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# Maths year 11

1. (Original post by K-Man_PhysCheM)
OK, if you have 64 100 000 to 3 significant figures, what other numbers would round to it to 3 significant figures?

For example, would 64 110 000 round to 64 100 000? What about 64 090 000?

An easier way to approach it might be to take off all the excess zeros for now, so you have the number 641. What is the lower bound of 641, if this is only accurate to 3 sig figs? Then, move that back up to the millions.

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2. (Original post by K-Man_PhysCheM)
Not quite.
Would 640.8 round up to 641?
Or 640.1?

What's the lowest number that would round up to 641?

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yes, yes, no
I got 640.95

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3. (Original post by z_o_e)
I got 640.95

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But does 640.7 round up to 641? And is 640.7 lower than 640.95?

Yes, it does, and yes, 640.7 is lower than 640.95. So can 640.95 really be the lower bound? Or is the answer lower?
4. (Original post by K-Man_PhysCheM)
But does 640.7 round up to 641? And is 640.7 lower than 640.95?

Yes, it does, and yes, 640.7 is lower than 640.95. So can 640.95 really be the lower bound? Or is the answer lower?
Think answer might be something lower.

No I think 640.95 is the lower bound as it it closer to 641

And 641.05 is close to 641 and is the UB

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5. (Original post by z_o_e)
Think answer might be something lower.

No I think 640.95 is the lower bound as it it closer to 641

And 641.05 is close to 641 and is the UB

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Remember, upper and lower bounds aren't about proximity: it's about what a measurement could have been before it was rounded to the stated degree of accuracy.

Say a tree is 240.5cm tall. I might round this height to three significant figures to get 241cm. If the tree is 240.49cm tall, I would have to round it down to 240cm (to 3sf). This means that the lower bound of 241cm must be 240.5cm. Now apply this to 641.
6. (Original post by K-Man_PhysCheM)
Remember, upper and lower bounds aren't about proximity: it's about what a measurement could have been before it was rounded to the stated degree of accuracy.

Say a tree is 240.5m tall. I might round this height to three significant figures to get 241m. If the tree is 240.49m tall, I would have to round it down to 240m (to 3sf). This means that the lower bound of 241m must be 240.5m. Now apply this to 641.
641.49?

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7. That would be the upper bound (though we would just write 641.5, because the upper bound is 641.49999999999999999999 =~ 641.5)

But the question asks for the minimum population. So find the lower bound and scale it up to the millions.
8. (Original post by K-Man_PhysCheM)
That would be the upper bound (though we would just write 641.5, because the upper bound is 641.49999999999999999999 =~ 641.5)

But the question asks for the minimum population. So find the lower bound and scale it up to the millions.
Lb =640.5

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9. (Original post by z_o_e)
Lb =640.5

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Correct. That is the lowest number that would round up to 641 (accurate te 3sf). And then you need to scale that back up to the millions, as expressed in the question.

Just to test that you really understand this, find the upper and lower bounds of the following:

13 (accurate to 2 sf)

990 (accurate to 3 sf)

10.34 (accurate to 2 decimal places)
10. (Original post by K-Man_PhysCheM)
Correct. That is the lowest number that would round up to 641 (accurate te 3sf). And then you need to scale that back up to the millions, as expressed in the question.

Just to test that you really understand this, find the upper and lower bounds of the following:

13 (accurate to 2 sf)

990 (accurate to 3 sf)

10.34 (accurate to 2 decimal places)

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12. The first one was correct, but this second one is sneaky.
A number has been rounded to three significant figures to give 990.
995 rounded to 3sf is 995, and 985 rounded to 3sf is 985 ---> not 990.

Your bounds need to be numbers which, when rounded to 3 sig figs, give 990. For example, 990.152 rounded to 3sf is 990. However, 992.3 rounded to 3sf is 992, NOT 990.

This question was sneaky, but you need to really be careful to note what the degree of accuracy is. In this case, 990 has been rounded to 3 sf, so the upper and lower bounds will have 4 significant figures, which will round to the 3 sig figs of 990.
13. (Original post by K-Man_PhysCheM)
The first one was correct, but this second one is sneaky.
A number has been rounded to three significant figures to give 990.
995 rounded to 3sf is 995, and 985 rounded to 3sf is 985 ---> not 990.

Your bounds need to be numbers which, when rounded to 3 sig figs, give 990. For example, 990.152 rounded to 3sf is 990. However, 992.3 rounded to 3sf is 992, NOT 990.

This question was sneaky, but you need to really be careful to note what the degree of accuracy is. In this case, 990 has been rounded to 3 sf, so the upper and lower bounds will have 4 significant figures, which will round to the 3 sig figs of 990.

Yeah i need to work on the second one more. I still don't understand It that clearly. Llike significant figures are after the decimal point so I could have 989.999 which would round to 990. And I could have 990.095 I think
14. (Original post by z_o_e)
Yeah i need to work on the second one more. I still don't understand It that clearly. Llike significant figures are after the decimal point so I could have 989.999 which would round to 990. And I could have 990.095 I think

Ahh, there's a bit of a misunderstanding there.

For significant figures, we count all the digits after (and including) the first non-zero digit.
Say we have 2508.715

This number to one significant figure is just the first non-zero figure, ie the first figure. The first non-zero figure in (000000)2508.715 is 2. Now you round the figure furthest to the right: in this case, just the 2. There's a 5 after the 2, so it rounds up to 3. Hence, 2508.715 to 1sf is 3000.

To 2sf, we take the first 2 numbers after the first non-zero number, including the first non-zero number. 2508.715 to 2 sf is 2500.

To 3sf, we look at the three digits after (including) the first non-zero digit. These are 2, 5 and 0. 0 is furthest to the right (most precise) and we need to decide to round it. To the right of the 0 is an 8, so we round the 0 up to a 1. So 2508.715 to 3sf is 2510.

To 4sf, we repeat this again. Take the first 4 digits and round the one furthest to the right. We get 2509, to 4sf.

The zero rule is important, because if we have a number like 0.0000060127, we only count the first non-zero digit as the first significant figure. So this to 1sf is 0.000006, to 2sf is 0.0000060, to 3sf: 0.00000601, etc...

What is 2047 rounded to 3 significant figures?
15. [QUOTE=K-Man_PhysCheM;67073246]Ahh, there's a bit of a misunderstanding there.

For significant figures, we count all the digits after (and including) the first non-zero digit.
Say we have 2508.715

This number to one significant figure is just the first non-zero figure, ie the first figure. The first non-zero figure in (000000)2508.715 is 2. Now you round the figure furthest to the right: in this case, just the 2. There's a 5 after the 2, so it rounds up to 3. Hence, 2508.715 to 1sf is 3000.

To 2sf, we take the first 2 numbers after the first non-zero number, including the first non-zero number. 2508.715 to 2 sf is 2500.

To 3sf, we look at the three digits after (including) the first non-zero digit. These are 2, 5 and 0. 0 is furthest to the right (most precise) and we need to decide to round it. To the right of the 0 is an 8, so we round the 0 up to a 1. So 2508.715 to 3sf is 2510.

To 4sf, we repeat this again. Take the first 4 digits and round the one furthest to the right. We get 2509, to 4sf.

The zero rule is important, because if we have a number like 0.0000060127, we only count the first non-zero digit as the first significant figure. So this to 1sf is 0.000006, to 2sf is 0.0000060, to 3sf: 0.00000601, etc...

What is 2047 rounded to 3 significant figures?[/QUOTE

GOT IT!!
2047= 2050
16. (Original post by z_o_e)

GOT IT!!
2047= 2050
Correct! To consolidate your understanding, try and answer these questions (progressively more difficult):

1) Express the following to 2 sig figs:
a) 431 208
b) 30 800
c) 0.05081

2) Express the following to 3 sig figs:
a) 3.14159
b) 500.6
c) 0.03006

3) The population of Andorra is 79 000, expressed to 2 significant figures. What are the upper and lower bounds?
17. (Original post by K-Man_PhysCheM)
Correct! To consolidate your understanding, try and answer these questions (progressively more difficult):

1) Express the following to 2 sig figs:
a) 431 208
b) 30 800
c) 0.05081

2) Express the following to 3 sig figs:
a) 3.14159
b) 500.6
c) 0.03006

3) The population of Andorra is 79 000, expressed to 2 significant figures. What are the upper and lower bounds?

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19. They are correct! Though if I'm really picky, in c) you shouldn't write the extra zeroes after the second significant figure, because that implies that you have rounded to 4sf instead of 2sf.

For example, 9.00001 to 3sf is 9.00, and to 2sf it is 9.0. If you were to write 9.00000, that would imply accuracy to 6sf instead of 2 or 3sf, and really it is 9.00001 to 6sf. They can sometimes be very picky with that, so just be very careful with not writing extra zeroes after the decimal point.
20. Again, kind of correct, but a) and b) have too many excess 0s.
3.14159 is itself expressed to 6sf, and 3.14000 would imply rounded to 6sf, which is wrong. 3.14 (without excess 0s) is better.

c) is completely correct.

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