# Jamb 2020 Mathematics Mock Expo Answer

Mathematics 2019 JAMB Past Questions

Mathematics 2019 JAMB Past Questions

1.Make q the subject of the formula in the equation mna2−pqb2=1

A. q=b2(mn−a2)a2p

B. q=m2n−a2p2

C. q=mn−2b2a2

D. q=b2(n2−ma2)n

Explanation

mna2−pqb2=1

mna2−1=pqb2

mn−a2a2=pqb2

pq=b2(mn−a2)a2

q=b2(mn−a2)a2p

2. The angle of elevation of the top of a tree from a point on the ground 6cm away from the foot of the tree is 78°. Find the height of the tree correct to the nearest whole number.

A. 148m

B. 382m

C. 282m

D. 248m

Explanation

tan78=h60

h=60tan78

h=60×4.705=282.27m

≊ 282m to the nearest whole number.

3. A binary operation ⊗

is defined by m⊗n=mn+m−n on the set of real numbers, for all m, n ∈ R. Find the value of 3 ⊗ (2 ⊗4).

A. 6

B. 25

C. 15

D. 18

Explanation

m⊗n=mn+m−n

3 ⊗ (2 ⊗ 4)

2 ⊗ 4 = 2(4) + 2 – 4 = 6

3 otimes 6 = 3(6) + 3 – 6 = 15

4

Age in years 7 8 9 10 11

No of pupils 4 13 30 44 9The table above shows the number of pupils in a class with respect to their ages. If a pie chart is constructed to represent the age, the angle corresponding to 8 years old is

A. 48.6°

B. 56.3°

C. 46.8°

D. 13°

Explanation

Total number of pupils : 4 + 13 + 30 + 44 + 9 = 100

The number of 8 – year olds = 13

The angle represented by the 8-year olds on the pie chart = 13100×360°

= 46.8°

5. In a class of 50 students, 40 students offered Physics and 30 offered Biology. How many offered both Physics and Biology?

A. 42

B. 20

C. 70

D. 54

Explanation

n(Total) = 50

n(Physics) = 40

n(Biology) = 30

Let n(Physics and Biology) = x

n(Physics only) = 40 -x

n(Biology only) = 30 – x

40 – x + 30 – x + x = 50

70 – x = 50

x = 20

6. Rationalize 2√+3√2√−3√

A. −5−26–√

B. −5+32–√

C. 5−23–√

D. 5+26–√

Explanation

2√+3√2√−3√

= (2√+3√2√−3√)(2√+3√2√+3√)

= 2+6√+6√+32−6√+6√−3

= 5+26√−1

= −5−26–√

7

Find the length of the chord |AB| in the diagram shown above.

A. 4.2 cm

B. 4.3 cm

C. 3.2 cm

D. 3.4 cm

Explanation

Length of chord = 2rsin(θ2)

= 2(3)sin(682)

= 6sin34

= 6×0.559

= 3.354 cm ≊ 3.4 cm

8.Given sin58°=cosp°

, find p.

A. 48°

B. 58°

C. 32°

D. 52°

Explanation

sinθ=cos(90−θ)

sinθ=cos(90−58)

= cos32

9

23÷4514+35−13

A. 3150

B. 2031

C. 3120

D. 5031

Explanation

23÷4514+35−13

23÷45=23×54

= 56

14+35−13=15+36−2060

= 3160

∴23÷4514+35−13=56÷3160

= 56×6031

= 5031

10

If 6×3+2×2−5x+1

divides x2−x−1

, find the remainder.

A. 9x + 9

A. 2x + 6

B. 6x + 8

C. 5x – 3

Explanation

11.If a fair coin is tossed 3 times, what is the probability of getting at least two heads?

A. 23

B. 45

C. 25

D. 12

Explanation

The outcomes are {HHH, HHT, HTT, HTH, THH, THT, TTH, TTT}

P(at least two heads) = 48

= 12

12. In how many ways can the word MATHEMATICIAN be arranged?

A. 6794800 ways

B. 2664910 ways

C. 6227020800 ways

D. 129729600 ways

Explanation

MATHEMATICIAN = 13 letters with 2M, 3A, 2T, 2I.

Hence, the word MATHEMATICIAN can be arranged in 13!2!3!2!2!

= 129729600 ways

13

Given matrix M = ∣∣∣∣−2050−16463∣∣∣∣

, find MT+2M

A. ∣∣∣∣−460206152∣∣∣∣

B. ∣∣∣∣−60140−31813189∣∣∣∣

C. ∣∣∣∣50321461−7∣∣∣∣

D. ∣∣∣∣−40100−2128−166∣∣∣∣

Explanation

M = ∣∣∣∣−2050−16463∣∣∣∣

MT

= ∣∣∣∣−2040−16563∣∣∣∣ 2M = ∣∣∣∣−40100−2128126∣∣∣∣

MT

+ 2M = ∣∣∣∣−60140−31813189∣∣∣∣

14

Score (x) 0 1 2 3 4 5 6

Freq (f) 5 7 3 7 11 6 7Find the mean of the data.

A. 3.26

B. 4.91

C. 6.57

D. 3.0

15

Score (x) 0 1 2 3 4 5 6

Freq (f) 5 7 3 7 11 6 7Find the variance

A. 3.42

B. 4.69

C.