Quantitative Aptitude Test - 1

Radius of big sphere \(\mathrm{R}=16 \mathrm{~cm}\)

Radius of smaller sphere \(r=4 \mathrm{~cm}\)

Ratio between radii of two spheres \(=16: 4=4: 1\)

Ratio between volumes of two spheres \(\frac{4}{3} \pi R^{3}: \frac{4}{3} \pi r^{3}\)

\(=4^{3}: 1^{3}\)

\(=64: 1\)

Therefore, 64 spheres can be made from one big sphere.

As we know,

Distance \(=\) Speed \(×\) Time

As Distance covered in both the cases is constant, Speed will be inversely proportional to time.

So, if Time by bus: Time by train \(= 2: 1\), Speed of bus: Speed of train \(= 1: 2\)

Speed of bus is \(40\) km/hr, so Speed of train is \(80\) km/hr.

Treat B and T as a single letter. Then the remaining letters \((5+1=6)\) can be arranged in \(6 !\) ways. 

Since, O is repeated twice, we have to divide by \(2\) and the B and T letters can be arranged in \(2 !\) ways.

Total no. of ways

\(=\frac{6 ! \times 2 !}{2}=720\)

Total number of students, n(U) = 100

Number of science students, n(U) = 35

Number of math students, n(M) = 45

Number of students who like both, n(M∩S) = 10

Number of students who like either of them,

n(MUS) = n(M) + n(S) – n(M∩S)

⇒ 45+35-10 = 70

Number of students who like neither = n(U) – n(MUS) = 100 – 70 = 30

Given,

The ratio of the cost price of two articles \(=4: 9\)

Let the cost price of the first article \(=4 x\)

The cost price of the second article \(=9 x\)

The articles are marked up by \(40 \%\) and \(15 \%\) respectively.

Marked price of first article \(=4 x \times\left(\frac{140}{100}\right)=5.6 x\)

Marked price of second article \(=9 x \times\left(\frac{115}{100}\right)=10.35 x\)

The ratio of their Marked price \(=112: 207\)

Marked price of first article \(=112 y\)

Marked price of second article \(=207 y\)

\(\therefore 5.6 x=112 y\)

\(x=20 y\)

According to question,

Selling price of first article \(=112 y \times\left(\frac{7}{8}\right)=98 y\)  .....\((12.5 \%=\frac{1}{8})\)

Profit \(=98 y-4 x=270\)

\(\Rightarrow 98 y-80 y=270\)

\(\Rightarrow 18 y=270\)

\(\Rightarrow y=15\)

\(x=20 \times 15=300\)

Cost price of second article \(=300 \times 9=2700\)

Marked price of second article \(=2700 \times\left(\frac{115}{100}\right)=3105\)

Selling price of second article \(=3105 \times\left(\frac{8}{9}\right)=2760\)     .....\((11.11 \%=\frac{1}{9})\)

Profit earned \(=2760-2700=60\)

\(\therefore\) The profit earned on second article is Rs. \(60\).

Given,

\(x=\tan ^{-1}(\frac{1 }{5})\)

\(\tan x=\frac{1 }{5}\)

As we know that, \(\cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\)

\(\cos 2 \mathrm{x}=\frac{1-\tan ^{2} \mathrm{x}}{1+\tan ^{2} \mathrm{x}}\)

\(=\frac{1-\left(\frac{1}{5}\right)^{2}}{1+\left(\frac{1}{5}\right)^{2}}\)

\(=\frac{\left(\frac{25-1}{25}\right)}{\left(\frac{25+1}{25}\right)}\)

\(=\frac{24}{26}\)

\(=\frac{12}{13}\)

Students passed in English = 80%

Students passed in Math's = 85%

Students passed in both subjects = 73%

Then, number of students passed in at least one subject

= (80+85)-73

= 92%. [The percentage of students passed in English and Maths individually, have already included the percentage of students passed in both subjects. So, We are subtracting percentage of students who have passed in both subjects to find out percentage of students at least passed in one subject.]

Thus, students failed in both subjects = 100-92 = 8%.

From the statement I, 3A = B + C ............ (i)

From the statement II, 5C = 3(A + B) ............. (ii)

From the question, A + B + C = 120 ........... (iii)

If we solve the three equation then A = 30, B = C = 45

In question, we need to conclude who among the three got the highest share here our answer can be B or C therefore we can not conclude the unique answer.

Let the speed of the train = x m/sec and the length of train = y meters 

From the question, 30 × x = y + 600 (distance = speed × time) 

From the statement I, y = (x + 20) × 6 

From the statement II, y = x × 10 

We have 2 variable and three equation, we can conclude the value of y by either of 2 equations. 

x = 30 and y = 300