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Q.1. The price of Darjeeling tea (in rupees per kilogram) is 100 + 0.10n, on the nth day of 2007 (n = 1, 2… 100), and then remains constant. On the other hand, the price of Ooty tea (in rupees per kilogram) is 89 + 0.15n, on the nth day of 2007 (n = 1, 2… 365). On which date in 2007 will the prices of these two varieties of tea be equal? (CAT 2007)
(1) May 21 (2) April 11 (3) May 20 (4) April 10 (5) June 30

Q.2 Let f(x) = max (2x + 1, 3 – 4x), where x is any real number. The minimum possible value of f(x) is (CAT 2006)
(1) 1/3 (2) 1/2 (3) 2/3 (4) 4/3 (5) 5/3

Q.3 The number of roots common between the two equations x^3 + 3x^2 + 4x + 5 = 0 and x^3 + 2x^2 + 7x + 3 = 0 is (CAT 2003)
(1) 0 (2) 1 (3) 2 (4) 3

Q.4. The remainder when 2^256 is divided by 17 is (CAT 2002)
(1) 7 (2) 13 (3) 11 (4) 1

Q.5 Consider a sequence where then nth term,
t_n=n/((n+2)) ,n=1,2,…
The value of t_3 * t_4 * t_5 * t_6 * … * t_53 equals: (CAT 2006)
(1) 2/495 (2) 2/477 (3) 12/55 (4) 1/1485 (5) 1/2970

Solutions:

Q.1. The price of Darjeeling tea (in rupees per kilogram) is 100 + 0.10n, on the nth day of 2007 (n = 1, 2… 100), and then remains constant. On the other hand, the price of Ooty tea (in rupees per kilogram) is 89 + 0.15n, on the nth day of 2007 (n = 1, 2… 365). On which date in 2007 will the prices of these two varieties of tea be equal? (CAT 2007)

Solution:

If we look at options, all are post April 10. Simple counting will show that January (31), February (28), March (31) add up to 90 days (n=90). On 10th April, price of Darjeeling Tea will be 100+0.1(100) = 110 and will remain constant for the year after that.

Clearly, looking at the equation, we can rule out April 10 and April 11 as they both will be lower. So we just need to find from the options, the value of n which when substituted in 89+0.15n gives the answer is 110. Without solving this equation, we must realize that n has to be an even number to give an integral answer.

n (May 21) = 141 Ruled out
n (May 20) = 140 Answer
n (June 30) = 181 Ruled out

Q.2 Let f(x) = max (2x + 1, 3 – 4x), where x is any real number. The minimum possible value of f(x) is (CAT 2006)

Solution:

Equate. 2x + 1 = 3 – 4x
6x = 2
X = 1/3. Hence f(x) = max (2/3 + 1, 3 – 4/3) = 5/3

Q.3 The number of roots common between the two equations x^3 + 3x^2 + 4x + 5 = 0 and x^3 + 2x^2 + 7x + 3 = 0 is (CAT 2003)

Solution:

If we equate these two, we get x^2 – 3x + 2 = 0
This gives us x as either 2 or 1. In both the equations, if we put these values we will never get zeroes as to get zero on RHS, we need to have negative value of x. The answer is 0.

Q.4. The remainder when 2^256 is divided by 17 is (CAT 2002)

Solution:

Use Fermat’s theorem for finding out the remainders easily.
Here 2 and 17 are coprimes. Hence Fermat theorem can be used.
Therefore, 216 divided by 17 gives a remainder of 1. Since 256 is a multiple of 16, the remainder will remain 1

Or you can simplify this further as (2^4)64 = 16^64 mod 17. 16 when divided by 17 gives -1 as the remainder. (-1)^64 = 1

Q.5 Consider a sequence where then nth term,
t_n=n/((n+2)) ,n=1,2,…
The value of t_3 * t_4 * t_5 * t_6 * … * t_53 equals: (CAT 2006)

Solution:

t_1=1/3
t_2=2/4
t_3=3/5
Every (n+2)th term numerator will get crossed out by denominator of nth term.
Alternate approach: The numerator is nothing but 53!/2!. The denominator is nothing but 55!/4!
(53!*4!)/(55!*2!) = 2/495

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