CBSE Class 10 Mathematics Basic: 2023 board paper, set 430/6/1

1. Athletes taking less than 17 s are those in classes 13-14, 14-15, 15-16 and 16-17.

2. Number $=2+4+5+71=82$, so option (C).

   1 mark

Answer: 82, option (C)

1. Distance from the origin $=\sqrt{5^2+0^2}=5$, so option (B).

   1 mark

Answer: 5, option (B)

1. $\tan A=\frac{BC}{AC}=\frac{8}{7}$.

2. $\cot B=\frac{\text{side adjacent to }B}{\text{side opposite to }B}=\frac{BC}{AC}=\frac{8}{7}$ (equivalently $\cot B=\tan (90^{\circ }-B)=\tan A$), so option (B).

   1 mark

Answer: $\frac{8}{7}$, option (B)

1. Area of quadrant $=\frac{1}{4}\pi r^2=\frac{1}{4}\times \frac{22}{7}\times 49=\frac{77}{2}$ cm${}^2$, so option (C).

   1 mark

Answer: $\frac{77}{2}$ cm${}^2$, option (C)

1. HCF × LCM = product of the numbers, so LCM $=\frac{72\times 120}{24}=360$, option (C).

   1 mark

Answer: 360, option (C)

1. There are 2 black kings (spades and clubs), so P $=\frac{2}{52}=\frac{1}{26}$, option (A).

   1 mark

Answer: $\frac{1}{26}$, option (A)

1. The zeroes of $f(x)$ are the x-coordinates of the points where the graph meets the x-axis.

2. The whole curve lies above the x-axis and never meets it, so there are 0 zeroes, option (A).

   1 mark

Answer: 0, option (A)

1. Substitute (6, k): $6-3k+6=0\Rightarrow 3k=12\Rightarrow k=4$, option (D).

   1 mark

Answer: k = 4, option (D)

1. $2304=2\times 1152=\cdots =2^8\times 9$, since $2304÷256=9$.

2. So $2304=2^8\times 3^2$, option (A).

   1 mark

Answer: $2^8\times 3^2$, option (A)

1. To end in 0, a number must have both 2 and 5 as prime factors.

2. $8^n=2^{3n}$ has no factor 5, so $8^n$ cannot end with 0, option (A).

   1 mark

Answer: 0, option (A)

1. The first seven primes are 2, 3, 5, 7, 11, 13, 17.

2. The median is the 4th term, which is 7, option (B).

   1 mark

Answer: 7, option (B)

1. By the mid-point formula, $\frac{6+a}{2}=2\Rightarrow 6+a=4\Rightarrow a=-2$, option (D).

   1 mark

Answer: a = −2, option (D)

1. For no solution, $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$, i.e. $\frac{k}{3}=\frac{2}{4}\ne \frac{5}{1}$.

2. $\frac{k}{3}=\frac{1}{2}\Rightarrow k=\frac{3}{2}$ (and $\frac{1}{2}\ne 5$ holds), so option (A).

   1 mark

Answer: $k=\frac{3}{2}$, option (A)

1. $\angle OQP=\angle ORP=90^{\circ }$ [tangent ⟂ radius].

2. In quadrilateral OQPR, $\angle QOR=360^{\circ }-90^{\circ }-90^{\circ }-65^{\circ }=115^{\circ }$, option (C).

   1 mark

Answer: 115°, option (C)

1. $16x^2-9=(4x-3)(4x+3)=0\Rightarrow x=\pm \frac{3}{4}$, option (B).

   1 mark

Answer: $-\frac{3}{4},\frac{3}{4}$, option (B)

1. For three consecutive terms of an A.P., $2x=-5+3=-2\Rightarrow x=-1$, option (D).

   1 mark

Answer: x = −1, option (D)

1. The odd primes on a die are 3 and 5, so there are 2 favourable outcomes out of 6.

2. P $=\frac{2}{6}=\frac{1}{3}$, option (D).

   1 mark

Answer: $\frac{1}{3}$, option (D)

1. Sum of the six numbers $=6\times 9=54$.

2. $6+7+8+14+x+y=54\Rightarrow 35+x+y=54\Rightarrow x+y=19$, option (B).

   1 mark

Answer: x + y = 19, option (B)

1. A leap year has 52 weeks + 2 extra days. These 2 days can be any of 7 equally likely pairs, and 2 of them (Sat-Sun, Sun-Mon) contain a Sunday, so P $=\frac{2}{7}$. A is true.

2. A non-leap year has 52 weeks + 1 extra day, so P(53 Sundays) $=\frac{1}{7}$. R is true, but it is about non-leap years and does not explain A, so option (B).

   1 mark

Answer: Option (B)

1. $(\mathrm{cosec}\theta -\cot \theta )(\mathrm{cosec}\theta +\cot \theta )={\mathrm{cosec}}^2\theta -{\cot }^2\theta =1$, so the two are reciprocals of each other. A is true.

2. The correct identity is ${\mathrm{cosec}}^2\theta -{\cot }^2\theta =1$, so ${\cot }^2\theta -{\mathrm{cosec}}^2\theta =-1$ and R is false. Option (C).

   1 mark

Answer: Option (C)

1. Use the standard values: $\mathrm{cosec}45^{\circ }=\sqrt{2}$, $\sin 90^{\circ }=1$, $\cos 0^{\circ }=1$. Substituting: $5(\sqrt{2}{)}^2-3(1{)}^2+5(1)$

   1½ marks

2. $=10-3+5=12$

   ½ mark

Answer: 12

1. Sum of zeroes $=6+(-3)=3$

   ½ mark

2. Product of zeroes $=6\times (-3)=-18$

   ½ mark

3. Required polynomial $=x^2-(\text{sum})x+(\text{product})=x^2-3x-18$, or more generally $k(x^2-3x-18)$, $k\ne 0$.

   1 mark

Answer: $x^2-3x-18$ (or $k(x^2-3x-18)$)

1. Split the middle term: $x^2+4x-12=x^2+6x-2x-12=x(x+6)-2(x+6)=(x+6)(x-2)$

   1 mark

2. Putting each factor equal to zero: $x=-6$ or $x=2$. Zeroes are $-6$ and $2$.

   1 mark

Answer: $-6$ and $2$

1. Comparing with $a{x}^2+bx+c=0$: $a=5$, $b=-10$, $c=k$.

   ½ mark

2. For real and equal roots, discriminant $D=b^2-4ac=0$.

   ½ mark

3. $(-10{)}^2-4(5)(k)=0\Rightarrow 100-20k=0$

   ½ mark

4. $k=5$

   ½ mark

Answer: k = 5

1. Let the roots be $\alpha$ and $7\alpha$.

   ½ mark

2. Sum of roots: $\alpha +7\alpha =-\frac{-8}{3}=\frac{8}{3}\Rightarrow 8\alpha =\frac{8}{3}\Rightarrow \alpha =\frac{1}{3}$

   ½ mark

3. Product of roots: $\alpha \cdot 7\alpha =\frac{-(2k+1)}{3}\Rightarrow 7{\alpha }^2=-\frac{2k+1}{3}$

   ½ mark

4. $7\cdot \frac{1}{9}=-\frac{2k+1}{3}\Rightarrow 2k+1=-\frac{7}{3}\Rightarrow 2k=-\frac{10}{3}\Rightarrow k=-\frac{5}{3}$

   ½ mark

Answer: $k=-\frac{5}{3}$

1. Total outcomes = 20. (i) 2-digit numbers are 10, 11, …, 20, i.e. 11 numbers. $P(\text{2-digit})=\frac{11}{20}$

   1 mark

2. (ii) Numbers less than 10 are 1, 2, …, 9, i.e. 9 numbers. $P(\text{less than }10)=\frac{9}{20}$

   1 mark

Answer: (i) 11/20; (ii) 9/20

1. Figure: circle with centre O, external point P, tangent PQ touching the circle at Q, with $PQ=24$ cm, $OP=25$ cm and OQ joined; $\angle OQP=90^{\circ }$ [tangent ⟂ radius at point of contact].

   ½ mark

2. In right $△OQP$, by Pythagoras theorem: $OQ=\sqrt{O{P}^2-P{Q}^2}=\sqrt{25^2-24^2}=\sqrt{625-576}=\sqrt{49}$

   1 mark

3. $OQ=7$ cm, so the radius is 7 cm.

   ½ mark

Answer: Radius = 7 cm

1. Let Varun's present age be $x$ years. Then his age 3 years ago was $x-3$ and 5 years from now will be $x+5$. According to the question: $\frac{1}{x-3}+\frac{1}{x+5}=\frac{1}{3}$

   1 mark

2. $\frac{(x+5)+(x-3)}{(x-3)(x+5)}=\frac{1}{3}\Rightarrow \frac{2x+2}{x^2+2x-15}=\frac{1}{3}\Rightarrow 6x+6=x^2+2x-15\Rightarrow x^2-4x-21=0$

   1 mark

3. $x^2-7x+3x-21=0\Rightarrow (x-7)(x+3)=0$

   ½ mark

4. $x=7$ or $x=-3$; age cannot be negative, so $x=-3$ is rejected. Varun's present age is 7 years.

   ½ mark

Answer: 7 years

1. Cumulative frequency table: 1–3: f = 7, cf = 7; 3–5: f = 8, cf = 15; 5–7: f = 2, cf = 17; 7–9: f = 2, cf = 19; 9–11: f = 1, cf = 20.

   1 mark

2. $N=20$, $\frac{N}{2}=10$. The cf just greater than 10 is 15, so the median class is 3–5.

   ½ mark

3. Here $l=3$, $cf=7$, $f=8$, $h=2$. Median $=l+\left(\frac{\frac{N}{2}-cf}{f}\right)\times h=3+\frac{10-7}{8}\times 2$

   1 mark

4. $=3+\frac{6}{8}=3+0.75=3.75$

   ½ mark

Answer: Median = 3.75

1. Figure: parallelogram ABCD with AD produced beyond D to E; BE drawn cutting CD at F.

   1 mark

2. In $△ABE$ and $△CFB$: $\angle A=\angle C$ [opposite angles of a parallelogram are equal].

   ½ mark

3. $\angle AEB=\angle CBF$ [alternate interior angles, since $AE\parallel BC$ (AD ∥ BC) and BE is a transversal].

   ½ mark

4. Therefore $△ABE\sim △CFB$ [AA similarity criterion].

   1 mark

Answer: ΔABE ~ ΔCFB by AA similarity

1. Since $△ABC\sim △PQR$, corresponding sides are proportional: $\frac{AB}{PQ}=\frac{AC}{PR}$.

   ½ mark

2. CM and RN are medians, so $AB=2AM$ and $PQ=2PN$. Hence $\frac{2AM}{2PN}=\frac{AC}{PR}\Rightarrow \frac{AM}{PN}=\frac{AC}{PR}$

   1 mark

3. Also $\angle A=\angle P$ [corresponding angles of similar triangles $ABC$ and $PQR$]. So in $△AMC$ and $△PNR$, $\frac{AM}{PN}=\frac{AC}{PR}$ and included $\angle MAC=\angle NPR$; therefore $△AMC\sim △PNR$ [SAS similarity].

   1½ marks

Answer: ΔAMC ~ ΔPNR by SAS similarity

1. Let A(5, 3) and B(4, 5), and let C and D be the points of trisection, so C divides AB in the ratio 1 : 2.

   ½ mark

2. By section formula: $C\left(\frac{1\times 4+2\times 5}{1+2},\frac{1\times 5+2\times 3}{1+2}\right)=C\left(\frac{14}{3},\frac{11}{3}\right)$

   1 mark

3. D divides AB in the ratio 2 : 1.

   ½ mark

4. $D\left(\frac{2\times 4+1\times 5}{2+1},\frac{2\times 5+1\times 3}{2+1}\right)=D\left(\frac{13}{3},\frac{13}{3}\right)$

   1 mark

Answer: $\left(\frac{14}{3},\frac{11}{3}\right)$ and $\left(\frac{13}{3},\frac{13}{3}\right)$

1. Assume, to the contrary, that $3-2\sqrt{5}$ is rational. Then $3-2\sqrt{5}=\frac{p}{q}$, where p, q are integers and $q\ne 0$.

   1 mark

2. Rearranging: $2\sqrt{5}=3-\frac{p}{q}\Rightarrow \sqrt{5}=\frac{3q-p}{2q}$

   1 mark

3. Since p, q are integers, $\frac{3q-p}{2q}$ is rational, so $\sqrt{5}$ would be rational. This contradicts the given fact that $\sqrt{5}$ is irrational. Hence our assumption is wrong, and $3-2\sqrt{5}$ is irrational.

   1 mark

Answer: $3-2\sqrt{5}$ is irrational (proved by contradiction)

1. LHS $=\frac{\cot A-\cos A}{\cot A+\cos A}=\frac{\frac{\cos A}{\sin A}-\cos A}{\frac{\cos A}{\sin A}+\cos A}$

   ½ mark

2. Taking $\frac{\cos A}{\sin A}$ common from numerator and denominator: $=\frac{1-\sin A}{1+\sin A}$

   1 mark

3. Multiplying numerator and denominator by $(1+\sin A)$: $=\frac{(1-\sin A)(1+\sin A)}{(1+\sin A)(1+\sin A)}$

   1 mark

4. $=\frac{1-{\sin }^{2}A}{(1+\sin A{)}^2}=\frac{{\cos }^{2}A}{(1+\sin A{)}^2}=$ RHS [since ${\sin }^{2}A+{\cos }^{2}A=1$]

   ½ mark

Answer: LHS = RHS, hence proved

1. LHS $=(\sec \theta +\tan \theta )(1-\sin \theta )=\left(\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }\right)(1-\sin \theta )$

   1 mark

2. $=\frac{(1+\sin \theta )(1-\sin \theta )}{\cos \theta }=\frac{1-{\sin }^2\theta }{\cos \theta }$

   1 mark

3. $=\frac{{\cos }^2\theta }{\cos \theta }=\cos \theta =$ RHS [since $1-{\sin }^2\theta ={\cos }^2\theta$]

   1 mark

Answer: LHS = RHS, hence proved

1. Figure: P is the point on the bridge, PQ = 3 m is perpendicular to the river AB, with A and B the banks on opposite sides. Angles of depression 30° and 45° give $\angle PAQ=30^{\circ }$ and $\angle PBQ=45^{\circ }$ [alternate angles].

   1 mark

2. In right $△APQ$: $\tan 30^{\circ }=\frac{PQ}{AQ}=\frac{3}{AQ}$

   1 mark

3. $\frac{1}{\sqrt{3}}=\frac{3}{AQ}\Rightarrow AQ=3\sqrt{3}$ m

   ½ mark

4. In right $△PBQ$: $\tan 45^{\circ }=\frac{PQ}{BQ}=\frac{3}{BQ}$

   1 mark

5. $1=\frac{3}{BQ}\Rightarrow BQ=3$ m

   ½ mark

6. Width $AB=AQ+BQ=3\sqrt{3}+3$

   ½ mark

7. $=3(1.73+1)=3\times 2.73=8.19$ m. Width of the river is 8.19 m.

   ½ mark

Answer: Width of river = $3(\sqrt{3}+1)$ m = 8.19 m

1. Figure: AB = 20 m is the building, BC = h m is the tower on top of it, D is the point on the ground with AD = x m; $\angle ADB=45^{\circ }$, $\angle ADC=60^{\circ }$, $\angle A=90^{\circ }$.

   1 mark

2. In right $△ABD$: $\tan 45^{\circ }=\frac{AB}{AD}=\frac{20}{x}$

   1 mark

3. $1=\frac{20}{x}\Rightarrow x=20$ m

   ½ mark

4. In right $△ACD$: $\tan 60^{\circ }=\frac{AC}{AD}=\frac{20+h}{x}$

   1 mark

5. $\sqrt{3}x=20+h\Rightarrow 20\sqrt{3}=20+h$

   ½ mark

6. $h=20(\sqrt{3}-1)$ m

   ½ mark

7. $h=20(1.73-1)=20\times 0.73=14.6$ m. Height of the tower is 14.6 m.

   ½ mark

Answer: Height of tower = $20(\sqrt{3}-1)$ m = 14.6 m

1. Given $a=22$, last term $l=a_n=-6$, $S_n=64$. Using $S_n=\frac{n}{2}(a+l)$: $64=\frac{n}{2}[22+(-6)]=\frac{n}{2}\times 16$

   2 marks

2. $8n=64\Rightarrow n=8$. So there are 8 terms.

   1 mark

3. Using $a_n=a+(n-1)d$: $-6=22+(8-1)d\Rightarrow 7d=-28$

   1 mark

4. $d=-4$

   1 mark

Answer: n = 8, d = −4

1. The 10 cm is the total height of the ice-cream (hemisphere top to cone tip), and the hemisphere has radius $r=5$ cm, so height of conical part $h=10-5=5$ cm.

   ½ mark

2. Volume of ice-cream in one cone = volume of cone + volume of hemisphere $=\frac{1}{3}\pi r^{2}h+\frac{2}{3}\pi r^3=\frac{1}{3}\pi r^2(h+2r)$

   1 mark

3. $=\frac{1}{3}\times \frac{22}{7}\times 5\times 5\times (5+10)$

   1½ marks

4. $=\frac{22\times 25\times 15}{21}$ cm³ $\left(=\frac{2750}{7}\approx 392.86{\text{ cm}}^3\right)$

   1 mark

5. Volume in 7 cones $=7\times \frac{22\times 25\times 15}{21}$

   ½ mark

6. $=\frac{8250}{3}\times \dots$ simplifies to $=\frac{22\times 25\times 15}{3}=2750$ cm³

   ½ mark

Answer: 2750 cm³

1. Figure: Draw $△ABC$ with a line DE parallel to BC, meeting AB at D and AC at E. Join BE and CD. Draw $EN\perp AB$ (N on AB) and $DM\perp AC$ (M on AC).

   1 mark

2. Given: In $△ABC$, $DE\parallel BC$, and DE meets AB at D and AC at E.

   ½ mark

3. To prove: $\frac{AD}{DB}=\frac{AE}{EC}$

   ½ mark

4. Construction: Join BE and CD. Draw $EN\perp AB$ and $DM\perp AC$.

   ½ mark

5. Proof: $△ADE$ and $△BDE$ have the same height EN on bases AD and DB:
   $\frac{ar(△ADE)}{ar(△BDE)}=\frac{\frac{1}{2}\times AD\times EN}{\frac{1}{2}\times DB\times EN}=\frac{AD}{DB}$ ...(i)

   1 mark

6. In the same way, $△ADE$ and $△CDE$ have the same height DM on bases AE and EC:
   $\frac{ar(△ADE)}{ar(△CDE)}=\frac{\frac{1}{2}\times AE\times DM}{\frac{1}{2}\times EC\times DM}=\frac{AE}{EC}$ ...(ii)

   ½ mark

7. $△BDE$ and $△CDE$ lie on the same base DE and between the same parallels DE and BC. So $ar(△BDE)=ar(△CDE)$ ...(iii)

   ½ mark

8. From (i), (ii) and (iii), the left-hand sides of (i) and (ii) are equal. Hence $\frac{AD}{DB}=\frac{AE}{EC}$. Proved.

   ½ mark

Answer: Proved: $\frac{AD}{DB}=\frac{AE}{EC}$ (Basic Proportionality Theorem).

1. In $△PQR$, $\angle 1=\angle 2$, i.e. $\angle PQR=\angle PRQ$.
   So $PQ=PR$ [sides opposite equal angles are equal].

   1 mark

2. Given $\frac{QR}{QS}=\frac{QT}{PR}$. Taking reciprocals gives $\frac{QS}{QR}=\frac{PR}{QT}$.
   Since $PR=PQ$, this becomes $\frac{QS}{QR}=\frac{PQ}{QT}$ ...(i)

   2 marks

3. In $△PQS$ and $△TQR$:

   • $\angle PQS=\angle TQR$ [common angle Q]
   • $\frac{PQ}{TQ}=\frac{QS}{QR}$ [from (i)]
   1½ marks

4. So $△PQS\sim △TQR$ [SAS similarity criterion]. Proved.

   ½ mark

Answer: $△PQS\sim △TQR$ by SAS similarity.

1. (i) Area of square ABCD $={\text{side}}^2=8^2=64{\text{cm}}^2$

   1 mark

2. (ii) By Pythagoras theorem in $△ABC$: $AC=\sqrt{8^2+8^2}=\sqrt{128}=8\sqrt{2}$ cm

   1 mark

3. (iii)(a) The radius of the inscribed circle is half the side: $r=\frac{8}{2}=4$ cm, and $\theta =90^{\circ }$.
   Area of sector $=\frac{\theta }{360^{\circ }}\times \pi r^2=\frac{90}{360}\times \frac{22}{7}\times 4\times 4$

   1 mark

4. $=\frac{88}{7}{\text{cm}}^2$ (about $12.57{\text{cm}}^2$)

   1 mark

5. (iii)(b) Area of circle $=\pi r^2=\frac{22}{7}\times 4\times 4=\frac{352}{7}{\text{cm}}^2$

   1 mark

6. Remaining area $=64-\frac{352}{7}=\frac{448-352}{7}=\frac{96}{7}{\text{cm}}^2$ (about $13.71{\text{cm}}^2$)

   1 mark

Answer: (i) 64 cm²; (ii) 8√2 cm; (iii)(a) 88/7 cm²; (iii)(b) 96/7 cm²

1. (i) Let the fixed charge be ₹ x and the charge per km be ₹ y.
   Then $x+10y=105$ ...(1) and $x+15y=155$ ...(2)

   ½ mark

2. Subtract (1) from (2): $5y=50$, so $y=10$.
   Put this in (1): $x=105-100=5$.
   Fixed charge = ₹ 5

   ½ mark

3. (ii) From the working above, $y=10$. Charge per km = ₹ 10

   1 mark

4. (iii)(a) Amount = fixed charge + 10 × charge per km $=20+10(10)$

   1 mark

5. $=₹120$

   1 mark

6. (iii)(b) Total amount $=(x+10y)+(x+25y)=2x+35y$

   1 mark

7. $=2(5)+35(10)=10+350=₹360$

   1 mark

Answer: (i) ₹ 5; (ii) ₹ 10 per km; (iii)(a) ₹ 120; (iii)(b) ₹ 360

1. (i) B is the mid-point of AC, so $BC=AB=20$ m.
   $AC=AB+BC=20+20=40$ m

   1 mark

2. (ii) The shortest distance from O to the road is OB, and $OB\perp AB$ [tangent ⟂ radius at the point of contact].
   In right $△OBA$: $OB=\sqrt{O{A}^2-A{B}^2}=\sqrt{25^2-20^2}=\sqrt{625-400}=\sqrt{225}=15$ m

   1 mark

3. (iii)(a) The radius is $r=OB=15$ m.
   Circumference $=2\pi r=2\pi (15)=30\pi$ m

   1 mark

4. $=30\times \frac{22}{7}=\frac{660}{7}$ m $\approx 94.29$ m

   1 mark

5. (iii)(b) Area $=\pi r^2=\pi (15{)}^2=225\pi {\text{m}}^2$

   1 mark

6. $=225\times \frac{22}{7}=\frac{4950}{7}{\text{m}}^2\approx 707.14{\text{m}}^2$

   1 mark

Answer: (i) 40 m; (ii) 15 m; (iii)(a) 30π m = 660/7 m; (iii)(b) 225π m² = 4950/7 m²