{"id":28152,"date":"2021-07-03T18:04:04","date_gmt":"2021-07-03T12:34:04","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=28152"},"modified":"2022-03-02T10:30:23","modified_gmt":"2022-03-02T05:00:23","slug":"ncert-solutions-for-class-8-maths-chapter-13-ex-13-1","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-8-maths-chapter-13-ex-13-1\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1"},"content":{"rendered":"

These NCERT Solutions for Class 8 Maths<\/a> Chapter 13 Direct and Inverse Proportions Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Exercise 13.1<\/h2>\n

Question 1.
\nFollowing are the car parking charges near a railway station upto<\/p>\n\n\n\n\n\n\n
4 hours<\/td>\n\u20b9 60<\/td>\n<\/tr>\n
8 hours<\/td>\n\u20b9 100<\/td>\n<\/tr>\n
12 hours<\/td>\n\u20b9 140<\/td>\n<\/tr>\n
24 hours<\/td>\n\u20b9 180<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Check if the parking charges are in direct proportion to the parking time.
\nAnswer:<\/p>\n\n\n\n\n\n\n\n
Parking time<\/td>\nParking charges<\/td>\nRatio<\/td>\n<\/tr>\n
4 hours<\/td>\n\u20b9 60<\/td>\n\\(\\frac { 60 }{ 4 }\\) = 15<\/td>\n<\/tr>\n
8 hours<\/td>\n\u20b9 100<\/td>\n\\(\\frac{100}{8}=\\frac{25}{2}\\)<\/td>\n<\/tr>\n
12 hours<\/td>\n\u20b9 140<\/td>\n\\(\\frac{140}{12}=\\frac{35}{3}\\)<\/td>\n<\/tr>\n
24 hours<\/td>\n\u20b9 180<\/td>\n\\(\\frac{180}{24}=\\frac{15}{2}\\)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Since \\(\\frac{15}{1} \\neq \\frac{25}{2} \\neq \\frac{35}{3} \\neq \\frac{15}{2}\\)
\n\u2234 The parking charges are not in direct proportion with the parking time.<\/p>\n

Question 2.
\nA mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.
\n\"NCERT
\nAnswer:
\nLet the number of parts of red pigment be x1<\/sub>, x2<\/sub>, x3<\/sub>, x4<\/sub> , x5<\/sub> and the number of parts of
\nthe base be y1<\/sub>, y2<\/sub>, y3<\/sub>, y4<\/sub>, y5<\/sub>
\nAs the number of parts of red pigment increases, number of parts of the base also increases in the same ratio It is a direct proportion
\n\"NCERT
\nThe required parts of bases are 32, 56, 96 and 160.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nIn question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?
\nAnswer:
\nWe have \\(\\frac{\\mathrm{x}_{1}}{\\mathrm{y}_{1}}=\\frac{1}{8}=\\frac{\\mathrm{x}_{2}}{\\mathrm{y}_{2}}\\)
\nHere x1<\/sub> = 1, y1<\/sub> = 75 and y2<\/sub> = 1800
\n\u2234 \\(\\frac{1}{75}=\\frac{x_{2}}{1800}\\)
\nx2<\/sub> = \\(\\frac{1 \\times 1800}{75}\\) = 24
\nThus, the required no. of red pigments = 24.<\/p>\n

Question 4.
\nA machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
\nAnswer:<\/p>\n\n\n\n\n\n
Number of bottles filled<\/td>\nNumber of hours<\/td>\n<\/tr>\n
840<\/td>\n6<\/td>\n<\/tr>\n
X<\/td>\n5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

More the number of hours, more the number of bottles that would be filled. Thus, given quantities vary directly.
\n\u2234 \\(\\frac{840}{x}=\\frac{6}{5}\\)
\nx = \\(\\frac{840 \\times 5}{6}\\) = 140 \u00d7 5 = 700
\n\u2234 The required number of bottles = 700<\/p>\n

Question 5.
\nA photograph of a bacteria enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?
\nAnswer:
\nActual length = \\(\\frac{5}{50000}=\\frac{1}{10000}\\) = 10-4<\/sup> cm<\/p>\n\n\n\n\n\n
Number of times pho\u00adtograph enlarged<\/td>\nLength (in cm)<\/td>\n<\/tr>\n
50,000<\/td>\n5<\/td>\n<\/tr>\n
20,000<\/td>\nx<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The length increases with an increment in the number of times the photograph enlarged.
\n\u2234 It is a case of direct proportion.
\n\\(\\frac{50000}{20000}=\\frac{5}{x}\\)
\n50,000 \u00d7 x = 5 \u00d7 20,000
\nx = \\(\\frac{5 \\times 20,000}{50,000}\\) = 2
\n\u2234The enlarged length = 2 cm.<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nIn a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship?
\nAnswer:
\nLet the required length of the model ship be ‘x’ cm
\n\"NCERT<\/p>\n\n\n\n\n\n
Length of the ship<\/td>\nHeight of the mast<\/td>\n<\/tr>\n
28<\/td>\n12<\/td>\n<\/tr>\n
X<\/td>\n9<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Since, more the length of the ship, more would be the length of its mast.
\n\u2234 It is a direct variation \\(\\frac{28}{x}=\\frac{12}{9}\\)
\n12 \u00d7 x = 28 \u00d7 9
\nx = \\(\\frac{28 \\times 9}{12}=\\frac{7 \\times 9}{3}\\) = 7 \u00d7 3 = 21
\nThe required length of the model = 21 cm.<\/p>\n

Question 7.
\nSuppose 2 kg of sugar contains 9 x 106 crystals. How many sugar crystals are there in (i) 5 kg of sugar? (ii) 1.2 kg of sugar?
\nAnswer:
\n(i) Let the required number of sugar crystals be Y.
\nSince more the amount of sugar, more would be the number of sugar crystals.
\n\u2234 It is a direct variation
\n\\(\\frac{2}{5}=\\frac{9 \\times 10^{6}}{\\mathrm{x}}\\)
\n2x = 5 \u00d7 9 \u00d7 106<\/sup>
\nx = \\(\\frac{5 \\times 9 \\times 10^{6}}{2}=\\frac{45 \\times 10^{6}}{2}\\) = 22.5 \u00d7 106<\/sup>
\n= 2.25 \u00d7 10 \u00d7 106<\/sup>
\n= 2.25 \u00d7 107<\/sup>
\n\u2234 Required number of sugar crystals = 2.25 x 107<\/sup><\/p>\n

(ii) Let the number of sugar crystals in a sugar be \u2018y\u2019
\nIt is a direct variation
\n\\(\\frac{2}{1.2}=\\frac{9 \\times 10^{6}}{\\mathrm{y}}\\)
\n2y = 9 \u00d7 106<\/sup> \u00d7 1.2
\ny = \\(\\frac{9 \\times 10^{6} \\times 1.2}{2}\\) = 9 \u00d7 106<\/sup> \u00d7 0.6
\n= 5.4 \u00d7 106<\/sup>
\nThe required number of sugar crystals = 5.4 \u00d7 106<\/sup><\/p>\n

\"NCERT<\/p>\n

Question 8.
\nRashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?
\nAnswer:
\nLet the required distance covered in the map be ‘x’ cm.<\/p>\n\n\n\n\n\n
Distance covered on the Road (in km)<\/td>\nDistance represented on the map (in cm)<\/td>\n<\/tr>\n
18<\/td>\n1<\/td>\n<\/tr>\n
72<\/td>\nx<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

It is a direct variation
\n\\(\\frac{18}{72}=\\frac{1}{x}\\)
\n18 \u00d7 x = 72 \u00d7 1
\nx = \\(\\frac{72 \\times 1}{18}\\) = 4 \u00d7 1 = 4
\n\u2234 The required distance on the map is 4 cm<\/p>\n

Question 9.
\nA 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time
\n(i) the length of the shadow cast by another pole 10 m 50 cm high
\n(ii) the height of a pole which casts a shadow 5 m long.
\nAnswer:
\nLet the required length of shadow be ‘x’ cm<\/p>\n\n\n\n\n\n
Height of the pole<\/td>\nLength of the shadow<\/td>\n<\/tr>\n
5 m 60 cm = 560 cm<\/td>\n3 m 20 cm = 320 cm<\/td>\n<\/tr>\n
10 m 50 cm = 1050 cm<\/td>\ncm<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

As the height of the pole increases, the length of the shadow also increases in the same ratio
\nIt is a direct variation.
\n\\(\\frac{560}{1050}=\\frac{320}{\\mathrm{x}}\\)
\n560 \u00d7 x = 1050 \u00d7 320
\nx = \\(\\frac{1050 \\times 320}{560}\\) = 600 cm
\n\u2234 Required length of the shadow = 600 cm (6 m)<\/p>\n

(ii) Let the required height of the pole be y\u2019 cm<\/p>\n\n\n\n\n\n
Height of the pole<\/td>\nLength of the shadow<\/td>\n<\/tr>\n
560 cm<\/td>\n320 cm<\/td>\n<\/tr>\n
y<\/td>\n5 m = 500 cm<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

It is a direct variation.
\n\\(\\frac{560}{y}=\\frac{320}{500}\\)
\ny \u00d7 320 = 560 \u00d7 500
\ny = \\(\\frac{560 \\times 500}{320}\\) = 125 \u00d7 7 = 875
\n7 320
\n\u2234 The height of the pole = 875 cm (or) 8 m 75 cm.<\/p>\n

\"NCERT<\/p>\n

Question 10.
\nA loaded truck travels 14 km in 25 minutes. If the speed remains the same, how for can it travel in 5 hours?
\nAnswer:
\nLet the distance travelled in 5 hours be ‘x’ km.<\/p>\n\n\n\n\n\n
Distance (km)<\/td>\nTime (minutes)<\/td>\n<\/tr>\n
14<\/td>\n25<\/td>\n<\/tr>\n
x<\/td>\n300<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

If the time increases, the distance also increases.
\n\u2234 It is a direct variation.
\n\\(\\frac{14}{x}=\\frac{25}{300}\\)
\n25 \u00d7 x = 300 \u00d7 14
\nx = \\(\\frac{300 \\times 14}{25}\\) = 12 \u00d7 14 = 168
\n\u2234 The required distance =168 km.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Exercise 13.1 Question 1. Following are the car parking charges near a railway station upto 4 hours …<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","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":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[7],"tags":[],"yoast_head":"\nNCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 - MCQ Questions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-8-maths-chapter-13-ex-13-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts. 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