CBSE Class 10 Mathematics Standard: 2025 board paper, set 30/6/1

1. $\sqrt{0.4}=\sqrt{\frac{4}{10}}=\frac{2}{\sqrt{10}}$. Since 10 is not a perfect square, $\sqrt{10}$ is irrational, so $\frac{2}{\sqrt{10}}$ is irrational. Option (D).

   1 mark

Answer: Irrational number, option (D)

1. Unit digits of $8^n$ cycle as 8, 4, 2, 6. For unit digit 0, $8^n=2^{3n}$ would need the prime factor 5, which it does not have. So 0 is impossible. Option (C).

   1 mark

Answer: 0, option (C)

1. (A) expands to $x^2+2x+1=2x+1$, i.e. $x^2=0$, so $D=0$: real and equal roots (both 0). (B) and (C) have $D>0$; (D) has $D=1-4<0$. Option (A).

   1 mark

Answer: $(x+1{)}^2=2x+1$, option (A)

1. If the zeroes are reciprocals, their product is 1. Product of zeroes $=\frac{2a/b}{a}=\frac{2}{b}=1$, so $b=2$. Option (A).

   1 mark

Answer: b = 2, option (A)

1. The distance of a point from the x-axis is |y-coordinate| $=|-4|=4$ units. Option (B).

   1 mark

Answer: 4, option (B)

1. In $△AXY$, $PQ\parallel XY$, so by BPT $\frac{AP}{PX}=\frac{AQ}{QY}$: $\frac{2}{1.5}=\frac{AQ}{0.75}$, giving $AQ=1$ cm.

   ½ mark

2. In trapezium $PXBC$-type section, the parallel lines cut the sides proportionally: $\frac{PX}{XB}=\frac{QY}{YC}$: $\frac{1.5}{4}=\frac{0.75}{CY}$, so $CY=2$ cm. Hence $AQ+CY=3$ cm. Option (C).

   ½ mark

Answer: 3 cm, option (C)

1. Similarity gives $\angle P=\angle A=30^{\circ }$, $\angle B=\angle Q=90^{\circ }$, so $\angle R=180^{\circ }-120^{\circ }=60^{\circ }$. Then $\angle R+\angle B=60^{\circ }+90^{\circ }=150^{\circ }$. Option (C).

   1 mark

Answer: 150°, option (C)

1. Outcomes: HH, HT, TH, TT (4). At least one head: HH, HT, TH (3). $P=\frac{3}{4}$. Option (C).

   1 mark

Answer: 3/4, option (C)

1. $OA\perp PA$, $OB\perp PB$ [tangent ⟂ radius] and $\angle P=90^{\circ }$, with $OA=OB$, so OAPB is a square with diagonal AB.

   ½ mark

2. Side $=\frac{AB}{\sqrt{2}}=\frac{3\sqrt{2}}{\sqrt{2}}=3$ cm = radius, so diameter = 6 cm. Option (D).

   ½ mark

Answer: 6 cm, option (D)

1. Parallel tangents touch the circle at ends of a diameter, so the distance between them is the diameter $=2\times 5=10$ cm. (S is false: a point at distance $5\sqrt{2}$ cm from O gives a 5 cm tangent.) Option (B).

   1 mark

Answer: Q, option (B)

1. $\angle OST=90^{\circ }$ [tangent ⟂ radius]. In $△OST$: $3x+x+90=180\Rightarrow x=22.5$, so $2x=45$. Option (B).

   1 mark

Answer: 45, option (B)

1. $\tan 30^{\circ }=\frac{1}{\sqrt{3}}$, ${\tan }^{2}30^{\circ }=\frac{1}{3}$. LHS bracket $=\frac{2/\sqrt{3}}{4/3}=\frac{\sqrt{3}}{2}$; RHS bracket $=\frac{2/\sqrt{3}}{2/3}=\sqrt{3}$.

   ½ mark

2. $x\cdot \frac{\sqrt{3}}{2}=y\sqrt{3}\Rightarrow x=2y$, so $x:y=2:1$. Option (C).

   ½ mark

Answer: 2 : 1, option (C)

1. Angle of depression = angle of elevation $\theta$ from the snake: $\tan \theta =\frac{10}{10\sqrt{3}}=\frac{1}{\sqrt{3}}\Rightarrow \theta =30^{\circ }$. Option (A).

   1 mark

Answer: 30°, option (A)

1. Greatest cone has same radius r and height h as the cylinder: $V_{cone}=\frac{1}{3}\pi r^{2}h$. Remaining $=\pi r^{2}h-\frac{1}{3}\pi r^{2}h=\frac{2}{3}\pi r^{2}h$.

   ½ mark

2. Ratio $=\frac{2}{3}:\frac{1}{3}=2:1$. Option (C).

   ½ mark

Answer: 2 : 1, option (C)

1. Empirical relation: $3\text{Median}=\text{Mode}+2\text{Mean}$. With Mean $=25-$Median: $3M=10+50-2M\Rightarrow M=12$.

   ½ mark

2. Mean $=25-12=13$. So mean 13, median 12. Option (B).

   ½ mark

Answer: 13 and 12, option (B)

1. The value occurring most often (highest frequency) is the mode, so 52 is the mode. Option (C).

   1 mark

Answer: 52 is the mode, option (C)

1. $x=-\frac{1}{2}$ (vertical line) and $y=\frac{5}{3}$ (horizontal line) intersect at exactly one point $(-\frac{1}{2},\frac{5}{3})$: unique solution. Option (A).

   1 mark

Answer: Unique solution, option (A)

1. $\cos B=\sqrt{1-\frac{1}{16}}=\frac{\sqrt{15}}{4}$ (B acute), so $\sec B=\frac{4}{\sqrt{15}}$. Option (D).

   1 mark

Answer: 4/√15, option (D)

1. For distinct primes p, q: HCF = 1 and LCM = pq, not p + q (e.g. 2, 3: LCM 6 ≠ 5). So A is false.

   ½ mark

2. HCF × LCM = product holds for any two natural numbers, so R is true. Option (D).

   ½ mark

Answer: Option (D)

1. $E_1=\{1,2\}$, $E_2=\{4,5,6\}$; outcome 3 is in neither, and $P(E_1)+P(E_2)=\frac{2}{6}+\frac{3}{6}=\frac{5}{6}\ne 1$. So A is false.

   ½ mark

2. For complementary events $P(E)+P(\bar{E})=1$, so R is true. Option (D).

   ½ mark

Answer: Option (D)

1. Adding the two equations: $203x+203y=609\Rightarrow x+y=3$ ...(1)

   ½ mark

2. Subtracting the first equation from the second: $x-y=1$ ...(2)

   ½ mark

3. Adding (1) and (2): $2x=4\Rightarrow x=2$

   ½ mark

4. Substituting in (1): $y=3-2=1$

   ½ mark

Answer: $x=2,y=1$

1. Let the smaller angle be $x^{\circ }$ and the greater angle be $y^{\circ }$. Since they are supplementary: $x+y=180$ ...(1)

   ½ mark

2. The greater exceeds the smaller by $50^{\circ }$: $y-x=50$ ...(2)

   ½ mark

3. Adding (1) and (2): $2y=230\Rightarrow y=115$

   ½ mark

4. From (1): $x=180-115=65$. So the angles are $65^{\circ }$ and $115^{\circ }$.

   ½ mark

Answer: Equations: $x+y=180$, $y-x=50$; angles $65^{\circ }$ and $115^{\circ }$

1. Squaring the first relation: $m^2=a^2{\sec }^2\theta +b^2{\tan }^2\theta +2ab\sec \theta \tan \theta$

   ½ mark

2. Squaring the second relation: $n^2=b^2{\sec }^2\theta +a^2{\tan }^2\theta +2ab\sec \theta \tan \theta$

   ½ mark

3. Subtracting: $m^2-n^2=a^2({\sec }^2\theta -{\tan }^2\theta )-b^2({\sec }^2\theta -{\tan }^2\theta )$

   ½ mark

4. Using ${\sec }^2\theta -{\tan }^2\theta =1$: $m^2-n^2=a^2-b^2\Rightarrow a^2+n^2=b^2+m^2$. Hence proved.

   ½ mark

Answer: $a^2+n^2=b^2+m^2$ (proved)

1. Dividing ${\sin }^{2}A+{\cos }^{2}A=1$ by ${\cos }^{2}A$ ($\cos A\ne 0$): $\frac{{\sin }^{2}A}{{\cos }^{2}A}+\frac{{\cos }^{2}A}{{\cos }^{2}A}=\frac{1}{{\cos }^{2}A}$

   ½ mark

2. $\Rightarrow {\tan }^{2}A+1={\sec }^{2}A$. Hence proved.

   ½ mark

3. Given $\sec A=\frac{5}{3}$: ${\tan }^{2}A={\left(\frac{5}{3}\right)}^2-1=\frac{25}{9}-1=\frac{16}{9}$

   ½ mark

4. Since A is acute, $\tan A>0$, so $\tan A=\frac{4}{3}$.

   ½ mark

Answer: $\tan A=\frac{4}{3}$

1. Let $P(x,y)$ be equidistant from A(7, 1) and B(3, 5). Then $PA=PB\Rightarrow P{A}^2=P{B}^2$.

   ½ mark

2. $(x-7{)}^2+(y-1{)}^2=(x-3{)}^2+(y-5{)}^2$ [distance formula]

   ½ mark

3. $x^2-14x+49+y^2-2y+1=x^2-6x+9+y^2-10y+25\Rightarrow -8x+8y+16=0$

   ½ mark

4. $\Rightarrow x=y+2$. So the abscissa of P is 2 more than its ordinate. Hence proved.

   ½ mark

Answer: $x=y+2$ (proved)

1. Figure: $△ABC$ with P on BC, AP joined; $\angle APC$ and $\angle BAC$ marked equal.

   ½ mark

2. In $△APC$ and $△BAC$: $\angle APC=\angle BAC$ (given) and $\angle ACP=\angle BCA$ (common). So $△APC\sim △BAC$ [AA similarity].

   1 mark

3. Corresponding sides are proportional: $\frac{AC}{BC}=\frac{CP}{CA}\Rightarrow A{C}^2=BC\cdot CP$. Hence proved.

   ½ mark

Answer: $A{C}^2=BC\cdot CP$ (proved)

1. Let black balls $=x$; red balls $=x+3$; total $=2x+3$.

   ½ mark

2. $P(\text{red})=\frac{x+3}{2x+3}=\frac{12}{23}$

   ½ mark

3. $23x+69=24x+36\Rightarrow x=33$

   ½ mark

4. Total number of balls $=2(33)+3=69$

   ½ mark

Answer: 69 balls

1. Assume, to the contrary, that $\sqrt{5}$ is rational.

   Then $\sqrt{5}=\frac{p}{q}$, where $p,q$ are integers, $q\ne 0$, and $p,q$ are co-prime.

   ½ mark

2. Squaring gives $5q^2=p^2$. So $p^2$ is divisible by 5.

   Since 5 is prime, $p$ is divisible by 5 [if a prime divides $p^2$, it divides $p$].

   So $p=5a$ for some integer $a$.

   1 mark

3. Substitute: $25a^2=5q^2\Rightarrow q^2=5a^2$. So $q^2$ is divisible by 5, and hence $q$ is divisible by 5.

   ½ mark

4. Now 5 is a common factor of $p$ and $q$. This contradicts the assumption that $p$ and $q$ are co-prime.

   Hence our assumption is wrong, and $\sqrt{5}$ is irrational.

   1 mark

Answer: $\sqrt{5}$ is irrational (proved by contradiction).

1. $p\cdot q\cdot r+q=q(pr+1)$

   ½ mark

2. Here $q>1$ and $pr+1>1$, so the number has factors other than 1 and itself; hence $pqr+q$ is a composite number.

   ½ mark

3. (i) Take $p=3,q=5,r=7$: $pqr+1=105+1=106=2\times 53$, a composite number.

   1 mark

4. (ii) Take $p=2,q=3,r=5$: $pqr+1=30+1=31$, a prime number.

   1 mark

Answer: $pqr+q$ is composite; (i) 3, 5, 7 give 106 (composite); (ii) 2, 3, 5 give 31 (prime)

1. $3x^2-4x-4=3x^2-6x+2x-4=(3x+2)(x-2)$, so the zeroes are $-\frac{2}{3}$ and $2$.

   1 mark

2. New zeroes (each 2 more): $-\frac{2}{3}+2=\frac{4}{3}$ and $2+2=4$.

   ½ mark

3. Sum of new zeroes $=\frac{4}{3}+4=\frac{16}{3}$

   ½ mark

4. Product of new zeroes $=\frac{4}{3}\times 4=\frac{16}{3}$

   ½ mark

5. Required polynomial $=x^2-\frac{16}{3}x+\frac{16}{3}$, i.e. $3x^2-16x+16$ (or any $k(3x^2-16x+16)$).

   ½ mark

Answer: Zeroes $-\frac{2}{3},2$; polynomial $3x^2-16x+16$

1. Write as $x+3y=6$ and $-2x+3y=-12$. Then $\frac{a_1}{a_2}=\frac{1}{-2}$, $\frac{b_1}{b_2}=\frac{3}{3}=1$.

   ½ mark

2. Since $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$, the lines intersect at one point, so the pair is consistent (unique solution).

   ½ mark

3. Tables: $x+3y=6$: (0, 2), (3, 1), (6, 0). $-2x+3y=-12$: (0, −4), (3, −2), (6, 0). Plot these points on graph paper and draw both lines; they meet at (6, 0) on the x-axis.

   1½ marks

4. Solution: $x=6,y=0$.

   ½ mark

Answer: Consistent; $x=6$, $y=0$

1. Diagonals of a parallelogram bisect each other, so midpoint of AC = midpoint of BD: $\left(\frac{6+9}{2},\frac{1+4}{2}\right)=\left(\frac{p+7}{2},\frac{2+q}{2}\right)$

   1 mark

2. $p+7=15\Rightarrow p=8$; $2+q=5\Rightarrow q=3$

   ½ mark

3. $AC=\sqrt{(9-6{)}^2+(4-1{)}^2}=\sqrt{18}=3\sqrt{2}$

   ½ mark

4. With B(8, 2), D(7, 3): $BD=\sqrt{(7-8{)}^2+(3-2{)}^2}=\sqrt{2}$

   ½ mark

5. Since $AC\ne BD$ (diagonals unequal), ABCD is not a rectangle.

   ½ mark

Answer: $p=8$, $q=3$; ABCD is not a rectangle

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

   ½ mark

2. Writing $1={\sin }^2\theta +{\cos }^2\theta$: $=\cot \theta \left[\frac{{\sin }^2\theta +{\cos }^2\theta -2{\cos }^2\theta }{{\sin }^2\theta +{\cos }^2\theta -2{\sin }^2\theta }\right]+\cot \theta$

   1 mark

3. $=\cot \theta \cdot \frac{{\sin }^2\theta -{\cos }^2\theta }{{\cos }^2\theta -{\sin }^2\theta }+\cot \theta$

   1 mark

4. $=-\cot \theta +\cot \theta =0=$ RHS. Hence proved.

   ½ mark

Answer: LHS = 0 = RHS (proved)

1. Squaring $\sin \theta +\cos \theta =x$: ${\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta =x^2\Rightarrow 2\sin \theta \cos \theta =x^2-1$

   1 mark

2. RHS $=\frac{2-(2\sin \theta \cos \theta {)}^2}{2}=\frac{2-4{\sin }^2\theta {\cos }^2\theta }{2}$

   ½ mark

3. $=1-2{\sin }^2\theta {\cos }^2\theta$

   ½ mark

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

   ½ mark

5. $={\sin }^4\theta +{\cos }^4\theta =$ LHS. Hence proved.

   ½ mark

Answer: ${\sin }^4\theta +{\cos }^4\theta =\frac{2-(x^2-1{)}^2}{2}$ (proved)

1. $\angle OPT=90^{\circ }$ [tangent ⟂ radius], so $\angle QPT=90^{\circ }-15^{\circ }=75^{\circ }$.

   ½ mark

2. $TP=TQ$ [tangents from an external point are equal], so $\angle PQT=\angle QPT=75^{\circ }$.

   ½ mark

3. In $△PTQ$: $\theta =180^{\circ }-75^{\circ }-75^{\circ }=30^{\circ }$

   1 mark

4. $\sin 2\theta =\sin 60^{\circ }$

   ½ mark

5. $=\frac{\sqrt{3}}{2}$

   ½ mark

Answer: $\sin 2\theta =\frac{\sqrt{3}}{2}$ ($\theta =30^{\circ }$)

1. Let $PB=x$ m; then $PA=(x+35)$ m.

   ½ mark

2. $\angle APB=90^{\circ }$ [angle in a semicircle], so by Pythagoras theorem: $(x+35{)}^2+x^2=65^2$

   1 mark

3. $2x^2+70x+1225-4225=0\Rightarrow x^2+35x-1500=0$

   2 marks

4. $(x+60)(x-25)=0$

   ½ mark

5. $x=25$ (rejecting $x=-60$ as distance cannot be negative)

   ½ mark

6. $PA=25+35=60$ m and $PB=25$ m.

   ½ mark

Answer: Distance from A = 60 m, from B = 25 m

1. For real roots, discriminant $D=b^2-4ac\ge 0$.

   ½ mark

2. Here $a=1,b=-2(p+1),c=p^2$: $[-2(p+1){]}^2-4p^2\ge 0$

   ½ mark

3. $4(p^2+2p+1)-4p^2\ge 0\Rightarrow 8p+4\ge 0\Rightarrow p\ge -\frac{1}{2}$

   1 mark

4. Smallest value of $p=-\frac{1}{2}$.

   ½ mark

5. At $p=-\frac{1}{2}$: $x^2-2\left(-\frac{1}{2}+1\right)x+{\left(-\frac{1}{2}\right)}^2=0$

   ½ mark

6. $x^2-x+\frac{1}{4}=0\Rightarrow 4x^2-4x+1=0$

   1 mark

7. $(2x-1)(2x-1)=0$

   ½ mark

8. Roots are $\frac{1}{2},\frac{1}{2}$.

   ½ mark

Answer: $p=-\frac{1}{2}$; roots $\frac{1}{2},\frac{1}{2}$

1. Statement (Basic Proportionality Theorem): If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

   1 mark

2. Given: $△ABC$ in which line $DE\parallel BC$ meets AB at D and AC at E. To prove: $\frac{AD}{DB}=\frac{AE}{EC}$. Construction: Join BE and CD; draw $EN\perp AB$ and $DM\perp AC$. Figure: triangle ABC with DE parallel to BC, BE, CD, EN, DM drawn.

   2 marks

3. Proof: $ar(ADE)=\frac{1}{2}\cdot AD\cdot EN$, $ar(BDE)=\frac{1}{2}\cdot DB\cdot EN$, so $\frac{ar(ADE)}{ar(BDE)}=\frac{AD}{DB}$ ...(1). Similarly $ar(ADE)=\frac{1}{2}\cdot AE\cdot DM$, $ar(DEC)=\frac{1}{2}\cdot EC\cdot DM$, so $\frac{ar(ADE)}{ar(DEC)}=\frac{AE}{EC}$ ...(2)

   1 mark

4. $△BDE$ and $△DEC$ are on the same base DE and between the same parallels DE and BC, so $ar(BDE)=ar(DEC)$ ...(3). From (1), (2), (3): $\frac{AD}{DB}=\frac{AE}{EC}$. Hence proved.

   1 mark

Answer: Converse = Basic Proportionality Theorem: a line parallel to one side divides the other two sides in the same ratio (proved by area method)

1. In $△ADB$ and $△CDA$: $\angle ADB=\angle CDA=90^{\circ }$ (given $AD\perp BC$).

   ½ mark

2. In right $△CAB$, $\angle B+\angle C=90^{\circ }$; in right $△ADB$, $\angle DAB+\angle B=90^{\circ }$. Hence $\angle DAB=\angle C=\angle DCA$.

   1 mark

3. So $△ADB\sim △CDA$ [AA similarity]. Hence proved.

   1 mark

4. Corresponding sides are proportional: $\frac{AD}{CD}=\frac{BD}{AD}$

   ½ mark

5. $\Rightarrow A{D}^2=BD\times CD$

   ½ mark

6. $BD=BC-CD=10-2=8$ cm, so $A{D}^2=8\times 2=16$

   1 mark

7. $AD=4$ cm

   ½ mark

Answer: $△ADB\sim △CDA$; $AD=4$ cm

1. Largest cone: diameter of base = side of cube = 14 cm, so radius $r=7$ cm.

   ½ mark

2. Height of cone $h=14$ cm.

   ½ mark

3. Slant height $l=\sqrt{h^2+r^2}=\sqrt{196+49}=7\sqrt{5}=7\times 2.2=15.4$ cm

   1 mark

4. Volume of remaining solid = volume of cube − volume of cone $=14^3-\frac{1}{3}\times \frac{22}{7}\times 7^2\times 14=2744-\frac{2156}{3}$

   1 mark

5. $=\frac{6076}{3}$ cm³ $\approx 2025.33$ cm³

   ½ mark

6. Surface area = TSA of cube − area of base circle + CSA of cone $=6\times 14^2-\frac{22}{7}\times 7^2+\frac{22}{7}\times 7\times 15.4=1176-154+338.8$

   1 mark

7. $=1360.8$ cm²

   ½ mark

Answer: Volume $=\frac{6076}{3}$ cm³ ($\approx 2025.33$ cm³); surface area $=1360.8$ cm²

1. Cumulative frequency table:

   Marks	$f$	cf
   10 – 20	12	12
   20 – 30	30	42
   30 – 40	$x$	$42+x$
   40 – 50	65	$107+x$
   50 – 60	$y$	$107+x+y$
   60 – 70	25	$132+x+y$
   70 – 80	18	$150+x+y$

   1 mark

2. Total frequency is 230, so $150+x+y=230\Rightarrow x+y=80$.

   1 mark

3. The median is 46, which lies in 40 – 50, so the median class is 40 – 50. Here $l=40$, $\frac{n}{2}=115$, $cf=42+x$, $f=65$, $h=10$.

   ½ mark

4. Median $=l+\left(\frac{\frac{n}{2}-cf}{f}\right)\times h$, so $46=40+\frac{115-(42+x)}{65}\times 10$.

   1 mark

5. $6=\frac{(73-x)\times 10}{65}\Rightarrow 73-x=39\Rightarrow x=34$.

   1 mark

6. From $x+y=80$: $y=80-34=46$.

   ½ mark

Answer: x = 34, y = 46

1. (i) The angle subtended by arc PB at the centre is twice the angle at the circumference: $\angle POB=2\angle PAB=60^{\circ }$. Since AOB is a straight line, $\angle POA=180^{\circ }-60^{\circ }=120^{\circ }$.

   1 mark

2. (ii) Radius $r=35$ m. Fence = semicircular arc + diameter $=\pi r+2r=\frac{22}{7}\times 35+70=110+70=180$ m.

   1 mark

3. (iii)(a) Mango region I is the segment cut off by chord PB. Its area = area of sector POB ($60^{\circ }$) − area of $△POB$. $△POB$ is equilateral because $OP=OB=35$ m and $\angle POB=60^{\circ }$. So area $=\frac{60}{360}\times \frac{22}{7}\times 35^2-\frac{\sqrt{3}}{4}\times 35^2$.

   1 mark

4. $=\left(\frac{1925}{3}-\frac{1225\sqrt{3}}{4}\right)$ m${}^2\approx 641.67-530.42\approx 111.2$ m${}^2$.

   1 mark

5. (iii)(b) $\angle APB=90^{\circ }$ [angle in a semicircle]. In right $△APB$, $\frac{AP}{AB}=\cos 30^{\circ }$, so $AP=70\times \frac{\sqrt{3}}{2}=35\sqrt{3}$ m.

   1 mark

6. Region III is bounded by chord AP and arc AP, and arc AP subtends $120^{\circ }$ at O. Arc AP $=\frac{120}{360}\times 2\times \frac{22}{7}\times 35=\frac{220}{3}$ m.

   ½ mark

7. Wire needed $=\left(\frac{220}{3}+35\sqrt{3}\right)$ m $\approx 73.33+60.62\approx 133.9$ m.

   ½ mark

Answer: (i) 120°; (ii) 180 m; (iii)(a) (1925/3 − 1225√3/4) m² ≈ 111.2 m²; (iii)(b) (220/3 + 35√3) m ≈ 133.9 m

1. The lane lengths form an AP: 400, 407.6, 415.2, … with $a=400$ and $d=7.6$.

   (i) $a_6=a+5d=400+5(7.6)=438$ m.

   1 mark

2. (ii) $a_8-a_4=(a+7d)-(a+3d)=4d=4(7.6)=30.4$ m.

   1 mark

3. (iii)(a) $S_6=\frac{6}{2}[2a+5d]=3[2(400)+5(7.6)]=3(838)$

   1 mark

4. $=2514$ m.

   1 mark

5. (iii)(b) Distance $=S_8-S_3=\frac{8}{2}[800+7(7.6)]-\frac{3}{2}[800+2(7.6)]=4(853.2)-1.5(815.2)$

   1 mark

6. $=3412.8-1222.8=2190$ m.

   1 mark

Answer: (i) 438 m; (ii) 30.4 m; (iii)(a) 2514 m; (iii)(b) 2190 m

1. (i) Diagram: a vertical line CQ represents the statue with its base, with C on the ground and Q the top. Point A is on the ground with $AC=80\sqrt{3}$ m. Join AQ and mark $\angle QAC=60^{\circ }$, with a right angle at C. Mark P on CQ with $CP=58$ m (the base).

   1 mark

2. (ii) Diagram: a vertical line CQ with $CQ=240$ m. A vertical platform EB stands with $BE=40$ m and E on the ground. Draw horizontal BR to meet CQ at R, so $RC=40$ m and $\angle QRB=90^{\circ }$. Join BQ and mark $\angle QBR=30^{\circ }$.

   1 mark

3. (iii)(a) In right $△ACQ$: $\frac{QC}{AC}=\tan 60^{\circ }=\sqrt{3}$, so $QC=80\sqrt{3}\times \sqrt{3}=240$ m. Height including the base = 240 m.

   1 mark

4. Height excluding the base $=240-58=182$ m.

   1 mark

5. (iii)(b) $QR=QC-RC=240-40=200$ m.

   ½ mark

6. In right $△QRB$: $\frac{QR}{RB}=\tan 30^{\circ }=\frac{1}{\sqrt{3}}$, so $RB=200\sqrt{3}$ m ($\approx 346.4$ m). This is the horizontal distance of B from the statue.

   ½ mark

7. Diagram: P is the top of the base on CQ, so $PC=58$ m and $PR=58-40=18$ m. Right $△PRB$ has the right angle at R and $\angle PBR=\alpha$.

   ½ mark

8. $\tan \alpha =\frac{PR}{RB}=\frac{18}{200\sqrt{3}}=\frac{3\sqrt{3}}{100}$.

   ½ mark