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...
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 R^{3}
Surface Area of the original bigger sphere S = 4 π R^{2}
Volume of the each smaller sphere v = 4/3 π r^{3}
Volume of the all the 1000 smaller spheres =1000*v = 4000/3 π
r^{3}
As the volume won’t change, 4000/3 π r^{3} = 4/3 π R^{3}
=> r^{3} = R^{3}/1000 => r = R/10
Surface Area of the smaller sphere s = 4 π r^{2}
Surface Area of all the 1000 smaller spheres
= 1000*4 π r^{2}
= 1000*4 π (R/10)^{2}
= 10*4 π R^{2}
= 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’
Answer (E)
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+2551200 = 4155

Answer (C)
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
Answer (A)
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
Answer (A)
Q5) Ram, a farmer managed to grow
shapedwatermelons inside glass cases of different shapes. The shapes he used
were: a perfect cube, hemispherical, 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) Hemisphere
(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 skinthickness 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.
Answer (E)
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
=> Answer is Bina
Answer (E)
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 NorthEast at exactly 45^{0} 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:
Fig1 shows the original path taken to travel and fig2 adds
the shortest distance path (line directly joining A and B).
AX = 50; YC = 50; BC = 50; XZ = 50;
ÐZXY = 45^{0}
Cos(ÐZXY) = cos(45^{0}) = XY/XZ
=> XY = XZ * cos(45^{0}) = 50(1/√2) = 25√2
AC = AX+XY+YC = 50+25√2+50 = 100+25√2
Similarly, BC = YZ = XZ * sin(45^{0}) = 25√2
Applying Pythagoras to the right triangle ABC,
AB^{2} = AC^{2} + BC^{2} = (100+25√2)^{2}
+ (25√2)^{2}
= 25^{2}{(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
Answer(E)
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.
(Equation1)  (Equation2) => 3x+y = 20 (3)
10*(Equation2)  7*(Equation1) => 2y+3z = 40 (4)
(Equation3) + (Equation4) => 3x+3y+3z = 60
=> x+y+z = 20 => 5x+5y+5z = 100
Answer (D)
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 5^{th}
to January 28^{th}, 2012. Last time CEO met the VPs on January 3, 2012.
When is the next time CEO will meet all the VPs?
Q10) A medical practitioner has created different potencies of a commonly used medicine by dissolving tablets in water and using the resultant solution.
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.
Q12) Lionel and Ronaldo had a discussion on the ages of Jose’s sons. Ronaldo made following sentences about Jose’s sons:
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?
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)?
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) 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 5^{th}
to January 28^{th}, He meets them on February 8
Answer (C)
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
Answer (B)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
The largest square from one end was sold at Rs. 1200/sqft
=> Area of b^{2} is sold at 1200
=> Amount received on this transaction = 1200 b^{2}
(1)
From the remaining rectangle, the largest square was sold at
Rs. 1150/sqft (2)
=> Area of (lb)^{2} is sold at 1150
=> Amount received on this transaction = 1150 (lb)^{2}
Area of the remaining part of the land = (2bl)(lb), let the
rate at which this shall be sold is Rs X /sqft
=> Amount received on this transaction = (2bl)(lb) X (3)^{}
Ratio of the perimeter of the remaining land to the perimeter
of the original land is 3:8
=> 2{(2bl)+(lb)} / 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 b^{2}+1150 (lb)^{2}+(2bl)(lb) X (6)
Equating (5) and (6),
1200 b^{2}+1150 (lb)^{2}+(2bl)(lb) X = 1100
lb
On Simplifying this using (4),
We get X = 550
Answer (B)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 : 138 = 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,
Answer (E)
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 r^{2}h = 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
Answer (E)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 => 64 = 2 tests failed in only
I and II
7 times failed in II and III => 74 = 3 tests failed in
only II and III
4 times failed in I and III => 44 = 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
Answer (B)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
x7 Km. As they have travelled for the same time,
7/a = (x7)/b (1)
Second meeting is at 4 km from B:
By that time first bus travels x+4 Km and second one travels x+(x4)
= 2x4 Km. As they have travelled for the same time,
(x+4)/a = (2x4)/b (2)
(1)/(2) gives,
7/(x+4) = (x7)/(2x4) => x^{2}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
Answer (D)
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%)

27x+x = 27

0.12*270.12x+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 (B)
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+105 = 50 litres
Total journey covered = 400 Km
=> Mileage = 400Km / 50 ltr = 8 Km per litre
Answer (A)
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 = 800650 = 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 = 650600 = 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 = 600400 = 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
Answer (D)
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

10x

x


Second iteration

to be removed

(10x)/10 * (x)

x/10 * (x)

to be added

0

x


Final volumes

(10x)  (10x)/10 * (x)

x  x/10 * (x) + x

Finally, the volumes of Orange juice
and Pineapple juice are equal
=> (10x) – (10x)x/10 = 2x – x^{2}/10
Multiplying on both sides with 10,
10010x10x+x^{2} = 20xx^{2}
2x^{2}40x+100 = 0 => x^{2}20x+50
= 0
=> x = {20±√(400200)}/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 (D)
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
Answer (E)
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*42(31) = +204 = +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 * 360^{0}
= 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
Answer (E)
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 = 6416 =
48
FI = (BI) SinB = 20 (3/5) = 12
From right triangle BEH,
BE = (BH) CosB = 40 (4/5) = 32 => AE = AB – BE = 6432 = 32
EH = (BH) SinB = 40 (3/5) = 24
From right triangle BDG,
BD = (BG) CosB = 60 (4/5) = 48 => AD = AB – BD = 6448 = 16
DG = (BG) SinB = 60 (3/5) = 36
Applying Pythagoras to the right triangle AIF,
AI^{2} = AF^{2} + FI^{2} = (48)^{2}
+ (12)^{2}
Applying Pythagoras to the right triangle AHE,
AH^{2} = AE^{2} + EH^{2} = (32)^{2}
+ (24)^{2}
Applying Pythagoras to the right triangle AGD,
AG^{2} = AD^{2} + DG^{2} = (16)^{2}
+ (36)^{2}
The sum of the squares of the lengths of the three paths
= AI^{2} + AH^{2} + AG^{2}(48)^{2}
+ (12)^{2} + (32)^{2} + (24)^{2}+ (16)^{2} + (36)^{2}
= 5600
Answer (C)
check solution for Question No. 7 the approach taken is completely wrong....I am talking about the question based on directions..
ReplyDeleteIf it is regarding 45 degrees, my point is like this:
DeleteI considered forward direction XY and not the reverse one XA as it is mentioned that he took 45 deg' turn to the earlier direction. That's why XZ is depicted at an angle 45 deg' with XY and not with XA. Pl comment if you contradict...
Howcome BC is 50??? If ZX is 50 and is the hypotenuse to the traingle then BC which is perpendicular cannot be 50.
ReplyDeletePlease clarify.
Its a typo, BC is not 50. But if you see the complete solution 25√2 is taken for BC. Thanks for thew information Rahul.
Deletevery helpful
ReplyDelete