Year : 
2016
Title : 
Mathematics (Core)
Exam : 
WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

21 - 30 of 49 Questions

# Question Ans
21.

\(\begin{array}{c|c}
Age(years) & 13 & 14 & 15 & 16 & 17 \\
\hline
Frequency & 10 & 24 & 8 & 5 & 3
\end{array}\)
Find the median age

A. 13

B. 14

C. 15

D. 16

Detailed Solution

Median = \(\frac{N}{2}\)th age

= \(\frac{50}{2}\)th age

= 25th age

= 14 years
22.

In the diagram, \(\bar{YW}\) is a tangent to the circle at X, |UV| = |VX| and < VXW = 50o. Find the value of < UXY.

A. 70o

B. 80o

C. 105o

D. 110o

Detailed Solution

In the diagram above, x1 = 50(angles in alternate segment)

x1 = x2(base angles of isos. \(\Delta\))

< UXY + x2 + 50o = 180o(sum of angles on a straight line)

< UXY + 50o + 50o = 180p

< UXY + 100o = 180o

< UXY = 180o - 100o

= 80o
23.

In the diagram, \(\bar{PF}\), \(\bar{QT}\), \(\bar{RG}\) intersect at S and PG||RG. If < SPQ = 113o and < RSt = 220, find < PSQ

A. 22o

B. 45o

C. 67o

D. 89o

Detailed Solution

In In the diagram given, \(\alpha\) = 22o (vertically opp. angles), \(\alpha\) = \(\beta\) (alternate angles)

< PSQ + 133o + \(\beta\) = 180o (sum of angles of a \(\Delta\))

< PSQ + 133o + 22o = 180o

< PSQ = 180o - (133 + 22)o

45o
24.

In the diagram, O is the centre of the circle, < XOZ = (10cm)o and < XWZ = mo. Calculate the value of m.

A. 30

B. 36

C. 40

D. 72

Detailed Solution

In the diagram above, \(\alpha\) = 2mo (angle at centre = 2 x angle at circumference)

\(\alpha\) + 10mo = 360o (angle at circumference)

\(\alpha\) + 10mo = 360o(angles round a point)

2mo + 10mo = 360o

12mo = 360o
25.

Kweku walked 8m up to slope and was 3m above the ground. If he walks 12m further up the slope, how far above the ground will he be?

A. 4.5m

B. 6.0m

C. 7.5m

D. 9.0m

Detailed Solution

By similar triangles, \(\frac{8}{3}\) = \(\frac{8 + 12}{h}\)

\(\frac{8}{3} = \frac{20}{h}\)

h = \(\frac{3 \times 20}{8}\)

= 7.5m
26.

In the diagram, TS is a tangent to the circle at S. |PR| and < PQR = 177o. Calculate < PST.

A. 54o

B. 44o

C. 34o

D. 27o

Detailed Solution

In the diagram above, x1 = 180o - 117o = 63o(opposite angles of a cyclic quad.)

x1 = x2 (base angles of isos. \(\Delta\))

x1 + x2 + \(\alpha\) = 180o (sum of angles of a \(\Delta\)

63o + 63o + \(\alpha\) = 180o<
27.

In the diagram, PR||SV||WY|, TX||QY|, < PQT = 48o and < TXW = 60o.Find < TQU.

A. 120o

B. 108o

C. 72o

D. 60o

Detailed Solution

In the diagram, < TUQ + 60o(corresp. angles)

< QTU = 48o (alternate angles)

< QU + 60o + 48o = 180o(sum of angles of a \(\Delta\))

< TQU = 180o - 108o

= 72o
28.

In the diagram, TX is perpendicular to UW, |UX| = 1cm and |TX| = |WX| = \(\sqrt{3}\)cm. Find UTW

A. 135o

B. 105o

C. 75o

D. 60o

Detailed Solution

In \(\Delta\) UXT, tan\(\alpha\) = \(\frac{1}{\sqrt{3}}\)

\(\alpha\) = tan-1(\(\frac{1}{\sqrt{3}}\))

= 30o

In \(\Delta\)WXT, tan\(\beta\) \(\frac{\sqrt{3}}{\sqrt{3}}\) = 1

\(\beta\) = tan-1(1) = 45o

Hence, < UTW = \(\alpha\) + \(\beta\)

= 30o + 45o = 75o
29.

The figure is a pie chart which represents the expenditure of a family in a year. If the total income of the family was Le 10,800,000.00, how much was spent on food?

A. Le 2,250,000.00

B. Le 22,700,000.00

C. Le 3,600,000.00

D. Le 4,500,000.00

Detailed Solution

sectoral angle representing food

= 360o - (80 + 70 + 90)o

= 120o

Amount spent on food

= \(\frac{\tect{sectoral angle}}{360^o}\) x Le 10,800,000

= Le 3,600,000
30.

A fair die is thrown two times. What is the probability that the sum of the scores is at least 10?

A. \(\frac{5}{36}\)

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

C. \(\frac{5}{18}\)

D. \(\frac{2}{3}\)

Detailed Solution

\(\begin{array}{c|c}
& 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
1 & 1,1 & 1,2 & 1,3 & 1,4 & 1,5 & 1,6 \\ \hline 2 & 2,1 & 2,2 & 2,3 & 2,4 & 2,5 & 2,6 \\ \hline 3 & 3,1 & 3,2 & 3,3 & 3,4 & 3,5 & 3,6 \\ \hline 4 & 4,1 & 4,2 & 4,3 & 4,4 & 4,5 & 4,6 \\ 5 & 5,1 & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \\ \hline 6 & 6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6\end{array}\)

From the table
21.

\(\begin{array}{c|c}
Age(years) & 13 & 14 & 15 & 16 & 17 \\
\hline
Frequency & 10 & 24 & 8 & 5 & 3
\end{array}\)
Find the median age

A. 13

B. 14

C. 15

D. 16

Detailed Solution

Median = \(\frac{N}{2}\)th age

= \(\frac{50}{2}\)th age

= 25th age

= 14 years
22.

In the diagram, \(\bar{YW}\) is a tangent to the circle at X, |UV| = |VX| and < VXW = 50o. Find the value of < UXY.

A. 70o

B. 80o

C. 105o

D. 110o

Detailed Solution

In the diagram above, x1 = 50(angles in alternate segment)

x1 = x2(base angles of isos. \(\Delta\))

< UXY + x2 + 50o = 180o(sum of angles on a straight line)

< UXY + 50o + 50o = 180p

< UXY + 100o = 180o

< UXY = 180o - 100o

= 80o
23.

In the diagram, \(\bar{PF}\), \(\bar{QT}\), \(\bar{RG}\) intersect at S and PG||RG. If < SPQ = 113o and < RSt = 220, find < PSQ

A. 22o

B. 45o

C. 67o

D. 89o

Detailed Solution

In In the diagram given, \(\alpha\) = 22o (vertically opp. angles), \(\alpha\) = \(\beta\) (alternate angles)

< PSQ + 133o + \(\beta\) = 180o (sum of angles of a \(\Delta\))

< PSQ + 133o + 22o = 180o

< PSQ = 180o - (133 + 22)o

45o
24.

In the diagram, O is the centre of the circle, < XOZ = (10cm)o and < XWZ = mo. Calculate the value of m.

A. 30

B. 36

C. 40

D. 72

Detailed Solution

In the diagram above, \(\alpha\) = 2mo (angle at centre = 2 x angle at circumference)

\(\alpha\) + 10mo = 360o (angle at circumference)

\(\alpha\) + 10mo = 360o(angles round a point)

2mo + 10mo = 360o

12mo = 360o
25.

Kweku walked 8m up to slope and was 3m above the ground. If he walks 12m further up the slope, how far above the ground will he be?

A. 4.5m

B. 6.0m

C. 7.5m

D. 9.0m

Detailed Solution

By similar triangles, \(\frac{8}{3}\) = \(\frac{8 + 12}{h}\)

\(\frac{8}{3} = \frac{20}{h}\)

h = \(\frac{3 \times 20}{8}\)

= 7.5m
26.

In the diagram, TS is a tangent to the circle at S. |PR| and < PQR = 177o. Calculate < PST.

A. 54o

B. 44o

C. 34o

D. 27o

Detailed Solution

In the diagram above, x1 = 180o - 117o = 63o(opposite angles of a cyclic quad.)

x1 = x2 (base angles of isos. \(\Delta\))

x1 + x2 + \(\alpha\) = 180o (sum of angles of a \(\Delta\)

63o + 63o + \(\alpha\) = 180o<
27.

In the diagram, PR||SV||WY|, TX||QY|, < PQT = 48o and < TXW = 60o.Find < TQU.

A. 120o

B. 108o

C. 72o

D. 60o

Detailed Solution

In the diagram, < TUQ + 60o(corresp. angles)

< QTU = 48o (alternate angles)

< QU + 60o + 48o = 180o(sum of angles of a \(\Delta\))

< TQU = 180o - 108o

= 72o
28.

In the diagram, TX is perpendicular to UW, |UX| = 1cm and |TX| = |WX| = \(\sqrt{3}\)cm. Find UTW

A. 135o

B. 105o

C. 75o

D. 60o

Detailed Solution

In \(\Delta\) UXT, tan\(\alpha\) = \(\frac{1}{\sqrt{3}}\)

\(\alpha\) = tan-1(\(\frac{1}{\sqrt{3}}\))

= 30o

In \(\Delta\)WXT, tan\(\beta\) \(\frac{\sqrt{3}}{\sqrt{3}}\) = 1

\(\beta\) = tan-1(1) = 45o

Hence, < UTW = \(\alpha\) + \(\beta\)

= 30o + 45o = 75o
29.

The figure is a pie chart which represents the expenditure of a family in a year. If the total income of the family was Le 10,800,000.00, how much was spent on food?

A. Le 2,250,000.00

B. Le 22,700,000.00

C. Le 3,600,000.00

D. Le 4,500,000.00

Detailed Solution

sectoral angle representing food

= 360o - (80 + 70 + 90)o

= 120o

Amount spent on food

= \(\frac{\tect{sectoral angle}}{360^o}\) x Le 10,800,000

= Le 3,600,000
30.

A fair die is thrown two times. What is the probability that the sum of the scores is at least 10?

A. \(\frac{5}{36}\)

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

C. \(\frac{5}{18}\)

D. \(\frac{2}{3}\)

Detailed Solution

\(\begin{array}{c|c}
& 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
1 & 1,1 & 1,2 & 1,3 & 1,4 & 1,5 & 1,6 \\ \hline 2 & 2,1 & 2,2 & 2,3 & 2,4 & 2,5 & 2,6 \\ \hline 3 & 3,1 & 3,2 & 3,3 & 3,4 & 3,5 & 3,6 \\ \hline 4 & 4,1 & 4,2 & 4,3 & 4,4 & 4,5 & 4,6 \\ 5 & 5,1 & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \\ \hline 6 & 6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6\end{array}\)

From the table