Basic Maths: Test 11

This is an assessment test.
These tests focus on the basics of Maths and are meant to indicate your preparation level for the subject.
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Basic Maths: Test 11

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Question 1

The value of

$\displaystyle 1+\frac{3}{1+\frac{2}{3+\frac{4}{5}}}$

A	$\displaystyle \frac{12}{29}$
B	$\displaystyle \frac{68}{19}$
C	$\displaystyle \frac{86}{29}$
D	$\displaystyle \frac{2}{19}$

Question 1 Explanation:

$\displaystyle \begin{array}{l}1+\frac{3}{1+\frac{2}{\frac{15+4}{5}}}\\=1+\frac{3}{1+\frac{2\times 5}{19}}\\=1+\frac{3}{\frac{19+10}{19}}\\=1+\frac{57}{29}=\frac{29+57}{29}=\frac{86}{29}\end{array}$

Question 2

Simplify

$\displaystyle \frac{19}{43}\div \frac{1}{2+\frac{1}{3+\frac{1}{1+\frac{1}{4}}}}$

A	$\displaystyle \frac{19}{43}$
B	$\displaystyle \text{1}$
C	$\displaystyle \frac{34}{19}$
D	$\displaystyle \frac{45}{43}$

Question 2 Explanation:

We can simplified this question as

$\displaystyle \begin{array}{l}\frac{19}{43}\div \frac{1}{2+\frac{1}{3+\frac{1}{1+\frac{1}{4}}}}\\=\frac{19}{43}\div \frac{1}{2+\frac{1}{3+\frac{4}{5}}}\\=\frac{19}{43}\div \frac{1}{2+\frac{5}{19}}\\=\frac{19}{43}\div \frac{19}{43}\\=\frac{19}{43}\times \frac{43}{19}=1\end{array}$
So the right answer for the question is option b

Question 3

The simplification of

$\displaystyle \frac{4}{3+\frac{3}{1-\frac{2}{3}}}$ gives

A	7
B	1/3
C	5/12
D	7/12

Question 3 Explanation:

We can simplified the equation as

$\displaystyle \begin{array}{l}\frac{4}{3+\frac{3}{\frac{3-2}{3}}}\\=\frac{4}{3+\frac{3}{\frac{1}{3}}}\\\frac{4}{3+3\times 3}\\=\frac{4}{3+9}\\=\frac{4}{12}\\=\frac{1}{3}\end{array}$
So the best option for the question is option b

Question 4

$\displaystyle 8=p+\frac{1}{1+\frac{1}{3+\frac{1}{4}}}$ , then the value of p is:

A	$\displaystyle \frac{191}{17}$
B	$\displaystyle \frac{123}{17}$
C	$\displaystyle \frac{141}{17}$
D	$\displaystyle \frac{133}{17}$

Question 4 Explanation:

$\displaystyle \begin{array}{l}8=p+\frac{1}{1+\frac{1}{3+\frac{1}{4}}}\\\Rightarrow 8=p+\frac{1}{1+\frac{1}{\frac{12+1}{4}}}\\\Rightarrow 8=p+\frac{1}{1+\frac{4}{13}}\\\Rightarrow 8=p+\frac{1}{\frac{13+4}{13}}\\\Rightarrow 8=p+\frac{1}{\frac{17}{13}}\\\Rightarrow 8=p+\frac{13}{17}\,\,\\\Rightarrow \,p=8-\frac{13}{17}\\=\frac{123}{17}\end{array}$

Question 5