Wednesday, 9 January 2013

XAT-2013 Detailed Solutions (Quantitative Ability)

Along with solutions, here I would like to give 'Notes' especially on some problems where I went in a wrong way or spent a little more time than usual while actually solving in the exam. And also for some problems, I summarized crucial clues and tagged those clues under 'Points'. I thought this approach is quite useful to solve the problems and get to know about the nitty-gritties.

Q1)Prof. Mandal walks to the market and comes back in an auto. It takes him 90 min to make the round trip. If he takes an auto both ways it takes him 30 minutes. On a Sunday he decides to walk both ways. How long would it take him?
(A)100 minutes  (B) 120 minutes        (C) 140 minutes        (D) 150 minutes
(E)None of the above
Sol:
one-way walk + one-way auto = 90 min  ---(1)
two-way auto = 30 min
=> one-way auto = 15 min                         ---(2)
From (1) and (2), one-way walk = 90-15 = 75
=>  two-way walk = 2*75 = 150  min

Q2)At the centre of a city’s municipal park there is a large circular pool. A fish is release in the water at the edge of the pool. The fish swims north for 300 feet before it hits the edge of the pool. It then turns east and swims for 400 feet before hitting the edge again. What is the area of the pool?
(A)62500π           (B) 125000π (C) 250000π (D) 500000π
(E)Cannot be answered from the given data
Sol:
“Travel of 300 m North and then 400 m East” creates a right-triangle on the circle
=> Funda here is “angle in semi circle is right angle”
=> hypt. of the right-triangle = diameter of the circle
=> diameter of the circle = √(3002+4002) = 500 m
=> rad. of the circle = 250 m
=> Area of the circle =

Q3)Mr.Mehra is planning for higher education expenses of his two sons aged 15 and 12.He plans to divide Rs 15 lakhs in two equal parts and invest in two different plans such that his sons may have access to Rs 21 lakhs each when they reach the age of 21. He is looking for plans that will give him a simple interest per annum. The rates of interest of the plans for his younger son and his elder son should be
(A) 5% and 7.5% respectively             (B) 8% and 12% respectively
(C) 10% and 15% respectively            (D) 15% and 22.5% respectively
(E) 20% and 30% respectively
Sol:
Points:
15 lakhs is divided in to two equal parts => 7.5 L each
Elder son aged 15 and younger one aged 12 years
Each one has to get 21 lakhs by the time they attain 21 yrs
=> Elder one after 6 years and younger one after 9 yrs
=> one sum of 7.5 L should grow to 21 L after 6 yrs
and other sum of 7.5 L should grow to 21 L after 9 yrs
=>Interest per annum for elder one = (21-7.5)*100/(7.5*6)
= 13.5*100/(7.5*6) = 30%
Interest per annum for younger one = (21-7.5)*100/(7.5*9)
= 13.5*100/(7.5*9) = 20%

Q4)Albela, Bob and Chulbul have to read a document of seventy eight pages and make a presentation next day. They realize that the article is difficult to understand and they would require team work to finish the assignment. Albela can read  a page in two minutes, Bob in 3 minutes and Chulbul in 4 minutes. If they divide the article in to 3 parts so that all three of them spend the equal amount of time on the article, the number of pages that Bob should read is
(A) 24        (B) 25             (C) 26             (D) 27 (E) 28
Sol:
Points:
Pages in the doc = 78
A reads a page in 2 min, B in 3 min and C in 4 min
=> in 1 min, A can read 1/2 of a page, B can read 1/3 of a page and C can read ¼ of a page
They have to divide the work such that all should spend equal amount of time. Let that be 't' min.
In 't' min, A can read t/2 , B can read t/3 and C can read t/4 pages
In total they read t/2+t/3+t/4 pages
=> t/2 + t/3 + t/4 = 78 => t = 72 min
So in his assignment B reads t/3 ie., 72/3 = 24 pages

Q5)The taxis plying in Wasseypur have the following fare structure: Rs 20 for the first two kilometers, Rs 5 for every km in excess of 2 km  and up to 10 km, and Rs 8 for every km in excess of 10 km. Bullock carts on the other hand charge Rs 2 per km. Sardar Khan takes a taxi from the Wasseypur Railway station to his home. On the way, at a distance of 14 km from the railway station, he meets Faizal Khan, and gets down from the taxi to talk to him. Later he takes a bullock cart to reach his home. He spends a total of Rs 102 to reach his home from the railway station. How far is his home from the railway station?
(A)17        (B) 18             (C) 19             (D) 20 (E) 21
Sol:
Points:
Taxi:
Rs 20 for first 2 km. Note here that it's not for each km, it is 20 for total 2 km
Rs 5 for every km from 3 up to 10 km
Rs 8 for every km for above 10 km range

Bullock:
It's a simple rate. Rs 2 per each km

First, Sardar travels 14 km on Taxi
Let us divide this in to three parts 2+8+4
=>Fare for taxi =   20 for first 2km + 8*5 for next 8 km + 4*8 for next 4 km
= 20+40+32 = 92
Afterwards he went by bullock. He spent a total amount of 102 for entire journey
=> He spent 102-92 = 10 for Bullock
=> He travels 10/2 = 5km on Bullock
Total distance of journey = 14km on Taxi + 5km on Bullock = 19 km
Note: While solving it in the exam, initially, I considered Rs 20 per each km of first 2km of taxi ride, and solved like this:
Fare of taxi = 2*20+8*5+4*8 = 112. As this figure is exceeding the total payment of Rs 102, I thought I made a mistake and rechecked. It took a little more time than usual.

Q6)Consider the expression: (a2+a+1)(b2+b+1)(c2+c+1)(d2+d+1)(e2+e+1)/abcde. The minimum value of the expression is:
(A)3          (B) 1   (C) 10             (D) 100          (E) 243
Sol:
(a2+a+1)(b2+b+1)(c2+c+1)(d2+d+1)(e2+e+1)/abcde
Take related terms first and simplify:
(a2+a+1)/a = a+1/a+1
we know that, minimum value of a+1/a is 2
=> minimum value of  (a+1/a+1) = 2+1 =3
=>  minimum value of  (a+1/a+1)(b+1/b+1)(c+1/c+1)(d+1/d+1)(e+1/e+1) = 35=243

Q7) The mean of six positive integers is 15. The median is 18, and the only mode of the integers is less than 18. The maximum possible value of the largest of the six integers is
(A)26        (B) 28             (C) 30             (D) 32 (E) 34
Sol:
If arranged in ascending order let the numbers be a,b,c,d,e,f. We want the largest number ie. 'f'.
Mean = 15 => (a+b+c+d+e+f)/6 = 15 =>  a+b+c+d+e+f = 90
median = 18 => (c+d)/2 = 18 => c+d = 36 => a+b+e+f = 90-36 = 54
The only mode is less than 18:
Mode is a number in the sequence which is repeated maximum number of times. To make the largest number maximum, the mode shall be minimum. The minimum possible mode is 1. so the cases may be
Case-I
a=b=1
Case-II:
a=b=c=1 : This case is ruled out. Follow the reason here:
if c=1, d=35 and
a+b+e+f = 2+e+f = 54 => e+f = 52.
'd' being 35, a value 52 is not possible for e+f because e and f shall be greater than 'd'
Case-III:
a=b=c=d=1: This case is ruled out as c+d shall be 36
So, there is only one case possible, ie., a=b=1
=>e+f = 52
so the sequence is 1,1,c,d,e,f
we have  c+d=36. In the sequence, to make f maximum, d shall be minimum. The least possible vale of 18 is not possible for d because in that case c is also equal to 18 in which case, there exists another mode which is not true as per the given condition. So we will go to the next possible minimum of 'd' ie., 19. So in this case c=17 and the sequence looks like this:
1,1,17,19,e,f.
And we have, e+f = 52. To make 'f' maximum, 'e' also should be as minmium as possible. Here the as the value of d is 19, the next minimum possible value for 'e' is 20 and in which case,
f= 52-20 = 32
Note: It took comparatively more time to solve in the exam.

Q8)Ramesh bought a total of 6 fruits (apples and oranges) from the market. He found that he required one orange less to extract the same quantity of juice as extracted from apples. If Ramesh had used the same number of apples and oranges o make the blend, then which of the following correctly represents the percentage of apple juice in the blend?
(A)25%    (B) 33.3%      (C) 60%         (D) 66.6%      (E) None of the above
Sol:
He has used same number of apples and oranges => A =3 and O =3
He found that he required one orange less to extract the same quantity of juice as extracted from apples
=> Had he used 2 Oranges and 3 Apples, it would have contributed equal sharing of juice
=> 2 Oranges and 3 Apples would have contributed equal volume of juice. Let that be ‘x’ each. I mean, 2 Oranges contribute ‘x’ volume of juice and 3 Apples also contribute ‘x’ volume of juice, making a total of ‘2x’ volume
But instead, he used 3 Oranges => Oranges contribute x+0.5x = 1.5x volume (x for 2-oranges and 0.5x for the extra orange)
=> Now the total volume of juice = 1.5x+x = 2.5x
(1.5x from Oranges and x from Apples)
=> The percentage of contribution of Apples = (x/2.5x)*100 = 40%

Q9) p and q are positive numbers such that pq = qp, and q=9p. The value of p is
(A)√9        (B) 91/6           (C) 91/9           (D) 91/8           (E) 91/3
Sol:
Let us try to solve the two equations:
p9p = (9p)p => p9p = 9p * pp => p9p-p = 9p => p8p = 9p
Let us check answer options one by one by substitution method.
(A) p = √9,
p8p = (91/2)8√9= 94√9
9p = 9√9
As the bases are same (ie., 9), just compare the powers.
Both are not equal and hence option A is out.
(B) p = 91/6,
p8p = (91/6)^(8*91/6)= 9^( 91/6*4/3)
9p = 9^(91/6)
Both are not equal and hence option B is out.
(C) p = 91/9,
p8p = (91/9)^(8*91/9)= 9^( 91/9*8/9)
9p = 9^(91/9)
Both are not equal and hence option C is out.
(D) p = 91/8,
p8p = (91/8)^(8*91/8)= 9^( 91/8*8/8) = 9^(91/8)
9p = 9^(91/8)
Both are equal and hence option D is correct.
And of course, we need not check option E.
Note: In the examination, first I have directly tried the substitution method without initially solving the two equations. It was a mess and then I have tried this way. It killed some time.

Q10) Sara has just joined Facebook. She has 5 friends. Each of her five friends has twenty five friends. It is found that at least two of Sara’s friends are connected with each other. On her birthday, Sara decides to invite her friends and the friends of her friends. How many people did she invite for her birthday party?
(A)≥ 105  (B) ≤ 123       (C) < 125       (D) ≥ 100 and ≤ 125 (E) ≥ 105 and ≤ 123
Sol:
This is a good problem and it troubled me a lot in exam.
Note: In exam, I made a mistake by not considering the ‘key point’ mentioned in the above solution. Then I got the range as ‘110 to 128’, which is not there in the answer options. I cross checked my way of approach several times but could not get the right one. Result is -it killed some valuable time.

Q11) 70% of the students who joined XLRI last year play football, 75% play cricket, 80% play basketball and 85% play carrom. The minimum percentage of people who play all games is:

(A) 5%     (B) 10%         (C) 15%         (D) 20% (E) None of the above

Sol:

70% play football => % not playing football = 30%

75% play cricket => % not playing cricket = 25%

Similarly, % not playing cricket = 20%, % not playing carrom = 15%

Max % of people not playing at least one game = 30+25+20+15 = 90%

A max of 90% do not play at least one game. This means, the remaining 10% play all the four games.

Q12) The central park is 40 meters long and 30 meters wide. The mayor wants to construct two roads of equal width in the park such that the roads intersect each other at right angles and the diagonals of the park are also the diagonals of the small rectangle formed at the intersection of the two roads. Further the mayor wants that the area of the two roads to be equal to the remaining area of the park. What should be the width of the roads?
(A)    10 meters     (B) 12.5 meters         (C) 14 meters            (D) 15 meters (E) 16 meters
Sol:
x(40-x)+ x(30-x)+x2 = 40*30-[ x(40-x)+ x(30-x)+x2]
=> 2[x(40-x)+ x(30-x)+x2] = 1200
=> 70x-x2 = 600 => x2-70x+600 = 0 => (x-10) (x-60) = 0
=> x = 10 or 60
As x must be less than 30, x = 10

Q13)How many whole numbers between 100 and 800 contain the digit 2?
(A)200     (B) 214           (C) 220           (D) 240          (E) 248
Here point is the number to be considered is of 3-digit. So, digit '2' can be in any of the three places- unit's ten's or hundred's position.
'2' in unit's position:
102,112,122,.....792=> by observing the first two digits (10 to 79), we can say that these are 70 in number.
'2' in ten's position:
120,121,122,...to 129 (122 should be excluded as it is already considered in the above list)
220,221,222,...to 229 (222 should be excluded as it is already considered in the above list)
….. such lists up to 729
=> 9*7 = 63
'2' in hundred's position:
200,202.......299: (100 numbers)
Out of all these, 220 to 229 shall be excluded. (10 numbers)
Out of all these, 202,212,222,232,... to 292 shall be excluded. (10-1 = 9 numbers)
100-10-9 = 81
Finally, the number of whole numbers meeting the criteria = 70+63+81 = 214
Note: My mistake in exam: For the case of '2' in ten's position, I had considered 9*8 instead of 9*7. I was contemplating that as the values range from 100 to 800 and so there are 8 such cases each of 9 numbers. But here the range stops at 800 and the numbers such as 820,821,...829 must not be considered. My answer was not matching with any of the given answer options. It killed some time.

Q14) A number is interesting if on adding the sum of the digits of the number, and the product of the digits of the number, the result is equal to the number. What fraction of numbers between 10 and 100 (both 10 and 100 included) is interesting?
(A)0.1       (B) 0.11          (C) 0.16          (D) 0.22         (E) None of the above
Sol:
As the number to be considered is in between 10 and 100, we can suppose it to be a two digit one 'xy'. The number 100 con be neglected as it does not meet the criteria, as
1+0+0+(1*0*0) = 1 100.
For a number 'xy', on adding the sum of digits of the number, and the product of the digits of the number yields: x+y+x*y
And if it is to meet the criteria, this sum shall be equal to the number 'xy', which is nothing but 10x+y.
x+y+x*y =  10x+y => x*y=9x => x(y-9) = 0 => x=0 or y=9
For the number to be two digit, x must be a non-zero number and hence, x must not be zero
=> y = 9.
All the two-digit numbers with y=9, ie., with 9 as unit digit are:
19,29,39,49,59,69,79,89,99
These are 9 in number. So, 9 numbers between 10 and 100 (both 10 and 100 included) are interesting
=> 9 out of 91 are interesting => fraction = 9/91 = 0.099  0.1

Q15)Consider the expression: (xxx)b = x3,where b is the base, and x is any digit of base b. Find the value of b:
(A)5          (B) 6   (C) 7   (D) 8   (E) None of the above
Sol:
It's all about the base system. For example, '234' in decimal system (base 10) can actually be written as 2*102+3*101+4*100.
Now I directly plunge in to the given equation:
(xxx)b = x3
=> x*b2+x*b1+x*b0 = x3
=> x(b2+b+1-x2)= 0
=> b2+b+1= x2
=>  b2+b+1 must be a perfect square
Now check the answer options one-by-one.
(A) b=5 =>  b2+b+1 = 25+5+1 = 31 ≠ a perfect square => option is out
(B) b=6 =>  b2+b+1 = 36+6+1 = 43 ≠ a perfect square => option is out
(C) b=7 =>  b2+b+1 = 49+7+1 = 57 ≠ a perfect square => option is out
(D) b=8 =>  b2+b+1 = 64+8+1 = 73 ≠ a perfect square => option is out
So the only option left out is (E)

Q16) Consider a function f(x) = x4+x3+x2+x+1, where x is a positive integer greater than 1. What will be the remainder if f(x5) is divided by f(x)?
(A)1          (B) 4               (C) 5   (D) a monomial in x (E) a polynomial in x
Sol:
Let x = 2.
f(x) = x4+x3+x2+x+1
f(x5) = x20+x15+x10+x5+1 =  x20-1+1+x15-1+1+x10-1+1+x5-1+1+1
=(x20-1)+(x15-1)+(x10-1)+(x5-1)+5
Now if we divide f(x5) with f(x), then we need to find the remainder.
But, just observe one thing here: all the terms of f(x5) except 5, ie., (x20-1), (x15-1), (x10-1) and (x5-1) are divisible by (x5-1). And (x5-1) can be split as (x-1)(x4+x3+x2+x+1).
=> all the terms of f(x5) except 5, ie., (x20-1), (x15-1), (x10-1) and (x5-1)
are divisible by (x4+x3+x2+x+1)
=> When we divide f(x5) with f(x), the only term left undivided is  '5'
=> remainder = 5
Note:
I have tried to get the remainder by directly dividing (x20+x15+x10+x5+1) with (x4+x3+x2+x+1) and got the answer. But as this involves a bit calculation oriented,  a little patience is required here, and there is a high chance of committing mistakes in this method.

Q17) p,q and r are three non-negative integers such that p+q+r=10. The maximum value of pq+qr+pr+pqr is

(A) ≥ 40 and ≤ 50           (B) ≥ 50 and ≤ 60     (C) ≥ 60 and ≤ 70

(D) ≥ 70 and ≤ 80           (E) ≥ 80 and ≤ 90

Sol:

This is a traditional type of problem, being touched in all types of competitive exams most frequently. The funda here is p=q=r.

p+q+r=10 =>3p=3q=3r=10 => p=q=r=10/3
pq+qr+pr+pqr = 3p2 +p3 = 3(10/3)2 +(10/3)3  = 100/3 + 1000/27 = 33.3+37.03 > 70

Q18) Ram, Shyam and Hari went out for a 100 km journey. Ram and Hari started the journey in Ram’s car at the rate of 25 kmph, while Shyam walked at 5 kmph. After some time Hari got off and started walking at the rate of 5 kmph and Ram went back to pick up Shyam. All three reached the destination simultaneously. The number of hours required for the trip was:
(A)8          (B) 7               (C) 6   (D) 5   (E) 4
Sol:
Key point to solve:
“If you break it in to pieces of each of the three friends individual journey, the puzzle will be resolved”
Let us take that 'R' and 'H' initially travels for distance 'x' km before 'H' gets off and starts walking and 'R' gets back in the car to pick up 'S'. And let us also consider 'y' km as the distance covered by 'R' before he meets 'S'. Then he pics up 'S' and both of them travel up to the end.
Now let us consider the journeys of the three one-by-one:
'H' travels in car with a speed of 25 kmph up to 'x' km and then he completes the remaining journey  of '100-x' km by walking at a speed of 5 kmph
=> the time taken by 'H' = (x/25) + (100-x/5) = (500-4x)/25

'R' travels in car with a speed of 25 kmph up to 'x' km and then he goes back in car a distance of 'y' km and pics up 'S' and again travels back 'y' km distance to reach the point of 'x' km and continues further to complete the remaining '100-x' km journey. So through out the journey, 'R' travels in the car a distance of x+y+y+(100-x) = 100+2y
=> the time taken by 'R' = (100+2y)/25

'S' walks at a speed of 5 kmph for a distance of (x-y) km, where he meets 'R' and continues the remaining journey of '100-(x-y)' km in the car at a speed of 25 kmph
=> the time taken by 'S' = (x-y)/5 + (100-(x-y))/25 = {100+4(x-y)}/25
Now the point is : 'All three reached the destination simultaneously '
=> all three took same time
=> (500-4x)/25 = (100+2y)/25 = {100+4(x-y)}/25
=> on solving we get y = 50 and x = 75
If we substitute these values in one of the times taken by R or S or H, we get the answer of 8 hrs

Q19) In a square PQRS, A and B are two points on PS and SR such that PA = 2AS and RB=2BS. If PQ=6, the area of the triangle ABQ is (is it a repeat)
(A)6          (B) 8   (C) 10             (D) 12 (E) 14
Sol:
'The mantra' => “Draw a figure and get the idea
We can trace out 4 triangles from the figure:
ABS, APQ, QRB and ABQ.
The point is - “summing up areas of these 4 triangles give us area of the square PQRS”.
And here, except ABQ all other are right-triangles and we also know the bases and heights of these right-triangles such that we can find their areas by using the formula 1/2*b*h.
Area of triangle ABS  = ½ * 2 * 2 = 2
Area of triangle APQ  = ½ * 6 * 4 = 12
Area of triangle QRB  = ½ * 4 * 6 = 12
Area of the square PQRS  = 6 * 6 = 36
Area of triangle ABQ
= Area of the square PQRS - (Area of triangle ABS + Area of triangle APQ + Area of triangle QRB)
= 36-(2+12+12) = 10

Q20) In the country of Four there are four cities, A,B,C and D. B is to the East of A, C is to the South of B, D is to the West of C, and A is to the North of D. The Government of Four is planning to connect these four cities by road such that it is possible for a person to go from a city to any of the other three cities. At the same time, the Government wants to ensure that the total road length is minimum. The distances between A to B, B to C, C to D and D to A are all equal to 10 km. What should be the total length of the road?
(A)26.64  (B) 27.32       (C) 28.30       (D) 30 (E) 36
Sol:
'The mantra' => “Draw a figure and get the idea
The points A,B,C and D are consecutive vertices of a square. A road with minimum length which connects each city with the remaining three cities can be built if the four cities are connected diagonally. There are two diagonals with each diagonal having a length of 10√2.
The total length of the road = 2 * 10√2 = 28.28

I – 103 and 7 are the only prime factors of 1000027
II -  6√6! > 7√7!
III – If I travel one half of my journey at an average speed of x km/h, it will be impossible for me to attain an average speed of 2x km/h for the entire journey.
(A)All the statements are correct        (B) Only statement II is correct
(C) Only statement III is correct          (D) Both statements I and II are correct
(E) Both statements I and III are correct
Sol:
Let us try to split 1000027:
1000027 = 7*142861 = 7*103*1387
Now we have to check for any factor of 1387:
First it is odd and hence 2 or any one of even numbers is not a factor.
Sum of digits = 1+3+8+7 = 19, not a multiple of 3 and hence 3 and 9 are not factors.
Unit digit is neither 0 nor 5 =>  5 or any one of numbers ending in 5 is not a factor
If we check, 7 is not a factor
Sum of alternate digits = 1+8 3+7 => 11 is not a factor
If we check, 13 and 17 are not factors
but, 19 is a factor
1000027 = 7*103*1387 =  7*103*19*73
So statement I is wrong => Answer options A,D and E can be eliminated. And the only left out answer options are B and C => only one of the options II and III is correct.
Statement III looks a simpler bet:
One half of the journey is completed x km/h. Let the other half is completed at an average speed y km/h. If the total distance is taken as 'd', then the overall average speed
= tot. distance / tot. time
= d/(d/2x + d/2y) = 4xy/(2x+2y) = 2xy/(x+y)
Now let us check whether this could be be equal to 2x:
2xy/(x+y) = 2x => y = x+y => x=0 which is wrong
=> it will be impossible to attain an average speed of 2x
=> statement III is correct
From the answer options, as we already know that only one of the statements II or III is correct, we can count on statement III and we need not check statement II.
Note:
Let us check option II:
Here a little manipulation is required.
6! = 6*5! = (6^6)(1/6^5)(5!)
=> 6√6! = (6) * 6√(5!/6^5)  ---(1)
7! = 6*7*5! = (6^7)(1/6^6)(7*5!)
=> 7√7! = (6) * 7√(7*5!/6^6) = (6)* 7√(5!/6^5) * 7√(7/6)          ---(2)
Now note the following two points:
5!/6^5 < 1 => 7√(5!/6^5) > 6√(5!/6^5) ---(3)
7/6 > 1 => 7√(7/6) > 1         ---(4)
From (3) and (4), we can say that,
(6) * 7√(5!/6^5) * 7√(7/6) >(6) * 6√(5!/6^5)
=> 7√7! > 6√6!
=> statement-II is wrong

Q22)Six playing cards are lying face down on a table, two of them are kings. Two cards are drawn at random. Let 'a' denote the probability that at least one of the cards drawn is a king, and 'b' denote the probability of not drawing a king. The ratio a/b is
(A)            ≥ 0.25 and < 0.5 (B) ≥ 0.5 and < 0.75  (C) ≥ 0.75 and < 1.0
(D) ≥ 1.0 and < 1.25 (E) ≥ 1.25
Sol:
There are three scenarios:
Either two are kings, only one card is king, none of the two is king
In all, the number of combinations of taking 2 cards out of 6 = 6C2 = 15
Split of this is like this:
The number of combinations of the two cards being kings = 2C2 = 1
The number of combinations of only one card being a king = 2C1 * 4C1 = 2*4 = 8
The number of combinations of neither card being a king =  4C2 = 6

Now, a = P(at least one king) = P(one king)+P(two kings) = (8+1)/15 = 3/5
b = P(no king) = 6/15 = 2/5
=> a/b = 3/2 = 1.5

Q23) Arun has to go to the country of Ten to work on a series of tasks for which he must get a permit from the Government of Ten. Once the permit is issued, Arun can enter the country within ten days of the date of issue of the permit. Once Arun enters Ten, he can stay for a maximum of ten days. Each of the tasks has a priority and takes certain number of days to complete. Arun can not work on more than one task at a time. The following table gives the details of the priority and the number of days required for each task.
Arun's first priority is to complete as many tasks as possible, and then try to complete the higher priority tasks. His last priority is to go back as soon as possible. The tasks that Arun should try to complete are:
(A)T1 and T2      (B) T1,T2 and T5      (C) T1,T4 and T5
(D) T1,T2 and T4           (E) T1,T3 and T4
Sol:
“Arun's first priority is to complete as many tasks as possible”
=>At the most, 3 tasks can be completed with in 10 days and all the combinations possible are:
T1,T2 and T5 => 3+5+2=10 days
T1,T3 and T4 => 3+3+4=10 days
T1,T4 and T5 => 3+4+2= 9 days
T2,T3 and T5 => 5+3+2=10 days
T3,T4 and T5 => 3+4+2= 9 days
To filter out further, let us see the second & third priorities:
“try to complete the higher priority tasks” and “to go back as soon as possible”
(T1,T2,T5) combination looks more suitable as it involves top two priorities.

More Problems in line.....

1. Hi Ravi, Thanks for the wonderful explanation.
Could I request you please do provide explanation for the rest of the questions as well.

Regards

1. Hi ak, they will be provided in a short span.

2. Great Blog..loved every line of this post..And would you be able to post such explanations for "Decision making" section?? That would be real helpful..

3. This is really helpful, thank you so much! Can you please take SNAP, TISS etc, too?

4. Thanks .. can i get for CAT as well for previous years 2013,14,