GCSE 9-1 maths: round, estimate, truncate, upper, lower bounds, error interval
The moon is an average of 384,400 km away from the Earth, but an answer of 'about 380,000 miles', is fine.
We can round off numbers to the nearest 10,000, nearest 1,000, nearest 100, nearest 10, nearest whole number, or any other specified number.
Example: Round 384,400 to the nearest thousand.
First, look at the digit in the thousands place. It is 4. So the number lies between 384,000 and 385,000.
Look at the digit to the right of the 4. It is 4. So 384,400 is closer to 384,000 than 385,000.
The rule for rounding
5 or more, we 'round up' , 4 or less, it stays as it is.
Q1: Round 151 to the nearest 100
Q2: Round 151 to the nearest 10
Q3: Round 151.4 to the nearest whole number
Q4: Round 155 to the nearest 10
We can round off the number up to a certain number of decimal places.
Be careful, this is not the same as rounding off using significant figures!
Example write 1.15 to 1 decimal place (dp)
Find the number in the 2nd decimal place - it is 5 so we round up. To 1 dp , 1.15 is 1.2
Q1: Round 3.738058 to 2 dp
Q2: Round 8.8762123 to 1 dp.
Truncation is a method of approximating a decimal number by dropping all decimal places past a certain point without rounding.
Truncate 2.718281828 to 4 dp
1. Find the 4th decimal place and ignore everything after this.
2.718281828 truncated to 4 dp is 2.7182
2. Truncate 345.6 to a whole number
Find the units place and ignore everything after this.
345.6 truncated to a whole number is 345
(note: correct to a whole number it is 346)
Q1: Truncate 12.099 to 1 dp
Q2: Truncate 1299.09 to the hundreds
Another method of approximating an answer is to round off using significant figures.
With the number 738,249, the 7 is the most significant digit, because it tells us that the number is 7 hundred thousand and something. Then the 3 is the next most significant, and so on.
With the number 0.00321, the 3 is the most significant digit, because it tells us that the number is 3 thousandth and something. Then the 2 is the next most significant, and so on.
Be careful with numbers with a zero, such as 90123, the 9 is the first significant figure and 0 the second, because of its value as a place holder.
Round off a number to a certain number of significant figures, usually to 1, 2 or 3 significant figures.
Q1: What is 156,456 correct to 1 sfs
Q2: What is 156,456 correct to 2 sfs?
Q3: What is 156,456 correct to 3 sfs?
Q4: What is 0.0015047 correct to 1, 2 and 3 sfs?
Q5: What is 1.99 correct to 2 sfs?
We can use significant figures to get an approximate answer to a problem. Look at these examples.
Look at the first example and round all the numbers to 1 sf to make 'easier' numbers.
It changes to: 200 × 40 / 0.5
Now the division is simpler and is: 8000 / 0.5 which is 16,000
Round the 2nd example to 1sf, and the 3rd example to 2sf
If a value has been rounded, it is useful to know what possible values the exact value could have been.
Example: An estimate of my height is 176 cm to the nearest cm. That means it could be any height from 175.5 cm to 176.5cm and these are called the lower and upper bounds.
The lower bound is the smallest value that would round up to the estimated value.
The upper bound is the smallest value that would round up to the next estimated value.
To calculate the upper and lower bounds half the degree of accuracy specified then add this to the rounded value for the upper bound and subtract it from the rounded value for the lower bound.
In my example, half the accuracy of 1cm is 0.5cm, so 0.5cm was added and subtracted to/from 176cm
Q1: What is the upper bound and lower bound of 33 kg, to the nearest kg
Q2: Find UB and LB of 190 cm,to the nearest mm
Q3: Find UB and LB of 10.12 seconds, to 2 dp
Q4: Find UB and LB of £2.50, to the 2 sfs
Q5: Find UB and LB of 100 to the 2 sfs
A piece of A4 paper is 11.3 inches by 15.7 inches.
What are the lower and upper bounds of its area, in square inches, correct to 1 dp?
Halve one decimal place is 0.05 inches
To find the lower bound for the area multiply the lower bounds together
i.e. 11.25 × 15.65 = 176.1 inches2 (1 dp)
To find the upper bound for the area multiply the upper bounds together
i.e. 11.35 × 15.75 = 178.7 inches2 (1 dp)
Q6: Henry ran 100m in 12.6 sec. both correct to 1 dp.
Find the UB and LB for his speed.
The Error interval is the range of values (between the upper and lower bounds) in which the precise value could be.
An estimate of my height is 176 cm to the nearest cm. That means it could be any height from 175.5 cm to 176.5cm and these are called the lower and upper bounds.
The error interval is 175.5 ≤ height < 176.5
i.e. Lower bound ≤ x < Upper bound
Q4: A number (n) rounded to the nearest 10 is 50. Which is the error interval
A. 40 ≤ n < 60 B. 40 < n ≤ 60
C: 45 ≤ n < 55 D: 45 < n ≤ 55
Q5: A number (n) rounded to the nearest whole number is 50. Which is the error interval
A. 49.9 ≤ n < 50.1 B. 49.5 < n ≤ 50.5
C: 45 ≤ n < 55 D: 49 < n ≤ 51
Find the error interval of 12.34 truncated to 2dp
The minimum number could have been 12.34, the maximum could have been 12.35
So Error interval is 12.34 < n ≤ 12.35
Q6: Find the error interval of 12.345 truncated to 3dp
Q7: Find the error interval of 123 truncated to a whole number
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