CAT 2007 Question Paper

Cost of 20 units = 240+20b+400c

Cost of 40 units = 240+40b+1600c = 5/3 * (240+20b+400c) => 720+120b+4800c = 1200+100b+2000c

=> 480 = 20b + 2800c => 120 = 5b + 700c

Cost of 60 units = 240+60b+3600c = 3/2 (240+40b+1600c) => 480 + 120b + 7200c = 720 + 120b + 4800c

=> 240 = 2400c => c = 1/10 and b = 10

Let the number of items needed for max profit be k

CP = $$240+10k+k^2/10$$

SP = 30k

Profit = SP - CP = $$30k - 240 - 10k - k^2/10$$ = $$20k - 240 - k^2/10$$

or Profit = $$\frac{1}{10} (-k^2 + 200k - 2400)$$

or, Profit = $$\frac{1}{10} (-(k^2 - 200k + 2400))$$

or, Profit = $$\frac{1}{10} (-(k^2 - 200k + 2400 + 7600 - 7600))$$

or, Profit = $$\frac{1}{10} (-(k^2 - 200k + 10000) + 7600)$$

or, Profit = $$\frac{1}{10} (-(k - 100)^2 + 7600)$$

To maximise the value of Profit, $$-(k - 100)^2$$ must be 0.

So, $$k$$ must be equal to 100.

Hence, option B is the correct answer.

Cost of 20 units = 240+20b+400c
Cost of 40 units = 240+40b+1600c = 5/3 * (240+20b+400c) => 720+120b+4800c = 1200+100b+2000c
=> 480 = 20b + 2800c => 120 = 5b + 700c
Cost of 60 units = 240+60b+3600c = 3/2 (240+40b+1600c) => 480 + 120b + 7200c = 720 + 120b + 4800c
=> 240 = 2400c => c = 1/10 and b = 10
Let the number of items needed for max profit be k
CP = $$240+10k+k^2/10$$
SP = 30k
Profit = SP - CP = $$30k - 240 - 10k - k^2/10$$ = $$20k - 240 - k^2/10$$
Maximum when 20 - k/5 = 0 or k = 100
Profit = 2000 - 240 - 1000 = 760

$$a_n + b_n$$ for even n = $$p*b_{n-1} + q*b_{n-1}$$

= $$(p+q)*b_{n-1}$$

$$b_{n-1} = q*a_{n-2} = qp*b_{n-3}$$

= $$q^2*p*a_{n-4} = q^2p^2*b_{n-5}$$

.

.

= $$(qp)^{n/2-1}*b_1 = (qp)^{n/2-1}*q$$

So, $$a_n + b_n$$ = $$q(pq)^{(n/2) -1}(p+q)$$

$$a_{n} + b_{n}$$ (n is odd) = $$p^{\frac{n+1}{2}}*q^{\frac{n-1}{2}} + p^{\frac{n -1}{2}}*q^{\frac{n+1}{2}}$$ = $$(p + q)pq^{\frac{n-1}{2}}$$

Substituting the values of p and q we get

$$a_{n} + b_{n}$$ = $$(\frac{2}{9})^{\frac{n-1}{2}}$$

Now substitute the values of n and check. 

We can see that the lowest value of n for which 

$$a_{n} + b_{n}$$ < .01 is 9

The odd numbers in the set are 3, 5, 7, ...2n+1

Sum of the odd numbers = 3+5+7+...+(2n+1) = $$n^2 + 2n$$

Average of odd numbers = $$n^2 + 2n$$/n = n+2

Sum of even numbers = 2 + 4 + 6 + ... + 2n = 2(1+2+3+...+n) = 2*n*(n+1)/2 = n(n+1)

Average of even numbers = n(n+1)/n = n+1

So, difference between the averages of even and odd numbers = 1

Ten years ago, the total age of the family is 231 years.

Seven years ago, (Just before the death of the first person), the total age of the family would have been 231+8*3 = 231+24 = 255.
This is because, in 3 years, every person in the family would have aged by 3 years,
Total change in age = 231+24 = 255

After the death of one member, the total age is 255-60 = 195 years.

Since a child takes birth in the same year, the number of members remain the same i.e. (7+1) = 8

Four years ago, (i.e. 6 years after start date) one of the member of age 60 dies, 
therefore, total age of the family is 195+24-60 = 159 years.

Since a child takes birth in the same year, the number of members remain the same i.e. (7+1) = 8

After 4 more years, the current total age of the family is = 8x4 + 159 = 191 years  

The average age is 191/8 = 23.875 years = 24 years (approx)


Alternatively,
Since the number of members is always the same throughout
The 2 older members dropped their age by 60
So, after 10yrs, total age = 231 + 8*10 - 2*60 = 191
Average age = 191/8 = 23.875 $$\simeq$$ 24

According to given conditions we get f(2)=f(1)/3 , then f(3)=f(1)/6, then  f(4)=f(1)/10 , then f(5)=f(1)/15 .

We can see the pattern here that the denominator goes on increasing from 3,3+3,6+4,10+5,15+6,.. so for the f(9) the denominator will be same as 15+6+7+8+9=45 .

So f(9)=3600/45 = 80

If two 50 Misos are used, the 107 can be paid in only 1 way.

If one 50 Miso is used, the number of ways of paying 107 is 6 - zero 10 Miso, one 10 Miso and so on till five 10 Misos.

If no 50 Miso is used, the number of ways of paying 107 is 11 - zero 10 Miso, one 10 Miso and so on till ten 10 Misos.

So, the total number of ways is 18

Let the value of cheque be x Rs and y ps and the amount she received is y Rs and x ps.

After 50 ps is deducted she has the amount which is 3 times the amount on cheque,

So 100y+x-50=3(100x+y)    (After converting the amount in paise)

y= (299x+50)/97 = 3x+ (8x+50)/97

Since x is an integer, 3x will definitely be an integer. Now, (8x+50)/97 has to be an integer for y to be an integer.

So, 8x+50 has to be a multiple of 97. 

8x+50= 97

$$\Rightarrow$$ x = 47/8, this value of x is not an integer. 

8x+50= 97*2 = 194

8x = 144

$$\Rightarrow$$ x = 144/8 = 18. Here, we got an integer.

So, the amount has to be 18 rupees and some paise. Hence, option D is correct.

1/m + 4/n = 1/12
So, 1/m = 1/12 - 4/n
So, m = 12n/(n-48)
Since m is positive, n should be greater than 48
Also, since n is an odd number, it can take only 49, 51, 53, 55, 57 and 59
If n = 49, 51, 57 then m is an integer, else it is not an integer
So, there are 3 pairs of values for which the equation is satisfied