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Three digits numbers which are divisible by 6 are 102, 108, 114....996This will form an arithmetic progression havingFirst term (a) = 102Common difference (d) = 6 and last term (a n ) = 996We know, a n = a + (n - 1) X dOr, 996 = 102 + (n - 1) X 6Or, (n - 1) = {(996 - 102) ÷ 6Or, (n - 1) = 149Or, 'n' = 150So, number of 3 digit numbers which are divisible by 6 = 150Similarly, number of 3-digit numbers divisible by 9 = {(999 - 108) /9} + 1 = 100Number of 3 digits numbers divisible by both 6 and 9 are the numbers divisible by LCM of (6, 9) , i.e., 18 = {(990 - 108) /18} + 1 = 50So, the required numbers are = 150 + 100 - 50 = 200
In the question, assuming the given statements to be true, find which of the following conclusion(s) among the two conclusions is/are true and then giv...
Statements: B ≥ C > D; B < E > J; G > A ≥ H > J
Conclusion:
I. D ≤ A
II. G > C
Statements: P % Q, Q & R, R $ S, S # Z
Conclusions:
I. P & R
II. R # Z
Statements: R = S ≥ T, U < Q < W = T, U = Z > V
Conclusions:
I. R > U
II. S ≥ Z
III. V < T
Statements: F > V > W ≥ L > G; F ≤ O = M < I
Conclusions: I. M > L II. V < I
Statement: D < F; D ≥ E > G; I ≥ H > F
Conclusion:
I. G ≥ F
II. H ≥ D
Statements:
I ≤ A ≤ S = X < L; N > W > G ≥ P ≥ S
Conclusions:
I. G ≥ A
II. N > L
If we arrange all the letters of the word ‘SUPPLIERS’ in alphabetical order from the left end, then, which letter will remain at its initial positio...
Statements: H + L & F; F + V $ U; A * U * R
Conclusions:
I. H + U
II. A * F
III. R + L
Statements: N ≥ M ≥ O; U < N; V < O ≤ R
Conclusions:
I. V < N
II. R ≥ N
III. O < U
