Mathematics Test with Solutions
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.
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.
Page 1 of 30
© www.testfunda.com
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
Page 2 of 30
© www.testfunda.com
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.