![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions.png?resize=159%2C13&ssl=1\")
<\/p>\nQuestion 1.<\/p>\n
(A) intersecting at exactly one point
\n(B) intersecting at exactly two points
\n(C) coincident
\n(D) parallel
\nAnswer:
\n(D) parallel<\/p>\n
Explanation:
\n\\(\\frac{a_{1}}{a_{2}}=\\frac{6}{2}=3\\) \\(\\frac{b_{1}}{b_{2}}=\\frac{-3}{-1}=3\\) \\(\\frac{c_{1}}{c_{2}}=\\frac{10}{9}\\)
\n\u21d2 \\(\\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}} \\neq \\frac{c_{1}}{c_{2}}\\)
\nSo, the system of linear equations is inconsistent (no solution) and graph will be a pair of parallel lines.<\/p>\n
<\/p>\n
Question 2.<\/p>\n
(A) a unique solution
\n(B) exactly two solutions
\n(C) infinitely many solutions
\n(D) no solution
\nAnswer:
\n(D) no solution<\/p>\n
Explanation:
\n\\(\\frac{a_{1}}{a_{2}}=\\frac{1}{-3}\\)\\(=-\\frac{1}{3},\\frac{b_{1}}{b_{2}}{a}\\)\\(\\frac{-1}{3},\\frac{c_{1}}{c_{2}}=\\frac{5}{1}\\)
\n\\(\\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}} \\neq \\frac{c_{1}}{c_{2}}\\)
\nSo, the system of linear equations has no solution<\/p>\n
Question 3.<\/p>\n
(A) parallel
\n(B) always coincident
\n(C) intersecting or coincident
\n(D) always intersecting E
\nAnswer:
\n(C) intersecting or coincident<\/p>\n
Explanation:
\nCondition for consistency:
\n\\(\\frac{a_{1}}{a_{2}} \\neq \\frac{b_{1}}{b_{2}}\\)has unique solution (consistent), i..e.,
\nintersecting at one point
\n\\(\\text { or } \\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}}=\\frac{c_{1}}{c_{2}}\\)
\n(Consistent lines, coincident or dependent)<\/p>\n
Question 4.<\/p>\n
(A) parallel
\n(B) intersecting at (b, a)
\n(C) coincident
\n(D) intersecting at (a, b)
\nAnswer:
\n(D) intersecting at (a, b)<\/p>\n
Explanation:
\n(x= a) is the equation of a straight line parallel to the y-axis at a distance ‘a’ from it. Again, y = b is the equation of a straight line parallel to the x-axis at a distance b from it. So, the pair of equations x = a and y = b graphically represents lines which are intersecting at (a,b).<\/p>\n
<\/p>\n
Question 5.<\/p>\n
(A) one solution
\n(B) two solutions
\n(C) infinitely many solutions
\n(D) no solution
\nAnswer:
\n(D) no solution<\/p>\n
Explanation:
\nWe know that equa tion of I he form y \u2014 a is a line parallel to r-axis at a distance V from it. y = 0 is the equation of the r-axis and y = -7 is the equation of the line parallel to the .v-axis. So, these two equations represent two parallel lines. Therefore, there is no solution<\/p>\n
Question 6.<\/p>\n
(A) 10x + 14y + 4 = 0
\n(B) -10x – 14y + 4 = 0
\n(C) -10x + 14y + 4 = 0
\n(D) 10x – 14y= – 4
\nAnswer:
\n(D) 10x – 14y= – 4<\/p>\n
Explanation:
\n\\(\\frac{a_{1}}{a_{2}}=\\)=\\(\\frac{b_{1}}{b_{2}}\\)=\\(\\frac{c_{1}}{c_{2}}\\)=\\(\\frac{1}{k}\\)
\nGiven equation of line is, – 5x + 7y – 2 = 0
\nHere,a1 = – 5, b1 = 7, c1 = – 2
\nFrom EQuestion (i), – \\(\\frac{5}{a_{2}}\\) = \\(\\frac{7}{b_{2}}\\) = \\(\\frac{2}{c_{2}}=\\) = \\(\\frac{1}{k}\\)
\n\u21d2 a2 = – 5k, b2 = 7k, c2 = – 2k
\nwhere k is any arbitrary constant.
\nputting k =2, then a2 = – 10, b2 = =14 and c2 = – 4
\n\u21d2 The required equation of line becomes,
\na2x + b2y + c2 = 0
\n\u21d2 – 10x + 14y – 4 = 0
\n\u21d2 10x – 14y + 4 = 0<\/p>\n
Question 7.<\/p>\n
(A) \\(\\frac{1}{2}\\)
\n(B) \\(\\frac{-1}{2}\\)
\n(C) 2
\n(D) -2
\nAnswer:
\n(C) 2
\nExplanation:
\n3x – y = – 18
\n6x – ky = – 16
\nFor coincident lince,
\n\u21d2 \\(\\frac{a_{1}}{a_{2}}\\)= \\(\\frac{b_{1}}{b_{2}}\\) = \\(\\neq \\frac{c_{1}}{c_{2}}\\)
\n\u21d2 \\(\\frac{3}{6}\\) = \\(\\frac{-1}{-k}\\) \\(\\frac{-8}{-16}\\)
\n\u21d2 \\(\\frac{1}{2}\\) = \\(\\frac{1}{k}\\) = \\(\\frac{1}{2}\\)<\/p>\n
Question 8.<\/p>\n
(A) \\(\\frac{-5}{4}\\)
\n(B) \\(\\frac{2}{5}\\)
\n(C) \\(\\frac{15}{4}\\)
\n(D) \\(\\frac{3}{2}\\)
\nAnswer:
\n(C) \\(\\frac{15}{4}\\)<\/p>\n
Explanation:
\nFor parallel lines (or no solution)
\n\\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{b_{1}}{b_{2}} \\neq \\frac{c_{1}}{c_{2}}\\)
\n\u21d2 \\(\\frac{3}{2}\\) = \\(\\frac{2 k}{5} \\neq \\frac{-2}{1}\\)
\n\u21d2 4k = 15
\n\u21d2 k = \\(\\frac{15}{4}\\)<\/p>\n
<\/p>\n
Question 9.<\/p>\n
(A) x + y = -1
\n2x – 3y = -5<\/p>\n
(B) 2x + 5y = -11
\n4x + 10y = -22<\/p>\n
(C) 2x-y = 1
\n3x + 2y = 0<\/p>\n
(D) x-4y-14 = 0
\nx -y-13=0
\nAnswer:
\nOption (B) & (D) is correct<\/p>\n
Explanation:
\n(B) 2x + 5y = \u201411 and 4,r + 10y = -22
\nPut x = 2 and y = -3 in both the equations,
\nLHS = 2x + 5y \u21d2 2 x 2 + (-3)
\n\u21d2 4- 15 = -11 = RHS
\nLHS = 4x + 10y -> 4(2) + 10(-3)
\n\u21d28 – 30 – 22 = RHS
\n(D) x – 4y – 14 = 0 and 5x – y – 13 = 0
\nx – 4y = 14 and 5x – y = 13
\nPut x = 2 and y = -3 in both the equations,
\nLHS = x – 4y \u21d2 2 – 4(- 3) \u21d2 2 + 12 = 14 = RHS
\nLHS = 5x – y\u00a0 \u21d2 5(2) – (- 3) \u21d2 10 + 3 = 13 = RHS<\/p>\n
Question 10.<\/p>\n
(A) 35 and 15
\n(B) 35 and 20
\n(C) 15 and 35
\n(D) 25 and 25
\nAnswer:
\n(D) 25 and 25<\/p>\n
Explanation:
\nLet the number of \u20b9 1 coins = x and the number of \u20b9 2 coins = y
\nSo, according to the question,
\nx + y = 50 ………(i)
\nx + 2y =75 ………(ii)
\nSubtracting equation (i) from (ii),
\ny = 25
\nSubstituting value of y in (i)x = 25
\nSo, y = 25 and x = 25<\/p>\n
Question 11.<\/p>\n
(A) 4 and 24
\n(B) 5 and 30
\n(C) 6 and 36
\n(D) 3 and 24
\nAnswer:
\n(C) 6 and 36
\nExplanation:
\nLet the present age of father be x years and the present age of son be y years.
\n\u2234 According to the question,
\nx = 6y …(i)
\nAge of the father after lour years = (x + 4) years
\nAge of son after four years = (y + 4) years Now, according to the question,
\nx + 4 = 4(y + 4)
\nx + 4 = 4(y + 4)
\nx +4 = 4y + 16
\n\u21d2 6y – 4y = 16 – 4
\n\u21d2 2y = 12
\n\u21d2 y = 6
\nx = 6 \u00d7 6 = 36 years
\ny = 6 years
\nSo, the present ages of the son and the father are 6 yeaers and 36 years, respectiveiy.<\/p>\n
<\/p>\n
Question 12.<\/p>\n
(A) 3 and 5
\n(B) 5 and 3
\n(C) 3 and 1
\n(D) – 1 and 3
\nAnswer:
\n(C) 3 and 1<\/p>\n
Explanation:
\nIf (a, b) is the solution of the given equations, then it must satisfy the given equations, so,
\na – b = 2 ………(i)
\n17 a + b = 4 ……. (ii)
\n\u21d2 2a = 6[Adding (i) and (ii) ]
\n\u21d2 a = 3
\nNow, 3 + 6 = 4
\n\u21d2 b = 1
\nSo, (a, b) = (3, 1).<\/p>\n
Question 13.<\/p>\n
(A) 99\u00b0, 81\u00b0
\n(B) 98\u00b0, 82\u00b0
\n(C) 97\u00b0, 83\u00b0
\n(D) None of these.
\nAnswer:
\n(A) 99\u00b0, 81\u00b0<\/p>\n
Explanation:
\nLet one angle be x.
\nThen, other angle (it’s supplementary angle)
\n= (180\u00b0 – x)
\nGiven, x + 18\u00b0 = (180\u00b0 – x)
\n2x = 180\u00b0 – 18\u00b0
\n2x = 162\u00b0
\nx = 81\u00b0
\nother angle = 180\u00b0 – 81\u00b0 = 99\u00b0
\nHence two required angles are 81\u00b0 and 99\u00b0<\/p>\n
<\/p>\n
Question 14.<\/p>\n
(A) \\(30, \\frac{5}{7}\\)
\n(B) \\(31, \\frac{-5}{7}\\)
\n(C) \\(32, \\frac{5}{7}\\)
\n(D) None of these.
\nAnswer:
\n(B) \\(31, \\frac{-5}{7}\\)
\nExplanation:
\nGiven,
\n2x + y = 23 …(i) and 4x – y = 19 …(ii)
\nOn adding eQuestion (i) and (ii), we get
\n6x = 42 \u21d2 x = 7
\nputting the value of x in eQuestion (i), we get
\n\u21d214 + y =23
\n\u21d2y = 23 \u201414 = 9
\nHence, 5y – 2x = 5 x 9 – 2 x 7 = 45 – 14 =31.
\nand \\(\\frac{y}{x}\\) -2 = \\(=\\frac{9}{7}\\) -2 = \\(\\frac{9-14}{7}\\) = \\(\\frac{-5}{7}\\)<\/p>\n
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: So, Here \\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{1}{2}\\),\\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{1}{k}\\) Question 2.<\/p>\n Answer: Explanation: 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. Situation 1: In city A, for a journey of 10 km, the charge paid is \u20b9 75 and for a journey of 15 km, the charge paid is \u20b9 110. Question 1.<\/p>\n (A) x + 10y = 110, x + 15y = 75 Answer: Explanation: Question 2.<\/p>\n (A) \u20b9 155 Explanation: Question 3.<\/p>\n (A) \u20b9 185 Question 4.<\/p>\n Question 1.<\/p>\n (A) (u + v) km\/hr Explanation: Question 2.<\/p>\n (A) (u + v) km\/hr Explanation: Question 3<\/p>\n (A) 60 km\/hr Explanation: Question 4.<\/p>\n (A) 60 km\/hr Explanation: Question 5.<\/p>\n (A) Pair of linear equations Explanation: III. Read the following text and answer the following questions on the basis of the same: Question 1.<\/p>\n (A) 15 Explanation: Question 2.<\/p>\n (A) 40 Explanation: Question 3.<\/p>\n (A) 15 Explanation: Question 4.<\/p>\n (A) 30 Explanation: Question 5.<\/p>\n (A) pair of linear equations Explanation: Pair of Linear Equations in Two Variables Class 10 MCQ Questions with Answers Question 1. Graphically, the pair of equations 6x – 3y + 10 = 0 2x-y + 9 = 0 represents two lines which are (A) intersecting at exactly one point (B) intersecting at exactly two points (C) coincident (D) parallel Answer: (D) …<\/p>\n
\n(A) Both A and R 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): If the pair of linear equations 3x 4 y = 3 and 6x 4 ky = 8 does not have a solution, then the value of k = 2.
\nReason (R): If the pair of linear equations x 4 y – 4 = 0 and 2x 4 ky = 3 does not have a solution, then the value of k = 2.<\/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:
\nGiven equation are:
\n3x + y – 3 = 0
\n6x + ky – 8 = 0
\nComparing eQuestion (i) with a1<\/sub> x + b1<\/sub>y + c1<\/sub> = 0 and eq. (ii) with a2<\/sub>x + b2<\/sub>y + c2<\/sub> = 0 we get,
\na1<\/sub> = 3, a2<\/sub> = 6, b1<\/sub> = 1 b2<\/sub> = k, c1<\/sub> = -3 and c2<\/sub> = – 8 Sine, given equation has solution.<\/p>\n
\n\\(\\frac{a_{1}}{a_{2}}\\) = \\(=\\frac{b_{1}}{b_{2}} \\neq \\frac{c_{1}}{c_{2}}\\)
\n\u21d2\\(\\frac{3}{6}\\) = \\(\\frac{1}{k} \\neq \\frac{-3}{-8}\\)
\n\u21d2\\(\\frac{1}{2}\\) = \\(\\frac{1}{k} \\neq \\frac{3}{8}\\)
\nEither \\(\\frac{1}{2}\\) = \\(\\frac{1}{k}\\) or \\(\\frac{1}{k} \\neq \\frac{3}{8}\\)
\n\u21d2k = 2 or \\(k \\neq \\frac{3}{8}\\)
\nhenece, the value of k is 2. Thus assertion is correct.
\nIn case of reason :
\nGiven equation :
\nx + y – 4 = 0
\n2 x z + ky – 3 = 0<\/p>\n
\nand \\(\\frac{c_{1}}{c_{2}}\\) = \\(\\frac{-4}{-3}\\) = \\(\\frac{4}{3}\\)
\nSystem has solution
\n\u21d2\\(\\frac{a_{1}}{a_{2}}\\) =\\(\\frac{b_{1}}{b_{2}} \\neq \\frac{c_{1}}{c_{2}}\\)
\n\\(\\frac{1}{2}\\) = \\(\\frac{1}{k} \\neq \\frac{4}{3}\\)
\nk =2 or k \u21d2\\(k \\neq \\frac{3}{4}\\)
\nHence, the value of k is 2.
\nThus, reason is correct. Hence both assetion and reason are correct but reason is not the correc t explanation for assertion.<\/p>\nAssertion (A): For all real values of c, the pair of equations x – 2y = 8 and 5x – 10y = c have a unique solution.
\nReason (R): Two lines are given to be parallel. The equation of one of the lines is 4x + 3y = 14,12x + 9y = 5.<\/h2>\n
\n(D) A is false and R is True<\/p>\n
\nIn case of assertion:
\nx -2y = 8
\n5x – 10 = c
\n\u21d2 \\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{1}{5}\\), \\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{-2}{10}\\) = \\(\\frac{1}{5}\\) and = \\(\\frac{c_{1}}{c_{2}}\\) = \\(\\frac{-8}{-c}\\)= \\(=\\frac{8}{c}\\)
\nAs \\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{b_{1}}{b_{2}}\\), so system of equations can never have unique solution.
\n\u2234 Assertion is incorrect.
\nIn case of reason:
\nThe equation of one line is 4x + 3y = 14. We know that if two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are parallel, then
\n\\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{c_{1}}{c_{2}}\\)
\nor \\(\\frac{4}{a_{2}}\\) = \\(\\frac{3}{b_{2}}\\) =\\(\\frac{-14}{c_{2}}\\)
\n\u21d2\\(\\frac{a_{2}}{b_{2}}\\) = \\(\\frac{4}{3}=\\) \\(\\frac{12}{9}\\)
\nhence one of the possible, second parallel line is 12 x + 9y = 5.
\n\u2234 Reason is correct.
\nHence assetin is in correct but reason is correct.<\/p>\nAssertion (A): If the equation 3x – y + 8 = 0 and 6x – ky = -16 represent coincident lines, then the value of k = 2.
\nReason (R): If the lines given by 3x + 2Ay = 2 and 2x + 5y + 1 = 0 are parallel, then the value of k is 15.<\/h2>\n
\n(C) A is true but R is false<\/p>\n
\nIn case of assertion:
\n3x – y = -18
\n6x – ky = -16
\nFor coincident lines,
\n\u21d2 \\(\\frac{a_{1}}{a_{2}}\\) =\\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{c_{1}}{c_{2}}\\)
\n\u21d2 \\(\\frac{3}{6}\\) = \\(\\frac{-1}{-k}\\) = \\(\\frac{-8}{-16}\\)
\n\u21d2 \\(\\frac{1}{2}\\) = \\(\\frac{1}{k}\\) = \\(\\frac{1}{2}\\)
\nSo, k = 2.
\n\u2234 Assertion is correct.
\nIn case of reason:
\nFor parallel lines ( or no solution)
\n\\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{c_{1}}{c_{2}}\\)
\n\u21d2 \\(\\frac{3}{2}\\) = \\(\\frac{2 k}{5}\\) = \\(\\frac{-2}{1}\\)
\n\u21d2 4k = 15
\n\u21d2 k = \\(\\frac{15}{4}\\)
\n\u2234 Reason is incorrect.
\nHence, assertion is but reason is incorrect.<\/p>\n<\/p>\nAssertion (A): If 4 chairs and 3 tables cost \u20b9 2100 and 5 chairs and 2 tables cost \u20b9 1750, then the cost of 1 chair is \u20b9 150.
\nReason (R): Sum of the ages of a father and the son is 40 years. If father’s age is 3 times that of his son, then the son’s age is 12 years.<\/h2>\n
\n(C) A is true but R is false<\/p>\n
\nIn case of assertion:
\nLet cost of 1 chair be \u20b9 x and cost of 1 table be \u20b9 y
\nAccording to the question,
\n4x + +3y = 2100
\n5x + 2y = 1750
\nMultiplying eqn. (i) by 2and enQuestion (ii) by 3, we get
\n8x + 6y = 4200
\n15x + 6y = 5250
\neqn. (iv) – eqn.(iii)
\n\u21d2 7x = 1050
\n\u21d2x = 150
\nSubstituting the value of x in (i) we gel y = 500 Thus, the cost of one chair and one table are \u2234 150 and \u2234 500 respectively.
\n\u2234 Assertion is correct. is
\nIn case of reason:
\nLet age of father and son be x and y respectively.
\nThen, x + y =40 …(i)
\nand x = 3y …(ii)
\nBy solving eqns. (i) and (ii), we get
\nx = 30 and y = 10
\nThus, the ages of father and son are 30 years and 10 years.
\n\u2234 Reason is incorrect.
\nHence, Assertion is correct but reason is incorrect.<\/p>\nAssertion (A): The solution of the pair of linear 19 equations x + y = 5 and 2x – 3y = 4 is x = \\(\\vec{a}\\) and \\(\\vec{a}\\)
\nReason (R): The solution of the pair of linear equations 3x + 4y = 10 and 2x – 2y = 2 is x = 2 and y = 1.<\/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:
\nBy elimination method,
\nx + y = 5 …(i)
\n2x – 3y \u20144 …(ii)
\nMultiplying equation (i) by (ii), we obtain
\n2x + 2y =10 …(iii)
\nSubtracting equation (ii) from equation (iii), we obtain
\n5y = 6
\ny = \\(\\frac{6}{5}\\)
\nSubstituting the value in equation (i), we obtain
\nx = 5 – \\(\\frac{6}{5}\\) = \\(\\frac{19}{5}\\)
\nx = \\(\\frac{19}{5}\\),y =\\(\\frac{6}{5}\\)
\n\u2234 Assertion is correct.
\nIn case of reason:
\nBy elimination method,
\n3x + 4y = 10
\n2x – 2y = 2
\nMultiplying equation (ii) by 2, we ob tain
\n4x – 4y = 4
\nAdding equstion (i) and (ii) , we obtain
\n7x = 14
\nx = 2
\nSubstituting in equstion (i), we obtain
\n6 + 4y = 10
\n4y = 4
\ny = 1
\nx = 2, y = 1
\n\u2234 Reason is correct.
\nHEnce ,both assertion and reason aer correct but reasn is not the correct explanation for assertion<\/p>\nAssertion (A): In a AABC, \u2220C = 3\u2220B = 2(\u2220A + \u2220B), then \u2220A = 20\u00b0.
\nReason (R): The angles of a triangle are x, y and 40\u00b0. The difference between the two angles x and y is 30\u00b0, then A = 85\u00b0 and y = 55\u00b0.<\/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:
\nGiven Aial,
\n\u2220C = 3\/B = 2(\u2220A + \u2220B) 3\u2220B = 2(\u2220A +\u2220B)
\n3\u2220B = 2(\u2220A + 2\u2220B)
\n3\u2220B =2 \u2220A + 2 \u2220B
\n\u2220B = 2 \u2220A
\n2 \u2220A -\u2220B = 0 …(i)
\nWe know that the sum ot the measures of all angles of a triangle is 180″. Therefore,
\n\u2220A + \u2220B + \u2220C = 180\u00b0
\n\u2220A + \u2220B + 3\u2220B = 180\u00b0
\n\u2220A + 4\u2220B = 180\u00b0 …(ii)
\nMultiplying equation (i) by 4, we obtain
\n8\u2220A – 4\u2220B = 0 ……….(iii)
\nAdding equations (ii) and (iii), we obtain
\n9\u2220A = 180\u00b0
\n\u2220A = 20\u00b0
\n\u2234 Assertion is correct.
\nIn case of reason:
\nGiven that, A, y and 40\u00b0are the angels of a triangle.
\nx + y + 40\u00b0 = 180\u00b0
\n[Since the sum of all the angels of a triangle is 180\u00b0]
\n\u21d2 x + y = 140\u00b0 …(i)
\nAlso, x – y =30\u00b0 …(ii)
\nOn adding Eqs. (i) and (ii), we get
\n2x = 170\u00b0
\n\u21d2x = \\(\\frac{170^{\\circ}}{2}\\)
\n\u21d2 x = 85\u00b0
\nOn putting x = 85\u00b0 in EQuestion (i), we get
\n85\u00b0 + y = 140\u00b0
\ny = 140\u00b0 -85\u00b0 = 55\u00b0
\n\u21d2 y = 55\u00b0
\nHence, the required values of x and y are 85\u00b0 and 55\u00b0, respectively.
\n\u2234 Reason is correct.
\nHence, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\n
\nI. Read the following text and answer the following questions on the basis of the same:
\nIt is common that Governments revise travel fares from time to time based on various factors such as inflation (a general increase in prices and fall in the purchasing value of money) on different types of vehicles like auto, Rickshaws, taxis, Radio cab etc. The auto charges in a city comprise of a fixed charge together with the charge for the distance covered. Study the following situations.
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-3-Pair-of-Linear-Equations-in-Two-Variables-1.png?resize=248%2C160&ssl=1\")
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-3-Pair-of-Linear-Equations-in-Two-Variables-2.png?resize=253%2C151&ssl=1\")
<\/p>\n\n\n
\n Name of the city<\/td>\n Distance travelled (km)<\/td>\n Amount paid (\u20b9)<\/td>\n<\/tr>\n \n City A<\/td>\n 10<\/td>\n 75<\/td>\n<\/tr>\n \n <\/td>\n 15<\/td>\n 110<\/td>\n<\/tr>\n \n City B<\/td>\n 8<\/td>\n 9<\/td>\n<\/tr>\n \n <\/td>\n 14<\/td>\n 145<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
\nSituation 2: In a city B, for a journey of 8 km, the charge paid is \u20b9 91 and for a journey of 14 km, the charge paid is \u20b9 145.
\nRefer situation 1<\/p>\nIf the fixed charges of auto rickshaw be \u20b9 x and the running charges be \u20b9 y km\/hr, the pair of linear equations representing the situation is<\/h2>\n
\n(B) x + 10y =75, x + 15y = 110
\n(C) 10x + y =110,15x + y = 75
\n(D) 10x + y = 75,15x + y =110<\/p>\n
\n(B) x + 10y =75, x + 15y = 110<\/p>\n
\nAccording to given situation, we have
\nx + 10y = 75 …(i)
\nx + 15y = 110 …(ii)<\/p>\nA person travels a distance of 50 km. The amount he has to pay is<\/h2>\n
\n(B) \u20b9255
\n(C) \u20b9 355
\n(D) \u20b9 455
\nAnswer:
\n(C) \u20b9 355<\/p>\n
\nSolving two equations,
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-3-Pair-of-Linear-Equations-in-Two-Variables-9.png?resize=124%2C101&ssl=1\")
\nNow, putting y= 7 in equation (i)
\nx – 10 x 7 =75′
\ny + 70 = 75
\nx =75 – 70
\nx = 5
\nNow, if a person travels a distance of 50 km then, amount
\n= x + 50y
\n= 5 + 50 x 7
\n= 5 + 350
\n= 355
\nRefer situation 2<\/p>\n<\/p>\nWhat will a person have to pay for travelling a distance of 30 km ?<\/h2>\n
\n(B) \u20b9 289
\n(C) \u20b9 275
\n(D) \u20b9 305
\nAnswer:
\n(B) \u20b9 289<\/p>\nThe graph of lines representing the conditions are:
\n(situation 2)<\/h2>\n![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-3-Pair-of-Linear-Equations-in-Two-Variables-3.png?resize=301%2C244&ssl=1\")
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-3-Pair-of-Linear-Equations-in-Two-Variables-4.png?resize=299%2C237&ssl=1\")
\n![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-3-Pair-of-Linear-Equations-in-Two-Variables-5.png?resize=290%2C239&ssl=1\")
\n![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-3-Pair-of-Linear-Equations-in-Two-Variables-6.png?resize=284%2C243&ssl=1\")
\nAnswer:
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-3-Pair-of-Linear-Equations-in-Two-Variables-5-2.png?resize=290%2C239&ssl=1\")
\nII. Read the following text and answer the following questions on the basis of the same:
\nPlaces A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour.
\n![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-3-Pair-of-Linear-Equations-in-Two-Variables-7.png?resize=289%2C210&ssl=1\")
<\/p>\nAssuming that the speed of first car and second car be u km\/h and v km\/h respectively.
\nWhat is the relative speed of both cars while they are travelling in the same direction ?<\/h2>\n
\n(B) (u – v) km\/hr
\n(C) (u\/v) km\/hr
\n(D) (uv) km\/hr
\nAnswer:
\n(B) (u – v) km\/hr<\/p>\n
\nRelative speed of both cars while they are travelling in same direction = (u – v) km\/hr.<\/p>\nWhat is the relative speed of both cars while they are travelling towards each other ?<\/h2>\n
\n(B) (u – v) km\/hr
\n(C) (u\/v) km\/hr
\n(D) (uv) km\/hr
\nAnswer:
\n(A) (u + v) km\/hr<\/p>\n
\nRelative speed of both cars while they are travelling in opposite directions i.c., travelling towards each other = (u + v) km\/hr.<\/p>\nWhat is the actual speed of one car?<\/h2>\n
\n(B) 40 km\/hr
\n(C) 100 km\/hr
\n(D) 20 km\/hr
\nAnswer:
\n(A) 60 km\/hr<\/p>\n
\nLet the speeds of first car and second car be u km\/hr and’v km\/hr respectively.
\nAccording to the given information.
\n5 (u – v) = 100
\nu – v = 20 \u201e.(i)
\nu + v = 100 ……..(ii)
\nSoliving eqs. (i) and (ii),
\nwe get u = 60 km\/hr.<\/p>\nWhat is the actual speed of pther car?<\/h2>\n
\n(B) 40 km\/hr
\n(C) 100 km\/hr
\n(D) 20 km\/hr
\nAnswer:
\n(B) 40 km\/hr<\/p>\n
\nForm above question 3, referring to the solution of both equations
\nv = 40 km\/hr<\/p>\nThe given problem is based on which mathematical concept<\/h2>\n
\n(B) Quadratic equations
\n(C) Polynomials
\n(D) none of the above
\nAnswer:
\n(A) Pair of linear equations<\/p>\n
\nThe given problem is based on pair of linear equations.<\/p>\n
\nJohn and Jivanti are playing with the marbles in the playground. They together have 45 marbles and John has 15 marbles more than Jivanti.
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-3-Pair-of-Linear-Equations-in-Two-Variables-8.png?resize=247%2C186&ssl=1\")
<\/p>\nThe number of marbles Jivanti had:<\/h2>\n
\n(B) 30
\n(C) 40
\n(D) 5
\nAnswer:
\n(A) 15<\/p>\n
\nLei the no. of marbles, John and Jivanti have, be x and y respectively. According to the given information,
\nx + y = 45 ……..(i)
\nx – y =15 …(ii)
\nSolving eqs. (i) and (ii), we get
\nx = 30 and y = 15<\/p>\nThe number of marbles John had:<\/h2>\n
\n(B) 30
\n(C) 15
\n(D) 20
\nAnswer:
\n(B) 30<\/p>\n
\nAccording to the solution of question 1, we get x = 30.<\/p>\n<\/p>\nIf 45 is replaced by 55 in the above case discussed in the question, then the number of marbles Jivanti have:<\/h2>\n
\n(B) 30
\n(C) 20
\n(D) 35
\nAnswer:
\n(C) 20<\/p>\n
\nAccording to given problem,
\nx + y = 55 …(i)
\nx – y = 15 …(ii)
\nSolving eqs.(i) and (ii), we get
\nx = 35 and y = 20.<\/p>\nAccording to the question 3, the number of marbles John have:<\/h2>\n
\n(B) 40
\n(C) 45
\n(D) 35
\nAnswer:
\n(D) 35<\/p>\n
\nFrom above question 3, we get x = 35. Hence, John had 35 marbles.<\/p>\nThe given problem is based on which mathematical concept ?<\/h2>\n
\n(B) Quadratic equations
\n(C) Polynomials
\n(D) None of the above
\nAnswer:
\n(A) pair of linear equations<\/p>\n
\nThe given problem is based on pair of linear equations.<\/p>\nMCQ Questions for Class 10 Maths with Answers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"