50 Must Solve Algebra Questions

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1 50 Must Solve Algebra Questions

2 50 Must Solve Algebra Questions 1. Classic furniture gallery employs male and female carpenters to create designer chairs for their stores. 5 males and 3 females can create as many chairs in 4 days as 3 males and 3 females in 5 days. If one man and one woman work together and complete an assignment of creating 400 chairs in 50 days, then how many chairs can be created by 6 males and 2 females in 7 days? 2. What is the minimum value of the function f(z) = (z 3) 2 + (5 z) 2? 3. By what minimum distance should the point (1, 3) be shifted so that it is collinear with the points ( 7, 1) and ( 3, 3)? A. 2 units B units C. 3 units D.3 2 units 4. There are 14 baskets of peaches and some baskets of plums. Each basket has only one kind of fruit. All the children of Montgomery School came and took some peaches and plums from the respective baskets. The number of students taking peaches from each basket is 6 and the number of students taking plums from each basket is 8. Also, each child took peaches from exactly 7 baskets and plums from exactly 4 baskets. How many baskets of plum were there? 5. In how many points do the line y = 2 and the curve y = x2 6x + 6 meet? A.2 B.3 C..4 D.5 6. If (a + c) 2 2b(a + c) + b 2 = 0, then which of the following is true? A. a = b + c B.. b = a + c C. c = a + b D. None of these 7. The sum of the roots of the equation x 2 17 x + 66 = 0 is. A..0 B.17 C.34 D What values of x satisfy this inequality A.x < 3 B..x < 3 4 4x 3 x 6x + 10 <0? C.x > 3 4 D.x > 3 9. If x, y > 0 and 3x 2 + 5y 2 = 10, then find the maximum value of 125x 4 y 6. A.144 B.288 C..384 D.768

3 10. Jacob has a certain number of 2-rupee and 1-rupee coins. Thickness of each 2-rupee coin is same as thickness of each 1-rupee coin. Inside a closed cylindrical container, he is able to exactly fit all 2- rupee coins and all but twenty 1-rupee coins when they are piled up one above the other. If the height of the container is increased by 20%, he is able to exactly fit all 1-rupee coins and all but ten 2-rupee coins when they are piled up one above the other. What is the total number of 2-rupee and 1-rupee coins available with Jacob? A. 50 B..70 C.80 D. Insufficient data 11. If 1 2x2 1 x2, then which of the following is true? A. x 2 3 B.. x 6 3 C. x 6 3 D. x In ABC the medians AD, BE and CF meet at G. The coordinates of A, B and G are (4, 7), (2, 3) and (4, 5) respectively. Find the length of the median CF. A.2 units B..3 units C.4 units D.5 units 13. In the year 2018, Raj's father was thrice as old as Raj was. In the year 2033, Raj's father will be twice as old as Raj will be then. In which Calendar year will Raj's father be 2.5 times as old as Raj will be then? A.2021 B.2022 C D A..40/41 B.36/37 C.44/45 D.32/ How many 2-digit numbers in base 4 are 2-digit numbers with the same digits (but not necessarily in the same order) in base 10? A.10 B.9 C..1 D X is p% more than Y and Z is p% less than Y. If X is 4p%more than Z, then find p. (Note: p 0) A.20 B.40 C.60 D If the roots of the equation x 3 px 2 + qx r = 0 are in the ratio 2 : 5 : 7 and r = 1890, then find the value of p + q.

4 18. A. B.. C. D. 19. If x, y > 0, x y and xy = 9, then find the minimum value of [x 2 ] + [y 2 ]. Note: [x] is the greatest integer less than or equal to x. A..17 B.16 C.19 D If A is a natural number less than 250, then for how many values of A is the inequality (A 100) (A 102) (A 104) (A 106)...(A 200) > 0 satisfied? A.25 B.49 C..74 D For a quadratic equation with roots a and b, the sum and the product of the roots is 4 and -3 respectively. Which of the following can be the sum of the roots of the quadratic equation whose product of the roots is given by a 2 (1 b) + 2ab + b 2 (1 a)? [Both the roots of the quadratic equation are positive integers.] A.11 B.7 C.4 D Consider the following two equations 2x 3 2y = 0 3x + 2y = 1 If points (x1, y1) and (x2, y2) are the two values of (x, y) respectively that satisfy both the equations, what is the value of x1 x2 y1 y2? A.0 B.77 C..154 D Given that a 3b =2. Which of the following is equal to A.2 B.9b C.12b D.6b 24. If x 2 5x + 6 < x 2 9, then which of the following is always true? A. 2x 2 + 3x + 1 < 2 B. 2x 2 + x + 1 > 3/2 C. Both the above D. None of the above 25. If p q 2 = 4 and p 2 + q 4 = 26, where p and q are positive and real, then the value of p is:

5 26. If (x + 3)(y 1) = 36, where x > 0 and y > 1, what is the minimum value of (x + y)? 27. The sum of a positive number i and its reciprocal is 8 less than thrice the sum of another 15 number j and the reciprocal of j. If i = 3j,then the value of j is: A. Between 8 and 12 B. Between 23 and 27 C. Between 33 and 37 D. Between 13 and Two friends tried solving a quadratic equations z 2 + bz + c = 0. One started with the wrong value of b and got the roots as 4 and 14 while the other started with the wrong value of c and got the roots as 17 and 2. Find the actual roots. A.7, 8 B.14, 1 C.19, 4 D.13, Given that a 2 xa b = 0 and b 2 a xb = 0. If a b, then what is the value of ab? Here x is a non-zero constant. A. x 1 B.x 2 1 C.1 x 2 D.1 x 30. Given that b is one of the roots of the equation p(p 2)(p 1) = a, where a and b are constants. Which of the following expressions is equal to 2b a 3 b? A. b 2 B. b 2 1 C.1 b 3 D. b A two-digit number is the sum of the product of its digits and the sum of its digits. The units digit of the number is 32. Four boys would like to buy a comic book which costs Rs.60. They have just enough money combined to pay for it. A has the same amount as B and C combined. D has Rs.10 less than A and C combined. The amount with A is greater than that with C by Rs.10. How much did D pay? 33. Find the remainder when x 3 5x 2 + 7x 9 is divided by x 2 + x 6. A.19x 45 B. 19x 38 C. 19x 49 D. 19x For a natural number i> 1, the number i 9 + 9i is definitely: I. prime II. even III. composite IV. odd A. I and II B. I and IV C. II and III D.. III and IV 35. Four people P, Q, R and S have together got Rs.100 with them. P and Q have got as much money as R and S put together. P has got more money than Q. R has half as money as S has. P has Rs.5 more than S.

6 Who has got the second biggest sum? A. P B. Q C. R D. S 36. Following is a computer program to calculate the value of B. I. B = AC II. A = A + 1 III. C = B A + 2 IV. GOTO STEP I The computer stops executing the program as soon as B becomes greater than or equal to Initially A = 1 and C = 2. Find the last value of C when the computer stops executing the program. A.1518 B.721 C.1026 D A and B solved a quadratic equation. In solving it, A made a mistake in the constant term and obtained the roots as (5, 3) while B made a mistake in the coefficient of x and obtained the roots as (1, 3). The correct roots of the equation are: A.. 1, 3 B.1, 3 C. 1, 3 D.1, For a quadratic equation ax 2 + bx + c = 0, which of the following statements is necessarily true, if the roots of the equation are opposite in sign, the root with the greater absolute value being negative? A. a and b have the same sign. B. a and c have the same sign. C. a and c have the same sign. D. c is necessarily positive. 39. Three friends A, B and C buy tickets for a Imagine dragons Rock Concert in Tomorrowland. The ticket agent accepts the currencies Nevers, Troopies and Pairsas, whose rates of exchange are 1 Pairsa = 135 Nevers and 1 Troopy = 136 Nevers. One ticket costs 1560 Nevers. After buying one ticket, the subsequent tickets are available at a discount of 30% each. B pays 6 Troopies, C pays 12 Troopies and 81 Nevers and A pays the remaining amount in Pairsas and then they decide to contribute an equal share later on. How many Nevers does B owe C? 40. How many Nevers does A owe C? 41. Nineteen years from now Gohan will be 3 times as old as Goten is now. Goten is three years younger than Gohan. I. Gohan s age now II. Goten s age now A. I > II B. I < II C. I = II D. Nothing can be said

7 42. Once I had been to the post-office to buy stamps of five rupees, two rupees and one rupee. I paid the clerk Rs 20, and since he did not have change, he gave me three more stamps of one rupee. If the number of stamps of each type that I had ordered initially was more than one, what was the total number of stamps that I bought? A. 10 B. 9 C. 12 D You can collect Sapphires and Diamonds as many as you can. Each Sapphire is worth Rs 4crores and each Diamond is worth of Rs 5crore. Each Sapphire weights 0.3 kg. and each Diamond weighs 0.4 kg. Your bag can carry at the most 12 kg. What you should collect to get the maximum wealth? A. 20 Sapphires and 15 Diamonds B. 40 Sapphires C. 28 Sapphires and 9 Diamonds D. None of these 44. If x = y = z y+z z +x x+y = r, then r cannot take any value except A. ½ B. 1 C. ½ or 1 D. ½ or A telecom service provider engages male and female operators for answering 1000 calls per day. A male operator can handle 40 calls per day whereas a female operator can handle 50 calls per day. The male and the female operators get a fixed wage of Rs. 250 and Rs. 300 per day respectively. In addition, a male operator gets Rs. 15 per call he answers and a female operator gets Rs. 10 per call she answer. To minimize the total cost, how many male operators should the service provider employ assuming he has to employ more than 7 of the 12 female operators available for the job? A. 15 B. 14 C. 12 D A confused bank teller transposed the rupees and paise when he cashed a cheque for Sharon, giving her rupees instead of paise and paise instead of rupees. After buying a toffee for 50 paise, Sharon noticed that she was left with exactly three times as much as the amount on the cheque. Which of the following is a valid statement about the cheque amount? A. Over Rupees 4 but less than Rupees 5 B. Over Rupees 13 but less than Rupees 14 C. Over Rupees 7 but less than Rupees 8 D. Over Rupees 18 but less than Rupees If i and j are integers then the equation 5i + 19j = 64 has: A. no solution for i < 300 and j < 0 B. no solution for i > 250 and j > 100

8 C. a solution for 250 < i < 300 D. a solution for 59 < j < There are fifty integers a1, a2,... a50, not all of them necessarily different. Let the greatest integer of these fifty integers be referred to as G, and the smallest integer be referred to as L. The integers a1 through a24 form sequence S1, and the rest form sequence S2. Each member of S1 is less than or equal to each member of S2. 7. All values in S1 are changed in sign, while those in S2remain unchanged. Which of the following statements is true? A. Every member of S1 is greater than or equal to every member of S2. B. G is in S1. C. If all numbers originally in S1 and S2 had the same sign, then after the change of sign, the largest number of S1 and S2 is in S1 D. None of the above 49. Elements of S1 are in ascending order, and those of S2 are in descending order. a24 and a25are interchanged then which of the following statements is true A. S1 continues to be in ascending order. B. S2 continues to be in descending order. C. S1 continues to be in ascending order and S2 in descending order. D. None of the above 50. Along a road lie an odd number of stones placed at intervals of 10 m. These stones have to be assembled around the middle stone. A person can carry only one stone at a time. A man carried out the job starting with the stone in the middle, carrying stones in succession, thereby covering a distance of 4.8 km. Then the number of stones is: A. 25 B. 10 C. 19 D. 21

9 Answers and Explanations: 1. [196] 2. [2] f(z) = (z 3) 2 + (5 z) 2 = z 2 6z z + z 2 = 2z 2 16z + 34 = 2(z 2 8z + 17) = 2(z 2 8z ) = 2[(z 4) 2 + 1] The minimum value of this function will be at z = 4 and the minimum value = 2 1 = 2. Therefore, the required answer is B 4. [6] Number of times peaches were picked = 14 6 = 84 Each child took peaches from exactly 7 baskets. Therefore, the total number of children = 84 7 = 12 Let the number of baskets of plum be n. Therefore, the number of times plums were picked = n 8 Each child took plums from exactly 4 baskets. Therefore, n 8 = Therefore, n = 6 Therefore, the required answer is 6.

10 5. C 6. B (a + c) 2 2b(a + c) + b 2 = 0 a 2 + c 2 + 2ac 2ab 2bc + b 2 = 0 (b a c) 2 = 0 b = a + c 7. A If x = 0, the equation becomes x 2 17x + 66 = 0. Therefore, (x 6)(x 11) = 0 Therefore, the roots in this case are 6& 11. If x < 0, the equation becomes x x + 66 = 0. Therefore, (x + 6)(x + 11) = 0 Therefore, the roots in this case are 6& 11. Required sum of roots = = 0 8. B x 6x + 10= x 6x =(x 3)

11 For all real values of x, the denominator will always be greater than 0. Thus the fraction will be less than 0 for 4x 3 < 0 i.e. x < C 10. B Let the number of 2-rupee and 1-rupee coins with Jacob be x and y respectively. The original height of the container can hold x 2-rupee coins and y 20 1-rupee coins. When the height of the container is increased by 20%, he is able to fit x 10 2-rupee coins and y 1-rupee coins. Therefore, we have: 1.2(x + y 20) = (x 10 + y) 0.2(x + y) = B x + y = B F is the mid pt of AB.

12 13. C 14. A Coordinates of F ( 4+2 2,7+3 2 ) (3, 5) GF = ((3-4) 2 + (5 5) 2 ) = 1 G intersects median in 2:1 ratio. CG : GF and GF = 1 CG = 2 CF = 3 units If Raj was x years old in 2018, his father was 3x years old in Raj s age in 2033 = x + 15 and his father s age in 2033 = 3x x + 15 = 2(x + 15). Solving for x, x = 15. That means in 2018, Raj was 15 years old and his father was 45 years old. If Raj s father will be 2.5 times as old as he will be y years after 2018, we get 2.5(15 + y) = 45 + y. Solving for y, we get y = 5. Therefore the required calendar year is C This is only possible if (ab) 4 = (ba) 10 Therefore, 4a + b = 10b + a. Therefore, a = 3b. The only possible solution is when a = 3 and b = 1. Therefore, there is only one 2-digit number.

13 16. D Let B = 100 Substitute values of x as given in answer options. Option1 : x = 25 A = 125 and C = is not (4 25)% more than 75. Option 2: x = 40 A = 140 and C = is not (4 40)% more than 60. Option 3: x = 60 A = 160 and C = is not (4 60)% more than 40. Option 4: x = 50 A = 150 and C = is (4 50)% more than [573]

14 18. B 19. A 20. C The inequality is a product of 51 natural numbers. (A 100)(A 102)(A 104) (A 106). (A 200) > 0 Out of these 51 terms, if odd number of terms are positive and even number of terms are negative, it will satisfy the inequality. If A< 100, there will be 51 negative terms and their product is negative. Therefore, A= 100, 102, 104,, 200 the product is zero and the inequality is not satisfied. So, if A = 101, there will be one positive number and 50 negative numbers and their product will be positive. If A = 103, there will be two positive numbers and 49 negative numbers, their product will be negative. if A = 105, there will be three positive number and 48 negative numbers and their product will

15 be positive. If A = 107, there will be four positive numbers and 47 negative numbers, their product will be negative. so on Thus, the inequality holds for 101, 105, 109,., 197 Number of terms: 197 = (n 1)4 Solving this, we get n = 25 All numbers greater than 200 and less than 250 will also satisfy the inequality and there are 49 such numbers. So, = 74 numbers will satisfy the inequality. 21. A a 2 (1 b) + 2ab + b 2 (1- a) =a 2 + 2ab + b 2 a 2 b ab 2 =(a + b)(a + b - ab) = 4 x 7 =28 = 1 x 28 = 2 x 14 = 4 x 7 sum of roots can be (1 + 28) 29, (2 + 14) 16 and (4 + 7) C 23. B let a = x and b = y

16 Therefore, 9b 24. D 25. [5] p q 2 = 4 (p q 2 ) 2 = p 2 2pq 2 + q 4 = 16 2(p 2 + q 4 ) (p 2 2pq 2 + q 4 ) = p 2 2pq 2 + q 4 = 2 x = 36 p + q 2 = ±6 But p an q are positive and real. p + q 2 = 6 Solving the equations, we get p= 5

17 26. [10] 27. D 28. A

18 29. D Given: a 2 xa b = 0 or a 2 = xa + b Given: b 2 a xb = 0 or b 2 = xb + a a 2 b 2 = (x 1)(a b) or (a + b)(a b) = (x 1)(a b) Since a b, (a b) 0 a + b = x 1 Also, a 2 + b 2 = (x 1)(a + b) = (x + 1)(x 1) = x 2 1 ab = [(a + b) 2 (a 2 + b 2 )]/2 = [(x 1) 2 x 2 + 1]/2 = 1 x 30. D 31. [9] 32. [30] A + B + C + D = (i) A = B + C... (ii) D + 10 = A + C... (iii) A = C (iv) Substituting (iv) in (ii), we get, B = 10. Then, substituting (iii) in (i), we get, 10 + D D = 60 i.e., D = 20.

19 33. A 34. D 35. D P + Q + R + S = Rs.50 R = of S => R = and S =, P = S + 5 = , Q = 50 P =. 3 3

20 36. B 37. A 38. A

21 39. [432] 40. [33] 41. D Let present age of Gohan be x years and present age of Goten be y years; then present age of Goten = x 3 After 19 years Gohan s age = (x + 19) years and Goten s age = (y + 19) years. According to the question: x + 19 = 3(y + 19) or x 3y = 38 So only from above equation x and y cannot be found. 42. A The number of stamps that were initially bought were more than one of each type. Hence, the total number of stamps = 2(5 rupee) + 2(2 rupee) + 3(1 rupee) + 3(1 rupee) = B Basically, the question is of weights, so let us analyse them.4 Sapphires weight as much as 3

22 Diamonds. Cost of 4 Sapphires = 16 crores. Cost of 3 Diamonds = 15 crores. All Sapphires, multiple of 4 allowed, is the best deal, so 40 Sapphires 4160 crores. 44. C There are 2 cases, first, the given condition will be satisfied only when x, y and z are equal and second, when x + y + z = 0. Case (i) : If x = y = z = 1, we get r = 1/2, Case (ii) : when x + y + z = 0, then we get the value of r = 1. There are no other values that r can take.

23 45. D

24 46. D Let the original amount of the cheque be x rupees and y paise. Hence, original cheque amount = (100x + y) paise. The amount paid by bank teller = (100y + x) paise According to the given question, 100y + x 50 = 3(100x + y) y= 299x x + 50 = = 3x + 8x On checking, we get if x = 18 (a positive integer), then only y has an positive integral value 56. Hence, original amount of the cheque = Rs i.e., over rupees18 but less than rupees C 5i + 19j = 64 We see that if j = 1, we get an integer solution i = 9, Now if j changes (increases or decreases) by 5, i will change (decrease or increase) by 19 Looking at options, if i = 256, we get, j = 64 Using these values we see option a, b and d which are eliminated and also that there exists a solution for 250 < i < D With the change in the sign of the members of S1 nothing definitely can be said because if both the sequences contain negative number then each term of S1 will be greater than each term of S2. So the answer must be none of these. 49. A With the interchanging of a24 and a25 still the sequence of S1 will be in ascending order because each term of S2 is either greater than or equal to each term of S D Suppose there are n stones are placed. So, there are (2n + 1) stones, on each side of middle stone To pick up and return at middle point the man will travel 20m for 1st, 40 m for 2nd, and so on. Therefore, total distance (2.2 km) is given by 2200 = 2[ n] 1100 = 20[ n] 55 = n(n + 1)/2 => n=10 No of stones = 2n +1 = =21

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