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2020 WAEC MATHEMATICS QUESTIONS AND ANSWERS

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MATHEMATICS OBJ

MATHS-OBJ

1 -10  CBCDACDCCD

11-20 AADBDACBBC

21-30 BDDABDADAD

31-40 CDACCCCCDA

41-50 BBBCDCACDB

Completed.

 

MATHEMATICS THEORY

WAEC MATHS ANSWERS

(1a)

Given A={2,4,6,8,…}

B={3,6,9,12,…}

C={1,2,3,6}

U= {1,2,3,4,5,6,7,8,9,10}

A’ = {1,3,5,7,9}

B’ = {1,2,4,5,7,8,10}

C’ = {4,5,7,8,9,10}

A’nB’nC’ = {5, 7}

(1b)

Cost of each premiere ticket = $18.50

At bulk purchase, cost of each = $80.00/50 = $16.00

Amount saved = $18.50 – $16.00

=$2.50

2ai).

p = (rk/Q -ms)2/3

p3/2 = (rk/Q -ms)2/8 * 3/2

p3/2 = rk/Q = ms

p3/2 + ms = rk/Q

q(p3/2 + ms ) = rk

Q =rk/p3/2 + ms

2aii)

Where :

r = 10, k = 4, s = 0.2

p = 3, m = 15

Q = 10 * 4/3^3/2 + (15 * 0.2)

Q = 40/5.2 + 3

Q = 40/8.2

Q = 4.88

(4a)

Total Surface Area = 224πcm²

r:l = 2:5

r/l = 2/5

Cross multiply

2l/2 = 5r/2

L = 5r / 2

Total surface = πrl + πr²

= πr (l + r)

24π/π = πr (5r/2 + r )/ π

224 = 5r²/2 + r²/1

L.c.m = 2

448 = 5r² + 2r²

448 / 7= 7r²/7

r² = 64

r = √64 = 8cm

L = 5*8/2 = 20cm

(4b)

Volume = 1/2πr²h

= 1/3 * 22/7 * 8 * 8 * 18.33

= 1228.98cm³

L² = h² + r ²

20² = h² + 8²

400 – 64 = h²

h² = 336

h = √ 336

h = 18.33cm

(5a)

Total income = 32+m+25+40+28+45

=170+m

PR(²)=m/170+m = 0.15/1

M=0.15(170+m)

M=25.5+0.15m

0.85m/0.85=25.5/0.85

M=30

(5b)

Total outcome = 170 + 30 = 200

(5c)

PR(even numbers) = 30+40+50/200

=115/200 = 23/40

Number: 8

Let ms Maureen income = x

1/4 x = shopping mall

1/3 x = at an open market

Hence shopping mall and open market = 1/4x. + 1/3x

=3x+4x /12. =.  7/12x

Hence de remaining amount

=X-7x/12.

=12x-7x/12.  = 5x/12

Then 2/5(5x/12) = mechanic workshop

=2x/12 *x/6

Amount left = #225,000

Total expenses

7/12 * x/6 *225,000. =#x

7x+2x+2,700,000/12 = #x

9x+2,700,000 /12 = #x

9x+2,700,000= 12x

2,700,000= 12x-9x

2,700,00 =3x

X= 2,700,000/3

=#900,000

8aii:) amount spent on open market

=1/3x=1/3*900,000

=#300,000

 

(9a)

Draw the triangle

(9b)

(i)Using cosine formulae

q² = x² + y² – 2xycosQ

q² = 9² + 5² – 2×9×5cos90°

q² = 81 + 25 – 90 × 0

q² = 106

q = square root 106

q = 10.30 = 10km/h

Distance = 10 × 2 = 20km

(ii)

Using sine formula

y/sin Y = q/sin Q

5/sin Y = 10.30/sin 90°

Sin Y = 5 × sin90°/10.30

Sin Y = 5 × 1/10.30

Sin Y = 0.4854

Y = sin‐¹(0.4854), Y = 29.04

Bearing of cyclist X from y

= 90° + 19.96°

= 109.96° = 110°

(9c)

Speed = 20/4, average speed = 5km/h

(10)

10 and graph complete

=======================================

11b)  8y = – 4x + 24

Y = – 4/8 x + 24/8

Y = – 1/2 x + 3

Gradient = m = – 1/2

:. Y- y1 = m(x-x)

Y- 12 = 1/2 (x + 8)

2(y-12) = 1 (x+8)

2y – 24 = -x-8

X + 2y – 24 + 8 = 0

X + 2y – 24 + 8 = 0

X + 2y – 16 = 0

Is the line above

12

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Posted by on August 18, 2020.

Categories: Waec, WAEC RUNS QUESTIONS AND ANSWERS

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One Response

  1. Good

    by thankgod on Oct 6, 2020 at 10:45 pm

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