Directions: Study the following information carefully and answer the questions. In a class there are 2 sections, section A and section B, All the students of section A and section B take part in an examination.<\/p>\n\n\n\n
The number of students in section B is 3 times the number of students in section A, in the exam the number of questions attempt by each students of section A is equal to the number of students in section A, and the number of questions attempted by each students of section B is equalto the total number of students in section B, 50% of total question attempted by all the students of section A is correct and rest are incorrect, and 70% of all the question attempted by the students of section B is correct and the rest are incorrect, each correct question carry 3 marks, and for each incorrect question 1 mark will be deducted. After the exam it is seen that the difference between the marks scored by all the students in section A and section B is 220.<\/p>\n\n\n\n
1) If the mark for correct answer is +4 and for the wrong answer is -2,then find the difference between the total students of both the sections.<\/p>\n\n\n\n
A. 40<\/p>\n\n\n\n
B. 50<\/p>\n\n\n\n
C. 80<\/strong><\/p>\n\n\n\n D. 70<\/p>\n\n\n\n E. 60<\/p>\n\n\n\n Sol:<\/p>\n\n\n\n Let the number of students in section A=x<\/p>\n\n\n\n So the number of students in section B= 3x<\/p>\n\n\n\n In section A, each students attempt x number of<\/p>\n\n\n\n question<\/p>\n\n\n\n And out of these 50% are correct= 0.5x<\/p>\n\n\n\n questions are correct by each students<\/p>\n\n\n\n And rest 0.5x are in correct by each students<\/p>\n\n\n\n In section B, each students attempt 3x number<\/p>\n\n\n\n of question<\/p>\n\n\n\n Out of which 70% are correct so 70% of 3x=2.1x<\/p>\n\n\n\n correct by each students<\/p>\n\n\n\n And the rest 30% is incorrect=0.9x is incorrect by<\/p>\n\n\n\n each students of section B<\/p>\n\n\n\n Correct question carry 3 marks<\/p>\n\n\n\n And for incorrect carry -1 marks<\/p>\n\n\n\n So marks of all students in A section get<\/p>\n\n\n\n =0.5x*3+0.5x*(-1)<\/p>\n\n\n\n And<\/p>\n\n\n\n marks of all students in section B get<\/p>\n\n\n\n =2.1x*3+0.9x*(-1)<\/p>\n\n\n\n According to the question<\/p>\n\n\n\n [2.1x*3+0.9x*(-1)]- [0.5x*3+0.5x*(-1)] =220<\/p>\n\n\n\n [6.3x-0.9x] -[1.5x-0.5x] =220<\/p>\n\n\n\n =>5.4x -x=220<\/p>\n\n\n\n =>4.4x=220<\/p>\n\n\n\n x= 220\/4.4<\/p>\n\n\n\n x=50<\/p>\n\n\n\n Number of students in section A= 50<\/p>\n\n\n\n And section B= 3* 50 =150<\/p>\n\n\n\n The number of students in section B= 150<\/p>\n\n\n\n According to the question correct question= +4<\/p>\n\n\n\n And incorrect question=-2<\/p>\n\n\n\n So<\/p>\n\n\n\n [2.1x*4+0.9x*(-2)]- [0.5x*4+0.5x*(-2)] =220<\/p>\n\n\n\n 8.4x-1.8x-2x+x=220<\/p>\n\n\n\n 5.6x=220<\/p>\n\n\n\n x=220\/5.6<\/p>\n\n\n\n x=39.28= 40 (Approximately)<\/p>\n\n\n\n The Number of students in section B= 120<\/p>\n\n\n\n The number of students in section A= 40<\/p>\n\n\n\n So the difference= 120-40 =80<\/p>\n\n\n\n 20) If 60% of total questions attempted by all the students in section A give correct answers and 40% of total question attempted by section B give correct questions, find the difference between the average marks of all the students in section A and section B.<\/p>\n\n\n\n A. 30<\/p>\n\n\n\n B. 20<\/strong><\/p>\n\n\n\n C. 10<\/p>\n\n\n\n D. 40<\/p>\n\n\n\n E. 60<\/p>\n\n\n\n Sol:<\/p>\n\n\n\n Let the number of students in section A=x<\/p>\n\n\n\n So the number of students in section B= 3x<\/p>\n\n\n\n In section A, each students attempt x number of<\/p>\n\n\n\n question<\/p>\n\n\n\n And out of these 50% are correct= 0.5x<\/p>\n\n\n\n questions are correct by each students<\/p>\n\n\n\n And rest 0.5x are in correct by each students<\/p>\n\n\n\n In section B, each students attempt 3x number<\/p>\n\n\n\n of question<\/p>\n\n\n\n Out of which 70% are correct so 70% of 3x=2.1x<\/p>\n\n\n\n correct by each students<\/p>\n\n\n\n And the rest 30% is incorrect=0.9x is incorrect by<\/p>\n\n\n\n each students of section B<\/p>\n\n\n\n Correct question carry 3 marks<\/p>\n\n\n\n And for incorrect carry -1 marks<\/p>\n\n\n\n So marks of all students in A section get<\/p>\n\n\n\n =0.5x*3+0.5x*(-1)<\/p>\n\n\n\n And<\/p>\n\n\n\n marks of all students in section B get<\/p>\n\n\n\n =2.1x*3+0.9x*(-1)<\/p>\n\n\n\n According to the question<\/p>\n\n\n\n [2.1x*3+0.9x*(-1)]- [0.5x*3+0.5x*(-1)] =220<\/p>\n\n\n\n [6.3x-0.9x] -[1.5x-0.5x] =220<\/p>\n\n\n\n =>5.4x -x=220<\/p>\n\n\n\n =>4.4x=220<\/p>\n\n\n\n x= 220\/4.4<\/p>\n\n\n\n x=50<\/p>\n\n\n\n Number of students in section A= 50<\/p>\n\n\n\n And section B= 3* 50 =150<\/p>\n\n\n\n The number of students in section B= 150<\/p>\n\n\n\n In section A 60% of all the question correct<\/p>\n\n\n\n In section B 40% of all the question correct<\/p>\n\n\n\n Number of students in section A=50 and the<\/p>\n\n\n\n number of students in section B=3x=150<\/p>\n\n\n\n In section B each students attempt 3x number of<\/p>\n\n\n\n question<\/p>\n\n\n\n 40% correct<\/p>\n\n\n\n So 1.2x correct and 1.8 x incorrect<\/p>\n\n\n\n In section A each students attempt x number of<\/p>\n\n\n\n question<\/p>\n\n\n\n 60% of all question correct<\/p>\n\n\n\n So 0.6x correct and 0.4x incorrect<\/p>\n\n\n\n The required difference<\/p>\n\n\n\n =[1.2*50*3+1.8*50*(-1)]- [0.6*50*3+0.4*50*(-1)]<\/p>\n\n\n\n =[60*3+90(-1)]-[30*3+20*(-1)]<\/p>\n\n\n\n =[180-90]-[90-20]<\/p>\n\n\n\n =90-70<\/p>\n\n\n\n =20<\/p>\n\n\n\n 21) Find the ratio of the total number of marks obtained by section A to section B.<\/p>\n\n\n\n A. 5:81<\/strong><\/p>\n\n\n\n B. 6:52<\/p>\n\n\n\n C. 7:63<\/p>\n\n\n\n D. 4:31<\/p>\n\n\n\n E. 9:82<\/p>\n\n\n\n Sol:<\/p>\n\n\n\n Let the number of students in section A=x<\/p>\n\n\n\n So the number of students in section B= 3x<\/p>\n\n\n\n In section A, each students attempt x number of<\/p>\n\n\n\n question<\/p>\n\n\n\n And out of these 50% are correct= 0.5x<\/p>\n\n\n\n questions are correct by each students<\/p>\n\n\n\n And rest 0.5x are in correct by each students<\/p>\n\n\n\n In section B, each students attempt 3x number<\/p>\n\n\n\n of question<\/p>\n\n\n\n Out of which 70% are correct so 70% of 3x=2.1x<\/p>\n\n\n\n correct by each students<\/p>\n\n\n\n And the rest 30% is incorrect=0.9x is incorrect by<\/p>\n\n\n\n each students of section B<\/p>\n\n\n\n Correct question carry 3 marks<\/p>\n\n\n\n And for incorrect carry -1 marks<\/p>\n\n\n\n So marks of all students in A section get<\/p>\n\n\n\n =0.5x*3+0.5x*(-1)<\/p>\n\n\n\n And<\/p>\n\n\n\n marks of all students in section B get<\/p>\n\n\n\n =2.1x*3+0.9x*(-1)<\/p>\n\n\n\n According to the question<\/p>\n\n\n\n [2.1x*3+0.9x*(-1)]- [0.5x*3+0.5x*(-1)] =220<\/p>\n\n\n\n [6.3x-0.9x] -[1.5x-0.5x] =220<\/p>\n\n\n\n =>5.4x -x=220<\/p>\n\n\n\n =>4.4x=220<\/p>\n\n\n\n x= 220\/4.4<\/p>\n\n\n\n x=50<\/p>\n\n\n\n Number of students in section A= 50<\/p>\n\n\n\n And section B= 3* 50 =150<\/p>\n\n\n\n The number of students in section B= 150<\/p>\n\n\n\n Total number obtain in section<\/p>\n\n\n\n A=x*[0.5x*3+0.5x*(-1)]<\/p>\n\n\n\n =50*[0.5*50*3+0.5*50*(-1)]<\/p>\n\n\n\n =50*[75-25] =50*[50] =2500<\/p>\n\n\n\n Total number obtain in section<\/p>\n\n\n\n B=3x*[2.1x*3+0.9x*(-1)]<\/p>\n\n\n\n =3*50[2.1*50*3+0.9*50*(-1)]<\/p>\n\n\n\n =3*50[105*3+45*(-1)]<\/p>\n\n\n\n =3*50[315-45]<\/p>\n\n\n\n =3*50*270<\/p>\n\n\n\n =40500<\/p>\n\n\n\n So the required ratio=2500:40500<\/p>\n\n\n\n =5:81<\/p>\n\n\n\n 22) If the difference between the marks of all the students together in section A and section B is 264, then find the number of students in section A.<\/p>\n\n\n\n A. 120<\/p>\n\n\n\n B. 160<\/p>\n\n\n\n C. 150<\/p>\n\n\n\n D. 140<\/p>\n\n\n\n E. 60<\/strong><\/p>\n\n\n\n Sol<\/strong>:<\/p>\n\n\n\n Let the number of students in section A=x<\/p>\n\n\n\n So the number of students in section B= 3x<\/p>\n\n\n\n In section A, each students attempt x number of<\/p>\n\n\n\n question<\/p>\n\n\n\n And out of these 50% are correct= 0.5x<\/p>\n\n\n\n questions are correct by each students<\/p>\n\n\n\n And rest 0.5x are in correct by each students<\/p>\n\n\n\n In section B, each students attempt 3x number<\/p>\n\n\n\n of question<\/p>\n\n\n\n Out of which 70% are correct so 70% of 3x=2.1x<\/p>\n\n\n\n correct by each students<\/p>\n\n\n\n And the rest 30% is incorrect=0.9x is incorrect by<\/p>\n\n\n\n each students of section B<\/p>\n\n\n\n Correct question carry 3 marks<\/p>\n\n\n\n And for incorrect carry -1 marks<\/p>\n\n\n\n So marks of all students in A section get<\/p>\n\n\n\n =0.5x*3+0.5x*(-1)<\/p>\n\n\n\n And<\/p>\n\n\n\n marks of all students in section B get<\/p>\n\n\n\n =2.1x*3+0.9x*(-1)<\/p>\n\n\n\n According to the question<\/p>\n\n\n\n [2.1x*3+0.9x*(-1)]- [0.5x*3+0.5x*(-1)] =220<\/p>\n\n\n\n [6.3x-0.9x] -[1.5x-0.5x] =220<\/p>\n\n\n\n =>5.4x -x=220<\/p>\n\n\n\n =>4.4x=220<\/p>\n\n\n\n x= 220\/4.4<\/p>\n\n\n\n x=50<\/p>\n\n\n\n Number of students in section A= 50<\/p>\n\n\n\n And section B= 3* 50 =150<\/p>\n\n\n\n The number of students in section B= 150<\/p>\n\n\n\n According to the question<\/p>\n\n\n\n [2.1x*3+0.9x*(-1)]- [0.5x*3+0.5x*(-1)] =264<\/p>\n\n\n\n [6.3x-0.9x] -[1.5x-0.5x] =264<\/p>\n\n\n\n =>5.4x -x=264<\/p>\n\n\n\n =>4.4x=264<\/p>\n\n\n\n x= 264\/4.4<\/p>\n\n\n\n x= 60<\/p>\n\n\n\n Students in section A= 60<\/p>\n","protected":false},"excerpt":{"rendered":" Directions: Study the following information carefully and answer the questions. In a class there are 2 sections, section A and section B, All the students of section A and section B take part in an examination. The number of students in section B is 3 times the number of students in section A, in the […]<\/p>\n","protected":false},"author":419,"featured_media":162961,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-4)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-4)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-4)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[5138],"tags":[5139],"class_list":["post-162960","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-quantitative-aptitude","tag-quantitative-aptitude"],"_links":{"self":[{"href":"https:\/\/www.gkseries.com\/blog\/wp-json\/wp\/v2\/posts\/162960","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.gkseries.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.gkseries.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.gkseries.com\/blog\/wp-json\/wp\/v2\/users\/419"}],"replies":[{"embeddable":true,"href":"https:\/\/www.gkseries.com\/blog\/wp-json\/wp\/v2\/comments?post=162960"}],"version-history":[{"count":1,"href":"https:\/\/www.gkseries.com\/blog\/wp-json\/wp\/v2\/posts\/162960\/revisions"}],"predecessor-version":[{"id":162962,"href":"https:\/\/www.gkseries.com\/blog\/wp-json\/wp\/v2\/posts\/162960\/revisions\/162962"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.gkseries.com\/blog\/wp-json\/wp\/v2\/media\/162961"}],"wp:attachment":[{"href":"https:\/\/www.gkseries.com\/blog\/wp-json\/wp\/v2\/media?parent=162960"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.gkseries.com\/blog\/wp-json\/wp\/v2\/categories?post=162960"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.gkseries.com\/blog\/wp-json\/wp\/v2\/tags?post=162960"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}