Mathematics Questions and Answers

\(3\sqrt{12} + 10\sqrt{3} - \frac{6}{\sqrt{3}}\)

= \(3\sqrt{4\times3} + 10\sqrt{3} - \frac{6}{\sqrt{3}}\)

= \(6\sqrt{3} + 10\sqrt{3} - \frac{6}{\sqrt{3}}\)

treating  like a fraction, then 

= \(\frac{18 + 30 - 6}{\sqrt{3}}\)

= \(\frac{42}{\sqrt{3}}\)

rationalizing 

= \(14\sqrt{3}\)

Reflex ∠MOR = 2 × 124° = 248° (angle at the centre is twice the angle at the circumference)
∠MOR = 360° - 248° = 112° (sum of angle at a point is 360°)
∠OMN = 360° - (124°+ 48° + 112°) (sum of angles in a quadrilateral is 360°)
= 360° - 284°
∴ ∠OMN = 76°

Let the two numbers be x and y 

\(\frac{1}{3}( x + y) = 12\)

then x + y = 36 ..........i

2( x - y) = 12

x - y = 6 ............ii

add equations i and ii 

2x = 42

x = 21, put x = 21 into equation i

x + y = 36

21 + y = 36 

y = 36 - 21 = 15

therefore the numbers are 21 and 15

From the diagram above; Tan\(\theta = \frac{opp}{adj}\)

tan50° = \(\frac{124}{d}\)

d = \(\frac{124}{tan50}\)

therefore, d = 104.05m

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.

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.

\(\theta = 115° , radius = 3.4cm^2\)

Area of a sector = \(\frac{\theta}{360} \times \pi r^2\)

= \(\frac{115}{360} \times \frac{22}{7} \times 3.4\times 3.4\)

= \(\frac{29246.4}{2520}\)

= \(11.6cm^2\)

\(1 + \sqrt [3]{x-3} = 4\)

= \(\sqrt[3]{x - 3} = 4 - 1\)

\(\sqrt[3]{x - 3} = 3 \)

take the cube of both sides

= x - 3 = 27
 x = 27 + 3
∴ x = 30

∠SOR = 2 × 28° = 56° (angle at the centre is twice the angle at the circumference)
From ∆SOR
|OS| = |OR| (radii)
So, ∆SOR is isosceles.

∠ORS = \(\frac{180^0 -  56^0}{2} = \frac{124^0}{2}\)    ( base angles of isosceles triangle are equal)

∴ ∠ORS = 62°

12 men working 8 hours a day can complete the work in 4 days

12 men working 1 hour a day will complete the same work in (8 x 4) days [working less hours a day means they have to work for more days]


1 man working 1 hour a day will complete same work in (8 x 4 x 12) days [fewer number of men working means the remaining ones will have to work for more days ]

So, it will take 1 man working 1 hour a day to complete a piece of work in (8 x 4 x 12) days = 384 days

Now,

1 man working 16 hours a day will complete the piece of work in (384/16) days = 24 days [working more hours a day means less days to complete the work]

∴ 4 men working 16 hours a day will complete the piece of work in (24/4) days = 6 days [more men working means fewer days to complete the work]

ALTERNATIVELY

12 x 8 x 4 = 4 x 16 x d

⇒ d = \(\frac{12 \times 8 \times 4}{4 \times 16}\)

⇒ d = \(\frac{384}{64}\)

⇒ d = 6

∴ 4 men working 16 hours a day will complete the piece of work in 6 days