IIFT 2022 Slot 1 Question Paper (18th Dec)

It is given that 10th July 2010 was a Thursday.

Let's calculate what day it is on 10th July 2003; from that; we can calculate 6th July 2003.

One year = 365 days = 52 weeks + 1day 

Hence, For any given date, the day will change by one place (i.e. Monday to Tuesday,...Wednesday to Thursday,...) for the succeeding year.

One Leap year = 366 days 52 weeks + 2 days.

Hence, If the year is a leap year, then the day will change by two places (i.e. Monday to Wednesday,...Wednesday to Friday,...) for the succeeding year.

We get the following table,

As 10th July 2003 is Tuesday, 6th July 2003 will be Friday.

Option (C) is the answer.

Let $$'y'$$ be the principal amount invested at a 6.5% per annum rate.

Let $$'x'$$ be the principal amount invested at a 7.5% per annum rate.

It is given that, $$\frac{y}{4}=\frac{x}{5}$$............(1)

It is also given that without reinvestment, his income from interest after 3 years is Rs. 400

i.e., $$y\times\ 3\times\ 0.065+x\times\ 3\times\ 0.075=400$$

$$\frac{4x}{5}\times\ 3\times\ 0.065+x\times\ 3\times\ 0.075=400$$

$$0.381x=400$$

Hence, $$x=Rs.\ 1049.87$$, then $$y=\frac{4x}{5}=\frac{1049.87\times\ 4}{5}=Rs.\ 839.90$$

Total invested sum = $$y+x=$$ = $$1889.77\approx Rs.1890$$

Option (A) is correct.

Let $$'x'$$ be the true time when the clock indicates 7 PM on 17th January 2022.

Let $$'y'$$ be the number of hours elapsed between 7 AM on 12th January, 2022 and $$'x'$$

The clock loses 18 minutes per 24 hours. 

For one hour, it loses$$=\frac{18}{24\ \times \ 60}=\frac{1}{80}\ \ hrs$$

For $$'y'$$ hours it loses = $$\frac{y}{80}\ \ hrs$$

Hence, the number of hours elapsed between 7 AM on 12th January 2022 and 7 PM on 17th January 2022 = $$y-\frac{y}{80}$$

$$y-\frac{y}{80}=5\times\ 24+12$$

$$\frac{79y}{80}=132$$

$$y=133.671\ hrs$$

$$y=24\times\ 5.5+1.67\ hrs$$

Hence, the approximate true time when the clock indicates 7 PM on 17th January 2022 = 7 AM, 12th January 2022 + $$'y' hrs$$

= 7 AM, 12th January 2022 + 5days +12 hrs +1.67 hrs

= 7 PM on 17th January 2022 + 1.67 hrs

= 08:40:15 PM 17th January 2022.

Option (D) is the answer.

Let the volume of the tank be L.C.M of (12,18) = $$72 m^3$$

The rate of inflow of the pipes are, $$\frac{72}{12}=6\ \frac{m^3}{hr}\ \&\ \frac{72}{18}=4\ \frac{m^3}{hr}$$

Let the rate of leakage be $$x\ \frac{m^3}{hr}$$

The total time it takes for both pipes to fill the tank = $$\frac{72}{6+4}=\frac{72}{10}=7.2\ hr$$

But due to leakage, the pipes took 36 min more to fill the tank, $$\frac{72}{6+4-x}=7.2\ +\frac{36}{60}$$

$$\frac{72}{10-x}=7.8$$

on solving, we get $$x=\frac{10}{13}\ \frac{m^3}{hr}$$

The total time taken by the leak to empty the full tank = $$\frac{72}{\frac{10}{13}}=93.6\ hr$$

The total marks = $$100\times\ 6=600$$

The recruit got 55% of the total marks, i.e. = $$55\%\times\ 600=330$$.

The recruit gets marks in the ratio of 10:9:8:7:6:5 in each parameter.

Let the marks in each parameter be $$10x,\ 9x,\ ...,\ 5x$$.

we have, $$10x+\ 9x+8x+7x+6x+5x=330$$

$$45x=330$$

$$x=\frac{22}{3}$$

Hence, the marks in each parameter are $$73.33,\ 66,\ 58.67,\ 51.33,\ 44,\ 36.67$$

The criteria for clearing the performance score is 55% of the maximum marks for first 3 and for the next 3 parameters is 50% of the maximum marks.

Hence the recruit cleared four parameters. Option (c) is the answer.

Let the lengths of BD, CD and AD be $$x,\ y\ \&\ h$$, respectively.

=> $$ BD+CD = x+y=21$$.....(1) 

As AD is altitude, using Pythagoras' theorem, we get

$$13^2=x^2+h^2$$ .....(2)

$$20^2=y^2+h^2$$.....(3)

Eq (3) - (2) gives 

$$20^2-13^2=y^2-x^2$$ 

$$231=\left(y-x\right)\ \left(y+x\right)$$

 $$\left(y-x\right)=\frac{231}{x+y}=\frac{231}{21}=\ 11$$......(4)

Adding Eq(1) & (4), we get $$2y=32$$  => $$y=16 cm$$ 

we get $$x+16=21$$ => $$x=5 cm$$

From Eq(2) => $$13^2=5^2+h^2$$ 

$$h^2=13^2-5^2$$ => $$h=12 cm$$

The area of the smaller triangle ABD = $$\ \frac{\ BD\times\ AD}{2}$$

= $$\ \frac{\ 5\times\ 12}{2}$$ = $$30 cm^2$$

30% of area of smaller triangle = $$30\%\ \times\ 30=9\ cm^2$$

Option (C) is the answer

It is given that 900 bags were made, and only 785 were sold.

Cost of manufacturing 900 bags $$=1200+\frac{125}{105}\times\ 900+0.80\times\ 900$$ $$=\ Rs. \frac{20940}{7}$$.....(1)

Revenue of the factory = Amount from sales + Amount from advertisements......(2)

Let the amount from advertisements be $$ Rs. 'y'$$

Amount from sales = $$785\times\ 3.25= Rs. 2551.25$$

Factory's profit percentage = 12%.....(3)

From Eq (1), (2) & (3)

We get, $$2551.25\ +y=\frac{20940}{7}\ \times\ 1.12$$

On solving, we get $$y=Rs.\ 799.15$$

Hence the approximate revenue from advertisements is = $$Rs.\ 799$$

Option (C) is the answer.

Let '$$p$$' be the principal amount and '$$r$$' be the interest rate per annum.

Simple Interest for t years = $$ptr$$

Compound Interest for t years = $$p\left(1+r\right)^t-p$$

From the given information,

$$\frac{\left(p\left(1+r\right)^4-p\right)-4pr}{\left(p\left(1+r\right)^3-p\right)-3pr}=\frac{20}{8}$$

$$\frac{\left(\left(1+r\right)^4-1\right)-4r}{\left(\left(1+r\right)^3-1\right)-3r}=\frac{5}{2}$$

$$2\times\ \left(\left(1+r\right)^4-1-4r\right)=5\times\ \left(\left(1+r\right)^3-1-3r\right)$$

$$2\times\ \left(1+r\right)^4-2-8r=5\times\ \left(1+r\right)^3-5-15r$$

$$2\times\ \left(1+r\right)^4=5\times\ \left(1+r\right)^3-3-7r$$

$$\ 2r^4+8r^3+12r^2+8r+2=\ 5r^3+15r^2+15r+5-3-7r$$

$$\ 2r^4+8r^3+12r^2=\ 5r^3+15r^2$$

$$\ 2r^2+3r-3=0$$

Hence, $$r=\frac{-3\pm\ \sqrt{\ 9+24}}{2\times\ 2}$$

$$r=\frac{-3\pm\ \sqrt{\ 33}}{4}$$

As interest rate cannot be negative,

$$r=\frac{-3+\ \sqrt{\ 33}}{4}$$ 

$$r=0.6861$$

Hence, the approximate value of $$r=69\%$$

Option (A) is correct.

The largest remainder we can get, = smallest of {largest remainder of '4', the largest remainder of '11', the largest remainder of '9'}

= smallest of {3,10,8} = 3.

The number which gives the remainder '3', when divided by 4,9& 11, will be of the form = L.C.M of (4,9&11) x n + 3.

L.C.M of (4,9&11) = 396.

The number which gives the remainder '3', when divided by 4,9& 11, will be of the form = 396n+3.

The largest 5-digit number of the form, 396n+3 = 396 x 252 + 3

=99792+3=99795.

Option (D) is correct.

From the above figure,

Using Pythagoras theorem

$$In\ \triangle\ AQO\ ,\ 55^2=a^2+c^2$$.....(1)

similarly. 

$$In\ \triangle\ BQO\ ,\ 65^2=a^2+d^2$$.....(2)

$$In\ \triangle\ CSO\ ,\ 75^2=b^2+d^2$$.....(3)

$$In\ \triangle\ DSO\ ,\ m^2=b^2+c^2$$.....(4)

From observation, Eq(1)-Eq(2)  = Eq(4)-Eq(3) = $$c^2-d^2$$

Hence, $$55^2-65^2=m^2-75^2$$

$$3025-4225=m^2-5625$$

$$m^2=4425$$

$$m=\pm\ 66.52$$

As 'm' cannot be negative 

$$m\approx\ \ 66.5$$

Option (A) is correct.