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Algebra Practice Quiz 2

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Algebra Practice Quiz 2

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Algebra is a division of mathematics designed to help solve certain types of problems quicker and easier. Algebra is based on the concept of unknown values called variables, unlike arithmetic which is based entirely on known number values.
The article Algebra Practice Quiz 2 is useful for candidates preparing for different competitive examinations like RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC, IBPS PO Exams and etc.

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In each of these questions, two equations (I) and (II) are given.You have to solve both the equations and give answer.
1. I. [latex]x^{3}[/latex] – 4913 = 0 II. [latex]y^{2}[/latex] – 361 = 0
    A. if x ≤ y B. if x = y or relationship between x and y can't be established C. if x ≥ y D. if x ≥ y E. if x ≤ y

Answer: Option B
Explanation: I. [latex]x^{3}[/latex] – 4913 = 0 or, [latex]x^{3}[/latex] = 4913 x = 17 II. [latex]y^{2}[/latex] = 361 or, y = ± 19 While comparing the values of x and y, one root value of y lies between the root values of x
2. I. [latex]x^{2}[/latex] = 361 II. [latex]y^{3}[/latex] = 7269 + 731
    A. if x ≤ y B. if x ≥ y C. if x ≥ y D. if x ≤ y E. if x = y or relationship between x and y can't be established

Answer: Option A
Explanation: I. [latex]x^{2}[/latex] = 361 x = ± 19 II. [latex]y^{3}[/latex] = 7269 + 731 [latex]y^{3}[/latex] = 8000 y = 20 x ≤ y
3. I. 15[latex]x^{2}[/latex] + x – 6 = 0 II. 5[latex]y^{2}[/latex] – 23y + 12 = 0
    A. if x ≥ y B. if x ≤ y C. if x ≥ y D. if x ≤ y E. if x = y or relationship between x and y can't be established

Answer: Option B
Explanation: I. 15[latex]x^{2}[/latex] + x – 6 = 0 15[latex]x^{2}[/latex] + 10x – 9x – 6 = 0 5x (3x + 2) – 3 (3x + 2) = 0 (5x – 3) (3x + 2) = 0 x = [latex]\frac{3}{5}[/latex], –[latex]\frac{2}{3}[/latex] II. 5[latex]y^{2}[/latex] – 23y + 12 = 0 5[latex]y^{2}[/latex] – 20y – 3y + 12 = 0 5y (y – 4) – 3 (y – 4) = 0 (y – 4) (5y – 3) = 0 y = 4, [latex]\frac{3}{5}[/latex] x ≤ y
4. I. [latex]x^{3}[/latex] – 2744 = 0 II. [latex]y^{2}[/latex] – 256 = 0
    A. if x ≥ y B. if x ≤ y C. if x ≥ y D. if x ≤ y E. if x = y or relationship between x and y can't be established

Answer: Option E
Explanation: I. [latex]x^{3}[/latex] – 2744 = 0 [latex]x^{3}[/latex] = 2744 x = 14 II. [latex]y^{2}[/latex] – 256 = 0 [latex]y^{2}[/latex] = 256 y = ± 16 While comparing the values of x and y, one root value of x lies between the root values of y.
5. I. [latex]x^{2}[/latex] – 8x – 20 = 0 II. 3[latex]y^{2}[/latex] – 60y + 297 = 0
    A. if x ≥ y B. if x ≤ y C. if x ≥ y D. if x ≤ y E. if x = y or relationship between x and y can't be established

Answer: Option B
Explanation: I. [latex]x^{2}[/latex] – 8x – 20 = 0 [latex]x^{2}[/latex] – 10x + 2x – 20 = 0 x (x – 10) + 2 (x – 10) = 0 (x – 10) (x + 2) = 0 Then, x = + 10 or x = – 2 II. 3[latex]y^{2}[/latex] – 60y + 297 = 0 [latex]y^{2}[/latex] – 20y + 99 = 0 [Dividing both sides by 3] [latex]y^{2}[/latex] – 11y – 9y + 99 = 0 y (y – 11) – 9 (y – 11) = 0 (y – 11) (y – 9) = 0 Then, y = + 11 or y = + 9 So, when x = + 10, x y for y = + 9 And when x = – 2, x ≤ y for y = + 11 and x ≤ y for y = + 9 So, we can observe that one root value of x lies between the root values of y. Therefore, the relation between x and y can't be determined.
1. The equations ax + b = 0 and cx + d = 0 are consistent, if
    A. ad = bc B. ad + bc = 0 C. ab - cd = 0 D. ab + cd = 0

Answer: Option B
Explanation: These equation are consistent if [latex]\frac{a}{c}[/latex] = [latex]\frac{b}{d}[/latex], i.e ad = bc
2. Points A and B are 60km apart. A bus starts from A and another from B at the same time. If they go in the same direction they meet in 6 hours and if they go in opposite direction they meet in 2 hours. The speed of the bus with greater speed is:
    A. 10 km/hr B. 20 km/hr C. 30 km/hr D. 40 km/hr

Answer: Option B
Explanation: Let their speeds be x km/hr and y km/hr. When they move in same direction, let them meet at M. Then, AM - BM = AB 6x - 6y = 60 or x - y = 10 ...(i) When they move in opposite direction, let them meet at N. Then, AN + BN = AB 2x + 2y = 60 or x + y = 30 ...(ii) Solving (i) and (ii) we get x = 20, y = 10.
3. A railway half ticket costs half the full fare but the reservation charges are the same for half ticket as well as for full ticket. one reserved first class ticket for a journey between two stations is Rs. 362 and one full and one half reserved first class tickets cost Rs. 554. The reservation charges are:
    A. Rs. 18 B. Rs. 22 C. Rs. 38 D. Rs. 46

Answer: Option A
Explanation: Let first class fare be Rs. x and reservation charges be Rs. y. Then x + y = 362 & (x + y) + ([latex]\frac{1}{2}[/latex]x+y)= 554 x + y = 362 & 3x + 4y = 1108. Solving these equations, we get: y = 22.
4. What is the value of x + y in the solution of the equations? [latex]\frac{x}{4}[/latex]+[latex]\frac{y}{3}[/latex]=[latex]\frac{5}{12}[/latex] and [latex]\frac{x}{2}[/latex]+ y= 1 is
    A. [latex]\frac{1}{3}[/latex] B. [latex]\frac{3}{2}[/latex] C. 2 D. [latex]\frac{5}{2}[/latex]

Answer: Option B
Explanation: Given equations, 3x + 4y = 5 ........ (i) x + 2y = 2 .........(ii) Solving (i) and (ii), we get y= [latex]\frac{1}{2}[/latex] and x = 1. so x + y =[latex]\frac{3}{2}[/latex]
5. The number of solutions of these equations x+[latex]\frac{1}{y}[/latex]= 2 and 2xy – 3y = – 2 is:
    A. 0 B. 1 C. 2 D. None of these

Answer: Option A
Explanation: First equation gives[latex]\frac{1}{y}[/latex] = 2 - x or y =[latex]\frac{1}{2 - x}[/latex] Second equation is: y(2x - 3) = -2 or[latex]\frac{2x - 3}{2 - x}[/latex] = – 2 2x – 3 = – 2(2 – x). This gives 1 = 0. This is impossible. So there is no solution at all.
1. L ≤ O ≥ V = E ≥ S Which of the following ones is correct? 1) E ≥ G 2) H ≥ G 3) H ≥ F 4) I ≥ G
    A. Only 1 B. Only 2 C. Only 3 & 4 D. Only 3 E. None of these

Answer: Option D
Explanation: E = F ≥ G ≤ H = I We can’t compare H and F because between H & F opposite symbol used. We know that the inequalities does not works between opposite symbol
2. L ≤ O ≥ V = E ≥ S Which of the following ones is correct? 1) L ≤ V 2) O = E 3) O ≥ S 4) S ≥ L
    A. Only 1 B. Only 2 C. Only 3 D. Only 3 & 4 E. None of these

Answer: Option C
Explanation: L ≤ O ≥ V = E ≥ S We can compare O and S. which shows that the option 3rd is correct because the common symbol between O and S is ‘≥’.
3. B ≥ E ≤ A = T ≥ S Which of the following ones is correct? 1) B ≥ S 2) E = T 3) E ≤ T 4) E ≤ S
    A. Only 1 B. Either 2 or 3 C. Only 2 D. Either 3 or 4 E. None of these

Answer: Option B
Explanation: B ≥ E ≤ A = T ≥ S We can compare E and T but either 2 or 3 equation is correct.
4. M = O ≤ N = K ≤ S Which of the following ones is correct? 1) M = S 2) O ≤ S 3) N ≥ S 4) O = K
    A. Only 1 B. Only 2 C. Only 2 & 3 D. Either 3 or 4 E. None of these

Answer: Option E
Explanation: M = O ≥ N = K ≤ S We can compare O & S. which shows that the option 2 is correct because the common symbol between O & S is ‘≤’.
5. C ≥ H = A ≥ T ≥ S Which of the following ones is correct? 1) S ≥ C 2) T = C 3) H ≤ T 4) H ≤ S
    A. Only 1 B. Only 2 C. Either 1 or 2 D. Only 4 E. None of these

Answer: Option A
Explanation: C ≥ H = A ≥ T ≥ S We can compare S & C. which shows that the option first one is correct because the common symbol between S & C is ‘≤’.

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