Algebra

  1. There were x pigeons and y mynahs in a cage. One fine morning p of them escaped to freedom. If the bird keeper, knowing only that p = 7, was able to figure out without looking into the cage that at least one pigeon had escaped, then which of the following does not represent a possible (x,y) pair?
    (a) (10,8)
    (b) (7,2)
    (c) (25,6)
    (d) (12,4)

2. What is the sum of the following series: 1/(1 × 2)+1/(2 × 3)+1/(3 × 4)+ …….. +1/(100 × 101) (a) 99/100           (b) 1/100             (c) 100/101             (d) 101/102

3. The roots of the equation ax^2 + 3x + 6 = 0 will be reciprocal to each other if the value of a is
(a) 3
(b) 4
(c) 5
(d) 6

4. N the set of natural numbers is partitioned into subsets S1 = (1), S2 = (2,3), S3 ={4,5,6), S4 = {7,8,9,10} and so on. The sum of the elements of the subset S50 is

(a) 61250

(b) 65525

(c) 42455

(d) 62525

Solutions:

1. For the bird keeper to figure out that at least 1 pigeon had escaped, the number of mynahs has to be less than 7. In other words, y < 7. Hence the pair (10,8) is not a valid one.

2. 1/(1× 2)+1/(2×3)+1/(3× 4)+−−−+1/(100 ×101)

=( 1−1/2) +( 1/2 −1/3)+ (1/3 −1/ 4)+ — + (1/99 – 1/100) + (1/100-1/101)

= 1 – 1/101

= 100/101.

3. If we are to take out the first elements of each set we find them as : 1, 2, 4, 7, 11, 16….. This series is neither an AP nor a GP, but the difference between the terms viz.1, 2, 3, 4, 5 ….. is in AP with a=1 and d=1. Hence to find the 50th term of the original series we have to add the sum of 49 terms of the second series to the first term of the original series. Since the difference series is the natural number series, the sum of first 49 terms = (49 x 50)/2 = 1225. Hence the 50th term of the original series = (1225 + 1) = 1226. This will be the first element of the set S50, which will have 50 elements ie. The last element will be 1275. So, the sum of the elements in this set is given as :   n(a + l)/2 = 50x(1226 + 1275)/2 = 62525.

4. If the roots are reciprocal of each other their product = 1. But product of roots in our equation = 6/a. Therefore 6/a = 1 a=6

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