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Updated on: 15 Nov 2023, 07:40
Originally posted by Sajjad1994 on 30 Apr 2021, 08:30.
Last edited by BottomJee on 15 Nov 2023, 07:40, edited 2 times in total.
Last edited by BottomJee on 15 Nov 2023, 07:40, edited 2 times in total.
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Dropdown 1: 46
Dropdown 2: greater than
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
54% (03:27) correct
46%
(03:24)
wrong
based on 721
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For each statement, select the option from the drop-down menu that completes the statement as accurately as possible according to the information provided.
1. is the smallest integer value for x that would result in a final value of 2 for d.
2. The final value of r when 15 is input for x is the final value for r when 55 is input for x.
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Dropdown 1: 46
Dropdown 2: greater than
04 Jul 2024, 22:05
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1. According to the question, initially the value of d is 1 and the value of y is 100. Now we have to find the smallest integer possible for x that will have the value of d as 2.
Now,
firstly imagine value of x as half of y, x=50. As we put it in the process we will get d as 2 and that sums up that x=50 is possible, so we will have to try a smaller value. Lets take
40. After applying it in the process we can see that d does not equal to 2, hence we take 45 as x. It does fulfill conditions of the question.
To check for a lower value put 44 in the process and we can see that it does not fulfill the condition of d as 2.
Hence 45 must be the smallest possible integer value for x when d will be 2.
(note: I have done this question thorugh trial and error method, because I find the other methods to be lengthy.)
2. Here, we have to compare the final values of r, when x is 15 and 55.
When xwill be 15, initial r will be 50. The value of d will be 4 hence we will have to divide the initial r with 2. Hence the final r will be 25.
when x will be 55, initial r will be 20. the value of d will be 2 hence we don't have to divide it with anything and our initial r is our final r.
Question asks about the difference of the r with respect to different r values.
r when x is 15 - r when x is 55 = 25-20 = 5
Now,
firstly imagine value of x as half of y, x=50. As we put it in the process we will get d as 2 and that sums up that x=50 is possible, so we will have to try a smaller value. Lets take
40. After applying it in the process we can see that d does not equal to 2, hence we take 45 as x. It does fulfill conditions of the question.
To check for a lower value put 44 in the process and we can see that it does not fulfill the condition of d as 2.
Hence 45 must be the smallest possible integer value for x when d will be 2.
(note: I have done this question thorugh trial and error method, because I find the other methods to be lengthy.)
2. Here, we have to compare the final values of r, when x is 15 and 55.
When xwill be 15, initial r will be 50. The value of d will be 4 hence we will have to divide the initial r with 2. Hence the final r will be 25.
when x will be 55, initial r will be 20. the value of d will be 2 hence we don't have to divide it with anything and our initial r is our final r.
Question asks about the difference of the r with respect to different r values.
r when x is 15 - r when x is 55 = 25-20 = 5
06 Jul 2024, 05:43
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We just need the value of d to be 2 right? So why is 46 the answer? D=2 at 45 and 46. 45 is the smallest integer and hence, the answer should be 45.
