Download Top-20 IBPS RRB PO Quadratic Equation Questions PDF. Quadratic Equation questions based on asked questions in previous year exam papers very important for the IBPS RRB PO (Officer Scale-I, II & III) exam.<\/p>\n
Download Quadratic Equation Questions For IBPS RRB PO<\/a><\/p>\n 35 IBPS RRB PO Mocks @ Rs. 149. Enroll Now<\/a><\/p>\n 70 IBPS RRB (PO + Clerk) Mocks @ Rs. 199<\/a><\/p>\n Take a free mock test for IBPS RRB PO<\/a><\/p>\n Download IBPS RRB PO Previous Papers PDF<\/a><\/p>\n Instructions<\/b><\/p>\n In the following questions 2 quantities X and Y are given. Select the option which best captures the relations between X and Y.<\/p>\n Question 1:\u00a0<\/b>3 dices are thrown. a)\u00a0X>Y<\/p>\n b)\u00a0X<Y<\/p>\n c)\u00a0X$\\geq$Y<\/p>\n d)\u00a0X$\\leq$Y<\/p>\n e)\u00a0X=Y or Cannot be determined<\/p>\n Question 2:\u00a0<\/b>Solution A has p and q in the ratio 1:2. Solution B has p and q in the ratio 5:4. Solution A and Solution B are mixed in the ratio 3:4 to form the solution C. Solution A and Solution B are mixed in the ratio 5:3 to form the solution D. a)\u00a0X>Y<\/p>\n b)\u00a0X<\/p>\n c)\u00a0X$\\geq$Y<\/p>\n d)\u00a0X$\\leq$Y<\/p>\n e)\u00a0X=Y or Cannot be determined<\/p>\n Question 3:\u00a0<\/b>Cost price of 6 articles is equal to the Selling price of 4 articles. The shopkeeper Marked up the price by 100%. X is the % discount offered. a)\u00a0X>Y<\/p>\n b)\u00a0X<Y<\/p>\n c)\u00a0X$\\geq$Y<\/p>\n d)\u00a0X$\\leq$Y<\/p>\n e)\u00a0X=Y or Cannot be determined<\/p>\n Question 4:\u00a0<\/b>2 dices are thrown. a)\u00a0X>Y<\/p>\n b)\u00a0X<\/p>\n c)\u00a0X$\\geq$Y<\/p>\n d)\u00a0X$\\leq$Y<\/p>\n e)\u00a0X=Y or Cannot be determined<\/p>\n Question 5:\u00a0<\/b>a and b are the roots of the quadratic equation $x^2-12x+25 = 0$ a)\u00a0X>Y<\/p>\n b)\u00a0X<\/p>\n c)\u00a0X$\\geq$Y<\/p>\n d)\u00a0X$\\leq$Y<\/p>\n e)\u00a0X=Y or Cannot be determined<\/p>\n Instructions<\/b><\/p>\n Calculate the quantity I and the quantity II on the basis of the given information then compare them and answer the following questions accordingly.<\/p>\n Question 6:\u00a0<\/b>Quantity 1: Selling price of an article on which a seller made a profit of 20% and he bought it for Rs. 800. a)\u00a0Quantity 1 > Quantity 2<\/p>\n b)\u00a0Quantity 1 $\\geq$ Quantity 2<\/p>\n c)\u00a0Quantity 1 < Quantity 2<\/p>\n d)\u00a0Quantity 1 $\\leq$ Quantity 2<\/p>\n e)\u00a0Quantity 1 = Quantity 2<\/p>\n Free Mock Test for IBPS RRB PO<\/a><\/p>\n IBPS RRB Clerk Previous Papers<\/a><\/p>\n Question 7:\u00a0<\/b>Quantity 1: Simple interest earned by Ramesh by lending Rs. 6000 for 3 years at 20% per annum. a)\u00a0Quantity 1 > Quantity 2<\/p>\n b)\u00a0Quantity 1 $\\geq$ Quantity 2<\/p>\n c)\u00a0Quantity 1 < Quantity 2<\/p>\n d)\u00a0Quantity 1 $\\leq$ Quantity 2<\/p>\n e)\u00a0Quantity 1 = Quantity 2<\/p>\n Question 8:\u00a0<\/b>Quantity 1: Volume of a cylinder whose height is 10 cm and perimeter of the base is 88 cm. a)\u00a0Quantity 1 > Quantity 2<\/p>\n b)\u00a0Quantity 1 $\\geq$ Quantity 2<\/p>\n c)\u00a0Quantity 1 < Quantity 2<\/p>\n d)\u00a0Quantity 1 $\\leq$ Quantity 2<\/p>\n e)\u00a0Quantity 1 = Quantity 2<\/p>\n Question 9:\u00a0<\/b>Quantity 1: The surface area of a cuboid whose side lengths are 20 cm, 30 cm and 50 cm. a)\u00a0Quantity 1 > Quantity 2<\/p>\n b)\u00a0Quantity 1 $\\geq$ Quantity 2<\/p>\n c)\u00a0Quantity 1 < Quantity 2<\/p>\n d)\u00a0Quantity 1 $\\leq$ Quantity 2<\/p>\n e)\u00a0Quantity 1 = Quantity 2<\/p>\n Question 10:\u00a0<\/b>Quantity 1: Cost price of an article on which a seller made a profit of 15% by selling it for Rs. 345. a)\u00a0Quantity 1 > Quantity 2<\/p>\n b)\u00a0Quantity 1 $\\geq$ Quantity 2<\/p>\n c)\u00a0Quantity 1 < Quantity 2<\/p>\n d)\u00a0Quantity 1 $\\leq$ Quantity 2<\/p>\n e)\u00a0Quantity 1 = Quantity 2<\/p>\n Instructions<\/b><\/p>\n Calculate the quantity I and the quantity II on the basis of the given information then compare them and answer the following questions accordingly.<\/strong><\/p>\n Question 11:\u00a0<\/b>Quantity 1: Simple interest charged by bank on a sum of Rs. 5000 at the rate of 23% annum for 3 years. a)\u00a0Quantity 1 > Quantity 2<\/p>\n b)\u00a0Quantity 1 $\\geq$ Quantity 2<\/p>\n c)\u00a0Quantity 1 < Quantity 2<\/p>\n d)\u00a0Quantity 1 $\\leq$ Quantity 2<\/p>\n e)\u00a0Quantity 1 = Quantity 2<\/p>\n Question 12:\u00a0<\/b>Quantity 1: Volume of a cube whose length of a side is 50 cm. a)\u00a0Quantity 1 > Quantity 2<\/p>\n b)\u00a0Quantity 1 $\\geq$ Quantity 2<\/p>\n c)\u00a0Quantity 1 < Quantity 2<\/p>\n d)\u00a0Quantity 1 $\\leq$ Quantity 2<\/p>\n e)\u00a0Quantity 1 = Quantity 2<\/p>\n Question 13:\u00a0<\/b>Quantity 1: Area of an equilateral triangle with side equals to 56 cm. a)\u00a0Quantity 1 > Quantity 2<\/p>\n b)\u00a0Quantity 1 $\\geq$ Quantity 2<\/p>\n c)\u00a0Quantity 1 < Quantity 2<\/p>\n d)\u00a0Quantity 1 $\\leq$ Quantity 2<\/p>\n e)\u00a0Quantity 1 = Quantity 2<\/p>\n Instructions<\/b><\/p>\n In the following questions 2 quantities X and Y are given. Select the option which best captures the relations between X and Y.<\/strong><\/p>\n Question 14:\u00a0<\/b>Ramesh has 400L of milk with him. He sells 20% of the mixture and replaces it water. He then sells 25% of the mixture and replaces it water. Finally, he sells 40% of the mixture and replaces it water. a)\u00a0X>Y<\/p>\n b)\u00a0X<Y<\/p>\n c)\u00a0X$\\geq$Y<\/p>\n d)\u00a0X$\\leq$Y<\/p>\n e)\u00a0X=Y or Cannot be determined<\/p>\n Question 15:\u00a0<\/b>Solution A has p and q in the ratio 3:5 whereas solution B has p and q in the ratio 4:3. a)\u00a0X>Y<\/p>\n b)\u00a0X<Y<\/p>\n c)\u00a0X$\\geq$Y<\/p>\n d)\u00a0X$\\leq$Y<\/p>\n e)\u00a0X=Y or Cannot be determined<\/p>\n Quantitative Aptitude formulas PDF<\/a><\/p>\n 520 Banking Mocks – Just Rs. 499<\/a><\/p>\n Question 16:\u00a0<\/b>A shopkeeper marks up the price of an article by 40%. X is the Cost price of 3 articles. a)\u00a0X>Y<\/p>\n b)\u00a0X<Y<\/p>\n c)\u00a0X$\\geq$Y<\/p>\n d)\u00a0X$\\leq$Y<\/p>\n e)\u00a0X=Y or Cannot be determined<\/p>\n Question 17:\u00a0<\/b>Solution A has p and q in the ratio 3:2 whereas solution B has p and q in the ratio 1:4. a)\u00a0X>Y<\/p>\n b)\u00a0X<Y<\/p>\n c)\u00a0X$\\geq$Y<\/p>\n d)\u00a0X$\\leq$Y<\/p>\n e)\u00a0X=Y or Cannot be determined<\/p>\n Question 18:\u00a0<\/b>5 men can complete a job in 6 days. 12 women and 6 children take $\\frac{20}{7}$ days to complete the same job instead if 6 women and 12 children work it will take them $\\frac{20}{91}$ more days to complete the work. a)\u00a0X>Y<\/p>\n b)\u00a0X<Y<\/p>\n c)\u00a0X$\\geq$Y<\/p>\n d)\u00a0X$\\leq$Y<\/p>\n e)\u00a0X=Y or Cannot be determined<\/p>\n Instructions<\/b><\/p>\n In the following questions, two equations numbered I and II are given. You have to establish a relation between the two variables and select the option accordingly:<\/p>\n Question 19:\u00a0<\/b>I: $x^2 – 11x + 30 = 0$ a)\u00a0$x > y$<\/p>\n b)\u00a0$x \\geq y$<\/p>\n c)\u00a0$y > x$<\/p>\n d)\u00a0$y \\geq x$<\/p>\n e)\u00a0$x = y$ or No relationship can be established<\/p>\n Question 20:\u00a0<\/b>I: $2x^2 – 33x + 133 = 0$ a)\u00a0$x > y$<\/p>\n b)\u00a0$x \\geq y$<\/p>\n c)\u00a0$y > x$<\/p>\n d)\u00a0$y \\geq x$<\/p>\n e)\u00a0$x = y$ or No relationship can be established<\/p>\n 18000 Free Solved Questions – Banking Study Material<\/a><\/p>\n 35 IBPS RRB PO Mocks @ Rs. 149. Enroll Now<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(E)<\/strong><\/p>\n 7 = 1 + 1 + 5 = 2 + 2 + 3 = 3 + 3 + 1 = 1 + 2 + 4 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the quantity of A and B be 9L. 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the CP be 10 Rs. 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Total possible combinations = 36. 5)\u00a0Answer\u00a0(E)<\/strong><\/p>\n a and b are the roots of the quadratic equation $x^2-12x+25 = 0$ 6)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Quantity 1: Selling price of the article = $\\dfrac{100+20}{100}\\times 800$ = Rs. 960 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Quantity 1: Simple interest earned by Ramesh = 6000*0.20*3 = Rs. 3600. 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Quantity 1: Radius of the cylinder = $\\dfrac{88}{2\\pi}$ = 14 cm 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Quantity 1: The surface area of a cuboid whose side lengths are 20 cm, 30 cm and 50 cm = 2(20*30 + 30*50 + 50*20) = 6200 $\\text{cm}^2$.<\/p>\n Quantity 2: The curved surface area of a hemisphere of 35 cm radius = $2\\pi*(35)^2$ = 7700 $\\text{cm}^2$.<\/p>\n Hence we can say that Quantity 1 < Quantity 2. Option C is the correct answer.<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Quantity 1: Cost price of the article = $\\dfrac{100}{100+15}\\times 345$ = Rs. 300 11)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Simple interest paid to bank = 5000*0.23*3 = Rs. 3450 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Volume of a cube whose length of a side is 50 cm = $50^3$ = $125000$ 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Quantity 1: Area of the equilateral triangle with side equals to 56 cm = $\\dfrac{\\sqrt{3}}{4}\\times 56^2$ = 1357.93 sq.cm 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n The amount of milk with Ramesh at the end = 400*0.8*0.75*0.6 = 144L= X 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the amount of solution A and solution B be 56L. 16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the CP of the article be x. 17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the amount of A and B be 5L each. 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n 5 men can complete a job in 6 days. Thus, 1 men will take 30 days to complete the work. In 1 day 1 man will complete $\\frac{1}{30}$ of the work. 19)\u00a0Answer\u00a0(A)<\/strong><\/p>\n On solving $x^2 – 11x + 30 = 0$, 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n On solving $2x^2 – 33x + 133 = 0$,
\nX is the probability that sum of the number on the faces of the dice is 7.
\nY is the probability that sum of the number on the faces of the dice is 14.<\/p>\n
\nX = the ratio of p and q in Solution C
\nY = the ratio of p and q in Solution D.<\/p>\n
\nY is the % profit obtained.<\/p>\n
\nX is the probability that the sum of the number on the faces of the dice is 7.
\nY is the probability that the sum of the number on the faces of the dice is 5.<\/p>\n
\nc and d are the roots of the equation $x^2-11x+15x = 0$
\nX = $a^3+b^3+4$
\nY = $c^3+d^3-4$<\/p>\n
\nQuantity 2: Cost price of an article on which a seller made a loss of 40% by selling it for Rs. 576.<\/p>\n
\nQuantity 2: Compound interest earned by Suresh by lending Rs. 5000 for 2 years at 25% per annum.<\/p>\n
\nQuantity 2: Volume of a sphere of 10.5 cm.<\/p>\n
\nQuantity 2: The curved surface area of a hemisphere of 35 cm radius.<\/p>\n
\nQuantity 2: Selling price of an article on which a seller made a loss of 20% and he bought it for Rs. 400.<\/p>\n
\nQuantity 2: Compound interest charged by another bank on a sum of Rs. 5000 at the rate of 30% annum for 2 years compounded annually.<\/p>\n
\nQuantity 2: Volume of a right-circular cone whose height is 50 cm and radius of the circular base is 50 cm.<\/p>\n
\nQuantity 2: Area of a square with side equals to 35 cm.<\/p>\n
\nSuresh has 500L of milk with him. He sells 30% of the mixture and replaces it water. He then sells 40% of the mixture and replaces it water. Finally, he sells 30% of the mixture and replaces it water.
\nX = The amount of milk Ramesh is left with
\nY = The amount of milk Suresh is left with<\/p>\n
\nSolution A and Solution B are mixed in the ratio m:n to form solution C. Solution C has p and q in ratio 53:59 . Solution A is mixed with solution B in ratio k:l to form solution D. Solution D has p and q and ratio 191:201
\nX is the ratio of m and n
\nY is the ratio of k and l<\/p>\n
\nHe gives 1 item free for every 2 items bought. Y is the revenue earned by selling 3 articles.<\/p>\n
\n3 parts of Solution A is mixed with 5 parts of solution B to form solution C and 2 parts of Solution A is mixed with 3 parts of solution B to form solution D.
\nX is the ratio of p and q in solution C.
\nY is the ratio of p and q in solution D.<\/p>\n
\nX is the number of it will take 8 women and 12 children to complete the work.
\nY is the number of days it will take 2 men and 15 children to complete the work.<\/p>\n
\nII: $y^2 + 6y – 16 = 0$<\/p>\n
\nII: $y^2 – 4y – 21 = 0$<\/p>\n
\nThus, the probability that sum of the number on the faces of the dice is 7 = (3C1+3C1+3C1+3!)\/216 = 15\/216 = X
\n10 = 6 + 3 + 1 = 6 + 2 + 2 = 5 + 3 + 2 = 4 + 4 + 2 = 4 + 3 + 3 = 1 + 4 + 5
\nThus, the probability that sum of the number on the faces of the dice is 14 = (3C1+3C1+3C1+3!)\/216 = 15\/216 = Y
\nHence, X=Y
\nHence, option E is the correct answer.<\/p>\n
\nThus, the amount of p and q in A is 3L and 6L respectively.
\nAlso, the amount of p and q in B is 5L and 4L respectively.
\nA and B are mixed in the ratio 3:4 to form solution C.
\nThus, the amount of p and q in solution C = (3*3+5*4):(6*3+4*4) = 29:34=X
\nSolution A and Solution B are mixed in the ratio 5:3 to form the solution D.
\nThus, the amount of p and q in solution D = (5*3+5*3):(6*5+4*3) = 30:42=Y
\nHence, X>Y
\nHence, option A is the correct answer.<\/p>\n
\nThus, 6*10 = 4*SP. Thus, SP = 15 Rs
\nMP = 2*10 = 20 Rs
\n% discount = $\\frac{20-15}{20}$ = 25% = X
\n% profit = $\\frac{15-10}{10}$ = 50% = Y
\nHence, option B is the correct answer.<\/p>\n
\nSum of the number on the faces of the dice is 7 will be for (1,6),(6,1),(2,5),(5,2),(3,4) and (4,3)
\nThus, X = 6\/36
\nSum of the number on the faces of the dice is 5 will be for (1,4),(4,1),(2,3) and (3,2).
\nThus, Y = 4\/36
\nHence, X>Y
\nHence, option A is the correct answer.<\/p>\n
\nThus, a+b = 12 and ab = 25
\nThus, $a^3+b^3+4$ = $(a+b)^3-3ab(a+b)+4 = 13^3-3*25*13 = 832$
\nc and d are the roots of the equation $x^2-11x+15x = 0$
\nThus, c+d = 11 and cd=15
\nThus, $c^3+d^3-4$ = $(c+d)^3-3cd(c+d)-4$ = $11^3-3*11*15-4 = 832$
\nThus, X=Y.
\nHence, option E is the correct answer.<\/p>\n
\nQuantity 2: Cost price of the article = $\\dfrac{100}{100-40}\\times 576$ = Rs. 960
\nWe can see that Quantity 1 = Quantity 2. Hence, option E is the correct answer.<\/p>\n
\nQuantity 2: Compound interest earned by Suresh = $5000(1 +\\dfrac{25}{100})^2 – 5000$ = Rs. 2812.50.
\nHence, we can say that Quantity 1 > Quantity 2. Option A is the correct answer.<\/p>\n
\nTherefore, the volume of the cylinder = $\\pi*14^2*10$ = 6160 $\\text{cm}^3$
\nQuantity 2: Volume of the sphere = $\\dfrac{4\\pi}{3}10.5^3$ = 4849 $\\text{cm}^3$
\nHence, we can say that Quantity 1 > Quantity 2. Option A is the correct answer.<\/p>\n
\nQuantity 2: Selling price of the article = $\\dfrac{100-20}{100}\\times 400$ = Rs. 320
\nWe can see that Quantity 1 < Quantity 2. Hence, option C is the correct answer.<\/p>\n
\nCompound interest paid to bank = $5000(1 + \\dfrac{30}{100})^2 – 5000$ = Rs. 3450
\nHence, we can say that Quantity 1 = Quantity 2. Option E is the correct answer.<\/p>\n
\nVolume of right-circular cone whose height is 100 cm and radius of circular base is 50 cm = $\\dfrac{\\pi}{3}\\times 50^2*50$ = $\\dfrac{125000\\pi}{3}$
\nHence, we can say that Quantity 1 < Quantity 2. Option C is the correct answer.<\/p>\n
\nQuantity 2: Area of square with side equals to 35 cm = $35^2$ = 1225 sq. cm
\nHence, we can say that Quantity 1 > Quantity 2. Option A is the correct answer.<\/p>\n
\nThe amount of milk Suresh is left with at the end = 500*0.7*0.6*0.7 = 147L = Y
\nHence, Y>X
\nThus, option B is the correct answer.<\/p>\n
\nIn solution A p is 21L and q is 35L.
\nIn solution B p is 32L and q is 24L.
\nIn solution C p:q = 53:59 = (21*m+32*n):(35*m+24*n)
\nSolving we get, m:n = 1:1 = X
\nIn solution D p:q = 191:201 = (21*k+32*l):(35*k+24*l)
\nSolving we get k:l = 3:4 = Y
\nThus, X > Y
\nHence, option A is the correct answer.<\/p>\n
\nThus CP of 3 articles = 3x.
\nMP = 1.4x
\nHe gives 1 item free for every 2 items bought. Y is the revenue earned by selling 3 articles.
\nY = 2.8x
\nThus, X>Y.
\nhence, option A is the correct answer.<\/p>\n
\nThus, the amount of p and q in A is 3L and 2L and the amount of p and q is 1L and 4L.
\n15L of A and 25L of B is taken to form C.
\nThus, in solution C the ratio of p:q = (3*3+1*5):(2*3+4*5) = 14:26 = 7:13 = X
\n10L of A and 15L of B is taken to form D
\nThus, in solution D the ratio of p:q = (3*2+1*3):(2*2+4*3) = 9:16= Y
\nThus, Y>X.
\nHence, option B is the correct answer.<\/p>\n
\nLet W be the number of days 1 women will take to complete the work and C be the number of days 1 children will take to complete the work
\nAs per the given Condition,
\n$\\frac{12}{W}+\\frac{6}{C} = \\frac{7}{20}$
\n$\\frac{20}{7}+\\frac{20}{91} = \\frac{13}{40}$ = number of days 6 women and 12 children will take.
\nand $\\frac{6}{W}+\\frac{12}{C} = \\frac{13}{40}$
\nSolving these 2 equation we get W = 48 and C = 60
\nThus, the number of it will take 8 women and 12 children to complete the work = $\\dfrac{1}{\\dfrac{8}{48}+\\dfrac{12}{60}} = \\dfrac{60}{22}$ = X
\nthe number of days it will take 2 men and 15 children to complete the work =
\n$\\dfrac{1}{\\dfrac{2}{30}+\\dfrac{15}{60}} = \\dfrac{60}{19}$ = Y
\nThus, Y>X
\nHence, option B is the correct answer.<\/p>\n
\n$x = 5$ or $x = 6$
\nOn solving $y^2 + 6y – 16 = 0$,
\n$y = 2$ or $y = -8$
\nWe can see that $x > y$ for all values of x and y.
\nHence, option A is the correct answer.<\/p>\n
\n$x = 7$ or $x = 8.5$
\nOn solving $y^2 – 4y – 21 = 0$,
\n$y = 7$ or $y = -3$
\nWe can see that $x \\geq y$ for all values of x and y.
\nHence, option B is the correct answer.<\/p>\n