Class 10 Maths Standard

(i) Answer: $2^2\times 3\times 7$

$84=2^2\times 3\times 7$.

(ii) Answer: $2\times 3^2\times 7$

$126=2\times 3^2\times 7$.

(iii) Answer: $42$ kits; $2$ pencils and $3$ erasers per kit.

HCF$(84,126)=2\times 3\times 7=42$. Thus each kit has $84/42=2$ pencils and $126/42=3$ erasers.

(i) Answer: $7$

$\alpha +\beta =-b/a=-(-7)/1=7$.

(ii) Answer: $12$

$\alpha \beta =c/a=12/1=12$.

(iii) Answer: $(x-3)(x-4)$; $(3,0)$ and $(4,0)$.

$x^2-7x+12=(x-3)(x-4)$. The zeroes are $3$ and $4$, giving x-intercepts $(3,0)$ and $(4,0)$.

(i) Answer: $a+s=20$

The group bought $20$ tickets in all, so $a+s=20$.

(ii) Answer: $100a+60s=1520$

Adult tickets cost Rs. $100$ each and student tickets cost Rs. $60$ each.

(iii) Answer: $a=8$, $s=12$

Divide $100a+60s=1520$ by $20$: $5a+3s=76$. Subtract $3(a+s)=60$, giving $2a=16$. Hence $a=8$ and $s=12$.

(i) Answer: $x+3$ m

The length is $3$ m more than the width, so it is $x+3$ m.

(ii) Answer: $x^2+3x-70=0$

Area gives $x(x+3)=70$, so $x^2+3x-70=0$.

(iii) Answer: $7$ m by $10$ m

$(x+10)(x-7)=0$, so $x=7$ or $-10$. Width cannot be negative, so width is $7$ m and length is $10$ m.

(i) Answer: $a=8$, $d=3$

The first-day reading is $8$, and each next day increases by $3$.

(ii) Answer: $35$ pages

$a_{10}=8+(10-1)3=35$.

(iii) Answer: $215$ pages

$S_{10}=\frac{10}{2}(8+35)=5(43)=215$.

(i) Answer: Basic Proportionality Theorem

A line parallel to one side of a triangle divides the other two sides proportionally.

(ii) Answer: $1/2$

$AD/DB=3/6=1/2$.

(iii) Answer: $EC=8$ m; $AC=12$ m

$AD/DB=AE/EC$, so $1/2=4/EC$. Thus $EC=8$ m and $AC=4+8=12$ m.

(i) Answer: $6$ units

$8-2=6$ units.

(ii) Answer: $10$ units

$AB=\sqrt{(8-2{)}^2+(11-3{)}^2}=\sqrt{36+64}=10$.

(iii) Answer: $P=(\frac{7}{2},5)$

$P=\left(\frac{1(8)+3(2)}{4},\frac{1(11)+3(3)}{4}\right)=(\frac{7}{2},5)$.

(i) Answer: $13$ m

By Pythagoras, ramp length $=\sqrt{5^2+12^2}=13$ m.

(ii) Answer: $5/12$

$\tan \theta =\frac{\text{opposite}}{\text{adjacent}}=5/12$.

(iii) Answer: $\sin \theta =5/13$, $\cos \theta =12/13$; identity $=1$

$\sin \theta =5/13$ and $\cos \theta =12/13$. Then $25/169+144/169=1$.

(i) Answer: Tangent

Tangent relates the opposite side (height) to the adjacent side (horizontal distance).

(ii) Answer: $\tan 30^{\circ }=h/18$

Opposite over adjacent gives $\tan 30^{\circ }=h/18$.

(iii) Answer: $h=6\sqrt{3}$ m; new tangent $=\sqrt{3}$

$h=18/\sqrt{3}=6\sqrt{3}$ m. The new distance is $6$ m, so tangent $=(6\sqrt{3})/6=\sqrt{3}$.

(i) Answer: $90^{\circ }$

The radius to a tangent at the point of contact is perpendicular to the tangent.

(ii) Answer: $PA=PB$

Tangents from the same external point to a circle are equal.

(iii) Answer: $PA=PB=12$ m

In right triangle $OAP$, $PA=\sqrt{O{P}^2-O{A}^2}=\sqrt{13^2-5^2}=12$ m. Hence both tangents are $12$ m.

(i) Answer: $1/4$

$90^{\circ }/360^{\circ }=1/4$.

(ii) Answer: $616$ m${}^2$

$\pi r^2=\frac{22}{7}\times 14^2=616$ m${}^2$.

(iii) Answer: Area $=154$ m${}^2$; perimeter $=50$ m

Sector area $=616/4=154$ m${}^2$. Arc length $=22$ m. Perimeter $=22+14+14=50$ m.

(i) Answer: $72\pi$ cm${}^3$

$\pi r^{2}h=\pi (3^2)(8)=72\pi$ cm${}^3$.

(ii) Answer: $18\pi$ cm${}^3$

$\frac{2}{3}\pi r^3=\frac{2}{3}\pi (3^3)=18\pi$ cm${}^3$.

(iii) Answer: Volume $=90\pi$ cm${}^3$; external area $=75\pi$ cm${}^2$

Total volume $=72\pi +18\pi =90\pi$. External area $=2\pi (3)(8)+2\pi (3^2)+\pi (3^2)=48\pi +18\pi +9\pi =75\pi$ cm${}^2$.

(i) Answer: $30$ students

$3+7+12+8=30$.

(ii) Answer: $20$-$30$

Cumulative frequencies are $3,10,22,30$. Since $N/2=15$, the first cumulative frequency above $15$ is $22$, so the median class is $20$-$30$.

(iii) Answer: $\frac{145}{6}\approx 24.17$ minutes

$l=20$, $cf=10$, $f=12$, $h=10$. Median $=20+\frac{15-10}{12}\times 10=\frac{145}{6}\approx 24.17$ minutes.

(i) Answer: $12$

$5+4+3=12$ counters.

(ii) Answer: $1/3$

$P(\text{blue})=4/12=1/3$.

(iii) Answer: $2/3$

Non-blue counters $=5+3=8$, so $P(\text{not blue})=8/12=2/3$. Complement check: $1-P(\text{blue})=1-1/3=2/3$.