{"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":"
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 \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 \nThe required parts of bases are 32, 56, 96 and 160.<\/p>\n
<\/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
Number of bottles filled<\/td>\n
Number of hours<\/td>\n<\/tr>\n
\n
840<\/td>\n
6<\/td>\n<\/tr>\n
\n
X<\/td>\n
5<\/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
Number of times pho\u00adtograph enlarged<\/td>\n
Length (in cm)<\/td>\n<\/tr>\n
\n
50,000<\/td>\n
5<\/td>\n<\/tr>\n
\n
20,000<\/td>\n
x<\/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
<\/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<\/p>\n
\n\n
\n
Length of the ship<\/td>\n
Height of the mast<\/td>\n<\/tr>\n
\n
28<\/td>\n
12<\/td>\n<\/tr>\n
\n
X<\/td>\n
9<\/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
<\/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
Distance covered on the Road (in km)<\/td>\n
Distance represented on the map (in cm)<\/td>\n<\/tr>\n
\n
18<\/td>\n
1<\/td>\n<\/tr>\n
\n
72<\/td>\n
x<\/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
Height of the pole<\/td>\n
Length of the shadow<\/td>\n<\/tr>\n
\n
5 m 60 cm = 560 cm<\/td>\n
3 m 20 cm = 320 cm<\/td>\n<\/tr>\n
\n
10 m 50 cm = 1050 cm<\/td>\n
cm<\/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
Height of the pole<\/td>\n
Length of the shadow<\/td>\n<\/tr>\n
\n
560 cm<\/td>\n
320 cm<\/td>\n<\/tr>\n
\n
y<\/td>\n
5 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
<\/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
Distance (km)<\/td>\n
Time (minutes)<\/td>\n<\/tr>\n
\n
14<\/td>\n
25<\/td>\n<\/tr>\n
\n
x<\/td>\n
300<\/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