Daily Quant Questions Set

Answer: Option C

First, note that by symmetry, ∠ACB = ∠ADB.

Next, since A, B, C, D lie on a circle, the quadrilateral ACBD is cyclic and hence opposite corners ∠ACB and ∠ADB sum to 180◦ .

Therefore, it follows that ∠ACB = ∠ADB = 90◦ so AB must be the diameter of the circle containing points A, B, C, D. Since this circle has radius 3, AB = 6.

Next, AC is a radius of circle A so AC = 2 and BC is a radius of circle B. Applying Pythagoras to the triangle ABC, we have

$A C^{2}+B C^{2}=A B^{2}$

${{2}^{2}}+B{{C}^{2}}={{6}^{2}}$

$B C^{2} =32$

$B C =4 \sqrt{2}$

Answer: Option D

Let the length of the tunnel x metres and let Anant have covered a distance of kx metres inside the tunnel when Amit is 100 m away from the tunnel. Here k is a proper fraction i.e 0 < k < 1.

By the problem, $\frac{(100+x)}{[80+(1-k) x]}=\frac{9}{8}$

$\Rightarrow 800+8 x=720+9(1-k) x$

$\Rightarrow x[8-9(1-k)]=-80$

That means $x[8-9(1-k)]<0$

since $x$ cannot be negative, $[8-9(1-k)]$ must be negative.

Therefore, $[8-9(1-k)]<0$ or, $k<\frac{1}{9}$

Answer: Option A

Let $P(x)=x^{3}+a x^{2}+b x+3$ and

$Q(x)=x^{3}+b x^{2}+a x+2$

$P(x)$ and $Q(x)$ have an intersection point.

$\Rightarrow$ At their intersection point, $P(x)=Q(x)$

$\Rightarrow x^{3}+a x^{2}+b x+3=x^{3}+b x^{2}+a x+2$

$\Rightarrow(a-b) x^{2}+(a-b) x+1=0$

There is only one intersection point, thus, discriminant $=0$.

$\Rightarrow a-b=4(a-b)=0$

$\Rightarrow a-b=4(\text { because a cannot be equal to b })$

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

$\Rightarrow x=0.5$

The intersection point lies on the x-axis.

$\Rightarrow$ at the intersection point, $\mathrm{P}(x)=\mathrm{Q}(\mathrm{x})=0$

$\Rightarrow \mathrm{P}(0.5)=0$ and $\mathrm{Q}(0.5)=0$

$\Rightarrow 0.125+0.25 \mathrm{a}+0.5 \mathrm{b}+3=0$ and

$0.125+0.25 \mathrm{b}+0.5 \mathrm{a}+2=0$

On solving, $\mathrm{a}=-1.5$ and $\mathrm{b}=-5.5$

$\Rightarrow \mathrm{a}+\mathrm{b}=-7$

Answer: Option 30

Let the names of the groups of 3 members, 2 members and 1 member be A , B and C respectively.

Let A can do the work in 60 days

B can do the work In 3x days

and C can do the work In 2x days

$\Rightarrow \frac{1}{6 x}+\frac{1}{3 x}+\frac{1}{2 x}=\frac{1}{12}$

$\rightarrow x=12 .$

If they work on alternate days for 6 days,

work done $=2\left(\frac{1}{6 x}+\frac{1}{3 x}+\frac{1}{2 x}\right)=\frac{2}{12}=\frac{1}{6}$

Remaining work $=\frac{5}{6}$ th

We know that $B$ can do complete the work in $3 x=36$ days

So, time taken by group B to complete $\frac{5}{6}$ th of the work $=\frac{5}{6} \times 36=30$ days.

Answer: Option B

Let a , ar, $a r ^ { 2 } , \ldots$

$a + a r = 12 \cdots (i)$

$a{{r}^{2}}+a{{r}^{3}}=48\cdots (ii)$

dividing (ii) by (i), we have

$\frac { \mathrm { ar } ^ { 2 } ( 1 + \mathrm { r } ) } { \mathrm { a } ( \mathrm { r } + 1 ) } = 4$

$\Rightarrow r ^ { 2 } = 4$ if $r \neq - 1$

Therefore, r=-2 and a=-12.