Question
In a shop, there are a total of 15,000 chip packets. Out
of these, 28% are Green Lays packets, and out of the remaining, 75% are Red Lays packets, and the rest are Blue Lays packets. If 20% of Green Lays packets, 10% of Red Lays packets, and 15% of Blue chip packets were rejected for quality, then how many chip packets, in total, were rejected for quality?Solution
ATQ, Number of Green lays packets in the shop = 15000 × 0.28 = 4200 Number of Red lays packets in the shop = (15000 – 4200) × 0.75 = 8100 Number of Blue lays packets in the shop = 10800 – 8100 = 2700 Number of Green lays packets rejected for quality = 4200 × 0.20 = 840 Number of Red lays packets rejected for quality = 8100 × 0.10 = 810 Number of Blue lays packets rejected for quality = 2700 × 0.15 = 405 So, total number of chips packets rejected for quality = (840 + 810 + 405) = 2055
Statements: M % C & G @ T $ D; W % M # PÂ
Conclusions :Â Â Â Â Â I. D % CÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â II. M % GÂ Â Â Â Â Â Â Â Â Â Â Â Â ...
Statements: G > N > P = E ≥ H < L; M < E < B < C = Q > X; U > W > Y = Q > H
Conclusions:
I). U > P
II). Y > P
...Statements: A > B; C > D; E ≥ A; F = C; C < B
Conclusions:
(i) B > D (ii) A > F (iii) F < E
...Which of the following symbols should replace the question mark in the given statement in order to make conclusion 'B>Z' as well as 'C>X' definitely tr...
Statements: F % W, W © R, R @ M, M $ D
Conclusions:
 I.D @ R                               II.M $ F�...
Statements: H > S ≥ F = B ≤ U≤ T; E ≤ B ≤ K
Conclusions:I. K > F II. K = F
Statements:  B > K < Y, E > C ≥ O = Y
Conclusions:
I. C > B
II. E ≤ Y
III. E > K
IV. O ≥ K
...Statements: B > D = C ≥ E ≥ G, C = H ≤ I < F
Conclusions:
I. B > H
II. I ≥ G
III. F > DStatement: E < F ≤ G = H, I ≥ G ≤ J ≤ K
Conclusion: I. K > EÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â II. H > K
...Statement: W>Y<X<Z=U>S; W<T ≥V
I. Y<T
II. X > V
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