XAT 2011 Question Paper - QUANT

Total value of assets in the year 2004 = 100 crore

Total value of sales in the year 2004 = 0.60*100 = 60 crore

CSR spending in the year 2004 = $$\dfrac{5/3}{100}\times 100$$ = 1 crore.

Similarly, we can calculate for the remaining years. Tabulating the same data, (All the values are in crores)

From the table, we can see that the increase in CSR spending 2006 is more than 1.5 times as compared to previous year whereas for none of the other values this is true. Hence, option B is the correct answer. 

Total value of assets in the year 2004 = 100 crore

Total value of sales in the year 2004 = 0.60*100 = 60 crore

CSR spending in the year 2004 = $${5/3}{100}\times 100$$ = 1 crore.

Similarly, we can calculate for the remaining years. Tabulating the same data, (All the values are in crores) 

CSR/ Assets ratio for the year 2004 = $$\dfrac{1}{100}$$ = 0.01

CSR/ Assets ratio for the year 2005 = $$\dfrac{1.15}{110}$$ = 0.01045

CSR/ Assets ratio for the year 2006 = $$\dfrac{2}{125}$$ = 0.016

CSR/ Assets ratio for the year 2007 = $$\dfrac{2}{135}$$ = 0.0148

CSR/ Assets ratio for the year 2008 = $$\dfrac{2.5}{150}$$ = 0.0166

We can see that CSR/Asset ratio is maximum for the year 2008. Hence, option E is the correct answer.

Total value of assets in the year 2004 = 100 crore

Total value of sales in the year 2004 = 0.60*100 = 60 crore

CSR spending in the year 2004 = $$\dfrac{5/3}{100}\times 100$$ = 1 crore.

Similarly, we can calculate for the remaining years. Tabulating the same data, (All the values are in crores) 

From the table we can see that maximum spending on CSR activities in the period 2004 - 2009 = 3.16 crore $$\approx$$ 3 crore. Hence, option D is the correct answer. 

Total value of assets in the year 2004 = 100 crore

Total value of sales in the year 2004 = 0.60*100 = 60 crore

CSR spending in the year 2004 = $$\dfrac{5/3}{100}\times 100$$ = 1 crore.

Similarly, we can calculate for the remaining years. Tabulating the same data, (All the values are in crores) 

From the table, we can see that the CSR spending never decreased as compared to previous year. Hence, option E is the correct answer. 

Percentage of Ex-defense serviceman in level 1 = $$\dfrac{6}{52}\times 100$$ = 11.54 percent

Percentage of Ex-defense serviceman in level 2 = $$\dfrac{8}{65}\times 100$$ = 12.31 percent

Percentage of Ex-defense serviceman in level 3 = $$\dfrac{30}{210}\times 100$$ = 14.28 percent

Percentage of Ex-defense serviceman in level 4 = $$\dfrac{25}{130}\times 100$$ = 19.23 percent

Percentage of Ex-defense serviceman in level 5 = $$\dfrac{60}{330}\times 100$$ = 18.18 percent

We can see that in level 4 the Ex-Defence Servicemen are the highest in percentage terms. Hence, option D is the correct answer. 

From the bar graph we can see that actual staff strength is 130 whereas the staff requirements is 255. So, the company has to abolish 125 vacant seats. we can see that the company has to abolish nearly 50 percent of the total strength which is greater than any other remaining levels. Hence, option D is the correct answer.  

Percentage representation of Ex- policemen in level 1 = $$\dfrac{4}{52}\times 100$$ = 7.69 percent

Percentage representation of Ex- policemen in level 2 = $$\dfrac{4}{65}\times 100$$ = 6.15 percent

Percentage representation of Ex- policemen in level 3 = $$\dfrac{9}{210}\times 100$$ = 4.29 percent

Percentage representation of Ex- policemen in level 4 = $$\dfrac{7}{130}\times 100$$ = 5.38 percent

Percentage representation of Ex- policemen in level 5 = $$\dfrac{15}{330}\times 100$$ = 4.54 percent

We can see that in level 3 the Ex- policemen are the lowest in percentage terms. Hence, option C is the correct answer. 

Total number of houses = 10

If first house is robbed, then II is not and if II house is robbed, then III is not and so on.

Thus 2 adjacent houses can never be chosen

So, number of ways in which three houses can be robbed such that no two of them are next to each other.

= $$C^{10 - 2}_3 = C^8_3$$

= $$\frac{8 \times 7 \times 6}{1 \times 2 \times 3}$$

= $$56$$

It is given that $$x=(9+4\sqrt{5})^{48}$$ ... (1)

Let us assume that $$y=(9-4\sqrt{5})^{48}$$ ... (2)

We can see that  0 < y < 1. 

Also, x + y = 2($$48C0*(9)^{48}$$+$$48C2*(9)^{46}*(4\sqrt{5})^2$$+$$48C4(9)^{44}*(4\sqrt{5})^4$$+...+$$48C48*(4\sqrt{5})^{48}$$)

We can see that x + y is an integer therefore we can say that y + f = 1. Hence, y = (1 - f)

We can see that = x(1 - f) = x*y

$$\Rightarrow$$ $$(9+4\sqrt{5})^{48}(9-4\sqrt{5})^{48}$$

$$\Rightarrow$$ $$(9^2-(4\sqrt{5})^2)^{48}$$

$$\Rightarrow$$ $$(81-80)^{48}$$ = 1

Hence, option A is the correct answer.

i. $$a_{741}$$ has 741 ones, it will be divisible by 3

ii. $$a_{534}$$ has 534 ones, it will be divisible by 3

iii. $$a_{123}$$ has 123 ones, it will be divisible by 3

iv. $$a_{77}$$ has 77 ones and each pair of 11 ones is divisible by 13.

Therefore, i,ii,iii and iv are not prime.

Ans -(D) All are correct.