CAT 1996 Question Paper

Break even implies that cost price is equal to selling price
Hence let's say in season x watches were sold
Cost price will be = $$1500 \times 150 + 30000 = 255000$$
So total selling price = $$250x + (1500 -x) 100 = 255000 $$
Or $$x = 700$$

 

As shopkeeper gave 3 one-rupee change for 20 rs. change, Buyer must have ordered for a total of 17 rs. stamps.
Now buyer ordered for at least more than 1 stamp for each type
Hence the minimum he bought was:

2 stamp for 5 rupees = 10 rs.
2 stamp for 2 rupees = 4 rs.
2 stamp for 1 rupee = 2 rs.

For the total to be seventeen, the buyer must have purchased 3 one rupee stamps.  
And 3 one rupee stamps were also there as changes given by shopkeeper.
So total number of stamps = 2+2+(3+3) = 10
 

A circle is circumscribed to triangle ABC.

For this circle D will be centre of circle, and AD , DC , BD will radius of this circle.

Hence AD=BD=DC=3 cm

Total sales value $$= 1148 = 4\cdot7\cdot41$$
Let AB be the actual price of the product and CD be the actual quantity.

Since the quantity sold reduced by 54 upon reversing
$$CD-DC\ =\ (10\cdot C+D)-\left(10\cdot D+C\right)=9\cdot\left(C-D\right)$$
wkt, $$9\cdot\left(C-D\right)=54\ \ \longrightarrow\ \ C-D=6$$

Now, wkt, $$AB\cdot CD=1148\ \ \&\ \ BA\cdot DC=1148$$
The possible values of CD are $$93$$, $$82$$, & $$71$$
Since only 82 divides 1148, $$CD=82$$, Therefore, $$AB=14$$
Actual price and Actual quantity are 14 and 82 respectively.

Alternatively,
Since final value $$= 1148 = 4\cdot7\cdot41$$ remains the same,
1148 can be represented as product of interchangeable numbers i.e. $$41 \times 28$$ and $$82 \times 14$$
As inventory sold reduced by 54, so, the entry of quantity sold should be 28 and price entry will be 41, and hence, actual price has to be 14 and actual sales volume has to be 82.

Total sales value $$= 1148 = 4\cdot7\cdot41$$
Let AB be the actual price of the product and CD be the actual quantity.

Since the quantity/inventory increased by 54 upon reversing
$$CD-DC\ =\ (10\cdot C+D)-\left(10\cdot D+C\right)=9\cdot\left(C-D\right)$$
wkt, $$9\cdot\left(C-D\right)=54\ \ \longrightarrow\ \ C-D=6$$

Now, wkt, $$AB\cdot CD=1148\ \ \&\ \ BA\cdot DC=1148$$
The possible values of CD are $$93, 82 $$ & $$ 71$$
Since only 82 divides 1148, $$CD=82$$, Therefore, $$AB=14$$
Actual price and Actual quantity are 14 and 82 respectively.

Alternatively,
Since final value $$= 1148 = 4\cdot7\cdot41$$ remains the same,
1148 can be represented as product of interchangeable numbers i.e. $$41 \times 28$$ and $$82 \times 14$$
As inventory decreased by 54, so, the quantity sold should be 82 and actual price will be 14.

Let the distribution of people having cable TV and VCR be as given in the diagram above.
Hence, $$a+c =\frac{2}{3}$$
$$b+c=\frac{1}{5}$$
and $$c=\frac{1}{10}$$

We need to find $$a+b+c=?$$
This equals $$(a+b)+(b+c)-c = \frac{2}{3}+\frac{1}{5}-\frac{1}{10}$$
Which equals $$\frac{20+6-3}{30}=\frac{23}{30}$$

$$\frac{1}{1 + \frac{1}{3-\frac{4}{2+\frac{1}{3-\frac{1}{2}}}}}$$

=$$\frac{1}{1 + \frac{1}{3-\frac{4}{2+\frac{1}{\frac{5}{2}}}}}$$

=$$\frac{1}{1 + \frac{1}{3-\frac{4}{2+\frac{2}{5}}}}$$

=$$\frac{1}{1 + \frac{1}{3-\frac{20}{12}}}$$

=$$\frac{1}{1 + \frac{1}{3-\frac{20}{12}}}$$

=$$\frac{1}{1 + \frac{12}{16}}$$

=$$\frac{1}{\frac{28}{16}}$$

=$$\frac{16}{28}$$

Similarly, Second term can be reduced to $$\frac{33}{21}$$ or $$\frac{11}{7}$$

Sum will be = $$\frac{11}{7}$$ + $$\frac{4}{7}$$ = $$\frac{15}{7}$$

For summation of square of roots to be zero, individual roots should be zero.
Hence summation should be zero i.e.  A-3=0 ; A = 3
And product of roots will also be zero i.e. A-2 = 0 ; A =2
So there is no unique value of A which can satisfy above equation.

Let the center be O and the point at which the semicircle intersects CD be P.

Let the radius of the semicircle be R and the circle be r.

OP = R and OC = R$$\sqrt{2}$$

OC - OT = CC' - TC'

$$R\sqrt{2} - R - 2r$$ = $$r\sqrt{2} - r$$

=> $$R\sqrt{2} - R$$ = $$r\sqrt{2} + r$$

=> r = $$\frac{(\sqrt{2}-1)R}{\sqrt{2}+1}$$

=> r = $$(\sqrt{2}-1)^2$$R

Ratio of areas will be $$r^2 : \frac{R^2}{2}$$ = $$2(\sqrt{2}-1)^4$$ : 1

Let the cost of pen, pencil and eraer be x,y,z respectively

5x+7y+4z = A

6x+8z+14y = 3A/2

4x + 16/3 z + 28/3 y = A

Comparing two equations

5x+7y+4z = 4x+16/3 z + 28/3 y

x = 7/3 y + 4/3 z

3x = 7y+4z

Now required percentage = $$\frac{5x}{5x+7y+4z}\times100=\frac{5x}{5x+3x}=62.5$$%