Quantitative Aptitude Test - 6

Let the CP of each of two cheapest articles = x and the CP of costliest article = x + 1

Then, x + x + x + 1 = 49,

x = 16

therefore, the CP of costliest article = 16 + 1 = 17

From the Statement II, we can say that the cost price of two articles is same

i.e., the cost price of first article = cost price of second article= x

And from Statement III, we can say that the cost price of costliest article is 6.25% more than the cost price of cheapest article

Therefore, the cost price of costliest article\( = x+x \times 6.25 \% \)

\(=1.0625x\)

According to question,

\(x+x+ 1.0625x=49\)

\(x=16\)

Therefore, cost price of costliest article\(=1.0625x\)

\(= 1.0625 \times 16=17\)

Therefore, by combining both the statement we can also get our answer.

If we have the dimensions, from Statement a,

Volume of cuboid \(=10 \times 5 \times 12=600 \mathrm{~cm}^{3}\)

If thickness is ' \(t\) ' and let side of square sheet be \(S\), then,

\(600=\left(\mathrm{S}^{2}\right) \times (\mathrm{t})\)

If \(t=1.5 \mathrm{~cm}\) is taken from Statement II,

\(\frac{600}{1.5}=\left(\mathrm{S}^{2}\right)=400\)

\(\mathrm{S}=20 \mathrm{~cm}\)

Height of cylinder \(=\mathrm{S}=20 \mathrm{~cm}\) [As square sheet is rolled so the side of the cylinder will be equal to side of square]

Outer circumference \(=\mathrm{S}=20 \mathrm{~cm}=2 \pi \mathrm{r}\)

Or, \(r=\frac{10}{\pi} \approx 3.185\)

Thickness taken, \(t=1.5 \mathrm{~cm}\)

So inner radius \(=3.185-1.5=1.685 \mathrm{~cm}\)

Whereas Statement III has no significance anywhere.

But none of the statement alone can answer the question individually.

Hence, answer is using statement I and II together is sufficient

Given:

There are a total of 7 red \(+4\) blue \(=11\) balls.

Probability of drawing 1 red ball \(=\frac{{ }^{7} \mathrm{C}_{1}}{{ }^{11} \mathrm{C}_{1}}=\frac{7}{11}\).

Probability of drawing 1 blue ball \(=\frac{{ }^{4} \mathrm{C}_{1}}{{ }^{11} \mathrm{C}_{1}}=\frac{4}{11}\).

Probability of drawing (1 red) AND ( 1 blue) ball \(=\frac{7}{11} \times \frac{4}{11}=\frac{28}{121}\).

Similarly, Probability of drawing (1 blue) AND (1 red) ball \(=\frac{4}{11} \times \frac{7}{11}=\frac{28}{121}\).

Probability of getting the balls of different colors \(=\frac{28}{121}+\frac{28}{121}=\frac{56}{121}\)

Given:

Percentage of passengers travelling from Haridwar to Delhi, Dehradun and Rishikesh = 20%, 15%, and 19%

Total percentage of passengers travelling from Haridwar to Dehradun and Rishikesh =15% + 19%

= 34%

Required percentage =$\frac{\mathrm{Passengers}\mathrm{travelling}\mathrm{from}\mathrm{Haridwar}\mathrm{to}(\mathrm{Dehradun}\mathrm{and}\mathrm{Rishikesh}-\mathrm{Delhi})}{\mathrm{Passengers}\mathrm{travelling}\mathrm{from}\mathrm{Haridwar}\mathrm{to}\mathrm{Dehradun}\mathrm{and}\mathrm{Rishikesh}}$ × 100

=$\frac{34-20}{34}\times 100$

=$\frac{14}{34}\times 100$

= 41.17%≈ 41%

∴ Required percentage will be 41%.

Given:

Percentage of passengers travelling from Haridwar to Dehradun = 15%

Percentage of passengers travelling from Haridwar to Jaipur = 10%

Required percentage =$\frac{15-10}{10}$ × 100

=$\frac{5}{10}\times 100$

= 50%

∴ Required percentage will be 50%.

Given:

Percentage of passengers travelling from Haridwar to Delhi, Rishikesh, Jaipur and Amritsar respectively = 20%, 19%, 10%, and 11%

Total numbers of passengers = 1500

Total percentage of passengers travelling from Haridwar to Delhi, Rishikesh, Jaipur and Amritsar = 20% + 19% + 10% + 11%

= 60%

So the number of passengers travelling from Haridwar to Delhi, Rishikesh, Jaipur and Amritsar = 60% of 1500

=$1500\times \frac{60}{100}$

= 900

Average =$\frac{\mathrm{Sum}\mathrm{of}\mathrm{Values}}{\mathrm{Number}\mathrm{of}\mathrm{values}}$

=$\frac{900}{4}$

= 225

∴ The average number of passengers travelling between Haridwar to Delhi, Rishikesh, Jaipur and Amritsar is 225.

Given:

Percentage of passengers travelling from Haridwar to Moradabad = 25%

Central angle for pie chart =$\frac{\mathrm{Value}\mathrm{of}\mathrm{the}\mathrm{component}}{100}\times 360°$

Central angle for passengers = 25% × 360°

=$\frac{25}{100}\times 360$

= 90°

∴ Central angle for passengers travelling from Haridwar to Moradabad = 90°

Given:

Total numbers of passengers = 1500

Percentage of children travelling between Haridwar to Delhi = 32%

Percentage of men travelling between Haridwar to Delhi = 24%

Percentage of woman travelling from Haridwar to Delhi =Total percentage passengers – (Percentage of children + Percentage of men)

=100% - (32% + 24%)

=100% - 56%

= 44%

We know that,

Percentage =$\frac{\mathrm{Actual}}{\mathrm{Total}}$ × 100

Total number of woman travelling from Haridwar to Delhi =1500 × 44% × 20%

=$1500\times \frac{44}{100}\times \frac{20}{100}$

=$\frac{1320000}{10000}$

= 132

∴ Number of woman travelling from Haridwar to Delhi = 132

Using statement I

Let the total students in the class be \(x\)

Male in the class \(=(\frac{60}{100}) \times x=\frac{3x}{5}\)

Females in the class \(=(\frac{40}{100}) \times x=\frac{2x}{5}\)

Let the average weight of female students be \(y\)

Then, the average weight of male students \(=y+15\)

Sum of the weight of male students \(=(\frac{3x}{5}) \times(y+15)=\frac{3xy}{5}+9 x\)

Sum of the weight of female students \(=(\frac{2x}{5})\times y=\frac{2xy}{5}\)

\(\therefore\) Average weight of the class \(=\frac{(\frac{3xy}{5}+9 x+\frac{2xy}{5})}{x}=\frac{(x y+9 x)}{x}=y+9\)

New average of the class \(=y+9+0.84=y+9.84 \quad \ldots\) (i)

Statement I alone is not sufficient to answer the question.

Using statement II

Total male students in the class is \(12\) more than female students.

\(\therefore\) If number of female students \(=x\)

Then, number of male students \(=x+12\)

Let average weight be male students be \(y\)

\(\therefore\) New average weight of male \(=y+1\)

Average weight of female students be \(z\)

\(\therefore\) New average weight of female \(=z+0.6\)

Statement II alone is not sufficient to answer the question.

Using statement I and statement II together,

Total male students in the class is \(12\) more than female students

\(\Rightarrow(\frac{3x}{5})-(\frac{2x}{5})=12\)

\(x=60\)

\(\therefore\) Male students \(=\frac{3x}{5}=\frac{180}{5}=36\)

Female students \(=\frac{2x}{5}=\frac{120}{5}=24\)

Avg weight of female and male students is \(y\) and \( (y+15)\) respectively

Adding \(9\) male students to the class increases the average weight of male students by \(1 \mathrm{~kg}\) and adding \(6\) female students increases the average weight of female students by \(0.6\)

Total male in the class \(=36+9=45\)

Total female in the class \(=24+6=30\)

\(\Rightarrow\) New average of male \(=y+16\), New average of female \(=y+0.6\)

\(\therefore\) New total average of class \(=\frac{(45({y}+16)+30({y}+0.6}{75}=\frac{(75 {y}+738)}{75} \quad-\)-(ii)

Equating (i) and (ii)

\(\frac{(75 {y}+738)}{75}={y}+9.84\)

\(y\) cannot be determined.

\(\therefore\) The data even in both the statements I and Il together are not sufficient to answer the question.