Question
If the sum of sweets in carton M and N is 88 and sum of
sweets in carton N and P is 66, then what is the probable number of sweets in carton L? Answer the questions based on the information given below. Seven cartons (J, K, L, M, N, O and P) have different numbers of sweets. M has more sweets than N but less than K. J has more sweets than P but less than O. L doesn’t have lowest number of sweets but less than N. N doesn’t have 4th highest number of sweets. O has more sweets than N but less than M. Carton, which has 2nd highest number of sweets, has 52 sweets.Solution
M + N = 88 52 + N = 88 N = 88 – 52 = 36. N + P = 66 36 + P = 66 P = 66 – 36 = 30. L will have number of sweets between 30 and 36. M has more sweets than N but less than K. L doesn’t have lowest number of sweets but less than N. N doesn’t have 4th highest number of sweets. K > M > N > L J has more sweets than P but less than O. O has more sweets than N but less than M. Carton, which has 2nd highest number of sweets, has 52 sweets. K > M (52) > O > N > J > L > P The final arrangement is as follow: K > M (52) > O > J > N > L > P
Choose the missing term.
3F, 6G, 11I, 18L,?
60, 100, 160, 240, ?, 460
1 12 31 58 93 ?
32, 48, 72, 108, ?, 243
36   37    33   42    ?      51    15
...1500 Â Â Â 150Â Â Â 30Â Â Â ? Â Â Â Â 3.6Â Â Â Â 1.8
9 28 83 250 ? 2248
4    9    28    99    ?   2105   12660
...12, 25, 51, ?, 207, 415Â
18    ?    20    13     22     11
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