Class 7: Maths Chapter 7 solutions. Complete Class 7 Maths Chapter 7 Notes.<\/p>\n\n\n\n
Selina 7th Maths Chapter 7, Class 7 Maths Chapter 7 solutions<\/p>\n\n\n\n
1. Weight of 8 identical articles is 4.8 kg. What is the weight of 11 such articles?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n Weight of 8 identical articles = 4.8 kg<\/p>\n\n\n\n So the weight of 1 article = 4.8\/8 kg<\/p>\n\n\n\n Here the weight of 11 such articles = 4.8\/8 \u00d7 11<\/p>\n\n\n\n We get<\/p>\n\n\n\n = 0.6 \u00d7 11<\/p>\n\n\n\n = 6.6 kg<\/p>\n\n\n\n 2. 6 books weigh 1.260 kg. How many books will weigh 3.150 kg?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n Weight of 6 books = 1.260 kg or 1 kg 260 g<\/p>\n\n\n\n Number of books = 6<\/p>\n\n\n\n So the number of books in 1 kg = 6\/ 1.260<\/p>\n\n\n\n Number of books in 3.150 kg = (6 \u00d7 3.150)\/ 1.260<\/p>\n\n\n\n Multiplying 1000 to numerator and denominator<\/p>\n\n\n\n = (6 \u00d7 3150)\/ 1260<\/p>\n\n\n\n = 3150\/ 210<\/p>\n\n\n\n = 15 books<\/p>\n\n\n\n 3. 8 men complete a work in 6 hours. In how many hours will 12 men complete the same work?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n Time taken by 8 men to complete a work = 6 hours<\/p>\n\n\n\n So the time taken by 1 man to complete the work = 6 \u00d7 8 hours<\/p>\n\n\n\n Time taken by 12 men to complete the work = (6 \u00d7 8)\/ 12 = 4 hours<\/p>\n\n\n\n 4. If a 25 cm long candle burns for 45 minutes, how long will another candle of the same material and same thickness but 5 cm longer than the previous one burn?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Time taken by a 25 cm long candle to burn = 45 minutes<\/p>\n\n\n\n Time taken by a 1 cm long candle to burn = 45\/ 25 minutes<\/p>\n\n\n\n So the time taken by a 25 + 5 = 30 cm long candle to burn = (45 \u00d7 30)\/ 25 = 54 minutes<\/p>\n\n\n\n 5. A typist takes 80 minutes to type 24 pages. How long will he take to type 87 pages?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Time taken by a typist to type 24 pages = 80 minutes<\/p>\n\n\n\n Time taken by a typist to type 1 page = 80\/24 minutes<\/p>\n\n\n\n Time taken by a typist to type 87 pages = (80 \u00d7 87)\/ 24 = 290 minutes<\/p>\n\n\n\n 6. \u20b9750 support a person for 15 days. For how many days will \u20b9 2,500 support the same person?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n \u20b9 750 can support a family for 15 days<\/p>\n\n\n\n So \u20b9 1 will support a family = 15\/ 750 days<\/p>\n\n\n\n Similarly \u20b9 2500 will support a family = 15\/ 750 \u00d7 2500 = 50 days<\/p>\n\n\n\n 7. 400 men have provisions for 23 weeks. They are joined by 60 men. How long will the provisions last?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n No. of weeks 400 men have provisions = 23 weeks<\/p>\n\n\n\n No. of weeks 1 man have provisions = 23 \u00d7 400 weeks<\/p>\n\n\n\n So the no. of weeks 400 + 60 = 460 men have provisions = (23 \u00d7 400)\/ 460 = 20 weeks<\/p>\n\n\n\n 8. 200 men have provisions for 30 days. If 50 men have left, for how many days the same provisions would last for the remaining men?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n No. of days 200 men have provisions = 30 days<\/p>\n\n\n\n No. of days 1 man have provisions = 30 \u00d7 200 days<\/p>\n\n\n\n So the no. of days 200 \u2013 50 = 150 men have provisions = (30 \u00d7 200)\/ 150 = 40 days<\/p>\n\n\n\n 9. 8 men can finish a certain amount of provisions in 40 days. If 2 more men join them, find for how many days will the same amount of provisions be sufficient.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n No. of days 8 men can finish a certain amount of provision = 40 days<\/p>\n\n\n\n No. of days 1 man can finish provision = 40 \u00d7 8 days<\/p>\n\n\n\n So the no. of days 8 + 2 = 10 men finish a provision = (40 \u00d7 8)\/ 10 = 32 days<\/p>\n\n\n\n 10. If the interest on \u20b9 200 be \u20b9 25 in a certain time, what will be the interest on \u20b9 750 for the same time?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n Interest on \u20b9 200 = \u20b9 25<\/p>\n\n\n\n Interest on \u20b9 1 = \u20b9 25\/200<\/p>\n\n\n\n So the interest on \u20b9 750 = (25 \u00d7 750)\/ 200 = 750\/8 = \u20b9 93.75<\/p>\n\n\n\n 11. If 3 dozen eggs cost \u20b9 90, find the cost of 3 scores of eggs. [1 score = 20]<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n We know that<\/p>\n\n\n\n 3 dozen = 3 \u00d7 12 = 36 eggs<\/p>\n\n\n\n 3 scores = 3 \u00d7 20 = 60<\/p>\n\n\n\n Cost of 36 eggs = \u20b9 90<\/p>\n\n\n\n Cost of 1 egg = \u20b9 90\/36<\/p>\n\n\n\n Cost of 60 eggs = (90 \u00d7 60)\/ 36 = \u20b9 150<\/p>\n\n\n\n 12. If the fare for 48 km is \u20b9 288, what will be the fare for 36 km?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n Fare for 48 km = \u20b9 288<\/p>\n\n\n\n So the fare for 1 km = (288 \u00d7 36)\/ 48 = \u20b9 216<\/p>\n\n\n\n 13. What will be the cost of 3.20 kg of an item, if 3 kg of it costs \u20b9 360?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n Cost of 3 kg of an item = \u20b9 360<\/p>\n\n\n\n Cost of 1 kg of an item = \u20b9 360\/3<\/p>\n\n\n\n So the cost of 3.20 kg of an item = (360 \u00d7 3.20)\/ 3 = \u20b9 384<\/p>\n\n\n\n 14. If 9 lines of a print, in a column of a book, contain 36 words, how many words will a column of 51 lines contain?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n No. of words in a 9 lines of a print = 36<\/p>\n\n\n\n No. of word in a 1 line of a print = 36\/9<\/p>\n\n\n\n No. of words in 51 lines of a print = 36\/9 \u00d7 51 = 204<\/p>\n\n\n\n 15. 125 students have food sufficient for 18 days. If 25 more students join them, how long will the food last now?<\/strong><\/p>\n\n\n\n What assumption have you made to come to your answer?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n No. of pupils in the beginning = 125<\/p>\n\n\n\n No. of pupils joined = 25<\/p>\n\n\n\n So the total no. of pupils = 125 + 25 = 150<\/p>\n\n\n\n No. of days food is sufficient for 125 pupils = 18 days<\/p>\n\n\n\n No. of days food is sufficient for 1 pupil = 18 \u00d7 125 days<\/p>\n\n\n\n No. of days food is sufficient for 150 pupils = (18 \u00d7 125)\/ 150<\/p>\n\n\n\n We get<\/p>\n\n\n\n = (18 \u00d7 5)\/ 6<\/p>\n\n\n\n = 15 days<\/p>\n\n\n\n 1. The cost of 3\/5 kg of ghee is \u20b9 96, find the cost of:<\/strong><\/p>\n\n\n\n (i) one kg ghee.<\/strong><\/p>\n\n\n\n (ii) 5\/8 kg ghee.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n Cost of 3\/5 kg of ghee = \u20b9 96<\/p>\n\n\n\n (i) Cost of 1 kg of ghee = 96 \u00d7 5\/3 = \u20b9 160<\/p>\n\n\n\n (ii) Cost of 5\/8 kg of ghee = 160 \u00d7 5\/8 = \u20b9 100<\/p>\n\n\n\n 2. 3 \u00bd m of cloth costs \u20b9 168, find the cost of 4 1\/3 m of the same cloth.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n Cost of 3 \u00bd m of cloth = \u20b9 168<\/p>\n\n\n\n So the cost of 1 m of cloth = 168 \u00d7 2\/7 = \u20b9 48<\/p>\n\n\n\n Similarly the cost of 4 1\/3 m of cloth = 48 \u00d7 13\/3 = \u20b9 208<\/p>\n\n\n\n 3. A wrist-watch loses 10 sec in every 8 hours. In how much time will it lose 15 sec?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n Time taken by a wrist watch to lose 10 sec = 8 hours<\/p>\n\n\n\n So the time taken by a wrist watch to lose 1 sec = 8\/10 hours<\/p>\n\n\n\n Similarly the time taken by a wrist watch to lose 15 sec = 8\/10 \u00d7 15 = 12 hours<\/p>\n\n\n\n 4. In 2 days and 20 hours a watch gains 20 sec. Find, how much time the watch will take to gain 35 sec.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n We know that<\/p>\n\n\n\n 2 days 20 hours = 2 \u00d7 24 + 20 = 48 + 20 = 68 hours<\/p>\n\n\n\n Time in which 20 sec are gained = 68 hours<\/p>\n\n\n\n So the time in which 1 sec will be gained = 68\/20 hours<\/p>\n\n\n\n Similarly the time in which 35 sec will be gained = 68\/20 \u00d7 35<\/p>\n\n\n\n By further calculation<\/p>\n\n\n\n = 119 hours<\/p>\n\n\n\n So we get<\/p>\n\n\n\n = 119 \u00f7 24 days<\/p>\n\n\n\n = 4 days 23 hours<\/p>\n\n\n\n 5. 50 men mow 32 hectares of land in 3 days. How many days will 15 men take to mow it?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Land is similar for both the case.<\/p>\n\n\n\n No. of days taken by 50 men to mow the land = 3 days<\/p>\n\n\n\n No. of days taken by 1 man to mow the land = 3 \u00d7 50 days<\/p>\n\n\n\n No. of days taken by 15 men to mow the land = (3 \u00d7 50)\/ 15 = 10 days<\/p>\n\n\n\n 6. The wages of 10 workers for six days week are \u20b9 1,200. What are the one day wages:<\/strong><\/p>\n\n\n\n (i) of one worker?<\/strong><\/p>\n\n\n\n (ii) of 4 workers?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n Wages of 10 workers for 6 days a week = \u20b9 1200<\/p>\n\n\n\n Wages of 10 workers per day = 1200\/6 = \u20b9 200<\/p>\n\n\n\n Wages of 1 worker per day = 200\/10 = \u20b9 20<\/p>\n\n\n\n Wages of 4 workers per day = 4 \u00d7 20 = \u20b9 80<\/p>\n\n\n\n 7. If 32 apples weigh 2 kg 800 g, how many apples will there be in a box, containing 35 kg of apples?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Weight of apples in a box = 35 kg<\/p>\n\n\n\n If the weight of apples is 2 kg 800 g (2.8 kg) then the number of apples = 32<\/p>\n\n\n\n No. of apples if the weight is 1 kg = 32\/2.800<\/p>\n\n\n\n No. of apples if the weight is 35 kg = (32 \u00d7 35)\/ 2.800<\/p>\n\n\n\n Multiplying both numerator and denominator by 1000<\/p>\n\n\n\n = (32 \u00d7 35 \u00d7 1000)\/ 2800<\/p>\n\n\n\n = 400<\/p>\n\n\n\n 8. A truck uses 20 litres of diesel for 240 km. How many litres will be needed for 1200 km?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Diesel needed for 240 km = 20 litres<\/p>\n\n\n\n Diesel needed for 1 km = 20\/ 240 litres<\/p>\n\n\n\n Diesel needed for 1200 km = 20\/240 \u00d7 1200 = 100 litres<\/p>\n\n\n\n 9. A garrison of 1200 men has provisions for 15 days. How long will the provisions last if the garrison be increased by 600 men?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n No. of days 1200 men has provisions = 15 days<\/p>\n\n\n\n No. of days 1 man has provisions = 15 \u00d7 1200 days<\/p>\n\n\n\n No. of days 1200 + 600 = 1800 men has provisions = (15 \u00d7 1200)\/ 1800 = 10 days<\/p>\n\n\n\n 10. A camp has provisions for 60 pupils for 18 days. In how many days, the same provisions will finish off if the strength of the camp is increased to 72 pupils?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n No. of days 60 pupil have provisions = 18 days<\/p>\n\n\n\n No. of days 1 pupil have provision = 18 \u00d7 60 days<\/p>\n\n\n\n No. of days 72 pupils have provision = (18 \u00d7 60)\/ 72 = 15 days<\/p>\n\n\n\n 1. A can do a piece of work in 6 days and B can do it in 8 days. How long will they take to complete it together?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n A can do a piece of work in 6 days<\/p>\n\n\n\n So A\u2019s one day work = 1\/6<\/p>\n\n\n\n B can do the same work in 8 days<\/p>\n\n\n\n So B\u2019s one day work = 1\/8<\/p>\n\n\n\n Here A and B one day work = 1\/6 + 1\/8<\/p>\n\n\n\n By taking LCM<\/p>\n\n\n\n = (4 + 3)\/ 24<\/p>\n\n\n\n = 7\/24<\/p>\n\n\n\n Similarly A and B can do the same work = 24\/7 = 3 3\/7 days<\/p>\n\n\n\n 2. A and B working together can do a piece of work in 10 days. B alone can do the same work in 15 days. How long will A along take to do the same work?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n A and B work together to do a piece of work in 10 days<\/p>\n\n\n\n B alone can do the same work in 15 days<\/p>\n\n\n\n So we get<\/p>\n\n\n\n A and B one day work = 1\/10<\/p>\n\n\n\n B\u2019s one day work = 1\/15<\/p>\n\n\n\n A\u2019s one day work = 1\/10 \u2013 1\/15<\/p>\n\n\n\n Taking LCM<\/p>\n\n\n\n = (3 \u2013 2)\/ 30<\/p>\n\n\n\n = 1\/30<\/p>\n\n\n\n No. of days A can do the same work = 30 days<\/p>\n\n\n\n 3. A can do a piece of work in 4 days and B can do the same work in 5 days. Find, how much work can be done by them working together in:<\/strong><\/p>\n\n\n\n (i) one day<\/strong><\/p>\n\n\n\n (ii) 2 days.<\/strong><\/p>\n\n\n\n What part of work will be left, after they have worked together for 2 days?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n A can do a piece of work in 4 days<\/p>\n\n\n\n B can do the same work in 5 days<\/p>\n\n\n\n A\u2019s one day work = \u00bc<\/p>\n\n\n\n B\u2019s one day work = 1\/5<\/p>\n\n\n\n (i) A and B both one day work = \u00bc + 1\/5<\/p>\n\n\n\n Taking LCM<\/p>\n\n\n\n = (5 + 4)\/ 20<\/p>\n\n\n\n = 9\/ 20<\/p>\n\n\n\n (ii) A and B two days\u2019 work = 9\/20 \u00d7 2 = 9\/10<\/p>\n\n\n\n So the work left after 2 days = 1 \u2013 9\/10<\/p>\n\n\n\n Taking LCM<\/p>\n\n\n\n = (10 \u2013 9)\/ 10<\/p>\n\n\n\n = 1\/10<\/p>\n\n\n\n 4. A and B take 6 hours and 9 hours respectively to complete a work. A works for 1 hour and then B works for two hours.<\/strong><\/p>\n\n\n\n (i) How much work is done in these 3 hours?<\/strong><\/p>\n\n\n\n (ii) How much work is still left?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n A takes 6 hours to finish the work<\/p>\n\n\n\n B takes 9 hours to finish the work<\/p>\n\n\n\n A\u2019s 1 hour work = 1\/6<\/p>\n\n\n\n B\u2019s 1 hour work = 1\/9<\/p>\n\n\n\n B\u2019s 2 hour work = 1\/9 \u00d7 2 = 2\/9<\/p>\n\n\n\n (i) A\u2019s 1 hour work + B\u2019s 2 hours\u2019 work = 1\/6 + 2\/9<\/p>\n\n\n\n By taking LCM<\/p>\n\n\n\n = (3 + 4)\/ 18<\/p>\n\n\n\n = 7\/18<\/p>\n\n\n\n (ii) The work still left = 1 \u2013 7\/18<\/p>\n\n\n\n Taking LCM<\/p>\n\n\n\n = (18 \u2013 7)\/ 18<\/p>\n\n\n\n = 11\/18<\/p>\n\n\n\n 5. A, B and C can do a piece of work in 12, 15 and 20 days respectively. How long will they take to do it working together?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n A can do a piece of work in 12 days<\/p>\n\n\n\n B can do a piece of work in 15 days<\/p>\n\n\n\n C can do a piece of work in 20 days<\/p>\n\n\n\n A\u2019s one day work = 1\/12<\/p>\n\n\n\n B\u2019s one day work = 1\/15<\/p>\n\n\n\n C\u2019s one day work = 1\/20<\/p>\n\n\n\n So A, B and C\u2019s together one day work = 1\/12 + 1\/15 + 1\/20<\/p>\n\n\n\n Taking LCM<\/p>\n\n\n\n = (5 + 4 + 3)\/ 60<\/p>\n\n\n\n = 12\/60<\/p>\n\n\n\n = 1\/5<\/p>\n\n\n\n Hence, they can do the work in 5 days.<\/p>\n\n\n\n 6. Two taps can fill a cistern in 10 hours and 8 hours respectively. A third tap can empty it in 15 hours. How long will it take to fill the empty cistern, if all of them are opened together?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n First tap can fill a cistern in 10 hours<\/p>\n\n\n\n Second tap can fill a cistern in 8 hours<\/p>\n\n\n\n Third tap can empty the cistern in 15 hours<\/p>\n\n\n\n First tap\u2019s one hour work = 1\/10<\/p>\n\n\n\n Second tap\u2019s one hour work = 1\/8<\/p>\n\n\n\n Third tap\u2019s one hour work = 1\/15<\/p>\n\n\n\n Here if all the taps are opened together then their one hour work = 1\/10 + 1\/8 \u2013 1\/15<\/p>\n\n\n\n Taking LCM<\/p>\n\n\n\n = (12 + 15 \u2013 8)\/ 120<\/p>\n\n\n\n So we get<\/p>\n\n\n\n = (27 \u2013 8)\/ 120<\/p>\n\n\n\n = 19\/120<\/p>\n\n\n\n All the taps can fill the cistern in = 120\/19 hours = 6 6\/19 hours<\/p>\n\n\n\n 7. Mohit can complete a work in 50 days, whereas Anuj can complete the same work in 40 days.<\/strong><\/p>\n\n\n\n Find:<\/strong><\/p>\n\n\n\n (i) work done by Mohit in 20 days.<\/strong><\/p>\n\n\n\n (ii) work left after Mohit has worked on it for 20 days.<\/strong><\/p>\n\n\n\n (iii) time taken by Anuj to complete the remaining work.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n Mohit can complete a work in 50 days<\/p>\n\n\n\n Anuj can complete the same work in 40 days<\/p>\n\n\n\n Mohit\u2019s one day work = 1\/50<\/p>\n\n\n\n Anuj\u2019s one day work = 1\/40<\/p>\n\n\n\n (i) Work done by Mohit in 20 days = 1\/50 \u00d7 20 = 2\/5<\/p>\n\n\n\n (ii) Work left after Mohit has worked on it for 20 days = 1 \u2013 2\/5 = (5 \u2013 2)\/ 5 = 3\/5<\/p>\n\n\n\n (iii) Time taken by Anuj to complete the remaining work = 40 \u00d7 3\/5 days = 24 days<\/p>\n\n\n\n 8. Joseph and Peter can complete a work in 20 hours and 25 hours respectively.<\/strong><\/p>\n\n\n\n Find:<\/strong><\/p>\n\n\n\n (i) work done by both together in 4 hrs.<\/strong><\/p>\n\n\n\n (ii) work left after both worked together for 4 hrs.<\/strong><\/p>\n\n\n\n (iii) time taken by Peter to complete the remaining work.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Time taken by Joseph to complete a work = 20 hours<\/p>\n\n\n\n Time taken by Peter to complete a work = 25 hours<\/p>\n\n\n\n Joseph\u2019s one hour work = 1\/20<\/p>\n\n\n\n Peter\u2019s one hour work = 1\/25<\/p>\n\n\n\n So both Joseph\u2019 and Peter\u2019s work in one hour = 1\/20 + 1\/25<\/p>\n\n\n\n Taking LCM<\/p>\n\n\n\n = (5 + 4)\/ 100<\/p>\n\n\n\n = 9\/100<\/p>\n\n\n\n (i) Work done by both together in 4 hrs = 9\/100 \u00d7 4 = 9\/25<\/p>\n\n\n\n (ii) Work left after both worked together for 4 hrs = 1 \u2013 9\/25<\/p>\n\n\n\n Taking LCM<\/p>\n\n\n\n = (25 \u2013 9)\/ 25<\/p>\n\n\n\n = 16\/25<\/p>\n\n\n\n (iii) Time taken by Peter to complete the remaining work = 25 \u00d7 16\/25 = 16 hours<\/p>\n\n\n\n 9. A is able to complete 1\/3 of a certain work in 10 hrs and B is able to complete 2\/5 of the same work in 12 hrs.<\/strong><\/p>\n\n\n\n Find:<\/strong><\/p>\n\n\n\n (i) how much work can A do in 1 hour?<\/strong><\/p>\n\n\n\n (ii) how much work can B do in 1 hour?<\/strong><\/p>\n\n\n\n (iii) in how much time will the work be completed, if both work together?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n A can complete 1\/3 of a certain work in 10 hours<\/p>\n\n\n\n Time in which A can do full work = (10 \u00d7 3)\/ 1 = 30 hours<\/p>\n\n\n\n B can complete 2\/5 of a certain work in 12 hours<\/p>\n\n\n\n Time in which B can do full work = (12 \u00d7 5)\/ 2 = 30 hours<\/p>\n\n\n\n (i) Work done by A in 1 hour = 1\/30<\/p>\n\n\n\n (ii) Work done by B in 1 hour = 1\/30<\/p>\n\n\n\n (iii) Work done by both in 1 hour = 1\/30 + 1\/30 = 2\/30 = 1\/15<\/p>\n\n\n\n Hence, both can finish the work in 15 hours.<\/p>\n\n\n\n 10. Shaheed can prepare one wooden chair in 3 days and Shaif can prepare the same chair in 4 days. If they work together, in how many days will they prepare:<\/strong><\/p>\n\n\n\n (i) one chair?<\/strong><\/p>\n\n\n\n (ii) 14 chairs of the same kind?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Work done by Shaheed in one day = 1\/3<\/p>\n\n\n\n Work done by Shaif in one day = \u00bc<\/p>\n\n\n\n So the work done by both in 1 day = 1\/3 + 1\/4<\/p>\n\n\n\n By taking LCM<\/p>\n\n\n\n = (4 + 3)\/ 12<\/p>\n\n\n\n = 7\/12<\/p>\n\n\n\n So both can prepare the chair in 12\/7 = 1 5\/7 days<\/p>\n\n\n\n We know that<\/p>\n\n\n\n One chair can be prepared in 1 5\/7 days<\/p>\n\n\n\n So 14 chairs can be prepared in = 12\/7 \u00d7 14 = 24 days<\/p>\n\n\n\n 11. A, B and C together finish a work in 4 days. If A alone can finish the same work in 8 days and B in 12 days, find how long will C take to finish the work.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n A, B and C together finish the work in 4 days<\/p>\n\n\n\n Work done by A, B and C together in 1 day = \u00bc<\/p>\n\n\n\n A\u2019s one day work = 1\/8<\/p>\n\n\n\n B\u2019s one day work = 1\/12<\/p>\n\n\n\n C\u2019s one day work = \u00bc \u2013 (1\/8 + 1\/12)<\/p>\n\n\n\n By taking LCM<\/p>\n\n\n\n = (6 \u2013 [3 + 2])\/ 24<\/p>\n\n\n\n = 1\/24<\/p>\n\n\n\n Hence, C can finish the work in 24 days.<\/p>\n\n\n\n 1. The cost of 3\/5 kg of ghee is \u20b9 96, find the cost of:<\/strong><\/p>\n\n\n\n (i) one kg ghee.<\/strong><\/p>\n\n\n\n (ii) 5\/8 kg ghee.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n Cost of 3\/5 kg of ghee = \u20b9 96<\/p>\n\n\n\n (i) Cost of 1 kg of ghee = 96 \u00d7 5\/3 = \u20b9 160<\/p>\n\n\n\n (ii) Cost of 5\/8 kg of ghee = 160 \u00d7 5\/8 = \u20b9 100<\/p>\n\n\n\n 2. 3 \u00bd m of cloth costs \u20b9 168, find the cost of 4 1\/3 m of the same cloth.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n Cost of 3 \u00bd m of cloth = \u20b9 168<\/p>\n\n\n\n So the cost of 1 m of cloth = 168 \u00d7 2\/7 = \u20b9 48<\/p>\n\n\n\n Similarly the cost of 4 1\/3 m of cloth = 48 \u00d7 13\/3 = \u20b9 208<\/p>\n\n\n\n 3. A wrist-watch loses 10 sec in every 8 hours. In how much time will it lose 15 sec?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n Time taken by a wrist watch to lose 10 sec = 8 hours<\/p>\n\n\n\n So the time taken by a wrist watch to lose 1 sec = 8\/10 hours<\/p>\n\n\n\n Similarly the time taken by a wrist watch to lose 15 sec = 8\/10 \u00d7 15 = 12 hours<\/p>\n\n\n\n 4. In 2 days and 20 hours a watch gains 20 sec. Find, how much time the watch will take to gain 35 sec.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n We know that<\/p>\n\n\n\n 2 days 20 hours = 2 \u00d7 24 + 20 = 48 + 20 = 68 hours<\/p>\n\n\n\n Time in which 20 sec are gained = 68 hours<\/p>\n\n\n\n So the time in which 1 sec will be gained = 68\/20 hours<\/p>\n\n\n\n Similarly the time in which 35 sec will be gained = 68\/20 \u00d7 35<\/p>\n\n\n\n By further calculation<\/p>\n\n\n\n = 119 hours<\/p>\n\n\n\n So we get<\/p>\n\n\n\n = 119 \u00f7 24 days<\/p>\n\n\n\n = 4 days 23 hours<\/p>\n\n\n\n 5. 50 men mow 32 hectares of land in 3 days. How many days will 15 men take to mow it?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Land is similar for both the case.<\/p>\n\n\n\n No. of days taken by 50 men to mow the land = 3 days<\/p>\n\n\n\n No. of days taken by 1 man to mow the land = 3 \u00d7 50 days<\/p>\n\n\n\n No. of days taken by 15 men to mow the land = (3 \u00d7 50)\/ 15 = 10 days<\/p>\n\n\n\n 6. The wages of 10 workers for six days week are \u20b9 1,200. What are the one day wages:<\/strong><\/p>\n\n\n\n (i) of one worker?<\/strong><\/p>\n\n\n\n (ii) of 4 workers?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n Wages of 10 workers for 6 days a week = \u20b9 1200<\/p>\n\n\n\n Wages of 10 workers per day = 1200\/6 = \u20b9 200<\/p>\n\n\n\n Wages of 1 worker per day = 200\/10 = \u20b9 20<\/p>\n\n\n\n Wages of 4 workers per day = 4 \u00d7 20 = \u20b9 80<\/p>\n\n\n\n 7. If 32 apples weigh 2 kg 800 g, how many apples will there be in a box, containing 35 kg of apples?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Weight of apples in a box = 35 kg<\/p>\n\n\n\n If the weight of apples is 2 kg 800 g (2.8 kg) then the number of apples = 32<\/p>\n\n\n\n No. of apples if the weight is 1 kg = 32\/2.800<\/p>\n\n\n\n No. of apples if the weight is 35 kg = (32 \u00d7 35)\/ 2.800<\/p>\n\n\n\n Multiplying both numerator and denominator by 1000<\/p>\n\n\n\n = (32 \u00d7 35 \u00d7 1000)\/ 2800<\/p>\n\n\n\n = 400<\/p>\n\n\n\n 8. A truck uses 20 litres of diesel for 240 km. How many litres will be needed for 1200 km?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Diesel needed for 240 km = 20 litres<\/p>\n\n\n\n Diesel needed for 1 km = 20\/ 240 litres<\/p>\n\n\n\n Diesel needed for 1200 km = 20\/240 \u00d7 1200 = 100 litres<\/p>\n\n\n\n 9. A garrison of 1200 men has provisions for 15 days. How long will the provisions last if the garrison be increased by 600 men?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n No. of days 1200 men has provisions = 15 days<\/p>\n\n\n\n No. of days 1 man has provisions = 15 \u00d7 1200 days<\/p>\n\n\n\n No. of days 1200 + 600 = 1800 men has provisions = (15 \u00d7 1200)\/ 1800 = 10 days<\/p>\n\n\n\n 10. A camp has provisions for 60 pupils for 18 days. In how many days, the same provisions will finish off if the strength of the camp is increased to 72 pupils?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n No. of days 60 pupil have provisions = 18 days<\/p>\n\n\n\n No. of days 1 pupil have provision = 18 \u00d7 60 days<\/p>\n\n\n\n No. of days 72 pupils have provision = (18 \u00d7 60)\/ 72 = 15 days<\/p>\n\n\n\n 1. A can do a piece of work in 6 days and B can do it in 8 days. How long will they take to complete it together?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n A can do a piece of work in 6 days<\/p>\n\n\n\n So A\u2019s one day work = 1\/6<\/p>\n\n\n\n B can do the same work in 8 days<\/p>\n\n\n\n So B\u2019s one day work = 1\/8<\/p>\n\n\n\n Here A and B one day work = 1\/6 + 1\/8<\/p>\n\n\n\n By taking LCM<\/p>\n\n\n\n = (4 + 3)\/ 24<\/p>\n\n\n\n = 7\/24<\/p>\n\n\n\n Similarly A and B can do the same work = 24\/7 = 3 3\/7 days<\/p>\n\n\n\n 2. A and B working together can do a piece of work in 10 days. B alone can do the same work in 15 days. How long will A along take to do the same work?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n It is given that<\/p>\n\n\n\n A and B work together to do a piece of work in 10 days<\/p>\n\n\n\n B alone can do the same work in 15 days<\/p>\n\n\n\n So we get<\/p>\n\n\n\n A and B one day work = 1\/10<\/p>\n\n\n\n B\u2019s one day work = 1\/15<\/p>\n\n\n\n A\u2019s one day work = 1\/10 \u2013 1\/15<\/p>\n\n\n\n Taking LCM<\/p>\n\n\n\n = (3 \u2013 2)\/ 30<\/p>\n\n\n\n = 1\/30<\/p>\n\n\n\n No. of days A can do the same work = 30 days<\/p>\n\n\n\n 3. A can do a piece of work in 4 days and B can do the same work in 5 days. Find, how much work can be done by them working together in:<\/strong><\/p>\n\n\n\n (i) one day<\/strong><\/p>\n\n\n\n (ii) 2 days.<\/strong><\/p>\n\n\n\n
\n\n\n\nExercise 7B page: 90<\/h4>\n\n\n\n
\n\n\n\nExercise 7C page: 92<\/h4>\n\n\n\n
Exercise 7B<\/h3>\n\n\n\n
Exercise 7C<\/h3>\n\n\n\n