**QUESTION: 1**

Two circles with centers A and B respectively intersect at two points C and D. Given that A, B, C, D lie on a circle of radius 3 and circle A has radius 2, what is the radius of circle B?

Level: Moderate

**OPTIONS**

- 4
- 5
- $4\sqrt{2}$
- $3\sqrt{2}$

**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

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

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

$B C^{2} =32$

$B C =4 \sqrt{2}$

**QUESTION: 2**

Amit is chasing Anant in his car on Mumbai-Pune Expressway. He observes that when he is just 100 m away from a dark tunnel, Anant has already entered the tunnel and covered some distance inside. When Amit just emerges from the tunnel, he can see Anant 80 m ahead of him. Amit is driving at a speed 12.5% faster than the speed at which Anant is driving. What fraction of the length of the tunnel Anant must have covered, when Amit was 100 m away from the tunnel?

Level: Difficult

**OPTIONS**

- more than 1/9th
- more than 1/9th but less than 8/9th
- more than 8/9th
- less than 1/9th

**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, $k<\frac{1}{9}$

**QUESTION: 3**

If $x^{3}+a x^{2}+b x+3$ and $x^{3}+b x^{2}+a x+2$ intersect at exactly one point, which lies on x axis, in the X -Y plane, what is the value of (a+b)?

Level: Difficult

**OPTIONS**

- -7
- -8
- -2
- 0

**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 $\mathrm{b}=-5.5$

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

**QUESTION: 4**

A family of comprising 6 members can do a task in 12 days. The time taken to complete the same work by 3 out of the 6 members is twice of the time taken by 2 other members of the family, and thrice of the time taken by the 6th member. If these three groups work alternatively for 6 days, then in how many days can the group containing two members working alone complete the remaining work?

Level: Easy

**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.

**QUESTION: 5**

The first two terms of a geometric progression add up to 12. The sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is

Level: Easy

**OPTIONS**

- -4
- -12
- 12
- None of these

**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.