Question
If ‘1x5629’ is a six digit number which is divisible
by 9, then which of the following can be the minimum value of ‘x’?Solution
For a number to be divisible by ‘9’, the sum of its digits must be divisible by ‘9’.So, (1 + x + 5 + 6 + 2 + 9) = (23 + x)So, number after 23 which is multiple of 9 is 27,Therefore, minimum value of ‘x’ = 27 – 23 = 4
Statements:
B < C ≤ I = O; C ≥ J ≥ K ≥ N; Z ≤ M ≤ N
Conclusions:
I) O > Z
II) O = Z
...Which of the following symbols should replace the question marks (?) in the given expression in order to make the expression ‘G ≤ J’ definitely tr...
Statements: X @ Y % M % N; M $ P $ Z
Conclusions : I. Y % Z II. X @ N �...
In the question, assuming the given statements to be true, find which of the conclusion (s) among given two conclusions is/are definitely true and then ...
In the following questions assuming the given statements to be true, find which of the conclusion among given conclusions is/are definitely true and the...
Statement- 1 - 6#≥9#≥10#≤5#
2 - 7*≤10#˃5*≥6*
Conclusions:
1) 7* > 6*
2) 6* > 7*
3) 6# ≤ 5#<...
Which of the following symbols should be placed in the blank spaces respectively (in the same order from left to right) in order to complete the given e...
Statements: Q $ Z % C # T @ H
Conclusions:
I. Z # H
II. Q © T
III. H % Z
Statement: P `>=` U > S `<=` F < Z; I `>=` F > M
Conclusion: I. P > M II. I `>=` Z
...Statements:
J $ R % U % C
Conclusions:
I. R © C
II. J * U
III. C % J
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