Year : 
2021
Title : 
Mathematics (Core)
Exam : 
WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

1 - 10 of 49 Questions

# Question Ans
1.

Correct 0.007985 to three significant figures.

A. 0.0109

B. 0.0800

C. 0.00799

D. 0.008

Detailed Solution

  1. To round to three significant figures, look at the fourth significant figure. It's a 5 , so round up or round down if below 5.
  2. To round to two significant figures, look at the third significant figure. It's an 8 , so round up.
2.

Simplify: (11\(_{two}\))\(^2\)

A. 1001\(_2\)

B. 1101\(_2\)

C. 101\(_2\)

D. 10001\(_2\)

Detailed Solution

(11\(_{two}\))\(^2\) = (11\(_2\) \(\times\) (11\(_2\))
= \({1 \times 2^1 + 1 \times 2^0} \times ({1 \times 2^1 + 1 \times 2^0})\)
= \({1 \times 2 + 1 \times 1} \times ({1 \times 2 + 1 \times 1})\)
= \({2 + 1} \times ({2 + 1})\)
= 3 \(\times\) 3
= 9\(_{10}\) or
1001\(_2\) from


3.

Solve: 2\(^{√2x + 1}\) = 32

A. 13

B. 24

C. 12

D. 11

Detailed Solution

2\(^{√2x + 1}\) = 32
2\(^{√2x + 1}\) = 2\(^5\)
√2x + 1 = 5
square both sides
2x + 1 = 5\(^2\)
2x + 1 = 25
2x = 25 - 1
2x = 24
x = \(\frac{24}{2}\)
x = 12
4.

If log\(_{10}\) 2 = m and log\(_{10}\) 3 = n, find log\(_{10}\) 24 in terms of m and n.

A. 3m + n

B. m + 3n

C. 4mn

D. 3mn

Detailed Solution

log\(_{10}\) 24 = log\(_{10}\) 8 \(\times\) log\(_{10}\) 3
where log\(_{10}\) 8 = 3 log\(_{10}\) 2 = 3 \(\times\) m
and log\(_{10}\) 3 = n
: log\(_{10}\) 24 = 3m + n
5.

Find the 5th term of the sequence 2,5,10,17....?

A. 22

B. 24

C. 36

D. 26

Detailed Solution

Simply add odd number starting from '3' to the next number
2
2 + 3 = 5
5 + 5 = 10
10 + 7 = 17
17 + 9 = 26

The fifth term = 26
6.

If p = {-3<x<1} and Q = {-1<x<3}, where x is a real number, find P n Q.

A. 0

B. -3, -2, -1, 0 and 1

C. -2, -1 and 0

D. -1, 0 and 1

Detailed Solution

p = {-3<x<1} = {-2,-1 and 0}
Q = {-1<x<3} = {0,1 and 2}
P n Q = {0} or {-1<x<1}
7.

Factorize 6pq-3rs-3ps+6qr

A. 3(r -p)(2q + s)

B. 3(p + r)( 2q - 2q - s)

C. 3(2q - s)(p + r)

D. 3(r - p)(s - 2q)

Detailed Solution

6pq-3rs-3ps+6qr = 3 (2pq - rs - ps + 2qr)
= 3 ({2pq + 2qr} {-ps - rs})
= 3 (2q{ p + r} -s{p + r})
= 3 ({2q - s}{p + r})
8.

What number should be subtracted from the sum of 2 \(\frac{1}{6}\) and 2\(\frac{7}{12}\) to give 3\(\frac{1}{4}\)?

A. \(\frac{1}{3}\)

B. 1\(\frac{1}{2}\)

C. 1\(\frac{1}{6}\)

D. \(\frac{1}{2}\)

Detailed Solution

The sum of 2 \(\frac{1}{6}\) and 2\(\frac{7}{12}\)
= \(\frac{13}{6}\) + \(\frac{31}{12}\)
= \(\frac{13 \times 2 + 31}{12}\)
= \(\frac{26 + 31}{12}\)
= \(\frac{57}{12}\)
What should be subtracted from \(\frac{57}{12}\) to give 3\(\frac{1}{4}\)
\(\frac{57}{12}\) - y = 3\(\frac{1}{4}\)
: y = \(\frac{57}{12}\) - 3\(\frac{1}{4}\) = \(\frac{57}{12}\) - \(\frac{13}{4}\)
y = \(\frac{57 - 3 \times 13}{12}\) = \(\frac{57 - 39}{12}\)
y = \(\frac{18}{12}\)
y = \(\frac{3}{2}\) or 1\(\frac{1}{2}\)
9.

Mensah is 5 years old and joyce is thrice as old as mensah. In how many years will joyce be twice as old as Mensah?

A. 3 years

B. 10 years

C. 5 years

D. 15 years

Detailed Solution

Mensah’s age is 5. Thus,
Joyce’s age is 15 (5*3=15)
The difference between their ages is 10 (15–5=10)
As we ought to find how many years Joyce’s age will be twice of Mensah’s age, we should write down the following :
15+X=2*(5+X)
15+X=10+2X lets add (-10-X) to both sides of the equation and
15+X-10-X = 10+2X-10-X
5=X —-> X=5
After 5 years Joyce’s age will be 20 (15+5=20)
After 5 years Mensah’s age will be 10 (5+5=10)
After 5 years Joyce will be twice as old as Mensah (10*2=20)
10.

If 16 * 2\(^{(x + 1)}\) = 4\(^x\) * 8\(^{(1 - x)}\), find the value of x.

A. -4

B. 4

C. 1

D. -1

Detailed Solution

16 * 2\(^{(x + 1)}\) = 4\(^x\) * 8\(^{(1 - x)}\)
= 2\(^4\) * 2\(^{(x + 1)}\) = 2\(^{2x}\) * 2\(^{3(1 - x)}\)
--> 4 + x + 1 = 2x + 3 - 3x
collect like terms
--> x - 2x + 3x = 3 - 1 - 4
--> 2x = -2
--> x = -1
1.

Correct 0.007985 to three significant figures.

A. 0.0109

B. 0.0800

C. 0.00799

D. 0.008

Detailed Solution

  1. To round to three significant figures, look at the fourth significant figure. It's a 5 , so round up or round down if below 5.
  2. To round to two significant figures, look at the third significant figure. It's an 8 , so round up.
2.

Simplify: (11\(_{two}\))\(^2\)

A. 1001\(_2\)

B. 1101\(_2\)

C. 101\(_2\)

D. 10001\(_2\)

Detailed Solution

(11\(_{two}\))\(^2\) = (11\(_2\) \(\times\) (11\(_2\))
= \({1 \times 2^1 + 1 \times 2^0} \times ({1 \times 2^1 + 1 \times 2^0})\)
= \({1 \times 2 + 1 \times 1} \times ({1 \times 2 + 1 \times 1})\)
= \({2 + 1} \times ({2 + 1})\)
= 3 \(\times\) 3
= 9\(_{10}\) or
1001\(_2\) from


3.

Solve: 2\(^{√2x + 1}\) = 32

A. 13

B. 24

C. 12

D. 11

Detailed Solution

2\(^{√2x + 1}\) = 32
2\(^{√2x + 1}\) = 2\(^5\)
√2x + 1 = 5
square both sides
2x + 1 = 5\(^2\)
2x + 1 = 25
2x = 25 - 1
2x = 24
x = \(\frac{24}{2}\)
x = 12
4.

If log\(_{10}\) 2 = m and log\(_{10}\) 3 = n, find log\(_{10}\) 24 in terms of m and n.

A. 3m + n

B. m + 3n

C. 4mn

D. 3mn

Detailed Solution

log\(_{10}\) 24 = log\(_{10}\) 8 \(\times\) log\(_{10}\) 3
where log\(_{10}\) 8 = 3 log\(_{10}\) 2 = 3 \(\times\) m
and log\(_{10}\) 3 = n
: log\(_{10}\) 24 = 3m + n
5.

Find the 5th term of the sequence 2,5,10,17....?

A. 22

B. 24

C. 36

D. 26

Detailed Solution

Simply add odd number starting from '3' to the next number
2
2 + 3 = 5
5 + 5 = 10
10 + 7 = 17
17 + 9 = 26

The fifth term = 26
6.

If p = {-3<x<1} and Q = {-1<x<3}, where x is a real number, find P n Q.

A. 0

B. -3, -2, -1, 0 and 1

C. -2, -1 and 0

D. -1, 0 and 1

Detailed Solution

p = {-3<x<1} = {-2,-1 and 0}
Q = {-1<x<3} = {0,1 and 2}
P n Q = {0} or {-1<x<1}
7.

Factorize 6pq-3rs-3ps+6qr

A. 3(r -p)(2q + s)

B. 3(p + r)( 2q - 2q - s)

C. 3(2q - s)(p + r)

D. 3(r - p)(s - 2q)

Detailed Solution

6pq-3rs-3ps+6qr = 3 (2pq - rs - ps + 2qr)
= 3 ({2pq + 2qr} {-ps - rs})
= 3 (2q{ p + r} -s{p + r})
= 3 ({2q - s}{p + r})
8.

What number should be subtracted from the sum of 2 \(\frac{1}{6}\) and 2\(\frac{7}{12}\) to give 3\(\frac{1}{4}\)?

A. \(\frac{1}{3}\)

B. 1\(\frac{1}{2}\)

C. 1\(\frac{1}{6}\)

D. \(\frac{1}{2}\)

Detailed Solution

The sum of 2 \(\frac{1}{6}\) and 2\(\frac{7}{12}\)
= \(\frac{13}{6}\) + \(\frac{31}{12}\)
= \(\frac{13 \times 2 + 31}{12}\)
= \(\frac{26 + 31}{12}\)
= \(\frac{57}{12}\)
What should be subtracted from \(\frac{57}{12}\) to give 3\(\frac{1}{4}\)
\(\frac{57}{12}\) - y = 3\(\frac{1}{4}\)
: y = \(\frac{57}{12}\) - 3\(\frac{1}{4}\) = \(\frac{57}{12}\) - \(\frac{13}{4}\)
y = \(\frac{57 - 3 \times 13}{12}\) = \(\frac{57 - 39}{12}\)
y = \(\frac{18}{12}\)
y = \(\frac{3}{2}\) or 1\(\frac{1}{2}\)
9.

Mensah is 5 years old and joyce is thrice as old as mensah. In how many years will joyce be twice as old as Mensah?

A. 3 years

B. 10 years

C. 5 years

D. 15 years

Detailed Solution

Mensah’s age is 5. Thus,
Joyce’s age is 15 (5*3=15)
The difference between their ages is 10 (15–5=10)
As we ought to find how many years Joyce’s age will be twice of Mensah’s age, we should write down the following :
15+X=2*(5+X)
15+X=10+2X lets add (-10-X) to both sides of the equation and
15+X-10-X = 10+2X-10-X
5=X —-> X=5
After 5 years Joyce’s age will be 20 (15+5=20)
After 5 years Mensah’s age will be 10 (5+5=10)
After 5 years Joyce will be twice as old as Mensah (10*2=20)
10.

If 16 * 2\(^{(x + 1)}\) = 4\(^x\) * 8\(^{(1 - x)}\), find the value of x.

A. -4

B. 4

C. 1

D. -1

Detailed Solution

16 * 2\(^{(x + 1)}\) = 4\(^x\) * 8\(^{(1 - x)}\)
= 2\(^4\) * 2\(^{(x + 1)}\) = 2\(^{2x}\) * 2\(^{3(1 - x)}\)
--> 4 + x + 1 = 2x + 3 - 3x
collect like terms
--> x - 2x + 3x = 3 - 1 - 4
--> 2x = -2
--> x = -1