Tuesday, February 26, 2013
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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...
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%
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
TCS latest pattern 2013
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...
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%
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
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
Answer: D
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
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
Answer: D
1. M, N, O and P are all different
individuals; M is the daughter of N; N is the son of O; O is the father
of P; Among the following statements, which one is true?
A. M is the daughter of P
B. If B is the daughter of N, then M and B are sisters
C. If C is the granddaughter of O, then C and M are sisters
D. P and N are bothers.
Answer: B
From the diagram it is clear that If B is the daughter of N, then M and B are sisters. Rectangle indicates Male, and Oval indicates Female.
12. In the adjoining diagram, ABCD and EFGH are squres of side 1 unit such that they intersect in a square of diagonal length (CE) = 1/2. The total area covered by the squares is
A. Cannot be found from the information
B. 1 1/2
C. 1 7/8
D. None of these
Answer: C
Let CG = x then using pythogerous theoremCG2+GE2=CE2
⇒ x2+x2=(1/2)2⇒2x2=1/4⇒x2=1/8
Total area covered by two bigger squares = ABCD + EFGE - Area of small square = 2 - 1/8 = 15/8
13. There are 10 stepping stones numbered 1 to 10 as shown at the side. A fly jumps from the first stone as follows; Every minute it jumps to the 4th stone from where it started - that is from 1st it would go to 5th and from 5th it would go to 9th and from 9th it would go to 3rd etc. Where would the fly be at the 60th minute if it starts at 1?
A. 1
B. 5
C. 4
D. 9
Answer : A
Assume these steps are in circular fashion.
Then the fly jumps are denoted in the diagram. It is clear that fly came to the 1st position after 5th minute. So again it will be at 1st position after 10th 15th .....60th. min.
So the fly will be at 1st stone after 60th min.
14. What is the remainder when617+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 x72 + 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
Answer: A
We know thatA1 = 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.
AssumeAn = an2+bn+c
NowA1 = 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 inAn = n2+bn+c = n2−n+2
Substitute 100 in the above equation we get 9902.
17.Find the number of rectangles from the adjoining figure (A square is also considered a rectangle)
A. 864
B. 3276
C. 1638
D. None
Answer: C
To form a rectangle we need two horizontal lines and two vertical lines. Here there are 13 vertical lines and 7 horizontal lines. The number of ways of selecting 2 lines from 13 vertical lines is13C2 and the number of ways of selecting 2 lines from 7 horizontals is 7C2 . So total rectangles = 7C2x13C2
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
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
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
Answer: A or C (But go for 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!
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
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
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
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
Answer: A
A. M is the daughter of P
B. If B is the daughter of N, then M and B are sisters
C. If C is the granddaughter of O, then C and M are sisters
D. P and N are bothers.
Answer: B
From the diagram it is clear that If B is the daughter of N, then M and B are sisters. Rectangle indicates Male, and Oval indicates Female.
12. In the adjoining diagram, ABCD and EFGH are squres of side 1 unit such that they intersect in a square of diagonal length (CE) = 1/2. The total area covered by the squares is
A. Cannot be found from the information
B. 1 1/2
C. 1 7/8
D. None of these
Answer: C
Let CG = x then using pythogerous theorem
Total area covered by two bigger squares = ABCD + EFGE - Area of small square = 2 - 1/8 = 15/8
13. There are 10 stepping stones numbered 1 to 10 as shown at the side. A fly jumps from the first stone as follows; Every minute it jumps to the 4th stone from where it started - that is from 1st it would go to 5th and from 5th it would go to 9th and from 9th it would go to 3rd etc. Where would the fly be at the 60th minute if it starts at 1?
A. 1
B. 5
C. 4
D. 9
Answer : A
Assume these steps are in circular fashion.
Then the fly jumps are denoted in the diagram. It is clear that fly came to the 1st position after 5th minute. So again it will be at 1st position after 10th 15th .....60th. min.
So the fly will be at 1st stone after 60th min.
14. What is the remainder when
A. 1
B. 6
C. 0
D. 3
Answer: C
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,
So
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
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
A. 9902
B. 9900
C. 10100
D. 9904
Answer: A
We know that
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
Now
Solving above equations we get a = 1, b = - 1 and C = 2
So substituting in
Substitute 100 in the above equation we get 9902.
17.Find the number of rectangles from the adjoining figure (A square is also considered a rectangle)
A. 864
B. 3276
C. 1638
D. None
Answer: C
To form a rectangle we need two horizontal lines and two vertical lines. Here there are 13 vertical lines and 7 horizontal lines. The number of ways of selecting 2 lines from 13 vertical lines is
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
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
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
Answer: A or C (But go for 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!
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
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
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
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
Answer: A
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