CAT 1990 Question Paper

$$\frac{-p}{1-p}$$ can be compared with sum of an infinite G.P. series i.e. $$\frac{a}{1-r}$$ (where a is first term and r is common ratio)
Hence here a=(-p)
and r = p
So k^th term will be = $$(-p) (p)^{(k-1)}$$

In second round, number of winners were odd hence 1 winner will be played in next round. Since in third round also number of winners are odd, then last round player will be played in this round and one another player will be played in next round.

Now in 4th round total winners will be (7+1), this 1 winner belongs to 3rd round where he was sent to play in next round.

Summation of all matches will be 115.

$$n^3-n$$ can be written as:
$$(n-1)n(n+1)$$ (where n is a positive integer)
i.e. product of three consecutive integers.
Hence for any number n=2 or >2 , product will have a factor of 6 in it.
When two numbers are prime in product, then third number will always be divisible by 6 
Or product will always have a factor of $$3\times2$$ into it.

$$\frac{(1-d^3)}{(1-d)}$$ = $$1+d^2+d$$ (where $$d \neq 1$$)
Let's say $$f(d)=1+d^2+d$$ 
Now $$f(d)$$ will always be greater than 0 and have its minimum value at d = -0.5. The value is $$\frac{3}{4}$$.
For $$d<-1$$ ; $$f(d) >1$$
$$-1<d<0$$ ; $$\frac{3}{4} <f(d)<1$$
$$0<d<1$$ ; $$1<f(d)<3$$
$$d>1$$ ; $$f(d)>3$$

So, for d > 1, f(d) > 3. Option b) is the correct answer.

Let's say total money was $$x$$ rs.
So cost price of 40 mango will be = $$x$$ ;
Hence cost price of 20 mangoes will be = $$\frac{x}{2}$$
Taxi fare = $$\frac{10x}{100}$$
Total expense = $$\frac{x}{2}$$ +  $$\frac{10x}{100}$$ = $$\frac{6x}{10}$$
Remaining money =$$ \frac{4x}{10}$$
Cost price of 1 orange will be = $$\frac{x}{50}$$
Hence in $$\frac{4x}{10}$$ rs. 20 oranges can be purchased.

It is given that the maximum cost is 12 lakhs
$$1200000 <= T \times 10000 + V \times 15000$$ (Where T is number of TV's and V is number of VCR's)
and as we know $$T+V = 100$$
By solving the above two equation, we will get v<=40

We have to keep the no. of VCRs as maximum, as they fetch higher profit than TV, so

$$T = 60$$
$$V = 40$$​​​​​

As we know for maximizing profit, Total Revenue = Marginal Cost
i.e. $$T \times 10000 + V \times 15000 = 1200000$$
and $$T+V = 120$$
By solving above two equations we will get
$$T=120$$
$$V=0$$
Hence ratio will be 0.​​​​​​

The total property consists of cash, gold coins and silver bars.
And ghosh babu gave equal parts to 4 daughters, hence they should have 25% of total property each.
As eldest daughter possess gold coins as 20% worth of total property, so 25000 cash should be equal to 5% of total property.
So total property will be =$$\frac{ 25000 \times 100}{5}$$ = 500000 
Hence property amounting only gold coins and silver bars will be = Total property - Total Cash
i.e. = 500000 - (25000 + 50000 + 75000 + 75000)
= 2,75,000 

The total property consists of cash, gold coins and silver bars.
And ghosh babu gave equal parts to 4 daughters, hence they should have 25% of total property each.
As eldest daughter possess gold coins as 20% worth of total property, so 25000 cash should be equal to 5% of total property.
So total property will be =$$\frac{ 25000 \times 100}{5}$$ = 500000