Year : 
2015
Title : 
Mathematics (Core)
Exam : 
JAMB Exam

Paper 1 | Objectives

11 - 20 of 62 Questions

# Question Ans
11.

Solve the equation \( 3x^2 − 4x − 5 = 0 \)

A. x = 1.75 or − 0.15

B. x = 2.12 or − 0.79

C. x = 1.5 or − 0.34

D. x = 2.35 or −1.23

Detailed Solution

Using the quadratic formula, we have
\(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)
From the equation \(3x^{2} - 4x - 5 = 0\), a = 3, b = -4 and c = -5.
\(\therefore x = \frac{-(-4) \pm \sqrt{(-4)^{2} - 4(3)(-5)}}{2(3)}\)
= \(\frac{4 \pm \sqrt{16 + 60}}{6}\)
= \(\frac{4 \pm \sqrt{76}}{6}\)
= \(\frac{4 \pm 8.72}{6}\)
= \(\frac{4 + 8.72}{6} \text{ or } \frac{4 - 8.72}{6}\)
= \(\frac{12.72}{6} \text{ or } \frac{-4.72}{6}\)
\(x = \text{2.12 or -0.79}\)
There is an explanation video available below.
12.

The first and last term of a linear sequence (AP) are 6 and 10 respectively. If the sum of the sequence is 40. Find the number of terms

A. nth = 3

B. nth = 4

C. nth = 5

D. nth = 6

Detailed Solution

nth term of a linear sequence (AP) = a+(n − 1)d

first term = 6, last term = 10 sum − 40

i.e. a = 6, l = 10, S = 40

S\(_{n}\)= n/2(2a + (n − 1)d or Sn = ÷2 (a + l)

S\(_{n}\) = n/2(a + l)

40 = n/2(6 + 10)

40 = 8n

8n = 40

8n = 40

n = 40/8

= 5

The number of terms = 5
There is an explanation video available below.
13.

Find the equation of a line which is form origin and passes through the point (−3, −4)

A. y = \( \frac{3x}{4} \)

B. y = \( \frac{4x}{3} \)

C. y = \( \frac{2x}{3} \)

D. y = \( \frac{x}{2} \)

Detailed Solution

The slope of the line from (0, 0) passing through (-3, -4) = \(\frac{-4 - 0}{-3 - 0}\)
= \(\frac{4}{3}\)
Equation of a line is given as \(y = mx + b\), where m = slope and b = intercept.
To get the value of b, we use a point on the line, say (0, 0).
\(y = \frac{4}{3} x + b\)
\(0 = \frac{4}{3}(0) + b\)
\(b = 0\)
The equation of the line is \(y = \frac{4}{3} x\)
There is an explanation video available below.
14.

The amount A to which a principal P amounts at r% compound interest for n years is given by the formula A = P(1 + (r ÷ 100)\(^n\). Find A, if P = 126, r = 4 and n = 2.

A. N132.50K

B. N136.30K

C. N125.40K

D. N257.42K

Detailed Solution

\( A = P \left(1 + \frac{r}{100}\right)^n \)

Where P = 126, r = 4,n = 2

A=126 \( \left(1 + \frac{4}{100}\right)^2 \text{Using LCM} \)

=126 \( \left(\frac{100+4}{100}\right)^2 = 126 \left(\frac{104}{100}\right)^2 \)


=126 \( \left(1.04^2 \right) \)

= 126 * 1.04 * 1.04

=136.28

A = 136.30 (approx.)

The Amount A, = N136.30k
There is an explanation video available below.
15.

Make x the subject of the equation
s = 2 + \(\frac{t}{5} \)(x + ⅗y)

A. x = 5[(s − 2) ÷ t] - 3/5y

B. x = 25[(s − 2) ÷ t] − 3ty

C. x = [1 ÷ (s − 2)3ty]

D. x = [5(s − 2) 2 ÷ 3ty] × t

Detailed Solution

There is an explanation video available below.
16.

If log520 = x, find x

A. 1.761

B. 1.354

C. 1.861

D. 2.549

Detailed Solution

There is an explanation video available below.
17.

Find at which rate per annum simple interest N525 will amount to N588 in 3 years.

A. 3%

B. 2%

C. 5%

D. 4%

Detailed Solution

There is an explanation video available below.
18.

Given that A = {1, 5, 7}

B = {3, 9, 12, 15}

C = {2, 4, 6, 8}

Find (A ∪ B) ∪ C

A. {1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15}

B. {1, 2, 3, 5, 6, 8, 12, 15}

C. {2, 4, 5, 9, 12, 15}

D. {1, 5, 6, 7, 8, 9, 12, 15}

Detailed Solution

There is an explanation video available below.
19.

The extension of a stretched string is directly proportional to its tension. If the extension produced by a tension of 8 Newton’s is 2cm, find the extension produced by a tension of 12 newton’s.

A. 2

B. 1

D. 3

Detailed Solution

Let the extension be E and the tension be T.
Then \(E \propto T\)
\(E = kT\)
when T = 8N, E = 2cm
\(2 = k \times 8\)
\(k = \frac{2}{8} = 0.25\)
\(\therefore E = 0.25T\)
when T = 12N, \(E = 0.25 \times 12 = 3cm\)
There is an explanation video available below.
20.

Factorize \( x^2 − 2x − 15 \)

A. (x + 3)2

B. (x + 5)(x − 3)

C. (x − 5)2

D. (x − 5)(x + 3)

Detailed Solution

There is an explanation video available below.
11.

Solve the equation \( 3x^2 − 4x − 5 = 0 \)

A. x = 1.75 or − 0.15

B. x = 2.12 or − 0.79

C. x = 1.5 or − 0.34

D. x = 2.35 or −1.23

Detailed Solution

Using the quadratic formula, we have
\(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)
From the equation \(3x^{2} - 4x - 5 = 0\), a = 3, b = -4 and c = -5.
\(\therefore x = \frac{-(-4) \pm \sqrt{(-4)^{2} - 4(3)(-5)}}{2(3)}\)
= \(\frac{4 \pm \sqrt{16 + 60}}{6}\)
= \(\frac{4 \pm \sqrt{76}}{6}\)
= \(\frac{4 \pm 8.72}{6}\)
= \(\frac{4 + 8.72}{6} \text{ or } \frac{4 - 8.72}{6}\)
= \(\frac{12.72}{6} \text{ or } \frac{-4.72}{6}\)
\(x = \text{2.12 or -0.79}\)
There is an explanation video available below.
12.

The first and last term of a linear sequence (AP) are 6 and 10 respectively. If the sum of the sequence is 40. Find the number of terms

A. nth = 3

B. nth = 4

C. nth = 5

D. nth = 6

Detailed Solution

nth term of a linear sequence (AP) = a+(n − 1)d

first term = 6, last term = 10 sum − 40

i.e. a = 6, l = 10, S = 40

S\(_{n}\)= n/2(2a + (n − 1)d or Sn = ÷2 (a + l)

S\(_{n}\) = n/2(a + l)

40 = n/2(6 + 10)

40 = 8n

8n = 40

8n = 40

n = 40/8

= 5

The number of terms = 5
There is an explanation video available below.
13.

Find the equation of a line which is form origin and passes through the point (−3, −4)

A. y = \( \frac{3x}{4} \)

B. y = \( \frac{4x}{3} \)

C. y = \( \frac{2x}{3} \)

D. y = \( \frac{x}{2} \)

Detailed Solution

The slope of the line from (0, 0) passing through (-3, -4) = \(\frac{-4 - 0}{-3 - 0}\)
= \(\frac{4}{3}\)
Equation of a line is given as \(y = mx + b\), where m = slope and b = intercept.
To get the value of b, we use a point on the line, say (0, 0).
\(y = \frac{4}{3} x + b\)
\(0 = \frac{4}{3}(0) + b\)
\(b = 0\)
The equation of the line is \(y = \frac{4}{3} x\)
There is an explanation video available below.
14.

The amount A to which a principal P amounts at r% compound interest for n years is given by the formula A = P(1 + (r ÷ 100)\(^n\). Find A, if P = 126, r = 4 and n = 2.

A. N132.50K

B. N136.30K

C. N125.40K

D. N257.42K

Detailed Solution

\( A = P \left(1 + \frac{r}{100}\right)^n \)

Where P = 126, r = 4,n = 2

A=126 \( \left(1 + \frac{4}{100}\right)^2 \text{Using LCM} \)

=126 \( \left(\frac{100+4}{100}\right)^2 = 126 \left(\frac{104}{100}\right)^2 \)


=126 \( \left(1.04^2 \right) \)

= 126 * 1.04 * 1.04

=136.28

A = 136.30 (approx.)

The Amount A, = N136.30k
There is an explanation video available below.
15.

Make x the subject of the equation
s = 2 + \(\frac{t}{5} \)(x + ⅗y)

A. x = 5[(s − 2) ÷ t] - 3/5y

B. x = 25[(s − 2) ÷ t] − 3ty

C. x = [1 ÷ (s − 2)3ty]

D. x = [5(s − 2) 2 ÷ 3ty] × t

Detailed Solution

There is an explanation video available below.
16.

If log520 = x, find x

A. 1.761

B. 1.354

C. 1.861

D. 2.549

Detailed Solution

There is an explanation video available below.
17.

Find at which rate per annum simple interest N525 will amount to N588 in 3 years.

A. 3%

B. 2%

C. 5%

D. 4%

Detailed Solution

There is an explanation video available below.
18.

Given that A = {1, 5, 7}

B = {3, 9, 12, 15}

C = {2, 4, 6, 8}

Find (A ∪ B) ∪ C

A. {1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15}

B. {1, 2, 3, 5, 6, 8, 12, 15}

C. {2, 4, 5, 9, 12, 15}

D. {1, 5, 6, 7, 8, 9, 12, 15}

Detailed Solution

There is an explanation video available below.
19.

The extension of a stretched string is directly proportional to its tension. If the extension produced by a tension of 8 Newton’s is 2cm, find the extension produced by a tension of 12 newton’s.

A. 2

B. 1

D. 3

Detailed Solution

Let the extension be E and the tension be T.
Then \(E \propto T\)
\(E = kT\)
when T = 8N, E = 2cm
\(2 = k \times 8\)
\(k = \frac{2}{8} = 0.25\)
\(\therefore E = 0.25T\)
when T = 12N, \(E = 0.25 \times 12 = 3cm\)
There is an explanation video available below.
20.

Factorize \( x^2 − 2x − 15 \)

A. (x + 3)2

B. (x + 5)(x − 3)

C. (x − 5)2

D. (x − 5)(x + 3)

Detailed Solution

There is an explanation video available below.