iii)(b) Volume of the ball $= \dfrac{4}{3}\pi R^3 - 315 \times \dfrac{2}{3}\pi r^3 = 38808 - 5280 = 33528$ mm³.