1. Year 10 Exam Revision
1. Round to the stated accuracy…
a) 3.4985 = (2dp)
b) 45300 = (1sf)
c) 0.03491 = (2sf)
2. How many sides are there on a…
a) Hexagon
b) Nonagon
c) Rhombus
3. Solve for x…
a) 4x – 21 = 479
b) 9x + 7 = 3x + 79
c) 3(x + 8) + 3 = 42
4. Find the area of the following shapes
430m
210m
60m
a)
b)
190m
5. Convert the areas from question 4 into
hectares.
6. Give the equations for the following
graphs…
●(0,6)
(9,0)
●
a)
b) (7,3)
●
7. What % did Gary get if he scored 83
right out of 112 questions.
3.50
50000
0.035
6
9
4
x = (479 + 21) ÷ 4 = 125
9x–3x=79–7 6x=72 x = 12
x = (42– 3)÷ 3 – 8 = 5
A = (430+210) ÷ 2 x 60
A = 19200m2
A = 0.5πr2
A = 0.5x 3.14159 x 952
A = 14176.4m2
(1dp)
A = 19200m2
A = 1.92ha
A = 14176.4m2
(1dp) A = 1.41764ha
y = mx + c
m = rise/run
c = y intercept
y = –6
/9x + 6
83/112 x 100 =74%

2. Year 10 Exam Revision
8. Simplify the following…
a) 3e + 7f + 8f – e =
b) (3e2
)3
=
c) 5ef x 2ef x 3f2
=
9. Increase 180 by 35%.
10. Convert the following measures…
a) 3400cm = m
b) 0.45km = m
c) 8000cm3
= litres
11. What is the scale factor of this
enlargement?
5cm
12. What is the length of the side marked x
in question 11.
13cm
object image
xcm
4cm
13. What is the shaded area of the following
shape
21m
14. Find the length of the missing side.
32m
75m
x
2e + 15f
3e2
x 3e2
x 3e2
= 27e6
30e2
f4
old x (1 ± %) = new
180 x 1.35 = 243
34
450
8
k = image/object k = 13/5 = 2.6
x = 4 x 2.6 = 10.4cm
area = π x r2
area = π x 212
area = 1385.44236…
area = 1385m2
(0dp)
a2
+ b2
= c2
c
a
b
322
+ x2
= 752
1024 + x2
= 5625
x2
= 5625 - 1024
x2
= 4601
x = 67.8m (1dp)

3. Year 10 Exam Revision
15. Simplify the following fractions
a) 9 + 7 =
20 10
b) 3 x 4 =
5 5
c) 6a – 3a =
5 4
16. Find the missing numbers…
a) 35, 42, 49, ___, ___, ___
b) 1, 4, 9, 16, ___, ___, ___
c) 1600, 1200, 900, 675, ___, ___
17. Calculate the size of the missing
angles…
123o
a
b
c
d
e
293o
18. From the following data calculate the
statistics required.
35, 42, 71, 33, 42, 63, 48, 54, 60, 39
a) mean
b) median
c) mode
d) range
e) upper quartile
f) lower quartile
19. Rewrite these standard forms into normal
numbers…
a) 4.9 x 10 6
=
b) 9.03 x 10 –4
=
20. Rewrite these normal numbers into standard
form…
a) 99,000,000 =
b) 0.00000057 =
21. Name these quadrilaterals…
9 +
20
14 =
20
23
20
12
25
24a –
20
15a =
20
9a
20
56 63 70
25 36 49
506.25, 379.6875
= 57o
= 123o
= 67o
= 67o
= 46o
= 487 ÷ 10 = 4.87
between 5th
& 6th
= 45
33, 35, 39, 42, 42, 48, 54, 60, 63, 71
= 42
high - low= 71 – 33 = 38
= 60
= 39
4,900,000
0.000903
9.9 x 107
5.7 x 10-7
parallelogram arrowhead
trapezium

4. Year 10 Exam Revision
22. The ratio of cordial to water should be 2:15.
If I have 70ml of cordial how much drink can I
make?
25. Calculate the missing side…
23. The car travelled 76km in 50minutes.
a) what is its speed in km/min?
b) what is its speed in km/hour?
c) how many hours and minutes would it take to
drive 150km?
24. Simplify the following algebraic
expressions…
a) 55e4
f9
= b) 8e3
x 4e5
=
5ef3
c) = d) (3e)0
=
4
9e
x
24o
170m
26. Factorise the following…
a) 3ab + 30bc =
b) e2
+ 19e + 90 =
c) e2
– 16 =
27. Find the missing angles…
a
●
b
156o
28. What percentage change is it to go from 59
up to 70?
1 to 7.5
70 x 7.5 = 525ml
speed = distance/time
76/50 =1.52km/min
50min = 0.833333hr
76÷0.833333hr = 91.2km/hr
150 ÷ 1.52 = 99mins (0dp)
= 1 hour and 39mins
11e3
f6
32e8
3e2 1
S
O
H C
A
H T
O
A
H
A x = 170 ÷ cos24
x = 186m (0dp)
3b( )a+ 10c
(e )(e )+ 9 + 10
(e )(e )– 4 + 4
(n-2)x180÷n
a = (5-2)x180÷5
a = 108o
b = 78o
(new-old)÷old x 100 = % change
(70-59)÷59 x 100 =19% (0dp)

5. Year 10 Exam Revision
1. Round to the stated accuracy…
a) 3.4985 = (2dp)
b) 45300 = (1sf)
c) 0.03491 = (2sf)
2. How many sides are there on a…
a) Hexagon
b) Nonagon
c) Rhombus
3. Solve for x…
a) 4x – 21 = 479
b) 9x + 7 = 3x + 79
c) 3(x + 8) + 3 = 42
4. Find the area of the following shapes
430m
210m
60m
a)
b)
190m
5. Convert the areas from question 4 into
hectares.
6. Give the equations for the following
graphs…
●(0,6)
(9,0)
●
a)
b) (7,3)
●
7. What % did Gary get if he scored 83
right out of 112 questions.

6. Year 10 Exam Revision
8. Simplify the following…
a) 3e + 7f + 8f – e =
b) (3e2
)3
=
c) 5ef x 2ef x 3f2
=
9. Increase 180 by 35%.
10. Convert the following measures…
a) 3400cm = m
b) 0.45km = m
c) 8000cm3
= litres
11. What is the scale factor of this
enlargement?
5cm
12. What is the length of the side marked x
in question 11.
13cm
object image
xcm
4cm
13. What is the shaded area of the following
shape
21m
14. Find the length of the missing side.
32m
75m
x

7. Year 10 Exam Revision
15. Simplify the following fractions
a) 9 + 7 =
20 10
b) 3 x 4 =
5 5
c) 6a – 3a =
5 4
16. Find the missing numbers…
a) 35, 42, 49, ___, ___, ___
b) 1, 4, 9, 16, ___, ___, ___
c) 1600, 1200, 900, 675, ___, ___
17. Calculate the size of the missing
angles…
123o
a
b
c
d
e
293o
18. From the following data calculate the
statistics required.
35, 42, 71, 33, 42, 63, 48, 54, 60, 39
a) mean
b) median
c) mode
d) range
e) upper quartile
f) lower quartile
19. Rewrite these standard forms into normal
numbers…
a) 4.9 x 10 6
=
b) 9.03 x 10 –4
=
20. Rewrite these normal numbers into standard
form…
a) 99,000,000 =
b) 0.00000057 =
21. Name these quadrilaterals…

8. Year 10 Exam Revision
22. The ratio of cordial to water should be 2:15.
If I have 70ml of cordial how much drink can I
make?
25. Calculate the missing side…
23. The car travelled 76km in 50minutes.
a) what is its speed in km/min?
b) what is its speed in km/hour?
c) how many hours and minutes would it take to
drive 150km?
24. Simplify the following algebraic
expressions…
a) 55e4
f9
= b) 8e3
x 4e5
=
5ef3
c) = d) (3e)0
=
4
9e
x
24o
170m
26. Factorise the following…
a) 3ab + 30bc =
b) e2
+ 19e + 90 =
c) e2
– 16 =
27. Find the missing angles…
a
●
b
156o
28. What percentage change is it to go from 59
up to 70?