## Tuesday, 10 January 2012

### XAT-2012 Detailed Solutions (Quantitative Ability)

Q1) A spherical metal of 10 cm radius is molten and made in to 1000 smaller spheres. In this process, the surface area of the metal is increased by:
(A)  1000
(B)   100
(C)   10
(D)  No change
(E)   None of the above
Solution follows here...
Solution:
Let the radius of the original bigger sphere be ‘R’ and that of the smaller sphere be ‘r’.
When the metal sphere is molten and made into 1000 smaller spheres, the volume won’t change bu the surface area changes.
Volume of the original bigger sphere V = 4/3 π R3
Surface Area of the original bigger sphere S = 4 π R2
Volume of the each smaller sphere v = 4/3 π r3
Volume of the all the 1000 smaller spheres =1000*v = 4000/3 π r3
As the volume won’t change, 4000/3 π r3 = 4/3 π R3
=> r3 = R3/1000 => r = R/10
Surface Area of the smaller sphere s = 4 π r2
Surface Area of all the 1000 smaller spheres
= 1000*4 π r2
= 1000*4 π (R/10)2
= 10*4 π R2
= 10*S
=> Total surface area is increased to 10 times
=> Total surface area is increased by 9 times, which is not available in the answer options.
‘Most of the exam takers commit mistake here’. We need to be careful at the point- whether the increase is asked for ‘to’ or ‘by’

Q2) A man borrows Rs 6000 at 5% interest, on reducing balance, at the start of the year. If he repays Rs 1200 at the end of each year, find the amount of loan outstanding at the beginning of the third year?
(A)  3162.75
(B)   4125.00
(C)   4155.00
(D)  5100.00
(E)   5355.00
Solution:
The following table explains step by step procedure:
 Interest @ 5% Amount paid at the end of Balance Amount at the start of 6000*5/100 = 300 1st year => 1200 2nd year => 6000+300 -1200 = 5100 5100*5/100 = 255 2nd year => 1200 3rd year => 5100+255-1200 = 4155

Q3) Ramesh analysed the monthly salary figures of five vice presidents of his company. All the salary figures are integers. The mean and median salary figures are Rs 5 lakh and the only mode is Rs 8 lakh. Which of the options below is the sum (in Rs lakh) of the highest and the lowest salaries?
(A)  9
(B)   10
(C)   11
(D)  12
(E)   None of the above
Solution:
Key here is  - “The mean and median salary figures are Rs 5 lakh and the only mode is Rs 8 lakh”
As there are five figures and given that the median is 5, the figures in increasing order would be as follows:
a, b, 5, c, d
Given that the only mode is 8 => 8 is the only repeated figure and the other figures occur only once. By this we can conclude that c = d = 8. Hence, the figures in increasing order would be as follows:
a, b, 5, 8, 8
Given that the mean is 5 => (a+b+5+8+8)/5 = 5 => a+b = 4
The possibilities are “a=1,b=3” and “a=b=2”
But, “a=b=2” possibility won’t work here in which case another mode 2 exists apart form ‘8’, which is wrong. Hence the only possibility is “a=1,b=3”, and hence, the figures in increasing order would be as follows:
1, 3, 5, 8, 8
=> Sum of the highest and the lowest salaries = 1+8 = 9

Q4) Nikhil’s mother asks him to buy 100 pieces of sweets worth Rs 100/-. The sweet shop has three kinds of sweets kajubarfi, gulabjamun and sandesh. Kajubarfi costs Rs 10/- per piece, Gulabjamun costs Rs 3/- per piece and Kajubarfi costs 50 paise per piece. If Nikhil decides to buy at least one sweet of each type, how many Gulabjamuns should he buy?
(A)  1
(B)   2
(C)   3
(D)  4
(E)   5
Solution:
Let the number of Kajubarfis, Gulabjamuns and Sandeshs be k,g and s respectively.
Total number of pieces should be 100 => k + g + s = 100           ---(1)
Net worth of all sweets to be bought is Rs 100 => 10k + 3g + 0.5s = 100
=> 20k + 6g + s = 200                                   ---(2)
(2) – (1) => 19k + 5g = 100
Here two terms are added to get 100. Second term 5g is a multiple of 5. Keeping this in mind, we will be cornered to one point:
“The only possible value of k is 5’’
For this value of k, we get g = 1

Q5) Ram, a farmer managed to grow shaped-watermelons inside glass cases of different shapes. The shapes he used were: a perfect cube, hemi-spherical, cuboid, cylindrical along with the normal spherical shaped watermelons. Thickness of the skin was same for all the shapes. Each of the glass cases was so designed that the total volume and the weight of all the watermelons would be equal irrespective of the shape.
A customer wants to buy watermelons for making juice, for which the skin of the watermelon has to be peeled off, and therefore is a waste. Which shape should the customer buy?
(A)  Cube
(B)   Hemi-sphere
(C)   Cuboid
(D)  Cylinder
(E)   Normal spherical
Solution:
This one can be easily dealt with commonsense. The customer’s perspective would be to minimise the waste => to minimise the skin of watermelon
As the skin-thickness of all the shapes is same, we need to select the one with minimum surface area.
Even though all the shapes have the same volume, we need to select the one with minimum surface area.
Thanks to the symmetry of shape and closed structure, it is nothing but the normal sphere, which has minimum surface area.

Q6) Tina, Mina, Gina, Lina and Bina are 5 sisters, aged in that order, with Tina being the eldest. Each of them had to carry a bucket of water from a well to their house. Their buckets’ capacities were proportional to their ages. While returning equal amount of water got splashed out of their buckets. Who lost maximum amount of water as a percentage of the bucket capacity?
(A)  Tina
(B)   Mina
(C)   Gina
(D)  Lina
(E)   Bina
Solution:
As Tina is eldest, capacity of Tina’s bucket is highest
=> % of water splashed = (water splashed) / (bucket capacity) * 100
As equal amount of water got splashed out of all the buckets,
Numerator is same for all the five cases and percentage depends inversely on the denominator (ie., the bucket capacity)
=> The bucket with minimum capacity splashes maximum % of water

Q7) Shyam, a fertilizer salesman sells directly to farmers. He visits two villages A and B. Shyam starts from A and travels 50 meters to the East, then 50 meters North-East at exactly 450 to his earlier direction, and then another 50 meters East to reach village B. If the shortest distance between villages A and B is in the form of a√(b + √c) meters, find the value of a+b+c?
(A)  52
(B)   54
(C)   58
(D)  59
(E)   None of the above
Solution:

Fig-1 shows the original path taken to travel and fig-2 adds the shortest distance path (line directly joining A and B).
AX = 50; YC = 50; BC = 50; XZ = 50;
ÐZXY = 450
Cos(ÐZXY) = cos(450) = XY/XZ
=> XY = XZ * cos(450) = 50(1/√2) = 25√2
AC = AX+XY+YC = 50+25√2+50 = 100+25√2
Similarly, BC = YZ = XZ * sin(450) =  25√2
Applying Pythagoras to the right triangle ABC,
AB2 = AC2 + BC2 = (100+25√2)2 + (25√2)2
= 252{(4+√2)2 + (√2)2}
=> AB = 25√(16+2+8√2+2) = 25√(20+8√2)
=> AB = 50√(5+2√2) = 50√(5+√8)
Given that the shortest distance AB can be represented as a√(b + √c)
=> 50√(5+√8) = a√(b + √c)
=> a = 50; b = 5; c = 8 => a+b+c = 63

Q8) Three truck drivers Amar, Akbar and Anthony stop at a road side eating joint. Amar orders 10 rotis, 4 plates of tadka and a cup of tea. Akbar orders 7 rotis, 3 plates of tadka and a cup of tea. Amar pays Rs 80 for the meal and Akbar pays Rs 60. Meanwhile, Anthony orders 5 rotis, 5 plates of tadka and 5 cups of tea. How much (in Rs) Anthony will pay?
(A)  75
(B)   80
(C)   95
(D)  100
(E)   None of the above
Solution:
This is a problem based on manipulation of linear equations.
Let the unit prices of roti, tadka and tea be x,y, and z respectively.
Converting the problem in to linear equations,
Amar’s payment => 10x+4y+z = 80           ---(1)
Akbar’s payment => 7x+3y+z = 60             ---(2)
We need to find (5x+5y+5z), the amount to be paid by Anthony.
(Equation-1) - (Equation-2) => 3x+y = 20                          --(3)
10*(Equation-2) - 7*(Equation-1) => 2y+3z = 40             --(4)
(Equation-3) + (Equation-4) => 3x+3y+3z = 60
=> x+y+z = 20 => 5x+5y+5z = 100

Q9)Three Vice Presidents (VP) regularly visit the plant on different days. Due to labour unrest, VP(HR) regularly visits the plant after a gap of 2 days. VP(operations) regularly visits the plant after a gap of 3 days. VP(Sales) regularly visits the plant after a gap of 5 days. The VPs do not deviate from their regular schedules. CEO of the company meets the VPs when all the three VPs come to plant together. CEO is on leave from January 5th to January 28th, 2012. Last time CEO met the VPs on January 3, 2012. When is the next time CEO will meet all the VPs?
(A)  February 6, 2012
(B)   February 7, 2012
(C)   February 8, 2012
(D)  February 9, 2012
(E)   None of the above
Solution follows here:
Solution:
This is a simple LCM problem.
VP(HR) regularly visits the plant after a gap of 2 days => VP(HR) visits the plant for every 3 days
VP(operations) regularly visits the plant after a gap of 3 days => VP(operations) visits the plant for every 4 days
VP(Sales) regularly visits the plant after a gap of 5 days => VP(Sales) visits the plant for every 6 days
They all meet for every ‘LCM(3,4,6)’ days => They all meet for every 12 days
=> CEO meets VPs for every 12 days
CEO met the VPs on January 3 => CEO shall meet on January 15, January 27, February 8
As the CEO was on leave from January 5th to January 28th, He meets them on February 8

Q10) A medical practitioner has created different potencies of  a commonly used medicine by dissolving tablets in water and using the resultant solution.
Potency 1 solution:  When 1 tablet is dissolved in 50 ml, the entire 50 ml solution is equivalent to one dose.
Potency 2 solution:  When 2 tablets are dissolved in 50 ml, the entire 50 ml solution is equivalent to 2 doses.
... and so on.
This way he can give fractions of tablets based on the intensity of infection and the age of the patient.
For a particular patient, he administers 10 ml of potency 1, 15 ml of potency 2 and 30 ml of potency 4. The dosage administered to the patient is equivalent to
(A)  > 2 and 3
(B)   > 3 and 3.25
(C)   > 3.25 and 3.5
(D)  > 3.5 and 3.75
(E)   > 3.75 and 4
Solution:
10 ml of potency 1, 15 ml of potency 2 and 30 ml of potency 4:
50 ml of potency 1 is equvalent to 1 tablet
=> 10 ml of potency 1 is equvalent to 10/50 = 1/5 tablet
50 ml of potency 2 is equvalent to 2 tablets
=> 15 ml of potency 2 is equvalent to (15/50) * 2 = 3/5 tablet
50 ml of potency 4 is equvalent to 4 tablets
=> 30 ml of potency 4 is equvalent to (30/50) * 4 = 12/5 tablet
Total dosage equivalent = 1/5 + 3/5 + 12/5 = 16/5 = 3.2

Q11)A property dealer bought a rectangular piece of land at Rs. 1000/sft. The length of the plot is less than twice its breadth. Due to its size, there were no buyers for the full plot. Hence he decided to sell it in smaller sized pieces as given below.
The largest square from one end was sold at Rs. 1200/sqft. From the remaining rectangle, the largest square was sold at Rs. 1150/sqft.
Due to crash in the property prices, the dealer found it difficult to make profit from the sale of the remaining part of the land. If the ratio of the perimeter of the remaining land to the perimeter of the original land is 3:8, at what price (in Rs) the remaining part of the land is to be sold such that the dealer makes an overall profit of 10%?
(A)  500 /sqft
(B)   550 /sqft
(C)   600 /sqft
(D)  650 /sqft
(E)   None of the above
Solution:
Diagrammatic representation:

The largest square from one end was sold at Rs. 1200/sqft => Area of b2 is sold at 1200
=> Amount received on this transaction = 1200 b2                     ---(1)
From the remaining rectangle, the largest square was sold at Rs. 1150/sqft  ---(2)
=> Area of (l-b)2 is sold at 1150
=> Amount received on this transaction = 1150 (l-b)2
Area of the remaining part of the land = (2b-l)(l-b), let the rate at which this shall be sold is Rs X /sqft
=> Amount received on this transaction = (2b-l)(l-b) X               ---(3)
Ratio of the perimeter of the remaining land to the perimeter of the original land is 3:8
=> 2{(2b-l)+(l-b)} /  2{l+b} = 3 / 8
=> l/b = 3/5               ---(4)
“Remaining part of the land is to be sold such that the dealer makes an overall profit of 10%”
=> Amount for which he purchased the land = (l*b)*1000 = 1000 lb
=> Amount for which he need to sell in order to get 10% profit = 1.1 * (1000lb) = 1100 lb ---(5)
From (1), (2) and (3), total amount he received on all the three transactions
= 1200 b2+1150 (l-b)2+(2b-l)(l-b) X             ---(6)
Equating (5) and (6),
1200 b2+1150 (l-b)2+(2b-l)(l-b) X = 1100 lb
On Simplifying this using  (4),
We get X = 550

Q12) Lionel and Ronaldo had a discussion on the ages of Jose’s sons. Ronaldo made following sentences about Jose’s sons:
i)                    Jose has three sons
ii)                  The sum of the ages of Jose’s sons is 13
iii)                The product of the ages of the sons is same as the age of Lionel
iv)                Jose’s eldest son Zizou weighs 32 kilos
v)                  The sum of the ages of younger sons of Jose is 4
vi)                Jose has fathered a twin
vii)              Jose is not the father of triple
viii)    The LCM of ages of Jose’s sons is more than the sum of their ages
Which of the following combinations gives information sufficient to determine the ages of Jose’s sons?
(A)  i), ii), iii) and iv)
(B)   i), ii), iv) and vi)
(C)   i), ii), iii) and v)
(D)  i), ii), v) and vii)
(E)   i), ii), v) and vi)
Solution:
Let us analyse the sentences one by one:
i)Jose has three sons – this is very much required without which we cannot proceed
vi)Jose has fathered a twin – this in combination with clue-(v) is quite useful to find the ages of the younger sons
v)The sum of the ages of younger sons of Jose is 4 – this in combination with clue-(iv) helps in finding the ages of two younger sons : 4 years each
ii)The sum of the ages of Jose’s sons is 13 - this in combination with clue-(v) helps in finding the age of the eldest son : 13-8 = 5 years
iii)The product of the ages of the sons is same as the age of Lionel – This is not a useful one as we do not the age of Lionel
iv)Jose’s eldest son Zizou weighs 32 kilos – not at all required as the it is asked for the ages but not the weights
vii)Jose is not the father of triple – Given the clue (vi), this is redundant and hence not required
viii)The LCM of ages of Jose’s sons is more than the sum of their ages- this is a redundant one and hence not required
Overall, clues i), ii), v) and vi) are enough to find the ages,

Q13) Suresh who runs a bakery, uses a conical shaped equipment to write decorative labels using cream. The height of this equipment is 7 cm and the diameter of the base is 5 mm. A full charge of this equipment will write 330 words on an average. How many words can be written using three fifth of a litre of cream?
(A)  45090
(B)   45100
(C)   46000
(D)  43200
(E)   None of the above
Solution:
Height of the cone = h = 7 cm
Radius of the cone = r = 5/2 mm = 2.5 mm = 0.25 cm
Volume of the cone = 1/3 π r2h = 1/3 * 22/7 * (0.25)2 * (7)
=11/24 cc = 11/24 * 10-3 ltr
11/24 * 10-3 ltr will write 330 words
=> 1 ltr will write 330*24*1000/11 = 720000 words
=> 3/5 th of litre will write 3/5 * (720000) = 432000 words
But this answer is not available in the options A to D

Q14) a computer program was tested 300 times before its release. The testing was done in three stages of 100 tests each. The software failed 15 times in Stage I, 12 times in Stage II, 8 times in Stage III, 6 times in both Stage I and Stage II, 7 times in both Stage II and Stage III, 4 times in both Stage I and Stage III, 4 times in all the three stages. How many times the software failed in a single stage only?
(A)  10
(B)   13
(C)   15
(D)  17
(E)   35
Solution:
Representing this in a Venn diagram:
To build the venn diagram, first start from ‘all the three stages’ case. It is failed 4 times in all three stages. Then go to ‘two stages’ cases.
6 times failed in I and II => 6-4 = 2 tests failed in only I and II
7 times failed in II and III => 7-4 = 3 tests failed in only II and III
4 times failed in I and III => 4-4 = 0 tests failed in only I and III

Then go to ‘one stage’ cases.
Only I
15 times failed in I => 15 – (numbers involving stage I)
= 15 –(I and II) – (I and III) – (I and II and III)
= 15 – 2 – 0 – 4 = 9
Only II
12 times failed in II => 12 – (numbers involving stage II)
= 12 –(I and II) – (II and III) – (I and II and III)
= 12 – 2 – 3 – 4 = 3
Only III
8 times failed in III => 8 – (numbers involving stage III)
= 8 –(I and III) – (II and III) – (I and II and III)
= 8 – 0 – 3 – 4 = 1
Cases failed in a single stage only = I only + II only + III only = 13

Q15) City bus corporation runs two buses from terminus A to terminus B, each bus making 5 round trips in a day. There are no stops in between. These buses ply back and forth on the same route at different but uniform speeds. Each morning the buses start at 7 AM from the respective terminuses. They meet for the first time at a distance of 7 Km from terminus A. Their next meeting is at a distance of 4 Km from terminus B, while travelling in opposite directions. Assuming that the time taken by the buses at the terminuses is negligibly small, and the cost of running the bus is Rs 20 per Km, find the daily cost of running the buses (in Rs)?
(A)  3200
(B)   4000
(C)   6400
(D)  6800
(E)   None of the above
Solution:
Let the distance between A and B be x. Let the speeds of the buses starting at A and B be by a and b respectively.
First meeting is at 7 km from A:
By that time first bus travels 7 Km and second one travels x-7 Km. As they have travelled for the same time,
7/a = (x-7)/b              ---(1)
Second meeting is at 4 km from B:
By that time first bus travels x+4 Km and second one travels x+(x-4) = 2x-4 Km. As they have travelled for the same time,
(x+4)/a = (2x-4)/b                 ---(2)
(1)/(2) gives,
7/(x+4) = (x-7)/(2x-4) => x2-17x = 0 => x =17
=> one round trip is equal to 2*17 = 34 Km
“Each bus makes 5 round trips in a day” => Each bus makes 5*34 Km per day
=> Total distance travelled by the two buses in a day = 2*5*34 Km
“the cost of running the bus is Rs 20 per Km”
=> Total cost = 2*5*34*20 = Rs 6800

Q16) Ram prepares solutions of alcohol in water according to customers’ needs. This morning Ram has prepared 27 litres of 12% alcohol solution and kept it ready in a 27 litres delivery container to be shipped to the customer. Just before delivery he finds out that the customer had asked 27 litres of 21% alcohol solution. To prepare what the customer wants, Ram replaces a portion of 12% solution with 39% solution. How many litres of 12% solution are replaced?
(A)  5
(B)   9
(C)   10
(D)  12
(E)   15
Solution:
Let the volume of alcohol replaced be ‘x’.
The following table gives step by step procedure:
 Tot. Vol Alcohol Vol Water Vol Original 12% solution 27 0.12*27 0.88*27 (-) removed 12% solution x 0.12x 0.88x (+) Added 39% solution x 0.39x 0.61x Final solution (21%) 27-x+x = 27 0.12*27-0.12x+0.39x = 0.12*27 + 0.27x 0.88*27- 0.88x + 0.61x = 0.88*27 - 0.27x

The volume of Alcohol in the final solution = 0.12*27 + 0.27x
It is 21% solution => 0.12*27 + 0.27x = 0.21*27
=> 0.27x = 0.09*27 => x = 0.09*27/0.27 = 9

Answer Questions 17 and 18 based on the following information:
Ramya, based in Shanpur took her car to a 400 Km trip to Rampur. She maintained a log of Odometer readings and the amount of petrol she purchased at different petrol pumps at different prices (given below). Her car already had 10 litres petrol at the start of the journey, and she first purchased petrol at the start of the journey, as given in table below, and she had 5 litres remaining at the end of the journey.
 Odometer Reading (Km) Petrol Purchased (Litre) Rate of Petrol (Rs per litre) Start of journey 400 20 30 600 15 35 650 10 40 End of journey 800

Q17) What has been the mileage (in kilometres per litre) of her car over the entire trip?
(A)  8.00
(B)   8.50
(C)   9.00
(D)  9.50
(E)   None of the above
Solution:
Petrol purchased in the trip = 20+15+10 = 45 litres
“Her car already had 10 litres petrol at the start of the journey”
“she had 5 litres remaining at the end of the journey”
=> Total petrol consumed = 45+10-5 = 50 litres
Total journey covered = 400 Km
=> Mileage = 400Km / 50 ltr = 8 Km per litre

Q18) Her car’s tank capacity is 35 litres. Petrol costs Rs 45 per litre in Rampur. What is the minimum amount of money she would need for purchasing petrol for the return trip from Rampur to Shanpur, using the same route? Assume that the mileage of the car remains unchanged throughout the route, and she did not use her car to travel around in Rampur?
(A)  1714
(B)   1724
(C)   1734
(D)  1744
(E)   None of the above
Solution:
Return Trip:
Distance to the next immediate petrol point = 800-650 = 150 Km
Petrol required to reach this point (mileage calculated at 8 km/ltr)
= 150/8 litres
Petrol left with at Rampur = 5 litres
=> Minimum amount of petrol to be purchased at Rampur = 150/8 - 5 = 110/8 litres
(this is at the rate of Rs 45/litre)
Distance to the next immediate petrol point = 650-600 = 50 Km
Petrol required to reach this point (mileage considered 8 km/ltr)
= 50/8 litres
(this is at the rate of Rs 40/litre)
Distance to the next immediate petrol point = 600-400 = 200 Km
Petrol required to reach this point (mileage considered 8 km/ltr)
= 200/8 litres
(this is at the rate of Rs 35/litre)
Hence minimum amount to be spent = (45*110/8)+(40*50/8)+(35*200/8)
= (50/8) (99+40+140) = 1743.75 ≈ 1744

Q19) Gopal sells fruit juice mixture using orange juice and pineapple juice. Gopal prepares this mixture by drawing out a jug of orange juice from a 10 litre container filled with orange juice and replacing it with pineapple juice. If Gopal draws out another jug of resultant mixture and replaces it with pineapple juice, the container will have equal volumes of orange juice and pineapple juice. The volume of the jug in litres is
(A)  2
(B)   > 2 and ≤ 2.5
(C)   2.5
(D)  > 2.5 and ≤ 3
(E)   ≥ 3
Solution:
Let the volume of jug be ‘x’.
The following table specifies the steps:
 Orange Pineapple Initial volumes 10 0 After First iteration 10-x x Second iteration to be removed (10-x)/10 * (x) x/10 * (x) to be added 0 x Final volumes (10-x) - (10-x)/10 * (x) x - x/10 * (x) + x

Finally, the volumes of Orange juice and Pineapple juice are equal
=> (10-x) – (10-x)x/10 = 2x – x2/10
Multiplying on both sides with 10,
100-10x-10x+x2 = 20x-x2
2x2-40x+100 = 0 => x2-20x+50 = 0
=> x = {20±√(400-200)}/2 = 10±5√2 = 10±(7.07) = 17.07 (or) 2.93
As the jug volume should not be greater than the volume of the mixture ie., 10,
17.07 value is not correct, the only answer is 2.93

Answer question numbers 20 and 21 based on the following information:
The following pie chart shows the percentage distribution of runs scored by a batsman in a test innings.

Q20) If the batsman has scored a total of 306 runs, how many 4’s and 6’s did he hit?
(A)  31 and 3 respectively
(B)   32 and 2 respectively
(C)   32 and 3 respectively
(D)  33 and 1 respectively
(E)   33 and 2 respectively
Solution:
In the pie chart, 4’s occupy 43.14% and 6’s occupy 3.92%.
=> Total score contribution by 4’s = 43.14% of 306
43.14% of 300 = 43.14/100  * 300 = 43.14 * 3 = 129.42
43.14% of 6 = 43.14/100  * 6 = 0.4314 * 6 = 2.5884
43.14% of 306 = 129.42 + 2.5884 = 132.xx
=> Total score contribution by 4’s = 132.xx
=> Number of 4’s = 132/4 = 33
Total score contribution by 6’s = 3.92% of 306
3.92% of 300 = 3.92/100  * 300 = 11.76
3.92% of 6 = 3.92/100  * 6 = 0.0392 * 6 = 0.2352
3.92% of 306 = 11.76 + 0.2352 = 12.xx
=> Total score contribution by 6’s = 12.xx
=> Number of 6’s = 12/6 = 2

Q21) If 5 of the dot balls had been hit for 4’s, and if two of the shots for which the batsman scored 3 runs each had fetched him one run instead, what would have been the central angle of the sector corresponding to the percentage of runs scored in 4’s?
(A)  160
(B)   163
(C)   165
(D)  167
(E)   170
Solution:
If 5 of the dot balls had been hit for 4’s, and if two of the shots for which the batsman scored 3 runs each had fetched him one run instead,
The change in total score = +5*4-2(3-1) = +20-4 = +16
The change in share of 4’s = +5*4 = +20 = +20
If the total score 306 runs given in the previous question is considered here as the initial value, the new score = 306+16 = 322
Initial share of 4’s = 43.14% of 306 = 132.xx
New share of 4’s = 132.xx+20 = 152.xx
New ratio share of 4’s = 152/322
=> Central Angle subtended by 4’s sector = 152/322 * 3600
= 76*360/161,
Now see the following procedure for simplification:
76*360/162 < 76*360/161 < 76*360/160
=> 1520/9 < 76*360/161 < 171
=> 168.89 < 76*360/161 < 171
The only answer satisfying this criterion is 170

Q22) A city has a park shaped as a right angled triangle. The length of the longest side of this triangle is 80 m. The mayor of the city wants to construct three paths from the corner point opposite to the longest side such that these three paths divide the longest side in to four equal segments. Determine the sum of the squares of the lengths of the three paths?
(A)  4000 m
(B)   4800 m
(C)   5600 m
(D)  6400 m
(E)   7200 m
Solution:
If right triangle triad 3,4,5 is considered and the proportionate values to suite length of diagonal to 80, the triad becomes: 48,64,80
=> The other two sides are 48 and 64
BC = 80, AB = 64, AC = 48
From trigonometric ratios,
ÐB = sin-1(48/80) = sin-1(3/5)
(or)
cos-1(64/80) = cos-1(4/5)
From right triangle BIF,
BF = (BI) CosB = 20 (4/5) = 16 => AF = AB – BF = 64-16 = 48
FI = (BI) SinB = 20 (3/5) = 12
From right triangle BEH,
BE = (BH) CosB = 40 (4/5) = 32 => AE = AB – BE = 64-32 = 32
EH = (BH) SinB = 40 (3/5) = 24
From right triangle BDG,
BD = (BG) CosB = 60 (4/5) = 48 => AD = AB – BD = 64-48 = 16
DG = (BG) SinB = 60 (3/5) = 36

Applying Pythagoras to the right triangle AIF,
AI2 = AF2 + FI2 = (48)2 + (12)2
Applying Pythagoras to the right triangle AHE,
AH2 = AE2 + EH2 = (32)2 + (24)2
Applying Pythagoras to the right triangle AGD,
AG2 = AD2 + DG2 = (16)2 + (36)2
The sum of the squares of the lengths of the three paths
= AI2 + AH2 + AG2(48)2 + (12)2 + (32)2 + (24)2+ (16)2 + (36)2 = 5600