IIFT 2009 Question Paper

Let the number of 1 rupee , 2 rupee and 5 rupee coins be x, y and z respectively

At the grocery store he gives half of his 5 Rupee coins and in return receives the same number of 1 Rupee coins i.e z now become 0.5z and x now becomes x+0.5z

At the dairy store he gives all 2 Rupee coins and in return gets thirty 5 Rupee coins which increases the number of 5 Rupee coins to 75% more than the original number i.e now y = 0 and 0.5z = 0.5z+30

Also given 0.5z+30 = 1.75z

i.e. z=24

Also given that number of 1 Rupee coins now is 50 i.e. x+0.5z = 50 

i.e. x = 38

Therefore the number of 1 Rupee and 5 Rupee coins originally were 38 and 24 respectively. 

Therefore our answer is Option 'D'

Sukriti covers a distance of 1 km in 5 minutes and 50 seconds therefore will cover 5 kms in 25 mins 250 secs = 29.16 mins

Saloni covers the 1 km in 6 minutes and 4 seconds therefore her speed is $$\frac{1}{6.06}$$ kms per minute

Therefore Distance Covered by Saloni in 29.16 mins is 29.16 mins*$$\frac{1}{6.06}$$ $$\approx$$ 4.8 kms 

Therefore the distance with which Sukriti will win a 5 km mini marathon is $$\approx$$ 200 meters

therefore our answer is option 'B'

Area of a equilateral triangle is $$\frac{\sqrt3}{4}$$*$$a^2$$

Area of a regular pentagon is $$\frac{5}{4}$$*$$\sqrt{1+\frac{2}{\sqrt{5}}}$$*$$a^2$$

Therefore ratio of shaded area to unshaded area = $$\frac{\frac{5}{4}*\sqrt{1+\frac{2}{\sqrt{5}}}}{\frac{5*\sqrt3}{4}}$$.

i.e. $$\frac{\sqrt{1+\frac{2}{\sqrt{5}}}}{{\sqrt3}}$$

therefore our answer is Option 'D'.

P1 can fill the tank in 8 hours while P2 can fill the tank in 12 hours therefore when Both the pumps are opened simultaneously they can fill $$\frac{1}{8}$$ + $$\frac{1}{12}$$ = $$\frac{5}{24}$$ this much of tank in an hour.Therefore Time required by P1 and P2 to fill full tank is $$\frac{24}{}$$ Hence they can fill half the tank in $$\frac{12}{5}$$ = 2.4 hours. After all the 3 pipes are opened the tank will be filled at the rate of $$\frac{1}{8}$$ + $$\frac{1}{12}$$ - $$\frac{1}{8}$$ = $$\frac{1}{12}$$ therefore for filling the remaining tank we will need 6 more hours. If supervisor was planning to come exactly when the tank is fully filled then he must come after 8.4 hours

 Time required by P1 and P2 to fill 1/3rd of the tank is $$\frac{24}{5*3}$$ = $$\frac{8}{5}$$ = 1.6 hours therefore supervisor will come after = 8.4 - 1.6 = 6.8 hours 

In 6.8 hours all the three pipe will fill $$\frac{6.8}{12}$$ = 0.56 of the capacity of tank

therefore remaining tank to be filled when the supervisor comes = 1- 0.56 - 0.33 $$\approx$$ 0.1 i.e. 10%

Therefore option 'C' is our answer

When the ball is dropped from 45m height it will cover a distance of 45m while going down after rebound it'll cover a distance of $$45*\frac{3}{5}$$ while going up and a distance of $$45*\frac{3}{5}$$ while going down again. after 2nd rebound it'll cover a distance of $$45*\frac{3}{5}*\frac{3}{5}$$ while going up and a distance of $$45*\frac{3}{5}*\frac{3}{5}$$ while going down again and so on 

i.e. total distance traveled by the ball is 45 + $$45*\frac{3}{5}$$ + $$45*\frac{3}{5}$$ + $$45*\frac{3}{5}*\frac{3}{5}$$ + $$45*\frac{3}{5}*\frac{3}{5}$$ +...........

this form 2 infinite GPs as 45 + $$45*\frac{3}{5}$$ + $$45*\frac{3}{5}*\frac{3}{5}$$+....... and $$45*\frac{3}{5}$$+$$45*\frac{3}{5}*\frac{3}{5}$$+....

For 1st GP a=45 and r=$$\frac{3}{5}$$ therefore sum of this infinite GP is $$\frac{45}{1-\frac{3}{5}}$$ = $$\frac{225}{2}$$

For 2nd GP a=45*$$\frac{3}{5}$$  and r=$$\frac{3}{5}$$ therefore sum of this infinite GP is  $$\frac{45*\frac{3}{5}}{1-\frac{3}{5}}$$ = $$\frac{135}{2}$$

Therefore our answer = $$\frac{225}{2}+\frac{135}{2}$$ = 180 m i.e. option 'B'

From 2nd customer for each customer the purity of oil becomes 0.9 of it's previous value i.e. 2nd customer will get only 40*0.9=36 liters of pure petrol 3rd customer will get 36*0.9= 32.4 liters of pure petrol and so on. Therefore the 7th customer will get $$40*(0.9)^6$$= 21.25 liters of pure petrol

Therefore our answer is option 'B'.

Given that probability of getting a heart attack before any drug = 0.80

Drug D1 reduces the probability of a heart attack by 35 %. Therefore, the probability of a patient getting heart attack after he has taken D1 = (1-0.35)*0.80 = 0.52

Drug D2 reduces the probability of a heart attack by 20 %. Therefore, the probability of a patient getting heart attack after he has taken D2 = (1-0.20)*0.80 = 0.64

It is given that both D1 and D2 work independently. Therefore, the probability of a patient getting heart attack after he has taken both D1 and D2 = (1-0.35)*(1-0.20)*0.80 = 0.416

A total of 100 patients have taken both the drugs whereas only 50-50 patients took drug D1 and D2.

Hence, the probability that the selected patient was given both the drug is = $$\dfrac{0.416}{0.416+0.5*0.52+0.5*0.64}$$ = 0.417 $$\approx$$ 0.42

Therefore, we can say that option A is the correct answer.

zone 1 will grow to become 193.8*1.0725= 207.85

zone 2 will grow to become 79.3*1.082=85.8

zone 3 will grow to become 57.5*1.0715=61.61

Total sales this year = 355.26

Total sales last year = 330.6

Growth = $$\frac{355}{330}$$ = 1.0746 i.e. growth of 7.46%

decrease in purchase price for first month was $$\frac{3000-2750}{3000}$$ = 0.083

decrease in purchase price for second month was $$\frac{2750-2500}{2750}$$ = 0.0909

decrease in purchase price for the third month was $$\frac{2500-2400}{2500}$$ = 0.04

decrease in purchase price for the fourth month was $$\frac{2400-2250}{2400}$$ = 0.0625

Average decrease = $$\frac{0.083+0.0909+0.04+0.0625}{4}$$ = 0.069 i.e 6.9%

Therefore our answer is option'C'

We need to find the coefficient of $$x^7$$ in the expansion $$(1-x^{2}+x^{3})(1 + x)^{10}$$ 

Now, $$(1 + x)^{10}$$ will have all the powers of x from 0 to 10. Multiplying these powers by 1, $$x^2$$ and $$x^3$$ will yield different results but we are interested in finding only the coefficient of $$x^7$$. When we multiply $$x^7$$ of $$(1 + x)^{10}$$ by 1, $$x^5$$ of $$(1 + x)^{10}$$ by $$x^2$$ and  $$x^4$$ of $$(1 + x)^{10}$$ by $$x^3$$ we will get $$x^7$$. coefficient of $$x^7$$ in $$(1 + x)^{10}$$ is $$10C_7$$ = 120, coefficient of $$x^5$$ in $$(1 + x)^{10}$$ is $$10C_5$$ = 252, coefficient of $$x^4$$ in $$(1 + x)^{10}$$ is $$10C_4$$ = 210 adding 120 and 210 and subtracting(since $$x^2$$ has a negative sign) 252 we get coefficient of $$x^7$$ as 78

Therefore our answer is option 'B'