Question 1
LEVEL 6
The sides of a cuboid are measured to be \text{length}=4 cm, \text{height} =5 cm, and \text{width} =6 cm, to the nearest cm. Find the interval for the volume, V, of this cuboid.
Select the correct answer from the list below:
A: 86.625 \leq V <160.875
B: 64.32 \leq V <65.82
C: 60 \leq V <210
D: 116.337\leq V <204.695
CORRECT ANSWER: A: 86.625 \leq V <160.875
WORKED SOLUTION:
To find the interval for volume we need to find the smallest and largest possible values for the volume of the cuboid. Because the volume of a cuboid is found by doing l \times h \times w, we find the lower bound by multiplying the smallest values for length, height and width; similarly for upper bound, we use the largest values.
We need to do \pm5 lots of the units to the right of where the value was rounded to. Here, out values were round to the nearest unit, so we need to look at the tenths.
\pm0.5
Lower bound
4-0.5=3.5
5-0.5=4.5
6-0.5=5.5
Upper Bound
4+0.5=4.5
5+0.5=5.5
6+0.5=6.5
We can now create our interval, remembering that the lower bound uses a non-strict inequality and the upper bounds uses a strict inequality.
86.625 \leq V <160.875
Question 2
LEVEL 6
The area of a square is measured to be 25 cm^2 to the nearest cm^2. Find the interval for length of the sides, l.
Select the correct answer from the list below:
A: 4 \leq l <6
B: 4.95 \leq l <5.05
C: 4.5 \leq l <5.5
D: 24.5 \leq l <25.5
CORRECT ANSWER: B: 4.95 \leq l <5.05
WORKED SOLUTION:
To find the interval for the side lengths, i.e. the smallest and largest lengths they can be, we have to find the smallest and largest possible values for the area of the square. We can find these values by seeing what the area might have been rounded from.
The area was rounded to the nearest cm^2, so we need to look at the tenths.
\pm0.5
Smallest value
25-0.5=24.5Largest value
25+0.5=25.5
The bounds for the side lengths can now be found from these areas.
Lower bound
\sqrt{24.5}=4.95Upper bound
\sqrt{25.5}=5.05
Giving an interval of:
4.95 \leq l <5.05
Question 3
LEVEL 6
A plane flies from Manchester to Chengdu. The distance between these cities is 8300km to the nearest 100 km. The average speed of the plane was 880 km/h to the nearest 10 km/h. Find an interval for the time in hours, t taken for this flight.
Use the formula speed=\frac{distance}{time}
Select the correct answer from the list below:
A: 9.21 \leq t <9.65
B: 10 \leq t <11.25
C: 8.53 \leq t <8.88
D: 9.32 \leq t <9.54
CORRECT ANSWER: D: 9.32 \leq t <9.54
WORKED SOLUTION:
To find the time taken for a journey we need to rearrange or speed/distance/time formula.
speed=\frac{distance}{time}
time =\frac{distance}{speed}
So, to find an interval for the time, we need to find the upper and lower bounds of the fraction \frac{distance}{speed}.
Lower Bound
To find the lower bound we need the fraction to be as small as possible, which we do by making the denominator as big as possible and the numerator as small as possible.
8300-50=8250
880+5=885
time =\frac{distance}{speed} =\frac{8250}{885}=9.32
Upper Bound
To find the upper bound we need the fraction to be as big as possible, which we do by making the denominator as small as possible and the numerator as big as possible.
8300+50=8350
880-5=875
time =\frac{distance}{speed} =\frac{8350}{875}=9.54
Putting lower and upper bounds together gives:
9.32 \leq t <9.54
Question 4
LEVEL 6
Umar buys 5 bags of gravel. Each bag of gravel weighs 5000 g to the nearest 100 g. Find an interval for the total weight in g, W, of the 5 bags of gravel.
Select the correct answer from the list below:
A: 24750 \leq W < 25250
B: 24975 \leq W < 25025
C: 4950 \leq W < 5050
D: 4995 \leq W < 5005
CORRECT ANSWER: A: 24750 \leq W < 25250
WORKED SOLUTION:
To find the interval for the total weight, we need to find the smallest and largest possible values for the weight of one bag of gravel. We then need to multiply the smallest value by 5 to find the smallest possible value for total weight, and similarly for the largest total weight.
We need to do \pm5 lots of the units to the right of where the value was rounded to. Here, our values were rounded to the nearest 100 g, so we need to look at 10 g.
\pm 50 g
Lower Bound
5000 - 50 = 4950 g
4950 \times 5 = 24750 g
Upper Bound
5000 + 50 = 5050 g
5050 \times 5 = 25250 g
Putting lower and upper bounds together gives:
24750 \leq W < 25250
Question 5
LEVEL 6
Adam has 11000 followers on his social media account, to the nearest 1000. Bethany has 5200 followers to the nearest 100. Find an interval for the difference, D, between the number of followers Adam has and the number of followers Bethany has.
Select the correct answer from the list below:
A: 5250 \leq D < 6350
B: 5350 \leq D < 6250
C: 5800 \leq D < 5900
D: 5200 \leq D < 6400
CORRECT ANSWER: A: 5250 \leq D < 6350
WORKED SOLUTION:
To calculate the lower bound for the difference, we need to subtract the largest number of followers Bethany has from the smallest number of followers Adam has.
To calculate the upper bound for the difference, we need to subtract the smallest number of followers Bethany has from the largest number of followers Adam has.
Adam:
Smallest possible value is 11000-500=10500
Largest possible value is 11000+500=11500
Bethany:
Smallest possible value is 5200-50=5150
Largest possible value is 5200+50=5250
Difference:
Lower bound = 10500-5250=5250
Upper bound = 11500-5150=6350
Putting lower and upper bounds together gives:
5250 \leq D < 6350