Find t in
the figure below.
Solution
4t + t + 900 = 1800
5t = 1800
- 900
5t = 900
5t = 900 [dividing
by 4 on both sides]
5 5
c = 180
Saturday, March 8, 2014
DEGREES B9
Find c in
the figure below.
Solution
2c + c + 900 = 1800
3c = 1800
- 900
3c = 900
3c = 900 [dividing
by 4 on both sides]
3 3
c = 300
DEGREES B8
Find a in
the figure below.
Solution
3a + 1230
= 1800
3a = 1800
- 1230
3a = 1170
3a = 1170 [dividing
by 4 on both sides]
3 3
a = 390
RATIOS C8
If
2: 8 = 10 : m; Find the value of m.
Solution
2 = 10 [cross multiply]
8 m
2
x m =
10 x
8
2m
= 80
2m = 80 [divide by 2 on both sides]
2 2
m
= 40
Hence
the value of m is 40
TRY THIS…………
If
8:3 = 11: w; Find the value of w.
DEGREES B7
Find m in
the figure below.
Solution
3m + m + 960
+ 1080 = 3600
4m + 2040
= 3600 [collecting like
terms]
4m = 3600
- 2040
4m = 1560
4m = 1560 [dividing
by 4 on both sides]
4 4
m = 390
RATIOS C7
If
7: 5 = 14:B; Find the value of B.
Solution
7 = 14 [cross multiply]
5 B
7
x B =
5 x 14
7B
= 70
7B = 70 [divide by 7 on both sides]
7 7
B
= 10
Hence
the value of B is 10
TRY THIS…………
If
7:2 = 14: c; Find the value of c.
DEGREES B7
Find a in
the figure below.
Solution
2a + 80 =
1800
2a = 1800
- 800
2a = 1000
2a = 1000 [dividing
by 4 on both sides]
2 2
a = 500
ALGEBRA B59
Simplify
20(4k)
Solution
=20(4k)
=20
x 4 x k
=(20
x 4) x k
=(80)
x k
=80k
Hence 20(4k) = 80k
TRY THIS…………
Simplify
18(5e)
RATIOS C6
If
3: 5 = 9:h; Find the value of h.
Solution
3 = 9 [cross multiply]
5 h
3
x h =
5 x 9
3h
= 45
3h = 45 [divide by 3 on both sides]
3 3
h
= 15
Hence
the value of h is 15
TRY THIS…………
If
8:4 = 14: c; Find the value of c.
DEGREES B5
Find a in
the figure below.
Solution
3a + 72 =
1800
3a = 1800
- 720
3a = 1080
3a = 1080 [dividing
by 4 on both sides]
3 3
a = 360
DEGREES B6
Find a in
the figure below.
Solution
3a + 81 =
1800
3a = 1800
- 810
3a = 990
3a = 990 [dividing
by 4 on both sides]
3 3
a = 330
RATIOS C5
If
7: 5 = 14 : B; Find the value of B.
Solution
7 = 14 [cross multiply]
5 B
7
x B =
5 x 14
7B
= 70
7B = 70 [divide by 7 on both sides]
7 7
B
= 10
Hence
the value of B is 10
TRY THIS…………
If
10:4 = 14: y; Find the value of y.
ALGEBRA B58
Simplify
20p + 6m + 5m + 7p
Solution
=20p
+ 6m + 5m + 7p
=20p
+ 7p + 6m + 5m ( collecting the like
terms)
=
27p + 11m ( adding the like terms)
Hence 10p + 5m + 2m + 7p = 27p
+ 11m
TRY THIS………………..
Simplify
20p + 7m + 8m + 9p
RATIOS C4
If
6: 5 = 18 : B; Find the value of B.
Solution
6 = 18 [cross multiply]
5 B
6
x B =
5 x 18
6B
= 90
6B = 90 [divide by 6 on both sides]
6 6
B
= 15
Hence
the value of B is 15
TRY THIS…………
If
7:4 = 14: c; Find the value of c.
DEGREES B4
Find a in
the figure below.
Solution
3a + a + 160
+ 40 = 3600
4a + 2000
= 3600 [collecting like
terms]
4a = 3600
- 2000
4a = 1600
4a = 1600 [dividing
by 4 on both sides]
4 4
a = 400
RATIOS C3
If
6: 5 = 10 : W; Find the value of W.
Solution
6 = 10 [cross multiply]
5 W
6
x W =
5 x 10
6W
= 50
6W = 50 [divide by 6 on both sides]
6 6
W
= 82/6
Hence
the value of W is 82/6
TRY THIS…………
If
9:3 = 14: c; Find the value of c.
BODMAS B5
Find
the value of 30 – 100 ÷ 20 x
2
Solution
Here
we use BODMAS
1st
we divide, then we multiply, lastly we subtract.
=
30 – 100 ÷ 20 x
2
= 30 – 5
x 2
(after dividing)
=
30 – 10
(after multiplication)
=
20 (then we get 20 after
subtraction)
TRY THIS…………………………………
Find
the value of 120 – 64 ÷ 8 x
8
DEGREES B3
Find a in
the figure below
Solution
3a + 72 + 30
= 1800
3a + 1020
= 1800 [collecting like
terms]
3a = 1800
- 1020
3a = 780
3a = 780 [dividing
by 4 on both sides]
3 3
a = 260
RATIOS C2
If
6: 8 = 11:B; Find the value of B.
Solution
6 = 11 [cross multiply]
8 B
6
x B =
11 x
8
6B
= 88
6B = 88 [divide by 6 on both sides]
6 6
B
= 144/6 = 142/3
Hence
the value of B is 142/3
TRY THIS…………
If
7:2 = 11: h; Find the value of h.
DEGREES B1
Find a in
the figure below
Solution
3a + a + 72
+ 20 = 1800
4a + 920
= 1800 [collecting like
terms]
4a = 1800
- 920
4a = 880
4a = 880 [dividing
by 4 on both sides]
4 4
a = 220
RATIOS C1
If
3:2 = 11:3B; Find the value of B.
Solution
3 = 11 [cross multiply]
2 3B
3x3B =
11 x
2
9B
= 22
9B = 22 [divide by 9 on both sides]
9 9
B
= 24/9
Hence
the value of B is 24/9
TRY THIS…………
If
7:2 = 11:4B; Find the value of B.
Thursday, March 6, 2014
PERIMETER A6
Find perimeter of the circle below
solution
P = ∏ x D D= diameter
= 22 x 840120
= 22 x 120
= 2640m
Hence perimeter = 2640metres.
PERIMETER A5
Find the
perimeter of the square below.
Solution
Perimeter =
4 x
side
= 4
x 35
= 140
hence perimeter 140 mm.
AREAS A5
Area of triangle is 150cm2.
Find its base if the height is 50cm.
Solution
150 = 1
x base x height
2
Let base = b
150 = 1 x b x 50
2
150 = 1 x b x 5025, canceling by 2
150 = 1 x b
x 25
150 = 25b
6150 = 125b
125 125
b = 6cm
hence base is 6cm
TRY THIS..............
Area of triangle is 180cm2. Find its base if the height is 60cm.
PERIMETER A4
Perimeter of a rectangle
is 260cm. If its width is 50cm, find its length.
Solution
Perimeter = 2(length +
width)
260= 2(L + 50)
260=2L + 100
260-100 = 2L + 100 – 100
160 = 2L+0
160 = 2L
80160
= 1 2L
12 1 2
L = 80
Hence Length is 80cm.
TRY THIS……………
Perimeter of a rectangle
is 340cm. If its width is 40cm, find its length.
FRACTIONS C2
Find 3/5 of
200 pupils.
Solution
=3/5
x 200
=3/5
x 20040
=3 x 40
=120
120 pupils answer
TRY THIS………….
Find 3/8 of 400
pupils
PERIMETER A3
Area of rectangle is 160cm2.
If its length is 20cm, find its width.
Solution
Area = Length x width
Let width=w,
160 = 20
x w
160 = 20w
8160 = 120w
1
20 120
w = 8cm
Hence width is 8cm.
FRACTIONS C1
Find 3/5 of 400
pupils.
Solution
=3/5
x 400
=3/5
x 40080
=3 x 80
=240
240 pupils answer
TRY THIS………….
Find 3/20
of 800 pupils
AREAS A4
Area of rectangle is 80cm2.
if its length is 16cm, find its width.
Solution
Area = Length x width
Let width=w,
80 = 16
x w
80 = 16w
580 = 116w
1 16 116
w = 5cm
Hence width is 5cm.
TRY THIS...............
Area of rectangle is 64cm2. if its length is 16cm, find its width.
BODMAS B4
Evaluate 50 ÷ 2 – 7.
Solution
We apply
BODMAS.
= 50 ÷ 2 – 7
= 25 - 7 [after division]
= 18
[after subtraction]
TRY THIS…………
Evaluate 90 ÷ 9 – 16
PERIMETER A2
Perimeter of a rectangle
is 200cm. If its width is 40cm, find its length.
Solution
200 = 2(length + width)
200= 2(L + 40)
200=2L + 80
200-80 = 2L + 80 – 80
120 = 2L+0
120 = 2L
60120
= 1 2L
12 1 2
L = 60
Hence Length is 60cm.
TRY THIS........................
Perimeter of a rectangle is 300cm. If its width is 50cm, find its length.
BODMAS B3
Evaluate 60 ÷ 2 – 7
Solution
We apply
BODMAS.
= 60 ÷ 2 – 7
= 30 - 7 [after division]
= 23
[after subtraction]
TRY THIS…………
Evaluate 93 ÷ 3 – 1
AREAS A3
Area of rectangle is 240cm2.
Find length if width is 10cm.
Solution
Area = Length x width
240
= L x 10
240 = 10L
24240 = 110L
1 10 110
L = 24cm
Hence length is 24cm
TRY THIS....................
Area of rectangle is 140cm2. Find length if width is 10cm.
ALGEBRA B57
Simplify
11(4k)
Solution
=11(4k)
=11
x 4 x k
=(11
x 4) x k
=(44)
x k
=44k
Hence 11(4k)=44k
TRY THIS…………
Simplify
18(2k)
BODMAS B2
Find
the value of 30 – 60 ÷ 20 x
2
solution
Here
we use BODMAS
1st
we divide, then we multiply, lastly we subtract.
=
30 – 60 ÷ 20 x
2
=
30 – 3 x 2 (after dividing)
=
30 – 6
(after multiplication)
=
24 (then we get 24 after
subtraction)
TRY THIS…………………………………
Find
the value of 100 – 60 ÷ 10 x
8
AREAS A3
Area of rectangle is 550cm2.
Find length if width is 5cm.
Solution
Area = Length x width
550 = L
x 5
550 = 5L
110550
= 15L
1 5 15
L = 110cm
Hence length is 110cm
TRY THIS..............
Area of rectangle is 90cm2. Find length if width is 5cm.
BODMAS B1
Find
the value of 11 – 7 + 3 – 1 + 5 – 4
solution
Here
we use BODMAS
There
is only addition and subtraction, We add first before we subtract
=
11 – 7 + 3 – 1 + 5 – 4
=
11 + 3 + 5 – 7– 1– 4 (rearranging
positives and negatives)
=
19 – 7– 1– 4 (dealing
with positives first we get 19)
=
19 – 12 (then we add the
negatives to get -12)
=
7 (then we get 7 after
subtraction)
Hence
11 – 7 + 3 – 1 + 5 – 4 = 7
TRY THIS............
Find the value of 16 – 7 + 3 – 1 + 3 – 6
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