Using LCM and ratios to find a minimum even distribution of items
Question
A person X wants to distribute some pens among six children A, B, C, D, E and F. Suppose A gets twice the number of pens received by B, three times that of C, four times that of D, five times that of E and six times that of F. What is the minimum number of pens X should buy so that the number of pens each one gets is an even number?
Options
147
150
294
300
Explanation
Let the number of pens A receives be k. Based on the conditions: A = k, B = k/2, C = k/3, D = k/4, E = k/5, F = k/6.
For everyone to receive whole pens, k must be a multiple of the Least Common Multiple (LCM) of the denominators \{2, 3, 4, 5, 6\}. LCM(2, 3, 4, 5, 6) = 60. Let k = 60m (where m is a positive integer).
If m = 1, the distribution is: A = 60, B = 30, C = 20, D = 15, E = 12, F = 10. Here, D receives 15 pens. The prompt strictly requires that everyone gets an even number of pens.
To make the odd number (15) even without violating the ratios, multiply the entire set by 2 (m = 2): A = 120, B = 60, C = 40, D = 30, E = 24, F = 20. All values are now even integers.
Calculate the total minimum pens required: Total = 120 + 60 + 40 + 30 + 24 + 20 = 294.
Answer: (c).
Question details
Year
2022
Paper
CSAT
Question
Q47
Section
Numerical Ability
Sub-topic
LCM & Ratios
Type
Factual single
Difficulty
Hard
Source hint
Distribute pens A, B, C, D, E, F
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