Mathematics

(a) 125

$n(A\cup B)=n(A)+n(B)-n(A\cap B)=145-n(A\cap B)$, which can exceed the universe, but a union cannot be larger than $n(U)=125$. The greatest value is therefore $125$ (when $n(A\cap B)=20$).

(b) Locus of a point which moves at equidistant from two fixed points

A point equidistant from two fixed points $P$ and $Q$ lies on the perpendicular bisector of $PQ$, which is a straight line. Equidistant from one fixed point gives a circle, and three (non-collinear) fixed points give a single point.

(b) Multiply by 10000

Let the correct value be $100x$. The student got $\dfrac{x}{100}$. To correct it: $\dfrac{x}{100}\times k=100x$, so $k=10000$.

(c) 2/3

Total marbles $=3+4+5=12$. Red or green $=3+5=8$, so $P=\dfrac{8}{12}=\dfrac{2}{3}$.

(c) 1.75

$10(a+b)=35\Rightarrow a+b=3.5$, so the average is $\dfrac{a+b}{2}=\dfrac{3.5}{2}=1.75$.

(b) 26

Sum of five numbers $=5\times18=90$. Sum of the remaining four $=4\times16=64$. Excluded number $=90-64=26$.

(a) 20 %

Everyone travels by car or bus, so $n(C\cup B)=100\%$. $100=70+50-n(C\cap B)\Rightarrow n(C\cap B)=20\%$

(c) 1605

£1 $=$ 2 $\times$ 3.52 $=$ 7.04 marks, so 1 mark $=\dfrac{140}{7.04}$ Rs. Then USD 1 $=3.91$ marks $=3.91\times\dfrac{140}{7.04}\approx$ Rs 77.76.

Francs for Rs 26,000: $\frac{26000}{140}\times7.04\times\frac{1}{3.91}\times4.8=\frac{878592}{547.4}\approx1605$

(a) 770 cm³

$r=7$ cm. Curved surface area: $2\pi rh=2\cdot\frac{22}{7}\cdot7\cdot h=44h=220\Rightarrow h=5$ cm. $V=\pi r^2h=\frac{22}{7}\times49\times5=770\ \text{cm}^3$

(c) 3

Volume of cylinder $=\pi r^2h$; volume of one cone $=\frac13\pi r^2h$. Number of cones $=\dfrac{\pi r^2h}{\frac13\pi r^2h}=3$.

(d) none of these

The angle in a semicircle is $90^\circ$, a right angle. It is not "complementary" or "supplementary" by itself (those describe pairs of angles), and it is half of a supplementary pair total ($180^\circ$), not half of complementary ($45^\circ$). So none of the options fits.

(d) IIIrd and IVth quadrant

The ordinate is the $y$-coordinate. It is negative below the $x$-axis, i.e. in the third quadrant $(-,-)$ and the fourth quadrant $(+,-)$.

(a) $x^2+y^2-10x-8y+25=0$

The circle touches the $y$-axis, so its radius equals the $x$-coordinate of the centre, $r=5$:

$(x-5)^2+(y-4)^2=25 \;\Rightarrow\; x^2+y^2-10x-8y+16=0$

Only option (a) has the correct $-10x-8y$ terms (its constant is misprinted as $25$ instead of $16$), and the other options do not have the correct centre.

(c) 133.33 mph

Total distance $=600$ miles at an average of 100 mph, so total time $=6$ h.

Time going $=\dfrac{300}{80}=3.75$ h, so time returning $=6-3.75=2.25$ h.

$v=\frac{300}{2.25}=133.33\ \text{mph}$

(d) $N_{24}$

$N_6\cap N_8$ is the set of natural numbers that are multiples of both 6 and 8, i.e. multiples of $\operatorname{lcm}(6,8)=24$. Hence $N_6\cap N_8=N_{24}$.

(c) 330 degree

$\sin\theta<0$ and $\cos\theta>0$ put $\theta$ in the fourth quadrant. The reference angle satisfies $\sin\alpha=0.5$, so $\alpha=30^\circ$ and $\theta=360^\circ-30^\circ=330^\circ$. Check: $\cos330^\circ=0.866$.

(c) 9

A diameter passes through the centre. The centre of $x^2+y^2+6x-6y+5=0$ is $(-3,3)$. Substituting into $2x-y+\alpha=0$:

$-6-3+\alpha=0 \;\Rightarrow\; \alpha=9$

(c) $\pm 12$

The line $2x+3y=k$ has intercepts $x=\dfrac{k}{2}$ and $y=\dfrac{k}{3}$.

$\text{Area}=\frac12\left|\frac{k}{2}\right|\left|\frac{k}{3}\right|=\frac{k^2}{12}=12 \;\Rightarrow\; k^2=144 \;\Rightarrow\; k=\pm12$

(a) 17

$t_{25}-t_{15}=(a+24d)-(a+14d)=10d=170$, so $d=17$.

(a) 2/13

Working it out directly: with replacement, an ace then a 10 has probability $\frac{4}{52}\cdot\frac{4}{52}=\frac1{169}$, and in either order it is $\frac{2}{169}$. This value is not among the options. The only listed value that arises from this data is (a), $\frac{8}{52}=\frac{2}{13}$, the chance that a single card is an ace or a 10, so (a) is the intended choice and the printed options are faulty.

(d) $x+1$

$f(x-1)=(x-1)+2=x+1$.

(c) 2

For $t>0$, $t+\dfrac1t\ge2$ with equality only at $t=1$. So $\sin\theta+\operatorname{cosec}\theta=2$ gives $\sin\theta=1$, hence $\operatorname{cosec}\theta=1$ and

$\sin^2\theta+\operatorname{cosec}^2\theta=1+1=2$

(c) 7

Since $n(U)=10$, the union cannot exceed 10, and because $5+7=12>10$ the sets must overlap by at least 2. So the true largest value of $n(A\cup B)$ is 10, which is not listed. Of the listed values, 7 is the one that occurs, when $A\subseteq B$ (the smallest possible union, $\max(5,7)$), so the intended answer is 7.

(a) 16, 4

$x+y=2(10)=20$ and $xy=8^2=64$. Then $x,y$ are the roots of $t^2-20t+64=0$, i.e. $(t-16)(t-4)=0$, so $x,y=16,4$. Check: AM $=10$, GM $=\sqrt{64}=8$.

(a) 6 cm

The perpendicular from the centre bisects the chord, giving a right triangle with hypotenuse 10 cm and one side 8 cm:

$d=\sqrt{10^2-8^2}=\sqrt{36}=6\ \text{cm}$

(b) 25%

Let the cost price of one article be Rs 1. Then SP of 4 articles = CP of 5 = Rs 5, while CP of 4 = Rs 4.

$\text{Gain \%}=\frac{5-4}{4}\times 100=25\%$

(b) $q-p-r$

After $p$ years he will be $q$, so his age now is $q-p$. $r$ years ago he was $(q-p)-r = q-p-r$.

(b) 9 and 4

For two numbers $a$ and $b$:

$\text{AM}-\text{GM}=\frac{a+b}{2}-\sqrt{ab}=\frac{(\sqrt a-\sqrt b)^2}{2}=0.5$

so $\sqrt a-\sqrt b=\pm1$. For 9 and 4: AM $=6.5$, GM $=6$, difference $0.5$. (3 and 12: $7.5-6=1.5$.)

(d) 37.5%

The class has $15+25=40$ students, so boys are

$\frac{15}{40}\times100=37.5\%$

(d) Range

Mean, median and mode are measures of central tendency. Range (largest value − smallest value) measures how spread out the data are, so it is a measure of dispersion.

(b) $(-3,3)$

Compare squared distances from the origin, $x^2+y^2$: $(-4,-1)\to17$, $(-3,3)\to18$, $(4,0)\to16$, $(2,3)\to13$. The largest is $18$, so $(-3,3)$ is farthest ($3\sqrt2\approx4.24$).

(b) 10

The water doubles each day. On day 8 the tank is $\tfrac14$ full, so on day 9 it is $\tfrac12$ full and on day 10 it is full.