I. 56x^2 – 127x + 72 = 0 II. 6y^2 – 17y + 12 = 0 1 : if x ≥ y 2 : if x ≤ y 3 : if x > y 4 : if x < y 5 : if x = y or relationship cannot be established

I. 56x^2 – 127x + 72 = 0

II. 6y^2 – 17y + 12 = 0

1 : if x ≥ y

2 : if x ≤ y

3 : if x > y

4 : if x < y

5 : if x = y or relationship cannot be established
by

1 Answer

Answer :

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4 : if x < y

by

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I. 63x^2 – 275x + 300 = 0 II. 18y^2 – 85y + 100 = 0 1 : if x ≥ y 2 : if x ≤ y 3 : if x > y 4 : if x < y 5 : if x = y or relationship cannot be established

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I. 3x^2 – 47x + 184 = 0 II. 3y^2 – 38y + 119 = 0 1 : if x ≥ y 2 : if x ≤ y 3 : if x > y 4 : if x < y 5 : if x = y or relationship cannot be established

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I. 6x^2 – 71x + 195 = 0 II. 12y^2 – 97y + 195 = 0 1 : if x ≥ y 2 : if x ≤ y 3 : if x > y 4 : if x < y 5 : if x = y or relationship cannot be established

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I. 2a^2 − 11a + 12 = 0 II. 35b^2 − 18b − 8 = 0 1 : if a > b 2 : if a < b 3 : if a ≥ b 4 : if a ≤ b 5 : if a = b or the relation between a and b cannot be established.

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Last Answer : 1 : if a > b

I. 32a^2 + 52a + 15 = 0 II. 8b^2 + 38b + 35 = 0 1 : if a > b 2 : if a < b 3 : if a ≥ b 4 : if a ≤ b 5 : if a = b or the relation between a and b cannot be established.

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I. 12a^2 − a − 6 = 0 II. 8b^2 − 26b + 15 = 0 1 : if a > b 2 : if a < b 3 : if a ≥ b 4 : if a ≤ b 5 : if a = b or the relation between a and b cannot be established.

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I. 28a^2 + 25a − 8 = 0 II. 5b^2 − 13b − 6 = 0 1 : if a > b 2 : if a < b 3 : if a ≥ b 4 : if a ≤ b 5 : if a = b or the relation between a and b cannot be established.

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Last Answer : 5 : if a = b or the relation between a and b cannot be established.  

I. 20a^2 + 51a + 28 = 0 II. 15b^2 − 61b + 56 = 0 1 : if a > b 2 : if a < b 3 : if a ≥ b 4 : if a ≤ b 5 : if a = b or the relation between a and b cannot be established.

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Naphthenic acid is represented by (A) CnH2n+2O2 (B) CnH2n-2O2 (C) CnH2n+2O2 (n ≥ 6) (D) CnH2n+6O2 (n ≤ 6)

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The grammar with production rules S → aSb |SS|λ generates language L given by: (A) L = {w∈{a, b}* | na(w) = nb(w) and na(v) ≥ nb(v) where v is any prefix of w} (B) L = {w∈{a, b}* | na(w) = nb(w) and na(v) ≤ nb(v) where v is any prefix of w} (C) L = {w∈{a, b}* | na(w) ≠ nb(w) and na(v) ≥ nb(v) where v is any prefix of w} (D) L = {w∈{a, b}* | na(w) ≠ nb(w) and na(v) ≤ nb(v) where v is any prefix of w}

Description : The grammar with production rules S → aSb |SS|λ generates language L given by: (A) L = {w∈{a, b}* | na(w) = nb(w) and na(v) ≥ nb(v) where v is any prefix of w} (B) L = {w∈{a, b}* | na(w) = nb(w) and na(v ... (D) L = {w∈{a, b}* | na(w) ≠ nb(w) and na(v) ≤ nb(v) where v is any prefix of w}

Last Answer : (A) L = {w∈{a, b}* | na(w) = nb(w) and na(v) ≥ nb(v) where v is any prefix of w} 

A set of processors P1, P2, ......, Pk can execute in parallel if Bernstein's conditions are satisfied on a pair wise basis; that is P1 || P2 || P3 || ..... || Pk if and only if: (A) Pi || Pj for all i ≠ j (B) Pi || Pj for all i = j+1 (C) Pi || Pj for all i ≤ j (D) Pi || Pj for all i ≥ j

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Last Answer : (A) Pi || Pj for all i ≠ j Explanation: Bernstein's Condition: 1. If process Pi writes to a memory cell Mi, then no process Pj can read the cell Mi. 2. If process Pi read from a memory ... Mi. 3. If process Pi writes to a memory cell Mi, then no process Pj can write to the cell Mi.

19/14 x 12/72 x 18/29 x 767 = ? (Take 2 digits after the decimal point) a) 110.69 b) 118.78 c) 111.56 d) None e) 108.98

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X says to Y,”I am 3 times as old as you were 3 years ago”.How old is X? I. Y ‘s age 17 years from now is same as X’s present age. II. X’s age nine years from now is 3 times Y’s present age. If the question can be answered with the help of statement I alone, If the question can be answered with the help of statement II alone, If both,statement I and statement II are needed to answer the question If the question cannot be answered even with the help of both the statements. If the data in either statement I or statement II alone is sufficient to answer the question

Description : X says to Y, I am 3 times as old as you were 3 years ago .How old is X? I. Y s age 17 years from now is same as X's present age. II. X's age nine years from now is ... with the help of both the statements. If the data in either statement I or statement II alone is sufficient to answer the question

Last Answer : Let the current ages of X and Y be a and b respectively From the question: A=3(a-b) From statement I: A=b + 17 B = a-17 Therefore, a = 3(a-17-3) a=3a-60 2a=60 A=30 Hence, X ... a+9)/3 Substituting this in first equation, A=a+9-9 Hence, no result Therefore, statement I alone is sufficient Answer :a

An electric field is given as E = 6y 2 z i + 12xyz j + 6xy 2 k. An incremental path is given by dl = -3 i + 5 j – 2 k mm. The work done in moving a 2mC charge along the path if the location of the path is at p(0,2,5) is (in Joule) a) 0.64 b) 0.72 c) 0.78 d) 0.80

Description : An electric field is given as E = 6y 2 z i + 12xyz j + 6xy 2 k. An incremental path is given by dl = -3 i + 5 j – 2 k mm. The work done in moving a 2mC charge along the path if the location of the path is at p(0,2,5) is (in Joule) a) 0.64 b) 0.72 c) 0.78 d) 0.80

Last Answer : b) 0.72

Sum of present age of A, B and C is 72 years. If 4 years ago, their ages were in the ratio 1 : 2 : 3, find A’s present age. a) 7 years b) 10 years c) 12 years d) 14 years e) None of these

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In how many different ways can the letters of the word DESIGN be arranged so that the vowels are at the two ends? a) 48 b) 66 c) 12 d) 72 e) 96

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187.5, 182, 171, ?, 132.5, 105, 72 a) 145.5 b) 155.5 c) 154.5 d) 144.5 e) None of these

Description : 187.5, 182, 171, ?, 132.5, 105, 72 a) 145.5 b) 155.5 c) 154.5 d) 144.5 e) None of these

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The sum of ages of family members (both children and parents) is 360 years.The total ages of children and parents are in the ratio 2:1 and the ages of wife and husband are in the ratio 5:7.What will be the age of husband? A.65 B.75 C.72 D.70

Description : The sum of ages of family members (both children and parents) is 360 years.The total ages of children and parents are in the ratio 2:1 and the ages of wife and husband are in the ratio 5:7.What will be the age of husband? A.65 B.75 C.72 D.70

Last Answer : Answer : D (70) ExplanationSum of ages is 360 years.The ratio of children and parents ages is 2:1. Total age of parents = 360 x 1 / 3 = 120 years Ratio of wife and husband age is 5:7. Therefore, the age of husband = 120 x 7 / 12 = 70 Hence the age of husband is 70 years.

The sum of ages of family members (both children and parents) is 360 years.The total ages of children and parents are in the ratio 2:1 and the ages of wife and husband are in the ratio 5:7.What will be the age of husband? A.65 B.75 C.72 D.70

Description : The sum of ages of family members (both children and parents) is 360 years.The total ages of children and parents are in the ratio 2:1 and the ages of wife and husband are in the ratio 5:7.What will be the age of husband? A.65 B.75 C.72 D.70

Last Answer : Answer : D (70) ExplanationSum of ages is 360 years.The ratio of children and parents ages is 2:1. Total age of parents = 360 x 1 / 3 = 120 years Ratio of wife and husband age is 5:7. Therefore, the age of husband = 120 x 7 / 12 = 70 Hence the age of husband is 70 years.

The average weight of four boys A, B, C, and D is 75 kg. The fifth boy E is included and the average weight decreases by 4 kg. A is replaced by F. The weight of F is 6 kg more than E. Average weight decreases because of the replacement of A and now the average weight is 72 kg. Find the weight of A. 1) 57 kg 2) 54 kg 3) 56 kg 4) 60 kg 5) 58 kg

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Last Answer : 3) 56 kg

24+(72 ÷ 48×8+14) × 36 ÷ (3×4) + 36 ÷ 3 ×4 a) 120 b) 140 c) 160 d) 180 e) None of these

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72× 25 + 45× 96 = 163 + ? a) 1912 b) 2024 c) 1844 d) 1908 e) 1960

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1824 ÷ 32 + 5040 ÷ ? – 5600 ÷ 35 = 37 a) 36 b) 48 c) 60 d) 72 e) 84

Description : 1824 ÷ 32 + 5040 ÷ ? – 5600 ÷ 35 = 37 a) 36 b) 48 c) 60 d) 72 e) 84

Last Answer : 57 + 5040 ÷ ? – 160 = 37 5040 ÷ ? = 37 – 57 + 160 = 140 = 5040 ÷ 140 = 36 Answer: a)

53 × 47 + 64 × 56 -72 × 68 = ? a) 1221 b) 1403 c) 1249 d) 1337 e) 1179

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Last Answer : Answer: e)

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?% of 725 + 14% of 2100 = 700 a) 64 b) 80 c) 72 d) 56 e) 48

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The average of weight of three men A,B and C is 84 kg. Another man D joins the group and the average now becomes 80 kg. If another man E, whose weight is 3 kg.more than that of D, replaces A, then the average weight of B,C,D and E becomes 79 kg. The weight of A is : a) 70 kg b) 72 kg c) 75 kg d) 80 kg e) 78 kg

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Last Answer : A+B+C = 3×84=252 A+B+C+D= (4×80)=320 D = (320-252)=68 and E = (68+3)=71 Now, B+C+D+E = (4×79)=316 B+C+D=(316-71)=245 kg So, A = (320-245)=75 kg Answer: c)

A vessel contains 100 litres of milk. From this, 10 litres of milk was taken out and replaced by water. The process was repeated further two times. How much milk is now contained by the vessel? a) 72.9 litres b) 60.4 litres c) 71.6 litres

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A man distributed 43 chocolates to his children. How many of his children are more than five years old? I. A child older than five years gets 5 chocolates II. A child 5 years or younger in age gets 6 chocolates. If the question can be answered with the help of statement I alone, If the question can be answered with the help of statement II alone, If both,statement I and statement II are needed to answer the question If the question cannot be answered even with the help of both the statements. If the data in either statement I or statement II alone is sufficient to answer the question

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Is it more profitable for Company M to produce Q? I. Product R sells at a price four times that of Q II. One unit of Q requires 2 units of labour, while one unit of R requires 5 units of labour. There is a no other constraint on production. If the question can be answered with the help of statement I alone, If the question can be answered with the help of statement II alone, If both,statement I and statement II are needed to answer the question If the question cannot be answered even with the help of both the statements. If the data in either statement I or statement II alone is sufficient to answer the question

Description : Is it more profitable for Company M to produce Q? I. Product R sells at a price four times that of Q II. One unit of Q requires 2 units of labour, while one unit of R requires 5 ... both the statements. If the data in either statement I or statement II alone is sufficient to answer the question

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A particle moves along the curve `6y = x^3 + 2`. Find the points on the curve at which y-co-ordinate is changing 8 times as fast as the x-co-ordinate.

Description : A particle moves along the curve `6y = x^3 + 2`. Find the points on the curve at which y-co-ordinate is changing 8 times as fast as the x-co-ordinate.

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The profit P, in dollars, of selling x widgets and y gadgets is given by the function P(x, y) = 7x 6y. Which corner point of the shaded region will maximize the profit?

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A rectangular surface 3' × 4', has the lower 3 edge horizontal and 6' below a free oil surface (sp. gr. 0.8). The surface inclination is 300 with the horizontal. The force on one side of the surface is (where, y = specific weight of water) (A) 39.6y (B) 48y (C) 49.2y (D) 58y

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A and B established a firm together. A’s investment was thrice that of B’s. A also kept the investment for twice as much time as B. If B got a profit of 4000, what was the total profit? A.30,000 B.28,000 C.40,000 D.45,000

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What is the distance from a point on the x axis to the centre of a circle when a tangent line at the point 3 4 meets the circle of x2 plus y2 -2x -6y plus 5 equals 0?

Description : What is the distance from a point on the x axis to the centre of a circle when a tangent line at the point 3 4 meets the circle of x2 plus y2 -2x -6y plus 5 equals 0?

Last Answer : Circle equation: x^2 +y^2 -2x -6y +5 = 0Completing the squares: (x-1)^2 +(y-3)^2 = 5Centre of circle: (1, 3)Tangent line meets the x-axis at: (0, 5)Distance from (0, 5) to (1, 3) = 5 units using the distanceformula

Francis and George started a business with investments of Rs.7000 and Rs.10000. After x months, Francis left. After two more months David joined the business with an investment of Rs.10000. If the annual profit is shared among David, Francis and George in the ratio 10:7:24, find the respective time periods of David, Francis and George for which they stayed that year. a)5:5:12 b)1:1:3 c)1:1:4 d)1:1:2

Description : Francis and George started a business with investments of Rs.7000 and Rs.10000. After x months, Francis left. After two more months David joined the business with an investment of Rs.10000. If the annual profit is shared among David, ... which they stayed that year. a)5:5:12 b)1:1:3 c)1:1:4 d)1:1:2

Last Answer : a)5:5:12

.12 X 13+105% of 993 +879/18 +15 (1)1150 (2)1170 (3)1185 (4)1200 (5) 1215 .12 X 13+105% of 993 +879/18 +15 (1)1150 (2)1170 (3)1185 (4)1200 (5) 1215

Description : .12 X 13+105% of 993 +879/18 +15 (1)1150 (2)1170 (3)1185 (4)1200 (5) 1215

Last Answer : (4)1200

√(4.9 − 0.49) = ? a) 10.42 b) 5.84 c) 12.84 d) 2.1 e) None of these

Description : √(4.9 − 0.49) = ? a) 10.42 b) 5.84 c) 12.84 d) 2.1 e) None of these

Last Answer : Answer: d)

If x:y =3:4,in the ratio of (7x-4y) : ( 3x + y) a) 2:11 b) 4:9 c) 5:13 d) 6:5 e) 13:5

Description : If x:y =3:4,in the ratio of (7x-4y) : ( 3x + y) a) 2:11 b) 4:9 c) 5:13 d) 6:5 e) 13:5

Last Answer : An easy way to solve this question is use x = 3, y=4 7x – 4y = 21 – 16 = 5 3x + y = 9+4 = 13 Required ratio = 5:13 Answer: c)

Three pipes P, Q and R can fill a tank from empty to full in 40 minutes, 15 minutes, and 30 minutes respectively. When the tank is empty, all the three pipes are opened. P, Q and R discharge chemical solutions X, Y and Z respectively. What is the proportion of the solution Y in the liquid in the tank after 5 minutes? a) 8 / 15 b) 7 / 15 c) 8 / 17 d) 6 / 13 e) 8 / 13

Description : Three pipes P, Q and R can fill a tank from empty to full in 40 minutes, 15 minutes, and 30 minutes respectively. When the tank is empty, all the three pipes are opened. P, Q and R discharge chemical solutions X, Y and Z respectively. ... 5 minutes? a) 8 / 15 b) 7 / 15 c) 8 / 17 d) 6 / 13 e) 8 / 13

Last Answer : a Part of the tank filled by pipe P in 1 minute = 1 / 40 Part of the tank filled by pipe Q in 1 minute = 1 / 15 Part of the tank filled by pipe R in 1 minute = 1/ 30 Here we have to find the proportion of ... together in 5 minute = 5 1/8 = 5/ 8 Required proportion = (1/3) / ( 5/8) = 8 / 15