SNAP Quant Questions

Total employees in Accounts department= 8% of 4600= 368

Ratio of Men: women= 1:3

So, total women in accounts department= $$\ \frac{\ 1}{4}\times\ 368=92$$

The total number of employees :4600
Now total percentage of employees working in IT and HR : 26+11 =37 %
Total number of employees woking in the two sectors = 4600(0.37)
=1702

Thus, the correct option is D.

Number of Men in HR= $$\ \frac{\ 1}{2}\times11\%\ of\ 4600=\ \ \frac{\ 1}{2}\times\ \ \frac{\ 11}{100}\times\ 4600=253\ $$

Number of Men in Accounts=$$\ \frac{\ 3}{4}\times8\%\ of\ 4600=\ \ \frac{\ 1}{2}\times\ \ \frac{\ 8}{100}\times\ 4600=276\ $$

Number of Men in Production= $$\ \frac{\ 3}{5}\times15\%\ of\ 4600=\ \ \frac{\ 1}{2}\times\ \ \frac{\ 15}{100}\times\ 4600=414\ $$

Number of Men in IT= $$\ \frac{\ 1}{4}\times26\%\ of\ 4600=\ \ \frac{\ 1}{2}\times\ \ \frac{\ 26}{100}\times\ 4600=299\ $$

Number of Men in Marketing= $$\ \frac{\ 1}{2}\times22\%\ of\ 4600=\ \ \frac{\ 1}{2}\times\ \ \frac{\ 22}{100}\times\ 4600=506\ $$

Number of Men in Merchandising= $$\ \frac{\ 5}{6}\times18\%\ of\ 4600=\ \ \frac{\ 1}{2}\times\ \ \frac{\ 18}{100}\times\ 4600=690\ $$

Total men= 253+276+414+299+506+690=2438.

Total Women= 4600-2438=2162.

Ratio of men and women= $$\ \frac{\ 2438}{2162}=\ $$= $$\ \frac{\ 53}{47}\ $$

Hence, Option C

Total Employees =4600.

Total employees in Merchandising= 18% of 4600= 828.

Ratio of women and men= 1/5. 

So, number of women= $$\ \frac{\ 1}{6}\times828=\ 138\ $$

% of women= $$\ \frac{\ 138}{4600}\times100=\ 3\%\ $$

Total employees given= 4600

Total employees in production = 15% of 4600= 690.

Ratio of men and women in Production= 3:2

Number of men= $$\ \ \frac{\ 3}{5}\times\ 690=3\times\ 138=414$$

Total employees in Marketing= 22% of 4600= 22$$\times\ $$46= 1012. 

Ratio of men and women in Production= 1:1

Therefore number of men= $$\ \frac{\ 1}{2}\times\ 1012=\ 506$$

So, Number of men in Production: Number of men in Marketing= $$\ \frac{\ 414}{506}=\ \frac{\ 9}{11}$$

From the above table, it is clear that the percentage increase in the consumption of rice over the previous year was highest in 2008

B is the correct answer.

Population = $$\ \frac{\ Consumption}{Percapita\ consumption}$$

In 2008, the population of A = $$\ \frac{\ 230-138}{38.7}\ $$ = 2.79 million

C is the correct answer.

Exports to consumption ratio was highest in the year 2006

Answer is option A.

Population = $$\ \frac{\ Consumption}{Percapita\ consumption}$$

The population of country A was the highest in the year 2010.

Option A is correct,

In the centre region, the dots and stripes are alternating, thus the centre will now be "dots".

Hexagon and circles will alternate in twos.

The sales of Voveran in 2006 = 23 Crore

The sales of Calpol in 2005 = 13 Crore

Difference= 10 Crore

i.e 1000 lakhs

Percentage of increase in sales from 2005 to 2006:

Voveran: (23-16)/16= 43.75

Volini: (9.5-6.5)/6.5=46.15

Dolonex:(10-6)/6= 66.66

Sumo:(7-5)/5=40

Option C is correct.

Percentage of increase in sales from 2005 to 2006:

Voveran: (23-16)/16= 43.75%

Volini: (9.5-6.5)/6.5=46.15%

Moov: 1/4= 25%

Nise: (18-15)/15=20%

Option D 

Percentage of increase in sales from 2005 to 2006:

Voveran: (23-16.5)/16.5= 39.39%  ..(nearly) 40%

Option B

Lower number is eight times the upper number, i.e.

6 $$\times\ $$ 8 = 48

12 $$\times\ $$ 8 = 96

y $$\times\ $$ 8 = 192

y = $$\frac{192}{8}$$ = 24

Answer is option D.

Images are shifting by one position to the left side. In each step, there are two shaded and one unshaded image. Therefore, shaded triangle with index C at upper vertex will fill the blank.

Image 6 is suitable image and answer is option B.

Total students in Arts faculty = 1049
Total non US students in arts faculty = 79+21+6+2+4 = 112
Now in percentage
We get :$$\frac{112}{1049}\times\ 100$$= 11%

Let the total number of students be x
Now it is given that 1049 students are in arts which is 23%
so we can say
x(23/100) = 1049
We get x =4560
Now Engineering faculty = 9% 
so we get number = (0.09)(4560) = 410

Let the total number of students be x
Now it is given that 1049 students are in arts which is 23%
so we can say
x(23/100) = 1049
We get x =4560

Let the total number of students be x
Now it is given that 1049 students are in arts which is 23%
so we can say
x(23/100) = 1049
We get x =4560
Now students studying science will be 4560(21/100) = 957
Now Asian students are studying Science will be 6% of total science 
= 0.06(957) 
= 57

Let the total number of students be x 
Now it is given that 1049 students are in arts which is 23%
so we can say 
x(23/100) = 1049
We get x =4560 
Now we know that medical students = 5% 
So number of medical students : 4560(0.05) = 228 students 
Now out of this 34 students are European 
So percentage of these with respect to total = $$\frac{34}{228}\times\ 100\ =\ 15\%$$

Total number of respondents under 31 = 66 
Total number of persons who liked blue = 3+2 =5
So in percentage we get (5/66 )(100) = 7.6

Total respondents in 21-30 =33
Respondents liking Rock = 12
So respondents liking other = 33-12 =21
Therefore in percentage : (21/33)*100 = 64%

Total number of people liking Jazz = 16
Total sample = 68+66 =134
So percentage = (16/134)(100) =12%

Now X = 8.7-2.5-0.5-1 =4.7
Y = 9.1-2.5-0.5-1.5 =4.6
Z = 5.6-1-1.5-0.5 =3.1
Now at least 1 will be 
4.7+4.6+3.1+2.5+1+1.5+0.5 = 17.4

As the graph representing production line 1 is showing a steady increase, the given graphs represent cumulative productions. The graph representing production line 2 is showing no change from the 4th month to 6th month so It was the period of breakdown.
As the graph representing production line 2 is not following any pattern, we cannot quantify the loss of production as the production in those months cannot be determined.
So Only A can be answered

Tabulating the percentage, we get

From the table, in case of Y, the percentage is 50% and rest of the hotels have less than 50%. Hence C is the answer.

In statistics, the coefficient of variation (COV) is a simple measure of relative event dispersion. It is equal to the ratio between the standard deviation and the mean. The most common use of the COV is to compare relative risk, although it can be applied to any type of quantitative likelihood or probability distribution.

There is another use and meaning of the COV. When interpreting mathematical models, COV is calculated as the ratio between root mean squared error and the mean of a separate dependent variable. This type of COV analysis is less common, but it can be constructive when determining if a model is a good fit for a specific task or type of analysis. It is used to measure disparity and consistency.

The total number of moviegoers in a city = Number of people who watch less than 1 movie per week + Number of people who watch one or more movie per week

For city A, the percentage of viewers who watch less than one movie a week = 60, hence the percentage of viewers who view one or more movies per week = 100-60 = 40

=> 40% of the total number of viewers in A = 2400    => The total number of viewers = $$\ \frac{\ 2400\times\ 100}{40}$$ = 6000

=> The number of viewers who watch less than one movie a week = $$6000\times\ \frac{\ 60}{100}$$ =3600

Similarly, we can tabulate the data for the rest of the cities:

The viewers in the city C watch less than one movie a week = 13600

The total number of moviegoers in a city = Number of people who watch less than 1 movie per week + Number of people who watch one or more movie per week

For city A, the percentage of viewers who watch less than one movie a week = 60, hence the percentage of viewers who view one or more movies per week = 100-60 = 40

=> 40% of the total number of viewers in A = 2400 => The total number of viewers = $$\ \frac{\ 2400\times\ 100}{40}$$ = 6000

=> The number of viewers who watch less than one movie a week = $$6000\times\ \frac{\ 60}{100}$$ =3600

Similarly, we can tabulate the data for the rest of the cities:

E has the highest number of viewers who watch less than one movie a week.

The total number of moviegoers in a city = Number of people who watch less than 1 movie per week + Number of people who watch one or more movie per week

For city A, the percentage of viewers who watch less than one movie a week = 60, hence the percentage of viewers who view one or more movies per week = 100-60 = 40

=> 40% of the total number of viewers in A = 2400    => The total number of viewers = $$\ \frac{\ 2400\times\ 100}{40}$$ = 6000

=> The number of viewers who watch less than one movie a week = $$6000\times\ \frac{\ 60}{100}$$ =3600

Similarly, we can tabulate the data for the rest of the cities:

Both A and D have second lowest number of movie watchers. B is the answer.

The total number of moviegoers in a city = Number of people who watch less than 1 movie per week + Number of people who watch one or more movie per week

For city A, the percentage of viewers who watch less than one movie a week = 60, hence the percentage of viewers who view one or more movies per week = 100-60 = 40

=> 40% of the total number of viewers in A = 2400    => The total number of viewers = $$\ \frac{\ 2400\times\ 100}{40}$$ = 6000

=> The number of viewers who watch less than one movie a week = $$6000\times\ \frac{\ 60}{100}$$ =3600

Similarly, we can tabulate the data for the rest of the cities:

The total number of all moviegoers in the five cities who watch less than one movie per week = 3600+750+13600+3300+24000 = 45250

Average Salary in Administration = $$\ \frac{\ Total\ Salary\ }{Total\ number\ of\ People}$$

Total number of people on Administration = $$22\%\ of\ 18,000\ =\ 3,960$$

Hence $$12,000\ =\ \frac{\ Total\ Salary}{3,960}$$

Total Salary = $$12,000\times\ 3960\ =\ 47520000$$ = $$4.7\ crores$$

No of the employees in Sales & Marketing in 2005 = $$18\%\ of\ 18,000\ =\ 3240$$

No of the employees in Sales & Marketing in 2006 = $$20\%\ of\ 20,000\ =\ 4000$$

Percent Increase = $$\ \frac{\ 4000-3240}{3240}\times\ 100\%\ =\ 23.45\%$$

From above chart we can get the following table.

Variation of A = |4800-3960| = 840

Variation of B = |5200-4140 | = 1060

Variation of C = |4000-3240 | = 760

Variation of D = | 4000 - 5220 | = 1220

Variation of E = |2000-1440| =560

D has max variation

From the above charts, the following table can be made.

Let $$N$$ be the number of people who joined in 2006 in Operations department, then 

$$\ 4140-300+N\ =5200$$

$$N\ =1360$$

We will go by the options to check each one of them

Option A says SIFY on BSE.

% increase= $$\ \frac{\text{ Closing Price-Opening Price}}{\text{Opening Price}}\cdot100\%=\ \frac{\ 247-232}{232}\cdot100\%=\ \ \frac{\ 15}{232}\cdot100\%=6.465\%$$

Option B says INFY on NQE.

% increase= $$\ \frac{\text{ Closing Price-Opening Price}}{\text{Opening Price}}\cdot100\%=\ \frac{\ 10.5-9.5}{9.5}\cdot100\%=\ \ \frac{\ 1}{9.5}\cdot100\%=10.52\%$$

Option C says Wipro on NQE.

% increase= $$\ \frac{\text{ Closing Price-Opening Price}}{\text{Opening Price}}\cdot100\%=\ \frac{\ 6.5-5.5}{5.5}\cdot100\%=\ \ \frac{\ 1}{5.5}\cdot100\%=18.18\%$$

Option D says TCS on NQE

% increase= $$\ \frac{\text{ Closing Price-Opening Price}}{\text{Opening Price}}\cdot100\%=\ \frac{\ 40-40.5}{40.5}\cdot100\%=\ \ \frac{\ -0.5}{40.5}\cdot100\%=-1.23\%$$

So, Option C shows the highest increase.

First we convert the currency of Kya Kya into Rs. by multiplying the values to 11 and we can get the following table:

In SIFY, profit made by him= $$\ \frac{\ 232-231}{231}\cdot100\%=0.43\%$$

In INFY, profit made= $$\ \frac{\ 105-104.5}{104.5}\cdot100\%=0.47\%$$

In WIPRO, profit made= $$\ \frac{\ 60.5-60}{60}\cdot100\%=0.83\%$$

In TCS, Profit made= $$\ \frac{\ 450-445.5}{445.5}\cdot100\%=1.01\%$$

Therefore, max profit is made in TCS share

Given, total shares of SIFY= 1000000

Shares purchased by SATYAM= 15% of 1000000= 150000.

Now, 50% of the shares were purchased at BSE's opening price i.e. Rs. 232 and 50% of the shares were purchased at BSE's closing price, Rs. 247.

So, total amount paid by SATYAM= 75000*(232)+ 75000(247)= Rs. 75000(479)= Rs. 35925000 which is close to Rs. 36 millions.

Considering statement X,

Percentage increase for Managerial candidates= $$\ \frac{\ \left(45292-19236\right)}{19236}\cdot100=\ \ \frac{\ 26056}{19236}\cdot100=135.45\%$$

Percentage increase in Technical candidates= $$\ \frac{\ \left(63298-61205\right)}{61205}\cdot100=3.3\%$$

Therefore statement X is true.

Considering statement Y,

Total registrations in 2004= 61205+19236= 80441.

Total registrations in 2005= 63298+45292= 108690.

So, growth in the overall number of registrations= $$\ \frac{\ \left(108690-80441\right)}{80441}\cdot100=35.11\%$$, which is greater than 25%.

So, both statement X and Y are true.

Statement X:

Percentage of drop-outs for Managerial category in 2004 = $$\ \frac{\ 19236\ -\ 15389}{19236}\times\ 100\ \approx\ 20\%$$

Percentage of drop-outs for Managerial category in 2005 = $$\ \frac{\ 45292\ -\ 40763}{45292}\times\ 100\ \approx\ 10\%$$

Therefore, percentage of drop-outs from 2004 to 2005 has decreased. Statement X is true.

Statement Y:

Percentage of drop-outs for Managerial category in 2005 = $$\ \frac{\ 45292\ -\ 40763}{45292}\times\ 100\ \approx\ 10\%$$

Percentage of drop-outs for Technical category in 2005 = $$\ \frac{\ 63298\ -\ 60133}{63298}\times\ 100\ \approx\ 5\%$$

Percentage of drop-out is lower for Technical category.

Statement Y is false

Answer is option A.

Evaluating statement X,

In the year 2005, In Managerial category, the number of candidates the number of candidates shortlisted based on CV=$$\ \frac{\ 399}{40763}=0.00978$$

In the year 2005, In Technical category, the number of candidates the number of candidates shortlisted based on CV=$$\ \frac{\ 637}{60133}=0.0106$$

So, statement X is false.

Evaluating statement Y,

In the year 2005, In Managerial category, the number of candidates the number of candidates shortlisted based on CV=$$\ \frac{\ 399}{40763}=0.00978$$

In the year 2004, In Managerial category, the number of candidates the number of candidates shortlisted based on CV=$$\ \frac{\ 138}{15389}=0.00896$$

So, statement Y is True. Hence, Option B.

Let's evaluate statement X first.

For technical jobs in 2004, the number of candidates offered jobs as a proportion of the number of CV’s posted=$$\ \frac{\ 181}{59981}=0.00301$$

For Managerial jobs in 2004, the number of candidates offered jobs as a proportion of the number of CV’s posted=$$\ \frac{\ 48}{15389}=0.00312$$

So, statement X is false.

Evaluating Y,

In technical category in 2004, the number of candidates offered jobs as a proportion of the candidates shortlisted=$$\ \frac{\ 181}{684}=0.2646$$

In Managerial category in 2004, the number of candidates offered jobs as a proportion of the candidates shortlisted=$$\ \frac{\ 48}{138}=0.3478$$

So, statement Y is also false.

Hence, option D.

In the year 2003, Wheat: Rice=1.2 and it is given that Rice production= 4 lac tons.

Let the wheat production during 2003 be x.

So, $$\ \frac{\ x}{4}=\ \frac{\ 1.2}{1}$$

=> x= 4.8 lac tons

Average of share price for 10/8 to 12/9 is  $$\ \frac{\ 483+475+467+513}{4}$$ = 483

Price at end of September (near to beginning of october)= 535

Price at the end October = 461

$$% decrease$$ = $$\frac{535-461\ }{535}\times\ 100\%$$

= $$13.83$$%

10% is the closest option 

Money spent to buy 100 shares on 31/08 is 100*461 = 46,100

Money Recieved after selling 100 shares on 10/10 = 538*100 = 53,800

Overall profit = 53,800-46,100 = 7,700

Engineering, Material, Services and CIS contributes around 81% of market share.

Let us represent the sales revenue in terms of 'k'.

Total sales revenue = 100k

Sales revenue from Engineering = 31k

Sales revenue from CIS = 19k

Remaining revenue from other sectors = 50k

Profit margin = Profit/Revenue

Let the profit margin of remaining sectors be x%.

So the eqn is:

10*100k = x*50k + 12*31k + 20*19k

after cancelling 'k' from both sides, we get:

=> x = 4.94% (answer)