Question 1
The students of Vidya Mandir Secondary school, Jalpaiguri went for an excursion. It was seen that a boat travels at 14.5 km per hour when it goes along with the stream. By the time he starts returning the speed of the river doubled than its original value due to a sudden storm. The speed of the boat is 7 km per hour when it goes against the river stream. What is the speed of boat per hour?
a) 10 km b) 8 km c) 12 km d) 7 km
Answer : c) 12 km
Solution :
Let the speed of the boat be B and the speed of stream be S.
Equation for travel along with Stream : B + S = 14.5 —> eq 1
During the travel against the stream, the speed of the stream temporarily doubled.
Therefore, Equation for travel against the stream = B – 2S = 7 —> eq 2 (NOTE: We are using 2S instead of S in the equation as speed of the stream has temporarily doubled when he travelled against the stream)
eq 1 – eq 2 => 3S = 7.5
or S = 2.5 kph
Substitute S = 2.5 in eq 1
B = 14.5 – 2.5 = 12
Question 2
A boat man was driving a boat during a cyclonic storm during which time the speed of the river was considerably high than normal times. He found that the boat travelled 222 km in three hours when he was driving along with the river. But when he drove the boat against the river the boat travelled 100 km in two hours. What is the speed of the river per hour during cyclonic storm?
a) 10 km b) 12 km c) 8 km d) 9 km
Answer : b) 12 km
Solution :
Let the speed of the boat be B and the speed of the river be S.
B + S = 222/3 —> eq 1
B – S = 100/2 —> eq 2
Adding equations 1 and 2 we get : 2B = 300 + 444 / 6
2B = 744/6
2B = 124
B = 62
Substituting B = 62 in eq 2 we get,
S = B – 100/2 = 62 – 50 = 12
Question 3
Maruti Cruisers sailed 132 km along with the river in six hours. Suddenly the boat had to return to the starting point and it started returning against the river and this time the Cruiser travelled at 128 km in eight hours. By how much percentage Cruiser’s speed exceeds the speed of the River ?
a) 533.33% b) 444.44% c) 267.67% d) none of these.
Answer : a) 533.33%
Solution :
Speed of boat along with the river per hour = 22 km (132 divided by 6)
Speed of boat against the river per hour = 16 km (128 divided by 8)
B + S = 22 —> eq 1
B – S = 16 —> eq 2
Adding equations 1 and 2, we get : B+SB+S = 22 16 = 6 =2S
S = 3 km/hour
Substituting S= 3, in eq 1 we get : B + 3 = 22
B = 22 – 3 = 19 km /hour
Cruiser’s speed exceeds the speed of the River By : Cruiser’s Speed – River Speed / River Speed x 100 = B – S / S X 100% = 19 – 3/ 3 X 100 = 16/3 X 100 = 533.33%
Question 1
Ramesh Khanna was five times old as his son Kishan Khanna eight years ago. Now he is three times as old as his son Kishan Khanna. Assume they both live in a country where people understand only binary numbers and they use symbol ‘0’ for binary 1 and ‘1’ for binary zero. Also they add 1 before any binary number. Express Ramesh Khanna’s present age in accordance to that world
a) 1001111 b) 1001111 c) 1100011 d) 1010000
Answer : a) 1001111
Solution :
Let Ramesh Khanna’s age be represented as ‘F’ eight years ago and the age of Kishan Khanna be represented as ‘S”.
F = 5 S
F + 8 = 3 (S + 8)
5S + 8 = 3S + 24
5S – 3 S = 24 – 8
2S = 16
S = 8
Now Ramesh Khanna’s age will be 5S + 8 = (5×8) + 8 = 48
Expressed as binary 110000 . However the people in their country use symbol 0 for binary one and 1 for binary zero. Replacing 1 with 0 and 0 with 1 we get 001111. Also those people usually add 1 before any binary number. Hence the answer becomes 1001111.
Question 2
Bhavna is presently three times as old as her daughter Anushka. Ten years ago Bhavna was five times as old as her daughter Anushka was. After how many years, sum of their ages will be 100.
a) 10 b) 20 c)15 d) none of these.
Answer : a) 10
Solution :
Let Bhavna’s present age be represented as B
Let her daughter Anushka’s age be represented as A.
B = 3 A
Ten years ago,
B – 10 = 5(A – 10)
3A – 10 = 5A – 50
10 + 50 = 5A – 3A = 2 A
40 = 2 A
A = 20.
B = 3 x A = 60.
Present sum of their ages = A + B = 80.
When 10 is added to both A and B, Sum of their ages will become 100. Hence, after 10 years, their sum of ages will be 100.
Question 3
Monisha was seven times as old as her daughter Mercy eight years ago. Now three times Mercy’s age is her mother age. Before how many years the ratio of their ages (ratio of mother’s age to Mercy’s age) will be increased by 1 from the current ratio
a) 5 b) 4 c) 8 d) none of these.
Answer : b) 4
Solution :
Eight years ago — let Monisha’s age be M and Mercy’s age be D.
M = 7 D
Now M + 8 = 3 (D + 8)
7 D + 8 = 3D + 24
7D – 3 D = 24 – 8 = 16
D = 4 = age of Mercy before 8 years
M = 28 = age of Mother before 8 years
Current age of Mercy = D + 8 = 12
Current age of Monisha = M + 8 = 36
Current ratio of mother’s age to Mercy’s age = 36/12 = 3.
Let before x years this ratio becomes 1 more than the current ratio. i.e x denotes the number of years before which the ratio becomes 3 + 1 = 4.
Then 36 – x / 12 – x = 4
36 – x = 48 – 4x
3x = 12
Or x = 4
i.e before 4 years, the ratio of their ages will be 4. Therefore answer = 4
Question 1
Ramakanth and his son went to a ball shop. There were 37 balls in the shop. Out of these one ball alone was weighing 6 gms whereas the other balls weighed 7 gms. How many minimum weighings are required to find out the 6 gm ball.( You are given a balance with two pans for weighing.)
a) 3 b) 4 c) 5 d) not possible to determine.
Answer : b) 4.
Solution :
Below let us discuss how we can handle the problem assuming three possible situations that can occur.
SITUATION ONE
1. Divide balls into three groups of 12, 12 and 13 balls each.Put one set of 12 balls into one pan and another set of 12 balls in another pan.Assume one pan shows lesser weight than the other.(a pan will go up and the other down).Discard the balls in the pan weighing more and the13 balls. .The ball weighing 6 gms has to be one among the 12 balls that were there in the pan that weighed less.
2.Now divide the 12 balls into 4,4 and 4. Put one set of 4 balls and another set of 4 balls in each pan. In case one of them weighs less –
3.Divide 4 into 2 and 2. Put them in two pans with 2 balls each from this lot. One pan will weight less.
4.Take the balls from the pan weighing less and then weigh one in each pan. The ball in the pan going up is the lesser weighing ball.(6 gm). So, in this case minimum 4 weighings are required.
SITUATION TWO
1. Divide balls into 12, 12 and 13. Put one set of 12 balls into one pan and another set of 12 balls in another pan.Consider both weigh same. NOW the ball weighing 6 gms has to be within 13 balls.
2. Divide 13 balls as 4,4 and 5. Put 4 balls each in the two pans.
3. Assume one pan goes down and the other goes up. Then the lighter ball has to be in the pan going up. Divide 4 into 2 and 2. Put them in two pans with 2 balls each from this lot. One pan will weight less.
4. Take the balls from the pan weighing less and then weigh one in each pan and the ball in the pan going up is the lesser weighing ball.(6 gm). So minimum 4 weighings are required.
SITUATION THREE
1 Divide balls into 12, 12 and 13. Put one set of 12 balls into one pan and another set of 12 balls in another pan. Consider both weigh same. NOW the ball weighing 6 gms has to be within 13 balls.
2. Divide 13 balls into 4,4 and 5. Put 4 balls each in the two pans. Assume both the pans weigh same. Then the lighter ball has to be one among the 5 balls.
3. Divide these 5 balls into 2,2 and 1. Weigh 2 and 2. If they weigh same then lighter ball is the fifth ball.
4. If one pan goes up and the other down take the two balls in the pan going up and then weigh as 1 and 1. Totally 4 weighings are required.
RESULT : In each of the above situations, the weighings required was 4. Hence 4 is the answer.
Question 2
England cricket captain and Indian cricket team captain went to a ball shop in Chennai. There they were told that there are 22 balls and out of them one weighs 15 gms whereas other balls weigh 16 gms. They were given a balance with two pans for weighing. How many minimum weighings are required by them to find out the ball weighing 15 gms.
a) 3 b) 4 c) 2 d) none of these.
Answer : a) 3
Solution :
Below let us discuss how we can handle the problem assuming three possible situations that can occur.
SITUATION ONE
1. Divide the balls into groups of 7, 7 and 8 balls each. Put the 7 balls in one pan and another set of 7 balls in another pan. Assume one pan goes up indicating that there is a ball which weighs 15 gms.(lesser weighing ball)
2. Discard the balls in the pan weighing more. Also discard the group of 8 balls set aside. Divide 7 balls in lesser weighing pan into 2,2 and 3. From these put 2 balls in one pan and another set of 2 balls in another pan.
3. Assume both weigh same. Then the lighter ball has to be one among the 3 balls set aside. So in third weighing put one ball in one pan and another ball from these in another pan. If one pan goes up then the ball in that pan is weighing 15 gms. If they weigh same then the third ball in this lot is the lighter ball.
SITUATION TWO
1. Divide the balls into 7, 7 and 8. Put the 7 balls in one pan and another set of 7 balls in another pan. Assume both them weigh same. Then the lighter ball has to be one among the 8 balls.
2. Divide 8 balls into 3,3 and 2. Weigh 3 balls in one pan and another set of 3 balls in another pan. Imagine they weigh same. Then the lighter ball has to be one among the two.
3. Do a third weighing putting one in each pan. The ball in the pan going up is the lighter ball.
SITUATION THREE
1. Divide the balls into 7, 7 and 8. Put the 7 balls in one pan and another set of 7 balls in another pan. Assume one pan goes up. Then the lighter ball has to be one among the balls in that pan.
2. Divide 7 balls into 2,2 and 3. Weigh 2 balls in one pan and 2 balls in another pan. Assume they weigh the same.
3. The lighter ball has to be one among the 3. Weigh 1 ball in one pan and another 1 ball in another pan. The ball in the pan going up is lighter one. If they weigh same then the third ball among these is the lighter ball.
RESULT : In each of the above situations, the weighings required was 3. Hence 3 is the answer.
Question 3
Ramesh Khanna wanted to buy 87 balls of the same weight – say 12 gms each. In a shop that he visited they were showing 88 balls. Out of them one weighed 14 gms and all others weighed 12 gms. How many minimum weighings are required to find out the heavier ball?
a) 4 b) 3 c) 5 d) none of these.
Answer : c) 5.
Solution :
Now in this problem we are going to use a shortcut which we haven’t used in the above two problems. We could had said this earlier, but knowing the formula after knowing the actual solution would make you more knowledgeable and equipped to solve these kinds of problems.
If the number of objects (balls in our example) is > 3^n1 and <= 3^n, then n weighings would be required to find the one odd object (ball) that will be weighing lesser or heavier than the rest.
Based on the above formula, for determining lighter or heavier one ball among 9 balls (3^2 balls) – 2 weighings are required. For determining lighter or heavier one ball from 10 to 27 balls (3^3 balls) – 3 weighings are required. From 28 to 81 – 4 weighings are required. From 82 to 243 balls 5 weighings are required. (follow the same principle of dividing into three parts.)
Question 1
A tourist from USA wanted to experiment different types of travel. He travelled 1/4 of the distance by car. He travelled by train 50% of the balance left. From out of the present balance 1/4^{th} was covered by him in his motor cycle. Thereafter he covered the balance of 108 km by boat. What is the total distance travelled by him?
a) 468 km b) 728 km c) 546 km d) 384 km
Answer : d) 384 km
Solution :
1/4^{ th} distance covered by car.
Out of the balance i.e. 3/4 ^{th} distance , 50% was covered by train.
i.e. 3/8^{th} of distance was covered by train.
Balance left was 3/8. Out of this 1/4^{th} of distance was covered by motor cycle.
Distance covered by motorcycle = 3/8 x 1/4^{ th} of distance = 3/32 ^{th} of distance.
Remaining Distance left out for boat = 3/8 – 3/32 = 9/32 = 9/32 ^{th} of distance
This portion 9/32 was covered by him in boat which is given as 108 km
Let total distance be D. Then 9/32 x D = 108. Or D = 384 km.
Question 2
A tourist company arranged 3/7^{th} of total travel by train. Of the balance half the distance was covered by Volvo luxury bus. The balance distance of 70 km was covered by call taxi. What is the total distance of travel arranged by the tourist company?
a) 145 km b) 245 km c) 345 km d) 445 km
Answer : b) 245 km
Solution :
3/7 ^{th} of distance covered by train , Balance – 4/7 ^{th} of distance
Out of 4/7 ^{th} balance , 1/2^{ th} of distance was covered by luxury bus.
i.e.4/14^{ th} of distance was covered by luxury bus. Balance left = 4/14 = 2/7 th of distance
This 2/7^{th} of distance is covered by call taxi i.e. 70 km
Let total distance be D. Then 2/7 x D = 245. Or D = 245 km
Question 3
An enterprising traveller covered 22% of the total distance by race car. Out of the balance he covered 50% by train. Now with the remaining balance he covered half of it by motor cycle and the balance of 156 km by cycle. What is the total distance travelled by him?
A) 600 km b) 900 km c) 700 km d) 800 km
Answer : d) 800 km
Solution :
By race car he travelled 22% of total distance
Balance distance to cover is 78% and out of this 1/2 that is 39% is covered by train.
In the remaining 50% i.e. 39% , he had covered half by motor cycle i.e. 19.5% of the distance.
Distance covered by cycle is 156 km which is the remaining 19.5% of the distance.
Hence 19.5% covers 156 km , then for 100% i.e. 15600 / 19.5 = 800
The total distance travelled will be 800 km
Question 1
Saikumar, Varun and Prasanth start from one of the points X and Y and move in their vehicles towards the other point. The distance between X and Y is 550 km. Saikumar and Varun start at 7 am from X and Y respectively whereas Prasanth starts from X at 8 am. Saikumar, Varun and Prasanth maintain constant speed of 60 km, 50 km and 70 km respectively. Which pair SaikumarVarun or VarunPrasanth or PrasanthSaikumar would meet first.
a) Saikumar – Varun b)VarunPrasanth c) PrasanthSaikumar d) cannot be determined
Answer : a) Saikumar – Varun
Solution :
Saikumar and Varun start at the same time, the initial distance between them is the distance XY = 550 km
They are moving towards each other i.e. in opposite directions and hence the relative speed = sum of the speeds = 60 + 50 = 110 km.
Hence the time taken for meeting = 550/110 = 5 hours.
i.e. Saikumar and Varun would meet 5 hours from 7 am or at 12 noon.
Saikumar and Prasanth starting from the same point begin their travel at different points of time. Since Prasanth is the late starter, the initial distance between them is the distance between them when Prasanth starts, namely at 8 am.
From 7 to 8 am Saikumar would have travelled 60 km and hence the distance between them at 8 am is 60 km.
Since they are moving in the same direction the relative speed is the difference between their
Speeds — 70 – 60 = 10 km/hr.
Therefore, time taken for meeting = 60/10 = 6 hours. This implies that Saikumar and Prasanth would meet 6 hours after 8 am. i.e. 2 pm
Varun and Prasnath start from opposite points at different points of time.
At 8 am. The distance between them is 54050 = 490 km . Since they are moving in opposite directions, the relative speed = 50 + 70 = 120 kmph
The time taken for meeting = 4.08 from 8 am i.e. 12.08 hrs.
Thus, Saikumar and Varun meet first at 12 noon, Varun and Prasanth meet at 12.08 pm and the last pair to meet Sai Kumar and Prasanth at 2 pm.
Question 2
Anna Salai, Madurai has 1100 buildings. A painter was engaged by the Madurai Corporation ( after calling for sealed tenders fixing the last date for receipt of tenders and the tender was opened in the presence of representatives of tenderers). How many zeros will he need to paint.?
a) 212 b) 192 c) 213 d) none of these.
Answer : a) 212.
Solution :
Divide 1100 building numbers into groups of 100 each as follow:
1) [(1..100)],
2) [(101..200), (201..300), …to (801…900) also (11011100)] …
3) [ (901..1000)]
For the first group, signmaker will need 11 zeroes.
For group numbers 201 to 900 and 1101 to 1100 , he will require 20 zeroes each.
And for group number 10 (from 901 to 1000), he will require 21 zeroes.
The total numbers of zeroes required are
= 11 + (9 x 20) + 21 = 212
Question 3
St. Xavier Matriculation School, Salem conducted sports meet and awarded prizes. In the sports contest there were m medals awarded on n successive days (n > 1).
1. On the first day 1 medal and 1/6 of the remaining m – 1 medals were awarded.
2. On the second day 2 medals and 1/6 of the now remaining medals was awarded; and so on.
3. On the nth and last day, the remaining n medals were awarded.
How many days did the contest last, and how many medals were awarded altogether?
a) 6 days , 36 medals b) 7 days, 49 medals c) 5 days, 25 medals d) none of these.
Answer : c) 5 days , 25 medals.
Solution :
For this question, we have to go by options. Only for option c) all the conditions in questions will be met as given below. (We recommend you to try with other options, to prove that they are wrong.)
On day 1: Medals awarded = (1 + 24/6) = 5 : Remaining 20 medals
On day 2: Medals awarded = (2 + 18/6) = 5 : Remaining 15 medals
On day 3: Medals awarded = (3 + 12/6) = 5 : Remaining 10 medals
On day 4: Medals awarded = (4 + 6/6) = 5 : Remaining 5 medals
On day 5: Medals awarded = 5
Question 1
A horse is tied to a pole by a rope of length 21m. If the length of the rope is increased to 28m, then how much excess area will it be able to graze?
a) 1386 m^{2} b) 1078 m^{2} c) 2464 m^{2} d) 1378 m^{2}
Answer : b) 1078 m^{2}
Solution :
Let r and R be the original and new proposed lengths of the rope respectively.
Given r = 21 m and R = 28 m.
Area of the surface with radius r = π x r^{2} = 22/7 * 21 * 21 m^{2} = 1386 m^{2}
Area of the surface with radius R = π * R^{2} = 22/7 * 28 * 28 m^{2} = 2464 m^{2}
Therefore, Additional area for grazing if length is increased = 2464 – 1386 m^{2} = 1078 m^{2}
Question 2
A cow was tied in the middle of the circular field by a rope of length 70 ft to graze the field. If the area of the circular field is 20000 sq.ft, then what is the area of the field that is not accessible to the cow?
a) 1540 sq.ft b) 6400 sq.ft c) 4600 sq.ft d) 1450 sq.ft
Answer : c) 4600sq.ft
Solution :
The length of the rope is 70 ft
The area grazed by the cow = π x 70 x 70 sq.ft = 15400sq.ft
Given that the area of the circular field = 20000ft
Then, the area of the field not accessible to the cow = 20000sq.ft – 15400sq.ft = 4600 sq.ft
Question 3
A man bought a field of length 250 m and breath 200 m. He planned to build a house of area 37500 m^{2}. And in the remaining area he planned to cultivate wheat crop. What is the length of the cultivating area if the breadth is 50m?
a) 150m b) 250m c) 350 m d) 200m
Answer : b) 250m
Solution :
Area of the field, he bought = 250 x 200 m^{2} = 50000 m^{2}
Given that the area of the house = 37500 m^{2}
Then the remaining area = 50000 m^{2} – 37500 m^{2} = 12500 m^{2}
Also given that the breadth of the remaining area = 50 m
Then the length of the cultivating area = 12500 / 50 = 250 m
Question 1
Priya wanted to mail 120 messages to her friend Banu. She mailed 1 message on the first day, 2 messages on the second day, 3 messages on the third day and so on. How many days she required to send all messages to Banu?
a)16 days b) 17 days c) 15 days d) 14 days.
Answer : c) 15 days.
Solution:
Total number of messages = 120
On the 1^{st} day, Priya mailed 1 message to Banu.
On the 2^{nd} day, she mailed 2 messages to Banu and so on…..
Let X denote the number of messages send on the X^{th} day.
Therefore, 1^{st} day messages + 2^{nd} day messages + ……. + X^{th} day messages = 120.
1 + 2 + 3 + …… + X = 120
X ( X + 1) / 2 = 120
X^{2} + X = 240
X^{2} + X – 240 = 0
By factoring the above eqn. we get
X ^{2} – 15 X + 16 X – 240 = 0 ( the middle term is obtained by the multiplicants of last term i.e 15 x 16 =240 and the subtracted value is 1 which is the middle term)
X(X – 15) + 16 ( X – 15) = 0 or (X + 16)(X15)
X = 15 0r X = 16
X = 15 ( since number of days cannot be negative).
So Priya required 15 days to send all messages.
Question 2
An IT company conducted interview for B.E Students. On the first day, they selected one student. On the second day, they selected 8 students and on the third day, they selected 27 students and so on. How many students will be selected if they conduct interview for 10 days?
a) 6050 b) 2530 c) 3025 d) 6025
Answer : c) 3025
Solution:
On the first day, they selected 1 student.
On the second day, they selected 8 students.
On the third day, they selected 27 students and so on.
Since they conduct interview for 10 days, the number of students selected on the 10^{th} day is 103 = 1000 students.
Total number of students selected = No. of students selected on 1^{st} day + No. of students selected on 2^{nd} day + ….. + No. of students selected on 10^{th} day.
Therefore, Total number of students selected = 1 + 8 + 27 + ….. + 1000.
= 1^{3} + 2^{3} + 3^{3} + …. + 10^{3}.
By using the formula,
1^{3} + 2^{3} + 3^{3} + …. + n^{3} = (n (n+1) / 2)^{2}, we get
Number of students selected = (10 (10+1) / 2)^{2}.
= (10 *11*10*11) / 4
= 3025.
Hence, the company has selected 3025 students in 10 days.
Question 3
Jeeva has a story book of 2047 pages. He read 1 page on the first day, 2 pages on the second day, 4 pages on the third day, 8 pages on the fourth day and so on. How many days it took for Jeeva to complete the book?
a)10 days b) 12 days c) 11days d) 13 days
Answer : c) 11 days
Solution:
Total number of Pages = 2047
On the 1^{st} day, Jeeva read 1 page.
On the 2^{nd} day, he read 2 pages
On the 3^{rd} day, he read 4 pages and so on…..
Let X denote the number of pages read on the X^{th} day.
Therefore, 1 + 2 + 4 + …. + X = 2047
Then by using the formula, the sum of X numbers in G.P = a ( r^{X} 1)/(r1) where a – First term & r – Common ratio, we get
1 (2^{x} – 1) / (2 – 1) = 2047
2^{X}1 = 2047
2^{X} = 2048
2^{X} = 2^{11}
X = 11.
So, Jeeva took 11 days to complete the book.
Question 1
If a Cone and a sphere have equal volumes and equal radius of 6 cm , then what is the height of the Cone and the diameter of the Sphere?
a) 12cm, 24cm b) 24cm, 6cm c) 24cm, 12cm d) 6cm, 24cm
Answer : c) 24cm, 12cm
Solution :
Let h be the height of the cone and r be the radius of the sphere as well as the radius of base of the cone.
Volume of the cone =1/3 x π x r^{2} x h cm^{3}
Volume of the Sphere = 4/3 x π x r^{3} cm^{3}
Given that r = 6 cm and Volume of the Cone= Volume of the Sphere
1/3 x π x r^{2} x h cm^{3} = 4/3 x π x r^{3} cm^{3}
h = 4 x r cm
h = 4 x 6 cm
h = 24 cm
Diameter of the sphere = 2r =12 cm
Question 2
A hemisphere bowl was filled with milk. The radius of the bowl was 20 cm. If the same quantity of milk was filled in another conical vessel of same radius, then what is the height of the new conical vessel?
a) 20cm b) 60cm c) 40cm d) 80cm
Answer : c) 40cm
Solution :
Given that the radius of the bowl = 20 cm
Since the quantity of the milk is same in both conical vessel and hemisphere bowl,
Volume of the hemisphere bowl = Volume of the conical vessel
2/3 x π x r^{3} = 1/3 x π x r^{2} x h
2 x r = h
h = 40 cm
Hence, the height of conical vessel is 40 cm.
Question 3
Nithesh built a wall around his house. After two years, he increased the length and the breadth of the wall by 20% and 15%. Then what is the consequent increase in area?
a) 23% b) 28% c) 38% d) 32%
Answer : c) 38%
Solution :
Let the length and breadth of the wall be L and B.
Increase in length = 20% = L x 20/100 =0.2 L
Then, New length = 1.2 L
Increase in breadth = 15% = B x 15/100 =0.15 B
Then, New breadth = 1.15 B
New area = 1.2L * 1.15B = 1.38 LB
Increase in area = 0.38 LB
Area increased in % = 0.38 LB / LB x 100 = 38 %
Question 1
In a conference hall, 56 members are seated by the following arrangements. After the first women, one man seated. After the second women, two men are seated. After the third women, three men are seated and so on. How many numbers of men are seated in the second half of the arrangements?
a) 27 b) 25 c) 24 d) 26
Answer : b) 25
Solution :
The seating arrangement will be
From the above diagram, we can see that the second half arrangement is :
M M M M M M M W M M M M M M M M W M M M M M M M M M W M
Here, there are 25 men in the above arrangement.
Question 2
In a bank, there are two counters namely counter1 and counter2. They will call the customers by a token number. If counter1 calls the token numbers from 40 in descending order and the counter2 calls only the token numbers which are prime numbers in ascending order. What token number will they call out at the same time if they were calling in the same speed?
a) 29 b) 31 c) 23 d) never call out the same number
Answer : d) never call out the same number
Solution :
Both counters will call in the following manner
So, both will never call out the same token number at a time.
Question 3
If the numbers from 1 to 100 which are exactly divisible by 4 and 6 are arranged in descending order, maximum number being on the top, which would come at the seventh place from the top?
a) 24 b) 12 c) 36 d) 18
Answer :a) 24
Solution :
From 1 to 100, the number which is divisible by 4 are 100, 96, 92, 88, 84, 80, 76, 72, 68, 64, 60, 56, 52, 48, 44, 40, 36, 32, 28, 24, 20, 16, 12, 8 and 4.
From 1 to 100, the numbers which is divisible by 6 are 96, 90, 84, 78, 72, 66, 60, 54, 48, 42, 36, 30, 24, 18, 12 and 6.
Therefore, the numbers which are divisible by 4 and 6 are 96, 84, 72, 60, 48, 36, 24 and 12.
Hence, the seventh place from the top is 24.
Question 1
Anantharama Iyer and Gopalakrishna Gandhi – two friends were meeting after a long time. Gandhi told he has two kids and their age is less than 10. The sum of the squares of their age is 130.Anantharama Iyer told that he has three kids and first two of them are of the age of Gandhi’s two children and the sum of the squares of their age is 139. What is the age of the third child of Anantharama Iyer?
a) 4 b)5 c)3 d)cannot be determined
Answer : c) 3.
Solution :
The age of the two kids of Anantharama Iyer is same as that of Gandhi’s children. The sum of the squares of age of Gandhi’s children is 130. The sum of squares of age of Anantharama Iyer’s children is 139. That means the square of the age of the third child is 9. Age is 3 years.
Question 2
Skilled Men, skilled women and unskilled men are employed to do a road construction work in the proportion of 1:2:3 and their wages in the proportion of 6:3:2. When 50 skilled men are employed total daily wages of all the hands amount to Rs. 45000. Find the total weekly wages to be paid to a skilled man, a skilled woman and an unskilled man.
a) Rs.3850 b) Rs.4850 c)Rs.5850 d)Rs.6850
Answer : a) Rs. 3850
Solution :
Number of men, women and unskilled men are 1: 2: 3.
But the skilled men are 50 in number. Therefore number of skilled women workers = 50 x 2 = 100 and the number of unskilled men = 50 x 3 = 150
Their wages are in the ratio 6:3:2.
Assuming a man gets paid Rs.6, woman will get Rs.3 and an unskilled man will get Rs. 2
Total wages paid to all the employees – 50 x 6 + 100 x 3 + 150 x 2 = Rs.900
If the total wages are Rs. 900, a skilled man gets Rs.6
If the total wages are Rs.45000, a skilled man gets (45000 /900 ) x 6 = Rs. 300
Weekly wage of a skilled man = daily wage x 7 = 300 x 7 = Rs.2100
The ratio of wages between men and women is 6:3. i.e Each woman will earn half as that of a skilled man.
Total weekly wages of a woman = 2100/2 = Rs.1050
The ratio of wages between skilled men and unskilled men is 6:2. i.e Each unskilled man will earn 1/3rd of that of a skilled man.
Total weekly wages of an unskilled man = 2100/3 = Rs.700
Total weekly wages of a man, a woman and an unskilled man = 2100 + 1050 + 700 = Rs.3850
Question 3
Fill the Sequence DI,MR,VA, ?
a) EJ b) SZ c) BK d) FR
Answer : a) EJ
Solution :
In the sequence M is the 9th letter from D and V is the 9th letter from M. So the next letter will be the 9th character from V i.e E.
In DI, I is the 5th letter from D. In MR, R is the 5th letter from M and in VA and so on.
The fifth character from E will be J. So the next sequence letters will be EJ
Question 1
Brindavan Express started from Chennai Central Station at 7.10 am. Karnatak express train for the same destination started from the same station on the same day at 8.40 am and ran with an average speed of 60 km/hour. Karnatak express crossed the Brindavan Express at 11.40 am. Find the average speed of Brindavan Express.
a) 75 km/hr b) 40 km/hr c) 30 km/hr d) 45 km/hr
Answer : b) 40 km/hr.
Solution :
Karnatak Express train crosses the Brindavan Express in three hours
Distance covered by Karnatak express when it crosses Brindavan Express = Speed of Karnatak Express x Time Until It Crosses Brindavan = 60 x 3 = 180 km
This means that, Brindavan express would had covered the same distance 180 Km from Chennai when it is crossed (Overtook) by Karnatak.
The distance covered by Brindavan Express from 7.10 am till 11.40 am = 180 km (in 4 1/2 hours)
So the average speed of Brindavan Express is = Distance covered by Brindavan till crossing / Time taken till crossing = 180 / 4 1/2
180 / 9 x 2 = 40 km/hour
Question 2
A Tata Indica car started from Mumbai towards Kolkata at 6 am and travelled at an average speed of 60 km/hr. It was found that an important material to be sent to Kolkata was omitted to be included in the packets sent in Tata Indica car. The omitted material was carried on an Innova car that started at 8.15 am and travelled with an average speed of 90 km/hr. At what time Innova car will cross Tata Indica car ?
a) 1.15 pm b) 3.15 PM c) 12.45 pm d) 4.15 pm
Answer : c) 12 . 45 p.m.
Solution :
Indica had started 2 hours 15 mins in advance to that of Innova Car
Before, Innova started, Tata Indica covers a distance of 135 km in 2 hours and 15 minutes. (you can check this by using the formula Speed x Time where speed of Indica = 60km)
Innova travels at 90 km/hr.
Relative speed of Innova with respect to Indica = Speed of Innova – Speed of Indica = 90 – 60 = 30 km/hr
Time that would be taken by Innova to catch up with Indica = Distance Difference at Start of Innova/ Relative Speed of Innova with respect to Indica = 135/30 = 4 1/2 hours for Innova to cross Tata Indica
Innova started at 8.15 am . It will cross Tata Indica at 12.45 pm.
Question 3
A Transport bus started Madurai Bus Terminal at 7 am and travelled at an average speed of 40 km per hour till 8.30 am. This bus stopped at the onway hotel for breakfast till 9 am and travelled at an average speed of 30 km per hour thereafter till its destination. Another express bus started from Madurai at 11 am and travelled at an average speed of X km per hour. The Express bus crossed the transport bus at 5 pm on the same day. Find the value of X?
a) 60 km/hr b) 40 km/hr c) 20 km/hr d) 50 km/hr
Answer : d) 50 km/hr
Solution :
Transport bus travelled at average speed of 40 km/hour
For 1 1/2 hours (i.e from 7 am till 8.20 am) Distance covered = 60 km.
Express bus starts at 11 am.
Till the express bus starts, transport bus would have covered further distance of
30 x 2 = 60 km.
Total distance covered upto 11 am by Transport bus is 60 + 60 = 120 km
Transport bus is travelling further at an average speed of 30 km per hour.
Express bus crossed Transport bus at 5 pm after travelling for 6 hours
Distance covered by the Transport bus till 5 pm = 120 + 180 = 300 km
This distance of 300 km is covered by express bus in 6 hours.
Hence average speed of Express bus = X = 300/6 = 50 km/hour
Question 1
Ashwin Kumar scored 32 marks in Geography, 29 marks in History, 40 marks in Science, 36 marks in mathematics , 33 marks in English and 35 marks in Hindi. Consider that pass mark is same across all subjects. Also assume that college hasn’t declared a pass mark before the exams are conducted. Now, the college administration sets a maximum pass mark that would take care that Ashwin Kumar JUST PASSES. Considering all the subjects, by what percentage Ashwin Kumar would have exceeded the pass marks ?
a) 15% b) 33.3% c) 17.8% d) 19.7%
Answer : c) 17.8%
Solution :
For Ashwin to just pass in all subjects, the minimum pass mark should be 29 (which is his lowest score.)
The average mark of Ashwin = His Total Marks / No of subjects
= 32+29+40+36+33+35 / 6 = 205/6
Average mark of Ashwin exceeds the pass percentage by = (Ashwin’s Average – Minimum Pass Mark ) / Minimum Pass Mark x 100 %
= (205/6 – 29)/29 x 100
= 31/174 * 100
= 17.8% (approximately)
Question 2
Average age of total 80 students from three classes V, VI and VII is 13 years. Average age of 30 students of class V is 11 1/2 years. If average age of another 30 students of class VI is 12 1/2 years, what is the average of the rest of the students who are in class VII?
a) 13 b) 14 c) 16 d) 15
Answer : c) 16
Solution :
Sum of ages of all the 80 students = 80 x 13 = 1040
Sum of ages of 30 students of class V = 30 x 11.5 = 345
Sum of ages of 30 students of class VI = 30 x 12.5 = 375
Total age of rest of the students (20) of class VII = 1040 – (345 + 375) = 320
Average age of these 20 students = 320/20 = 16
Question 3
Assume a project team of 4 students. The project guide (their professor) had estimated that every one of the students would take approximately 200 working hours to complete the project. But, the first two students lagged by 20 hours each while the last two students completed their tasks 10 hours each in advance. By what percentage is the actual working hours exceeds the actual estimate ?
a) 2.5% b) 5% c) 4% d) 8.5%
Answer : a) 2.5%
Solution :
Estimated Total Hours to complete the project = Working Hours Estimate Per Student x No of Students = 200 x 4 = 800.
Case I : Consider first two students.
They lagged by 20 hours each. This means each took 220 hours to complete their tasks.
Actual hours of work by first two students = actual working hours per student x no of students = 220 x 2 = 440.
Case II : Consider last two students.
They completed in advance by 10 hours each. This means each took 190 hours to complete their tasks.
Actual hours of work by last two students = actual working hours per student x no of students = 190 x 2 = 380.
Conclusion :
Actual Total Working hours of all 4 students = Working hours of first 2 students + Working hours of last 2 students = 440 + 380 = 820 hours.
Percentage increase in actual working hours compared to estimate = (Actual total working hours – Estimated total working hours) / Estimated total working hours x 100%
= ((820 – 800) / 800) x 100
= 2000/800 %
= 2.5%
Question 1
Karthikeyan covers a certain distance by car.Karthikeyan divided the complete distance into four equal parts and after finishing each part he increased his speed to twice the previous one. If, in the first part, his speed was 30 kmph then what is the average speed of the complete journey.
a) 48 km/hour b)56 km/hour c)64 km/hour d) 72 km/hour
Answer : c) 64 km/hour
Solution :
Let Karthikeyan cover a distance of 4X divided into four equal parts of X each
Let the initial speed be V
Then total time taken for the journey = X/V + X/2V + X /4V + X /8V
= 15X / 8V
Therefore Average speed = 4X / (15X/8V) = 32V/15
But V is given to be, V = 30 km per hour
Therefore Average speed = 32 x 30/15 = 64 km per hour
Question 2
Two Express trains started at the same time from two stations – Chennai Central and Secunderabad and proceed towards each other at the speed of 75 kmph and 45 kmph respectively. When they met, it was found that one express train had travelled 240 km more than the other. What is the distance between Secunderabad and Chennai Central, as per the problem?
a) 1200 km b) 11140 km c) 1050 km d) 960 km
Answer : d) 960 km
Solution :
The relative speed of the faster train = Speed of the faster train – Speed of the slower train = 75 – 45 = 30 kmph
In the question it is given that the faster train had covered 240 Km more when compared to the slower train when they meet each other.
Time taken for the trains to meet = Additional distance covered by faster train from chennai station compared to the slower train from Secunderabad (during the meeting) / Relative speed of faster train with respect to slower train
= (x + 240) – x / 30
= 240 / 30 = 8 hours.
Hence, the trains will cross each other in 8 hours.
So the distance between the two stations = Distance covered by the faster train in 8 hours + Distance covered by the slower train in 8 hours
= Speed of the faster train x 8 + Speed of the slower train x 8
= 75 x 8 + 45 x 8 = (75 + 45) x 8 = 960 km
Question 3
Two trains of length 200 m and 160 m respectively run on parallel lines of rails. When running in the same direction the faster train passes the slower train in 36 seconds. When they run in opposite directions with the same speeds as earlier, they pass each other in 18 seconds. Find the speed of the faster train?
a) 15 m/sec b)10 m/sec c)5 m/sec d) 20 m/sec
Answer : a) 15 m/sec
Solution :
Length of the two trains – 200 m and 160 m.
Let the speed of the faster train be A and that of the slower train be B.
The distance to be crossed by the faster train = sum of the lengths of two trains = 360 m
Case I : Trains travelling in same direction :
Time taken while running in same direction – 36 sec.
Relative speed of faster train with respect to the slower train when travelling in same direction = A – B = 360 / 36 = 10 m/sec >equation 1
Case II : Trains travelling in opposite direction :
When they travel in opposite direction, again the distance to be crossed is 360 m
But time taken is 18 seconds.
Relative speed of faster train with respect to the slower train when travelling in opposite direction = A + B = 360/18 = 20 m/sec >equation 2
Adding equations 1 and 2,
2A = (20 + 10 ) = 30 m / sec
Or, A = 30/2 = 15 m/sec.
(slower train will run at 5 m/sec)
Question 1
25^{2.7} x 5^{4.2}÷ 5^{6.4} = 25^{(?)}
a) 1.7 b) 3.2 c) 1.6 d) 3.6
Answer : c)1.6
Solution :
(5^{2})^{2.7} x 5^{4.2}÷ 5 ^{6.4} =(5^{2}) ^{?}
(5^{5.4} x 5^{4.2})/ 5^{6.4} = (5^{2})^{?}
5^{3.2} = (5^{2})^{?}
2? = 3.2
? = 3.2 / 2 = 1.6
Question 2
When a number is divided by 138 the remainder is 26. What will be the remainder if the same number is divided by 23?
a) 1 b) 3 c) 2 d) 4
Answer : b)3
Solution :
138 is a multiple of 23. (23 x 6 = 138).
So when the number is divided by 23 the remainder will be 2623 = 3.
Question 3
Mr. Mukhambo has two numbers and say that their difference, their sum and their product are to one another as 1:7:24. Mukhambo wants to find out whether you can get him the product of those two numbers:
a) 24 b) 6 c) 48 d) 12
Answer : c)48
Question 1
Anand is the brother of Bachan. Coolie is the son of Bachan. Divya, the daughter of anand, is married to Elgin. Fox and Bachan are sisters.
Now consider the below proposed conclusions:
Conclusion 1) Divya is the daughter of Fox.
Conclusion 2) Coolie is brother of Divya.
Conclusion 3) Elgin is soninlaw of Anand.
Conclusion 4) Fox is sister of Coolie.
Which of the following statements is true.
a) Conclusions 1) and 2) are true
b) Conclusions 1) and 3) are true
c) None of the above.
d) Conclusion 3) alone is true
Answer : d) Conclusion 3) alone is true:
Reason :
Based on the information given in the question, one can draw a diagrammatic representation as given below.
(In the above diagram, a forward arrow from a person A to person B indicates that A is related to B as given in the label near that particular arrow.)
Comparing this with the conclusions 1,2,3 and 4, One can easily say that only conclusion 3 is true. Therefore option d is true.
Anand’s daughter Divya and her husband is Elgin. Hence Elgin is soninlaw of Anand.
Question 2
P,Q,R,S,T, U, V and W are eight classmates studying in Padma Seshadri School, KK Nagar, Chennai. They are standing along the four sides of a square, two on each side. P is oppoisite V, R is opposite T, S is on the same side as V, W is opposite Q and T is on the side as Q.
If the above information is true, which of the following is true?
a) T is on the side of P.
b) U is on the side of P.
c) R is on the side of Q.
d) P is on the side of S.
Answer : b) U is on the side of P.
Reason :
The information provided may be drawn in
Clockwise – TQ,VS,RW,PU
Comparing the above diagram with the options given, one can easily say that option b is the correct one.
Hence choice b) is correct.
Question 3
Six paintings – F, G, H, J, K and L – are to be sold at a three day auction.
The paintings are to be divided into three groups – group 1, group 2 and group 3 – and each group is to be sold on one of the days of the auction. The paintings to be included in each group are to be selected according to the following conditions.
1.Group 2 must contain at least as many paintings as Group 1 and Group 3 must contain at least as many paintings as Group 2.
2.H and K, paintings by the same artist, must be in the same group as each other.
3.F and L paintings of similar subjects, must be in different groups from each other.
4. G and H estimated to be the two most valuable paintings, must be in different groups from each other.
5. If J is in group 3, K must also be in group 3 because of a request from the auctioneer for the third day.
If H is in group 1, which of the following must be true?
a) F is in group 2.
b) G is in group 2.
c) J is in group 2
d) L is in group 2
Answer : c) J is in group 2.
Reason :
Rule 1 states that Group 2 must contain at least as many paintings as Group 1 and Group 3 must contain at least as many paintings as Group 2 i.e. G1 ≤ G2 ≤ G3 where G1, G2, and G3 are the number of paintings of Group 1, Group 2 and Group 3 respectively. As the total number of paintings is 6, they can be divided into three groups in three possible ways
In the question it is given that H is in group 1.
From rule 2, K should be in the same group as H.
Hence H and K are in G1. (There can be no more paintings in group 1 as the diagram above clearly indicates that there is no way for group 1 or group 2 to have more than 2 paintings.)
Rule 5 states that if J has to be in group 3, K also has to be in group 3. But as per last paragraph, we know that K is in group 1 and not in group 3. Hence J cannot be in group 3. Therefore it needs to be in group 2.
Since J has to be in group 2, option C is certainly true without doubt.
Question 1
A box contains 4 blue and 5 white balls and another box contains 5 blue and 4 white balls. One ball is to be drawn from either of the two boxes. What is the probability of drawing a blue ball?
a) 2/9 b) 1/9 c) 4/9 d) 1/2
Answer : d) 1/2
Solution :
Probability of choosing first box = 1/2 and
Probability of choosing second box = 1/2
Imagine there is no second box at all :
P1 = Probability of choosing one blue ball from first box when there is no second box at all
= No of ways of choosing 1 blue ball from among 4 blue balls / Total number of ways of choosing 1 ball from all 9 balls.
= 4C_{1} / 9C_{1}
Imagine there is no first box at all :
P2 = Probability of choosing one blue ball from second box when there is no first box at all
= No of ways of choosing 1 blue ball from among 5 blue balls / Total number of ways of choosing 1 ball from all 9 balls.
= 5C_{1} / 9C_{1}
Now, consider our real scenario of having both the boxes :
P3 = Probability of choosing one blue ball from the first box = P1 x Probability of choosing first box = 4C_{1} / 9C_{1} x 1 / 2 = 2/9
P4 = Similarly probability of choosing one blue ball from the second box = P2 x Probability of choosing second box = 5C_{1} / 9C_{1} x 1 / 2 = 5/18
P (E) = Probability of choosing a blue ball from either of the boxes =
P3 + P4 = 2/9 + 5/18 = 1/2 –> eq 1
Note : Since the question reads we have to choose one blue ball from first OR second box we have used ‘+’ sign in eq 1. If the question were choosing one blue ball from first AND second box, we should had used multiplication i.e ‘ x ‘.
Question 2
An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four marbles are picked at random, what is the probability that at least one is blue?
a) 4/15 b) 69/91 c) 11/15 d) 22/91
Answer : b) 69/91 .
Solution :
Total possible outcomes = n(S)
= Selection of 4 marbles out of 15 marbles
= 15C_{4 }= (15 x 14 x 13 x 12) divided by 1 x 2 x 3 x4 = 1365
When no marble is blue, favourable number of cases n(E)
= Selection of 4 marbles out of 11 marbles
= 11C_{4} = 11 x 10 x 9 x 8 divided by 1 x 2 x 3 x 4 = 330
Probability that atleast one ball is blue + Probability that no ball is blue = 1
Therefore, Probability that atleast one ball is blue = 1 – Probability that no ball is blue
Therefore, Probability that atleast one ball is blue = 1 – n(E) / n(S)
= 1 – 330/1365 = 1 – 22/91 = 69 /91
Question 3
A committee of 4 members is to be selected from a group of 4 women and 3 men. What is the probability that the committee has at least one man.?
a) 1/35 b) 3/7 c) 34/35 d) 4/7
Answer : c) 34/35
Solution :
Probability of there being all women in the committee = 4C_{4} divided by 7C_{4} = 1/35
But, Probability of there being all women in the committee + Probability of there being at least one man = Probability of formation of a committee with no restriction of gender = 1
Probability of there being at least one man = 1 – 1/35 = 34/35
Question 1
Ansar can do a work of making a sofa set in 40 hours and with the help of Nayeem, he can complete the same work in 24 hours. If they get Rs.1200 for the work, then what could be Nayeem’s share?
a) Rs. 360 b)Rs.420 c)Rs.480 d) Rs.540
Answer : c) Rs.480
Solution :
Ansar can do in one hour 1/40 of the sofa set work
Both Ansar and Nayeem perform in one hour 1/24 of the sofa set work.
Nayeem’s one hour work = One hour work of Both Ansar and Nayeem – One hour work of Ansar = 1/24 – 1/40 = 1/60
So, Nayeem will finish the work in 60 hours.
Ansar’s revenue share / Nayeem’s revenue share = Time taken for Ansar to complete the work / Time taken for Nayeem to complete the work = 60 / 40 = 3 / 2
Ansar’s revenue share : Nayeem’s revenue share = 3 : 2
Let Ansar’s share be 3X and let Nayeem’s share be 2X. Therefore 3X + 2X = 1200 (Total revenue of both)
Or X = 1200/5 = 240.
Therefore Nayeem’s share = 2X = 2 x (240) = Rs. 480
Question 2
8 men and 12 women together can complete construction of a house in 8 days. 6 men and 14 women can complete the construction of the same house in 10 days. Suppose 20 women alone work for construction of this house, how many days will they take to complete the same?
a) 36 days b)40 days c)32 days d)24 days
Answer : b) 40 days
Solution :
Work done by 8 men and 12 women for 8 days is equal to
Work done by 6 men and 14 women in 10 days.
Let m be the amount of work done by one man and let w be the amount of work done by one woman. Then
(8m + 12 w) 8 = (6m + 14 w) 10
64 m + 96 w= 60 m + 140 w
64 m – 60 m = 140 w – 96 w
4 m = 44 w
1 m = 11 w
Construction of the house requires
(8 x 11w + 12 w)8 = 800 women days.
So if the construction is to be done by 20 women only
It will take – 800/20 = 40 days.
Question 3
Bharath and Rajani together can complete a piece of work in 12 days and Rajani and Kamal together in 15 days. If Bharath is twice as good a workman as Kamal, then in how many days will Rajani along complete the same work?
a) 30 b)24 c)25 d) 20
Answer : d) 20
Solution :
Assume Bharath completes in B days
Rajani completes in R days
Kamal completes in K days
Then 1/B + 1/R = 1/12 –>eq 1 (Bharath and Rajani take 12 days to complete the work)
1/R + 1/K = 1/15 –> eq 2
eq 1 – eq 2 gives, 1/B – 1/K = 1/12 – 1/15 = 1/60 –> eq 3
From the question it is given that Bharath is twice as efficient as Kamal. That is, if Bharath takes B days Kamal will take 2B days to complete the work. Therefore eq 3 becomes,
1/B – 1/2B = 1/60
1/2B = 1/60
B = 30 days
1/R = 1/12 – 1/30
= 3/60 = 1/20
So Rajani will take 20 days to complete the work.
1. Choose the odd one from among the given options.
A) mercury B) Cast Iron c) Antimony d) Chromium
Answer : b) Cast Iron
Reason : All other choices are pure metals while cast iron is a compound.
2. Choose the odd one from among the given options.
A) Furnace oil b) petrol c) cow dung d) mustard oil
Answer : d) mustard oil
Reason : All others are fuels. Mustard oil is used for cooking.
3. Fill the blank with appropriate word.
The salaries and perks of the employees were not in ________ with their status in this industry.
(a) Value (b) Conformity (c) Accordance (d) Possession
Answer : c) Accordance
Reason : By using the word Accordance in the sentence there is a sense of Comparison is made here. Other choices do not fit in properly.
4. Fill the blank with appropriate word.
The recent spate of riots has proved that biases ______ in childhood are strengthened in ______
a. gathered, school b. learned, riots
c. inculcated, adulthood d. imbibed, force
Answer : c) inculcated, adulthood
Reason : When option c is used the sentence becomes The recent spate of riots has proved that biases inculcated (means implanted) in childhood are strengthened in adulthood.
In choice a) while gathered may fit in –the word school will not be suitable for the second blank. Same way in choice b) also the word riots is not relevant. Also in choice d) the word imbibed will be suitable for the first blank but the word force will not fit in the context of the sentence.Correct choice is c) as for both the blanks the words fit in aptly.
5. Fill the blank with appropriate word.
The recent resolution in the Assembly demands ______ and trial under POTA.
a. incident b. result c. extradition d. focus
Answer : c) extradition
Reason : extraditio can mean surrender of an accused person (through court of law) by one state to another.
Answer choices a), b) and d) are less relevant to the context of the sentence.
Question 1
Nine letters from English alphabet from B to J are arranged in three rows with three letters in each row.H is above I, which is right of D. E is above F, which is left of G. C is below B, which right of I.
Which of the following is true?
A)Letter G is at the extreme right of the first row and letter I is the middle of second row.
B)Letter J is at the extreme right of the first row and letter I is in the middle of second row.
C)Letter J is at the extreme right of the first row and letter B is in the middle of second row.
Answer : b) Letter J is at the extreme right of the first row and letter I is in the middle of second row
Reason :
Based on the arguments in the question, the letters are arranged as
E H J
D I B
F G C
Now by comparing this arrangement with the options we can conclude that answer choice b) is correct.
Question 2
There was a musical concert in Mumbai. A famous musician had sung six classical ragas viz. Kedar, Shivranjani, Todi, Bhairavi and Malkauns. The musician had sung Todi immediately before Malkauns. There were as many ragas between Todi and Kedar as after Kedar and before Shivranjani. Shivranjani had been sung before Durbari.
If the information above is true, which of the following is true?
a) Bhairavi was sung second from the beginning.
b) Todi rag was sung as the last raga.
c) Bhairavi was sung fourth from the beginning
d) Kedar was sung as the last raga.
Answer : c). Bhairavi was sung fourth from the beginning
Reason :
Based on the information provided — Todi – Malkauns – Kedar – Bhairavi – Shivranjani and Darbari is the order in which ragas were sung by the musician.
Question 3
Mallikarjun and Nagaraja are good at hockey and volleyball. Orange and Mallikarjun are good at hockey and baseball. Purushotham and Nagaraja are good at cricket and volleyball. Orange, Purushotham and Quicky are good at football and baseball.
If the above information is true, which of the following is true.
a) Quicky plays more number of games.
b) Mallikarjun plays less number of games
c) Purushotham plays maximum number of games
d) Nagaraja plays lesser number of games than others.
Answer : c) Purushotham plays maximum number of games.
Reason :
Based on information in the question, we can tabulate the games played by the players as below. (Y indicates that a player plays the corresponding game.)
Name Hockey Volleyball Baseball Cricket football
Mallikarjun Y Y Y
Nagaraja Y Y Y
Orange Y Y Y
Purushotham Y Y Y Y
Quicky Y Y
Observing the table above, option C is correct.
1. a) Anger b)Sorrow c)Feeling d) Joy
Answer : c)Feeling
Reason :
Choice a), b) and d) are different forms of feelings. Hence feelings is a superset and rightly the odd option.
2. a) Pointed – Blunt b)Sweet – Sour c)Hard – soft d)Long – High
Answer : d)Long – High
Reason :
All other choices refer synonym – antonym. Choice d) differs.
3. A) Sly B) Virtuous C) Pensive D) Cowardly
Answer : c) Pensive.
Reason :
Pensive refers to mood. All others are traits of an individual.
4. a) Instruct b) Explain C)Teach D)Train
Answer : b)Explain
Reason
All other choices indicate the presence of trainees/students.
—
Question 1
A contractor was engaged to construct a residential complex along with a water tank. The water tank can filled up by Pipe One in 6 hours and by Pipe Two in 8 hours. If the two pipes are opened one after the other each for one hour, how long will it take for the two pipes, pipe One and pipe Two to fill the water tank?
a) 8 hours 20 min b)7 hours 40 min c) 6 hours 45 min d) none of these.
Answer : c) 6 hours 45 min.
Solution :
Pipe one will fill 1/6 of the tank in one hour. Pipe two will fill 1/8 of the tank in one hour.
These pipes are opened one after the other, each run for one hour.
In one turn – that is one hour of pipe one and one hour of pipe two, 1/6 + 1/8 = 7/24 of the tank will be filled in.
So in 6 hours 21/24 of the tank will be filled in and balance will be 3/24 = 1/8 of the tank.
Now in the seventh hour Pipe one will be run. Pipe one can fill 1/6 of the tank in 1 hour or 60 minutes. We can calculate the time taken by Pipe 1 to fill 1/8th of the tank as shown below :
Tank Filled Time Taken
1/6  60 min
1/8  t min
t = 1/8 x 6/1 x 60 = 45 minutes.
So the tank will be filled in 6 hours 45 minutes.
Question 2
SivaChrist constructions constructed a community water tank for Pozhichalur village. The water tank can be filled by Karuna Pipe in 18 hours and Pandian pipe in 24 hours. Karuna Pipe will run for one hour and closed. Pandian pipe will run for the next one hour and closed. This process continues. Which pipe will be running when the tank is filled in fully? How many hours will it take for the tank to be filled in the above process?
a) Pandian, 30 hours 30 min b) Karuna ,10 hours 30 min c) Pandian,5 hours 30 min d) Karuna, 20 hours 30 min
Answer : d) Karuna, 20 hours 30 min
Solution :
Karuna Pipe can fill the tank in 18 hours.
In one hour Karuna Pipe will fill 1/18 of the tank
Pandian Pipe will fill 1/24 of the tank in one hour.
If both the pipes run one after the other once –
1/18 + 1/24 = 7/72 of the tank will be filled.
So, in 20 hours – that after 10 such running – 70/72 of the tank will be filled in and 1/36 of the tank need to be filled in.
Karuna Pipe can fill 1/18th of the tank in 60 minutes. Then the time taken to fill 1/36th of the tank can be calculated as below :
Tank Filled Time Taken
1/18  60 min
1/36  t min
t = 1/36 x 60 x 18 = 30 minutes
So Karuna pipe will be running when the tank is filled in and it will take 20 hours 30 min. for the tank to be filled in completely.
Question 3
Two pipe manufacturers were asked to show demonstration about the performance of their pipes. Pipe X and Pipe Y can fill a tank in four hours. Had they been opened separately Pipe Y will take 6 hours more than Pipe X. How much time will Pipe X take to fill the tank separately?
a) 2 hours b) 1 hour c) 6 hours d) 8 hours
Answer : c) 6 hours.
Solution :
Let the tank be filled in completely by Pipe X in x hours. Therefore in 1 hour, Pipe X will fill 1/x of the tank
Then Pipe Y will take x + 6 hours. Therefore in 1 hour, Pipe Y will fill 1/(x + 6) of the tank.
When Pipe X and Pipe Y are opened together they will take 4 hours to fill the entire tank. In other words, if Pipe X and Pipe Y are opened together for 1 hour, they would had filled 1/4th of the tank. Putting this argument in equation form we get :
1/x + 1/ x+6 = 1/4
x + 6 + x divided by x(x+6) = 1/4
8x + 24 = x^{2} + 6x
x^{2} – 2x – 24 = 0
(x6) (x+4) = 0
So x = 6 (neglecting the negative value of x)
Question 1
Two goats were standing 700 metres apart. One goat is standing at the western gate and another is standing at eastern gate. They move forward for two minutes at the rate of 50 metres per minute initially. Then they move backward for two minutes at the rate of 25 metres per minute. The process continues. When will the two goats meet/hit the other?
a) 20 min b) 22 min c) 26 min d) 28 sec
Answer : b) 22 min.
Solution :
For every 2 min. the two goats will come nearer by 200 metres. In the next 2 min. the two goats will go backwards 50 metres back and hence after every 4 minutes goats would have come nearer by 100 metres. The process continues. Hence after 20 min the two goats would have come nearer by 500 metres. At the end of 22 mins the two goats will meet each other.
Question 2
A new type of study about movement of trains was conducted by South Central Railway. Two trains are standing 2520 metres apart in two adjacent railway lines. Both the trains move for five seconds at 60 m/sec. They halt for 12 seconds. Then again they start moving for five second at 60 m/sec. The process continues. After how many seconds will the engines of the two trains cross in the adjacent railway lines?
a) 50 sec b) 60 sec c) 70 sec d) none of these.
Answer : c) 70 sec.
Solution :
The two trains are standing 2520 metres apart in two adjacent railway lines. After every five second they come closer by 600 metres.(300 + 300). They halt for 12 seconds and then start moving. So after every 17 seconds they come closer by 600 metres. The process continues. After 68 second they would have covered 2400 metres. So in the next two seconds that is after 70 second the engines of the two trains will cross each other in adjacent line having covered 2520 metres.
Question 3
Kumaresan Environment Industries Limited , planted a tree. The tree was growing at 6 cm every day. The tree was allowed to grow for ten days. On the 11th day the tree will be cut to 50% of its size. From the next day (12th day) for the next ten days it was allowed to grow at the rate of 6 cm every day. On the 11th day the tree is cut to 50% of its size. The process continued. Assuming the tree was planted afresh as a seed after how many days the tree would have reach a height of 105 cms?
a) 32 days b) 26 days c) 18 days d) none of these.
Answer : a) 32 days
Solution :
For ten days the tree grows at 6 cm per day and it will reach a height of 60 cm.
On the 11th day its height will be reduced to 50% of its size – 30 cm.
Till 21st day the tree will grow further 60 cm and reach a height of 90 cms.
On 22nd day its height would have been reduced to 45 cm.(50%)
It grows at 6 cm per day and on 32nd day it would have grown further 60 cm and reached height of 105 cms.
1. 
A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train? 

Answer: Option D Explanation:

2. 
A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is: 

Answer: Option B Explanation:
= 45 km/hr. Let the speed of the train be x km/hr. Then, relative speed = (x – 5) km/hr. x – 5 = 45 x = 50 km/hr. 
3. 
The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is: 

Answer: Option C Explanation:
Time = 30 sec. Let the length of bridge be x metres.
2(130 + x) = 750 x = 245 m. 
4. 
Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is: 

Answer: Option B Explanation: Let the speeds of the two trains be x m/sec and y m/sec respectively. Then, length of the first train = 27x metres, and length of the second train = 17y metres.
27x + 17y = 23x + 23y 4x = 6y

5. 
A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform? 

Answer: Option B Explanation:
Length of the train = (15 x 20)m = 300 m. Let the length of the platform be x metres.
x + 300 = 540 x = 240 m. 1. A shopkeeper arranges his fruits in such a way that putting say either 3,or 9,or 7 apples in a row each time one apple is left. but when he puts say 11 apples in a row no apple is left.find no. of apples. 2. A problem of staircase:if i start going down, having gone down 4 steps i see x coming up.i meet x in way going down.when i had still 7 steps to go down x had gone up .find no. of steps in the staircase.condition: for each of my one step down x goes up 2 steps.ans:22 3. Problem of merrygoround.tom on a merrygoround finds that one third of people ahead of him and three fourth of people behind him is equal to the no. of people on the merrygoround.find the no. of people on the merrygoround. 4. A problem of cards .four persons sitting on a table.infront of each one of them one card is lying not having same colours on both sides.2 blue,2 red,2 green faces. each one guesses the color of unseen face of card .(guesses i don?t remember) exactly 2 are lying.the front faces of card are red,blue,green,red respectively.find the colors of unseen faces of card. 5. five persons in a conference.a & b can communicate in english.when d joins the only coomon language of communication is spanish. a ,b,e can talk to each other in french only.exactly 3 people know portugese. the number of languages known by different persons are 1,2,3,4,5. some other similar conditions . you have to find answers to four questions based n these facts. 6. say ona sacle of 100:85 have phones,80 cars,75 married,70 have houses. how many minimum persons are married,have phone car and houses on a scale of 100. 7. some series given3 10 20 27 54 61 _ans .162some other seriestrick: find suare root then add some number etc. 8. some conditions on xz etc. 9. afternoon temperatures of five days are recorded.each temperature is different.multiple is 12. 
Sheila Vanconey said,
May 22, 2012 @ 12:39 pm
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