CAT 2007 Question Paper with Solutions PDF, Download PDF

Instructions

Directions for the following two questions:

Mr. David manufactures and sells a single product at a fixed price in a niche market. The selling price of each unit is Rs. 30. On the other hand, the cost, in rupees, of producing x units is $$240 + bx + cx^2$$ , where b and c are some constants. Mr. David noticed that doubling the daily production from 20 to 40 units increases the daily production cost by 66.67%. However, an increase in daily production from 40 to 60 units results in an increase of only 50% in the daily production cost. Assume that demand is unlimited and that Mr. David can sell as much as he can produce. His objective is to maximize the profit.

Question 11

How many units should Mr. David produce daily?

Video Solution

Solution

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.

Question 12

What is the maximum daily profit, in rupees, that Mr. David can realize from his business?

Video Solution

Solution

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

Instructions

Directions for the following two questions:

Let $$a_1= p$$ and $$b_1 = q$$, where p and q are positive quantities.

Define $$a_n = pb_{n-1} , b_n = qb_{n-1}$$ , for even n > 1. and $$a_n = pa_{n-1} , b_n = qa_{n-1}$$ , for odd n > 1.

Question 13

Which of the following best describes $$a_n + b_n$$ for even n?

Video Solution

Solution

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

Question 14

If p = 1/3 and q = 2/3 , then what is the smallest odd n such that $$a_n+b_n < 0.01$$?

Video Solution

Solution

$$a_{n} + b_{n}$$ (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

Instructions

For the following questions answer them individually

Question 15

Consider the set S = {2, 3, 4, ...., 2n+1}, where n is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of X - Y ?

Video Solution

Solution

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

Question 16

Ten years ago, the ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born during the same year. After another three years, one more member died, again at 60, and a child was born during the same year. The current average age of this eight-member joint family is nearest to

[CAT 2007]

Video Solution

Solution

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

Question 17

A function $$f (x)$$ satisfies $$f(1) = 3600$$, and $$f (1) + f(2) + ... + f(n) =n^2f(n)$$, for all positive integers $$n > 1$$. What is the value of $$f (9)$$ ?

Video Solution

Solution

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

Question 18

Suppose you have a currency, named Miso, in three denominations: 1 Miso, 10 Misos and 50 Misos. In how many ways can you pay a bill of 107 Misos?

Video Solution

Solution

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

Question 19

A confused bank teller transposed the rupees and paise when he cashed a cheque for Shailaja, giving her rupees instead of paise and paise instead of rupees. After buying a toffee for 50 paise, Shailaja noticed that she was left with exactly three times as much as the amount on the cheque. Which of the following is a valid statement about the cheque amount?

Video Solution

Solution

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.

Question 20

How many pairs of positive integers m, n satisfy 1/m + 4/n = 1/12 , where n is an odd integer less than 60?

Video Solution

Solution

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

CAT Content

CAT 2007 Question Paper

How many units should Mr. David produce daily?

What is the maximum daily profit, in rupees, that Mr. David can realize from his business?

Which of the following best describes $$a_n + b_n$$ for even n?

If p = 1/3 and q = 2/3 , then what is the smallest odd n such that $$a_n+b_n < 0.01$$?

Consider the set S = {2, 3, 4, ...., 2n+1}, where n is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of X - Y ?

A function $$f (x)$$ satisfies $$f(1) = 3600$$, and $$f (1) + f(2) + ... + f(n) =n^2f(n)$$, for all positive integers $$n > 1$$. What is the value of $$f (9)$$ ?

Suppose you have a currency, named Miso, in three denominations: 1 Miso, 10 Misos and 50 Misos. In how many ways can you pay a bill of 107 Misos?

How many pairs of positive integers m, n satisfy 1/m + 4/n = 1/12 , where n is an odd integer less than 60?

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CAT Content

CAT 2007 Question Paper

How many units should Mr. David produce daily?

What is the maximum daily profit, in rupees, that Mr. David can realize from his business?

Which of the following best describes $$a_n + b_n$$ for even n?

If p = 1/3 and q = 2/3 , then what is the smallest odd n such that $$a_n+b_n < 0.01$$?

Consider the set S = {2, 3, 4, ...., 2n+1}, where n is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of X - Y ?

A function $$f (x)$$ satisfies $$f(1) = 3600$$, and $$f (1) + f(2) + ... + f(n) =n^2f(n)$$, for all positive integers $$n > 1$$. What is the value of $$f (9)$$ ?

Suppose you have a currency, named Miso, in three denominations: 1 Miso, 10 Misos and 50 Misos. In how many ways can you pay a bill of 107 Misos?

How many pairs of positive integers m, n satisfy 1/m + 4/n = 1/12 , where n is an odd integer less than 60?

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