2015 - JAMB Mathematics Past Questions and Answers - page 2
11
Simplify √30 × √40
A
10 √3
B
5 √3
C
20 √3
D
15 √3
correct option: c
√30 × √40
√(30 × 40)
√(3 × 10 × 4 × 10)
√(400 × 3)
√400 × √3
20 × √3
20√3
Users' Answers & Comments√(30 × 40)
√(3 × 10 × 4 × 10)
√(400 × 3)
√400 × √3
20 × √3
20√3
12
If √ (2x + 2) − √x =1 ,find x.
A
x = 1 twice
B
x = 0 or 1
C
x = 3 or 1
D
x = 2 twice
correct option: a
√(2x + 2 ) − √x = 1
√(2x + 2) = 1 + √x
Square both sides
√(2x + 2)2 = (1 + √x)2
((2x + 2)2)½ = 1 + √x + √x+ x
2x + 2 = 1 + 2√x +x
Collect the like term together
2x − x + 2 − 1 = 2 √x
x + 1 = 2√x
Square both sides
(x + 1)2 = (2√x)2
x2 + 2x + 1 = 4x
x2 + 2x − 4x + 1 = 0
(x2− x)(x + 1)= 0
x(x − 1) −1(x − 1)
(x − 1)(x −1)
Either (x − 1) = 0 or (x − 1) = 0
x = 1 or 1
x = 1 twice
Users' Answers & Comments√(2x + 2) = 1 + √x
Square both sides
√(2x + 2)2 = (1 + √x)2
((2x + 2)2)½ = 1 + √x + √x+ x
2x + 2 = 1 + 2√x +x
Collect the like term together
2x − x + 2 − 1 = 2 √x
x + 1 = 2√x
Square both sides
(x + 1)2 = (2√x)2
x2 + 2x + 1 = 4x
x2 + 2x − 4x + 1 = 0
(x2− x)(x + 1)= 0
x(x − 1) −1(x − 1)
(x − 1)(x −1)
Either (x − 1) = 0 or (x − 1) = 0
x = 1 or 1
x = 1 twice
13
The sum of two numbers is 5; their product is 14. Find the numbers.
A
x = 5 or 1
B
x = 7 or 1
C
x = 7 or 0
D
x = 7 or − 2
correct option: d
Let x represent the first number;
Then, the other is (5 − x) , since their sum is 5 and their product is 14
x(5 − x) = 14
5x − x2 = 14
x2 − 5x − 14 = 0
(x2 − 7x) + (2x − 14) = 0
x(x − 7) + 2(x − 7) = 0
(x − 7)(x + 2) = 0
Either (x − 7) = 0 or (x + 2) = 0
x = 7 or x = − 2
The two numbers are 7 and − 2
Users' Answers & CommentsThen, the other is (5 − x) , since their sum is 5 and their product is 14
x(5 − x) = 14
5x − x2 = 14
x2 − 5x − 14 = 0
(x2 − 7x) + (2x − 14) = 0
x(x − 7) + 2(x − 7) = 0
(x − 7)(x + 2) = 0
Either (x − 7) = 0 or (x + 2) = 0
x = 7 or x = − 2
The two numbers are 7 and − 2
14
Solve the equation
\( 5^{(x − 2)} = (1 ÷ 125)^{(x +3)}\)
\( 5^{(x − 2)} = (1 ÷ 125)^{(x +3)}\)
A
3/2
B
− 7/4
C
4/7
D
7/4
correct option: b
\( 5^{(x − 2)} = (125^{−1} )^{(x + 3)}\\
5^{(x − 2)} = (5− 3)^{x + 3}\\
5^{(x − 2)} = 5^{− 3x − 9}\\
x + 3x = − 9 + 2\\
4x = − 7\\
x = \frac{−7}{4} \)
Users' Answers & Comments5^{(x − 2)} = (5− 3)^{x + 3}\\
5^{(x − 2)} = 5^{− 3x − 9}\\
x + 3x = − 9 + 2\\
4x = − 7\\
x = \frac{−7}{4} \)
15
How many sides has a regular polygon whose interior angles are 120°each
A
4
B
5
C
6
D
3
correct option: c
Each exterior angle = 180 − 120=60
Exterior angle =360/n
60 = 360/n
n =360/6
= 6
Users' Answers & CommentsExterior angle =360/n
60 = 360/n
n =360/6
= 6
16
If duty is levied at 25%, find the duty to be added to a bill of N80.
A
N100
B
N20
C
N80
D
N150
17
Simplify 2.04 × 3.7 (Leave your answer in 2 decimal place)
A
7.51
B
7.71
C
5.5
D
7.55
18
A public car dealer marked up the cost of a car at 30% in an attempt to make 20% gross profit. Due to the value of dollar, he now placed 20% discount on the car. What profit or loss will he make?
A
3%
B
2%
C
4%
D
1%
correct option: c
Let assume the cost price is 100%
Marked up price + cost price = 20 + 100 = 120%
Discount at 20% = 20/100 × 120 % of cost price
Selling price = cost price − gain
= (120 − 24)% of cost price
= 96% of cost price
Loss = (100− 96)% of cost price
= 4% of cost price
∴ He will wake 4% loss
Users' Answers & CommentsMarked up price + cost price = 20 + 100 = 120%
Discount at 20% = 20/100 × 120 % of cost price
Selling price = cost price − gain
= (120 − 24)% of cost price
= 96% of cost price
Loss = (100− 96)% of cost price
= 4% of cost price
∴ He will wake 4% loss
19
Rationalize 5 ÷ (2 − √3)
A
3(2 + √5)
B
2(3 + √5)
C
5(2 + √3)
D
3 √2 + 1
correct option: c
5÷ (2 − √3)
Using conjugate surds
[5 ÷ (2 − √3)]×(2 + √3) ÷ (2 + √3)]
[5(2+√3)÷((2− √3))2]
[5(2 +√3) ÷ (2)2− (√3)2]
[5(2 +√3) ÷(4− 3)]
= 5(2 + √3)
Users' Answers & CommentsUsing conjugate surds
[5 ÷ (2 − √3)]×(2 + √3) ÷ (2 + √3)]
[5(2+√3)÷((2− √3))2]
[5(2 +√3) ÷ (2)2− (√3)2]
[5(2 +√3) ÷(4− 3)]
= 5(2 + √3)
20
Factorize \( a^2 − b^2 − 4a + 4 \)
A
(a + b)(a − b)
B
(a − 2 + b)
C
(a + 1)(a − 2 + b)
D
(a + b) 2
correct option: b
The trinomial = \( a^2 − 4a + 4 \\
a^2 − b^2− 4a + 4 = (a^2 − 4a + 4) − b^2 \\
(a^2 − 2a − 2a + 4) − b^2 \\
a(a − 2)− 2(a − 2)] − b^2 \\
(a − 2)2 − b^2 \\
(a − 2 + b)(a − 2 − b )\)
Users' Answers & Commentsa^2 − b^2− 4a + 4 = (a^2 − 4a + 4) − b^2 \\
(a^2 − 2a − 2a + 4) − b^2 \\
a(a − 2)− 2(a − 2)] − b^2 \\
(a − 2)2 − b^2 \\
(a − 2 + b)(a − 2 − b )\)