\n\n\n\n
Step 1: Define variables<\/h3>\n\n\n\n\n- Let the motorcyclist’s original speed be xxx km\/h.<\/li>\n\n\n\n
- Distance between City A and City B: 225 km.<\/li>\n\n\n\n
- The motorcyclist rides at speed xxx for 1.5 hours.<\/li>\n\n\n\n
- Then stops for 0.5 hours (lunch).<\/li>\n\n\n\n
- After lunch, the motorcyclist increases speed by 10 km\/h, so the new speed is x+10x + 10x+10 km\/h.<\/li>\n\n\n\n
- The motorcyclist wants to reach City B on time<\/strong> (as if no stop was made).<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nStep 2: Calculate distances and times<\/h3>\n\n\n\n\n- Distance covered before lunch: distance1=x\u00d71.5=1.5x\u00a0km\\text{distance}_1 = x \\times 1.5 = 1.5x \\text{ km}distance1\u200b=x\u00d71.5=1.5x\u00a0km<\/li>\n\n\n\n
- Remaining distance after lunch: distance2=225\u22121.5x\u00a0km\\text{distance}_2 = 225 – 1.5x \\text{ km}distance2\u200b=225\u22121.5x\u00a0km<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nStep 3: Calculate total time without stopping<\/h3>\n\n\n\n
If the motorcyclist did not stop<\/strong>, the total time to cover 225 km at speed xxx would be: t=225x hourst = \\frac{225}{x} \\text{ hours}t=x225\u200b hours<\/p>\n\n\n\n
\n\n\n\nStep 4: Calculate total actual time with stopping and speed increase<\/h3>\n\n\n\n
The motorcyclist actually spends:<\/p>\n\n\n\n
\n- 1.5 hours riding before lunch<\/li>\n\n\n\n
- 0.5 hours stopping for lunch<\/li>\n\n\n\n
- Time to ride the remaining distance at speed x+10x + 10x+10: t2=225\u22121.5xx+10t_2 = \\frac{225 – 1.5x}{x + 10}t2\u200b=x+10225\u22121.5x\u200b<\/li>\n<\/ul>\n\n\n\n
So, total actual time: 1.5+0.5+225\u22121.5xx+10=2+225\u22121.5xx+101.5 + 0.5 + \\frac{225 – 1.5x}{x + 10} = 2 + \\frac{225 – 1.5x}{x + 10}1.5+0.5+x+10225\u22121.5x\u200b=2+x+10225\u22121.5x\u200b<\/p>\n\n\n\n
\n\n\n\nStep 5: Set actual time equal to time without stopping (arrive on time)<\/h3>\n\n\n\n
2+225\u22121.5xx+10=225×2 + \\frac{225 – 1.5x}{x + 10} = \\frac{225}{x}2+x+10225\u22121.5x\u200b=x225\u200b<\/p>\n\n\n\n
\n\n\n\nStep 6: Solve the equation<\/h3>\n\n\n\n
Multiply both sides by x(x+10)x(x + 10)x(x+10) to clear denominators: 2x(x+10)+x(225\u22121.5x)=225(x+10)2x(x + 10) + x(225 – 1.5x) = 225(x + 10)2x(x+10)+x(225\u22121.5x)=225(x+10)<\/p>\n\n\n\n
Expand terms: 2×2+20x+225x\u22121.5×2=225x+22502x^2 + 20x + 225x – 1.5x^2 = 225x + 22502×2+20x+225x\u22121.5×2=225x+2250<\/p>\n\n\n\n
Simplify left side: (2×2\u22121.5×2)+(20x+225x)=225x+2250(2x^2 – 1.5x^2) + (20x + 225x) = 225x + 2250(2×2\u22121.5×2)+(20x+225x)=225x+2250 0.5×2+245x=225x+22500.5x^2 + 245x = 225x + 22500.5×2+245x=225x+2250<\/p>\n\n\n\n
Bring all terms to one side: 0.5×2+245x\u2212225x\u22122250=00.5x^2 + 245x – 225x – 2250 = 00.5×2+245x\u2212225x\u22122250=0 0.5×2+20x\u22122250=00.5x^2 + 20x – 2250 = 00.5×2+20x\u22122250=0<\/p>\n\n\n\n
Multiply entire equation by 2 to eliminate decimals: x2+40x\u22124500=0x^2 + 40x – 4500 = 0x2+40x\u22124500=0<\/p>\n\n\n\n
\n\n\n\nStep 7: Solve quadratic equation<\/h3>\n\n\n\n
Use quadratic formula x=\u2212b\u00b1b2\u22124ac2ax = \\frac{-b \\pm \\sqrt{b^2 – 4ac}}{2a}x=2a\u2212b\u00b1b2\u22124ac\u200b\u200b, with a=1a = 1a=1, b=40b = 40b=40, and c=\u22124500c = -4500c=\u22124500: x=\u221240\u00b1402\u22124(1)(\u22124500)2x = \\frac{-40 \\pm \\sqrt{40^2 – 4(1)(-4500)}}{2}x=2\u221240\u00b1402\u22124(1)(\u22124500)\u200b\u200b x=\u221240\u00b11600+180002x = \\frac{-40 \\pm \\sqrt{1600 + 18000}}{2}x=2\u221240\u00b11600+18000\u200b\u200b x=\u221240\u00b1196002x = \\frac{-40 \\pm \\sqrt{19600}}{2}x=2\u221240\u00b119600\u200b\u200b x=\u221240\u00b11402x = \\frac{-40 \\pm 140}{2}x=2\u221240\u00b1140\u200b<\/p>\n\n\n\n
Two possible solutions:<\/p>\n\n\n\n
\n- x=\u221240+1402=1002=50x = \\frac{-40 + 140}{2} = \\frac{100}{2} = 50x=2\u221240+140\u200b=2100\u200b=50<\/li>\n\n\n\n
- x=\u221240\u22121402=\u22121802=\u221290x = \\frac{-40 – 140}{2} = \\frac{-180}{2} = -90x=2\u221240\u2212140\u200b=2\u2212180\u200b=\u221290 (not possible, speed can’t be negative)<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nFinal answer:<\/h3>\n\n\n\n
The motorcyclist\u2019s original speed<\/strong> is 50 km\/h<\/strong>.<\/p>\n\n\n\n
\n\n\n\nExplanation <\/h2>\n\n\n\n
This problem involves calculating the motorcyclist\u2019s original speed given a certain delay and an increased speed after a break to arrive on time. The total distance between City A and City B is 225 km, and the motorcyclist initially travels at speed xxx for 1.5 hours before stopping for lunch. The lunch break lasts 0.5 hours, causing a delay that the motorcyclist compensates for by increasing their speed by 10 km\/h after lunch.<\/p>\n\n\n\n
First, we express the distance traveled before lunch as 1.5×1.5×1.5x km and the remaining distance as 225\u22121.5×225 – 1.5×225\u22121.5x km. If the motorcyclist did not stop, the total time to reach City B would be 225x\\frac{225}{x}x225\u200b hours.<\/p>\n\n\n\n
Since the motorcyclist stops for lunch and then travels faster, the total time consists of three parts: 1.5 hours riding, 0.5 hours resting, and the time taken to cover the remaining distance at the increased speed x+10x + 10x+10 km\/h. This actual time must equal the no-stop time for the motorcyclist to arrive on schedule.<\/p>\n\n\n\n
Setting these equal, we derive an equation relating xxx, the original speed. By algebraic manipulation, we obtain a quadratic equation x2+40x\u22124500=0x^2 + 40x – 4500 = 0x2+40x\u22124500=0. Solving this yields two solutions: 50 km\/h and -90 km\/h. The negative speed is invalid physically, so the motorcyclist\u2019s original speed is 50 km\/h.<\/p>\n\n\n\n
This speed means that before lunch, the motorcyclist covers 1.5\u00d750=751.5 \\times 50 = 751.5\u00d750=75 km. After lunch, the remaining 150 km must be covered at 50+10=6050 + 10 = 6050+10=60 km\/h, which allows the rider to compensate for the 0.5-hour break and still reach City B on time. This problem showcases how speed, time, and distance interact and how to adjust for delays by changing speed.<\/p>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner10-55.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"A motorcyclist leaves City A and rides at a constant speed towards City B, which is 225 km away. After 1.5 hours, the motorcyclist stops for half an hour for lunch. To reach City B on time, the motorcyclist increases his speed after lunch by 10 km\/hour. Find the motorcyclist’s original speed. The Correct Answer […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","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-5)","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-5)","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-5)","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":[25],"tags":[],"class_list":["post-230433","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/230433","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=230433"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/230433\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=230433"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=230433"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=230433"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n
Step 2: Calculate distances and times<\/h3>\n\n\n\n\n- Distance covered before lunch: distance1=x\u00d71.5=1.5x\u00a0km\\text{distance}_1 = x \\times 1.5 = 1.5x \\text{ km}distance1\u200b=x\u00d71.5=1.5x\u00a0km<\/li>\n\n\n\n
- Remaining distance after lunch: distance2=225\u22121.5x\u00a0km\\text{distance}_2 = 225 – 1.5x \\text{ km}distance2\u200b=225\u22121.5x\u00a0km<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nStep 3: Calculate total time without stopping<\/h3>\n\n\n\n
If the motorcyclist did not stop<\/strong>, the total time to cover 225 km at speed xxx would be: t=225x hourst = \\frac{225}{x} \\text{ hours}t=x225\u200b hours<\/p>\n\n\n\n
\n\n\n\nStep 4: Calculate total actual time with stopping and speed increase<\/h3>\n\n\n\n
The motorcyclist actually spends:<\/p>\n\n\n\n
\n- 1.5 hours riding before lunch<\/li>\n\n\n\n
- 0.5 hours stopping for lunch<\/li>\n\n\n\n
- Time to ride the remaining distance at speed x+10x + 10x+10: t2=225\u22121.5xx+10t_2 = \\frac{225 – 1.5x}{x + 10}t2\u200b=x+10225\u22121.5x\u200b<\/li>\n<\/ul>\n\n\n\n
So, total actual time: 1.5+0.5+225\u22121.5xx+10=2+225\u22121.5xx+101.5 + 0.5 + \\frac{225 – 1.5x}{x + 10} = 2 + \\frac{225 – 1.5x}{x + 10}1.5+0.5+x+10225\u22121.5x\u200b=2+x+10225\u22121.5x\u200b<\/p>\n\n\n\n
\n\n\n\nStep 5: Set actual time equal to time without stopping (arrive on time)<\/h3>\n\n\n\n
2+225\u22121.5xx+10=225×2 + \\frac{225 – 1.5x}{x + 10} = \\frac{225}{x}2+x+10225\u22121.5x\u200b=x225\u200b<\/p>\n\n\n\n
\n\n\n\nStep 6: Solve the equation<\/h3>\n\n\n\n
Multiply both sides by x(x+10)x(x + 10)x(x+10) to clear denominators: 2x(x+10)+x(225\u22121.5x)=225(x+10)2x(x + 10) + x(225 – 1.5x) = 225(x + 10)2x(x+10)+x(225\u22121.5x)=225(x+10)<\/p>\n\n\n\n
Expand terms: 2×2+20x+225x\u22121.5×2=225x+22502x^2 + 20x + 225x – 1.5x^2 = 225x + 22502×2+20x+225x\u22121.5×2=225x+2250<\/p>\n\n\n\n
Simplify left side: (2×2\u22121.5×2)+(20x+225x)=225x+2250(2x^2 – 1.5x^2) + (20x + 225x) = 225x + 2250(2×2\u22121.5×2)+(20x+225x)=225x+2250 0.5×2+245x=225x+22500.5x^2 + 245x = 225x + 22500.5×2+245x=225x+2250<\/p>\n\n\n\n
Bring all terms to one side: 0.5×2+245x\u2212225x\u22122250=00.5x^2 + 245x – 225x – 2250 = 00.5×2+245x\u2212225x\u22122250=0 0.5×2+20x\u22122250=00.5x^2 + 20x – 2250 = 00.5×2+20x\u22122250=0<\/p>\n\n\n\n
Multiply entire equation by 2 to eliminate decimals: x2+40x\u22124500=0x^2 + 40x – 4500 = 0x2+40x\u22124500=0<\/p>\n\n\n\n
\n\n\n\nStep 7: Solve quadratic equation<\/h3>\n\n\n\n
Use quadratic formula x=\u2212b\u00b1b2\u22124ac2ax = \\frac{-b \\pm \\sqrt{b^2 – 4ac}}{2a}x=2a\u2212b\u00b1b2\u22124ac\u200b\u200b, with a=1a = 1a=1, b=40b = 40b=40, and c=\u22124500c = -4500c=\u22124500: x=\u221240\u00b1402\u22124(1)(\u22124500)2x = \\frac{-40 \\pm \\sqrt{40^2 – 4(1)(-4500)}}{2}x=2\u221240\u00b1402\u22124(1)(\u22124500)\u200b\u200b x=\u221240\u00b11600+180002x = \\frac{-40 \\pm \\sqrt{1600 + 18000}}{2}x=2\u221240\u00b11600+18000\u200b\u200b x=\u221240\u00b1196002x = \\frac{-40 \\pm \\sqrt{19600}}{2}x=2\u221240\u00b119600\u200b\u200b x=\u221240\u00b11402x = \\frac{-40 \\pm 140}{2}x=2\u221240\u00b1140\u200b<\/p>\n\n\n\n
Two possible solutions:<\/p>\n\n\n\n
\n- x=\u221240+1402=1002=50x = \\frac{-40 + 140}{2} = \\frac{100}{2} = 50x=2\u221240+140\u200b=2100\u200b=50<\/li>\n\n\n\n
- x=\u221240\u22121402=\u22121802=\u221290x = \\frac{-40 – 140}{2} = \\frac{-180}{2} = -90x=2\u221240\u2212140\u200b=2\u2212180\u200b=\u221290 (not possible, speed can’t be negative)<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nFinal answer:<\/h3>\n\n\n\n
The motorcyclist\u2019s original speed<\/strong> is 50 km\/h<\/strong>.<\/p>\n\n\n\n
\n\n\n\nExplanation <\/h2>\n\n\n\n
This problem involves calculating the motorcyclist\u2019s original speed given a certain delay and an increased speed after a break to arrive on time. The total distance between City A and City B is 225 km, and the motorcyclist initially travels at speed xxx for 1.5 hours before stopping for lunch. The lunch break lasts 0.5 hours, causing a delay that the motorcyclist compensates for by increasing their speed by 10 km\/h after lunch.<\/p>\n\n\n\n
First, we express the distance traveled before lunch as 1.5×1.5×1.5x km and the remaining distance as 225\u22121.5×225 – 1.5×225\u22121.5x km. If the motorcyclist did not stop, the total time to reach City B would be 225x\\frac{225}{x}x225\u200b hours.<\/p>\n\n\n\n
Since the motorcyclist stops for lunch and then travels faster, the total time consists of three parts: 1.5 hours riding, 0.5 hours resting, and the time taken to cover the remaining distance at the increased speed x+10x + 10x+10 km\/h. This actual time must equal the no-stop time for the motorcyclist to arrive on schedule.<\/p>\n\n\n\n
Setting these equal, we derive an equation relating xxx, the original speed. By algebraic manipulation, we obtain a quadratic equation x2+40x\u22124500=0x^2 + 40x – 4500 = 0x2+40x\u22124500=0. Solving this yields two solutions: 50 km\/h and -90 km\/h. The negative speed is invalid physically, so the motorcyclist\u2019s original speed is 50 km\/h.<\/p>\n\n\n\n
This speed means that before lunch, the motorcyclist covers 1.5\u00d750=751.5 \\times 50 = 751.5\u00d750=75 km. After lunch, the remaining 150 km must be covered at 50+10=6050 + 10 = 6050+10=60 km\/h, which allows the rider to compensate for the 0.5-hour break and still reach City B on time. This problem showcases how speed, time, and distance interact and how to adjust for delays by changing speed.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"A motorcyclist leaves City A and rides at a constant speed towards City B, which is 225 km away. After 1.5 hours, the motorcyclist stops for half an hour for lunch. To reach City B on time, the motorcyclist increases his speed after lunch by 10 km\/hour. Find the motorcyclist’s original speed. The Correct Answer […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","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-5)","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-5)","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-5)","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":[25],"tags":[],"class_list":["post-230433","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/230433","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=230433"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/230433\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=230433"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=230433"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=230433"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n
Step 3: Calculate total time without stopping<\/h3>\n\n\n\n
If the motorcyclist did not stop<\/strong>, the total time to cover 225 km at speed xxx would be: t=225x hourst = \\frac{225}{x} \\text{ hours}t=x225\u200b hours<\/p>\n\n\n\n The motorcyclist actually spends:<\/p>\n\n\n\n So, total actual time: 1.5+0.5+225\u22121.5xx+10=2+225\u22121.5xx+101.5 + 0.5 + \\frac{225 – 1.5x}{x + 10} = 2 + \\frac{225 – 1.5x}{x + 10}1.5+0.5+x+10225\u22121.5x\u200b=2+x+10225\u22121.5x\u200b<\/p>\n\n\n\n 2+225\u22121.5xx+10=225×2 + \\frac{225 – 1.5x}{x + 10} = \\frac{225}{x}2+x+10225\u22121.5x\u200b=x225\u200b<\/p>\n\n\n\n Multiply both sides by x(x+10)x(x + 10)x(x+10) to clear denominators: 2x(x+10)+x(225\u22121.5x)=225(x+10)2x(x + 10) + x(225 – 1.5x) = 225(x + 10)2x(x+10)+x(225\u22121.5x)=225(x+10)<\/p>\n\n\n\n Expand terms: 2×2+20x+225x\u22121.5×2=225x+22502x^2 + 20x + 225x – 1.5x^2 = 225x + 22502×2+20x+225x\u22121.5×2=225x+2250<\/p>\n\n\n\n Simplify left side: (2×2\u22121.5×2)+(20x+225x)=225x+2250(2x^2 – 1.5x^2) + (20x + 225x) = 225x + 2250(2×2\u22121.5×2)+(20x+225x)=225x+2250 0.5×2+245x=225x+22500.5x^2 + 245x = 225x + 22500.5×2+245x=225x+2250<\/p>\n\n\n\n Bring all terms to one side: 0.5×2+245x\u2212225x\u22122250=00.5x^2 + 245x – 225x – 2250 = 00.5×2+245x\u2212225x\u22122250=0 0.5×2+20x\u22122250=00.5x^2 + 20x – 2250 = 00.5×2+20x\u22122250=0<\/p>\n\n\n\n Multiply entire equation by 2 to eliminate decimals: x2+40x\u22124500=0x^2 + 40x – 4500 = 0x2+40x\u22124500=0<\/p>\n\n\n\n Use quadratic formula x=\u2212b\u00b1b2\u22124ac2ax = \\frac{-b \\pm \\sqrt{b^2 – 4ac}}{2a}x=2a\u2212b\u00b1b2\u22124ac\u200b\u200b, with a=1a = 1a=1, b=40b = 40b=40, and c=\u22124500c = -4500c=\u22124500: x=\u221240\u00b1402\u22124(1)(\u22124500)2x = \\frac{-40 \\pm \\sqrt{40^2 – 4(1)(-4500)}}{2}x=2\u221240\u00b1402\u22124(1)(\u22124500)\u200b\u200b x=\u221240\u00b11600+180002x = \\frac{-40 \\pm \\sqrt{1600 + 18000}}{2}x=2\u221240\u00b11600+18000\u200b\u200b x=\u221240\u00b1196002x = \\frac{-40 \\pm \\sqrt{19600}}{2}x=2\u221240\u00b119600\u200b\u200b x=\u221240\u00b11402x = \\frac{-40 \\pm 140}{2}x=2\u221240\u00b1140\u200b<\/p>\n\n\n\n Two possible solutions:<\/p>\n\n\n\n The motorcyclist\u2019s original speed<\/strong> is 50 km\/h<\/strong>.<\/p>\n\n\n\n This problem involves calculating the motorcyclist\u2019s original speed given a certain delay and an increased speed after a break to arrive on time. The total distance between City A and City B is 225 km, and the motorcyclist initially travels at speed xxx for 1.5 hours before stopping for lunch. The lunch break lasts 0.5 hours, causing a delay that the motorcyclist compensates for by increasing their speed by 10 km\/h after lunch.<\/p>\n\n\n\n First, we express the distance traveled before lunch as 1.5×1.5×1.5x km and the remaining distance as 225\u22121.5×225 – 1.5×225\u22121.5x km. If the motorcyclist did not stop, the total time to reach City B would be 225x\\frac{225}{x}x225\u200b hours.<\/p>\n\n\n\n Since the motorcyclist stops for lunch and then travels faster, the total time consists of three parts: 1.5 hours riding, 0.5 hours resting, and the time taken to cover the remaining distance at the increased speed x+10x + 10x+10 km\/h. This actual time must equal the no-stop time for the motorcyclist to arrive on schedule.<\/p>\n\n\n\n Setting these equal, we derive an equation relating xxx, the original speed. By algebraic manipulation, we obtain a quadratic equation x2+40x\u22124500=0x^2 + 40x – 4500 = 0x2+40x\u22124500=0. Solving this yields two solutions: 50 km\/h and -90 km\/h. The negative speed is invalid physically, so the motorcyclist\u2019s original speed is 50 km\/h.<\/p>\n\n\n\n This speed means that before lunch, the motorcyclist covers 1.5\u00d750=751.5 \\times 50 = 751.5\u00d750=75 km. After lunch, the remaining 150 km must be covered at 50+10=6050 + 10 = 6050+10=60 km\/h, which allows the rider to compensate for the 0.5-hour break and still reach City B on time. This problem showcases how speed, time, and distance interact and how to adjust for delays by changing speed.<\/p>\n\n\n\n A motorcyclist leaves City A and rides at a constant speed towards City B, which is 225 km away. After 1.5 hours, the motorcyclist stops for half an hour for lunch. To reach City B on time, the motorcyclist increases his speed after lunch by 10 km\/hour. Find the motorcyclist’s original speed. The Correct Answer […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","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-5)","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-5)","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-5)","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":[25],"tags":[],"class_list":["post-230433","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/230433","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=230433"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/230433\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=230433"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=230433"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=230433"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nStep 4: Calculate total actual time with stopping and speed increase<\/h3>\n\n\n\n
\n
\n\n\n\nStep 5: Set actual time equal to time without stopping (arrive on time)<\/h3>\n\n\n\n
\n\n\n\nStep 6: Solve the equation<\/h3>\n\n\n\n
\n\n\n\nStep 7: Solve quadratic equation<\/h3>\n\n\n\n
\n
\n\n\n\nFinal answer:<\/h3>\n\n\n\n
\n\n\n\nExplanation <\/h2>\n\n\n\n
<\/figure>\n","protected":false},"excerpt":{"rendered":"