Quantitative Aptitude Test - 3

Given,radius \(=22\) unit

Height=16 unit

Curved surface area of the cone \(=\pi r L\)

Where, \(L=\sqrt{\left(r^{2}+h^{2}\right)}\)

\(L=\sqrt{\left(22^{2}+16^{2}\right)} \)

\(L=\sqrt{(484+256)} \)

\(L=\sqrt{740}\)

\(\text { L=27.20 }\)

\(\text { Cured surface area of the cone }=\pi r L\)

\(=\frac{22}{7} \times 22 \times 27.20 \)

\(1880.6 \text { square unit }\)

Given:

A trip from Mumbai to Pune takes \(4 ~\text{hours 30 minutes}\) at a speed of \(60~\mathrm{km/hr}\)

Distance will be the same. Only speed and time change.

We know that:

Distance \(=\) Speed \(\times\) Time

Time \(=4\) \(\mathrm{hours}\) \(30\) \(\mathrm{minutes}\) \(=4.5\) \(\mathrm{hours}\)

Speed \(=60 \mathrm{~km} / \mathrm{hr}\)

Distance \(=4.5 \times 60=270 \mathrm{~km}\)

Now, if the speed is \(15 \mathrm{~m} / \mathrm{sec}\):

\(15 \mathrm{~m} / \mathrm{sec}=15 \times(\frac{18}{5}) \mathrm{~km} / \mathrm{hr}\)

\(\Rightarrow 54 \mathrm{~km} / \mathrm{hr}\)

Time to travel \(270 \mathrm{~km}\) with speed \(54 \mathrm{~km} / \mathrm{hr}\):

Time \(=\frac{270}{54}=5\) \(\mathrm{hours}\)

All the letters starting with A and B will comes before the given word.

Total number of words starting with 'A' = 4! = 24

Total number of words starting with 'B' = 4! = 24

Total number of words starting with 'CA' = 3! = 6

Total number of words starting with 'CB' = 3! = 6

After 'CB' the word will start with 'CD' and the first word will be 'CDABE'.

∴ Position of the given word = 24 + 24 + 6 + 6 + 1 = 61

Let the tens and unit digits be \(x\) and \(y\) respectively.

Statement I do not give the answer.

From statement II, we have,

\((x+y)=5(x-y) \)

\((x+y)=5 x-5 y\)

\(4 x=6 y \)

\(\frac{x}{y}=\frac{6}{4}=\frac{3}{2}\)

Therefore the required ratio \(=\frac{x}{y}=x y=3: 2\)

Here, the data in Statement II alone are sufficient to find the answer, while the data in Statement I alone are not sufficient.

In order to find out the rate of interest, time for which the sum has been deposited is needed.

But this has provided neither in statement I or statement II.

Statement III is also not an informative statement because it states about a general case which is true for any Sum. This is applicable for Simple Interest illustrated below:-

Suppose \(x\) is sum, it will double in 25 years at the rate of \(4 \%\) per annum.

\(S t=\frac{(x \times 4 \times 25)}{100}=x\)

Therefore, it doubles in 25 years which is true for any sum.

From statement I,

This statement gives us idea about the possibility of number (average)

The sum of the middle numbers = 128, so possible prime numbers = 59, 61, 67, 71

From statement II,

The sum of the 1st and last no. = 130, which is 59, 61, 67, 71

From statement III,

The difference between 1st and last in the above statements is 12, but there are other numbers also viz. (29, 41), (31, 43) etc.

So, either I or II is sufficient to answer the question.

Given,

Rajiv sold to Richa at Profit \(=17 \%\)

Richa sold to Shivani at Profit \(=9 \%\)

Shivani sold to Chetna at loss \(=5 \%\)

Selling price \(=\) Cost price \(+\) Profit

Let the cost price of the item for Rajiv be Rs. \(x\)

Profit \(=17 \%\) or Rs. \(0.17 x\)

Selling Price for Rajiv \(=\) Cost price for Richa

\(=\) Rs. \(1.17 x\)

Profit earned by Richa \(=9 \%\) of \(1.17 x\)

\(=\) Rs. \(0.1053 x\)

Selling Price for Richa \(=\) Cost Price for Shivani

\(=1.17 x+0.1053 x\)

\(=\) Rs. \(1.2753 x\)

Loss incurred by Shivani \(=5 \%\) of \(1.2753 x\) \(=0.063 x\)

Selling Price for Shivani \(=1.2753 x-0.063 x\)

\(=\) Rs. \(1.211 x\)

Total profit earned \(=0.17 x+0.1053 x-0.063 x\) \(=\) Rs. \(0.2123 x\) or \(21.23 \%\)

\(\therefore\) Total profit earned on the article is \(21.23 \%\)

\(a^{2}-b^{2}=(a-b)(a+b)\)

\(\sec ^{2} x-\tan ^{2} x=1\)

\(\sec ^{4} x-\tan ^{4} x\)

\(=\left(\sec ^{2} x-\tan ^{2} x\right)\left(\sec ^{2} x+\tan ^{2} x\right) \quad\left(\because a^{2}-b^{2}=(a-b)(a+b)\right)\)

\(=1 \times\left(1+\tan ^{2} x+\tan ^{2} x\right)\)

\(=1+2 \tan ^{2} x\)

Hence, the correct option is (C).

Let the price of a machine be Rs. $100$ and let original sales be$100$ pieces.

As the price of the machine is increased by $20\%$.

And the number sold is decreased by $25\%$.

New price $=100+\frac{20}{100}\times 100=120$

Number of sales after price rise $=100-\frac{25}{100}\times 100=75$

Initial revenue$=100\times 100=10000$

New revenue $=120\times 75=9000$

Decrease in revenue

$=\left(\frac{(10000-9000)}{10000}\right)\times 100$

$\therefore$ Decrease in revenue by $10\%$.