Year : 
2010
Title : 
Mathematics (Core)
Exam : 
JAMB Exam

Paper 1 | Objectives

21 - 30 of 49 Questions

# Question Ans
21.

Find r, if 6r7\(_8\) = 511\(_9\)

A. 3

B. 2

C. 6

D. 5

Detailed Solution

6r7\(_8\) = 511\(_9\)

6 x 8\(^2\) + r x 8\(^1\) + 7 x 8\(^0\) = 5 x 9\(^2\) + 1 x 9\(^1\) + 1 x 9\(^0\)

6 x 64 + 8r + 7 x 1 = 5 x 81 + 9 + 1 x 1

384 + 8r + 7 = 405 + 9 + 1

8r = 415 - 391
8r = 24

r = \(\frac{24}{8}\)

= 3
There is an explanation video available below.
22.

Simplify (\(\frac{3}{4}\) of \(\frac{4}{9}\) \(\div\) 9\(\frac{1}{2}\)) \(\div\) 1\(\frac{5}{19}\)

A. \(\frac{1}{5}\)

B. \(\frac{1}{4}\)

C. \(\frac{1}{36}\)

D. \(\frac{1}{25}\)

Detailed Solution

(\(\frac{3}{4}\) of \(\frac{4}{9}\) \(\div\) 9\(\frac{1}{2}\)) \(\div\) 1\(\frac{5}{19}\)

Applying the rule of BODMAS, we have:

(\(\frac{3}{4}\) x \(\frac{4}{9}\) \(\div\)\(\frac{19}{2}\)) \(\div\)\(\frac{24}{19}\)


(\(\frac{1}{3}\) x \(\frac{2}{19}\)) \(\div\)\(\frac{24}{19}\)

\(\frac{1}{3}\) x \(\frac{2}{19}\) x \(\frac{19}{24}\)

= \(\frac{1}{36}\)
There is an explanation video available below.
23.

A student measures a piece of rope and found that it was 1.26m long. If the actual length of the rope was 1.25m, what was the percentage error in the measurement?

A. 0.25%

B. 0.01%

C. 0.80%

D. 0.40%

Detailed Solution

Actual length of rope = 1.25m

Measured length of rope = 1.26m

error = (1.26 - 1.25)m - 0.01m

Percentage error = \(\frac{error}{\text{actual length}}\) x 100%

= \(\frac{0.01}{1.25}\) x 100%

= 0.8%
There is an explanation video available below.
24.

At what rate will the interest on N400 increases to N24 in 3 years reckoning in simple interest?

A. 3%

B. 2%

C. 5%

D. 4%

Detailed Solution

Using simple interest = \(\frac{P \times T \times R}{100}\),

where: P denoted principal = N400
T denotes time = 3 years
R denotes interest rate = ?

24 = \(\frac{400 \times 3 \times R}{100}\)

24 x 100 = 400 x 3 x R

R = \(\frac{24 \times 100}{400 \times3}\)

= 2%
There is an explanation video available below.
25.

If p : q = \(\frac{2}{3}\) : \(\frac{5}{6}\) and q : r = \(\frac{3}{4}\) : \(\frac{1}{2}\), find p : q : r

A. 12 : 15 : 10

B. 12 : 15 : 16

C. 10 : 15 : 24

D. 9 : 10 : 15

Detailed Solution

If p : q = \(\frac{2}{3}\) : \(\frac{5}{6}\), then the sum S1 of ratio = \(\frac{2}{3}\) + \(\frac{5}{6}\) = \(\frac{9}{6}\)

If q : r = \(\frac{3}{4}\) : \(\frac{1}{2}\), then the sum S2 of ratio = \(\frac{3}{4}\) + \(\frac{1}{2}\) = \(\frac{5}{4}\)

Let p + q = T1, then

q = (\(\frac{5}{6} \div \frac{9}{6}\))T1 = (\(\frac{5}{6} \times \frac{6}{9}\))T1 = \(\frac{5}{9}\)T1

Again, let q + r = T2, then

q = (\(\frac{3}{4} \div \frac{5}{4}\))T2 = (\(\frac{3}{4} \times \frac{4}{5}\))T2 = \(\frac{3}{5}\)T2

Using q = q

\(\frac{5}{9}\)T1 = \(\frac{3}{5}\)T2
26.

The 3rd term of an arithmetic progression is -9 and the 7th term is -29. Find the 10th term of the progression

A. -44

B. -165

C. 165

D. 44

Detailed Solution

3rd term : a + 2d = -9 .......(1)
7th term : a + 6d = -29 ......(2)
(2) - (1): 4d = -20
:. d = -20/4 = -5
From (1) : a + 2(-5) = -9
a - 10 = -9
:. a = -9 + 10 = 1
:. 10th term of A.P is a + 9d = 1 + 9 (-5)
= 1 - 45 = -44
There is an explanation video available below.
27.

At what value of X does the function y = -3 - 2x + X2 attain a minimum value?

A. -1

B. 14

C. 4

D. 1

Detailed Solution

Given that y = -3 - 2x + X2

then, \(\frac{dy}{dx}\) = -2 + 2x

At maximum value, \(\frac{dy}{dx}\) = O

therefore, -2 + 2x

2x = 2

x = 2/2 = 1
There is an explanation video available below.
28.

Find the equation of a line parallel to y = -4x + 2 passing through (2,3)

A. y + 4x + 11 = 0

B. y - 4x - 11 = 0

C. y + 4x - 11 = 0

D. y - 4x + 11 = 0

Detailed Solution

By comparing y = mx + c

with y = -4x + 2,

the gradient of y = -4x + 2 is m1 = -4

let the gradient of the line parallel to the given line be m2,

then, m2 = m1 = -4

(condition for parallelism)

using, y - y1 = m2(x - x1)

Hence the equation of the parallel line is

y - 3 = -4(x-2)

y - 3 = -4 x + 8

y + 4x = 8 + 3

y + 4x = 11

y + 4x - 11 = 0
There is an explanation video available below.<
29.

Make Q the subject of formula if p = \(\frac{M}{5}\)(X + Q) + 1

A. \(\frac{5P - MX + 5}{M}\)

B. \(\frac{5P - MX - 5}{M}\)

C. \(\frac{5P + MX + 5}{M}\)

D. \(\frac{5P + MX - 5}{M}\)

Detailed Solution

p = \(\frac{M}{5}\)(X + Q) + 1

P - 1 = \(\frac{M}{5}\)(X + Q)

\(\frac{5}{M}\)(p - 1) = X + Q

\(\frac{5}{M}\)(p - 1)- x = Q

Q = \(\frac{5(p -1) - Mx}{M}\)

= \(\frac{5p - 5 - Mx}{M}\)

= \(\frac{5p - Mx - 5}{M}\)
There is an explanation video available below.
30.

If 9x2 + 6xy + 4y2 is a factor of 27x3 - 8y3, find the other factor.

A. 2y + 3x

B. 2y - 3x

C. 3x + 2y

D. 3x - 2y

Detailed Solution

27x3 - 8y3 = (3x - 2y)3

But 9x2 + 6xy + 4y2 = (3x +2y)2

So, 27x3 - 8y3 = (3x - 2y)(3x - 2y)2

Hence the other factor is 3x - 2y
There is an explanation video available below.
21.

Find r, if 6r7\(_8\) = 511\(_9\)

A. 3

B. 2

C. 6

D. 5

Detailed Solution

6r7\(_8\) = 511\(_9\)

6 x 8\(^2\) + r x 8\(^1\) + 7 x 8\(^0\) = 5 x 9\(^2\) + 1 x 9\(^1\) + 1 x 9\(^0\)

6 x 64 + 8r + 7 x 1 = 5 x 81 + 9 + 1 x 1

384 + 8r + 7 = 405 + 9 + 1

8r = 415 - 391
8r = 24

r = \(\frac{24}{8}\)

= 3
There is an explanation video available below.
22.

Simplify (\(\frac{3}{4}\) of \(\frac{4}{9}\) \(\div\) 9\(\frac{1}{2}\)) \(\div\) 1\(\frac{5}{19}\)

A. \(\frac{1}{5}\)

B. \(\frac{1}{4}\)

C. \(\frac{1}{36}\)

D. \(\frac{1}{25}\)

Detailed Solution

(\(\frac{3}{4}\) of \(\frac{4}{9}\) \(\div\) 9\(\frac{1}{2}\)) \(\div\) 1\(\frac{5}{19}\)

Applying the rule of BODMAS, we have:

(\(\frac{3}{4}\) x \(\frac{4}{9}\) \(\div\)\(\frac{19}{2}\)) \(\div\)\(\frac{24}{19}\)


(\(\frac{1}{3}\) x \(\frac{2}{19}\)) \(\div\)\(\frac{24}{19}\)

\(\frac{1}{3}\) x \(\frac{2}{19}\) x \(\frac{19}{24}\)

= \(\frac{1}{36}\)
There is an explanation video available below.
23.

A student measures a piece of rope and found that it was 1.26m long. If the actual length of the rope was 1.25m, what was the percentage error in the measurement?

A. 0.25%

B. 0.01%

C. 0.80%

D. 0.40%

Detailed Solution

Actual length of rope = 1.25m

Measured length of rope = 1.26m

error = (1.26 - 1.25)m - 0.01m

Percentage error = \(\frac{error}{\text{actual length}}\) x 100%

= \(\frac{0.01}{1.25}\) x 100%

= 0.8%
There is an explanation video available below.
24.

At what rate will the interest on N400 increases to N24 in 3 years reckoning in simple interest?

A. 3%

B. 2%

C. 5%

D. 4%

Detailed Solution

Using simple interest = \(\frac{P \times T \times R}{100}\),

where: P denoted principal = N400
T denotes time = 3 years
R denotes interest rate = ?

24 = \(\frac{400 \times 3 \times R}{100}\)

24 x 100 = 400 x 3 x R

R = \(\frac{24 \times 100}{400 \times3}\)

= 2%
There is an explanation video available below.
25.

If p : q = \(\frac{2}{3}\) : \(\frac{5}{6}\) and q : r = \(\frac{3}{4}\) : \(\frac{1}{2}\), find p : q : r

A. 12 : 15 : 10

B. 12 : 15 : 16

C. 10 : 15 : 24

D. 9 : 10 : 15

Detailed Solution

If p : q = \(\frac{2}{3}\) : \(\frac{5}{6}\), then the sum S1 of ratio = \(\frac{2}{3}\) + \(\frac{5}{6}\) = \(\frac{9}{6}\)

If q : r = \(\frac{3}{4}\) : \(\frac{1}{2}\), then the sum S2 of ratio = \(\frac{3}{4}\) + \(\frac{1}{2}\) = \(\frac{5}{4}\)

Let p + q = T1, then

q = (\(\frac{5}{6} \div \frac{9}{6}\))T1 = (\(\frac{5}{6} \times \frac{6}{9}\))T1 = \(\frac{5}{9}\)T1

Again, let q + r = T2, then

q = (\(\frac{3}{4} \div \frac{5}{4}\))T2 = (\(\frac{3}{4} \times \frac{4}{5}\))T2 = \(\frac{3}{5}\)T2

Using q = q

\(\frac{5}{9}\)T1 = \(\frac{3}{5}\)T2
26.

The 3rd term of an arithmetic progression is -9 and the 7th term is -29. Find the 10th term of the progression

A. -44

B. -165

C. 165

D. 44

Detailed Solution

3rd term : a + 2d = -9 .......(1)
7th term : a + 6d = -29 ......(2)
(2) - (1): 4d = -20
:. d = -20/4 = -5
From (1) : a + 2(-5) = -9
a - 10 = -9
:. a = -9 + 10 = 1
:. 10th term of A.P is a + 9d = 1 + 9 (-5)
= 1 - 45 = -44
There is an explanation video available below.
27.

At what value of X does the function y = -3 - 2x + X2 attain a minimum value?

A. -1

B. 14

C. 4

D. 1

Detailed Solution

Given that y = -3 - 2x + X2

then, \(\frac{dy}{dx}\) = -2 + 2x

At maximum value, \(\frac{dy}{dx}\) = O

therefore, -2 + 2x

2x = 2

x = 2/2 = 1
There is an explanation video available below.
28.

Find the equation of a line parallel to y = -4x + 2 passing through (2,3)

A. y + 4x + 11 = 0

B. y - 4x - 11 = 0

C. y + 4x - 11 = 0

D. y - 4x + 11 = 0

Detailed Solution

By comparing y = mx + c

with y = -4x + 2,

the gradient of y = -4x + 2 is m1 = -4

let the gradient of the line parallel to the given line be m2,

then, m2 = m1 = -4

(condition for parallelism)

using, y - y1 = m2(x - x1)

Hence the equation of the parallel line is

y - 3 = -4(x-2)

y - 3 = -4 x + 8

y + 4x = 8 + 3

y + 4x = 11

y + 4x - 11 = 0
There is an explanation video available below.<
29.

Make Q the subject of formula if p = \(\frac{M}{5}\)(X + Q) + 1

A. \(\frac{5P - MX + 5}{M}\)

B. \(\frac{5P - MX - 5}{M}\)

C. \(\frac{5P + MX + 5}{M}\)

D. \(\frac{5P + MX - 5}{M}\)

Detailed Solution

p = \(\frac{M}{5}\)(X + Q) + 1

P - 1 = \(\frac{M}{5}\)(X + Q)

\(\frac{5}{M}\)(p - 1) = X + Q

\(\frac{5}{M}\)(p - 1)- x = Q

Q = \(\frac{5(p -1) - Mx}{M}\)

= \(\frac{5p - 5 - Mx}{M}\)

= \(\frac{5p - Mx - 5}{M}\)
There is an explanation video available below.
30.

If 9x2 + 6xy + 4y2 is a factor of 27x3 - 8y3, find the other factor.

A. 2y + 3x

B. 2y - 3x

C. 3x + 2y

D. 3x - 2y

Detailed Solution

27x3 - 8y3 = (3x - 2y)3

But 9x2 + 6xy + 4y2 = (3x +2y)2

So, 27x3 - 8y3 = (3x - 2y)(3x - 2y)2

Hence the other factor is 3x - 2y
There is an explanation video available below.