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1. If 3y + x > 2 and x + 2y≤3, What can be said about the value of y?
A. y = -1
B. y >-1
C. y <-1
D. y = 1
Answer: C
If a certain number is increased by x% then decreased by x% or vice versa, the net change is always decrease. This change is given by a simple formula −(x10)2= −(1010)2= −1%. Negitive sign indicates decrease.
3. If m is an odd integer and n an even integer, which of the following is definitely odd?
A. (2m+n)(m-n)
B. (m+n2)+(m−n2)
C. m2+mn+n2
D. m +n
Answer: C and D (Original Answer given as D)
You just remember the following odd ± odd = even; even ± even = even; even ± odd = odd
Also odd x odd = odd; even x even = even; even x odd = even.
4. What is the sum of all even integers between 99 and 301?
A. 40000
B. 20000
C. 40400
D. 20200
Answer: D
The first even number after 99 is 100 and last even number below 301 is 300. We have to find the sum of even numbers from 100 to 300. i.e., 100 + 102 + 104 + ............... 300.
Take 2 Common. 2 x ( 50 + 51 + ...........150)
There are total 101 terms in this series. So formula for the sum of n terms when first term and last term is known is n2(a+l)
So 50 + 51 + ...........150 = 1012(50+150)
So 2 x 1012(50+150) = 20200
5. There are 20 balls which are red, blue or green. If 7 balls are green and the sum of red balls and green balls is less than 13, at most how many red balls are there?
A. 4
B. 5
C. 6
D. 7
Answer: B
Given R + B + G = 17; G = 7; and R + G < 13. Substituting G = 7 in the last equation, We get R < 6. So maximum value of R = 6
6. If n is the sum of two consecutive odd integers and less than 100, what is greatest possibility of n?
A. 98
B. 94
C. 96
D. 99
Answer : C
We take two odd numbers as (2n + 1) and (2n - 1).
Their sum should be less than 100. So (2n + 1) + (2n - 1) < 100 ⇒ 4n < 100.
The largest 4 multiple which is less than 100 is 96
7. x2 < 1/100, and x < 0 what is the highest range in which x can lie?
A. -1/10 < x < 0
B. -1 < x < 0
C. -1/10 < x < 1/10
D. -1/10 < x
Answer: 5
Total ways of selecting two boxes out of 4 is 4C2 = 6. Now, the number of ways of selecting two boxes where none of the green or red box included is only 1 way. (we select yellow and blue in only one way). If we substract this number from total ways we get 5 ways.
9. All faces of a cube with an eight - meter edge are painted red. If the cube is cut into smaller cubes with a two - meter edge, how many of the two meter cubes have paint on exactly one face?
A. 24
B. 36
C. 60
D. 48
Answer : A
If there are n cubes lie on an edge, then total number of cubes with one side painting is given by 6×(n−2)2. Here side of the bigger cube is 8, and small cube is 2. So there are 4 cubes lie on an edge. Hence answer = 24
10. Two cyclists begin training on an oval racecourse at the same time. The professional cyclist completes each lap in 4 minutes; the novice takes 6 minutes to complete each lap. How many minutes after the start will both cyclists pass at exactly the same spot where they began to cycle?
Answer: D
The faster cyclyst comes to the starting point for every 4 min so his times are 4, 8, 12, ......... The slower cyclist comes to the starting point for every 6 min so his times are 6, 12, 18, ......... So both comes at the end of the 12th min.
14. What is the remainder when 617+1176 is divided by 7?
A. 1
B. 6
C. 0
D. 3
Answer: C
617 = (7−1)17 =
17C0.717−17C1.716.11.....+17C16.71.116−17C17.117
If we divide this expansion except the last term each term gives a remainder 0. Last term gives a remainder of - 1.
Now From Fermat little theorem, [ap−1p]Rem=1
So [1767]Rem=1
Adding these two remainders we get the final remainder = 0
15. In base 7, a number is written only using the digits 0, 1, 2, .....6. The number 135 in base 7 is 1 x 72+ 3 x 7 + 5 = 75 in base 10. What is the sum of the base 7 numbers 1234 and 6543 in base 7.
A. 11101
B. 11110
C. 10111
D. 11011
Answer: B
Answer: A
We know that A1 = 2 so A2=A1+1=A1+2(1)=4
A3=A2+1=A2+2(2)=8
A4=A3+1=A3+2(3)=14
So the first few terms are 2, 4, 8, 14, 22, ......
The differences of the above terms are 2, 4, 6, 8, 10...
and the differences of differences are 2, 2, 2, 2. all are equal. so this series represents a quadratic equation.
Assume An = an2+bn+c
Now A1 = a + b + c = 2
A2 = 4a + 2b + c = 4
A3 = 9a + 3b + c = 8
Solving above equations we get a = 1, b = - 1 and C = 2
So substituting in An = n2+bn+c = n2−n+2
Substitute 100 in the above equation we get 9902.
A. Statement X alone is enough to get the answer
B. Both statements X and Y are needed to get the answer
C. Statement Y alone is enough to get the answer
D. Data inadequate
Ans: Assume tank capacity is 45 Liters. Given that the first pipe fills the tank in 9 hours. So its capacity is 45 / 9 = 5 Liters/ Hour. Second pipe fills the tank in 5 hours. So its capacity is 45 / 5 = 9 Liters/Hour. If both pipes are opened together, then combined capacity is 14 liters/hour. To fill a tank of capacity 45 liters, Both pipes takes 45 / 14 = 3.21 Hours.
2) If 75 % of a class answered the first question on a certain test correctly, 55 percent answered the second question on the test correctly, and 20 percent answered neither of the questions correctly, what percentage answered both correctly?
It is a problem belongs to sets. We use the following formula n(A∪B) = n(A) + n(B) - n(A∩B)
Here n(A∪B) is the people who answered atleast one of the questions.
It was given that 20% answered neither question then the students who answered atleast one question is 100% - 20% = 80%
Now substituting in the formula we get 80% = 75% + 55% - n(A∩B)
⇒ n(A∩B) = 50%
3) A student's average ( arithmetic mean) test score on 4 tests is 78. What must be the students score on a 5th test for the students average score on the 5th test to be 80?
Ans: We know that Average =Sum of the observations No of observations
So Sum of 4 test scores = 78×4=312
Sum of 5 tests scores = 80×5=400
⇒ 5th test score=400-312=88
Alternative method: If the student scores 78 in the fifth test also, what could be his average? No change. Is it not?
But to bring the average to 80, he must have scored enough marks extra so that each of the five subject scores increase upto 80. i.e., he should have scored 2 x 5 = 10 runs extra in the fifth subject. So 5th subject score is 78 + 10 = 88
4) Rural households have more purchasing power than do urban households at the same income level, since some of the income urban and suburban households use for food and shelter can be used by the rural households for other needs. Which of the following inferences is best supported by the statement made above?
(A) The average rural household includes more people than does the average urban or suburban household.
(B) Rural households have lower food and housing costs than do either urban or suburban households.
(C) Suburban households generally have more purchasing power than do either rural or urban households.
(D) The median income of urban and suburban households is generally higher than that of rural households.
(E) All three types of households spend more of their income on housing than on all other purchases combined.
As the cylinder is filled to initially exactly half of the capacity, When this cylinder is placed on its side, Water comes upto the height of the radius.
So water comes upto 3 ft.
7) The present ratio of students to teachers at a certain school is 30 to 1. If the student enrollment were to increase by 50 students and the number of teachers were to increase by 5, the ratio of the teachers would then be 25 to 1 What is the present number of teachers?
Assume the present students and teachers are 30K, K
After new recruitments of students and teachers the strength becomes 30K + 50, K + 5 respectively. But given that this ratio = 25 : 1
⇒30K+50K+5=251
Solving we get K = 15
So present teachers are 15.
8) College T has 1000 students. Of the 200 students majoring in one or more of the sciences,130 are majoring in Chemistry and 150 are majoring in Biology. If at least 30 of the students are not majoring in either Chemistry or Biology, then the number of students majoring in both Chemistry and Biology could be any number from
If we assume exactly 30 students are not majoring in any subject then the students who take atleast one subject = 200 - 30 = 170
We know that n(A∪B) = n(A) + n(B) - n(A∩B)
⇒ 170 = 130 + 150 - n(A∩B)
Solving we get n(A∩B) = 110.
i.e., Students who can take both subjects are 110
But If more than 30 students are not taking any subject, what can be the maximum number of students who can take both the subjects?
As there are 130 students are majoring in chemistry, assume these students are taking biology also. So maximum students who can take both the subjects is 130
So boxes loaded in day shift = 4 x 5 = 20, and boxes loaded in night shift = 3 x 4 = 12
so fraction of boxes loaded in day shift = 2020+12=58
22) A bakery opened yesterday with its daily supply of 40 dozen rolls. Half of the rolls were sold by noon and 80 % of the remaining rolls were sold between noon and closing time. How many dozen rolls had not been sold when the bakery closed yesterday?
Ans: If half of the rolls were sold by noon, the remaining are 50 % (40) = 20.
Given 80% of the remaining were sold after the noon to closing time
⇒ 80% (20) = 16
Unsold = 20 - 16 = 4
23) If N=4P, where P is a prime number greater than 2, how many different positive even divisors does n have including n?
Ans: N = 22×P1
We know that total factors of a number which is in the format of aP×bQ×cR... = (P + 1). (Q + 1). (R + 1) .... = (2 + 1).(1 + 1) = 6
Also odd factors of any number can be calculated easily by not taking 2 and its powers.
So odd factors of 22×P1 = the factors of P1 = (1 + 1) = 2
Even factors of the number = 6 - 2 = 4
24) A dealer originally bought 100 identical batteries at a total cost of q rupees. If each battery was sold at 50 percent above the original cost per battery, then, in terms of q, for how many rupees was each battery sold?
Ans: Per battery cost = q / 100
If each battery is sold for 50% gain, then selling price = CostPrice×(100+Gain100)
⇒ q100×(100+50100)=3q200
25) The price of lunch for 15 people was 207 pounds, including a 15 percent gratuity of service. What was the average price per person, EXCLUDING the gratuity?
Ans: Let the net price excluding the gratuity of service = x pounds
Then, total price including 15% gratuity of service = x×(100+15100) = 1.15 x pounds
So, 1.15 x = 207 pounds
⇒ x = 207 / 1.15 = 180 pounds
Net price of lunch for each person = 180 / 15 = 12 pounds
1. Adam sat with his friends in the Chinnaswamy stadium at Madurai to watch the 100 metres running race organized by the Asian athletics Association. Five rounds were run. After every round half the teams were eliminated. Finally, one team wins the game. How many teams participated in the race?
Ans: Total five rounds were run. So in the final round 2 teams must have participated. In the penultimate round 4 teams, and 3rd round 8, 2nd round 16 and in the first round 32 teams must have participated as in each round half of the teams got eliminated.
3. 49 members attended the party. In that 22 are males, 27 are females. The shake hands are done between males, females, male and female. Total 12 people given shake hands. How many such kinds of such shake hands are possible?
Ans: If only 12 people shaked their hands, then total hand shakes are 12C2 = 66
4. Ferrari S.P.A is an Italian sports car manufacturer based in Maranello, Italy. Founded by Enzo Ferrari in 1928 as Scuderia Ferrari, the company sponsored drivers and manufactured race cars before moving into production of street-legal vehicles in 1947 as Ferrari S.P.A. Throughout its history, the company has been noted for its continued participation in racing, especially in Formula One where it has employed great success. Rohit once bought a Ferrari. It could go 4 times as fast as Mohan’s old Mercedes. If the speed of Mohan’s Mercedes is 35 km/hr and the distance traveled by the Ferrari is 490 km, find the total time taken for Rohit to drive that distance.
Ans: As Ferrari's speed is four times that of the mercedes, Its speed is 35 x 4 = 140
So time taken by the ferrari = 490 / 140 = 3.5 Hours
5. A sheet of paper has statements numbered from 1 to 40. For all values of n from 1 to 40, statement n says: ‘Exactly n of the statements on this sheet are false.’ Which statements are true and which are false?
a) The even numbered statements are true and the odd numbered statements are false.
b) The odd numbered statements are true and the even numbered statements are false.
c) All the statements are false.
d) The 39th statement is true and the rest are false
Ans: Assume there is only one statement is there. The statement should read "Exactly 1 statement on this sheet is false" . If the truth value of the statement is true, then given statement should be false. This is contradiction. If the statement is false, Then the given statement is true. but there is not other true statement.
Remember, while moving from east to west countries lag in time. Remember when Test cricket starts in England? 3. 30 in afternoon. Right? ie., We are in after noon means they are in morning.
If the coordinates change from 10 E to 70W, the plane has moved a total of 80 degrees. We know that with each degree time increases by 4 minutes while going from east to west. (How? 24 x 60 min / 360 degrees, So 1 degree = 4 min)
So total time change = 4 x 80 = 320 min = 5 hrs + 20 minutes.
After 10 hours local time is (2 am + 10 - 5.20 hrs) = 6.40 AM.
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Note :
As you already aware of the fact that TCS has changed the database of questions for its aptitude test. The questions below give you an overview of the models to be prepared. But don't depend on these models only. We solved these questions only as an indicative purpose. You are requested to go through all the arithmetic topics given in this site so that you become confident of sitting for TCS or any other written test. All the best... The question below have been taken from recent TCS Placement papers of 2016.
As you already aware of the fact that TCS has changed the database of questions for its aptitude test. The questions below give you an overview of the models to be prepared. But don't depend on these models only. We solved these questions only as an indicative purpose. You are requested to go through all the arithmetic topics given in this site so that you become confident of sitting for TCS or any other written test. All the best... The question below have been taken from recent TCS Placement papers of 2016.
1. If 3y + x > 2 and x + 2y≤3, What can be said about the value of y?
A. y = -1
B. y >-1
C. y <-1
D. y = 1
Answer: B
Multiply the second equation with -1 then it will become - x - 2y≥ - 3. Add the equations. You will get y > -1.
2. If the price of an item is decreased by 10% and then increased by 10%, the net effect on the price of the item is
A. A decrease of 99%
B. No change
C. A decrease of 1%
D. An increase of 1%
Multiply the second equation with -1 then it will become - x - 2y≥ - 3. Add the equations. You will get y > -1.
2. If the price of an item is decreased by 10% and then increased by 10%, the net effect on the price of the item is
A. A decrease of 99%
B. No change
C. A decrease of 1%
D. An increase of 1%
Answer: C
If a certain number is increased by x% then decreased by x% or vice versa, the net change is always decrease. This change is given by a simple formula −(x10)2= −(1010)2= −1%. Negitive sign indicates decrease.
3. If m is an odd integer and n an even integer, which of the following is definitely odd?
A. (2m+n)(m-n)
B. (m+n2)+(m−n2)
C. m2+mn+n2
D. m +n
Answer: C and D (Original Answer given as D)
You just remember the following odd ± odd = even; even ± even = even; even ± odd = odd
Also odd x odd = odd; even x even = even; even x odd = even.
4. What is the sum of all even integers between 99 and 301?
A. 40000
B. 20000
C. 40400
D. 20200
Answer: D
The first even number after 99 is 100 and last even number below 301 is 300. We have to find the sum of even numbers from 100 to 300. i.e., 100 + 102 + 104 + ............... 300.
Take 2 Common. 2 x ( 50 + 51 + ...........150)
There are total 101 terms in this series. So formula for the sum of n terms when first term and last term is known is n2(a+l)
So 50 + 51 + ...........150 = 1012(50+150)
So 2 x 1012(50+150) = 20200
5. There are 20 balls which are red, blue or green. If 7 balls are green and the sum of red balls and green balls is less than 13, at most how many red balls are there?
A. 4
B. 5
C. 6
D. 7
Answer: B
Given R + B + G = 17; G = 7; and R + G < 13. Substituting G = 7 in the last equation, We get R < 6. So maximum value of R = 6
6. If n is the sum of two consecutive odd integers and less than 100, what is greatest possibility of n?
A. 98
B. 94
C. 96
D. 99
Answer : C
We take two odd numbers as (2n + 1) and (2n - 1).
Their sum should be less than 100. So (2n + 1) + (2n - 1) < 100 ⇒ 4n < 100.
The largest 4 multiple which is less than 100 is 96
7. x2 < 1/100, and x < 0 what is the highest range in which x can lie?
A. -1/10 < x < 0
B. -1 < x < 0
C. -1/10 < x < 1/10
D. -1/10 < x
Answer: A
Remember:
(x - a)(x - b) < 0 then value of x lies in between a and b.
(x - a)(x - b) > 0 then value of x does not lie inbetween a and b. or ( −∞, a) and (b, −∞) if a < b
x2 < 1/100 ⇒
(x2−1/100)<0⇒(x2−(1/10)2)<0⇒(x−1/10)(x+1/10)<0
So x should lie inbetween - 1/10 and 1/10. But it was given that x is -ve. So x lies in -1/10 to 0
8. There are 4 boxes colored red, yellow, green and blue. If 2 boxes are selected, how many combinations are there for at least one green box or one red box to be selected?
A. 1
B . 6
C. 9
D. 5
Remember:
(x - a)(x - b) < 0 then value of x lies in between a and b.
(x - a)(x - b) > 0 then value of x does not lie inbetween a and b. or ( −∞, a) and (b, −∞) if a < b
x2 < 1/100 ⇒
(x2−1/100)<0⇒(x2−(1/10)2)<0⇒(x−1/10)(x+1/10)<0
So x should lie inbetween - 1/10 and 1/10. But it was given that x is -ve. So x lies in -1/10 to 0
8. There are 4 boxes colored red, yellow, green and blue. If 2 boxes are selected, how many combinations are there for at least one green box or one red box to be selected?
A. 1
B . 6
C. 9
D. 5
Answer: 5
Total ways of selecting two boxes out of 4 is 4C2 = 6. Now, the number of ways of selecting two boxes where none of the green or red box included is only 1 way. (we select yellow and blue in only one way). If we substract this number from total ways we get 5 ways.
9. All faces of a cube with an eight - meter edge are painted red. If the cube is cut into smaller cubes with a two - meter edge, how many of the two meter cubes have paint on exactly one face?
A. 24
B. 36
C. 60
D. 48
Answer : A
If there are n cubes lie on an edge, then total number of cubes with one side painting is given by 6×(n−2)2. Here side of the bigger cube is 8, and small cube is 2. So there are 4 cubes lie on an edge. Hence answer = 24
10. Two cyclists begin training on an oval racecourse at the same time. The professional cyclist completes each lap in 4 minutes; the novice takes 6 minutes to complete each lap. How many minutes after the start will both cyclists pass at exactly the same spot where they began to cycle?
A. 10
B. 8
C. 14
D. 12
B. 8
C. 14
D. 12
Answer: D
The faster cyclyst comes to the starting point for every 4 min so his times are 4, 8, 12, ......... The slower cyclist comes to the starting point for every 6 min so his times are 6, 12, 18, ......... So both comes at the end of the 12th min.
14. What is the remainder when 617+1176 is divided by 7?
A. 1
B. 6
C. 0
D. 3
Answer: C
617 = (7−1)17 =
17C0.717−17C1.716.11.....+17C16.71.116−17C17.117
If we divide this expansion except the last term each term gives a remainder 0. Last term gives a remainder of - 1.
Now From Fermat little theorem, [ap−1p]Rem=1
So [1767]Rem=1
Adding these two remainders we get the final remainder = 0
15. In base 7, a number is written only using the digits 0, 1, 2, .....6. The number 135 in base 7 is 1 x 72+ 3 x 7 + 5 = 75 in base 10. What is the sum of the base 7 numbers 1234 and 6543 in base 7.
A. 11101
B. 11110
C. 10111
D. 11011
Answer: B
In base 7 there is no 7. So to write 7 we use 10.
for 8 we use 11...... for 13 we use 16, for 14 we use 20 and so on.
So from the column d, 4 + 3 = 7 = 10, we write 0 and 1 carried over. now 1 + 3 + 4 = 8 = 11, then we write 1 and 1 carried over. again 1 + 2 + 5 = 8 = 11 and so on
16. The sequence {An} is defined by A1 = 2 and An+1=An+2n what is the value of A100
A. 9902
B. 9900
C. 10100
D. 9904
So from the column d, 4 + 3 = 7 = 10, we write 0 and 1 carried over. now 1 + 3 + 4 = 8 = 11, then we write 1 and 1 carried over. again 1 + 2 + 5 = 8 = 11 and so on
16. The sequence {An} is defined by A1 = 2 and An+1=An+2n what is the value of A100
A. 9902
B. 9900
C. 10100
D. 9904
Answer: A
We know that A1 = 2 so A2=A1+1=A1+2(1)=4
A3=A2+1=A2+2(2)=8
A4=A3+1=A3+2(3)=14
So the first few terms are 2, 4, 8, 14, 22, ......
The differences of the above terms are 2, 4, 6, 8, 10...
and the differences of differences are 2, 2, 2, 2. all are equal. so this series represents a quadratic equation.
Assume An = an2+bn+c
Now A1 = a + b + c = 2
A2 = 4a + 2b + c = 4
A3 = 9a + 3b + c = 8
Solving above equations we get a = 1, b = - 1 and C = 2
So substituting in An = n2+bn+c = n2−n+2
Substitute 100 in the above equation we get 9902.
18. A, B, C and D go for a picnic. When A stands on a
weighing machine, B also climbs on, and the weight shown was 132 kg. When
B stands, C also climbs on, and the machine shows 130 kg. Similarly the
weight of C and D is found as 102 kg and that of B and D is 116 kg. What
is D's weight
A. 58kg
B. 78 kg
C. 44 kg
D. None
A. 58kg
B. 78 kg
C. 44 kg
D. None
Answer : C
Given A + B = 132; B + C = 130; C + D = 102, B + D = 116
Eliminate B from 2nd and 4th equation and solving this equation and 3rd we get D value as 44.
19. Roy is now 4 years older than Erik and half of that amount older than Iris. If in 2 years, roy will be twice as old as Erik, then in 2 years what would be Roy's age multiplied by Iris's age?
A. 28
B. 48
C. 50
D. 52
Given A + B = 132; B + C = 130; C + D = 102, B + D = 116
Eliminate B from 2nd and 4th equation and solving this equation and 3rd we get D value as 44.
19. Roy is now 4 years older than Erik and half of that amount older than Iris. If in 2 years, roy will be twice as old as Erik, then in 2 years what would be Roy's age multiplied by Iris's age?
A. 28
B. 48
C. 50
D. 52
Answer: 48
20. X, Y, X and W are integers. The expression X - Y - Z is even and the expression Y - Z - W is odd. If X is even what must be true?
A. W must be odd
B. Y - Z must be odd
C. W must be odd
D. Z must be odd
20. X, Y, X and W are integers. The expression X - Y - Z is even and the expression Y - Z - W is odd. If X is even what must be true?
A. W must be odd
B. Y - Z must be odd
C. W must be odd
D. Z must be odd
Answer: C
21. Mr and Mrs Smith have invited 9 of their friends and their spouses for a party at the Waikiki Beach resort. They stand for a group photograph. If Mr Smith never stands next to Mrs Smith (as he says they are always together otherwise). How many ways the group can be arranged in a row for the photograph?
A. 20!
B. 19! + 18!
C. 18 x 19!
D. 2 x 19!
21. Mr and Mrs Smith have invited 9 of their friends and their spouses for a party at the Waikiki Beach resort. They stand for a group photograph. If Mr Smith never stands next to Mrs Smith (as he says they are always together otherwise). How many ways the group can be arranged in a row for the photograph?
A. 20!
B. 19! + 18!
C. 18 x 19!
D. 2 x 19!
Answer: C
22. In a rectanglular coordinate system, what is the area of a triangle whose vertices whose vertices have the coordinates (4,0), (6, 3) adn (6 , -3)
A. 6
B. 7
C. 7.5
D. 6.5
22. In a rectanglular coordinate system, what is the area of a triangle whose vertices whose vertices have the coordinates (4,0), (6, 3) adn (6 , -3)
A. 6
B. 7
C. 7.5
D. 6.5
Answer: A
23. A drawer holds 4 red hats and 4 blue hats. What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer and immediately returning every hat to the drawer before taking out the next?
A. 1/2
B. 1/8
C. 1/4
D. 3/8
23. A drawer holds 4 red hats and 4 blue hats. What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer and immediately returning every hat to the drawer before taking out the next?
A. 1/2
B. 1/8
C. 1/4
D. 3/8
Answer: B
24. In how many ways can we distribute 10 identical looking pencils to 4 students so that each student gets at least one pencil?
A. 5040
B. 210
C. 84
D. None of these
24. In how many ways can we distribute 10 identical looking pencils to 4 students so that each student gets at least one pencil?
A. 5040
B. 210
C. 84
D. None of these
Answer: C
25. The prime factorization of intezer N is A x A x B x C, where A, B and C are all distinct prine intezers. How many factors does N have?
A. 12
B. 24
C. 4
D. 6
25. The prime factorization of intezer N is A x A x B x C, where A, B and C are all distinct prine intezers. How many factors does N have?
A. 12
B. 24
C. 4
D. 6
Answer: A
26. Tim and Elan are 90 km from each other.they start to move each other simultanously tim at speed 10 and elan 5 kmph. If every hour they double their speed what is the distance that Tim will pass until he meet Elan
A. 45
B. 60
C. 20
D. 80
26. Tim and Elan are 90 km from each other.they start to move each other simultanously tim at speed 10 and elan 5 kmph. If every hour they double their speed what is the distance that Tim will pass until he meet Elan
A. 45
B. 60
C. 20
D. 80
Answer: B
27. A father purchases dress for his three daughter. The dresses are of same color but of different size .the dress is kept in dark room .What is the probability that all the three will not choose their own dress.
A. 2/3
B. 1/3
C. 1/6
D. 1/9
27. A father purchases dress for his three daughter. The dresses are of same color but of different size .the dress is kept in dark room .What is the probability that all the three will not choose their own dress.
A. 2/3
B. 1/3
C. 1/6
D. 1/9
Answer: B
28. N is an integer and N>2, at most how many integers among N +
2, N + 3, N + 4, N + 5, N + 6, and N + 7 are prime integers?
A. 1
B. 3
C. 2
D. 4
A. 1
B. 3
C. 2
D. 4
Answer: C
29. A turtle is crossing a field. What is the total distance (in meters) passed by turtle? Consider the following two statements
(X) The average speed of the turtle is 2 meters per minute
(Y) Had the turtle walked 1 meter per minute faster than his average speed it would have finished 40 minutes earlier
29. A turtle is crossing a field. What is the total distance (in meters) passed by turtle? Consider the following two statements
(X) The average speed of the turtle is 2 meters per minute
(Y) Had the turtle walked 1 meter per minute faster than his average speed it would have finished 40 minutes earlier
A. Statement X alone is enough to get the answer
B. Both statements X and Y are needed to get the answer
C. Statement Y alone is enough to get the answer
D. Data inadequate
Answer: B
30. Given the following information, who is youngest?
C is younger than A; A is talled than B
C is older than B; C is younger than D
B is taller than C; A is older than D
A. D
B. B
C. C
D. A
30. Given the following information, who is youngest?
C is younger than A; A is talled than B
C is older than B; C is younger than D
B is taller than C; A is older than D
A. D
B. B
C. C
D. A
Answer: B
31. If P(x) = ax4+bx3+cx2+dx+e has roots at x = 1, 2, 3, 4 and P(0) = 48, what is P(5)
A. 48
B. 24
C. 0
D. 50
31. If P(x) = ax4+bx3+cx2+dx+e has roots at x = 1, 2, 3, 4 and P(0) = 48, what is P(5)
A. 48
B. 24
C. 0
D. 50
Answer: A
TCS latest Pattern Questions with Explanations - 2
1) The water from one outlet, flowing at a constant rate, can fill
the swimming pool in 9 hours. The water from second outlet, flowing at a
constant rate can fill up the same pool in approximately in 5 hours. If both
the outlets are used at the same time, approximately what is the number of
hours required to fill the pool?
Ans: Assume tank capacity is 45 Liters. Given that the first pipe fills the tank in 9 hours. So its capacity is 45 / 9 = 5 Liters/ Hour. Second pipe fills the tank in 5 hours. So its capacity is 45 / 5 = 9 Liters/Hour. If both pipes are opened together, then combined capacity is 14 liters/hour. To fill a tank of capacity 45 liters, Both pipes takes 45 / 14 = 3.21 Hours.
2) If 75 % of a class answered the first question on a certain test correctly, 55 percent answered the second question on the test correctly, and 20 percent answered neither of the questions correctly, what percentage answered both correctly?
It is a problem belongs to sets. We use the following formula n(A∪B) = n(A) + n(B) - n(A∩B)
Here n(A∪B) is the people who answered atleast one of the questions.
It was given that 20% answered neither question then the students who answered atleast one question is 100% - 20% = 80%
Now substituting in the formula we get 80% = 75% + 55% - n(A∩B)
⇒ n(A∩B) = 50%
3) A student's average ( arithmetic mean) test score on 4 tests is 78. What must be the students score on a 5th test for the students average score on the 5th test to be 80?
Ans: We know that Average =Sum of the observations No of observations
So Sum of 4 test scores = 78×4=312
Sum of 5 tests scores = 80×5=400
⇒ 5th test score=400-312=88
Alternative method: If the student scores 78 in the fifth test also, what could be his average? No change. Is it not?
But to bring the average to 80, he must have scored enough marks extra so that each of the five subject scores increase upto 80. i.e., he should have scored 2 x 5 = 10 runs extra in the fifth subject. So 5th subject score is 78 + 10 = 88
4) Rural households have more purchasing power than do urban households at the same income level, since some of the income urban and suburban households use for food and shelter can be used by the rural households for other needs. Which of the following inferences is best supported by the statement made above?
(A) The average rural household includes more people than does the average urban or suburban household.
(B) Rural households have lower food and housing costs than do either urban or suburban households.
(C) Suburban households generally have more purchasing power than do either rural or urban households.
(D) The median income of urban and suburban households is generally higher than that of rural households.
(E) All three types of households spend more of their income on housing than on all other purchases combined.
Ans: If average rural household includes more people, then how come they have
more purchasing power? Infact, they have less purchasing power as they have to
feed more people. Option A ruled out.
Option C does not explain why rural households have more purchasing power than urban. Ruled out.
If median income of urban and suburban households is generally higher than rural households they are likely to have more purchasing power, assuming other parameters constant. But this does not explain why rural households have more purchasing power. Options D ruled out.
Option E does not provide any explanation why rural households have more purchasing power. Ruled out.
Option B is correct as, If rural households spend less income on food and shelter due to less prices they definitely have more disposable income to spend.
5) Jose is a student of horticulture in the University of Hose. In a horticultural experiment in his final year, 200 seeds were planted in plot I and 300 were planted in plot II. If 57% of the seeds in plot I germinated and 42% of the seeds in plot II germinated, what percent of the total number of planted seeds germinated?
Ans: Total seeds germinated in Plot I = 57% of 200 = 114
Total seeds germinated in Plot II = 42% of 300 = 126
Total germinated seeds = 114 + 126 = 240
The percentage of germinated seeds of the total seeds = 240500×100 = 48%
6) A closed cylindrical tank contains 36π cubic feet of water and its filled to half its capacity. When the tank is placed upright on its circular base on level ground, the height of water in the tank is 4 feet. When the tank is placed on its side on level ground, what is the height, in feet, of the surface of the water above the ground?
Ans: We know that the volume of cylinder = πr2h
Given tank hight = 4ft.
⇒ πr24 = 36π
⇒ r = 3
So the radius is 3 which means the diameter is 6.
Option C does not explain why rural households have more purchasing power than urban. Ruled out.
If median income of urban and suburban households is generally higher than rural households they are likely to have more purchasing power, assuming other parameters constant. But this does not explain why rural households have more purchasing power. Options D ruled out.
Option E does not provide any explanation why rural households have more purchasing power. Ruled out.
Option B is correct as, If rural households spend less income on food and shelter due to less prices they definitely have more disposable income to spend.
5) Jose is a student of horticulture in the University of Hose. In a horticultural experiment in his final year, 200 seeds were planted in plot I and 300 were planted in plot II. If 57% of the seeds in plot I germinated and 42% of the seeds in plot II germinated, what percent of the total number of planted seeds germinated?
Ans: Total seeds germinated in Plot I = 57% of 200 = 114
Total seeds germinated in Plot II = 42% of 300 = 126
Total germinated seeds = 114 + 126 = 240
The percentage of germinated seeds of the total seeds = 240500×100 = 48%
6) A closed cylindrical tank contains 36π cubic feet of water and its filled to half its capacity. When the tank is placed upright on its circular base on level ground, the height of water in the tank is 4 feet. When the tank is placed on its side on level ground, what is the height, in feet, of the surface of the water above the ground?
Ans: We know that the volume of cylinder = πr2h
Given tank hight = 4ft.
⇒ πr24 = 36π
⇒ r = 3
So the radius is 3 which means the diameter is 6.
As the cylinder is filled to initially exactly half of the capacity, When this cylinder is placed on its side, Water comes upto the height of the radius.
So water comes upto 3 ft.
7) The present ratio of students to teachers at a certain school is 30 to 1. If the student enrollment were to increase by 50 students and the number of teachers were to increase by 5, the ratio of the teachers would then be 25 to 1 What is the present number of teachers?
Assume the present students and teachers are 30K, K
After new recruitments of students and teachers the strength becomes 30K + 50, K + 5 respectively. But given that this ratio = 25 : 1
⇒30K+50K+5=251
Solving we get K = 15
So present teachers are 15.
8) College T has 1000 students. Of the 200 students majoring in one or more of the sciences,130 are majoring in Chemistry and 150 are majoring in Biology. If at least 30 of the students are not majoring in either Chemistry or Biology, then the number of students majoring in both Chemistry and Biology could be any number from
If we assume exactly 30 students are not majoring in any subject then the students who take atleast one subject = 200 - 30 = 170
We know that n(A∪B) = n(A) + n(B) - n(A∩B)
⇒ 170 = 130 + 150 - n(A∩B)
Solving we get n(A∩B) = 110.
i.e., Students who can take both subjects are 110
But If more than 30 students are not taking any subject, what can be the maximum number of students who can take both the subjects?
As there are 130 students are majoring in chemistry, assume these students are taking biology also. So maximum students who can take both the subjects is 130
So the number of students who can take both subjects can be any
number from 110 to 130.
9) Kelly and Chris are moving into a new city. Both of them love books and thus packed several boxes with books. If Chris packed 60% of the total number of boxes, what was the ratio of the number of boxes Kelly packed to the number of boxes Chris packed?
Simple questions. If chris packs 60% of the boxes, kelly packs remaining 40%
So Kelly : Chris = 40% : 60% = 2 : 3
9) Kelly and Chris are moving into a new city. Both of them love books and thus packed several boxes with books. If Chris packed 60% of the total number of boxes, what was the ratio of the number of boxes Kelly packed to the number of boxes Chris packed?
Simple questions. If chris packs 60% of the boxes, kelly packs remaining 40%
So Kelly : Chris = 40% : 60% = 2 : 3
11) Machine A produces bolts at a uniform rate of 120 every 40
second, and Machine B produces bolts at a uniform rate of 100 every 20 seconds.
If the two machines run simultaneously, how many seconds will it take for them
to produce a total of 200 bolts?
Ans: Machine A produces 120/40 = 3 bolts in 1 second and machine B produces 100/20 = 5 bolts in one second.
Hence, both of them will produce 8 bolts per second.
Hence, they wil take 200/8 = 25 seconds to produce 200 bolts.
12) How many prime numbers between 1 and 100 are factors of 7150?
Ans: 7, 150 = 2×52×11×13
So there are 4 distinct prime numbers that are below 100
13) Analysing the good returns that Halocircle Insurance Pvt Ltd was giving, Ratika bought a 1-year, Rs 10,000 certificate of deposit that paid interest at an annual rate of 8% compounded semi-annually.What was the total amount of interest paid on this certificate at maturity?
This is a question on compound interest to be calculated semi annually.
In the case of semi annual compounding, Interest rate becomes half and Number of periods becomes 2 per year.
So A = P(1+R100)n
⇒A=10,000(1+4100)2=10,000×2625
= 10,816
Interest = A - P = 10, 816 - 10,000 = 816
14) Juan is a gold medalist in athletics. In the month of May, if Juan takes 11 seconds to run y yards, how many seconds will it take him to run x yards at the same rate?
Ans: If juan takes 11 seconds to run Y yards, for 1 yard he will take 11 / y seconds. To run x yards his time will be 11 / y × x = 11x/ y
15) A certain company retirement plan has a rule of 70 provision that allows an employee to retire when the employee's age plus years of employment with the company total at least 70. In what year could a female employee hired in 1986 on her 32nd birthday first be eligible to retire under this provision?
Assume it has taken x years to the female employee to reach the rule of 70.
So her age should be 32 + x. Also she gains x years of experience.
⇒ (32 + x) + x = 70
⇒ x = 19.
Her age at the time of retirement = 1986 + 19 = 2005
16) Of the following, which is the closest approximation of (50.2*0.49)/199.8 ?
ans: For approximation (50.2×0.49)/199.8 can be taken as
50×0.5/200 = 25/200 = 1/8 = 0.125
17) Andalusia has been promoting the importance of health maintenance. From January 1,1991 to January 1,1993, the number of people enrolled in health maintenance organizations increased by 15 percent. The enrollment on January 1,1993 was 45 million. How many million people(to the nearest million) was enrolled in health maintenance organizations on January 1,1991?
Ans: If a number K is to be increased by x % it should be multiplied by (100+x)100
So When the enrollment in January 1, 1991 is multiplied by (100+x)100 we got 45 million.
K×(100+15)100=45
K = 45×100115 = 39.13
18) What is the lowest possible integer that is divisible by each of the integers 1 through 7, inclusive?
Ans: If a number has to be divisible by each number from 1 to 7, that number should be L.C.M of(1,2,3,4,5,6,7) = 420
19) If the area of a square region having sides of length 6 cms is equal to the area of a rectangular region having width 2.5 cms, then the length of the rectangle, in cms, is
Ans: Given Area of the square = Area of rectangle
⇒a2=l.b
Substituting the above values in the formula
⇒62=l.2.5
⇒ l = 14.4 cm
20) A tank contains 10,000 gallons of a solution that is 5 percent sodium chloride by volume. If 2500 gallons of water evaporate from the tank, the remaining solution will be approximately what percentage of sodium chloride?
Ans: Sodium chloride in the original solution = 5% of 10,000 = 500
Water in the original solution = 10,000 - 500 = 9,500
If 2,500 Liters of the water is evaporated then the remaining water = 9,500 - 2,500 = 7,000
Sodium chloride concentration = 500500+7000×100 = 6.67 %
(concentration should be calculated always on the total volume)
21) After loading a dock, each worker on the night crew loaded 3/4 as many boxes as each worker on the day of the crew. If the night crew has 4/5 as many workers as the day crew, what fraction of all the boxes loaded by two crews did the day crew load?
Assume the number of boxes loaded in dayshift is equal to 4, then the number of boxed loaded in night shift = 3
Assume the worked on dayshift = 5, then workers on night shift = 4
Ans: Machine A produces 120/40 = 3 bolts in 1 second and machine B produces 100/20 = 5 bolts in one second.
Hence, both of them will produce 8 bolts per second.
Hence, they wil take 200/8 = 25 seconds to produce 200 bolts.
12) How many prime numbers between 1 and 100 are factors of 7150?
Ans: 7, 150 = 2×52×11×13
So there are 4 distinct prime numbers that are below 100
13) Analysing the good returns that Halocircle Insurance Pvt Ltd was giving, Ratika bought a 1-year, Rs 10,000 certificate of deposit that paid interest at an annual rate of 8% compounded semi-annually.What was the total amount of interest paid on this certificate at maturity?
This is a question on compound interest to be calculated semi annually.
In the case of semi annual compounding, Interest rate becomes half and Number of periods becomes 2 per year.
So A = P(1+R100)n
⇒A=10,000(1+4100)2=10,000×2625
= 10,816
Interest = A - P = 10, 816 - 10,000 = 816
14) Juan is a gold medalist in athletics. In the month of May, if Juan takes 11 seconds to run y yards, how many seconds will it take him to run x yards at the same rate?
Ans: If juan takes 11 seconds to run Y yards, for 1 yard he will take 11 / y seconds. To run x yards his time will be 11 / y × x = 11x/ y
15) A certain company retirement plan has a rule of 70 provision that allows an employee to retire when the employee's age plus years of employment with the company total at least 70. In what year could a female employee hired in 1986 on her 32nd birthday first be eligible to retire under this provision?
Assume it has taken x years to the female employee to reach the rule of 70.
So her age should be 32 + x. Also she gains x years of experience.
⇒ (32 + x) + x = 70
⇒ x = 19.
Her age at the time of retirement = 1986 + 19 = 2005
16) Of the following, which is the closest approximation of (50.2*0.49)/199.8 ?
ans: For approximation (50.2×0.49)/199.8 can be taken as
50×0.5/200 = 25/200 = 1/8 = 0.125
17) Andalusia has been promoting the importance of health maintenance. From January 1,1991 to January 1,1993, the number of people enrolled in health maintenance organizations increased by 15 percent. The enrollment on January 1,1993 was 45 million. How many million people(to the nearest million) was enrolled in health maintenance organizations on January 1,1991?
Ans: If a number K is to be increased by x % it should be multiplied by (100+x)100
So When the enrollment in January 1, 1991 is multiplied by (100+x)100 we got 45 million.
K×(100+15)100=45
K = 45×100115 = 39.13
18) What is the lowest possible integer that is divisible by each of the integers 1 through 7, inclusive?
Ans: If a number has to be divisible by each number from 1 to 7, that number should be L.C.M of(1,2,3,4,5,6,7) = 420
19) If the area of a square region having sides of length 6 cms is equal to the area of a rectangular region having width 2.5 cms, then the length of the rectangle, in cms, is
Ans: Given Area of the square = Area of rectangle
⇒a2=l.b
Substituting the above values in the formula
⇒62=l.2.5
⇒ l = 14.4 cm
20) A tank contains 10,000 gallons of a solution that is 5 percent sodium chloride by volume. If 2500 gallons of water evaporate from the tank, the remaining solution will be approximately what percentage of sodium chloride?
Ans: Sodium chloride in the original solution = 5% of 10,000 = 500
Water in the original solution = 10,000 - 500 = 9,500
If 2,500 Liters of the water is evaporated then the remaining water = 9,500 - 2,500 = 7,000
Sodium chloride concentration = 500500+7000×100 = 6.67 %
(concentration should be calculated always on the total volume)
21) After loading a dock, each worker on the night crew loaded 3/4 as many boxes as each worker on the day of the crew. If the night crew has 4/5 as many workers as the day crew, what fraction of all the boxes loaded by two crews did the day crew load?
Assume the number of boxes loaded in dayshift is equal to 4, then the number of boxed loaded in night shift = 3
Assume the worked on dayshift = 5, then workers on night shift = 4
So boxes loaded in day shift = 4 x 5 = 20, and boxes loaded in night shift = 3 x 4 = 12
so fraction of boxes loaded in day shift = 2020+12=58
22) A bakery opened yesterday with its daily supply of 40 dozen rolls. Half of the rolls were sold by noon and 80 % of the remaining rolls were sold between noon and closing time. How many dozen rolls had not been sold when the bakery closed yesterday?
Ans: If half of the rolls were sold by noon, the remaining are 50 % (40) = 20.
Given 80% of the remaining were sold after the noon to closing time
⇒ 80% (20) = 16
Unsold = 20 - 16 = 4
23) If N=4P, where P is a prime number greater than 2, how many different positive even divisors does n have including n?
Ans: N = 22×P1
We know that total factors of a number which is in the format of aP×bQ×cR... = (P + 1). (Q + 1). (R + 1) .... = (2 + 1).(1 + 1) = 6
Also odd factors of any number can be calculated easily by not taking 2 and its powers.
So odd factors of 22×P1 = the factors of P1 = (1 + 1) = 2
Even factors of the number = 6 - 2 = 4
24) A dealer originally bought 100 identical batteries at a total cost of q rupees. If each battery was sold at 50 percent above the original cost per battery, then, in terms of q, for how many rupees was each battery sold?
Ans: Per battery cost = q / 100
If each battery is sold for 50% gain, then selling price = CostPrice×(100+Gain100)
⇒ q100×(100+50100)=3q200
25) The price of lunch for 15 people was 207 pounds, including a 15 percent gratuity of service. What was the average price per person, EXCLUDING the gratuity?
Ans: Let the net price excluding the gratuity of service = x pounds
Then, total price including 15% gratuity of service = x×(100+15100) = 1.15 x pounds
So, 1.15 x = 207 pounds
⇒ x = 207 / 1.15 = 180 pounds
Net price of lunch for each person = 180 / 15 = 12 pounds
TCS latest Pattern Questions with Explanations - 3
1) Of the following, which is the closest approximation of
(50.2*0.49)/199.8 ?
Ans: For approximation (50.2×0.49)/199.8 can be taken as
50×0.5/200 = 25/200 = 1/8 = 0.125
2) How many prime numbers between 1 and 100 are factors of 7150?
Ans: 7, 150 = 2×52×11×13
So there are 4 distinct prime numbers that are below 100
Ans: For approximation (50.2×0.49)/199.8 can be taken as
50×0.5/200 = 25/200 = 1/8 = 0.125
2) How many prime numbers between 1 and 100 are factors of 7150?
Ans: 7, 150 = 2×52×11×13
So there are 4 distinct prime numbers that are below 100
4) Country Club has an indoor swimming club. Thirty percent of the
members of a swim club have passed the lifesaving test. Among the members who
have not passed the test, 12 have taken the preparatory course and 30 have not
taken the course. How many members are there in the swim club?
Ans: 30 + 12 = 42 did not pass the test. This is equal to 70 % of the total
members. So total members = 100/ 70 x 42 = 60
5) A necklace is made by stringing N individual beads together in the repeating pattern red bead, green bead, white bead, blue bead and yellow bead. If the necklace begins with a red bead and ends with a white bead, then N could be:
Ans: The pattern is R G W B Y R G W B Y R .......
So, White bead comes at these positions 3rd, 8th, 13th, 18th...
If we take this as a arithmetic progression, then this series can be expressed as 3 + (n - 1) 5. ( From the formula for general term of AP = a + (n-1)d).
This can be expressed as 5n - 2
We check the answer options so only 68 satisfy the condition.
6) A dog taken four leaps for every five leaps of hare but three leaps of the dog is equal to four leaps of the hare. Compare speed?
Ans: In terms of number of leaps, the ratio of the Dog and hare speeds are 4 : 5
But Given that 3 leaps of dog = 4 leaps of hare,. i.e., Leap lengths = 4 : 3 (If Dog is covering in 3 leaps what hare as covered in 4 leaps then Leap lengths are inversely proportional)
So Dog speed = 4 x 4 = 16
Hare speed = 5 x 3 = 15
So speeds ratio = 16 : 15
7) There are two boxes,one containing 39 red balls & the other containing 26 green balls.you are allowed to move the balls b/w the boxes so that when you choose a box random & a ball at random from the chosen box,the probability of getting a red ball is maximized.this maximum probability is
Ans: Very interesting question.
As we are allowed to move the balls, we keep only one red ball in first box and move all the remaining balls to the second box
So fist box contains 1 redball, second box contains 38 red + 26 green = 64 balls
Probability of choosing any box is 1/ 2.
So probability of taking one red ball = 12×(1)+12(3864)≃0.8
8) In how many ways can 3 postcards can be posted in 5 postboxes?
Ans: First card can go into any of the five boxes, Second can go into any of the five boxes, Third can go into any of the five boxes = 5×5×5=125
9) Apple costs L rupees per kilogram for first 30kgs and Q rupees per kilogram for each additional kilogram. If the price of 33 kilograms is 11.67and for 36kgs of Apples is 12.48 then the cost of first 10 kgs of Apples is
Ans: By framing equations we get
30L+3Q=11.67
30L+6Q=12.48
Eliminate Q by multiplying the first equation by 2 and subtracting second equation from the first
Then we get L = 0.362
Cost of 10 kgs of apples = 0.362 x 10 = 3.62
10) letters in the word ABUSER are permuted in all possible ways and arranged in alphabetical order then find the word at position 49 in the permuted alphabetical order?
a) ARBSEU
b) ARBESU
c) ARBSUE
d) ARBEUS
5) A necklace is made by stringing N individual beads together in the repeating pattern red bead, green bead, white bead, blue bead and yellow bead. If the necklace begins with a red bead and ends with a white bead, then N could be:
Ans: The pattern is R G W B Y R G W B Y R .......
So, White bead comes at these positions 3rd, 8th, 13th, 18th...
If we take this as a arithmetic progression, then this series can be expressed as 3 + (n - 1) 5. ( From the formula for general term of AP = a + (n-1)d).
This can be expressed as 5n - 2
We check the answer options so only 68 satisfy the condition.
6) A dog taken four leaps for every five leaps of hare but three leaps of the dog is equal to four leaps of the hare. Compare speed?
Ans: In terms of number of leaps, the ratio of the Dog and hare speeds are 4 : 5
But Given that 3 leaps of dog = 4 leaps of hare,. i.e., Leap lengths = 4 : 3 (If Dog is covering in 3 leaps what hare as covered in 4 leaps then Leap lengths are inversely proportional)
So Dog speed = 4 x 4 = 16
Hare speed = 5 x 3 = 15
So speeds ratio = 16 : 15
7) There are two boxes,one containing 39 red balls & the other containing 26 green balls.you are allowed to move the balls b/w the boxes so that when you choose a box random & a ball at random from the chosen box,the probability of getting a red ball is maximized.this maximum probability is
Ans: Very interesting question.
As we are allowed to move the balls, we keep only one red ball in first box and move all the remaining balls to the second box
So fist box contains 1 redball, second box contains 38 red + 26 green = 64 balls
Probability of choosing any box is 1/ 2.
So probability of taking one red ball = 12×(1)+12(3864)≃0.8
8) In how many ways can 3 postcards can be posted in 5 postboxes?
Ans: First card can go into any of the five boxes, Second can go into any of the five boxes, Third can go into any of the five boxes = 5×5×5=125
9) Apple costs L rupees per kilogram for first 30kgs and Q rupees per kilogram for each additional kilogram. If the price of 33 kilograms is 11.67and for 36kgs of Apples is 12.48 then the cost of first 10 kgs of Apples is
Ans: By framing equations we get
30L+3Q=11.67
30L+6Q=12.48
Eliminate Q by multiplying the first equation by 2 and subtracting second equation from the first
Then we get L = 0.362
Cost of 10 kgs of apples = 0.362 x 10 = 3.62
10) letters in the word ABUSER are permuted in all possible ways and arranged in alphabetical order then find the word at position 49 in the permuted alphabetical order?
a) ARBSEU
b) ARBESU
c) ARBSUE
d) ARBEUS
Ans: The best way to solve this problems is Just ask how many words starts with
A. If we fix A, then the remaining letters can be arranged in 5! ways = 120. So
the asked word must start with A.
Arrange all the given letters in alphabetical order. ABERSU
Let us find all the words start with AB. AB**** = 4!= 24 ways
Now we find all the words start wit AE. AE****= 4!= 24 ways
So next word start with AR and remaining letters are BESU
So option B
11) A is twice efficient than B. A and B can both work together to complete a work in 7 days. Then find in how many days A alone can complete the work?
Ans: Let us assume A can do 2 units of work each day, then B can do only 1 unit a day. If both can complete the work in 7 days, total work done by these two togeter = (2 + 1 ) x 7 = 21 units
If these 21 units to be done by A alone, then he will take 21 / 2 = 10.5 days.
12) In a 8 x 8 chess board what is the total number of squares.
Ans: The total number of squares in a n x n chess board is equal to "the sum of first n natural number squares"
i.e., n(n+1)(2n+1)6
So Substituting 8 in the above formula we get 204
13) X, Y, W and Z are intezers and the expressing X - Y - Z is even and Y - W - Z is odd. If X is even then which of the following is true?
(a) Y must be odd
(b) Y-Z must be odd
(c) W must be odd
(d) Z must be odd
Arrange all the given letters in alphabetical order. ABERSU
Let us find all the words start with AB. AB**** = 4!= 24 ways
Now we find all the words start wit AE. AE****= 4!= 24 ways
So next word start with AR and remaining letters are BESU
So option B
11) A is twice efficient than B. A and B can both work together to complete a work in 7 days. Then find in how many days A alone can complete the work?
Ans: Let us assume A can do 2 units of work each day, then B can do only 1 unit a day. If both can complete the work in 7 days, total work done by these two togeter = (2 + 1 ) x 7 = 21 units
If these 21 units to be done by A alone, then he will take 21 / 2 = 10.5 days.
12) In a 8 x 8 chess board what is the total number of squares.
Ans: The total number of squares in a n x n chess board is equal to "the sum of first n natural number squares"
i.e., n(n+1)(2n+1)6
So Substituting 8 in the above formula we get 204
13) X, Y, W and Z are intezers and the expressing X - Y - Z is even and Y - W - Z is odd. If X is even then which of the following is true?
(a) Y must be odd
(b) Y-Z must be odd
(c) W must be odd
(d) Z must be odd
Ans. If X is even and X - Y - Z is even then Y and Z both should be odd or both
should be even.
If Y - W - Z is odd, and Y and Z are also odd W should be odd
If Y - W - Z is even, and Y and Z are even then W should be odd.
So option C is correct. i.e., W must be ODD
14) The remainder when 1!+2!+3!...+50! divided by 5! will be
The remainder when the terms greater than 5! are divided by 5! becomes 0 so we need to consider the terms upto 4!.
So remainder will be whatever is obtained by dividing 1!+2!+3!+4! with 5!.
So remainder is obtained by dividing (1+2+6+24)= 33 with 5! ( 120)
So remainder is 33.
15) If there are Six periods in each working day of a school, In how many ways can one arrange 5 subjects such that each subject is allowed at least one period?
Ans. To arrange 6 periods with 5 subjects, then one subject can be arranged in two slots.
5 Subjects can be arranged in 6 periods in 6P5 ways and now we have 1 period which we can fill with any of the 5 subjects in 5 ways. so 6P5×5=3600
If Y - W - Z is odd, and Y and Z are also odd W should be odd
If Y - W - Z is even, and Y and Z are even then W should be odd.
So option C is correct. i.e., W must be ODD
14) The remainder when 1!+2!+3!...+50! divided by 5! will be
The remainder when the terms greater than 5! are divided by 5! becomes 0 so we need to consider the terms upto 4!.
So remainder will be whatever is obtained by dividing 1!+2!+3!+4! with 5!.
So remainder is obtained by dividing (1+2+6+24)= 33 with 5! ( 120)
So remainder is 33.
15) If there are Six periods in each working day of a school, In how many ways can one arrange 5 subjects such that each subject is allowed at least one period?
Ans. To arrange 6 periods with 5 subjects, then one subject can be arranged in two slots.
5 Subjects can be arranged in 6 periods in 6P5 ways and now we have 1 period which we can fill with any of the 5 subjects in 5 ways. so 6P5×5=3600
Alternate method:
Assume the subjects are X1, X2, A, B , C, D,. Here X is the subject which repeats. So arranging 6 objects in 6 places will be equal to 6! = 720 (here no need to divide this number with 2! as even though the subject is same, but not identical)
But this repeated subect can be any of the five. So total arrangements are 720 x 5 = 3600
16) An article manufactured by a company consists of two parts X and Y. In the process of manufacturing of part X, 9 out 100 parts many be defective. Similarly , 5 out of 100 are likely to be defective in the manufacturer of Y. Calculate the probability that the assembled product will not be defective?
a) 0.6485
b) 0.6565
c) 0.8645
d) none of these
Assume the subjects are X1, X2, A, B , C, D,. Here X is the subject which repeats. So arranging 6 objects in 6 places will be equal to 6! = 720 (here no need to divide this number with 2! as even though the subject is same, but not identical)
But this repeated subect can be any of the five. So total arrangements are 720 x 5 = 3600
16) An article manufactured by a company consists of two parts X and Y. In the process of manufacturing of part X, 9 out 100 parts many be defective. Similarly , 5 out of 100 are likely to be defective in the manufacturer of Y. Calculate the probability that the assembled product will not be defective?
a) 0.6485
b) 0.6565
c) 0.8645
d) none of these
Ans: Probability that the part X is nondefective is = 1 - 9/100=.91
Probablity that the part Y is nondefective is = 1 - 5/100=.95
so, Probablity of nondefective product=0.91×0.95=0.8645
Probablity that the part Y is nondefective is = 1 - 5/100=.95
so, Probablity of nondefective product=0.91×0.95=0.8645
TCS Latest Placement Paper Questions - 2016 (4)
1. Adam sat with his friends in the Chinnaswamy stadium at Madurai to watch the 100 metres running race organized by the Asian athletics Association. Five rounds were run. After every round half the teams were eliminated. Finally, one team wins the game. How many teams participated in the race?
Ans: Total five rounds were run. So in the final round 2 teams must have participated. In the penultimate round 4 teams, and 3rd round 8, 2nd round 16 and in the first round 32 teams must have participated as in each round half of the teams got eliminated.
3. 49 members attended the party. In that 22 are males, 27 are females. The shake hands are done between males, females, male and female. Total 12 people given shake hands. How many such kinds of such shake hands are possible?
Ans: If only 12 people shaked their hands, then total hand shakes are 12C2 = 66
4. Ferrari S.P.A is an Italian sports car manufacturer based in Maranello, Italy. Founded by Enzo Ferrari in 1928 as Scuderia Ferrari, the company sponsored drivers and manufactured race cars before moving into production of street-legal vehicles in 1947 as Ferrari S.P.A. Throughout its history, the company has been noted for its continued participation in racing, especially in Formula One where it has employed great success. Rohit once bought a Ferrari. It could go 4 times as fast as Mohan’s old Mercedes. If the speed of Mohan’s Mercedes is 35 km/hr and the distance traveled by the Ferrari is 490 km, find the total time taken for Rohit to drive that distance.
Ans: As Ferrari's speed is four times that of the mercedes, Its speed is 35 x 4 = 140
So time taken by the ferrari = 490 / 140 = 3.5 Hours
5. A sheet of paper has statements numbered from 1 to 40. For all values of n from 1 to 40, statement n says: ‘Exactly n of the statements on this sheet are false.’ Which statements are true and which are false?
a) The even numbered statements are true and the odd numbered statements are false.
b) The odd numbered statements are true and the even numbered statements are false.
c) All the statements are false.
d) The 39th statement is true and the rest are false
Ans: Assume there is only one statement is there. The statement should read "Exactly 1 statement on this sheet is false" . If the truth value of the statement is true, then given statement should be false. This is contradiction. If the statement is false, Then the given statement is true. but there is not other true statement.
Assume there are two statements. By the above logic, 2nd statement should
not be true. But 1st statement is true as it truthfully says the
truthfulness. By this logic we know that If there are "n"
statements, (n-1)th statement is the only true statement And all other are
false
6. If there are 30 cans out of them one is poisoned if a person tastes very little he will die within 14 hours so if there are mice to test and 24 hours to test, what is the minimum no. of mice’s required to find poisoned can?
Ans:
6. If there are 30 cans out of them one is poisoned if a person tastes very little he will die within 14 hours so if there are mice to test and 24 hours to test, what is the minimum no. of mice’s required to find poisoned can?
Ans:
If only 3 person are used, by giving wine drops suggested by the diagram, we
can find the poisoned casks upto 8.
for example, If the 2nd and 3rd persons die, then 7th cask is poisoned. As a rule of thumb, If we have n mice, we can easily find the poison casks upto 2n. As the number of casks are less than 32 we can use only 5 mice.
7. How many 9 digit numbers are possible by using the digits 1, 2, 3, 4, 5 which are divisible by 4 if the repetition is allowed?
Ans: If A number has to be divisible by 4, the last two digits must be divisible by 4. So possibilities are, 12, 24, 32, 44, 52. And the of the remaining 7 places, each place got filled by any of the five digits. So these 7 places got filled by 5 x 5 x .....(7 times) = 57 ways. So total ways are 5 x 57 = 58
8. A hare and a tortoise have a race along a circle of 100 yards diameter. The tortoise goes in one direction and the hare in the other. The hare starts after the tortoise has covered 1/5 of its distance and that too leisurely. The hare and tortoise meet when the hare has covered only 1/8 of the distance. By what factor should the hare increase its speed so as to tie the race
for example, If the 2nd and 3rd persons die, then 7th cask is poisoned. As a rule of thumb, If we have n mice, we can easily find the poison casks upto 2n. As the number of casks are less than 32 we can use only 5 mice.
7. How many 9 digit numbers are possible by using the digits 1, 2, 3, 4, 5 which are divisible by 4 if the repetition is allowed?
Ans: If A number has to be divisible by 4, the last two digits must be divisible by 4. So possibilities are, 12, 24, 32, 44, 52. And the of the remaining 7 places, each place got filled by any of the five digits. So these 7 places got filled by 5 x 5 x .....(7 times) = 57 ways. So total ways are 5 x 57 = 58
8. A hare and a tortoise have a race along a circle of 100 yards diameter. The tortoise goes in one direction and the hare in the other. The hare starts after the tortoise has covered 1/5 of its distance and that too leisurely. The hare and tortoise meet when the hare has covered only 1/8 of the distance. By what factor should the hare increase its speed so as to tie the race
Assume the circumference of the circle is 200 meters. Hare
and tortoise started at the same point but moves in the opposite direction. It
is given that by that time tortoise covered 40 m (1/5th of the distance), Hare
started and both met after hare has covered 25. This implies, in the time hare
has covered 25m, hare has covered 200 - 40 - 25 = 135 meters.
So Hare : tortoise speeds = 25 : 135 = 5 : 27
Now Hare and tortoise has to reach the starting point means, Hare has to cover 175 meters and Tortoise has to cover only 25 meters in the same time.
As time =DistanceSpeed=2527=1755×K
Ie., Hare has to increase its speed by a factor K. Solving we get K = 37.8
9. For the FIFA world cup, Paul the octopus has been predicting the winner of each match with amazing success. It is rumored that in a match between 2 teams A and B, Paul picks A with the same probability as A’s chances of winning. Let’s assume such rumors to be true and that in a match between Ghana and Bolivia; Ghana the stronger team has a probability of 2/3 of winning the game. What is the probability that Paul will correctly pick the winner of the Ghana-Bolivia game?
a) 1/9
b) 4/9
c) 5/9
d) 2/3
The probability that Paul correctly picks the winner = (A's Chances of winning)x(Pauls picking the winner corectly) + (A's chances of loosing) x (Paul picks wrongly) = 23×23+13×13=59
10. 36 people {a1, a2… a36} meet and shake hands in a circular fashion. In other words, there are totally 36 handshakes involving the pairs, {a1, a2}, {a2, a3}, …, {a35, a36}, {a36, a1}. Then size of the smallest set of people such that the rest have shaken hands with at least one person in the set is
a) 12
b) 11
c) 13
d) 18
Ans: {a1, a2}, {a2, a3},{a3, a4}, {a4, a5},{a5, a6}, {a6, a7} …, {a35, a36}, {a36, a1}
From the above arrangement, If we separate a3, a6, a9, .....a36. Total 12 persons the reamining persons must have shaked hand with atleast one person. So answer is 12.
11. There are two boxes, one containing 10 red balls and the other containing 10 green balls. You are allowed to move the balls between the boxes so that when you choose a box at random and a ball at random from the chosen box, the probability of getting a red ball is maximized. This maximum probability is
If rearrangement is not allowed, then actual probability of picking up a red ball = 12(10)+12(0)=12
As we are allowed to move the ball, we keep only 1 red in the first box, and shirt the remaining 9 to the second.
So = 12(1)+919(0)=1419
12. The difference between two no is 9 and the product of the two is 14. What is the square of their sum?
We know that (a+b)2=(a−b)2 + 4ab
Substituting a - b = 9, and ab = 14, (a+b)2=(9)2+4(14)=137
13. There are two water tanks A and B, A is much smaller than B. While water fills at the rate of one liter every hour in A, it gets filled up like 10, 20, 40, 80, 160 in tank B. (At the end of first hour, B has 10 liters, second hour it has 20, third hour it has 40 and so on). If tank B is 1/32 filled after 21 hours, what is the total duration required to fill it completely?
Ans: The data related to the first tank A is not necessary. As you can see, the capacity that gets filled in the tank B after each hour is doubled. So If the tank is 1/32nd part is full after 21 hours, it is 1/16th part full after 22 hours, 1/8th part full after 23 hours, 1/4th part full after 24 hours, 1/2 full after 25 hours, completely full after 26 hours.
14. 3 friends A, B, C went for week end party to McDonald’s restaurant and there they measure there weights in some order In 7 rounds. A, B, C, AB, BC, AC, ABC. Final round measure is 155kg then find the average weight of all the 7 rounds?
Average weight = [(a + b + c + (a+b) + (b+c) + (c+a)+(a+b+c)] / 7 = 4 (a+b+c) /7 = 4 x 155/7 = 88.5 kgs
15. A grand father has 3 grand children. Age difference of two children among them is 3. Eldest child age is 3 times the youngest child’s age and the eldest child age is two year more than the sum of age of other two children. What is the age of the eldest child?
Ans: As the eldest son's age is 3 times that of the youngest, eldest son's age should be a multiple of 3. From the given options take 15 as the eldest son's age. Then youngest son's age becomes 5. But Eldest sons age is 2 more than the sum of the remaining two sons. So Sum of the remaining two sons is 13. So the age of the middle son is 13 - 5 = 8. Which satisfies another condition in the question that the difference between the two sons age is 3. So answer is 15.
16. In a mixture of a, b and c, if a and b are mixed in 3:5 ratio and b and c are mixed in 8:5 ratio and if the final mixture is 35 liters, find the amount of b?
Ans: As b is common in both ratios, we should equate b in both ratios by multiplying suitable numbers.
a:b = 3 : 5 = 24 : 40
b:c = 8 : 5 = 40 : 25
Now a : b : c = 24 : 40 : 25.
Amount of b in the mixture = 4089×35 = 15.73
17. After the typist writes 12 letters and addresses 12 envelopes, she inserts the letters randomly into the envelopes (1 letter per envelope). What is the probability that exactly 1 letter is inserted in an improper envelope?
Ans: Tricky one but simple. How do you put exactly 1 letter in the wrong envelope? we need minimum two. So answer is 0.
18. 10 suspects are rounded by the police and questioned about a bank robbery. Only one of them is guilty. The suspects are made to stand in a line and each person declares that the person next to him on his right is guilty. The rightmost person is not questioned. Which of the following possibilities are true?
A. All suspects are lying.
B. leftmost suspect is innocent.
C. leftmost suspect is guilty
a) A only
b) A or C
c) A or B
d) B only
There are only 2 cases. Either left one is guilty or one of the remaining 9 to his right is guilty.
So If the left most is guilty, All the statements including the guilty one are lies. A and C are correct.
Or If Any one except left most one is guilty, Then one of the statements given by the person should be true. In this case all the suspects are lying does not hold. So If B is correct, A is not correct. i.e., only A or B is correct. Option C is correct.
19. A hollow cube of size 5 cm is taken, with a thickness of 1 cm. It is made of smaller cubes of size 1 cm. If 1face of the outer surface of the cube are painted, totally how many faces of the smaller cubes remain unpainted?
The Hallow cube volume = n3−(n−2)2, Here n is the number of small cubes lie on the big cube edge.
Now n = 5 so Hallow cube volume = 53−(5−2)2=125−27=98
So 98 small cubes required to make a hallow cube of size 5 cm. Now total surfaces = 6 x 98 = 588
Now if the bigger cube is painted 4 sides, total 4 x 25 small faces got paint. So remaining small faces which does not have paint after cutting is 588 - 100 = 488
20. My flight takes of at 2am from a place at 18N 10E and landed 10 Hrs later at a place with coordinates 36N70W. What is the local time when my plane landed?
So Hare : tortoise speeds = 25 : 135 = 5 : 27
Now Hare and tortoise has to reach the starting point means, Hare has to cover 175 meters and Tortoise has to cover only 25 meters in the same time.
As time =DistanceSpeed=2527=1755×K
Ie., Hare has to increase its speed by a factor K. Solving we get K = 37.8
9. For the FIFA world cup, Paul the octopus has been predicting the winner of each match with amazing success. It is rumored that in a match between 2 teams A and B, Paul picks A with the same probability as A’s chances of winning. Let’s assume such rumors to be true and that in a match between Ghana and Bolivia; Ghana the stronger team has a probability of 2/3 of winning the game. What is the probability that Paul will correctly pick the winner of the Ghana-Bolivia game?
a) 1/9
b) 4/9
c) 5/9
d) 2/3
The probability that Paul correctly picks the winner = (A's Chances of winning)x(Pauls picking the winner corectly) + (A's chances of loosing) x (Paul picks wrongly) = 23×23+13×13=59
10. 36 people {a1, a2… a36} meet and shake hands in a circular fashion. In other words, there are totally 36 handshakes involving the pairs, {a1, a2}, {a2, a3}, …, {a35, a36}, {a36, a1}. Then size of the smallest set of people such that the rest have shaken hands with at least one person in the set is
a) 12
b) 11
c) 13
d) 18
Ans: {a1, a2}, {a2, a3},{a3, a4}, {a4, a5},{a5, a6}, {a6, a7} …, {a35, a36}, {a36, a1}
From the above arrangement, If we separate a3, a6, a9, .....a36. Total 12 persons the reamining persons must have shaked hand with atleast one person. So answer is 12.
11. There are two boxes, one containing 10 red balls and the other containing 10 green balls. You are allowed to move the balls between the boxes so that when you choose a box at random and a ball at random from the chosen box, the probability of getting a red ball is maximized. This maximum probability is
If rearrangement is not allowed, then actual probability of picking up a red ball = 12(10)+12(0)=12
As we are allowed to move the ball, we keep only 1 red in the first box, and shirt the remaining 9 to the second.
So = 12(1)+919(0)=1419
12. The difference between two no is 9 and the product of the two is 14. What is the square of their sum?
We know that (a+b)2=(a−b)2 + 4ab
Substituting a - b = 9, and ab = 14, (a+b)2=(9)2+4(14)=137
13. There are two water tanks A and B, A is much smaller than B. While water fills at the rate of one liter every hour in A, it gets filled up like 10, 20, 40, 80, 160 in tank B. (At the end of first hour, B has 10 liters, second hour it has 20, third hour it has 40 and so on). If tank B is 1/32 filled after 21 hours, what is the total duration required to fill it completely?
Ans: The data related to the first tank A is not necessary. As you can see, the capacity that gets filled in the tank B after each hour is doubled. So If the tank is 1/32nd part is full after 21 hours, it is 1/16th part full after 22 hours, 1/8th part full after 23 hours, 1/4th part full after 24 hours, 1/2 full after 25 hours, completely full after 26 hours.
14. 3 friends A, B, C went for week end party to McDonald’s restaurant and there they measure there weights in some order In 7 rounds. A, B, C, AB, BC, AC, ABC. Final round measure is 155kg then find the average weight of all the 7 rounds?
Average weight = [(a + b + c + (a+b) + (b+c) + (c+a)+(a+b+c)] / 7 = 4 (a+b+c) /7 = 4 x 155/7 = 88.5 kgs
15. A grand father has 3 grand children. Age difference of two children among them is 3. Eldest child age is 3 times the youngest child’s age and the eldest child age is two year more than the sum of age of other two children. What is the age of the eldest child?
Ans: As the eldest son's age is 3 times that of the youngest, eldest son's age should be a multiple of 3. From the given options take 15 as the eldest son's age. Then youngest son's age becomes 5. But Eldest sons age is 2 more than the sum of the remaining two sons. So Sum of the remaining two sons is 13. So the age of the middle son is 13 - 5 = 8. Which satisfies another condition in the question that the difference between the two sons age is 3. So answer is 15.
16. In a mixture of a, b and c, if a and b are mixed in 3:5 ratio and b and c are mixed in 8:5 ratio and if the final mixture is 35 liters, find the amount of b?
Ans: As b is common in both ratios, we should equate b in both ratios by multiplying suitable numbers.
a:b = 3 : 5 = 24 : 40
b:c = 8 : 5 = 40 : 25
Now a : b : c = 24 : 40 : 25.
Amount of b in the mixture = 4089×35 = 15.73
17. After the typist writes 12 letters and addresses 12 envelopes, she inserts the letters randomly into the envelopes (1 letter per envelope). What is the probability that exactly 1 letter is inserted in an improper envelope?
Ans: Tricky one but simple. How do you put exactly 1 letter in the wrong envelope? we need minimum two. So answer is 0.
18. 10 suspects are rounded by the police and questioned about a bank robbery. Only one of them is guilty. The suspects are made to stand in a line and each person declares that the person next to him on his right is guilty. The rightmost person is not questioned. Which of the following possibilities are true?
A. All suspects are lying.
B. leftmost suspect is innocent.
C. leftmost suspect is guilty
a) A only
b) A or C
c) A or B
d) B only
There are only 2 cases. Either left one is guilty or one of the remaining 9 to his right is guilty.
So If the left most is guilty, All the statements including the guilty one are lies. A and C are correct.
Or If Any one except left most one is guilty, Then one of the statements given by the person should be true. In this case all the suspects are lying does not hold. So If B is correct, A is not correct. i.e., only A or B is correct. Option C is correct.
19. A hollow cube of size 5 cm is taken, with a thickness of 1 cm. It is made of smaller cubes of size 1 cm. If 1face of the outer surface of the cube are painted, totally how many faces of the smaller cubes remain unpainted?
The Hallow cube volume = n3−(n−2)2, Here n is the number of small cubes lie on the big cube edge.
Now n = 5 so Hallow cube volume = 53−(5−2)2=125−27=98
So 98 small cubes required to make a hallow cube of size 5 cm. Now total surfaces = 6 x 98 = 588
Now if the bigger cube is painted 4 sides, total 4 x 25 small faces got paint. So remaining small faces which does not have paint after cutting is 588 - 100 = 488
20. My flight takes of at 2am from a place at 18N 10E and landed 10 Hrs later at a place with coordinates 36N70W. What is the local time when my plane landed?
a) 12 noon
b) 6: 40 AM
c) 5: 20 PM
d) 6:50 AM
b) 6: 40 AM
c) 5: 20 PM
d) 6:50 AM
Remember, while moving from east to west countries lag in time. Remember when Test cricket starts in England? 3. 30 in afternoon. Right? ie., We are in after noon means they are in morning.
If the coordinates change from 10 E to 70W, the plane has moved a total of 80 degrees. We know that with each degree time increases by 4 minutes while going from east to west. (How? 24 x 60 min / 360 degrees, So 1 degree = 4 min)
So total time change = 4 x 80 = 320 min = 5 hrs + 20 minutes.
After 10 hours local time is (2 am + 10 - 5.20 hrs) = 6.40 AM.
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