suminha wrote:
nick1816 wrote:
\(5*10^{20} ≤ N_h < 5*10^{21}\)
\(5*10^{20} ≤ N_o < 5*10^{21}\)
Statement 1-\(5*10^{20} ≤ N_h < 3*10^{21}\)......
(1)\(5*10^{20} ≤ N_o < 3*10^{21}\)......
(2)Combining (1) and (2), wen can get the range of \(N_h\)+\(N_o\).
\(10^{21} ≤ N_h+N_o < 6*10^{21}\)
Case 1- If \(10^{21} ≤ N_h+N_o < 5*10^{21}\)
\(N_h\)+\(N_o\)is rounded off to \(10^{21}\)
Case 2- \(5*10^{21} ≤ N_h+N_o < 6*10^{21}\)
\(N_h\)+\(N_o\) is rounded off to \(10^{22}\)
InsufficientStatement 2-
\(10^{21} ≤ N_h < 5*10^{21}\)......
(1)\(0.5*10^{21} ≤ N_o < 2.5*10^{21}\)......
(2)Hence,
\(1.5*10^{21} ≤ N_h+N_o < 7.5*10^{21}\)
Case 1- If \(1.5*10^{21} ≤ N_h+N_o < 5*10^{21}\)
\(N_h\)+\(N_o\) is rounded off to \(10^{21}\)
Case 2- \(5*10^{21} ≤ N_h+N_o < 7.5*10^{21}\)
\(N_h\)+\(N_o\) is rounded off to \(10^{22}\)
InsufficientCombining both statements\(10^{21} ≤ N_h < 3*10^{21}\)......
(1)\(0.5*10^{21} ≤ N_o < 1.5*10^{21}\)......
(2)\(1.5*10^{21} ≤ N_h+N_o < 4.5*10^{21}\)
Hence, \(N_h\)+\(N_o\) rounded off to \(10^{21}\)
Sufficient gmatt1476 wrote:
Given a positive number N, when N is rounded by a certain method (for convenience, call it Method Y), the result is \(10^n\) if and only if n is an integer and \(5*10^{n − 1} ≤ N < 5*10^n\). In a certain gas sample, there are, when rounded by Method Y, \(10^{21}\) molecules of \(H_2\) and also \(10^{21}\) molecules of \(O_2\). When rounded by Method Y, what is the combined number of \(H_2\) and \(O_2\) molecules in the gas sample?
(1) The number of \(H_2\) molecules and the number of \(O_2\) molecules are each less than \(3*10^{21}\).
(2) The number of \(H_2\) molecules is more than twice the number of \(O_2\) molecules.
DS36141.01
I did understand until «Combining (1) and (2), wen can get the range of \(N_h\)+\(N_o\). \(10^{21} ≤ N_h+N_o < 6*10^{21}\)», but I don’t understand after that.
How did you arrive to make a conclusion that “If \(10^{21} ≤ N_h+N_o < 5*10^{21}\), \(N_h\)+\(N_o\)is rounded off to \(10^{21}\), and If \(5*10^{21} ≤ N_h+N_o < 6*10^{21}\), \(N_h\)+\(N_o\) is rounded off to \(10^{22}\).”
And at the statement 2, \(10^{21} ≤ N_h < 5*10^{21}\), where did 5 go??? Wasn’t it
\(5\)\(*10^{20} ≤ N_h < 3*10^{21}\)??? How did you arrive to \(10^{21} ≤ N_h < 5*10^{21}\), and \(0.5*10^{21} ≤ N_o < 2.5*10^{21}\)[/b]. Hence, \(1.5*10^{21} ≤ N_h+N_o < 7.5*10^{21}\) from \(N_h>2*N_o\)???
Seems like everyone understood what your saying but me....
Please help me 🙏🏻
Thank you in advance!
Posted from my mobile deviceHi
I know i'm very late , you probably don't need it anymore but this might help someone. (PS- i tried my best explaining the answer as it took me also a very long time to understatnd it)
This question has been made this way so that most of the test takers skip it
If you dive deep into its logic, it's pretty simple
I'll try to explain
Normal approximation: Lets take a number , for example- 1.7
In normal approximation , you would round it off to 2 because it lies in the range 1.5
< NUMBER GIVEN < 2.5 and the numbers from 1.5 to 2 would be rounded up to 2 and numbers greater than 2 to numbers < 2.5 would be rounded down to 2.
The numbers in the given range after approximation will become equal to the mid value of the range i.e. 2
BACK TO THE ORIGINAL QUESTION the result is 10^n if and only if n is an integer and 5∗10^n−1≤N<5∗10^n
This is the same kind of approxmation I have explained above
The numbers in the given range after approximation will become equal to the mid value of the range i.e. 10^n
The range is a complicated because it involves exponents and variables but look through it, it's just a number with many many zeros in it and range for our approximation is large
the method Y is nothing but our usual approximation but with a very large range
Now , THE EXACT NUMBER OF oxygen and hydrogen molecules lies in this given range ( 5∗10^n−1≤N<5∗10^n )
THE QUESTION WANTS US TO ( STEP 1) FIND OUT THE EXACT NUMBER OF O2 AND H2 molecules----->2 ADD THEM UP ---->3 Find combined number of H2 and O2 molecules in the gas sample AFTER APPROXIMATION/ROUNDING UP
since we want a unique answer and the answer is a rounded up value, we don't really need an exact value of number of molecules, WE WANT VALUE THAT GIVES A UNIQUE ANSWER
STATEMENT 1: The number of H2 molecules and the number of O2 molecules are each less than 3∗10^21
CASE 1Step 1: if No. EXACT NUMBER of O2 and H2 molecules is 2.5*10^21 each
Step 2: ADDING THEM UP : TOTAL= 5*10^21
Step 3: Round them--> since 5*10^21 lies outside the range given in the question and it actually lies in the next possible range i.e 5∗10^21≤N<5∗10^22 whose mid-value is 10^22 hence after rounding up the value of the number of molecules willl be 10^22
CASE 2Step 1: if No. OF EXACT NUMBER of O2 and H2 molecules is 2*10^21 each
Step 2: ADDING THEM UP : TOTAL= 4*10^21
Step 3: Round them--> since 4*10^21 lies inside the range given in the question , approxmated value is 10^21
2 POSSIBLE ANSWERS , HENCE NOT SUFFICIENT
STATEMENT 2 The number of H2 molecules is more than twice the number of O2 molecules.
CASE 1Step 1: if No. OF EXACT NUMBER of O2 molecules is 1*10^21 and H2 molecules is 2.1*10^21 ( 2.1 is more than 2*1=2)
( Also notice, the no of molecules of O2 and H2 satisfies the range given for each of them)
Step 2: ADDING THEM UP : TOTAL= 3.1*10^21
Step 3: Round them--> since 3.1*10^21 lies intside the range given in the question, approximated value will be 10^21
CASE 2Step 1: if No. EXACT NUMBER of O2 molecules is 2.3*10^21 and H2 molecules is 4.7*10^21 ( 4.7 is more than 2*2.3=4.6)
Step 2: ADDING THEM UP : TOTAL= 7*10^21
Step 3: Round them--> since 7*10^21 lies outside the range given in the question and it actually lies in the next possible range i.e 5∗10^21≤N<5∗10^22 whose mid-value is 10^22 hence after rounding up the value of the total number of molecules willl be 10^22
2 POSSIBLE ANSWERS , HENCE NOT SUFFICIENT
COMBINING 1 & 2Step 1: if EXACT NUMBER of O2 molecules is 1.4*10^21 and H2 molecules is 2.9*10^21 ( 2.9 is more than 2*1.4=2.8)
( Also notice, we are taking the maximum permissible values)
Step 2: ADDING THEM UP : TOTAL= 4.3*10^21
Step 3: Round them--> since 4.3*10^21 lies inside the range given in the question, approximated value will be 10^21
we can try the same with other values as well but we won't get another answer as we have already tried out the maximum possible values . Taking minumum values will complicate the calculations so i just took the maximum possible values
WE ARE GETTING A UNIQUE ANSWER HENCE ---> answer : C Hope it helps !!!