SECTION B: DATA INTERPRETATION AND QUANTITIATIVE ABILITY

 

 

In the questions 64-65, one statement is followed by three conclusions. Section the appropriate answer from the options given below.

(A) Using the given statement, only conclusion I can be derived.

(B) Using the given statement, only conclusion II can be derived.

(C) Using the given statement, only conclusion

III can be derived.

(D) Using the given statement, conclusions I, II and III can be derived.

(E) Using the given statement, none of the three conclusions I, II and III can be derived.

 

64. A0, A1, A2, …… is a sequence of numbers with A0 = 1, A1 = 3,

      and At = (t + 1)At-1 – t At-2 for t = 2, 3, 4, ….

      Conclusion I.   A8 = 77

      Conclusion II.   A10 = 121

      Conclusion III.  A12 = 145.

 

Solution:

At = (t + 1) At –1 – tAt –2

It'll be convenient to arrange the terms in two columns as shown below.

A0 = 1                        A1 = 3

A2 = 3 (3) – 2 (1) = 7   A3 = 4 (7) – 3 (3) = 19

A4 = 5 (19) – 4 (7) = 67    

A5 = 6 (67) – 5(19) = 307

A6 = 7(307) – 6(67) = 1747

We can see that none of the conclusions follow.                                                                                                Choice (E)

 

65. A, B, C be real numbers satisfying A < B < C, A + B + C = 6 and AB + BC + CA = 9

      Conclusion I.   1 < B < 3

      Conclusion II.   2 < A < 3

      Conclusion III.  0 < C < 1

 

Solution:

A + B + C = 6  ®  (1) (given)

      AB + BC + CA = 9      ®  (2) (given)

      Squaring (1) we get

      (A + B + C)2 = 62

      A2 + B2 + C2 + 2AB + 2BC + 2CA = 36 ® (3)

      Þ from (2) and (3) we get

      A2 + B2 + C2 + 2(9) = 36

      A2 + B2 + C2 = 18  ®  (4)

If A = 1, B = 1 and C = 4, the equations (1) and (4) will be satisfied.

      Conclusion I: 1 < B < 3.

Suppose B £ 1, then A must be decreased by some value (say x)

\ C must be increased by at least x so that A + B + C = 6 to be satisfied.

Now, if we consider the new value of A2 + B2 + C2, it will be greater than 18, as the value of C (greatest of the three) is increasing.

For example, if A = 0.9, B = 1 and C = 4.1, then A2 + B2 + C2 > 18.

\ B > 1.

If B > 3, then A + B + C will be greater than 6 as A < B < C.

\ 1 < B < 3 is always true.

      Conclusion II: 2 < A < 3.

If A > 2, then A + B + C > 6 as A < B < C.

Conclusion III: 0 < C < 1.

If C < 1, then A + B + C < 3 as A < B < C.

\ Only conclusion I can be derived.                                                                                                                   Choice (A)