CBSE Class 10 Mathematics Standard: 2023 board paper, set 30/1/1

1. The number of zeroes of $p(x)$ equals the number of points where the graph of $y=p(x)$ meets the $x$-axis.

2. The downward-opening curve touches the $x$-axis at exactly one point, its vertex. It does not cross the $x$-axis anywhere else.

3. So $p(x)$ has exactly 1 zero. Option (B).

   1 mark

Answer: 1, option (B)

1. Write both equations in standard form: $kx-y-2=0$ and $6x-2y-3=0$.

2. For infinitely many solutions we need $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$, that is $\frac{k}{6}=\frac{-1}{-2}=\frac{-2}{-3}$.

3. Here $\frac{b_1}{b_2}=\frac{1}{2}$ but $\frac{c_1}{c_2}=\frac{2}{3}$, and $\frac{1}{2}\ne \frac{2}{3}$ whatever $k$ is. So no value of $k$ works: such a $k$ does not exist. Option (B).

   1 mark

Answer: k does not exist, option (B)

1. For three terms in A.P., twice the middle term equals the sum of the other two: $2(p+1)=(p-1)+(2p+3)$.

2. $2p+2=3p+2\Rightarrow p=0$. Option (C).

   1 mark

Answer: p = 0, option (C)

1. Let the $x$-axis divide AB in the ratio $k:1$. Any point on the $x$-axis has $y$-coordinate 0.

2. By the section formula, $y=\frac{k(-3)+1(6)}{k+1}=0\Rightarrow -3k+6=0\Rightarrow k=2$.

3. So the ratio is $2:1$. Option (D).

   1 mark

Answer: 2 : 1, option (D)

1. In $△AOB$, $OA=OB$ (radii), so $\angle OBA=\angle OAB=\frac{180^{\circ }-95^{\circ }}{2}=42.5^{\circ }$.

2. $\angle OBQ=90^{\circ }$ [tangent ⟂ radius at the point of contact].

3. $\angle ABQ=90^{\circ }-42.5^{\circ }=47.5^{\circ }$. Option (A).

   1 mark

Answer: 47.5°, option (A)

1. From $2\tan A=3$ we get $\tan A=\frac{3}{2}$.

2. Divide the numerator and denominator by $\cos A$: $\frac{4\tan A+3}{4\tan A-3}=\frac{6+3}{6-3}=3$. Option (C).

   1 mark

Answer: 3, option (C)

1. For $x^2+x-1$, the sum of zeroes is $\alpha +\beta =-\frac{b}{a}=-1$ and the product is $\alpha \beta =\frac{c}{a}=-1$.

2. $\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{-1}{-1}=1$. Option (A).

   1 mark

Answer: 1, option (A)

1. The roots are rational when the discriminant is a perfect square: $D=k^2-4(2)(-4)=k^2+32$.

2. Test the options:

   • $k=\pm 2\sqrt{2}$ gives $D=40$, not a perfect square.
   • $k=\sqrt{2}$ gives $D=34$, not a perfect square.
   • $\pm 2$ is not a single positive value.
   • $k=2$ gives $D=36=6^2$, a perfect square.

3. So the required least positive value is $k=2$ (the roots are then $1$ and $-2$). Option (B).

   1 mark

Answer: k = 2, option (B)

1. Use the standard values: ${\tan }^{2}30^{\circ }=\frac{1}{3}$, ${\sec }^{2}45^{\circ }=2$, ${\sin }^{2}60^{\circ }=\frac{3}{4}$.

2. $\frac{3}{4}\cdot \frac{1}{3}-2+\frac{3}{4}=\frac{1}{4}-2+\frac{3}{4}=1-2=-1$. Option (A).

   1 mark

Answer: −1, option (A)

1. Curved surface area of a cylinder $=2\pi rh$.

2. $2\times 3.14\times r\times 5=94.2\Rightarrow 31.4r=94.2\Rightarrow r=3$ cm. Option (B).

   1 mark

Answer: 3 cm, option (B)

1. Convert the 'less than' cumulative frequencies into class frequencies:

   • 0–10: 3
   • 10–20: $12-3=9$
   • 20–30: $27-12=15$
   • 30–40: $57-27=30$
   • 40–50: $75-57=18$
   • 50–60: $80-75=5$

2. The highest frequency is 30, for the class 30–40. So the modal class is 30–40. Option (C).

   1 mark

Answer: 30 – 40, option (C)

1. Slant height $l=\sqrt{h^2+r^2}=\sqrt{24^2+7^2}=\sqrt{625}=25$ cm.

2. CSA $=\pi rl=\frac{22}{7}\times 7\times 25=550\mathrm{c}{\mathrm{m}}^2$. Option (C).

   1 mark

Answer: 550 cm², option (C)

1. By the distance formula, $d=\sqrt{(-2\sqrt{5}-0{)}^2+(0-2\sqrt{5}{)}^2}$.

2. $d=\sqrt{20+20}=\sqrt{40}=2\sqrt{10}$ units. Option (A).

   1 mark

Answer: $2\sqrt{10}$ units, option (A)

1. Sum of zeroes $=0$ and product $=-\frac{4}{9}$. So the polynomial is $k\left(x^2-\frac{4}{9}\right)=\frac{k}{9}(9x^2-4)$ for any non-zero constant $k$.

2. Taking $k=45$ gives $5(9x^2-4)$. Check: $9x^2=4\Rightarrow x=\pm \frac{2}{3}$. The other options have zeroes $\pm \frac{3}{2}$ or no real zeroes. Option (D).

   1 mark

Answer: $5(9x^2-4)$, option (D)

1. New mean $=\frac{\sum (x_i+3)}{n}=\frac{\sum x_i+3n}{n}=\bar{x}+3$.

2. So the mean increases by 3. Option (B).

   1 mark

Answer: Increases by 3, option (B)

1. An event and its complement together cover all outcomes: $P(E)+P(\bar{E})=1$.

2. Hence $p+q=1$. Option (A).

   1 mark

Answer: p + q = 1, option (A)

1. $P(\text{win})=\frac{\text{tickets she bought}}{\text{total tickets}}\Rightarrow 0.08=\frac{n}{6000}$.

2. $n=0.08\times 6000=480$. Option (C).

   1 mark

Answer: 480, option (C)

1. Number who can swim $=20-5=15$.

2. $P(\text{can swim})=\frac{15}{20}=\frac{3}{4}$. Option (A).

   1 mark

Answer: 3/4, option (A)

1. Assertion: on the $y$-axis, $x=0$, so $2y=4\Rightarrow y=2$. The line meets the $y$-axis at $(0,2)$, so A is true.

2. Reason: the distance of $(0,2)$ from the $x$-axis is $|y|=2$ units, so R is true.

3. R only gives a distance. It does not explain why $P$ lies on the line $3x+2y=4$. So both are true but R is not the correct explanation of A. Option (B).

   1 mark

Answer: Option (B)

1. By Pythagoras, $AC=\sqrt{2^2+3^2}=\sqrt{13}$ cm, which is irrational.

2. Perimeter $=2+3+\sqrt{13}=5+\sqrt{13}$, which is irrational. So A is false.

3. R: if $a,b$ are rational, then $a^2+b^2$ is rational (rationals are closed under multiplication and addition). So R is true. Option (D).

   1 mark

Answer: Option (D)

1. Draw both lines on the same axes:

   • $x=3$ is a vertical line through $(3,0)$, passing through points such as $(3,0),(3,2),(3,-4)$.
   • $y=-4$ is a horizontal line through $(0,-4)$, passing through points such as $(0,-4),(2,-4),(3,-4)$.
   1 mark

2. The two lines intersect at $(3,-4)$. Hence the solution is $x=3,y=-4$.

   1 mark

Answer: x = 3, y = −4 (lines meet at (3, −4))

1. Draw both lines on the same axes:

   • $x=0$ is the $y$-axis itself.
   • $y=-7$ is a horizontal line through $(0,-7)$, passing through points such as $(1,-7),(2,-7)$.
   1 mark

2. The lines intersect at exactly one point, $(0,-7)$. A unique solution exists, so the system is consistent.

   1 mark

Answer: Consistent; unique solution x = 0, y = −7

1. In $△ABC$, $XZ\parallel BC$, so by the Basic Proportionality Theorem $\frac{AX}{XB}=\frac{AZ}{ZC}=\frac{3}{2}$.

   ½ mark

2. In $△AXY$ and $△ABM$, $\angle A$ is common and $\angle AXY=\angle ABM$ (corresponding angles, $XY\parallel BM$). So $△AXY\sim △ABM$ (AA similarity).

   ½ mark

3. Hence $\frac{XY}{BM}=\frac{AX}{AB}$. From the first step, $\frac{AX}{AB}=\frac{3}{3+2}=\frac{3}{5}$, so $\frac{XY}{3}=\frac{3}{5}$.

   ½ mark

4. $XY=\frac{9}{5}=1.8$ cm.

   ½ mark

Answer: XY = 9/5 cm = 1.8 cm

1. Square both sides of $\sin \theta +\cos \theta =\sqrt{3}$: ${\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta =3$.

   1 mark

2. Use ${\sin }^2\theta +{\cos }^2\theta =1$: $1+2\sin \theta \cos \theta =3$.

   ½ mark

3. $2\sin \theta \cos \theta =2\Rightarrow \sin \theta \cos \theta =1$.

   ½ mark

Answer: sin θ · cos θ = 1

1. Since $\mathrm{cosec}\alpha =\frac{1}{\sin \alpha }$, we get $\mathrm{cosec}\alpha =\frac{1}{1/\sqrt{2}}=\sqrt{2}$.

   ½ mark

2. Use the identity ${\mathrm{cosec}}^2\beta =1+{\cot }^2\beta$. Then $\mathrm{cosec}\beta =\sqrt{1+(\sqrt{3}{)}^2}=\sqrt{4}=2$.

   1 mark

3. Therefore $\mathrm{cosec}\alpha +\mathrm{cosec}\beta =\sqrt{2}+2$, which can also be written as $\sqrt{2}(1+\sqrt{2})$.

   ½ mark

Answer: $\sqrt{2}+2$

1. The required number leaves remainder 1 when it divides 85, so it divides $85-1=84$ exactly. It leaves remainder 2 when it divides 72, so it divides $72-2=70$ exactly. Hence the required number is the HCF of 84 and 70.

   1 mark

2. Write the prime factorisations: $84=2^2\times 3\times 7$ and $70=2\times 5\times 7$.

   So HCF $=2\times 7=14$.

   The greatest such number is 14.

   1 mark

Answer: 14

1. Total number of balls $=4+3+2=9$, so there are 9 equally likely outcomes.

2. (i) Number of red balls = 4, so $P(\text{red})=\frac{4}{9}$.

   1 mark

3. (ii) Number of yellow balls = 2, so $P(\text{yellow})=\frac{2}{9}$.

   1 mark

Answer: (i) $\frac{4}{9}$; (ii) $\frac{2}{9}$

1. Let the greater number be $x$ and the smaller number be $y$.

   Half of their difference is 2:
   $\frac{1}{2}(x-y)=2\Rightarrow x-y=4$ ...(i)

   1 mark

2. The greater number plus twice the smaller number is 13:
   $x+2y=13$ ...(ii)

   1 mark

3. Subtract (i) from (ii): $3y=9\Rightarrow y=3$.

   Then $x=4+3=7$.

   The numbers are 7 and 3.

   1 mark

Answer: 7 and 3

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. Let $A(-5,3)$, $B(5,3)$ and let the third vertex be $C(x,y)$.

   Side $AB=\sqrt{(5+5{)}^2+0^2}=10$.

   Since the triangle is equilateral, $AC=BC=AB=10$.

   1 mark

2. From $A{C}^2=B{C}^2$:
   $(x+5{)}^2+(y-3{)}^2=(x-5{)}^2+(y-3{)}^2$

   $\Rightarrow 20x=0\Rightarrow x=0$.

   1 mark

3. From $A{C}^2=100$ with $x=0$:
   $25+(y-3{)}^2=100\Rightarrow (y-3{)}^2=75\Rightarrow y-3=\pm 5\sqrt{3}$.

   ½ mark

4. $AB$ lies on the line $y=3$, above the origin. For the origin to lie inside the triangle, $C$ must lie below the x-axis, so we take the negative sign.

   $y=3-5\sqrt{3}=3-5(1.7)=3-8.5=-5.5$.

   The third vertex is $(0,-5.5)$.

   ½ mark

Answer: $(0,-5.5)$, i.e. $(0,3-5\sqrt{3})$

1. $TP=TQ$ [tangents drawn from an external point are equal].

   So in $△TPQ$, $\angle TPQ=\angle TQP$ [angles opposite equal sides].

   1 mark

2. Let $\angle PTQ=\theta$. By the angle sum property of $△TPQ$:
   $\angle TPQ=\angle TQP=\frac{180^{\circ }-\theta }{2}=90^{\circ }-\frac{\theta }{2}$.

   1 mark

3. $\angle OPT=90^{\circ }$ [tangent $\perp$ radius at the point of contact].

   So $\angle OPQ=90^{\circ }-\angle TPQ=90^{\circ }-\left(90^{\circ }-\frac{\theta }{2}\right)=\frac{\theta }{2}$.

   Hence $\angle PTQ=\theta =2\angle OPQ$. Hence proved.

   1 mark

Answer: $\angle PTQ=2\angle OPQ$ (proved)

1. $DR=DS=3$ cm [tangents from an external point are equal].

   ½ mark

2. $AR=AD-DR=17-3=14$ cm.

   1 mark

3. $AQ=AR=14$ cm [tangents from A].

   ½ mark

4. $QB=AB-AQ=20-14=6$ cm.

   ½ mark

5. In quadrilateral $OQBP$:

   • $\angle OQB=\angle OPB=90^{\circ }$ [tangent $\perp$ radius]
   • $\angle B=90^{\circ }$
   • $OQ=OP=r$

   So $OQBP$ is a square, and $r=OP=QB=6$ cm.

   The radius is 6 cm.

   ½ mark

Answer: Radius = 6 cm

1. Replace 1 in the numerator by ${\sec }^2\theta -{\tan }^2\theta$ [since ${\sec }^2\theta -{\tan }^2\theta =1$]:

   LHS $=\frac{(\tan \theta +\sec \theta )-({\sec }^2\theta -{\tan }^2\theta )}{\tan \theta -\sec \theta +1}$

   1 mark

2. Factorise ${\sec }^2\theta -{\tan }^2\theta =(\sec \theta +\tan \theta )(\sec \theta -\tan \theta )$:

   $=\frac{(\tan \theta +\sec \theta )[1-\sec \theta +\tan \theta ]}{\tan \theta -\sec \theta +1}$

   1 mark

3. The bracket $1-\sec \theta +\tan \theta$ equals the denominator, so it cancels:

   $=\tan \theta +\sec \theta$

   ½ mark

4. $=\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta }=\frac{1+\sin \theta }{\cos \theta }=$ RHS. Hence proved.

   ½ mark

Answer: LHS = $\tan \theta +\sec \theta =\frac{1+\sin \theta }{\cos \theta }$ = RHS (proved)

1. Let $r$ be the radius of the hemisphere (and of the cylinder), and $h$ the height of the cylindrical part.

   Given $r=\frac{h}{2}$, so $h=2r$.

   ½ mark

2. Volume of room = volume of cylinder + volume of hemisphere:

   $\pi r^2(2r)+\frac{2}{3}\pi r^3=2\pi r^3+\frac{2}{3}\pi r^3=\frac{8}{3}\pi r^3$

   So $\frac{8}{3}\pi r^3=\frac{1408}{21}$.

   1 mark

3. $\frac{8}{3}\times \frac{22}{7}\times r^3=\frac{1408}{21}\Rightarrow \frac{176}{21}{r}^3=\frac{1408}{21}\Rightarrow r^3=8\Rightarrow r=2$ m.

   ½ mark

4. Height of the cylindrical part: $h=2r=4$ m.

   ½ mark

5. Total height of room $=h+r=4+2=6$ m.

   ½ mark

Answer: 6 m

1. Volume of cone $=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\times \pi \times 3^2\times 12=36\pi$ cm³.

   1 mark

2. $\frac{1}{6}$ of the cone is unfilled, so the ice-cream inside the cone is
   $\frac{5}{6}\times 36\pi =30\pi$ cm³.

   ½ mark

3. Volume of the hemisphere on top (radius 3 cm):
   $\frac{2}{3}\pi (3{)}^3=18\pi$ cm³.

   1 mark

4. Total ice-cream $=30\pi +18\pi =48\pi =48\times 3.14=150.72$ cm³.

   ½ mark

Answer: 150.72 cm³

1. Given: In $△ABC$, a line parallel to $BC$ meets $AB$ at $D$ and $AC$ at $E$, so $DE\parallel BC$.

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

   1 mark

2. Figure and construction: Draw $△ABC$ with $DE\parallel BC$, where $D$ is on $AB$ and $E$ is on $AC$. Join $BE$ and $CD$. Draw $EN\perp AB$ (N on AB) and $DM\perp AC$ (M on AC).

   1 mark

3. Proof: Using area $=\frac{1}{2}\times$ base $\times$ height:

   • $\mathrm{ar}(ADE)=\frac{1}{2}\cdot AD\cdot EN$ and $\mathrm{ar}(BDE)=\frac{1}{2}\cdot DB\cdot EN$
   • So $\frac{\mathrm{ar}(ADE)}{\mathrm{ar}(BDE)}=\frac{AD}{DB}$ ...(1)

   Similarly:

   • $\mathrm{ar}(ADE)=\frac{1}{2}\cdot AE\cdot DM$ and $\mathrm{ar}(DEC)=\frac{1}{2}\cdot EC\cdot DM$
   • So $\frac{\mathrm{ar}(ADE)}{\mathrm{ar}(DEC)}=\frac{AE}{EC}$ ...(2)
   1 mark

4. $△BDE$ and $△DEC$ stand on the same base $DE$ and lie between the same parallels $DE$ and $BC$.

   So $\mathrm{ar}(BDE)=\mathrm{ar}(DEC)$ ...(3)

   1 mark

5. From (1), (2) and (3): $\frac{AD}{DB}=\frac{AE}{EC}$.

   Hence the other two sides are divided in the same ratio. Hence proved.

   1 mark

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

1. Figure: Let $AB$ be the first tower with $AB=24$ m (foot A, top B). Let $CD=h$ be the second tower (foot C, top D). Let $AC=x$ be the distance between the towers.

   • $\angle BCA=60^{\circ }$ (elevation of B from C)
   • $\angle DAC=30^{\circ }$ (elevation of D from A)

   Draw $DE\perp AB$ (E on AB), so $DE=x$ and $AE=h$.

   1 mark

2. In right $△DCA$: $\tan 30^{\circ }=\frac{CD}{AC}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x}\Rightarrow x=h\sqrt{3}$ ...(i)

   1 mark

3. In right $△BAC$: $\tan 60^{\circ }=\frac{AB}{AC}\Rightarrow \sqrt{3}=\frac{24}{x}\Rightarrow x=\frac{24}{\sqrt{3}}=8\sqrt{3}$ ...(ii)

   1 mark

4. From (i) and (ii): the distance between the towers is $x=8\sqrt{3}$ m (≈ 13.86 m).

   The height of the other tower is $h=\frac{x}{\sqrt{3}}=8$ m.

   1 mark

5. The wire joins tops B and D. In right $△BED$:

   • $BE=24-8=16$ m
   • $ED=x=8\sqrt{3}$ m

   $BD=\sqrt{B{E}^2+E{D}^2}=\sqrt{256+192}=\sqrt{448}=8\sqrt{7}$ m (≈ 21.17 m).

   1 mark

Answer: Distance $=8\sqrt{3}$ m; height of other tower = 8 m; wire $=8\sqrt{7}$ m

1. Figure: Let $B$ be the observer's eye and $O$ the centre of the balloon, with radius $r$. $BQ$ and $BP$ are the two tangents from $B$ to the balloon, touching it at $Q$ and $P$. Join $OP$, so $OP=r$ and $OP\perp BP$. Draw $OA\perp$ the horizontal line through $B$. Then $\angle QBP=60^{\circ }$, $\angle ABO=45^{\circ }$, and $OA$ is the height of the centre.

   1 mark

2. The tangents from the external point $B$ are equally inclined to $BO$, so $BO$ bisects $\angle QBP$:
   $\angle PBO=\frac{1}{2}\times 60^{\circ }=30^{\circ }$.

   1 mark

3. In right $△BPO$ [tangent $\perp$ radius, so $\angle BPO=90^{\circ }$]:
   $\sin 30^{\circ }=\frac{OP}{OB}=\frac{r}{OB}$, which gives $\frac{1}{2}=\frac{r}{OB}$, so $OB=2r$ ...(i)

   1 mark

4. In right $△BAO$ ($\angle BAO=90^{\circ }$):
   $\sin 45^{\circ }=\frac{OA}{OB}$, which gives $\frac{1}{\sqrt{2}}=\frac{OA}{OB}$, so $OB=\sqrt{2}OA$ ...(ii)

   1 mark

5. From (i) and (ii): $\sqrt{2}OA=2r$, so $OA=\frac{2r}{\sqrt{2}}=\sqrt{2}r$.
   Hence the height of the centre of the balloon is $\sqrt{2}$ times its radius. Hence proved.

   1 mark

Answer: Height of centre $OA=\sqrt{2}r$ (proved)

1. Here $r=14$ cm and $\theta =60^{\circ }$.
   Area of sector $=\frac{\theta }{360^{\circ }}\pi r^2=\frac{60}{360}\times \frac{22}{7}\times 14\times 14=\frac{308}{3}{\text{cm}}^2$

   1 mark

2. The two radii are equal and the angle between them is $60^{\circ }$, so the triangle formed by the chord and the radii is equilateral with side 14 cm.
   Area of triangle $=\frac{\sqrt{3}}{4}\times 14^2=49\sqrt{3}{\text{cm}}^2$ (the same as $\frac{1}{2}\times 14\times 14\times \sin 60^{\circ }$).

   1 mark

3. Area of minor segment $=$ area of sector $-$ area of triangle
   $=\left(\frac{308}{3}-49\sqrt{3}\right){\text{cm}}^2\approx 102.67-84.87\approx 17.8{\text{cm}}^2$ (about $17.9{\text{cm}}^2$ if $\sqrt{3}=1.73$ is used).

   1 mark

4. Area of circle $=\frac{22}{7}\times 14\times 14=616{\text{cm}}^2$
   Area of major segment $=$ area of circle $-$ area of minor segment $=616-\left(\frac{308}{3}-49\sqrt{3}\right)$

   1 mark

5. $=\left(\frac{1540}{3}+49\sqrt{3}\right){\text{cm}}^2\approx 513.33+84.87\approx 598.2{\text{cm}}^2$ (about $598.1{\text{cm}}^2$ if $\sqrt{3}=1.73$ is used).

   1 mark

Answer: Minor segment $=\left(\frac{308}{3}-49\sqrt{3}\right){\text{cm}}^2\approx 17.8{\text{cm}}^2$; Major segment $=\left(\frac{1540}{3}+49\sqrt{3}\right){\text{cm}}^2\approx 598.2{\text{cm}}^2$

1. Let the first term be $a$ and the common difference $d$. Using $a_n=a+(n-1)d$:
   $\frac{a_{11}}{a_{17}}=\frac{a+10d}{a+16d}=\frac{3}{4}$

   1 mark

2. Cross-multiplying: $4a+40d=3a+48d$, so $a=8d$ ...(i)
   (Here $d\ne 0$; otherwise the ratio would be $1$, not $\frac{3}{4}$.)

   1 mark

3. $\frac{a_5}{a_{21}}=\frac{a+4d}{a+20d}=\frac{8d+4d}{8d+20d}=\frac{12d}{28d}=\frac{3}{7}$
   So $a_5:a_{21}=3:7$.

   1 mark

4. Using $S_n=\frac{n}{2}[2a+(n-1)d]$:
   $\frac{S_5}{S_{21}}=\frac{\frac{5}{2}(2a+4d)}{\frac{21}{2}(2a+20d)}=\frac{5(16d+4d)}{21(16d+20d)}=\frac{5\times 20d}{21\times 36d}$

   1 mark

5. $=\frac{100}{756}=\frac{25}{189}$
   So $S_5:S_{21}=25:189$.

   1 mark

Answer: $a_5:a_{21}=3:7$; $S_5:S_{21}=25:189$

1. The numbers of logs in the rows (from the bottom) form an A.P. $22,21,20,\dots$ with $a=22$, $d=-1$. Let the number of rows be $n$, so $S_n=250$.

   1 mark

2. $S_n=\frac{n}{2}[2a+(n-1)d]$, which gives $250=\frac{n}{2}[44+(n-1)(-1)]=\frac{n}{2}(45-n)$

   1 mark

3. $500=45n-n^2$, so $n^2-45n+500=0$

   1 mark

4. $(n-20)(n-25)=0$, so $n=20$ or $n=25$.
   For $n=25$: $a_{25}=22+24(-1)=-2$. A row cannot have a negative number of logs, so $n=25$ is rejected. Hence $n=20$.

   1 mark

5. Logs in the top row: $a_{20}=a+19d=22+19(-1)=3$.
   So the 250 logs are placed in 20 rows and the top row has 3 logs.

   1 mark

Answer: 20 rows; 3 logs in the top row

1. (I) New length $=(18+x)$ cm and new width $=(12+x)$ cm. The new area must be double the original area $18\times 12=216{\text{cm}}^2$:
   $(18+x)(12+x)=2(18\times 12)$

   1 mark

2. (II) Expanding: $216+30x+x^2=432$, so $x^2+30x-216=0$

   1 mark

3. (III)(a) Solve $x^2+30x-216=0$ by splitting the middle term: $x^2+36x-6x-216=0$, so $(x+36)(x-6)=0$.
   This gives $x=6$ or $x=-36$. A length increase cannot be negative, so $x=6$.

   1 mark

4. (III)(a) New dimensions: length $=18+6=24$ cm, width $=12+6=18$ cm. The enlarged photo is $24\text{ cm}\times 18\text{ cm}$ (area $432=2\times 216$ ✓).

   1 mark

5. (III)(b) For a new area of $220{\text{cm}}^2$: $(18+x)(12+x)=220$, so $x^2+30x+216=220$, giving $x^2+30x-4=0$.

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6. (III)(b) Discriminant $D=b^2-4ac=30^2-4(1)(-4)=900+16=916$. Since 916 is not a perfect square, $\sqrt{D}$ is irrational and so the roots are irrational. Hence **no rational value of $x$** can make the new area $220{\text{cm}}^2$.

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