Quantitative Aptitude (QA) Test - 26

Angle around a point add up to \(360^{\circ}\)

Sum of all angles of a triangle is \(180^{\circ}\)

\(\Rightarrow \angle \mathrm{y}=180^{\circ}-(\angle \mathrm{a}+\angle \mathrm{b}) \ldots \ldots \ldots . .(\mathrm{i})\)

\(\Rightarrow \angle \mathrm{X}=180^{\circ}-(\angle \mathrm{C}+\angle \mathrm{d}) \ldots \ldots \ldots \text { (ii) } \)

\(\Rightarrow \angle \mathrm{p}=180^{\circ}-(\angle e+\angle \mathrm{f}) \ldots \ldots \ldots \ldots .(i i i) \)

\(\Rightarrow \angle \mathrm{q}=180^{\circ}-(\angle \mathrm{h}+\angle \mathrm{g}) \ldots \ldots \ldots \ldots \text { (iv) }\)

\(\Rightarrow \angle \mathrm{Z}=180^{\circ}-(\angle \mathrm{i}+\angle \mathrm{j}) \ldots \ldots \ldots(\mathrm{v})\)

\(\text { also, } 2(\angle \mathrm{y}+\angle \mathrm{x}+\angle \mathrm{p}+\angle \mathrm{q}+\angle \mathrm{Z})=360^{\circ} \)

\(\Rightarrow(\angle \mathrm{y}+\angle \mathrm{x}+\angle \mathrm{p}+\angle \mathrm{q}+\angle \mathrm{z})=180^{\circ}\)

adding all equation we have

\(\Rightarrow(\angle \mathrm{y}+\angle \mathrm{x}+\angle \mathrm{p}+\angle \mathrm{q}+\angle \mathrm{Z})=900^{\circ}-(\angle \mathrm{a}+\angle \mathrm{b}+\angle \mathrm{C}+\angle \mathrm{d}+\angle \mathrm{e}+\angle \mathrm{f}+\angle \mathrm{g}+\angle \mathrm{h}+\angle \mathrm{i} \)

\(+\angle \mathrm{j}) \)

\(\Rightarrow 180^{\circ}=900^{\circ}-(\angle \mathrm{a}+\angle \mathrm{b}+\angle \mathrm{c}+\angle \mathrm{d}+\angle \mathrm{e}+\angle \mathrm{f}+\angle \mathrm{g}+\angle \mathrm{h}+\angle \mathrm{i}+\angle \mathrm{j}) \)

\(\Rightarrow(\angle \mathrm{a}+\angle \mathrm{b}+\angle \mathrm{C}+\angle \mathrm{d}+\angle \mathrm{e}+\angle \mathrm{f}+\angle \mathrm{g}+\angle \mathrm{h}+\angle \mathrm{i}+\angle \mathrm{j})=900^{\circ}-180^{\circ}=720^{\circ}\)

\(\therefore\) required value of angle \(=720^{\circ}\)

Hence, the correct answer is 720.

Given,

The total number of students is 50 and the number of selected students should be either multiple of 2 or \(5 .\)

The average weight of students who were not selected \(=56 ~kg\)

The average weight of students who were selected \(=58~ kg\)

Total students whose number was multiple of \(2=25\)

Total students whose number was multiple of \(5=10\)

Total students whose number was multiple of 5 and 2 both i.e., of \(10=5\)

Total students whose number was either multiple of 2 or multiple of \(5=(25+10-5)\) \(=30\)

It means, 30 students were selected and 20 students were not selected.

According to the question,

The sum of the weight of all the students \(=(56 \times 20+58 \times 30) kg\) \(=2860~ kg\)

The average weight of all the students \(=\frac{2860}{50}\) \(=57.2~ kg\)

Hence, the correct answer is57.2.

Let in A, quantity of liquid X, Y and Z is ‘5x’, ‘4x’ and ‘3x’ litres respectively

And in B, quantity of liquid Y and Z is ‘16y’ and ‘15y’ respectively

According to the question:

(5x) : (4x + 16y) : (3x + 15y) = 25:68:60

So,$\left(\frac{5x}{(4x+16y)}\right)=\frac{25}{68}$

$\left(\frac{5x}{x+4y}\right)=\frac{25}{17}$

85x = 25x + 100y

60x = 100y

3x = 5y

And,$\frac{(4x+16y)}{(3x+15y)}=\frac{98}{60}$

$\left(\frac{(x+4y}{(x+5y}\right)=\frac{17}{20}$

20x + 80y = 17x + 85y

3x = 5y

So, x=5, y=3

Quantity in container A = 12x = 60 litres

Quantity in container B = 31y = 93 litres

Desired percentage =$\left[\frac{(93-60)}{\left(60\times 100\right)}\right]=55\%$

Total number of letters in the word \(=10\)

Position of \(A\) and \(T\) are fixed.

Remaining places \(=8\)

\(=8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\)

Total number of ways of writing remaining 8 places \(=8\) !

\(=40320\)

Letters I and 0 are repeated two times, So divide by 2 ! two times.

Letter \(\mathrm{N}\) is repeated three times, So divide by 3 !

The required total number of ways \(=\frac{(8 \times 7 \times 6 \times 5 \times 4 \times 3 !)}{(3 ! \times 2 ! \times 2 !)}\)

\(=8 \times 7 \times 6 \times 5 \)

\(=1680\)

Hence, the corret answer is 1680.

Let the given triangle be ∆ABC, as shown below:

\(\Rightarrow B C=\text { Base }=\left(u^{2}-v^{2}\right) \)

\(\Rightarrow A C=\text { Hypotenuse }=\left(u^{2}+v^{2}\right)\)

According to Pythagoras theorem,

\((A C)^{2}=(A B)^{2}+(B C)^{2} \)

\(\Rightarrow\left(u^{2}+v^{2}\right)^{2}=(A B)^{2}+\left(u^{2}-v^{2}\right)^{2} \)

\(\Rightarrow u^{4}+v^{4}+2 u^{2} v^{2}=(A B)^{2}+u^{4}+v^{4}-2 u^{2} v^{2} \)

\(\Rightarrow(A B)^{2}=4 u^{2} v^{2} \)

\(\Rightarrow A B=2 u v\)

Now,

Area of triangle \(=\frac{1}{2} \times \text { Base } \times \text { Height } \)

\( \frac{1 }{ 2} \times B C \times A B=2016 \)

\(\Rightarrow\left(u^{2}-v^{2}\right) \times u v=2016 \)

\(\Rightarrow u v(u+v)(u-v)=32 \times 7 \times 9\)

We can write,

\( u v(u+v)(u-v)=16 \times 9 \sqrt{2} \times 7 \sqrt{2}\)

Thus, we get, \(u=8 \sqrt{2}\) and \(v=\sqrt{2}\)

\(\therefore\) Perimeter of triangle:

\(=A B+B C+A C=2 u v+u^{2}-v^{2}+u^{2}+v^{2}\)

\(=2 u v+2 u^{2}=32+256=288\) units

Given:

Cost of per articles \(=\) Rs. 5

Profit \(\%=30 \%\)

Profit percent \(=\left(\frac{S P-C P}{C P}\right) \times 100\)

\(\mathrm{CP}=\) cost price

\(\mathrm{SP}=\) selling price

CP of 600 articles \(=\) Rs. \((600 \times 5)\)

\(\Rightarrow\) Rs. 3000

SP of 500 articles \(=\) Rs. \(\left(3000 \times \frac{13}{10}\right)\)

\(\Rightarrow\) Rs. 3900

SP per articles \(=\) Rs. \(\left(\frac{3900}{500}\right)\)

\(\Rightarrow\) Rs. \(7.8\)

SP of 540 articles \(=\) Rs. \((540 \times 7.8)\)

\(\Rightarrow\) Rs. 4212

Profit \(=\) Rs. \((4212-3000)=\) Rs. 1212

Profit percent \(=\left(\frac{1212}{3000}\right) \times 100\)

\(\Rightarrow 40.4 \%\)

\(\therefore\) Profit percent is \(40.4 \%\)

Given:

Numbers from 200 to 800 .

Total number from 200 and \(800=800-200+1=601\) (including 200)

Total number from 1 to 200 (200 excluding) which are divisible by \(5=\frac{199}{5} \approx 39\)

Total number from 1 to 800 (800 including) which are divisible by 5 \(=\frac{800}{5}=160\)

Total number from 200 to 800 which are divisible by \(5=160-39\) \(=121\)

Total number from 1 to 200 (200 excluding) which are divisible by \(7=\frac{199}{7} \approx 28\)

Total number from 1 to 800 ( 800 including) which are divisible by 7 \(=\frac{800}{7} \approx 114\)

Total number from 200 to 800 which are divisible by \(7=114-28\) \(=86\)

Similarly, Total number from 200 to 800 which are divisible by \(35=\) \(\frac{800}{35}-\frac{200}{35}=22-5=17\)

Number neither divisible by 5 nor \(7=601-121-86+17=411\)

\(\therefore\) Number neither divisible by 5 nor 7 is 411 .

Hence, the correct answer is 411.

As we know,

If \(z=x+i y \ldots(1)\) be any complex number, then its modulus is given by,

\(|z|=\sqrt{x^2+y^2}\)

Let,

\(z=\frac{\sqrt{2}+i}{\sqrt{2}-i}\)

Multiplying by \((\sqrt{2}+i)\) in numerator and denominator, we get

\(z=\frac{\sqrt{2}+i}{\sqrt{2}-i} \times \frac{\sqrt{2}+i}{\sqrt{2}+i} \)

\(\Rightarrow z=\frac{2+2 \sqrt{2} i+i^2}{2-i^2} \)

\(\Rightarrow z=\frac{2+2 \sqrt{2} i-1}{2-(-1)} \quad\left[\because i^2=-1\right] \)

\(\Rightarrow z=\frac{1+2 \sqrt{2} i}{3}\)

\(\Rightarrow z=\frac{1}{3}+i \frac{2 \sqrt{2}}{3} \ldots(2)\)

Comparing equation (1) and (2), we get \(x=\frac{1}{3}\) and \(y=\frac{2 \sqrt{2}}{3}\)

\(\text { As, }|z|=\sqrt{x^2+y^2} \)

\(\therefore|z|=\sqrt{\left(\frac{1}{3}\right)^2+\left(\frac{2 \sqrt{2}}{3}\right)^2} \)

\(\Rightarrow|z|=\sqrt{\frac{1}{9}+\frac{8}{9}} \)

\(\Rightarrow|z|=1\)

So, the modulus of \(\frac{\sqrt{2}+i}{\sqrt{2}-i}\) is \(1\)

Hence, the correct answer is 1.

Given,

Number of days by Rakesh to complete the project \(=10\) days

Number of days taken by Ravi to complete a project \(=15\) days

\(\because\) Rakesh takes 10 days to complete a project.

\(\therefore\) Work done by Rakesh in a day \(=\frac{1}{10}\)

\(\because\) Ravi takes 15 days to complete a project.

\(\therefore\) Work done by Ravi in a day \(=\frac{1}{15}\)

Let Anas takes \(X\) days to complete the project.

Work done by Rakesh in 4days + work done by Rakesh and Ravi in

2 days \(+\) work done by all three persons in \(\frac{16}{13}\) days \(=1\)

\(\Rightarrow \frac{4}{10}+2 \times\left(\frac{1}{10}+\frac{1}{15}\right)+\frac{16}{13} \times\left(\frac{1}{10}+\frac{1}{15}+\frac{1}{X}\right)=1\)

\(\Rightarrow \frac{2}{5}+2 \times\left(\frac{5}{30}\right)+\frac{16}{13} \times\left(\frac{5}{30}+\frac{1}{X}\right)=1\)

\(\Rightarrow \frac{1}{3}+\frac{16}{13} \times\left(\frac{1}{6}+\frac{1}{X}\right)=1-\frac{2}{5}\)

\(\Rightarrow \frac{16}{13} \times\left(\frac{1}{6}+\frac{1}{X}\right)=\frac{3}{5}\)

\(\Rightarrow \frac{16}{13} \times\left(\frac{1}{6}+\frac{1}{X}\right)=\frac{3}{5}-\frac{1}{3}\)

\(\Rightarrow \frac{1}{6}+\frac{1}{X}=\frac{13}{16} \times \frac{4}{15}\)

\(\Rightarrow \frac{1}{X}=\frac{13}{60}-\frac{1}{6}\)

\(\Rightarrow \frac{1}{X}=\frac{1}{20}\)

\(\Rightarrow X=20\)

\(\therefore\) Number of days taken by Anas to complete the project \(=20\) days.

Hence, the correct answer is 20.

Let the total amount which is divided be ' \(x\) '

\(\Rightarrow\) Harsh's money \(=\frac{8 x}{25}\) \(\Rightarrow\) Nikhil's money \(=\frac{17 x}{25}\)

Harsh spent \(\frac{3}{4}^{\text {th }}\) of his money

\(\Rightarrow\) Money left with harsh \(=\frac{1}{4}^{\text {th }}\) of \(\frac{8 x}{25}=\frac{2 x}{25}\)

Nikhil spent \(20 \%\) of his money; \(80 \%\) is thus left.

\(\Rightarrow\) Money left with Nikhil \(=6528\)

\(\Rightarrow 6528=\left(\frac{80}{100}\right) \times\left(\frac{17 x}{25}\right)\)

\(\Rightarrow x=12000\)

\(\therefore\) Money spent by Harsh \(=\frac{3}{4}^{\text {th }}\) of \(\frac{8}{25}=3 \times 8 \times \frac{12000}{100}=\) Rs.

And, Money spent by Harsh \(=20 \%\) of

\(\frac{17}{25}=20 \times 17 \times \frac{12000}{2500}=R s .1632\)

\(\Rightarrow\) Total money spent by both

\(=R s .2880+R s .1632=R s .4512\)