Quantitative Aptitude Test - 5

As we know the sum of opposite angles of a cyclic quadrilateral is \(180^{\circ}\)

By, angles on a straight line \(\angle C E D=120^{\circ}\) By, angles on a straight line

By, angles in a straight line \(\angle B E D=60^{\circ}\) By, angles in a straight line

\(\therefore \angle E D B=90^{\circ}\)

\(\therefore \frac{B D}{B E}=\cos 30^{\circ}\)

\(\frac{6}{B E}=\frac{\sqrt{3}}{2}\)

\(\Rightarrow B E=4 \sqrt{3} \mathrm{~cm}\)

\(\therefore B C=B E+C E\)

\(=4 \sqrt{3}+5 \sqrt{3}=9 \sqrt{3} \mathrm{~cm}\)

Now, since \(A B\) and \(C B\) are the secants of the circle

\(\therefore B D \times B A=B E \times B C\)

\(6 \times B A=4 \sqrt{3} \times 9 \sqrt{3}\)

\(\Rightarrow B A=18 \mathrm{~cm}\)

Again, \(\triangle A B C\) is a right-angled triangle \(\left(\therefore \angle C=90^{\circ}\right)\)

\(A C=9 \mathrm{~cm}\) (alternatively apply Pythagoras theorem)

and \(A D=A B-B D=12 \mathrm{~cm}\)

\(\frac{A C}{A D}=\frac{9}{12}=\frac{3}{4}\)

Given:

A person covers \((\frac{1}{5})^{{tn}}\) of the distance with the speed \(=20 {~km} / {h}\)

A person covers \((\frac{1}{4})^{{th}}\) of the distance with the speed \(=10 {~km} / {hr}\)

We Know that:

Average Speed \(=\) \(\frac{\text{Total Distance}}{\text{Total time taken}}\)

Let the total distance \(={LCM}(4,5)=20 {unit}\)

A person cover distance with speed of \(20 {~km} / {hr}=\frac{1}{5} \times 20=4 {unit}\)

A person cover distance with speed of \(10 {~km} / {hr}=\frac{1}{4} \times 20=5 {unit}\)

Remaining Distance \(=20-4-5=11 {unit}\)

Average Speed \(=20\) unit \(\times[(\frac{4}{20})+(\frac{5}{10})+(\frac{11}{60})]\)

\(\Rightarrow 20\) unit \(\times \frac{60}{(12+30+11)}=22.64 {~km} / {hr}\)

\(\therefore\) The required answer is \(22.64 {~km} / {hr}\)

Given:

The given word = SOFTWARE

Formula used:

Arrangement of \(\mathrm{n}\) letter of a word

Case 1:- If there were no repeating letters \(=\mathrm{n} !\)

Case 2:- If there were "r" letter of one kind \(=\frac{\mathrm{n} !}{ \mathrm{r} !}\)

Keeping all the vowel at one place and consider it as one vowel

\(\Rightarrow\) SFTWR(OAE)

Here, \(n=6\) and number of vowel \(=3\)

All letter can be arranged in \(6 !\) ways and Vowel can be arranged in 3! ways

Total number of ways to arrange the word \(=6 ! \times 3 !=6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 3 \times 2 \times 1=4320\)

\(\therefore\) Total number of ways to arrange the word \(=4320\)

Statement I:

The age of Rahul at the time of marrige = 25 years.

Statement I alone is not sufficient to answer the question.

Statement II:

Total age of Rahul and his wife at the time of marrige = 24 × 2 = 48 years.

Statement II alone is not sufficient to answer the question.

Statement III:

Difference between the ages of Rahul and his son = 24 years.

Statement III alone is not sufficient to answer the question.

All the statements are not sufficient to answer the question because in these statements it is not given that when Rahul got married.

Taking all statement together,

Let the profit earned by company in \(2001=\) Rs. \(x\) and in \(2002=\) Rs. \(y\)

Profit earned in \(2003=1.4 x\)

\(x+y=\) Rs. 20 crore .... (i)

From statement (III),

\(1.4 x=y \times \frac{80}{100}\)

\(x=\frac{4}{7} y \ldots . . \text { (ii) }\)

From equation (i) and (ii), we can get the required profit.

So all the statements are required to find profit in the year 2002 .

Let the total students \(=x\)

Statement I:

Failed students \(=x \times 35 \%\), Passed students \(=x \times 65 \%\)

Statement II:

Difference between failed and passed students \(=240\)

Statement III:

Ratio of Boys and Girls \(=4: 5\)

Statement I + Statement II:

\(x \times 65 \%-x \times 35 \%=240\)

\(x \times 30 \%=240\)

\(x=800\)

Students who cleared the exam \(=800 \times 65 \%=520\)

Statement I and II together is sufficient to answer the question.

From Statement I:

Let Anuj \(=4 x\), Vipin \(=5 x\)

Sanjeev \(=4 x+4\)

From Statement II:

Anuj \(=20\) years, Sanjeev \(=x\), Vipin \(=x+1\)

Statement III:

Let age of Mohan \(=m\) years

Sanjeev \(=m-5\) years

From Statement \(I+\) Statement II:

Anuj \(=20\) years, Vipin \(=\frac{20}{4 x} \times 5 x=25\) years

Sanjeev \(=25-1=24\) years

Statement I and II both are necessary to answer the question.

External radius \(=11 \mathrm{~cm}\)

Internal radius \(=7 \mathrm{~cm}\)

Volume of iron \(=\frac{22}{7} \times\left(11^{2}-7^{2}\right) \times 63\)

\(=22 \times 72 \times 9=14256 \mathrm{~cm}^{3}\)

Weight of iron \(=(14256 \times 4)=57024 \mathrm{gms}=57.024 \mathrm{~kg}\)

Given:

Articles price \(=\) Rs. \(9520\) each

Profit \(=12 \%\)

Loss \(=15 \%\)

Total SP of both articles \(=2 \times\) Articles price

Total SP of both articles \(=2 \times 9520\)

\(=19040\)

CP of first articles \(=\) Articles price \(\times \frac{100}{100 \pm \text{Profit/Loss}}\)

CP of first articles \(=9520 \times \frac{100}{112}\)

\(=8500\)

CP of second articles \(=9520 \times \frac{100}{85}\)

\(=11200\)

Total CP of both articles \(=8500+11200\)

\(=19,700\)

Loss \(=19700-19040=660\)

Loss percentage \(=\frac{\text{Loss}}{\text{Total CP of both articles}}\)

\(\therefore\) Loss percentage \(=\frac{660}{19700} \times 100\)

\(=3.4 \%\)

Given:

\(\cos 20^{\circ}+\cos 100^{\circ}+\cos 140^{\circ}\)

\(=2 \cos \frac{20^{\circ}+100^{\circ}}{2} \cos \frac{100^{\circ}-20^{\circ}}{2}+\cos 140^{\circ}\)

\(=2 \cos 60^{\circ} \cos 40^{\circ}+\cos 140^{\circ}\)

\(=2 \times \frac{1}{2} \cos 40^{\circ}+\cos 140^{\circ}\)

\(=\cos 40^{\circ}+\cos 140^{\circ}\)

\(=2 \cos \frac{140^{\circ}+40^{\circ}}{2} \cos \frac{140^{\circ}-40^{\circ}}{2}\)

\(=2 \cos 90^{\circ} \cos 50^{\circ}\)

\(=0 \quad\left(\because \cos 90^{\circ}=0\right)\)