SNAP 2014 Question Paper

We always follow BODMAS 
Now first taking numerator 
which $$\frac{1}{2}divided\ by\ \frac{1}{2\ }\ of\ \frac{1}{2}$$
so we get $$\frac{\frac{1}{2}}{\frac{1}{2}}\times\ \frac{1}{2}=\frac{1}{2}$$
Now taking denominator we get $$\frac{1}{2}+\frac{1}{2}\times\ \frac{1}{2}=\frac{3}{4}$$
Taking ratio 
we get : $$\frac{\frac{1}{2}}{\frac{3}{4}}=\frac{2}{3}$$

Let the total distance covered by the taxi be 100 kms.

For 60% of the distance, i.e. 60 Kms, Speed of the taxi was 40 kmph. Time taken to cover 60 kms= $$\ \frac{\ Dis\tan ce}{Speed}\ =\ \frac{\ 60}{40}=\ \frac{\ 3}{2}\ hrs.$$

For 20% of the distance, i.e. 20 Kms, Speed of the taxi was 30 kmph. Time taken to cover 20 kms= $$\ \frac{\ Dis\tan ce}{Speed}\ =\ \frac{\ 20}{30}=\ \frac{\ 2}{3}\ hrs.$$

For remaining, i.e. 20% of the distance, i.e. 20 Kms, Speed of the taxi was 10 kmph. Time taken to cover 20 kms= $$\ \frac{\ Dis\tan ce}{Speed}\ =\ \frac{\ 20}{10}=\ \ 2\ hrs.$$

So, total distance= 100 kms.

Total time= $$\ \ \frac{\ 3}{2}+\ \frac{\ 2}{3}+2=\ \frac{\ 9+4+12}{6}=\ \frac{25\ }{6}hrs$$

.'. Average speed= $$\ \ \ \frac{\ Total\ Dis\tan ce}{Total\ Time}=\ \frac{\ 100}{\ \frac{\ 25}{6}}=\ \frac{\ 100\cdot6}{25}=4\times\ 6=24\ kmph$$

Let the two numbers be 3x and 4x.

According to the question,

$$\ \ \ \left(4x\right)^2=8\left(3x^2\right)-504$$

=> $$\ \ \ 16x^2=72x^2-504$$

=>$$\ \ \ 56x^2=504$$

=>$$\ \ \ x^2=9$$

Or, x=$$\pm\ 3$$

Since, x cannot be -3, it is 3.

Therefore the numbers are 3x and 4x, which is 9 and 12.

We will first look at the numbers that have 2 in the unit's digit.

So, the numbers between 1 to 100 having 2 in unit's place are 2, 12, 22, 32... 92. So, there are 10 such numbers.

For numbers having 2 at ten's place, we have numbers from 20 to 29. In total we have 10 such numbers.

But, 22 is common to both the groups, and hence, 

Total numbers with between 1 and 100 with digit 2 on it is 10+10-1=19.

Total sample space= 100.

So, required probability= $$\ \frac{\ 19}{100}$$

We have two parameters to look at in this question:

1) Length of the train. Let this be assumed to be a metres.

2) Length of the platform. Let us assume it as b metres.

It is given that speed of the train is 54 km/h Or, 54$$\ \times\ \frac{\ 5}{18}$$= 15 m/s

So, when it passes the man, distance covered will be the length of the train itself= a metres.

So, time taken= $$\frac{Dis\tan ce}{Speed}=\ \ \frac{\ a}{15}$$, which is given to be 20 seconds.

So, $$\frac{\ a}{15}=20$$. And a= 300 metres.

Now, to pass the entire platform, we need to account for both the length of the platform and that of the train.

So, effective distance to be covered= a+b= 300+b metres.

Speed of the train = 15 m/s.

So, time taken= $$\frac{\ 300+b}{15}$$, which is given to be 36 seconds.

So, $$\frac{\ 300+b}{15}=36$$

=>$$\ 300+b=36\times\ 15$$

=>b=540-300

So, b= Length of the train= 240 metres.

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= $$\ \frac{\ 414}{506}=\ \frac{\ 9}{11}$$