Mathematics Test with Solutions

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CAT 2007 Solutions

Section I 1. Note that the price of Darjeeling tea remains constant after the 100th day (n=100). If the prices of the two varieties of tea become equal before n = 100, then 100 + 0.1n = 89 + 0.15n ∴ n = 220, which is not possible. (Since n has been assumed to be less than 100) ∴ The prices of the two varieties will be equal after n = 100, i.e., when the price of Darjeeling tea = 100 + 0.1 × 100 = 110 ∴ 89 + 0.15n = 110 ∴ n = 140 2007 is not a leap year. Number of days till 30th April = 31 + 28 + 31 + 30 = 120 The prices of the two varieties will be equal on 20th May. Hence, option 3. 2. Let f(x) = px2 + qx + k, where p, q and k are integers, p, 0 ∴ f(0) = k = 1 ∴ f(x) = px2 + qx + 1 f(x) = px2 + qx + k f’(x) = 2px + q When f’(x) = 0, x = −q/2p = 1 f(x) attains maximum at x = 1 ∴ q = −2p f(1) = p + q + 1 = 3 ∴1–p=3 ∴ p = −2 ∴q=4 ∴ f(x) = −2x2 + 4x + 1 ∴ f(10) = −200 + 40 + 1 = −159 Hence, option 2.

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CAT 2007 Solutions

3.

P and Q do not lie within the intersection of the two circles. So they lie on the circumferences or outside the circumferences. If they lie on the circumferences, ∠ AQP = 60o From the diagram, if they lie outside the circumferences, ∠ AQ’P’ < 60o Also, ∠ AQP would be 0o if A, Q and P were collinear. But as P and Q cut each other in two distinct points, A, Q and P cannot be collinear. ∴ ∠ AQP > 0o ∴ ∠ AQP lies between ∠0o and ∠60o. Hence, option 3. 4. Enemies of every pair are the pairs formed with all numbers other than the two in the member itself. ∴ If there are n elements then each member has

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CAT 2007 Solutions

Hence, option 4. 5. Two members are friends if they have one element in common. ∴ All the members having one constituent as the common element are common friends. There are (n – 3) such friends. Also, one pair formed by the uncommon constituents of the two friends is a common friend. ∴ There are n – 3 + 1 = n – 2 common friends. Hence, option 4. 6. Let Shabnam have Rs. 100 to invest. Let Rs. x, Rs. y and Rs. z be invested in option A, B and C respectively. ∴ x + y + z = 100 ... (I) If there is a rise in the stock market, returns = 0.001x + 0.05y – 0.025z If there is a fall in the stock market, returns = 0.001x – 0.03y + 0.02z Now, x, y and z are such that regardless of whether the market rises or falls, they give the same return, which is the maximum guaranteed return. ∴ 0.001x + 0.05y – 0.025z = 0.001x – 0.03y + 0.02z ∴ y/z = 9/16 Now, consider different possible values of x, y and z. The returns are as follows: x 75 50 25 0 y 9 18 27 36 z 16 32 48 64 Returns = 0.001x + 0.05y – 0.025z 0.125 0.15 0.175 0.2

We see that as the values of y and z increase, the returns increase. ∴ The returns are maximum when x = 0%, y = 36% and z = 64%

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CAT 2007 Solutions

The maximum returns are 0.2%. Hence, option 3. 7. As shown by the table formulated in the first question, maximum returns are guaranteed by investing 36% in option B and 64% in option C. Hence, option 2. 8. Let the speed of the plane be x kmph. Then the speed from B to A is (x – 50) kmph and that from A to B is (x + 50) kmph. Note that the plane travels from B to A, halts for 1 hour and travels back to B, all in 12 hrs. ∴ 3000/(x – 50) + 1 + 3000/(x + 50) = 12 Now consider options for this question. We can easily see that x= 550 satisfies the above expression. Speed of plane = 550 kmph Now, the plane takes 3000/500 = 6 hrs to travel from B to A. It reaches A when the time at B is 8:00 am + 6 hrs = 2:00 p.m. => The time difference between A and B is 1 hour. Hence, option 1. 9. As calculated in the first question, the cruising speed of the plane is 550 kmph. Hence, option 2. 10. Let aabb (a ≠ 0, a and b being single digits) be a perfect square. aabb = 1100a + 11b = 11(100a + b) Also, as aabb is a perfect square, it is a multiple of 121. ∴ aabb = 121K, where K is also perfect square. K ≥ 9 121...
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