Mr Manu is 4 times as old as his son, Adu. 7 years ago the sum of their ages was 76. How old is Adu?
22years
12years
18years
15 years
Correct answer is C
Let Mr Manu be x years old and Adu be y years old.
But Mr Manu is four times as old as Adu then, x = 4y.
7 years ago, the sum of their ages was 76.
( x - 7) + ( y - 7) = 76
x + y - 14 = 76
x + y = 76 + 14 = 90
But x = 4y
Therefore, 4y + y = 90
5y = 90
y = \(\frac{90}{5}\)
y = 18
Therefore Adu is 18 years old.
13
11
5
9
Correct answer is B
Using the venn diagram above
μ = 30
n(W) = 15
n(M) = 13
\(n(W ∪ M)^1 = 6\)
Let x = number of students that study both woodwork and metalwork
i.e. n(W ∩ M) = x
Number of students that study only woodwork,\(n(W ∩ M^1)\) = \(15 - x\)
Number of students that study only metalwork, \(n(W^1 ∩ M)\) = \(13 - x \)
Bringing all together,
\(n(W ∩ M^1)\) +\( n(W^1 ∩ M)\) + \(n(W ∩ M)\) + \(n(W ∪ M)^1\) = \(μ\)
∴ (15 - x) + (13 - x) + x + 6 = 30
⇒ 34 - x = 30
⇒ 34 - 30 = x
∴ x = 4
\(n(W ∩ M^1)\) = \(15 - 4 = 11\)
∴ The number of students that study woodwork but not metalwork is 11.
\(11.6cm^2\)
\(12.7cm^2\)
\(10.2cm^2\)
\(9.4cm^2\)
Correct answer is A
\(\theta = 115° , radius = 3.4cm^2\)
Area of a sector = \(\frac{\theta}{360} \times \pi r^2\)
= \(\frac{115}{360} \times \frac{22}{7} \times 3.4\times 3.4\)
= \(\frac{29246.4}{2520}\)
= \(11.6cm^2\)
If goods are returned, the seller issues
a debit notes
a consignment note
an advice note
a credit note
Correct answer is D
Credit note is also known as credit memo is a commercial document issued by a seller to a buyer. Credit notes act a s a source document for the sales return journal
legible
concise
accurate
flexible
Correct answer is C
A meeting is accurate when it's clearly present the proceedings and outcomes of the meeting