\n(B) -2x2<\/sup> = (5 – x) \\(\\left(2 x-\\frac{2}{5}\\right)\\)
\n(C) (k + 1)x2<\/sup> +\\(\\frac{3}{2}\\)x = 7, where = -1
\n(D) x3<\/sup> \u2014 x2<\/sup> \u2014 (x – 1)3<\/sup>
\nAnswer:
\n(D) x3<\/sup> \u2014 x2<\/sup> \u2014 (x – 1)3<\/sup><\/p>\n
Explanation: Question 2.<\/p>\n (A) 2(x – 1)2<\/sup> = 4x2<\/sup> – 2x + 1 Explanation: Question 3.<\/p>\n (A) x2<\/sup> – 4x + 5 = 0 Explanation: Question 4<\/p>\n (A) 2 Explanation: Question 5.<\/p>\n (A) 2x2<\/sup>-3x + 6 = 0 Explanation: Question 6.<\/p>\n (A) -5, -2 Explanation: Question 7.<\/p>\n (A) 2x2<\/sup> -3\u221a2x + \\(\\frac{9}{4}\\) = 0 Explanation: Question 8.<\/p>\n (A) 0 only Explanation: Question 9.<\/p>\n (A) \\(\\frac{1}{8}\\) Explanation: Question 10.<\/p>\n (A) two distinct real roots Explanation: Question 11.<\/p>\n (A) x2<\/sup> -4x + 3\u221a2 =0 Explanation: Question 12.<\/p>\n (A) four real roots Explanation: Assertion and reason Based MCQs<\/span><\/p>\n Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as. Question 1.<\/p>\n Answer: Explanation: Question 2.<\/p>\n Answer: Explanation: \u21d2 x2<\/sup> + [(a + 3) \u2014 (a \u2014 2)]x – (a + 3)(a – 2) = 0 Question 3.<\/p>\n Answer: Explanation: Question 4.<\/p>\n Answer: Explanation: Question 5.<\/p>\n Answer: Explanation: Question 6.<\/p>\n Answer: Explanation: Case -Based MCQs<\/span><\/p>\n Attempt any four sub-parts from each question. Each sub-part carries 1 mark. Question 1.<\/p>\n (A) 2(x + 5) km Explanation: Question 2.<\/p>\n (A) x2<\/sup> – 5x – 500 = 0 Question 3.<\/p>\n (A) 20 km\/hour Question 4.<\/p>\n (A) 20 hour II. Read the following text and answer the following questions on the basis of the same: The speed of a motor boat is 20 km\/hr. For covering the distance of 15 km the boat took 1 hour more for upstream than downstream. Question 1.<\/p>\n (A) 20 km\/hr Explanation: Question 2.<\/p>\n (A) speed = (distance )\/time Question 3.<\/p>\n Which is the correct quadratic equation for the speed of the current ?<\/p>\n (A) x2<\/sup>+ 30x – 200 = 0 Question 4.<\/p>\n (A) 20 km\/hour Question 5.<\/p>\n (A) 90 minute III. Read the following text and answer the following questions on the basis of the same: Question 1.<\/p>\n (A) x-45 Explanation: Question 2.<\/p>\n (A) x-45 Explanation: Question 3.<\/p>\n (A) x2<\/sup> – 45x+ 324 = 0 Explanation: Question 4.<\/p>\n (A) 10 Explanation: Question 5.<\/p>\n (A) 10 Explanation: IV. Read the following text and answer the following questions on the basis of the same: There is a triangular playground as shown in the below figure. Many Children and people are playing and walking in the ground. As we see in the above figure of right angled triangle playground, the length of the sides are 5x cm and (3x – 1) cm and area of the triangle is 60 cm2. Question 1.<\/p>\n (A) 8 Explanation: Question 2.<\/p>\n (A) 8 cm Question 3.<\/p>\n (A) 17 cm Explanation: Question 4.<\/p>\n (A) 35 cm Explanation: Question 5.<\/p>\n (A) AP Explanation: Quadratic Equations Class 10 MCQ Questions with Answers Question 1. Which of the following is a quadratic equation ? (A) x2 + 2x + 1 = (4 – x)2 + 3 (B) -2×2 = (5 – x) (C) (k + 1)x2 +x = 7, where = -1 (D) x3 \u2014 x2 \u2014 (x – 1)3 …<\/p>\n
\nx3<\/sup> – x2<\/sup> = x3<\/sup> -1 – 3x(x -1)
\nx3<\/sup> – x2<\/sup> = x3<\/sup> -1 – 3x2<\/sup>+ 3x -x2<\/sup>+ 3x2<\/sup> +1 – 3x = 0
\n2x2<\/sup> – 3x +1 = 0
\nIt is of the form of ax2<\/sup> +bx +c=0.<\/p>\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions.png?resize=159%2C13&ssl=1\")
<\/p>\nWhich of the following is not a quadratic equation?<\/h2>\n
\n(B) 2x – x2<\/sup>= x2<\/sup> + 5
\n(C) (\u221a2x + \u221a3)2<\/sup> + x2<\/sup> = 3x2<\/sup> \u2014 5x
\n(D) (x2<\/sup> + 2x)2<\/sup> = x4<\/sup>+ 3 + 4x3<\/sup>
\nAnswer:
\n(C) (\u221a2x + \u221a3)2<\/sup> + x2<\/sup> = 3x2<\/sup> \u2014 5x<\/p>\n
\n(\u221a2x)2<\/sup> + (\u221a3)2<\/sup> + 2 x \u221a2x x \u221a3 + x2<\/sup> = 3x2<\/sup>-5x
\n2x2<\/sup> 3 + 2\u221a6x + x2<\/sup> = 3x2<\/sup> – 5x
\n3x2<\/sup> + 2\u221a6x + 3 = 3x2<\/sup> – 5x
\nx(5 + 2\u221a6) + 3 = 0
\nIt is not of the form of ax2<\/sup> +bx +c = 0.<\/p>\nWhich of the following equations has 2 as a root ?<\/h2>\n
\n(B) x2<\/sup> + 3x-12 = 0
\n(C) 2x2<\/sup>-7x + 6 = 0
\n(D) 3x2<\/sup>-6x-2 = 0 0
\nAnswer:
\n(C) 2x2<\/sup>-7x + 6 = 0<\/p>\n
\nPut the value of x = 2 in 3x(C) 2x2<\/sup>-6x – 2 = 0
\n3(2)2<\/sup> – 6(2) – 2 = 0
\n12 – 12 – 2 =0
\n12 – 14 =0
\n-2 \u2260 0
\nSo, x = 2 is not a root of 3x2<\/sup> – 6x – 2 = 0<\/p>\nIf \\(\\frac{1}{2}\\) is a root of the equation x2<\/sup> + kx – \\(\\frac{5}{4}\\) = 0 then,the value of k is<\/h2>\n
\n(B) -2
\n(C) \\(\\frac{1}{2}\\)
\n(D) \\(\\frac{1}{2}\\)
\nAnswer:
\n(A) 2<\/p>\n
\nSince,\\(\\frac{1}{2}\\) is a root of the equation
\nx2<\/sup> + kx\u2014\\(\\frac{5}{4}\\) = 0
\nThen, \\(\\left(\\frac{1}{2}\\right)^{2}\\) + k \\(\\left(\\frac{1}{2}\\right)\\) – \\(\\frac{5}{4}\\) = 0
\n\\(\\frac{1}{4}\\) + \\(\\frac{k}{2}\\) – \\(\\frac{5}{4}\\) = 0
\n\\(\\frac{k}{2}\\) = \\(\\frac{5}{4}\\) – \\(\\frac{1}{4}\\)
\n\\(\\frac{k}{2}\\) = 1
\nk = 1<\/p>\nWhich of the following equations has the sum of its roots as 3 ?<\/h2>\n
\n(B) \u2014x2<\/sup> + 3x-3 = 0
\n(C) \u221a2x2<\/sup>–\\(\\frac{3}{\\sqrt{2}}\\) = x + 1 = 0
\n(D) 3x2<\/sup> – 3x + 3 = 0
\nAnswer:
\n(B) \u2014x2<\/sup> + 3x-3 = 0<\/p>\n
\n-x2<\/sup> – + 3x – 3 = 0
\nOn comparing with ax2<\/sup> +bx +c = 0
\na = -1, b= 3, c = – 3 -b -3
\n\u2234Sum of the roots = \\(\\frac{-b}{a}\\) = \\(\\frac{-3}{-1}\\) = 3<\/p>\n<\/p>\nThe rooks of the equation x2<\/sup>2 + 7x + 10 = 0 are:<\/h2>\n
\n(B) 5, 2
\n(C) 5, -2
\n(D) -5, 2
\nAnswer:
\n(A) -5, -2<\/p>\n
\nGwen, .x2<\/sup> + 7x + 10
\nOn Comparing with ax2<\/sup> + bx + c = 0 we get a = 1, b = 7 and c = 10
\n\u21d2 x = \\(\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}\\) – \\(-\\frac{1}{2}\\)
\n\u21d2 x = \\(\\frac{-7 \\pm \\sqrt{(7)^{2}-4 \\times 1 \\times 10}}{2 \\times 1}\\)
\n\u21d2 x = \\(\\frac{-7 \\pm \\sqrt{9}}{2}\\)
\n\u21d2 x= \\(\\frac{-7 \\pm 3}{2}\\)
\n\u21d2 x = \\(\\frac{-7+3}{2}\\) or \\(\\frac{-7-3}{2}\\)
\n=> x= – 2 or -5
\nHence, the roots of the given equation are – 2 or – 5.<\/p>\nWhich of the following equations has two distinct real roots?<\/h2>\n
\n(B) x2<\/sup> + x- 5 = 0
\n(C) x2<\/sup> +3x + 2\u221a2 = 0
\n(D) 5x2<\/sup>-3x + 1 = 0
\nAnswer:
\n(B) x2<\/sup> + x- 5 = 0<\/p>\n
\nx2<\/sup> + x – 5 = 0
\nOn comparing with ax2<\/sup> +bx +c = 0
\na = 1, b= 1, c = – 5
\nb2<\/sup> – 4ac = 0
\n(1)-4(1) (-5) =1+20 =21 >0
\nHence, the equation has two distinct real roots.<\/p>\nValues of k for which the quadratic equation 2x2<\/sup> – kx + k = 0 has equal roots is:<\/h2>\n
\n(B) 4
\n(C) 8 only
\n(D) 0, 8
\nAnswer:
\n(B) 4<\/p>\n
\nGiven equation is 2x2<\/sup> – kx + k = 0 On comparing with ax2<\/sup> +bx +c =0
\na = 2, b=-k c=k
\nFor equal roots b2<\/sup> – 4ac = 0
\n(- k)2<\/sup> – 4 (2)(k) = 0
\nk2 \u2014 8k = 0
\nk(k-8) = 0
\nk = 0, 8
\nHence, the required values of k are 0 and 8.<\/p>\nWhich constant must be added and subtracted to solve the quadratic equation 9x2<\/sup> + \\(\\frac{3}{4}\\)x-\u221a2 = 0 by the method of completing the square?<\/h2>\n
\n(B) \\(\\frac{1}{64}\\)
\n(C) \\(\\frac{1}{4}\\)
\n(D) \\(\\frac{9}{64}\\)
\nAnswer:
\n(B) \\(\\frac{1}{64}\\)<\/p>\n
\nGiven equation is
\n9x2<\/sup> + \\(\\frac{3}{4}\\)x-\u221a2 = 0
\n(3x)2<\/sup>+\\(\\frac{1}{4}\\)(3x) = \u221a2
\n(3x)2<\/sup> + \\(\\frac{1}{4}\\) (3x) + \\(\\left(\\frac{1}{8}\\right)^{2}\\) = \\(\\left(\\frac{1}{8}\\right)^{2}\\) + \u221a2
\n\\(\\left(3 x+\\frac{1}{8}\\right)^{2}\\) = \\(\\frac{1}{64}\\) + \u221a2
\nThus, \\(\\frac{1}{64}\\) must be added and subtracted to solve the quadratic equation.<\/p>\n<\/p>\nThe quadratic equation 2x2<\/sup> – \u221a5x + 1 = 0 has<\/h2>\n
\n(B) two equal real roots
\n(C) no real roots
\n(D) more than 2 real roots
\nAnswer:
\n(C) no real roots<\/p>\n
\n2x2<\/sup> – \u221a5x + 1 = 0
\nOn comparing with ax2<\/sup> +bx +c =0
\na = 2, b = – \u221a5 , c= 1
\nDiscriminant = b2<\/sup> – 4ac = ( – \u221a5 )2<\/sup> – 4(2)(1)
\n= 5 – 8 = -3<0
\nTherefore, the equation has no real roots.<\/p>\nWhich of the following equations has no real roots?<\/h2>\n
\n(B) x2<\/sup> + 4x-3\u221a2 =0
\n(C) x2<\/sup> -4x-3\u221a2 =0
\n(D) 3x2<\/sup> + 4\u221a3x + 4 = 0
\nAnswer:
\n(A) x2<\/sup> -4x + 3\u221a2 =0<\/p>\n
\nx2<\/sup> – 4x + 3\u221a2 = 0
\nOn comparing with ax2<\/sup> +bx +c =0
\na = 1, b= – 4, c = 3\u221a2
\nDiscriminant = b2<\/sup> – 4ac = (- 4 )2<\/sup> – 4 (1)(3\u221a2)
\n= 16-12 \u221a2
\n= 16-12 \u00d7 1.41 = 16 – 16.92 = – 0.92 < 0
\nTherefore, the equation has no real roots.<\/p>\n(x2<\/sup> + 1)2<\/sup> – x2<\/sup> = 0 has<\/h2>\n
\n(B) two real roots
\n(C) no real roots
\n(D) one real root.
\nAnswer:
\n(C) no real roots<\/p>\n
\n(x2<\/sup>+ 1)2<\/sup>-x2<\/sup> = 0
\nx4<\/sup>+1+2x2<\/sup> – x2<\/sup> = 0
\nx4<\/sup> + x2<\/sup> + 1 = 0
\nLet x2<\/sup> = y
\ny2<\/sup> + y + 1 = 0
\nOn comparing with ay2<\/sup> + by + c = 0
\na = 1, b = 1, c = 1
\nDiscriminant = b2<\/sup> – 4ac
\n= (1)2<\/sup>-4(1)(1)
\n= 1- 4 = -3 < 0
\nTherefore, the equation has no real roots.<\/p>\n
\n(A) BothAandR are true and R is the correct explanation of A
\n(B) Both A and R are true but R is NOT the correct explanation of A
\n(C) A is true but R is false
\n(D) A is false and R is True<\/p>\nAssertion (A): The positive root of \\(\\sqrt{3 x^{2}+6}\\) =9 is 5.
\nReason (R): If x =\\(\\frac{2}{3}\\)and x = -3 are roots of the 3 quadratic equation ax2<\/sup> + 7x + b = 0, then the value of a and b are 3 and -6.<\/h2>\n
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
\nIn case of assertion:
\n\\(\\sqrt{3 x^{2}+6}\\) = 9
\n3x2<\/sup> + 6 = 81
\n3x2<\/sup> = 81 – 6 = 75
\nx2<\/sup> = \\(\\frac{75}{3}\\) = 25
\nx = \u00b1 5
\nHence, positive root = 5.
\n\u2234Assertion is correct.
\nIn case of reason:
\nSubstituting x =\\(\\frac{2}{3}\\) in ax2<\/sup> + 7x + b = 0
\n\\(\\frac{4}{9} a\\) + \\(\\frac{14}{3}\\) + b = 0
\n\u21d2 4a + 42 + 9b = 0
\n\u21d24a + 9b = – 42 …(i)
\nagain\/substituting x = – 3 in ax2<\/sup> + 7x + b = 0
\n9a – 21 + b = 0
\n\u21d29a + b = 21 …(ii)
\nSolving (i) and (ii), we get
\na = 3 and b = – 6
\n\u2234 Reason is correct
\nHence, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\nAssertion (A): The solution of the quadratic equation (x – 1)2<\/sup> – 5 (x – 1) – 6 = 0 is 0 or 7.
\nReason (R): The solution of the equation x2<\/sup> + 5x – (a2<\/sup> + a – 6) = 0 is a + 3.<\/h2>\n
\n(C) A is true but R is false<\/p>\n
\nIn case of assertion:
\nGiven, (x-1)2<\/sup>– -5(x -1) – 6 = 0
\n\u21d2 x2<\/sup> – 2x + 1 – 5x + 5 – 6 = 0
\n\u21d2 x2<\/sup> – 7x+ 6 – 6 = 0
\n\u21d2 x2<\/sup> – 7x = 0
\n\u21d2 x(x – 7) = 0
\nx = 0 or 7
\n\u2234 Assertion is correct.
\nIn case of reason:
\nx2<\/sup>+ 5x – (a2<\/sup> + a – 6) = 0
\n\u21d2 x2<\/sup> – 5x – (a2 + a – 6) = 0
\n\u21d2 x2<\/sup> + 5x – [a(a + 3) – 2 (a + 3)] = 0
\n\u21d2 x2<\/sup> + 5x – (a + 3)(a – 2) = 0<\/p>\n
\n\u21d2 x2<\/sup> + (a + 3)x -(a- 2)x – (a + 3 )(a – 2) = 0
\n\u21d2 x[x + (a + 3)] – (a – 2) [x + (a + 3)] = 0
\n\u21d2 [x + (a + 3)][x-(a – 2)] =0
\n\u21d2 x = – (a + 3) or x = a – 2
\nHence, roots of given equation are – (a + 3) and a – 2.
\nReason is incorrect
\nHence, assertion is correct but reason is incorrect.<\/p>\n<\/p>\nAssertion (A): A two digit number is four times the sum of the digits. If it is also equal to 3 times the product of the digits, then the number is 25.
\nReason (R): The denominator of a fraction is one more than twice its numerator. If the sum of the 16 fraction and its reciprocal is \\(2\\frac{16}{21}\\), then the fraction is \\(\\frac{3}{7}\\)<\/h2>\n
\n(D) A is false and R is True<\/p>\n
\nIn case of assertion:
\nLet unit’s digit and ten’s digit of the two digit number be x and y respectively .\u2019. Number is 10y + x According to question,
\n10y + x = 4 (y + x)
\n\u21d2 10y + x = 4y + 4x
\n\u21d2 10y – 4y = 4x – x
\n\u21d2 6y = 3x
\n\u21d2 2y = x …(i)
\nAlso, 10y + x = 3xy …(ii)
\n\u21d2 10y + 2y = 3(2y)y [From eq (i)]
\n\u21d2 12y = by2<\/sup>
\n\u21d2 6y2<\/sup> – 12y = 0
\n\u21d2 6y(y \u2014 2) = 0
\n\u21d2 y=0 or y = 2
\nRejecting y = 0 as tens digit should not be zero for a two digit number.
\n\u21d2 x = 4
\n.’. Required number = 10y + x
\n\u21d2 10 x 2 + 4 = 24.
\nAssertion is incorrect.
\nIn case of reason:
\nLet numerator be x.
\nThen, the fraction = \\(\\frac{x}{2 x+1}\\)
\nAgain,\\(\\frac{x}{2 x+1}\\) + \\(\\frac{2 x+1}{x}\\) = \\(\\frac{58}{21}\\)
\n\u21d2 21 [x2<\/sup> + (2x + 1)2<\/sup>] = 58 (2x2<\/sup> + x
\n\u21d2 11x2<\/sup> – 26x – 21 = 0
\n11x2<\/sup> – 33x + 7x – 21 = 0
\n(x- 3)(11x + 7) = 0
\nx = 3 or – \\(-\\frac{7}{11}\\)
\n(rejected negative value)
\nHence, fraction = \\(\\frac{3}{7}\\)
\nReason is correct Hence, assertion is correct incorrect but reason is correct<\/p>\nAssertion (A): If the coefficient of x2<\/sup> and the constant term have the same sign and if the coefficient of x term is zero then the quadratic equation has no real roots.
\nReason (R): The equation 13\u221a3x2<\/sup> + 10x + \u221a3 = 0 has not real roots.<\/h2>\n
\n(A) BothAandR are true and R is the correct explanation of A<\/p>\n
\nIn case of assertion:
\nSince, in this case discriminant is always negative, so it has no real roots, that is, if b = 0, then b2 – 4ac => – 4ac < 0 and ac > 0.
\n\u2234 Assertion is correct.
\nIn case of reason:
\nHere, a = 13 \u221a3 , b = 10 and c = \u221a3
\nThen, b2<\/sup> – 4ac = (10)2<\/sup> – 4(13 \u221a3)(\u221a3 )
\n= 100 -156
\n= -56
\nAs D < 0.
\nSo, the equation has not real roots.
\n\u2234 Reason is correct.
\nHence, assertion and reason are correct and reason is the correct explanation for assertion.<\/p>\nAssertion (A): Every quadratic equation has exactly one root.
\nReason (R): Every quadratic equation has almost two roots.<\/h2>\n
\n(D) A is false and R is True<\/p>\n
\nIn case of assertion:
\nSince, a quadratic equation has two and only two roots.
\n\u2234 Assertion is incorrect.
\nIn case of reason:
\nBecause every quadratic polynomial has almost two roots.
\n\u2234Reason is correct.
\nHence, assertion is incorrect but reason is : correct.<\/p>\nAssertion (A): If the coefficient of x2 and the constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots.
\nReason (R): Every quadratic equation has at least two roots.<\/h2>\n
\n(C) A is true but R is false<\/p>\n
\nIn case of assertion:
\nSince, in this case discriminant is always positive, so it has always real roots, that is, ac < 0 and so, b2<\/sup>– 4ac > 0.
\n\u2234 Assertion is correct.
\nTn case of reason:
\nSince, a quadratic equation has two and only two roots.
\n\u2234 Reason is incorrect.
\nHence, assertion is incorrect but reason is incorrect.<\/p>\n
\nI. Read the following text and answer the following questions on the basis of the same:
\nRaj and Ajay are very close friends. Both the families decide to go to Ranikhet by their own cars. Raj’s car travels at a speed of x km\/h while Ajay’s car travels 5 km\/h faster than Raj’s car. Raj took 4 hours more than Ajay to complete the journey of 400 km.
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-4-Quadratic-Equations-1.png?resize=337%2C178&ssl=1\")
<\/p>\nWhat will be the distance covered by Ajay’s car in two hours ?<\/h2>\n
\n(B) (x – 5) km
\n(C) 2(x + 10) km
\n(D) (2x + 5) km
\nAnswer:
\n(A) 2(x + 5) km<\/p>\n
\nSpeed of Raj’s car = x km\/hr
\nSpeed of Ajay’s car = (x + 5) km\/hr
\nDistance covered by Ajay in 2 hours
\n= [(x + 5) x 2] km
\n= 2(x + 5) km.<\/p>\n<\/p>\nWhich of the following quadratic equation describe the speed of Raj’s car?<\/h2>\n
\n(B) x2<\/sup> + 4x – 400 = 0
\n(C) x2<\/sup> + 5x – 500 = 0
\n(D) x2<\/sup> – 4x + 400 = 0
\nAnswer:
\n(C) x2<\/sup> + 5x – 500 = 0<\/p>\nWhat is the speed of Raj’s car<\/h2>\n
\n(B) 15 km\/hour
\n(C) 25 km\/hour
\n(D) 10 km\/hour
\nAnswer:
\n(A) 20 km\/hour<\/p>\nmuch time took Ajay to travel 400 km?<\/h2>\n
\n(B) 40 hour
\n(C) 25 hour
\n(D) 16 hour
\nAnswer:
\n(D) 16 hour<\/p>\n
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-4-Quadratic-Equations-2.png?resize=351%2C362&ssl=1\")
<\/p>\nLet speed of the stream be x km\/hr. then speed of the motorboat in upstream will be<\/h2>\n
\n(B) (20 + x) km\/hr
\n(C) (20 – x) km\/hr
\n(D) 2 km\/hr
\nAnswer:
\n(C) (20 – x) km\/hr<\/p>\n
\nSpeed of motorboat in upstream = Speed of motorboat – Speed of stream = (20 – x) km\/hr<\/p>\nWhat is the relation between speed .distance and time?<\/h2>\n
\n(B) distance = (speed )\/time
\n(C) time = speed x distance
\n(D) speed = distance x time
\nAnswer:
\n(B) distance = (speed )\/time<\/p>\n
\n(B) x2<\/sup> + 20x – 400 = 0
\n(C) x2<\/sup> + 30x – 400 = 0
\n(D) x2<\/sup> – 20x – 400 = 0
\nAnswer:
\n(C) x2<\/sup> + 30x – 400 = 0<\/p>\nWhat is the speed of current?<\/h2>\n
\n(B) 10 km\/hour
\n(C) 15 km\/hour
\n(D) 25 km\/hour
\nAnswer:
\n(B) 10 km\/hour<\/p>\nHow much time boat took in downstream?<\/h2>\n
\n(B) 15 minute
\n(C) 30 minute
\n(D) 45 minute
\nAnswer:
\n(C) 30 minute<\/p>\n
\nJohn and Jivanti are playing with the marbles. They together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124.
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-4-Quadratic-Equations-3.png?resize=241%2C188&ssl=1\")
<\/p>\nIf John had x number of marbles, then number of marbles Jivanti had:<\/h2>\n
\n(B) 45-x
\n(C) 45x
\n(D) x – 5
\nAnswer:
\n(B) 45-x<\/p>\n
\nIf John had x number of marbles, then Jivanti had (45 – x) marbles, because there are total 45 marbles.<\/p>\nNumber of marbles left with Jivanti, when she lost 5 marbles:<\/h2>\n
\n(B) 40-x
\n(C) 45 -x
\n(D) x- 40
\nAnswer:
\n(B) 40-x<\/p>\n
\nNumber of marbles left with Jivanti, when she lost 5 marbles = (45 – x – 5) = (40 – x)<\/p>\n<\/p>\nThe quadratic equation related to the given problem is:<\/h2>\n
\n(B) x2<\/sup> + 45x + 324 = 0
\n(C) x2<\/sup> – 45x – 324 = 0
\n(D) -x2<\/sup> – 45x + 324 = 0
\nAnswer:
\n(A) x2<\/sup> – 45x+ 324 = 0<\/p>\n
\nAccording to question,
\n(x-5)(40-x) = 124
\n\u21d2 -x2<\/sup> – 200 + 40x + 5x – 124 = 0
\n\u21d2 x2<\/sup> – 45x + 324 = 0<\/p>\nNumber of marbles John had:<\/h2>\n
\n(B) 9
\n(C) 35
\n(D) 30
\nAnswer:
\n(B) 9<\/p>\n
\nx2<\/sup> – 45x + 324 = 0
\n\u21d2 x2<\/sup>– 9x – 36x + 324 = 0
\n\u21d2 x(x – 9) – 36(x – 9) = 0
\n\u21d2 (x – 9)(x – 36) = 0
\nEither x = 9 or x = 36.
\nTherefore, the number of marbles John had 9 1 or 36.<\/p>\nIf John had 36 marbles, then number of marbles Jivanti had:<\/h2>\n
\n(B) 9
\n(C) 36
\n(D) 35
\nAnswer:
\n(B) 9<\/p>\n
\nIf John had 36 marbles, then I Jivanti had (45 – 36) = 9 marbles.<\/p>\n
\n![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-4-Quadratic-Equations-4.png?resize=274%2C104&ssl=1\")
<\/p>\nThe value of x is:<\/h2>\n
\n(B) 3
\n(C) 4
\n(D) 5
\nAnswer:
\n(B) 3<\/p>\n
\nGiven, area of triangle = 60 cm2<\/sup>
\n\u21d2 \\(\\frac{1}{2}\\) \u00d7 AB \u00d7BC = 0
\n\u21d2 (5x)(3x – 1) =120
\n\u21d2 3x2<\/sup>-x-24 = 0
\n\u21d2 3x2<\/sup>– 9x + 8x – 24 = 0
\n\u21d2 3x(x – 3) + 8(x – 3) =0
\n\u21d2 (x – 3)(3x + 8) = 0
\nEither x = 3 or x = \\(-\\frac{8}{3}\\)
\nSince length can’t be negative, then x = 3.<\/p>\nThe length of AB is:<\/h2>\n
\n(B) 10 cm
\n(C) 15 cm
\n(D) 17 cm
\nAnswer:
\n(C) 15 cm
\nExplanation:
\nThe length of AB = 5x cm
\n= 5 x 3 cm
\n= 15 cm<\/p>\nThe length of AC is:<\/h2>\n
\n(B) 15 cm
\n(C) 21 cm
\n(D)20 cm
\nAnswer:
\n(A) 17 cm<\/p>\n
\nAB = 15 cm and BC = (3x – 1) cm
\n= (3 x 3-1) cm
\n= 8 cm
\nNow, in right angled AABC,
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\n(By using Pythagoras theorem)
\n= (15)2<\/sup>+ (8)2<\/sup> = 225 + 64 = 289 = (17)2<\/sup>
\nHence, AC = 17 cm.<\/p>\n<\/p>\nThe perimeter of \u2206ABC is:<\/h2>\n
\n(B) 45 cm
\n(C) 30 cm
\n(D) 40 cm
\nAnswer:
\n(D) 40 cm<\/p>\n
\nHere, AB = 15 cm, BC = 8 cm and AC = 17 cm.
\nThen, the perimeter of \u2206ABC = (AB + BC + CA) cm
\n= (15 + 8 + 17) cm = 40 cm.<\/p>\nThe given problem is based on which mathematical concept?<\/h2>\n
\n(B) Linear equation in one variable
\n(C) Quadratic Equations
\n(D) None of these
\nAnswer:
\n(C) Quadratic Equations<\/p>\n
\nThe given problem is based on the concept of quadratic equations.<\/p>\nMCQ Questions for Class 10 Maths with Answers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"