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Bunuel
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Bunuel
150 = 2*3*5^2. Hence, for 150x to be a multiple of 5^2, 3^2, and 2^2, x must contribute a 2 and a 3. Thus, the smallest possible value for x is 2*3 = 6. Answer: B.
How on the exam situation, you can quickly differentiate X as a number that will multiply 150 instead of a single or 2 digit number.

Or the question will be more precise on that case
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Bunuel
150 = 2*3*5^2. Hence, for 150x to be a multiple of 5^2, 3^2, and 2^2, x must contribute a 2 and a 3. Thus, the smallest possible value for x is 2*3 = 6. Answer: B.
How on the exam situation, you can quickly differentiate X as a number that will multiply 150 instead of a single or 2 digit number.

Or the question will be more precise on that case
Exactly. ­If "150x" were a four-digit number, it would have been explicitly mentioned. Without such clarification, "150x" can only represent "150*x", as the multiplication sign (*) is typically omitted.
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Hello everyone! Does someone has reminder type excercises?
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Thank you very much Bunuel
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Is there any short method to this question? Thanks!
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Is there any short method to this question? Thanks!
Only lcm method may be helpful if think its a shortcut
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PrabhatKC
Is there any short method to this question? Thanks!
Take total work as 60.
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As the work remaining will be = 1 - “Work completed”, subtract all the reciprocals by 1 and multiply them to get the remaining fraction of work
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Remaining work is 24?/60?
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i did assuming 15 as total then a completed 2/3rd remained only 5 now b completed 1/5 of remaining now remained only 4 same way c completed 1/4 finally only 3 remained so 3/15 that is 1/5
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This is the short cut my teacher said but I don’t get why we are subtracting 1 from the individual work done and multiplying them
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PrabhatKC
This is the short cut my teacher said but I don’t get why we are subtracting 1 from the individual work done and multiplying them
Yes, this can be a direct formula. Reason we are subtracting from 1 is to get the remaining work of each of the worker A, B, C
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Why not subtracting
1. /1-2/3= 1/3
2. 1/3-1/5=2/15
3. 2/15-1/4=7/60
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A did 2/3 of the work, so remaining work of A = (1-2/3). The remaining work of A is done by B but only 1/5 of (1-2/3) is completed so the remaining work of B = (1-1/5)(1-2/3) and the remaining work of B is to be done by C but she does 1/4*(1-2/3)(1-1/5) and the remaining work of C is then (1-1/4)(1-1/5)(1-2/3). Since we don’t have anyone after C, so the remaining work is (1-1/4)(1-1/5)(1-2/3)

Hope it is clear!
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Extension-Oil7460
Why not subtracting 1. /1-2/3= 1/3 2. 1/3-1/5=2/15 3. 2/15-1/4=7/60
Because each person after A is doing a "fraction" of the remaining work by the person immediately before. What you said would be true if A,B,C are independently working on the same task. Example if you will, if A,B,C work "together" to create a house, then the method you propose is applicable. However, if A works on the house first and completes 2/3 of it and then B works on the remaining house = (1-2/3) and completes a fraction of the remaining house 1/5*(1-2/3) so remaining house is whatever B has left (1-1/5)*(1-2/3), and then C comes in to complete his fraction of the house which is 1/4, so finally remaining portion of the house in absolute terms is (1-1/4)(1-1/5)(1-2/3)
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Hi

Good morning sir and Ma
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