Here you can download a free Linear Equations questions PDF with answers for MAH MBA CET 2022 by Cracku. These are some tricky questions in the MAH MBA CET 2022 exam that you need to find the Linear Equations of answers for the given questions. These questions will help you to make practice and solve the Linear Equations questions in the MAH MBA CET exam. Utilize this best PDF practice set<\/strong> which is included answers in detail. Click on the below link to download the Linear Equations MCQ<\/strong> PDF for MBA-CET 2022 for free.<\/p>\n Download Linear Equations Questions for MAH-CET<\/a><\/p>\n Enroll to MAH-CET 2022 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>Rubina could get equal number of Rs. 55, Rs. 85 and Rs. 105 tickets for a movie. She spends Rs. 2940 for all the tickets. How many of each did she buy?<\/p>\n a)\u00a012<\/p>\n b)\u00a014<\/p>\n c)\u00a016<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let x be the number of tickets bought by tickets. Question 2:\u00a0<\/b>In a test, a candidate secured 468 marks out of maximum marks \u2018A\u201d. Had the maximum marks \u2018A\u2019 converted to 700, he would have secured 336 marks. What was the maximum marks of the test?<\/p>\n a)\u00a0775<\/p>\n b)\u00a0875<\/p>\n c)\u00a0975<\/p>\n d)\u00a01075<\/p>\n e)\u00a0None of these<\/p>\n 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n When the maximum marks were 700 the secured marks would be\u00a0336 Question 3:\u00a0<\/b>Six – elevenths of a number is equal to twentytwo percent of the second number. The second number is equal to one-fourth of the third number. The value of the third number is 2400. What is 45% of the first number?<\/p>\n a)\u00a0109.8<\/p>\n b)\u00a0111.7<\/p>\n c)\u00a0117.6<\/p>\n d)\u00a0123.4<\/p>\n e)\u00a0None of these<\/p>\n 3)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the 1st number be x, 2nd number be y Question 4:\u00a0<\/b>Rachita enters a shop to buy ice-creams, cookies and pastries. She has to buy at least 9 units of each. She buys more cookies than ice-creams and more pastries than cookies. She picks up a total of 32 items. How many cookies does she buy?<\/p>\n a)\u00a0Either 12 or 13<\/p>\n b)\u00a0Either 11 or 12<\/p>\n c)\u00a0Either 10 or 11<\/p>\n d)\u00a0Either 9 or 11<\/p>\n e)\u00a0Either 9 or 10<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number of Ice-creams < Cookies < Pastries <\/p>\n Question 5:\u00a0<\/b>The fare of a bus is Rs.X for the first five kilometres and Rs.13 per kilometer thereafter. If a passenger pays Rs.2,402 for a journey of 187 kilometres, what is the value of X?<\/p>\n a)\u00a0Rs. 29<\/p>\n b)\u00a0Rs. 39<\/p>\n c)\u00a0Rs. 36<\/p>\n d)\u00a0Rs. 31<\/p>\n e)\u00a0None of these<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total cost of the journey = 2402 Question 6:\u00a0<\/b>The sum of the ages of 4 members of a family 5 years ago was 94 years. Today, when the daughter has been married off and replaced by a daughter-in-law the sum of their ages is 92. Assuming that there has been no other change in the family structure and all the people are alive, what is the different in the age of the daughter and the daughter-in-law?<\/p>\n a)\u00a022 years<\/p>\n b)\u00a011 years<\/p>\n c)\u00a025 years<\/p>\n d)\u00a019 years<\/p>\n e)\u00a015 years<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Sum of the ages five years ago = 94 Question 7:\u00a0<\/b>Akash scored 73 marks in subject A. He scored 56% marks in subject B and X marks in subject C. Maximum marks in each subject were 150. The overall percentage marks obtained by Akash in the three subjects together were 54%. How many marks did he scored in subject C<\/p>\n a)\u00a084<\/p>\n b)\u00a086<\/p>\n c)\u00a079<\/p>\n d)\u00a073<\/p>\n e)\u00a0None of these<\/p>\n Take Free MAH-CET mock tests here<\/a><\/strong><\/span><\/p>\n Enrol to 5 MAH CET Latest Mocks For Just Rs. 299<\/a><\/strong><\/span><\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Percentage of marks obtained in A =\u00a0$\\frac{73}{150}$ x 100 = 48.66 Question 8:\u00a0<\/b>The present age of Romil is one fourth of that of her father. After 6 years the father\u2019s age will be twice the age of Kapil. If Kapil celebrated fifth birth day 8 years ago. What is Romil\u2019s percent age?<\/p>\n a)\u00a07 years<\/p>\n b)\u00a07.5 years<\/p>\n c)\u00a08 years<\/p>\n d)\u00a08.5 years<\/p>\n e)\u00a0None of these<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Kapil’s present age = 8 + 5 = 13 Question 9:\u00a0<\/b>In an examination, 30% of total students failed in Hindi, 45% failed in English and 20% failed in both subjects. Find the percentage of those who passed in both the subjects.<\/p>\n a)\u00a035.7 %<\/p>\n b)\u00a035%<\/p>\n c)\u00a040%<\/p>\n d)\u00a045%<\/p>\n e)\u00a044%<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n 45% of the total students failed in Hindi Question 10:\u00a0<\/b>Rita’s present age is four times her daughter’s present age and two-thirds of her mother’s present age. The total of the present ages of all of them is 154 years. What is the difference between Rita’s and her mohter’s present age?<\/p>\n a)\u00a028 years<\/p>\n b)\u00a034 years<\/p>\n c)\u00a032 years<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let Rita’s age be R. Question 11:\u00a0<\/b>In the following question, two equations are given. a)\u00a0x < y<\/p>\n b)\u00a0x > y<\/p>\n c)\u00a0x = y<\/p>\n d)\u00a0x \u2265 y<\/p>\n e)\u00a0x \u2264 y<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Solving 10x2<\/sup> – 7x + 1 = 0 Solving 35y2<\/sup> – 12y + 1 = 0 x $\\geq$ y<\/p>\n Question 12:\u00a0<\/b>In the following question, two equations are given. a)\u00a0x < y<\/p>\n b)\u00a0x > y<\/p>\n c)\u00a0x = y<\/p>\n d)\u00a0x \u2265 y<\/p>\n e)\u00a0x \u2264 y<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n x2<\/sup> – 6x – 7 = 0 Question 13:\u00a0<\/b>In the following question, two equations are given. a)\u00a0x < y<\/p>\n b)\u00a0x > y<\/p>\n c)\u00a0cannot be determined<\/p>\n d)\u00a0x \u2265 y<\/p>\n e)\u00a0x \u2264 y<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n 6x2<\/sup> + 77x + 121 = 0 Thus, the relationship between x and y cannot be determined.<\/p>\n Question 14:\u00a0<\/b>In the following question, two equations are given. a)\u00a0x < y<\/p>\n b)\u00a0x > y<\/p>\n c)\u00a0cannot be determined<\/p>\n d)\u00a0x \u2265 y<\/p>\n e)\u00a0x \u2264 y<\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n 2x2<\/sup> + 3x – 20 = 0 x \u2265 y<\/p>\n Question 15:\u00a0<\/b>In the following question, two equations are given. a)\u00a0x < y<\/p>\n b)\u00a0x > y<\/p>\n c)\u00a0cannot be determined<\/p>\n d)\u00a0x \u2265 y<\/p>\n e)\u00a0x \u2264 y<\/p>\n 15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n x2<\/sup> – 24x + 95 = 0 x < y or x>y<\/p>\n
\nTotal money spent\u00a0= 55x + 85x + 105x
\n245x = 2940
\nx = $\\frac{2940}{245}$ = 12<\/p>\n
\n% of marks out of 700 = $\\frac{336}{700}$ x 100 =\u00a048%
\nNow 468 is 48% of maximum marks \u2018A\u2019
\n468 = $\\frac{48}{100}$ x A
\nA = 468 x $\\frac{100}{48}$
\n= 975<\/p>\n
\nThird number 2400
\n$\\frac{6}{11}$ of x = 22% of y
\ny = $\\frac{1}{4}$th of the third number
\n2nd No. = y = 600
\n22% of 2nd No. = \u00a0$\\frac{22}{100}$ x 600 = 132
\n$\\frac{6}{11}$ of x = 132
\nx = 242
\n45% of 1st number = $\\frac{45}{100}$ x 242 = 108.9<\/p>\n
\nSum must be 32
\nIf ice-creams = 9
\nCookies > Ice-creams= 10 or 11
\nPastries > Ice-creams = 11 or 12<\/p>\n
\nX + 182 x 13 = 2402
\nX = 2402 -2366 = 36<\/p>\n
\nSum of their ages now = 94 + 20 = 114
\nSum of their ages after the daughter is left and replaced by daughter-in-law = 92
\nDifference = 114 – 92 = 22<\/p>\n
\nOverall percentage = 54%
\n$\\frac{A + B + C}{3}$ = 54%
\n$\\frac{48.66 + 56 + C}{3}$ = 54%
\nC = 57.34%
\nMarks = 57.34% of 150 = 86<\/p>\n
\nAfter 6 years Kapil’s age = 19
\nFather’s age after six years = Twice of Kapil’s age = 2 x 19 =38
\nFather’s present age = 32
\nRomil’s age =\u00a0$\\frac{1}{4}$th of 32 = 8<\/p>\n
\n30% failed in English
\n20% failed in both
\nTherefore 55% of the total students failed in atleat one subject.
\nHence 45% have passed in both<\/p>\n
\nMother’s age = 3R\/2
\nDaughter’s age = R\/4
\nTotal = R + 3R\/2 + R\/4 = 11R\/4 = 154 => R = 56
\nSo, difference of ages = R\/2 = 28 years
\nInstructions<\/b>Instructions: In the following question, two equations are given.\u00a0You have to solve those equations\u00a0and give your answer<\/p>\n
\nI. 10x2<\/sup> – 7x + 1 = 0
\nII. 35y2<\/sup> – 12y + 1 = 0<\/p>\n
\nx = $\\frac{7\\pm\\sqrt{49-40}}{20}$ = $\\frac{7 \\pm 3}{20}$ = 1\/5 or 1\/2<\/p>\n
\ny = $\\frac{12\\pm\\sqrt{144-140}}{70}$ = $\\frac{12 \\pm 2}{70}$ = 1\/5 or 1\/7<\/p>\n
\nI. x2<\/sup> – 6x – 7 = 0
\nII. 2y2<\/sup> + 13y + 15 = 0<\/p>\n
\nx = 7 , -1
\n2y2<\/sup> + 13y + 15 = 0
\ny = -5 , -3\/2
\nThus, x > y<\/p>\n
\nI. 6x2<\/sup> + 77x + 121 = 0
\nII. y2<\/sup> + 9y – 22 = 0<\/p>\n
\nx = -11 , -11\/6
\ny2<\/sup> + 9y – 22 = 0
\ny = -11 , 2<\/p>\n
\nI. 2x2<\/sup> + 3x – 20 = 0
\nII. 2y2<\/sup> + 19y + 44 = 0<\/p>\n
\nx = -8 , 5
\n2y2<\/sup> + 19y + 44 = 0
\ny = -11 , -8<\/p>\n
\nI. x2<\/sup> – 24x + 95 = 0
\nII. y2<\/sup> – 26y + 169 = 0<\/p>\n
\nx = 19 , 5
\ny2<\/sup> – 26y + 169 = 0
\ny = 13, 13<\/p>\n