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1.	The sum of a two digit number and the number obtained by reversing the order of the digits is 187. If the digits differ by 1, then what will be the number?(a) 67(b) 54(c) 89(d) 67I have been asked this question during an online interview.Enquiry is from Solution of Two Linear Equations in Two Variables in Different Methods in division Pair of Linear Equations in Two Variables of Mathematics – Class 10
Answer» Correct OPTION is (c) 89 The best I can explain: Let the two digit number be 10x+y The number obtained after reversing the digits will be 10y+x 10x+y+10y+x=187 11x+11y=187 x+y=17(1) ALSO, x-y=1(2) Adding (1) and (2) 2x=18 x=9 Substituting in EQUATION (1), 9+y=17 y=8 The number is 89 or 98.

2.	What will be the value of k, if the lines given by 3x+ky-4 and 5x+(9+k)y+41 represent two lines intersecting at a point?(a) k≠\(\frac {7}{2}\)(b) k≠\(\frac {27}{8}\)(c) k=\(\frac {27}{2}\)(d) k≠\(\frac {27}{2}\)I had been asked this question in an interview for job.The origin of the question is Solution of Two Linear Equations in Two Variables in Different Methods topic in section Pair of Linear Equations in Two Variables of Mathematics – Class 10
Answer» The correct answer is (d) k≠\(\frac {27}{2}\) To EXPLAIN I would say: The GIVEN equations are 3x+ky-4 and 5x+(9+k)y+41 . Here, a1=3, b1=k, c1=-4 and a2=5, b2=9+k, c2=41 Lines are intersecting at a point, so \(\frac {a_1}{a_2} \ne \frac {b_1}{b_2} \) Now, \(\frac {a_1}{a_2} =\frac {3}{5}, \frac {b_1}{b_2} = \frac {k}{9+k}, \frac {c_1}{c_2} =\frac {-4}{41}\) \(\frac {3}{5} \ne \frac {k}{9+k}\) 3(9+k)≠5k 27+3k≠5k 27≠5k-3k 2k≠27 k≠\(\frac {27}{2}\)

3.	The lines 5x-7y=13 and 10x-14y=15 are inconsistent.(a) True(b) FalseThe question was posed to me during an interview.I'd like to ask this question from Solution of Two Linear Equations in Two Variables in Different Methods in portion Pair of Linear Equations in Two Variables of Mathematics – Class 10
Answer» CORRECT choice is (a) True For explanation I would say: A system of linear equations is SAID to be inconsistent if it has no solution at all. The given equations are 5x-7y=13 and 10x-14y=15 Here, a1=5, b1=-7, c1=-13 and a2=10, b2=-14, c2=-15 Now, \(\frac {a_1}{a_2} = \frac {5}{10}=\frac {1}{2}, \frac {b_1}{b_2} =\frac {-7}{-14}=\frac {1}{2}, \frac {c_1}{c_2} =\frac {-13}{-15}\) CLEARLY, \(\frac {a_1}{a_2} =\frac {b_1}{b_2} \ne \frac {c_1}{c_2}\) Hence, it will have no solution. The given equations are inconsistent.

4.	The lines 2x+5y=17 and 5x+3y=14 are consistent.(a) False(b) TrueI got this question in an interview.The question is from Solution of Two Linear Equations in Two Variables in Different Methods topic in portion Pair of Linear Equations in Two Variables of Mathematics – Class 10
Answer» The correct ANSWER is (b) True Best explanation: A system of LINEAR equations is said to be consistent if it has at least one solution. The given equations are 2x+5y=17 and 5x+3y=14 Here, a1=2, b1=5, c1=-17 and a2=5, b2=3, c2=-14 Now, \(\frac {a_1}{a_2} =\frac {2}{5}, \frac {b_1}{b_2} =\frac {5}{3}, \frac {c_1}{c_2} =\frac {-17}{-14}\) Clearly, \(\frac {a_1}{a_2} \ne \frac {b_1}{b_2} \) Hence, it will have UNIQUE solution. The given equations are consistent.

5.	What will be the value of k, if the lines given by x+ky+3 and 2x+(k+2)y+6 are coincident?(a) 4(b) 2(c) 6(d) 8The question was asked during an interview for a job.My doubt is from Solution of Two Linear Equations in Two Variables in Different Methods in chapter Pair of Linear Equations in Two Variables of Mathematics – Class 10
Answer» The correct option is (b) 2 To explain: The given EQUATIONS are x+ky+3 and (k-1)x+4y+6. Here, a1=1, b1=k, c1=3 and a2=k-1, b2=4, c2=6 Lines are COINCIDENT, so \(\frac {a_1}{a_2} =\frac {b_1}{b_2} =\frac {c_1}{c_2}\) Now, \(\frac {a_1}{a_2} = \frac {1}{k-1}, \frac {b_1}{b_2} =\frac {k}{4}, \frac {c_1}{c_2} =\frac {3}{6}\) \(\frac {1}{k-1}=\frac {k}{4}=\frac {1}{2}\) 2k=4 k=2

6.	What will be the value of k, if the lines given by (5+k)x-3y+15 and (k-1)x-y+19 are parallel?(a) 5(b) 4(c) 6(d) 7I had been asked this question during an interview.The query is from Solution of Two Linear Equations in Two Variables in Different Methods in portion Pair of Linear Equations in Two Variables of Mathematics – Class 10
Answer» CORRECT option is (b) 4 For EXPLANATION: The GIVEN equations are (5+k)x-3y+15 and (k-1)x-y+19. Here, a1=5+k, b1=-3, c1=15 and a2=k-1, b2=-1, c2=19 Lines are parallel, so \(\frac {a_1}{a_2} = \frac {b_1}{b_2} \NE \frac {c_1}{c_2} \) Now, \(\frac {a_1}{a_2} = \frac {5+k}{k-1}, \frac {b_1}{b_2} =\frac {-3}{-1}\) = 3, \(\frac {c_1}{c_2} = \frac {15}{19} \) \(\frac {5+k}{k-1}\) = 3 5+k=3(k-1) 5+k=3k-3 5+3=3k-k 2k=8 k=4

7.	What will be the nature of the graph lines of the equations x+3y-2 and 2x-y+5?(a) Parallel(b) Coincident(c) Intersecting(d) Perpendicular to each otherThis question was addressed to me by my college professor while I was bunking the class.My question is from Solution of Two Linear Equations in Two Variables in Different Methods topic in chapter Pair of Linear Equations in Two Variables of Mathematics – Class 10
Answer» Correct answer is (C) Intersecting Easy EXPLANATION: The given equations are x+3y-2 and 2x-y+5. Here, a1=1, b1=3, c1=-2 and a2=2, b2=-1, c2=5 Now, \(\frac {a_1}{a_2} = \frac {1}{2}, \frac {b_1}{b_2} = \frac {3}{1}\) = 3, \(\frac {c_1}{c_2} = \frac {-2}{5} \) CLEARLY, \(\frac {a_1}{a_2} \ne \frac {b_1}{b_2} \) Therefore, the graph lines of the equations will intersect at a point.

8.	10 years ago, a woman was thrice the age of her daughter. Two years later her daughter’s age will be 30 more than the age of the mother. What are the present ages of the woman and the daughter?(a) 70 years, 40 years(b) 60 years, 40 years(c) 55 years, 25 years(d) 45 years, 20 yearsI got this question in an online interview.This is a very interesting question from Solution of Two Linear Equations in Two Variables in Different Methods in division Pair of Linear Equations in Two Variables of Mathematics – Class 10
Answer» Correct OPTION is (c) 55 years, 25 years The best explanation: Let the PRESENT ages of mother be x years and daughter be y years. 10 years ago, Age of mother = x-10 years Age of daughter = y-10 years Age of mother = 3(age of daughter) x-10=3(y-10) x-10=3y-30 x-3y+20=0 x=3y-20(1) Two years LATER, Age of mother = x+2 years Age of daughter = y+2 years Age of mother will be 30 more than the age of daughter x+2=y+2+30 x=y+30(2) From (1) and (2), we GET, y+30=3y-20 30+20=3y-y 50=2y y=25 Substituting y=25 in equation (1) we get, x=3(25)-20 x=55 The present age of mother is 55 years and that of daughter is 25 years.

9.	In a piggy bank the total number of coins of Rs. 5 and Rs. 1 is 100. If the total coins amount is 300, then what is the number of coins of each denomination?(a) 30, 70(b) 50, 50(c) 45, 55(d) 60, 40The question was asked in an internship interview.This intriguing question originated from Solution of Two Linear Equations in Two Variables in Different Methods topic in section Pair of Linear Equations in Two Variables of Mathematics – Class 10
Answer» RIGHT choice is (b) 50, 50 For explanation I WOULD say: LET the coins of RS. 5 be x and that of Rs. 1 be y Total number of coins is 100 x+y=100 y=100-x Total amount of coins is 300 Also, 5x+y=300(1) Substituting y=100-x in equation (1) we get, 5x+100-x=300 4x=200 x=50 y=100-x=100-50=50 The coins of each denomination are 50.

10.	What will be the nature of the graph lines of the equations 5x-2y+9 and 15x-6y+1?(a) Parallel(b) Coincident(c) Intersecting(d) Perpendicular to each otherI had been asked this question in unit test.The above asked question is from Solution of Two Linear Equations in Two Variables in Different Methods in chapter Pair of Linear Equations in Two Variables of Mathematics – Class 10
Answer» Right choice is (a) PARALLEL To explain: The given equations are 5x-2y+9 and 15x-6y+1. Here, a1=5, b1=-2, c1=9 and a2=15, b2=-6, c2=1 Now, \(\frac {a_1}{a_2} = \frac {5}{15} = \frac {1}{3}, \frac {b_1}{b_2} = \frac {-2 }{-6}=\frac {1}{3}, \frac {c_1}{c_2} = \frac {9}{1} \) Clearly, \(\frac {a_1}{a_2} =\frac {b_1}{b_2} \ne \frac {c_1}{c_2} \) THEREFORE, the graph lines of the equations will be parallel.

11.	It takes 10 men and 6 women to finish a piece of work in 4 days, while it takes 5 men and 7 women to finish the same job in 6 days. What will be the time taken by 1 man and 1 woman to finish the job?(a) Man = 34 days, Woman = 45 days(b) Man = 45 days, Woman = 34 days(c) Man = 53 days, Woman = 96 days(d) Man = 54 days, Woman = 96 daysI got this question by my school principal while I was bunking the class.The question is from Solution of Two Linear Equations in Two Variables in Different Methods in portion Pair of Linear Equations in Two Variables of Mathematics – Class 10
Answer» The correct option is (d) MAN = 54 days, Woman = 96 days The explanation is: Let 1 man TAKE x days to finish the job and 1 woman take y days to finish the same job. Then, 1 man’s 1 day work will be \(\frac {1}{x}\) days 1 woman’s 1 day work will be \(\frac {1}{y}\) days 10 men and 6 women can finish the job in 6 days (10 men’s 1 day work + 6 women’s 1 day work = \(\frac {1}{4}\)) \(\frac {10}{x}+\frac {6}{y}=\frac {1}{4}\) Let, \(\frac {1}{x}\) = u, \(\frac {1}{y}\) = v 10u+6v=\(\frac {1}{4}\)(1) 5 men and 7 women can finish the job in 6 days. (5 men’s 1 day work + 7 women’s 1 day work = \(\frac {1}{6}\)) \(\frac {5}{x}+\frac {7}{y}=\frac {1}{6}\) Let, \(\frac {1}{x}\) = u, \(\frac {1}{y}\) = v 5u + 7v = \(\frac {1}{6}\)(2) Multiplying equation by 2 and then subtracting both the equations we get, 10u + 14v = \(\frac {1}{6}\) -10u + 6v = \(\frac {1}{4}\) 8v = \(\frac {1}{3}-\frac {1}{4}\) 8v = \(\frac {1}{12}\) v = \(\frac {1}{96}\) v = \(\frac {1}{y}=\frac {1}{96}\) y = 96 Substituting the value of v in equation (1) we get, 10u + 6\((\frac {1}{96})=\frac {1}{4}\) 10u + \(\frac {1}{16}=\frac {1}{4}\) 10u = \(\frac {3}{16}\) u = \(\frac {3}{160}\) u = \(\frac {1}{x}=\frac {3}{160}\) x = \(\frac {160}{3}\) ≈ 54 Hence, a man alone can finish the job in 54 days and a woman can finish the job in 96 days.

12.	What will be the nature of the graph lines of the equations 2x+5y+15 and 6x+15y+45?(a) Parallel(b) Coincident(c) Intersecting(d) Perpendicular to each otherI got this question in an interview for job.This intriguing question comes from Solution of Two Linear Equations in Two Variables in Different Methods in division Pair of Linear Equations in Two Variables of Mathematics – Class 10
Answer» The CORRECT answer is (b) Coincident Explanation: The given EQUATIONS are 2x+5y+15 and 6x+15y+45. Here, a1=2, b1=5, c1=15 and a2=6, b2=15, c2=45 Now, \(\frac {a_1}{a_2} = \frac {2}{6} = \frac {1}{3}, \frac {b_1}{b_2} = \frac {5}{15} =\frac {1}{3}, \frac {c_1}{c_2} = \frac {15}{45} = \frac {1}{3} \) Clearly, \(\frac {a_1}{a_2} =\frac {b_1}{b_2} = \frac {c_1}{c_2} \) Therefore, the graph lines of the equations will be coincident.

13.	A father gives Rs. 500 to his children every month. If the boy gets Rs. 100 then, the girl gets Rs. 200 and if the boy gets Rs. 100 the girl gets Rs. 150. How many children does he have?(a) 0(b) 3(c) 2(d) 1I had been asked this question by my college professor while I was bunking the class.The doubt is from Solution of Two Linear Equations in Two Variables in Different Methods in section Pair of Linear Equations in Two Variables of Mathematics – Class 10
Answer» The CORRECT option is (b) 3 Easy explanation: Let the number of BOYS be x and number of GIRLS be y Now, if the boys get 100, the girls get 200 100x+200y=500 x+2y=5 x=5-2y If the boys get 200, the girls get 150 200x+150y=500 4x+3y=10(1) Substituting x=5-2y in equation 1 we get, 4(5-2y)+3y=10 20-8y+3y=10 -5y=-10 y=2 x=5-2y=5-2(2)=1 The FATHER has THREE children.