Class 9 Maths_ Manual

97 (iii) (a) Consumed water = Present reading – Previous reading (b) Consumed water in Jesth = 1013 – 1000 = 13 units (c) Consumed water on Asadh = 1037 – 1013 = 24 units = 10 units + 14 units Bill amount = 100 + 14 × 32 = Rs. 548 Drain charge = 50% of 548 = Rs. 274 Final bill of Asadh = bill + drain + meter rental charge = 548 + 274 + 5 = Rs. 827. (d) Consumed water in Shrawan = 1074 – 1037 = 37 units Hence, Jestha month has economy than Ashadh. 9. (i) A family used 358 units in a month and the payment was made. [Use the water charge and rules of payment from the page no. 114 and 115] (a) How much money should the family pay if payment was made within 62ndday? (b) How much more money should be paid if the payment was made on 1stmonth? (c) Compare the difference in percentage when payment is made on 62ndand on 108thday. (ii) The water meter in a house showed the following units used within the fourth month. 02578 units (At the end of Shrawan), 02986 units (At the end of Bhadra) (a) How can we find the consumed water units of Bhadra? Find it. (b) If the pipe line is of 3 4 inch whose minimum charge for 27 units is Rs. 1910 and charge of extra units = 71 per unit. Find the total bill paid by the family if the bill is paid on the 9th day of billing. (c) Compare the difference in percentage when payment is made on 65thand on 128thday. Solution: (i) (a) Consumed water = 358 = 10 + 348 Bill amount = 100 + 348 × 32 = Rs. 11,236 Bill amount within 62 days = Rs. 11,236. (b) On 1st month, bill amount = 11,236 – 3% of 11,236 = 11,236 – 337.08 = Rs. 10,898.92 (c) On 108th day, bill amount = 11,236 + 10% of 11,236 = 11,236 + 1123.6 = Rs. 12,359.6 Difference on 62th and 108th day = 12,359.6 – 11,236 = Rs. 1,123.60 Difference % = 1123.6 11236 × 100% = 10% more (ii) (a) Consumed water unit on Bhadra = 02986 – 02578 (b) Now, 408 units = 408 = 27units + 381units Bill amount = 1910 + 381 × 71 = Rs. 28,961 Bill on 9th day = 28961 – 3% of 28961 = 28961 – 868.83 = Rs. 28,092.17 (c) On 65th day, bill = 28961 + [according to bill] On 128th bill, 28961 + 20% of 28961 = 28961 + 5792.2 = Rs. 34,753.20 Difference in payment % = 34753.20 – 28961 28961 × 100% = 5792.2 28961 × 100% = 20% more.

98 4.3 Household Expenses for Use of Telephone PRACTICE4.3 Read/Understand/Think/Do Keeping Skill Sharp 1. (a) Define telephone. (b) Define unit call in landline phone. (c) How can you find the telephone calls? (d) Which formula do you use to calculate the bill amount of the telephone calls when the minimum charge rate and rate of extra call are given? (e) Write the formula to calculate the subtotal amount of telephone calls. (f) Write the formula to calculate the grand total amount of telephone calls. Solution: (a) A telephone is a telecommunications device that permits two or more users to conduct a conversation when they are too far apart to be easily heard directly. (b) Talking of 2 minutes between two persons by landline telephone is called a unit call. (c) Telephone calls = Present calls – Previous calls (d) Bill amount = Minimum charge + rate rate of extra call (e) Subtotal amount of telephone calls = Minimum charge + Rate Extra calls (f) Grand total amount of telephone calls = Total + VAT% of Total. 2. (a) In a telephone bill of a house, the previous reading 0952 and the present reading 1095 are shown. How many calls are made there? (b) In a telephone bill of an office, the previous reading is 12345 calls and the total calls is 543 calls. What is the present reading? (c) In a telephone bill of a house, the present reading is 98765 calls and the total calls is 1234 calls. What is the previous reading ? (d) In a telephone bill of a house, 7080 calls is shown on the last of Asar, 2076. If the total calling during the month of Asar is 524 calls, what was the appearing reading on the first of Asar, 2076? Solution: (a) Calls = Present calls – Previous calls = 1095 – 0952 = 143 calls. (b) Present calls = Previous + total calls = 12345 + 543 = 12888. (c) Previous calls = Present calls – total calls = 98765 – 1234 = 97531 calls. (d) On 1st of Asar = 7080 – 524 = 6556 calls. Check Your Performance 3. (i) The minimum charge for first 175 telephone calls is Rs. 200. If the charge for each additional call is Re. 1. (a) What percent of Telephone Service Charge (TSC) is imposed by NTC? (b) How much will be the charge for 792 calls? (c) Find the total payment charge including TSC and 13% VAT. (d) Find out the percentage of difference of answers from b and c.

99 (ii) The current and previous calls of the local telephone of a household are 12645 and 11265 respectively. (Minimum charge for first 175 calls = Rs. 200, Rate of additional call = Re. 1) (a) How can we calculate the telephone call? (b) What is the total number of calls done by the household? (c) Find out the total bill with 10% TSC and 13% VAT. (d) By how much the VAT amount is more than TSC amount? Solution: (i) (a) TSC = 10% of sub total bill (b) 792 calls = 175 calls + 617calls Call charge = 200 + 617 × 1 = Rs. 817 (c) Including TSC = 817 + 10% of 817 = 817 + 81.7 = Rs. 898.7 with VAT = 898.7 + 13% of 898.7 = 898.7 + 116.83 = Rs. 1015.53 (d) Difference in (b) and (c) = 1015.53 – 817 = Rs. 198.53 Difference % = 198.53 817 × 100% = 24.3% (ii) (a) Total call = Total call charge – Minimum charge Rate of additional call + Minimum call (b) Telephone call = 12645 – 11265 = 1380 calls (c) 1380 calls = 175 + 1205 Sub. total = 200 + 1205 × 1 = Rs. 1405 TSC = 10% of 1405 = 140.5 Total = 1405 + 140.5 = Rs. 1545.5 VAT amount = 13% of 1545.5 = 13 100 × 1545.5 = Rs. 200.92 Ground total = 1545.5 + 200.92 = Rs. 1746.42 (d) More percentage in VAT = 200.92 – 140.5 140.5 × 100 = 43% 4. (i) The minimum charge for the first 175 telephone calls is Rs. 200 and then additional charge per call is Re. 1. Moreover, 10% TSC and VAT are included in the bill. (a) What percent of VAT is levied on telephone bill with service charge? (b) How much should pay for 2368 calls without TSC and VAT? (c) Find the total bill amount to be paid with TSC and VAT. (d) By how much percent the final bill increased from the bill without TSC and VAT? (ii) The previous calls and current calls of the local telephone of a company in a month are 12689 calls and 13965 calls respectively (Minimum charge for first 175 calls = Rs. 200, Rate of additional call = Re. 1, TSC = 10% and VAT = 13%) (a) Write down the formula for calculating total bill amount if subtotal, TSC% and VAT% are given. (b) How many calls did the family make in the given month? (c) Find the grand total amount with 10% TSC and 13% VAT by the household. (d) Compare the TSC amount and VAT amount. Solution: (i) (a) VAT rate = 13%

100 (b) 2368 calls = 175 calls + 2193calls Sub bill = 200 + 2193 × 1 = Rs. 2393 (c) TSC = 10% of 2393 = 239.3 Total bill = 2393 + 239.3 = Rs. 2632.3 VAT = 13% of 2632.3 = Rs. 342.20 Grand total = 2632.3 + 342.2 = Rs. 2974.50 (d) Difference in subtotal and grand total = 2974.50 – 2393 = Rs. 581.50 Difference % = 581.50 2393 × 100% = 24.30% (ii) (a) Total bill amount = Subtotal + 10% of subtotal + 13% of (subtotal + 10% of subtotal) (b) Calls made = 13965 – 12689 = 1276 calls. (c) Subtotal = 175 + 1101 = 200 + 1101 × 1 = Rs. 1301. TSC = 10% of 1301 = Rs. 130.10 Total = 1301 + 130.10 = Rs. 1431.10 VAT amount = 13% of 1431.10 = Rs. 186.04 Grand total = 1431.10 + 186.04 = Rs. 1617.14 (d) More in VAT = 186.04 – 130.10 = Rs. 55.94 5. (i) A telephone booth charges Rs. 15 for the first minute and Rs. 1.5 for every additional 10 seconds for ISD call from Kathmandu to Canada. (a) What does ISD call stands for? (i) International Sub-Dialing (ii) Internal Subscriber Dialing (iii) International Subscriber Dial (iv) International Subscriber Dialing (b) If a girl made the telephone call of 4 minutes and 40 seconds from telephone booth, find the amount to be paid by the girl. (c) If the actual charge of ISD call from Kathmandu to Canada is Rs. 25 for first 2 minutes and Rs. 2 for every additional 20 seconds, find the actual telephone bill of booth owner. (d) Which was better for booth owner, the modified charging system or the actual charging system? (ii) The charge of ISD Telephone Booth, Jomsom are given below: ISD Telephone Booth, Jomsom Call London Call USA Minimum Charge Rs. 45 for a minute Minimum Charge Rs. 25 for a minute Additional Charge Rs. 4.25 per 15 seconds Additional Charge Rs. 2.5 per 10 seconds (a) If a person needs to call Paris from Jomsom which call he need to make? (i) Local (ii) STD (iii) ISD (iv) none of them (b) If a man made the telephone call for 8 minutes 45 seconds to London, then he made another call for 10 minutes and 30 second to Washington DC, how much amount did he pay altogether? (c) Which call was cheaper for him by how much? Compare them. (iii) The charge of ISD call for USA from Pokhara is Rs. 12 for the first minute and Rs. 2 for every additional 15 seconds. A boy made the telephone call and paid Rs. 24 for the bill. (a) Which of them is ISD call?

101 (i) Pokhara−Pokhara (ii) Pokhara−Kathmandu (iii) Pokhara−New York (iv) Pokhara−Dhunche (b) How long did the boy made the call? (c) If the rate was modified as Rs. 24 for first two minutes and Rs. 1.5 for every additional 15 second, then find the difference. (d) Which was best for the boy? Give reason. Solution: (i) (a) (iv) (b) We have, 4 min. 40 sec= 1 min + 3 min 40 sec. = 1 min + 220 sec = 1 min + 220 10 times = 1 min + 22 time Charge for 4 min 40 sec = 15 + 22 × 1.5 = 15 + 33 = Rs. 48 (c) We have, 4 min 40 sec = 2 min + 2 min 40 sec = 2 min + 160 sec = 2 min = 160 20 time = 2 min + 8 times Charge for 4 min 40 sec = 25 + 8 × 2 = 25 + 16 = Rs. 41 (d) The first system is better for booth owner as it generates more bill than modified system. (ii) (a) (iii) (b) For London; Total time = 8 min 45 sec = 1 + 7 min 45 sec = 1 min + 465 sec = 1 min + 465 15 times = 1 min + 31 times = Rs. 45 + 31 × Rs. 4.25 = Rs. 45 + Rs. 131.75 = Rs. 176.75 For USA; Total time = 10 min 30 sec = 1 + 9 min 30 sec = 1 + 570 sec = 1 min + 570 10 times = 1 min + 57 times = Rs. 25 + 57 × Rs. 2.5 = Rs. 25 + Rs. 142.5 = Rs. 167.50 Now, total payment = Rs. 176.75 + Rs. 167.5 = Rs. 344.25. (c) The call in London is cheaper for him by Rs. 344.25−Rs. 176.75 = Rs. 167.50 (iii) (a) (iii) (b) Total paid by boy = Rs. 24 24 = minimum + extra = 1 min + extra = Rs. 12 + Rs 12 Now, Rs. 2 can call 15 sec Rs. 1 can call 15 2 sec Rs. 12 can call 15 2 × 12 = 90 sec. Total call made = 1 min + 90 sec = 1 min + 1 min 30sec = 2 min 30 sec (c) According to the question. Time = 2 min 30 sec = 2 + 30 15 times = 2 min + 2 times

102 = Rs. 24 + 2 × 1.5 = Rs. 27 Now, difference = Rs. 27 – Rs. 24 = Rs. 3 (d) 1st option because, he can save Rs. 3 for the same time. 6. (i) The minimum charge for 175 local telephone calls is Rs. 200 and Re. 1 is charged for each extra call. (a) How many calls do we need to maintain if we want to pay only Rs. 200 each month? (b) If a person makes 738 calls in a month of Shrawan, then find out the sub-total charge of his local telephone call. (c) If the sub-total charge of Bhadra is Rs. 763, how many calls were made? (d) Compare the calls made by him in two different months mentioned above. (ii) The minimum charge of telephone call up to 175 calls is Rs. 200 and the charge of each additional call is Rs. 1.5. If a costumer paid Rs. 747.50 in a month, find the number of additional calls in the month. (a) Write down the minimum charge of telephone call up to 100 calls. (b) If a costumer paid Rs. 747.50 in a month, find the number of additional calls in the month. (c) Later, the owner realized that the actual rate was Rs. 1 for each additional call, then find the amount that the costumer get back. (d) What is the percentage of amount that he get back to amount he paid at first? (iii) The minimum charge for the first 175 telephone calls is Rs. 200 and then the additional charge per call is Re. 1. 10% TSC and 13% VAT are included in the bill. (a) Write down the formula to calculate the total charge of local telephone call if subtotal, TSC% and VAT% are. (b) Using the formula find how many calls can be used by Rs. 1602.227. (c) Find out the TSC and VAT amount separately. (d) Compare TSC and VAT in percentage. (iv) A man pays Rs. 423.50 in a month for local telephone calls. The charges are indicated below: Minimum rental charge (with 100 free calls) = Rs. 150, Each extra call = Re. 1, Telephone service charge = 10%, (imposed on the telephone call charges) and VAT = 13% (imposed on the telephone call charges with telephone service charges). (a) If the total calls are limited to 86 calls only, how much charge the man had to pay? (b) Find the total number of local calls made by that man. (c) Estimate the number of calls he needs to reduce if he wants to pay only minimum rental charge? Solution: (i) (a) Maximum 175 calls per month (b) Consumed telephone calls = Calls 738 = 175 Calls + 563Calls Now, subtotal = Rs. 200 + 563 × Rs. 1 = Rs. 763 (c) Subtotal for Bhadra is Rs. 763. Now, 763 = min. charge + extra or, x = 763 – 200 = 563 calls Total call = 175 + 563 = 738 calls (d) There are equal calls or same bill.

103 (ii) (a) Minimum charge for 175 calls is Rs. 200. (b) Total bill = Rs. 747.50 Let extra call be x, then 747.50 = 200 + x × 1.50 or, 747.50 – 200 = 1.50x or, x = 547.50 1.50 = 365 calls. Hence, total call = 175 + 365 = 540 calls. (c) If Rs. 1 for extra calls he has to pay of 540 = 175calls + 365calls Charge for 540 calls = Rs. 200 + 365 × Rs. 1 = Rs. 565 He will get back = 747.50 – 565 = Rs. 182.50 (d) Paid back% = 182.5 747.5 × 100% = 24.41% (iii) (a) See in 4. (ii)(a). (b) Let sub total bill be Rs. x. Then, TSC = 10% of x = x 10 Total = x + x 10 = 11x 10 VAT = 13% of 11x 10 = 13 100 × 11x 10 = 143x 1000 Now, Grand total = 11x 10 + 143x 1000 or, 1602.227 = 1100x + 143x 1000 or, 1602227 = 1243x or, x = 1602227 1243 = Rs. 1289. Now, call made = min + extra = 200 + 1089 = 175 calls + 1089 calls = 1264 calls (c) TSC = x 10 = 1289 10 = Rs. 128.90 VAT = 143x 1000 = 143 × 1289 1000 = Rs. 184.33 (d) More VAT % = 184.33 – 128.90 128.90 × 100% = 43% more (iv) (a) For 86 calls, the man need to pay Rs. 150. (b) Grand total = Rs. 423.50 Let subtotal be Rs. x, then TSC = 10% of x = x 10 Total = x + x 10 = 11x 10 VAT = 13% of 11x 10 = 143x 1000 Now, grand total = 11x 10 + 143x 1000

104 or, 423.50 = 1100x + 143x 1000 or, 423500 = 1243x or, x = 423500 1243 = Rs. 340.70. Now, Rs. 340.70 = Rs. 150 + Rs. 190.70 or, No. of calls = 100 calls + 191 calls = 291 calls [Rs.190.70 can make 191 extra calls.] (c) He needs to reduce 191 calls to pay only minimum rental charge. 7. (i) The minimum charge for the first 175 telephone calls Rs. 200 and then additional charge per call is Re. 1. 10% TSC and 13% VAT are included in the bill. (Rules of payment: Within 7th day according to the bill. After 7thday 10% late fee. (a) What percent of late fee is charged by NTC if the payment was done after 7thday of billing? (b) How much should pay for 1368 calls with TSC and VAT? (c) If the bill is to be paid within 15thof the billing, find the amount of late fee according to rule of payment. (d) If the bill was paid on 5thday, what amount would be saved by the man? (ii) The current calls and previous calls of the local telephone of a company in a month are 926 calls and 682 calls respectively. (Minimum charge for first 175 calls = Rs. 200, Rate of additional call = Re. 1) (a) On which amount will be the TSC imposed? (b) Find the payment amount with 10% TSC and 13% VAT. (c) Find the total bill of the household if 10% fine is added when the bill is to be paid at the end of second month after the billing month. (d) Compare TSC amount, VAT amount and late fee amount. Solution: (i) (a) 10% fee time on grand total (b) Calls 1368 Sub bill = 175 + 1193 = Rs. 200 + 1193 × 1 = Rs. 1393 TSC = 10% of 1393 = Rs. 139.30 Total = 1393 + 139.30 = Rs. 1532.30 VAT = 13% of 1532.30 = Rs. 199.12 Ground total = Rs. 1532.30 + Rs. 199.12 = Rs. 1731.42 (c) Late fee = 15% of 1731.42 = Rs. 259.71 (d) On 5th day, he saves = Rs. 259.71 (ii) (a) TSC imposed on sub total amount. (b) Call made = 926 –682 = 244 calls. Sub total for 244 calls = 175 + 69 = 200 + 69 × 1 = Rs. 269. TSC = 10% of 269 = 26.90 Total = 269 + 26.90 = Rs. 295.90 VAT = 13% of 295.90 = Rs. 38.467 Grand total = 295.90 + 38.47 = Rs. 334.37 (c) If 10% is fined, then Ground total = 334.37 + 10% of 334.37 = 334.37 + 33.44 = Rs. 367.81 (d) VAT > late fee > TSC.

105 4.4 Calculation of Taxi Fare PRACTICE4.3 Read/Understand/Think/Do Keeping Skill Sharp 1. (a) What type of device is called taxi meter? (b) How can you find the taxi fare when the minimum fare, rate of fare per 200m an travelled distance are given ? (c) What is the formula to find the rate of taxi fare? Write it. Solution: (a) A mechanical or electronic device that measured the travelling fare of taxi is called taxi meter. (b) Taxi fare = min. fare + R × Distance 200 m (c) Rate of taxi fare = (Total fare – min. fare) ÷ distance 2. (a) What is the taxi fare to travel 8.4 km at the rate of Rs. 125 per km? (b) Kirtipur is 6.2 km far from the Bhaisepati. How much the taxi fare should be paid at the rate of Rs. 25 per 200 m? (c) Deira City Centre, Dubai is 140.25 km far from Grand Sheikh Mosque, Abu Dhabi. How much taxi fare is paid at the rate of Rs. 244 per km? Solution: (a) Taxi fare = 8.4 × 125 = Rs. 1050 (b) Total distance = 6.2 km = 6.2 × 1000 m = 6200 m Total fare = 6200 200 × 25 = Rs. 775 (c) Taxi fare = 140.25 × 244 = Rs. 34,221 Check Your Performance 3. (i) A meter taxi shows Rs. 60 as the initial fare and Rs. 49 in each km. (a) What is the formula to calculate the taxi fare if initial fare, travelled distance and the rate of fare per km are given? (b) How much amount is paid by a passenger for travelling in the taxi for 20 km? (c) How much more money did he need to travel 4 additional km? (ii) Mr. Gyan hired a metered taxi and travelled 8.6 km. If the minimum fare is Rs. 200 and the fare goes on at the rate of Rs. 15 per 200 m, how much taxi fare is required to him? (a) What is the formula to calculate the taxi fare if initial fare, travelled distance in km and the rate of fare per 200 m are given? (b) How much amount is paid by Mr. Gyan as a taxi fare? (c) How much more distance can he travel with extra Rs. 75? (iii) Mr. Ram Ghimire travelled Purano Bus Park, Ratnapark to Naya Bus Park, Gongabu by a metered taxi at 11 pm. Find the total taxi fare paid by Mr. Ram Ghimire on the basis of the following information: Distance between Purano Bus Park, Ratnapark and Naya Bus Park, Gongabu = 6.2 km, minimum fare (from 9pm to 6am) = Rs. 90, fare for each 200 m = Rs. 14.70.

106 (a) What is the minimum fare of taxi from 9pm to 6am? (b) How much money did Mr Ram Ghimire pay as taxi fare? (c) He had returned from Gongabu to Ratnapark at 8am. Then did he paid the same fare? Give reason? Solution: (i) (a) Taxi fare = initial fare + distance × Rate (b) Taxi fare = 60 + 20 × 49 = 60 + 980 = Rs. 1040. (c) He need Rs. 196 more to travel additional 4 km. (ii) (a) Taxi fare = initial fare + R × Distance in km 200 m (b) Fare for 8.6 km = 8.6 × 1000 = 8600 m Taxi fare = initial + R × 8600 200 = 200 + 15 × 43 = 200 + 645 = Rs. 845 (c) With extra Rs. 75 he travel 1 km more distance. (iii) (a) Rs. 90 (b) Distance = 6.2 km = 6.2 × 1000 = 6200 m. For night = initial + R × D 200 = 90 + 14.70 × 6200 200 = 90 + 455.7 = Rs. 545.70 (c) No, he did not pay the same fare because night and day taxi fare are different. 4. (i) Daya travelled from Balaju to Kakani at 8 am by a metered taxi. The meter showed the fare of Rs. 432.40. (Minimum taxi fare (from 6am to 9pm)= Rs. 16, fare per 200 m = Rs. 9.80) (a) What is the minimum fare of taxi from 6am to 9pm? (b) Find the distance of Balaju to Kakani? (c) He had returned from Kakanni to Balaju at 8pm. Then did he paid the same fare? Give reason? (ii) Miss Sun Keshari paid Rs. 839.70 to the metered taxi driver by travelling some distance at 9:10 pm. (Minimum taxi fare = Rs. 90, fare per 200 m = Rs. 14.70) (a) What is the per 200m fare at night time? (b) Find the travelled distance by her. (c) If the taxi got punctured before 2km from her destination and she left the taxi, how much money would be less? Solution: (i) (a) Minimum fare = Rs. 16 (b) Total fare = Rs. 432.40 = 16 + 416.40 Fare of 200 m = 9.8 1 km = 9.8 × 5 = Rs. 49 So, distance = 416.4 49 = 8.49 km (c) No, as the rate of taxi fare is higher in night time than in day time. (ii) (a) Per 200 m, fare is Rs. 14.70. (b) Taxi fare = Rs. 839.70 Fare for 1 km = 5 × 14.7 = Rs. 73.5 Fare for distance = 839.70 – 90 = Rs. 749.7 Distance = 749.7 73.5 = 10.2 km (c) Less money for 2 km = 2 × 73.5 = Rs. 147

107 5. (i) The taxi meter shows Rs. 60 at first and then charges Rs. 9.80 every 200 meters and waiting charge for 2 minutes is Rs. 9.80. (a) What is the rate of waiting charge of taxi? (b) A girl travelled 15.4 km, calculate the fare. (c) If the girl had waited for 16 minutes on the travelling and gave one thousand rupee note, how much should she get back? (ii) A tourist travelled 28.6 km distance using metered taxi with the rate of Rs. 11.10 per 200 m plus minimum charge at the start Rs. 21 and Rs. 11.10 per 2 minutes waiting where total waiting time is 32 minutes. (a) Write down the formula to calculate the taxi fare when minimum charge, rate of travel, travelled distance, waiting time and rate of waiting charge are given? (b) What fare should the tourist had to pay? (c) If taxi had not stopped anywhere, how much money would be saved? (d) If he has given one thousand notes, how much should he add to? Solution: (i) (a) Rs. 9.80 per 2 minutes. (b) Travel distance 15.4 km Fare for 1 km = 9.80 × 5 = Rs. 49 Fare = initial + R × Distance = 60 + 15.4 × 49 = 60 + 754.6 = Rs. 814.60 (c) Waiting charge = 16 2 × 9.8 = 8 × 9.8 = Rs. 78.4 Taxi fare with 16 min. waiting charge = 814.60 + 78.4 = Rs. 893 So, she will get back = 1000 – 893 = Rs. 107 (ii) (a) Taxi fare = min. charge + Rate × Travelled distance + Rate Waiting time (b) Fare for 1 km = 11.1 × 5 = Rs. 55.5 Initial charge = Rs. 21 Waiting charge for 32 min = 32 2 × 11.1 = Rs. 177.6 Tourist had to pay = 21 + 28.6 × 55.5 + 177.6 = 21 + 1587.3 + 177.6 = Rs. 1785.90 (c) She saves Rs. 177.6. (d) She will get back = 2000 – 1785.90 = Rs. 214.10 6. (i) Mr. Kiran hired a metered taxi and travelled some distance at 8:00 pm. (Minimum taxi fare = Rs. 90, fare per 200 m = Rs. 15.20, waiting charge for each 2 minutes = Rs. 15.20) (a) Write down the formula to calculate the taxi fare when minimum charge, rate of travel and travelled distance are given. (b) If he paid the total fare of Rs. 804.40 including waiting charge of 8 minutes, calculate the travelled distance by him. (c) If taxi had not stopped anywhere, how far could he travel with the same money? (ii) Mrs. Shila travelled some distance by a metered taxi at 12:15 pm. She spent 28 minutes at shopping mall during her travel. If the taximeter showed the fare of Rs. 443.20, find the distance travelled by her. (Minimum taxi fare = Rs. 100, fare per 200 m = Rs. 16.50, waiting charge for each 2 min. = Rs. 16.50) (a) Which algorithm does a taxi meter use to calculate the taxi fare when minimum charge, rate of travel, travelled distance, waiting time and rate of waiting charge are given? (b) If the taxi meter shows the total fare of Rs. 727 including waiting charge of 28 minutes, calculate the travelled distance by her. (c) If Mrs Shila ignored shopping, how far could she travel with the same money?

108 Solution: (i) (a) Taxi fare = min. charge + Rate × Travelled distance (b) Minimum fare = Rs. 90 Fare for 1 km = 15.2 × 5 = Rs. 76 Waiting charge = 8 2 × 15.2 = Rs. 60.80 Now, fare for distance = 804.40 – 90 – 60.80 = Rs. 653.60 Hence, distance = 653.60 76 = 8.6 km (c) If taxi had not stopped, he can travel more distance of Rs. 60.80. i.e. 1000 m 76 × 60.80 = 800 m (ii) (a) Taxi fare = min. charge + Rate × Travelled distance + Rate Waiting time (b) Minimum fare = Rs. 100 Fare for 1 km = 5 × 16.5 = Rs. 82.5 Waiting change = 28 2 × 16.5 = Rs. 231 Fare for distance = 727 – 100 – 231 = Rs. 396 Now, travelled distance = 396 82.5 km = 4.8 km (c) If not stopped, she can travel more = 1 82.5 × 231 km = 2.8 km 7. (i) The taximeter shows Rs. 14 at first and then charges Rs. 7.40 every 200 meters and waiting charge for 2 minutes is Rs. 7.40. A tourist waited for 28 minutes on the travelling. (a) Define taxi fare. (b) Find the fare if the tourist travelled for 26.2 km. (c) If he adds 10% tips of the fare, what amount will the driver get in total? (ii) An American tourist travelled from Tribhuwan International Airport to Thamel by the metered taxi with the rate of Rs. 10 per 200 metres plus minimum charge at the start Rs. 50 and Rs. 10 per 2 minutes waiting where total waiting time is 12 minutes. Distance is 8.6 km. (a) Write down the rate of waiting charge of taxi. (b) Find the fare with 15% tips if the taxi travelled through the blue line route. (c) The taxi took long route (grey line). Tourist realized that he is being cheated by driver, so he refused to give tips and paid as per meter. How much did he save or loss? Solution: (i) (a) The money a customer pays for a journey in a taxi is called taxi fare. (b) Fare for 1 km = 5 × 7.4 = Rs. 37 Taxi fare = 14 + 37 × 26.2 = 14 + 969.4 = Rs. 983.40 Waiting change = 28 2 × 7.4 = Rs. 103.6 Total fare = 983.40 + 103.6 = Rs. 1087 (c) With 10% tips = 1087 + 10% of 1087 = 1087 + 108.7 = Rs. 1195.70 (ii) (a) Waiting charge = Rs. 10 for 2 mins. (b) Minimum charge = Rs. 50 8.6 km = 8600 m Rate of 200 m = Rs. 10 Charge for 8.6 km = Rs. 10 8600 m 200 m = Rs. 430

109 Waiting charge = 12 2 × 10 = Rs. 60 Taxi Fare = 50 + 430 + 60 = Rs. 540 Tips amount = 15% of Rs. 540 = Rs. 81 Total fare with tips = Rs. 540 + Rs. 81 = Rs. 621 (c) Loss amount of the driver = Tips amount = Rs. 81. 8. (i) A man hired a metered taxi and travelled from Prithvi Chowk to Hemja of Pokhara. (Minimum taxi fare = Rs. 50, fare/200 m = Rs. 10, waiting charge for each 2 minutes = Rs. 10) (a) Write down the rate of distance travel by taxi. (b) If the fare with 30 minutes waiting charge and 20% tips is Rs. 924, find the distance between Prithvi Chowk to Hemja. (c) If the driver had not get any tips, how much less amount will he get? (ii) A tourist travelled from Pokhara to Lumbini at 7:15 am by the metered taxi with the rate of Rs. 20 per 200 metres plus minimum charge at the start Rs. 100 and Rs. 20 per 2 minutes waiting where the taxi had waited 1 hour at hotel and 20 minutes at cafeteria. He gave Rs. 34892 to the taxi driver after arrived Lumbini by adding 30% tips. (a) Write down the initial charge of taxi fare. (b) How far is Lumbini from Pokhara ? (c) If the taxi ran non-stop, how much less amount will he get? Solution: (i) (a) Rate of distance travel by taxi = Rs. 10 per 200 m (b) Minimum charge = Rs. 50 Fare for 1 km = 10 × 5 = Rs. 50 Waiting charge = 30 2 × 10 = Rs. 150 Fare without tip is Rs. x, then x + 20% of x = 924 or, 6x 5 = 924 or, x = 924 × 5 6 = 770 Fare for distance = 770 – 50 – 150 = Rs. 570 Distance = 570 50 = 11.40 km (c) Tip = 20% of 770 = Rs. 154 Less amount = Rs. 154 (ii) (a) The initial charge of taxi fare is Rs. 100. (b) Minimum change = Rs. 100 Fare for 1 km = 5 × 20 = Rs. 100 Waiting charge = 80 2 × 20 = Rs. 800 Late fare before tips be Rs. x, then x + 30% of x = 34892 or, 1.3x = 34892 or, x = Rs. 26840 Now, fare for distance = 26840 – 100 – 800 = Rs. 25,940. Distance = 25940 100 km = 259.4 km (c) Less waiting charge = Rs. 800

110 Additional Practice – II 1. Ramesh is an unmarried secondary level teacher. His monthly salary is Rs. 43689. He gets Rs. 2000 per month dearness allowance and once a year one-month salary for festival expenses. A total of the 10% salary of each month is deposited in the employee's provident fund and the same amount is deposited by the government in the same fund. Rate of tax applicable to unmarried persons for the fiscal year 2078/79. Income Rate of tax Income Rate of tax Up to 5 lakhs 1 percentage 10- 20 lakhs up to 30 percentage 5 - 7 lakhs up to 10 percentage more than 20 lakhs 36 percentage 7 - 10 lakhs up to 20 percentage (a) What do you mean by income tax? write it. (b) Find the income of Ramesh in 13 months including festival expenses and EPF. (c) Find out how much income tax he has to pay annually if he also deposits Rs 2000 per month in citizen investment fund from his income. Solution: (a) The tax levied directly on personal income above some certain minimum amount fixed by government is called the income tax. (b) Income = 13 × 43689 + 2000 × 12 + 10% of (12 × 43689) = 5,67,957 + 24000 + 52426.8 = Rs. 6,44,383.80 (c) Taxable amount = 644,383.80 – 2 × 52,426.8 – 12 × 2000 = Rs. 5,15,530.20 Tax of 5,15,530.20 = 1% of 5,00,000 + 10% of 15,530.20 = 5000 + 1553.02 = 6553.02 2. The commission rate given by a company to an agent is mentioned below. Selling Goods Commission Rate Selling Goods Commission Rate Up to 15 lakhs 0.5% 25 lakhs - 54 lakhs 1.5% 15 lakhs - 25 lakhs 1% Above 54 lakhs 2% (a) Write the formula for finding rate of commission. (b) If an agent is selling goods worth Rs.28 lakhs, what amount does he get as commission? (c) If the amount of commission is Rs 25000, calculate the minimum value of goods to be sold. (d) The monthly salary of an agent is Rs. 15000 and he gets commission more than 5 lakhs book sold at 2%, what amounts of goods to be sold for being monthly income Rs. 19000? Solution: (a) Rate of commission = Commission Amount Total Sells × 100% (b) 28,00,000 = 15,00,000 + 10,00,000 + 3,00,000 = 0.5% of 15,00,000 + 1% of 1,00,000 + 1.5% of 3,00,000 = Rs. 7500 + 10000 + 4500 = Rs. 22,000 He gets Rs. 22,000 as commission. (c) Increases commission = 25000 – 22000 = Rs. 3000 Let x be increased – sales amount Then, 1.5% of x = 3000

111 or, 1.5 100 × x = 3000 or, x = Rs. 2,00,000 Hence, 28,00,000 + 2,00,000 = 30,00,000 value of goods to be sold. (d) Commission = 19000 – 15000 = Rs. 4000 Let sold amount be x, then 2% of x = 4000 or, 2 100 × x = 4000 or, x = 4000 × 100 2 = Rs. 2,00,000 Hence, total sold = 5,00,000 + 2,00,000 = Rs. 7,00,000 3. A man paid Rs. 15,750 for a palmtop with 12.5% VAT. (a) Find the listed price of the palmtop. (b) If 15% discount is given, find the price with 13% VAT. Solution: (a) Let listed price be Rs. x. Then, x + 12.5% of x = 15,750 or, x + 0.125x = 15,750 or, 1.125x = 15750 or, x = 15750 1.125 = Rs. 14,000 Listed price of palmtop is Rs. 14,000 (b) Discount amount =15% of 14000 = Rs. 2100 SP = 14000 – 2100 = Rs. 11900 VAT = 13% of 11,900 = Rs. 1547 SP with VAT = 11,900 + 1547 = Rs. 13,447. 4. A cost of an article with 15% discount and 10% VAT is Rs. 5610. (a) Find the marked price. (b) Find the difference between discount and VAT. (c) If the marked price is 20% above the cost price, find CP. Solution: (a) Let MP be Rs. x, then by question Discount = 15% of x = 0.15x SP = x – 0.15x = 0.85x VAT = 10% of 0.85x = 0.085x Now, SP with VAT = 0.85x + 0.085x or, 5610 = 0.935x or, x = 5610 0.935 = Rs. 6000 (b) Discount – VAT = 0.15x – 0.085x = 0.065x = 0.065 × 6000 = Rs. 390 (c) Let CP by Rs. y, then

112 MP = CP + 20% of CP or, 6000 = y + 20 100 × y or, 6000 = 1.2y or, y = 6000 1.2 = Rs. 5000 Hence, CP is Rs. 5000. 5. Mr. Lichwa bought a cycle of Rs. 35,000 with 12% discount and some percent of VAT at Rs. 34,804. (a) Find VAT percentage. (b) Find the actual discount percentage in the transaction for the customer. (c) If the discount is double percent of VAT, find the price of the cycle with the same VAT rate. Solution: (a) Here, MP of cycle = 35000 Discount = 12% of 35000 = Rs. 4200 SP = 35000 – 4200 = Rs. 30,800 Now, VAT = SP with VAT – SP = 34,804 – 30,800 = Rs. 4004 VAT % = 4004 30800 × 100% = 13% (b) Actual discount = 35,000 – 34,804 = Rs. 196 Actual discount % = 196 3500 × 100% = 0.56% (c) Discount is 2 × VAT rate = 2 × 13 = 26% Now, Discount = 26% of 35000 = Rs. 9100 SP = 35,000 – 9,100 = Rs. 25,900 VAT = 13% of 25,900 = Rs. 3367 SP with VAT = 25,900 + 3367 = Rs. 29,267. 6. The marked price of a jacket is Rs. 5000. If the discount and VAT rate are the same and is sold at Rs. 4950. Find (a) Rate of VAT. (b) Price of the jacket without bill. Solution: (a) Here, MP = 5000 Let discount & VAT rate be x%. then, discount = x% of 5000 = 50x SP = 5000 – 50x VAT = x% of (5000 – 50x) = 0.01x(5000 – 50x) = 50x – 0.5x2 Now, SP with VAT = SP + VAT = 5000 – 50x + 50x – 0.5x2 or, 4950 = 5000 – 0.5x2 or, 0.5x2 = 5000 – 4950 or, x2 = 50 0.5 = 100

113 or, x = 100 = 10% Hence, VAT rate = 10% (b) Price of jacket without bill = 5000 – 10% of 5000 = 5000 – 500 = Rs. 4500 7. A businessman imports EV car from Indra at Rs. 8,00,000 without 13% VAT. He pays 150% tax at Bhairahawa custom office and Rs. 15,000 for transportution to Kathmandu. The businessman sells the EV car to Laxmi Motors at a profit of Rs. 50,000. And Laxmi Motor spend Rs. 15000 for showroom rent, 20,000 for Metropolitan city and made 30,000 profit on it. If a customer buys the EV car, (a) Find the cost of the EV car. (b) What is the total VAT amount? (c) Find the VAT paid by Laxmi motors. Solution: (a) Cost of EV for businessman = 800000 + 150% of 800000 + 15000 = 800000 + 1200000 + 15000 = Rs. 20,15,000 Profit = Rs. 5000 SP of businessman = 20,15,000 + 50,000 = 20,65,000 Now, selling price of B. Man is CP for Laxmi Motors. Total CP for Laxmi motors = 20,65,000 + 15,000 + 20,000 = Rs. 21,00,000 SP of EV = 21,00,000 + 30,000 = Rs. 21,30,000 SP with VAT = 21,30,000 + 13% of 21,30,000 = 21,30,000 + 2,76,900 = Rs. 24,06,900 (b) VAT amount = Rs. 2,76,900 (c) VAT paid by Laxmi motors = 13% of SP = 13% of 20,65,000 = Rs. 2,68,450 8. Shrestha Carpet Udhog sold a carpet of amount Rs. 75,000 to a dealer with 20% profit and adding 13% VAT. The dealer paid 1% local tax, Rs. 2,000 for transportation and 10% profit is added on it. If the rate of VAT is 13%, find (a) VAT paid by the customer. (b) Cost of the carpet. (c) How much percent is more than actual price for customer? Solution: Here, SP of carpet = 75,000 + 20% of 75,000 = 75,000 + 15,000 = Rs. 90,000 SP with VAT = 90,000 + 13% of 90,000 = 90,000 + 11,700 = 1,01,700 Now, for dealer CP of carpet is Rs. 1,01,700 Dealer's CP = 1,01,700 + 1% of 1,01,700 + 2,000 = Rs. 1,04,717 SP with profit = 1,04,717 + 10% of 1,04,717 = 1,04,717 + 10,471.70 = Rs. 1,15,188.70 (a) VAT paid by customers = 11,700 + 13% of (1017 + 2000 + 10,471.70) = 11,700 + 13% of 13,488.7 = 11,700 + 1753.53 = Rs. 13,453.53 (b) Cost of carpet = 1,15,188.70 + 13% of 10,471.70 = 1,15,188.70 + 1,361.32 = Rs. 1,16,550.02

114 (c) Increased % in price = 116550.02 – 70000 70000 × 100% = 66.5% 9. Swikriti sold the photocopy machine to Hari for Rs. 92095 with 13 % VAT. Hari paid Rs. 2000 as transportation cost, Rs. 500 as local tax and Rs 5000 as a profit and delivered it to Shyam's house. (a) Write the formula for finding Value Added Tax (VAT). (b) How much did Hari pay for that machine without VAT? Find it. (c) How much did the machine cost when it reached Shyam's house? Find it. Solution: (a) VAT % of added amount (b) Let CP of machine for Hari be Rs. x. then, x + 13% of x = 92095 or, x + 13 100 x = 92095 or, 113x = 92095 × 100 or, x = 92095 × 100 113 = Rs. 81,500 Hence, Hari paid Rs. 81,500. (c) Cost price for Shyam = 92095 + 2000 + 500 + 5000 + 13% of (2000 + 500 + 5000) = 92095 + 7500 + 13% of 7500 = 99,595 + 975 = Rs. 1,00,570 10. There are 22,500 shares at the rate of Rs. 100 in Sanjeewani Co-operative. The co-operative earned Rs. 15,00,000 in a year and decided to distribute 15% cash dividend of total profit to the shareholders. Sunil is a member and he has 840 shares. (a) Find the cash dividend amount. (b) How much cash dividend amount will Sunil get? (c) If half of the profit amount is distributed as cash dividend to the shareholders, how much additional cash dividend amount would Sunil get? Solution: Here, No. of share = 22,500 (a) Cash dividend amount = 15% of 15,00,000 = Rs. 2,25,000 (b) Per share dividend = 225000 22500 = Rs. 10 Sunil get dividend = 840 × 10 = Rs. 8400 (c) If half of profit is distributed, the dividend amount is 1 2 of 15,00,000 = 7,50,000 So, each share get 750000 22500 = Rs. 100 3 . Now, Sunil will get = 840 × 100 3 = 28000 Additional cash dividend = 27,997.20 – 8400 = Rs. 19,597.20

115 11. In a water processing industry, water comes from a 1 2 inch pipe. To do water business, a person has also connected a telephone set for business communication. He uses a taxi to come and go to the water industry which is 20 km away from his house. Answer the following questions based on the rules given below: Calculation of taxi meter Rate of per month for 1 2'' water pipe Telephone Starting taxi fare is Rs. 60 after that additional charge per km. is Rs, 49 minimum use in liter minimum maintained price The additional charge of per 1 unit water when its use is more than the minimum maintained The minimum charge of first 175 calls is Rs. 200 and then additional charge Rs. 1 per call 10,000 liter (1 unit = 1000 liter) Rs. 100 Rs. 32 (a) How many rupees does the businessman have to pay for using a taxi every day just to go to the water processing industry from home? (b) How much rupees should the company pay for the water tariff, which uses 85,000 liters of water in a month? (c) How many maximum call can he do in one month without increasing the telephone expenses from Rs. 1000? Mention with calculation. Solution: (a) Taxi fare = minimum + rate × distance = 60 + 20 × 49 = Rs. 1040 (b) Uses of water = 85,000 = 85000 1000 = 85 units Now, 85 = (10 + 75) units Hence, company pay = Rs. 100 + 75 × 32 = 100 + 2400 = Rs. 2500 (c) Max. budget = Rs. 1000 Now, 1000 = min + additional cost = 200 + 800 = 175 calls + 800 calls = 975 calls. 975 calls can made by Rs. 1,000. 12. Ramesh has been travelling in a taxi from Kalanki to Koteshwor at 9:30 am. The distance between Kalanki and Koteshwor is 10.7 km. The taxi charges are as follows: 6 AM to 9 PM 9 PM to 6 AM Starting rate = Rs. 50 Starting rate = Rs. 75 The rate per 200 meters = Rs. 10 The rate per 200 meters = Rs. 15 (a) How much amount should be paid for travelling only1 kilometer? (b) How much total cost does Ramesh pay for travelling 10.7 km in a taxi at night time? Calculate and mention. (c) Ramesh gave Rs. 1000 to the driver, how much amount does driver return back to Ramesh? Calculate and mention. Solution: (a) Fare for 1 km = 5 × 15 = Rs. 75

116 (b) 10.7 km fare = min. + distance fare = 75 + 10.7 × 75 = 75 + 802.5 = Rs. 877.50 (c) If Ramesh gave one thousand note, he will get back = 1000 – 877.50 = Rs. 122.50. 13. The Electricity Consumer Tariff rate for 5 amperes on date 2080-07-25 by Nepal Electricity Authority is as follows. Consumed Unit 5 Ampere Service Charge (Rs.) Energy Charge (Rs.) 0 - 20 30 3.00 21 - 30 50 6.50 31 - 50 50 8.00 51 - 100 75 9.50 101-250 100 9.50 A 5 ampere meter has been installed in Ramesh's house. The electricity bill of the house for the month of Kartik shown as follows. Previous RDG 4283 Present RDG 4523 (a) How much electricity consumption in the month of Kartik? (b) How much rupees should he pay in a month of Kartik? (c) What is the maximum number of units that he can consume to reduce the tariff amount to Rs, 1501 for the month of mangsir? Calculate and mention with reason. Solution: (a) Electricity consumed on Kartik = 4523 – 4283 = 240 units. (b) Charge of 240 units = 20 + 10 + 20 + 50 + 140 = 20 × 3 + 10 × 6.5 + 20 × 8 + 50 × 9.5 + 140 × 9.5 = 60 + 65 + 160 + 475 + 1330 = Rs. 2090 Now, bill with service charge = 2090 + 100 = Rs. 2,190. (c) Units amount to Rs. 1501, we do, 1501 = 100 + 60 + 65 + 160 + 475 + 641 = 60 3 + 65 6.5 + 160 8 + 475 9.5 + 641 9.5 units = 20 + 10 + 20 + 50 + 67.47 = 146.47 units. [The first Rs. 100 is service charge.] 14. A pen drive was sold at a discount of the 10%. If 13% VAT was added to the price and the customer got Rs. 600 as discount. (a) How much was paid on VAT? (b) If VAT amount is the same as discount find the VAT%. Solution: Let MP of pendrive = Rs. x. then, discount = 10% of x = x 10 SP = x – x 10 = 9x 10 VAT = 13% of 9x 10 = 13 100 × 9x 10 = 117x 1000 Now, SP with VAT = SP + VAT

117 or, 600 = 9x 10 + 117x 1000 or, 600 = 900x + 117x 1000 or, 600000 = 1017x or, x = 600000 1017 = Rs. 589.97 (a) VAT amount = 117 × 589.97 1000 = Rs. 69.03 (b) Discount amount = 589.97 10 = Rs. 58.997 Now, SP = 9x 10 = 9 × 589.97 10 = Rs. 530.97 VAT % = 58.997 530.97 × 100% = 11.10% 15. A tourist paid Rs. 5610 for decorative window. If the shopkeeper has given 15% discount and 10% VAT was levied on it. (a) Find the marked price of the window. (b) Find the amount when he goes bact his to his country. (c) Also find the discount amount. Solution: (a) Let MP be Rs. x, then Discount = 15% of x = 3x 20 SP = x – 3x 20 = 17x 20 VAT = 10% of 17x 20 = 10 100 × 17x 20 = 17x 200 Now, SP with VAT = 5610 or, 17x 20 + 17x 200 = 5610 or, 170x + 17x 200 = 5610 or, 187x = 5610 × 200 or, x = 5610 × 200 187 = Rs. 6000. (b) VAT amount = 17x 200 = 17 × 6000 200 = Rs. 510. (c) Discount amount = 3x 20 = 3 × 6000 20 = Rs. 900 16. A telephone set has marked Rs. 2100. Including 20% discount and certain percent VAT added the price reached to Rs. 1848. Find, (a) VAT amount. (b) VAT percentage.

118 Solution: (a) Here, Marked price = Rs. 2100 Discount = 20% Discount amount = 20% of 2100 = Rs. 420 SP = 2100 – 420 = Rs. 1680 Now, VAT = SP with VAT – SP = 1848 – 1680 = Rs. 168 (b) VAT % = 168 1680 × 100% = 10% 17. The discount rate is twice of VAT rate. If a watch is bought at Rs. 1672.40 of Rs. 2000, find (a) VAT rate. (b) Discount rate. (c) Actual discount amount and rate. Solution: (a) Let VAT rate = x%, then discount rate = 2x% Now, discount = 2x% of 2000 = 2x 100 × 2000 = 40x and SP = (2000 – 40x) Now, VAT = x% of (2000 – 40x) = x 100 (2000 – 40x) = 20x – 2 5 x2 Hence, SP with VAT = 1672.40 SP + VAT = 1672.40 or, (2000 – 40x) + 20x – 2 5 x2 = 1672.40 or, 2000 – 20x – 2 5 x2 = 1672.40 or, 10000 – 100x – 2x2 = 1672.40 × 5 or, 2x2 + 100x – 10000 + 8362 = 0 or, 2(x2 + 50x – 819) = 0 or, x2 + 5x – 819 = 0 or, x2 + 63x – 13x – 819 = 0 or, x(x + 63) – 13(x + 63) = 0 or, (x + 63) (x – 13) = 0 Either, x + 63 = 0 OR, x – 13 = 0 or, x = – 63 (impossible) or, x = 13 Hence, VAT rate = 13% (b) Discount rate = 2x = 2 × 13 = 26% (c) Actual discount amount = 2000 – 1672.40 = Rs. 327.60 Discount rate = 327.60 2000 × 100 = 16.38%

119 UNIT III MENSURATION CHAPTER AREA 5 5.1 Area of Scalene Triangle PRACTICE 5.1 Read / Understand / Think / Do Keeping Skill Sharp 1. (a) Write the definition of scalene triangle. (b) Write the formula to compute the area of a scalene triangle with the sides a, b and c. (c) What is the formula to compute the area of a right-angled triangle with hypotenuse h, perpendicular p and base b? (d) What is the area of a parallelogram with base 14 cm and height 8 cm? (e) Write down the area of a rhombus having diagonals 10 cm and 16 cm. (f) Write down the area of a kite having diagonals 12 cm and 15 cm. Solution: (a) A triangle having no equal sides is called a scalene triagle. (b) Area of a scalene triangle = s(s – a) (s – b) (s – c) (c) Area of a right angled = 1 2 × p × b (d) Area of a parallelogram = base × height = 14 cm × 8 cm = 112 cm2 (e) Area of a rhombus = 1 2 × d1 × d2 = 1 2 × 10 cm × 16 cm = 80 cm2 (f) Area of a kite = 1 2 × d1 × d2 = 1 2 × 12 cm × 15 cm = 90 cm2 2. Circle the correct answer. (a) Which is the correct formula to calculate the area (A) of a scalene triangle with sides a, b and c? (i) A = s(s + a) (s + b) (s + c) (ii) A = s(s – a + s – b + s – c) (iii) A = s(s + a + s + b + s + c) (iv) A = s(s – a) (s – b) (s – c)

( ( ( ( ( Soluti ( ( Chec 3. O ( Soluti ( ( (b) Which is th (i) P = p + (iii) P = p + (c) What is the (i) 30 cm (d) What is the (i) 30.05 c (e) What is the (i) 14 cm2 (f) What is the (i) 26.43 c ion: (a) (iv) (e) (ii) ck Your Perfo Observe the follo (i) (a) Write (b) Find th (c) Find th (d) Find th ion: (i) (a) A close (b) Diagon (c) Area o Area o (d) Area o (ii) (a) See i. ( (b) Diagon he perimeter (P) + b + h + b + h2 e area of the give (ii) 30 e area of the give cm2 (ii) 0.3 e area of the give 2 (ii) 24 e area of the give cm2 (ii) 26 (b) (iv (f) (iv ormance owing polygons: the definition of he length of the he area of all tri he area of the po ed plane figure fo nal of polygon (d f A1 = 1 2 × 4 cm × f A2 = 1 2 × 3 cm × f polygons = A1 (a) nal of quad. (A2A 120 of a right-angle en ΔPQR below? 0 cm2 en ΔABC below? 3977 cm2 en ΔLMN below 4 cm2 en ΔCDE below? 6.83 cm3 v) v) : f polygon. diagonals in eac iangles in each p olygons. ormed by three or d) = 32 + 42 = × 3 cm = 6 cm2 × 52 – 32 = 1 2 × + A2 = 6 cm2 + 6 A3) = Hypotenu = 100 = 1 ed triangle with l (ii) P = a + b + (iv) P = p + b + ? (iii) 43.30 cm ? (iii) 0.1503 cm w? (iii) 24 cm ? (iii) 26.83 cm (c) (iv) (ii) ch polygon. polygon. r more straight li 9 + 16 = 25 = 5 3 cm × 25 – 9 = 6cm2 = 12 cm2 use of rt. A3 = 0 cm legs p and b? + c + p2 + b2 (iv) 43 m2 (iv) 10 (iv) 34 (iv) 26 (d) (iv ine segments is ca 5 cm = 1 2 × 3 × 4 = 6 cm 82 + 62 = 64 + 3.30 cm2 0.04 cm2 4 cm2 6.83 cm2 v) alled a polygon. m2 36

121 34 ft 20 ft 42 ft and Diagonal of quad. (A1A2) = Perpendicular of rt. A2 = 102 – 82 = 100 – 64 = 36 = 6 cm (c) Area of A3 = 1 2 × 8 × 6 = 24 cm2 Area of A2 = 1 2 × 8 × 6 = 24 cm2 For area of A1, s = 7 + 5 + 3 2 = 16 2 = 8 cm Area (A1) = 8(8 – 7) (8 – 6) (8 – 3) = 8 × 1 × 2 × 5 = 80 = 8.94 cm2 (d) The area of the polygons (A) = A1 + A2 + A3 = 8.94 + 24 + 24 = 56.94 cm2 4. (i) Observe a triangular land ABC with sides 5 m, 12 m and 13 m and answer the following questions. (a) Define a right-angled triangle. (b) Find the area of ΔABC. (c) How much cost is required to plough the land at the rate of Rs. 25 per square meter? Find it. (d) Is ΔABC a right-angled triangle? Justify. (ii) Observe a triangular pond with sides 42 ft, 34 ft and 20 ft whose bottom is congruent to its surface and answer the following questions. (a) Define a scalene triangle. (b) Find the area of the pond. (c) How much cost is needed to pave the tiles on its bottom at the rate of Rs. 85 per square ft? Find it. (d) If the height of the water level from the bottom is 15 ft, find the quantity of water in litre. Solution: (i) (a) A triangle having are angle 90° is called a right angled triangle. (b) Given: Three sides of ABC are 5 m, 12 m and 13 m Suppose a = 5 m, b = 12m and c = 13 m Now, s = a + b + c 2 = 5 + 12 + 13 2 = 30 2 = 15 m Area of ABC = s(s – a) (s – b) (s – c) = 15(15 – 5) (15 – 12) (15 – 13) = 15 × 10 × 3 × 2 = 900 = 30 m2 (c) Cost of requires to plough the land at the rate of Rs. 25 per m2 . So, Total cost to plough = Rs. 25 × 30 = Rs. 750. (d) Here, 132 = 122 + 52 or, 169 = 144 + 25 or, 169 = 169 Since, h2 = p2 + b2 is satisfied, ABC is a right angled triangle. (ii) (a) A triangle whose all sides are not equal is called a scalene triangle. (b) For the area of pond. Let a = 42 ft, b = 20 ft and c = 34 ft Now, Semi-perimeter (s) = a + b + c 2 = 42 + 20 + 34 2 = 48 ft

122 Area = s(s – a) (s – b) (s – c) = 48(48 – 42) (48 – 20) (48 – 34) = 48 × 6 × 28 × 14 = 112896 = 336 ft2 (c) Also, rate to pane the tiles on bottom (c) = Rs. 85 Total cost (T) = Rs. 336 × 85 = Rs. 28,560 (d) Here, height of the water level from bottom (h) = 15 ft Quantity of water (V) = ? Now, V = A × h = 336 × 15 = 5040 ft3 = 5040 × 28.317 liters = 142,717.68 liter 5. (i) The perimeter of a triangular tarpaulin is 900 cm and its edges are in the ratio 5:12:13. (a) Write the formula to find the area of a scalene triangle with sides a, b and c. (b) Find the edges of the tarpaulin. (c) Find the area of the tarpaulin. (d) Find the cost for waterproof painting on it at Rs. 5.25 per cm2 . (ii) The perimeter of a triangular pond is 540 ft and its sides are in the ratio 25:17:12. (a) Write the formula to find the area of a scalene triangle with base b and height h. (b) Find the length of sides of the pond. (c) Find the area of the pond. (d) Calculate the total cost for paving slates on its floor at the rate of Rs. 325/ft2 . (e) If the pond is empty and you fill it with water of a height of 7 ft, find the cost of water at Rs. 2.15 per liter. Solution: (i) Given, perimeter of a triangular tarpaulin (P) = 900 cm Ratio of it's edges is 5 : 12 : 13. (a) Formula to find the area of a scalene with sides a, b and c is s(s – a) (s – b) (s – c) (b) For the edges of tarpaulin, let it's edges be 5x, 12x and 13x, then 5x + 12x + 13x = 900 or, 30x = 900 or, x = 900 30 = 30 5x = 5 × 30 = 150 cm 12x = 12 × 30 = 360 cm 13x = 13 × 30 = 390 cm Hence, the edges of tarpaulin are 150 cm, 360 cm and 390 cm. (c) For the area of tarpaulin, s = a + b + c 2 = 150 + 360 + 390 2 = 450 cm Now, area = s(s – a) (s – b) (s – c) = 450(450 – 150) (450 – 360) (450 – 390) = 450 × 300 × 90 × 60 = 729000000 = 27000 cm2 (d) Cost of 1 cm2 for water proof painting (c) = Rs. 5.25 Total cost (T) = Rs. 27,000 × 5.25 = Rs. 1,41,750 (ii) Given, perimeter of a triangular pond (P) = 540 ft Ratio of it's sides = 25 : 17 : 12 (a) Formula to find the area of a scalene with base b and height h = 1 2 × b × h.

123 (b) For the length of sides of pond let it's sides be 25x, 17x and 12x. Then, 25x + 17x + 12x = 540 or, 54x = 540 x = 10 First side = 25x = 25 × 10 = 250 ft Second side = 17x = 17 × 10 = 170 ft Third side = 12x = 12 × 10 = 120 ft (c) For the area of pond, semi-perimeter (s) = a + b + c 2 = 250 + 170 + 120 2 = 270 ft Area (A) = s(s – a) (s – b) (s – c) = 270(270 – 250) (270 – 170) (270 – 120) = 270 × 20 × 100 × 150 = 81000000 = 9,000 sq. ft (d) Cost of 1 square ft for paving slates on floor (C) = Rs. 325 Total cost (T) = A × C = Rs. 9,000 × 325 = Rs. 29,25,000 (e) Here, height of pond (h) = 7 ft quantity of mater (V) = ? Now, V = A × h = 9000 × 7 ft3 = 63,000 ft3 = 63,000 × 28.317 liter = 17,83,971 liter Now, cost of 1 litre water = Rs. 2.15 Total cost of 1783971 litre water = Rs. 1783971 × 2.15 = Rs. 38,35,537.65 6. Observe the mentioned triangular shapes and answer the following question. (i) An isosceles triangular garden has an area 72 m2 and one of the equal sides is 12 m. (a) Find the length of unknown side of each triangular garden. (b) Find the perimeter of the garden. (c) Calculate the cost for fencing 5 times around its boundaries at the rate of Rs. 276/m. (ii) The area of an isosceles triangular island of the road is 192 m2 and the length of base is 24 m. (a) Calculate the length of equal sides. (b) Find the perimeter of each triangular shape. (c) Calculate the total cost for colouring on its boarders around it at the rate of Rs. 25.25/m. (iii) The length of the sum of three sides of an equilateral triangular paper is 18 cm. (a) Calculate the length of its sides. (b) Find the area of the paper. (c) Calculate the total cost for colouring on its surface at the rate of 60 paisa/cm2 . Solution: (i) Here, Area of isosceles garden (A) = 72 m2 one side of the equal side (a) = 12 m (a) Length of unknown side (b) = ?

124 Now, A = b 4 4a2 – b2 or, 72 = b 4 4 × 122 – b2 or, 288 = b 576 – b2 or, 82,944 = b2 (576 – b2 ) or, 82,944 = 576b2 – b4 or, b4 – 576b2 + 82,944 = 0 or, (b2 ) – 2 . b2 . 288 + (288)2 = 0 or, (b2 – 288)2 = 0 or, b2 – 288 = 0 or, b2 = 288 b = 288 = 16.97 m (b) Perimeter of garden (P) = 2a + b = 2 × 12 + 16.97 = 24 + 1697 = 40.97 m (c) Here, length in 1 round = 40.97 m length in 5 rounds = 5 × 40.97 m = 204.85 m cost of 1 m for fencing = Rs. 276 Total cost (T) = Rs. 204.85 × 276 = Rs. 56,538.60 (ii) Here, Area of an isosceles island of the round (A) = 192 m2 Length of base (b) = 24 m (a) Length of equal sides (a) = ? Now, area of isosceles = b 4 4a2 – b2 or, 192 = 24 4 4a2 – (24)2 or, 192 = 6 4a2 – 576 or, 32 = 4a2 – 576 or, 32 = 4(a2 – 144) or, 32 = 2 a 2 – 144 or, 16 = a 2 – 144 or, 256 = a2 – 144 or, 400 = a2 a = 400 = 20 m Hence, the length of equal side = 20 m. (b) Here, perimeter of triangle = 2a + b = 2 × 20 m = 24 m = 40 m + 24 m = 64 m (c) Here, cost of 1 m for colouring on boarder = Rs. 25.25 Total cost for colouring = Rs. 64 × 25.25 = Rs. 1616 (iii) Here, sum of three sides of an equilateral triangle (P) = 18 cm (a) Length of it's sides = ? Now, 3a = 18 cm a = 18 3 cm = 6 cm

125 P Q S R 45 m 45 m 30 m 20 m the length of it's side = 6 cm (b) Area of paper (A) = 3 4 a2 = 3 4 × (6 cm)2 = 3 4 × 36 cm2 = 9 3 cm2 = 15.59 cm2 (c) Cost of 1 cm2 for colouring on it's surface = 60 paisa Total cost (T) = 15.59 × 60 paisa = 935.4 paisa = Rs. 9.35 7. (i) Observe the plot of land in the form of a quadrilateral PQRS with dimensions; PQ = 45 m, QR = 30 m, RS = 45 m, PS = 21 m, and QS = 44 m (a) Find the area of ΔPQS and ΔQRS. (b) Calculate the area of the quadrilateral land PQRS in bighaha system. Use 1 kattha = 338.62 m2 . (c) If an agent takes 4% commission on selling the land, how much does she/he take commission when it sells on Rs. 1250000 per kattha? Find it. (d) How much money does the land owner get from it after commission ? Find it. (ii) Observe the plot of land in the form of a quadrilateral ABCD alongside. (a) State the Pythagoras' theorem. (b) If, in the land, ADC = 90°, how can you find the area of the land ? Write. (c) Find the area of the land in ropani system. Use 1 aana = 342.25 ft2 . (d) If an agent takes 5% commission on selling the land, how much does she/he take commission when it sells on Rs. 4525000 per aana ? Find it. (e) How much money does the land owner get from it after commission ? Find it. Solution: (i) Here, for area of PQS (a) s = PQ + QS + PS 2 = 45 m + 44 m + 21 m 2 s = 55 m Now, area of PQS = s(s – a) (s – b) (s – c) = 55(55 – 45) (55 – 44) (55 – 21) = 55 × 10 × 11 × 34 = 205700 = 453.54 m2 Again, for QRS s = QR + RS + QS 2 = 30 m + 45 m + 44 m 2 = 59.5 m Now, area of QRS = 59.5(59.5 – 30) (59.5 – 45) (59.5 – 44) = 59.5 × 29.5 × 14.5 × 15.5 = 394492.4375 = 628.08 m2 (b) Here, Area of ~PQRS = area of PQS + area of QRS = 453.54 m2 + 628.08 m2 = 1081.62 m2 = 1081.62 338.62 Katha = 3.1941999882 Katha = (3 + 0.1941999882) Katha = 3 Katha + 0.1941999882 × 20 Dhur = 3 Katha + 3.883999764 Dhur = 3 Katha 3.88 Dhur

126 a a a 2 m a + 2 (c) Here, selling price of land per Katha = Rs. 1250000 selling price of land 3.19 Katha = Rs. 1250000 × 3.19 = Rs. 3987,500 Commission amount = 4% of Rs. 3987,500 = 4 100 × 3987,500 = Rs. 159,500. (d) The land owner gets from it after commission = Rs. 3987500 – Rs. 159500 = Rs. 3828,000 (ii) (a) Statement of Pythagoras theorem In a right angled triangle, the square of hypotenuse side is equal to the sum of the squares of base and perpendicular sides. (b) Here, If ADC = 90°, then we find first AC diagonal (AC) by using Pythagoras theorem. After that find ADC and ABC. At last we find the sum of ADC and ABC. (c) To find the area of land AC (h) = AD2 + CD2 = (62.30)2 + (46.30)2 = 3881.29 + 2143.69 = 6024.98 = 77.62 ft. Now, area of ADC = 1 2 × 62.30 ft × 46.30 ft = 1442.245 sq.ft. Also, for area of ABC, s = AC + BC + AB 2 = 77.62 + 74 + 33.45 2 = 92.535 ft. Area of ABC = 92.54(92.54 – 77.62) (92.54 – 74) (92.54 – 33.45) = 92.54 × 14.92 × 18.54 × 59.09 = 1512592.83 = 1229.87 sq.ft. Total area of land ABCD = 1442.245 + 1229.87 = 2672.12 sq.ft. = 2672.12 342.25 Aana [1 ana = 342.25 ft2 ] = 7.81 Ana. (d) SP of land for 1 Aana = Rs. 45,25,000 SP of land for 7.81 Aana = Rs. 7.81 × 4525000 = Rs. 3,53,40,250 Commission amount = 5% of 35340250 = Rs. 17,67,012.50 (e) The land owner gets from it after commission = Rs. 35340250 – Rs. 1767012.50 = Rs. 33573237.50 8. (i) If each side of an equilateral triangular design on a garden is increased by 2 cm, its area increases by 5 3 cm2 . (a) Find the length of each side of the triangular design. (b) Find the area of the triangular design. (ii) The perimeter of a right-angled triangle is 60 cm and its hypotenuse is 26 cm. (a) Find the length of other two sides of the triangle. (b) Find the area of the triangle. Solution: (i) (a) Here, let the length of equlateral be 'a' then it's area is A1 = 3 4 a2

127 b p If each side of equilateral is increased by 2 cm, then length = (a + 2) cm. Then it's new area (A2) = 3 4 (a + 2)2 By question, A2 – A1 = 5 3 or, 3 4 (a + 2)2 – 3 4 a2 = 5 3 or, 3 4 {(a + 2)2 – a2 } = 5 3 or, a2 + 4a + 4 – a2 = 20 or, 4a = 20 – 4 or, 4a = 16 a = 16 4 = 4 Hence, length of each side of the triangular design = 4 cm. (b) Area of the triangle (A) = 3 4 a2 = 3 4 × 42 = 3 × 4 = 4 3 cm2 . (ii) (a) Here, perimeter of right angled (P) = 60 cm hypotenuse (p) = 26 cm base (b) = ? perpendicular (p) = ? Now, p + b + h = 60 or, p + b + 26 = 60 or, p + b = 60 – 26 or, p + b = 34 p = 34 – b ................... (i) Again, h2 = p2 + b2 or, 262 = (34 – b)2 + b2 or, 676 = 1156 – 68b + b2 + b2 or, 2b2 – 68b + 1156 – 676 = 0 or, 2b2 – 68b + 480 = 0 or, b2 – 34b + 240 = 0 or, b(b – 24) – 10(b – 24) = 0 or, (b – 24) (b – 10) = 0 Either, b – 24 = 0 Or, b – 10 = 0 b = 24 b = 10 when b = 24, p = 34 – 24 = 10 when b = 10, p = 34 – 10 = 24 Hence, other two sides are 10 cm and 24 cm. (b) Area of = 1 2 × p × b = 1 2 × 10 × 24 = 120 cm2

128 12 ft 9. (i) An umbrella is made by stitching 8 triangular pieces of canvas of four different colors, each piece measuring 15 cm, 45 cm and 45 cm. (a) How much canvas is required for stitching the umbrella? (b) If the cost of 1 cm2 is Rs. 0.65, find the total cost of the canvas. (ii) A regular octagonal court-yard has a side 12 ft and a main diagonal 20 ft. (a) What is the area of the court-yard? (b) If the cost of 1 sq. ft is Rs. 265 for paving stones, find the total cost of the stones. Solution: (i) Here, An umbrella is made by stitching 8 triangular pieces of canvas of form different colors. each piece measuring a = 15 cm, b = 45 cm and c = 45 cm (a) For area of one triangular piece, Semi-perimeter (s) = a + b + c 2 = 15 + 45 + 45 2 = 52.5 cm Now, area of one piece = 52.5(52.5 – 15) (52.5 – 45) (52.5 – 45) = 52.5 × 37.5 × 7.5 × 7.5 = 110742.1875 = 332.78 cm2 Also, area of 8 triangular piece = 8 × 332.78 cm2 = 2662.24 cm2 Hence, the required canvas for stiching the umbrella = 2662.24 cm2 . (b) The cost of 1 cm2 canvas = Rs. 0.65 The cost of 2662.24 cm2 canvas = Rs. 2662.24 × 0.65 = Rs. 1730.45 (ii) (a) Here, in a regular octagum court-yard side (b) = 12 ft length of main diagonal = 20 ft. Now, side of one triangle (issosceles) (a) = 20 ft 2 = 10 ft. Now, area of one triangle = b 4 4a2 – b2 = 12 4 4 × 102 – 122 = 3 4 × 100 – 144 = 3 400 – 144 = 3 256 = 3 × 16 = 48 sq. ft. Area of the court yard = 8 × 48 sq.ft. = 384 sq.ft. (b) Cost of 1 sq.ft. for paving stones = Rs. 265 The total cost of the stones = Rs. 384 × 265 = Rs. 1,01,760. 10. (i) If a, b and c are the length of three sides and s is the semi-perimeter of a triangle, find the area of the triangle in terms of a, b, c and s. (ii) If a and b are the lengths of two adjacent sides and d is the length of a diagonal of a parallelogram, find the area of the parallelogram in terms of a, b and d. Solution: See in the textbook. 10 ft 10 ft 10 ft 10 ft

129 5.2 Surface Area of Room and Cost Estimation PRACTICE 5.2 Read / Understand / Think / Do Keeping Skill Sharp 1. (a) What is the formula to calculate the area of the floor of a rectangular room having length , breadth b and height h ? (b) What is the area of the four walls of a rectangular room having length a cm, breadth b cm and height c cm ? (c) Write down the formula to find the area of the four walls and ceiling of a rectangular room having the length , breadth b and height h. (d) What is the formula to calculate the area of the four walls, ceiling and floor of a rectangular room having the length , breadth b and height h ? Solution: (a) Area of floor = × b (b) Area of 4 walls = 2c(a + b) cm2 (c) Area of 4 walls and ceiling = 2h( + b) + × b (d) Area of 4 walls, ceiling and floor = 2h( + b) + × b + × b 2. (a) N tiles are needed to pave the floor of a room. If the area of each tile is a sq. units, what is the area of the room? (b) What is the cost of N paving bricks at the rate of Rs. R to pave on the floor of the room ? (c) The total cost of paving stones on the floor of a room at Rs. R is Rs. T. What is the area of the floor? Solution: (a) Area of the room (A) = N × a sq. units. (b) Cost of N paving bricks = Rs. N × R (c) Area of floor = T R 3. (a) What is the area of the ceiling of a rectangular room having length 4 m, breadth 3 m and height 2.5 m? (b) What is the area of the four walls of a cubical room having the length 4 m ? Solution: (a) Area of ceiling = × b = 4 m × 3 m = 12 m2 (b) Area of 4 walls of cubical room = 4 2 = 4 × (4 m)2 = 4 × 16 m2 = 64 m2 Check Your Performance Observe the given situation and answer the following questions for each situation. 4. (i) A man wants to paste the carpet of 80 cm wide on the floor of a room 10 m long and 6 m broad. (a) Write the relation between the area of room and carpet. (b) Calculate the area of the room.

130 (c) What is the area of the carpet to be pasted on the room. (d) Find the length of the carpet to be needed to paste on the room. (e) If the cost of 1 meter carpet is Rs. 645, calculate the cost of carpet used in the room? (ii) When a girl pasted the carpet on the floor of a squared room of the length 15 ft, a 15 m long carpet used. (a) What formula is used to find the area of the square floor ? (b) Calculate the area of the room. (c) What is the area of the carpet to be pasted on the room. (d) Find the breadth of the carpet. Solution: (i) Given, length of room () = 10 m, breadth (b) = 6 m wide of capet = 80 cm (a) The relation between the area of room nd carpet is equal in area. (b) Area of the room (A) = × b = 10 m × 6 m = 60 m2 (c) The area of carpet to be pasted on the room is 60 m2 (d) Area of carpet = area of room or, length of carpet × wide of carpet = 60 m2 or, length of carpet × 80 cm = 60 m2 length of carpet = 60 m2 0.8 m = 75 m. (e) The cost of 1 meter carpet = Rs. 645. The cost of carpet used in the room = Rs. 75 × 645 = Rs. 48,375. (ii) Here, length of the squared room () = 15 ft. length of carpet (1) = 15 m (a) Formula to find the area of square floor = 2 (b) the area of the room = 2 = (15 ft)2 = 225 sq.ft. (c) The area of the carpet to be pasted in the room = 225 sq.ft. (d) Breadth of carpet (b1) = ? Now, area of carpet = area of floor. 1 × b1 = 225 ft2 or, 15 m × b1 = 225 ft2 b1 = 225 ft2 15 m = 225 × 0.305 × 0.305 15 m = 1.40 m Hence, the breadth of carpet = 1.4 m. 5. (i) The stones of each 30 cm long and 25 cm broad are required to pave the floor of a rectangular room 20 m by 18 m. (ii) The carpets of each 5 m long and 3 m wide will be required to cover the floor of a room measuring 15 m by 12 m. (a) Write the formula to find the number of materials paving on plane floor. (b) What are the area of the room and each material? (c) How many materials are needed to pave the floor of the above mentioned rooms? (d) If the cost of each material is Rs. 15, calculate the cost of all required materials.

131 Solution: (i) Here, length of stone (1) = 5 m, breadth (b1) = 3 m. size of rectangular court = 20 m × 18 m (a) The formula to find the no. of material paving on plane floor. (N) = Area of floor (A) Area of each material (b) Area of floor of a court = × b (A) = 20 m × 18 m = 360 m2 Area of each stones (a) = 30 cm × 25 m = 750 cm2 (c) No. of materials (stones) to pave the floor = A a = 360 m2 750 cm2 = 360 × 100 × 100 cm2 750 cm2 = 4800 (d) The cost of 1 stone (material) = Rs. 15 The cost of all materials = Rs. 15 × 4800 = Rs. 72,000. (ii) Here, length of carpet (1) = 5 m wide of carpet (b1) = 3 m size of room (A) = 15 m by 12 m = 15 m × 12 m = 180 m2 (a) Formula to find the no. of material paving on plane floor (N) = Area of floor (A) Area of carpet (a) (b) Area of floor of room (A) = 180 cm2 Area of each carpet (a) = 5 m × 3 m = 15 m2 (c) No. of carpet (N) = A a = 180 m2 15 m2 = 12 (d) The cost of 1 material (carpet) = Rs. 15 The cost 12 carpet = Rs. 15 × 12 = Rs. 180. 6. (i) The cost of carpeting on the floor of a room at Rs. 430/m2 was Rs. 17200. If the length of the room had been 2 m more, the expense would have been Rs. 19780. (ii) The cost of carpeting a room at Rs. 4.75 per sq. metre is Rs. 646. If the length of the room had been 3 m less, the cost would have been Rs. 532. (a) What is the cost of paving marbles on the floor of the area 'a' sq. m at Rs. 'b' per sq. m? (b) Find the area of the carpet used in the room. (c) What is the area of the room. (d) What is the length and breadth of the room? Find it. Solution: (i) Here, cost of carpeting on the floor (c) = Rs. 430 /m2 total cost (T) = Rs. 17200 length of room () = 2 m more, then total cost = Rs. 19780. (a) The cost of paving marbles on the floor = Rs. b × a (b) The area of the carpet used in the room (A) = T c = 17200 430 = 40 m2 (c) The area of room = area of carpet = 40 m2

132 (d) If length = and breadth = b, then Also, × b = 40 m2 or, ( + 2) × b = 19780 430 or, × b + 2b = 46 or, 40 + 2b = 46 or, 2b = 46 – 40 or, 2b = 6 b = 6 2 = 3 m. Hence, breadth = 3 m and length = 40 3 = 13.33 m. (ii) Here, the rate of carpeting a room (c) = Rs. 4.75 m2 total cost of carpeting (T) = Rs. 646 length of the room had been 3 m less then total cost = Rs. 532 (b) The area of carpet used in the room (A) = T c = Rs. 646 Rs. 4.75 = 136 m2 (c) The area of room = area of carpet = 136 m2 . (d) Where length = and breadth = b, then × b = 136 m2 Also, when length is reduced by 3 m, then ( – 3) × b = Rs. 532 Rs. 4.75 or, × b – 3b = 112 or, 136 – 112 = 3b or, 24 = 3b b = 8 m Hence, breadth (b) = 8 m and length () = 136 8 = 17 m. 7. (i) A rectangular room has the length 8 m, breadth 6 m and height 4 m. (ii) A square room has the length 10 m and height 3 m. (a) Write the formula to find the area of 4 walls of a rectangular/square room. (b) Find the area of 4 walls of the above room. (c) Find the area of the 4 walls and ceiling of the room. (d) What percent of the area of the 4 walls and ceiling is more than its 4 walls? Find it. Solution: (i) Here, length of room () = 8 m, breadth (b) = 6 m and height (h) = 4 m (a) The formula to find the area of 4 walls = 2h( + b) (b) The area of 4 walls = 2h( + b) = 2 × 4 m(8 m + 6 m) = 8 m × 14 m = 112 m2 (c) The area of 4 walls and ceiling = 2h( + b) + × b = 112 + 8 × 6 = 112 + 48 = 160 m2 (d) Percentage increasd the area of 4 walls and ceiling by 4 walls = 160 – 112 112 × 100% = 42.85% (ii) Here, length of square room () = 10 m (a) The formula to find the area of 4 walls of a square room = 4h (b) Area of 4 walls = 4 × 10 m × 3 m2 = 120 m2 (c) Area of 4 walls and ceiling = 4h + 2 = 120 + 102 = 120 + 100 = 220 m2

133 (d) Percentage increased the area of 4 walls and ceiling by 4 walls = 220 – 120 120 × 100% = 83.33% 8. (i) A rectangular room is 12 m long, 6 m broad and 4 m high. (a) How many metres of wall-paper of 80 cm wide will be required to paste on its walls? Find it. (b) Calculate the cost of paper at the rate of Rs. 50 per sq. m. (ii) A rectangular room is 6.5 m long, 5.5 m broad and 3.75 m high. (a) How many pieces of tiles 0.25 sq. meters each will be required to paste on the four walls and ceiling of the room? Find it. (b) Calculate the cost of tiles at the rate of Rs. 245 per piece. Solution: (i) Here, length of room () = 12 m breadth (b) = 6 m height (h) = 4 m (a) Area of 4 walls (A) = 2h( + b) = 2 × 4 m (12 m + 6 m) = 8 m × 18 m = 144 m2 wide of wall paper (b1) = 80 cm = 80 100 m = 0.8 m length of wall paper (1) = ? Now, area of wall paper = area of 4 walls or, 1 × b1 = 144 m2 or, 1 × 0.8 m = 144 m2 1 = 144 m2 0.8 m = 180 m The required length of wall paper = 180 m. (b) Rate of paper (C) = Rs. 50/m2 Total cost of paper (T) = Rs. 50 × 144 = Rs. 7200 (ii) Here, length () = 6.5 m breadth (b) = 5.5 m height (h) = 3.75 m (a) The area of 4 walls and ceiling = 2h( + b) + × b = 2 × 3.75(6.5 + 5.5) + 6.5 × 5.5 = 7.5 × 12 + 35.75 = 90 + 35.75 = 125.75 m2 Also, area of 1 tile (a) = 0.25 m2 Now, no. of tiles to paste in the 4 walls and ceiling (N) = A a = 125.75 m2 0.25 m2 = 503 (b) Rate of tiles per piece = Rs. 245 The total cost of tiles = Rs. 503 × 245 = Rs. 123,235. 9. (i) The area of four walls of a room is 77 m2 . The length and breadth of the room are 7.5 m and 3.5 m respectively. (ii) A room 6 m long and 5 m wide and the area of whose floor and ceiling together is equal to the area of the four walls. (a) Find the height of the room. (b) Find the four walls and floor of the room. (c) Calculate the cost of plastering on its four walls and floor at Rs. 125/m2 .

134 Solution: (i) Here, area of 4 walls (A) = 77 m2 length () = 7.5 m breadth (b) = 3.5 m (a) height of room = ? Now, A = 2h( + b) or, 77 = 2 × h(7.5 + 3.5) or, 77 = 2h × 11 or, 77 = 22 h h = 77 22 = 3.5 m (b) Area of 4 walls and floor of the room = 2h( + b) + × b = 77 + 7.5 × 3.5 = 77 + 26.25 = 103.25 m2 (c) Rate of plastering on 4 walls and floor = Rs. 125/m2 The total cost of plastering (T) = A × C = Rs. 103.25 × 125 = Rs. 12,906.25 (ii) Here, length of room () = 6 m breadth of room (b) = 5 m area of floor and ceiling together = area of 4 walls height (h) = ? (a) Now, Area of 4 walls = Area of floor & ceiling together or, 2h( + b) = × b + × b or, 2h(6 + 5) = 2 × 6 × 5 or, 2h × 11 = 60 h = 60 22 = 2.72 m. (b) Area of 4 walls and floor = 2h( + b) + × b = 2 × 2.72(6 + 5) + 6 × 5 = 5.44 × 11 + 30 = 59.84 + 30 = 89.84 m2 (c) Cost of plastering on 4 walls and floor = Rs. 89.84 × 125 = Rs. 11,230. 10. (i) A room is 12 m long, 8 m broad and 4 m high. It has two doors each 2 m high and 1.5 m wide and 4 windows each 2 m wide and 2.5 m high. Calculate the total cost of papering the walls and ceiling at the rate of Rs. 5.50 /m2 . (ii) A room is 9 m long, 8 m broad and 6.5 m high. It has one door 3 m × 1.5 m and three windows each of dimensions 1.5 m and 1 m. Find the cost of white washing the walls at Rs. 3.75 per m2 . Solution: (i) Here, length () = 12 m broad (b) = 8 m height (h) = 4 m area of two doors = 2 × 2 m × 1.5 m = 6 m2 area of 4 windows = 4 × 2 m × 2.5 m = 20 m2 Now, Area of 4 walls and ceiling exluding doors and windows = 2h( + b) – 6 m2 – 20 m2 + 12 × 8 = 2 × 4(12 + 8) – 6 – 20 + 96 = 8 × 20 – 26 + 96 = 160 – 26 + 96 = 230 m2

135 Rate of papering the walls and ceiling (C) = Rs. 5.50/m2 Total cost of papering = Rs. 230 × Rs. 5.50 = Rs. 1265 (ii) Here, length of room () = 9 m broad of room (b) = 8 m height of room (h) = 6.5 m area of one door = 3 m × 1.5 m = 4.5 m2 area of 3 windows = 3 × 1.5 × 1 = 4.5 m2 Now, the area of the walls to be washed = 2h( + b) – 4.5 – 4.5 = 2 × 6.5(9 + 8) – 9 = 13 × 17 – 9 = 221 – 9 = 212 m2 Rate of white washing the walls (C) = Rs. 3.75 / m2 Cost of white washing the walls (T) = Rs. 212 × 3.75 = Rs. 795. 11. (i) The length of a room is three times its breadth. The cost of carpeting its floor at Rs. 175 per sq. metre is Rs. 13125. (a) Find the area of the floor of the room. (b) Find the length and breadth of the room. (ii) The length of a room is twice its breadth and the area of its 4 walls is 120 m2 . The height of the room is 4 m. (a) Find the length and breadth of the room. (b) Compute the area of the ceiling of the room. (c) Calculate the cost of carving on the ceiling at Rs. 550 per m2 . Solution: (i) Here, let the breadth be b then it's length = 3b rate of carpeting the floor (C) = Rs. 175 / m2 total cost of carpeting (T) = Rs. 13125 (a) The area of floor of the room (A) = T C = Rs. 13125 Rs. 175 = 75 m2 (b) Area of floor (A) = 75 or, × b = 75 or, 3b × b = 75 or, b2 = 25 b = 25 = 5 3b = 3 × b = 15 m. Hence, length () = 15 m and breadth (b) = 5 m. (ii) Here, let the breadth of a room be x then it's length = 2x. area of 4 walls (A) = 120 m2 height (h) = 4 m (a) Now, A = 2h( + b) or, 120 = 2 × 4(2x + x) or, 120 = 8 × 3x or, 120 = 24 x x = 120 24 = 5 m length () = 2x = 2 × 5 m = 10 m. and breadth (b) = 5 m

136 (b) Area of the ceiling of the room (A) = × b = 10 m × 5 m = 50 m2 (c) Rate of carving on the ceiling (C) = Rs. 550 / m2 Total cost of carving (T) = ? Now, T = A × C = 50 × Rs. 550 = Rs. 27,500. 12. (i) A hall is 36 m long and 24 m broad. Allowing 40 sq. metres for doors and windows the cost of papering the walls and ceiling at Rs. 8.40 per square metre is Rs 9441.60. Find the height of the hall. (ii) A room is 7 m long and 5 m broad. It has one door measuring 2 m by 1.5 m and two windows, each measuring 1.5 m by 1 m. The cost of painting the walls at Rs. 7.50/m2 is Rs. 495. Find the height of the room. Solution: (i) Here, length of hall () = 36 m broad of hall (b) = 24 m rate of papering the walls and ceiling (C) = Rs. 8.40 total cost of papering (T) = Rs. 94441.60 Area of 4 walls & ceiling (A) = T C = 9441.60 8.40 = 1124 m2 Again by question, Area of 4 walls and ceiling = 2h( + b) + × b – 40 or, 1124 = 2h(36 + 24) + 36 × 24 – 40 or, 1124 + 40 = 2h × 60 + 864 or, 1164 – 864 = 120h or, 300 = 120h h = 300 120 = 2.5 m Hence, height of the hall is 2.5 m. (ii) Here, length of room () = 7 m broad of room (b) = 5 m measurement of 1 door = 2 m × 1.5 m = 3 m2 measurement of 2 windows = 2 × 1.5 m × 1 m = 3 m2 Now, the area of walls to be painted (A) = 2h( + b) – area of 1 door – area of 2 windows or, A = 2h(7 + 5) – 3 – 3 = 2h × 12 – 6 or, A = 24h – 6 ............. (i) Again, rate of painting the walls = Rs. 7.50/m2 Total cost (T) = Rs. 495 Area of walls to be painted (A) = T C = 495 7.50 = 66 m2 ............. (ii) From (i) and (ii), 24h – 6 = 66 or, 24h = 66 + 6 or, 24h = 72 h = 72 24 = 3 m Hence, the height of the room is 3 m.

137 13. (i) The cost of carpeting a square room at Rs. 160 per sq. metre is Rs. 5760. The cost of painting the four walls at Rs. 10 per sq. metre is Rs. 840. Find the height of the room. (ii) The total cost of painting the four walls of a room at the rate of Rs. 6 per m2 is Rs. 432. Find the cost of carpeting the room at Rs. 850 per sq. metre if the room is 4 m wide and 3 m high. Solution: (i) Here, Rate of carpeting a square room (C) = Rs. 160 / m2 total cost of carpeting (T) = Rs. 5760 Area of a square room (A) = T C = 5760 160 = 36 m2 2 = 36 = 36 = 6 m Again, rate of painting the 4 walls (C) = Rs. 10 / m2 and total cost of painting (T) = Rs. 840 Area of 4 walls (A) = T C = Rs. 840 Rs. 10 = 84 m2 Now, area of 4 walls = 4h or, 84 = 4 × h × 6 or, 84 = 24h h = 84 24 = = 3.5 m Hence, the height of the room is 5.5 m. (ii) Here, rate of painting the 4 walls of a room (C) = Rs. 6 / m2 total cost (T) = Rs. 432 Area of 4 walls (A) = T C = Rs. 432 Rs. 6 = 72 m2 Again, rate of carpeting the room (C) = Rs. 850 / m2 wide of the room (b) = 4 m height of the room (h) = 3 m total cost of carpeting (T) = ? Now, A = 2h( + b) or, 72 = 2 × 3( + 4) or, 72 = 6( + 4) or, 12 = + 4 = 12 – 4 = 8 m Again, area of room for carpeting (A) = × b = 8 m × 4 m = 32 m2 Total cost (T) = A × C = Rs. 32 × 850 = Rs. 27,200.

138 CHAPTER PRISM AND CYLINDER 6 6.1 Surface Area and Volume of Prism PRACTICE 6.1 Read / Understand / Think / Do Keeping Skill Sharp 1. Define the following terms related to prism: (a) Prism (b) Right Prism (c) Cross-section (d) Base (e) LSA (f) TSA Solution: (a) A solid geometric figure whose two ends are congruent and parallel, and whose sides are parallelograms, is called a prism. (b) A prism whose sides are perpendicular to its bases is called a right prism. (c) The shape obtained by the intersection of solid by a plane, is called cross-section. (d) The parallel polygons of a prism are called bases. (e) The area of the parallelogram sides of a prism is called LSA (Lateral Surface Area). (f) The area of the all visualized faces of a prism is called TSA (Total Surface Area). 2. (a) If the length of a cube is a cm, what is the area of cross-section of the cube? (b) What is the formula to compute the LSA of the prism having perimeter of the base P cm and height h cm ? (c) What is the formula to compute the TSA of the prism in which its LSA and base area A are given ? (d) Write the height of a prism with volume a cm3 and base area b cm2 . Solution: (a) CSA = a2 cm2 (b) LSA = p × h cm2 (c) TSA = LSA + A (d) height = a b 3. (a) What is the area of the cross-section of a cube having length of the side 3.5 cm ? (b) What is the volume of a cube having the length of the cross-section 5 cm ? (c) What is the TSA of a prism with LSA 96 cm2 and base area 16 cm2 ? Solution: (a) Cross-section area = 3.52 = 12.25 cm2 (b) V = 53 = 125 cm3 . (c) TSA = LSA + Base area = 96 + 16 = 112 cm2

4. W Soluti ( ( Chec Answ 5. F Soluti ( ( ( ( 6. F Soluti ( ( ( What is the area ion: (a) Area of cros (b) Area of cros ck Your Perfo er the given que Find the area of ion: (a) Area of cros (b) Area of cros (c) Area of cros (d) Area of cros Find the total su ion: (a) TSA = = (b) TSA = (c) Area of base To find the u h = 6 Now LSA a of the cross-sec ss-section = 10 ss-section = 82 = ormance estions for each s the cross-section ss-section = = ss-section = = ss-section = = ss section = = urface area of the = 2(b + bh + h) = 2(70 + 42 + 60) = 6a2 = 6 × (8 cm) e = 1 2 h(a + b) = 1 2 unknown side of 62 + 42 = 36 + 1 = p × h = (6 + 9 = 27.2 × 12 = 32 139 ction of the given 5 = 50 cm2 64 cm2 . situation. n of the followin = Base area = 14 = 140 cm2 – 20 c = 13 cm × 3 cm + = 39 cm2 + 39 cm = 12 cm × 3 cm + = 36 cm2 + 15 cm = 16 cm × 4 cm + = 64 cm2 + 24 cm e following prism = 2(10 × 7 + 7 × = 2 × 172 = 344 ) 2 = 6 × 64 = 384 1 2 × 6(9 + 5) = 3 × f base 6 = 52 = 7.2 + 7.2 + 5) × 12 26.4 cm2 n prism : ng prisms: cm × 10 cm – 5 cm2 + 13 cm × 3 cm + m2 + 15 cm2 = 93 + 5 cm × 3 cm + m2 + 9 cm2 = 60 c + 6 cm × 4 cm + m2 + 24 cm2 = 72 ms: × 6 + 10 × 6) cm2 cm2 × 14 = 42 cm2 cm × 4 cm + 5 cm × 3 cm cm2 3 cm × 3 cm cm2 6 cm × 4 cm cm2

( 7. F Soluti ( ( ( ( 8. ( ( Soluti ( ( TSA (d) Here, in the p = 8 c Now, h Now, area o LSA Now, TSA Find the volume ion: (a) Here, base a Volum (b) Here, Area o Volum (c) Here, area o Again, volum (d) Here, area o Now, volum (i) The dimens (a) Define (b) Find th (c) Find th (ii) A box conta (a) Is the s (b) Find th (c) Find th ion: (i) Here, dimen (a) Definit (b) Volum (c) Volum (ii) Here, no. of length = LSA + 2A = 3 = 326.4 + 84 = 4 base, cm, b = 10 cm, h = 82 + 102 = of base (A) = 1 2 × = p × h = (8 + 10 = LSA + 2A = 3 e of the following area (A) = 3 4 a2 = me of the prism = of base (A) = = me of prism (V) = of base (A) = 1 2 × me (V) = A × h = of base (A) = base me (V) = A × h = sion of a Sancho e a prism. he volume of the he volume of a b ains 20 cubical s soap a prism ? G he volume of the he volume of the nsion of a Sancho tion write you me of box (V) = me of a box which f cubical soaps = of each cube () 140 26.4 + 2 × 42 410.40 cm2 = ? 64 + 100 = 164 10 cm × 8 cm = 4 0 + 12.80) × 12 = 69.6 cm2 + 2 × 4 g prisms: = 3 4 × 62 = 3 4 × A × h = 15.58 × b 4 4a2 – b2 = 8 4 2 80 = 8.94 × 2 A × h = 17.88 × 6 × 6 = 18 cm2 = 18 cm2 × 9 cm e × height = 8 cm 48 cm2 × 12 cm = o box is 6 cm × 4 e Sancho box. box which conta soaps of the leng Give reason. e soap. e box. o box = 6 cm × 4 urself. × b × h = 6 cm × h contains such 8 20 = 15 cm 4 = 12.80 40 cm2 = 30.80 × 12 = 36 40 cm2 = 369.6 cm × 36 = 15.59 cm2 10 = 155.89 cm3 4 × 62 – 82 = 2 = 17.88 cm2 14 = 250.32 cm3 = 162 cm3 m × 6 cm = 48 cm = 576 cm3 cm × 2 cm. ins such 8 Sanch gth 15 cm each. cm × 2 cm × 4 cm × 2 cm = 4 boxes = 8 × 48 c 69.6 cm 9 2 m2 + 80 cm2 = 44 144 – 64 3 m2 ho boxes. 48 cm3 cm3 = 384 cm3 5 cm 9 cm 49.6 cm2

141 (a) Yes, the soap is a prism because it's cross section is parallel and congrucuit to it's bases. (b) The volume of 1 soap = 3 = (15 cm)3 = 3375 cm3 (c) The volume of 20 soap = 20 × 3375 cm3 = 67,500 cm3 . 9. (i) A cubical wooden block has a length 12 cm. (ii) A cuboid wooden block has the length 12 cm, breadth 8 cm and height 5 cm. (a) What are cube and cuboid ? Define them. (b) Find the cross-section area and lateral surface area of the block. (c) How much wood does each block contain ? Find it. Solution: (i) Here, length of cubical wooden block () = 12 cm. (a) Cube: A three dimensional solid whose all sides are equal is called a cube. Cuboid: A three dimensional solid vluse all sides are not equal is called a cuboid. (b) Cross section area of cube (A) = 2 = (12 cm)2 = 144 cm2 lateral surface are (LSA) = 4 2 = 4 × 144 = 576 cm2 . (c) Quactity of wood (V) = 3 = (12 cm)3 = 1728 cm3 (ii) Here, length of cuboid () = 12 cm breadth of cuboid (b) = 8 cm height of cuboid (h) = 5 m (b) Area of cross section (A) = × b = 12 cm × 8 cm = 96 cm2 Lateral surface area (LSA) = 2h( + b) = 2 × 5(12 + 8) = 10 × 20 = 200 cm2 (c) Quantity of wood (V) = × b × h = 12 cm × 8 cm × 5 cm = 480 cm3 10. (a) What is the volume of a cuboid having the area of the cross-section 25 cm2 and height 4 cm? (b) What is the volume of a squared base prism having the perimeter of the cross-section 144 cm ? (c) If the height of a squared based prism having side 6 cm is 12 cm, find its total surface area. (d) The area and perimeter of the cross-section of a cuboid are 125 cm2 and 60 cm respectively. If the height of the cuboid is 7 cm, find its total surface area. Solution: (a) Here, area of cross section (A) = 25 cm2 height (h) = 4 cm volume (V) = × b × h = A × h = 25 cm2 × 4 cm = 100 cm3 . (b) Here, perimeter of cross section of squared base prism (P) = 144 m or, 4 = 144 m = 36 m Now, volume of prism (V) = 3 = (36 m)3 = 46,656 m3 (c) Here, height of squared based prism (h) = 12 cm side of base () = 6 cm = b TSA = ?

142 Now, TSA = 2(b + h + bh) = 2(6 × 6 + 6 × 12 + 6 × 12) = 2(36 + 72 + 72) = 2 × 180 = 360 cm2 (d) Here, the area of cross section (A) = 125 cm2 perimeter of cross section (P) = 60 cm height of cuboid (h) = 7 cm Now, TSA = LSA + 2A = 2h( + b) + 2 × 125 = P × h + 250 = 60 × 7 + 250 = 420 + 250 = 670 cm2 11. (i) The length, breadth and height of a box of the volume 82944 m3 are in the ratio 3:2:1. (ii) The ratio of the length, breadth and height of a cuboid is 6:4:3. If the volume of the cuboid is 1944 cm3 , find (a) Find the its length, breadth and height. (b) Find its LSA and TSA. Solution: (i) Here, ratio of length, breadth and height of cuboid = 3 : 2 : 1. volume of cuboid (V) = 82944 m3 . (a) Let length, breadth and height be 3x, 2x and x. then, V = × b × h or, 82944 = 3x × 2x × x or, 82944 6 = x3 or, 13,824 = x3 or, 243 = x3 x = 24 length = 3x = 3 × 24 = 72 m. breadth = 2x = 2 × 24 = 48 m and height = x = 24 m. (b) LSA = 2h( + b) = 2 × 24(72 + 48) = 48 × 120 = 5760 m2 TSA = 2h( + b) + × b = 5760 + 72 × 48 = 5760 + 3456 = 9216 m2 (ii) Here, ratio of length, breadth and height of cuboid = 6 : 4 : 3. volume (V) = 1944 cm3 (a) Let the length, breadth and l\height be 6x, 4x and 3x. then, V = × b × h or, 1944 = 6x × 4x × 3x or, 1944 72 = x3 or, 27 = x3 x = 3 cm length = 6x = 6 × 3 = 18 cm breadth = 4x = 4 × 3 = 12 cm and height = 3x = 3 × 3 = 9 cm (b) LSA = 2h( + b) = 2 × 9(18 + 12) = 18 × 30 = 540 cm2 TSA = 2h( + b) + × b = 540 cm2 + 18 × 12 = 540 cm2 + 216 cm2 = 756 cm2

143 12. (i) A closed wooden box 15 ft long, 10 ft wide and 6 ft high, externally is made up of 1.5 inches thick wood. (a) Find the outer volume of the box. (b) Find the capacity of the box. (c) Find the weight of the box, if 1 ft3 of wood weighs 6 kg. (d) Calculate the cost of the wood at Rs. 1225 per cu. ft. (ii) An open rectangular cistern when measured from outside is 1.35 m long, 1.08 m broad and 90 cm deep and is made of 2.5 cm thick iron. (a) Find the outer volume of the cistern. (b) Find the capacity of the cistern. (c) Find the volume of the iron used. (d) Calculate the cost of the iron at Rs. 105 per kg if the cost of 1 cc iron is 50 gm. Solution: (i) Here, length of box () = 15 ft breadth of box (b) = 10 ft height of box (h) = 6 ft thickness of wood = 1.5 inch = 1.5 12 ft = 0.125 ft (a) volume of outer of box = × b × h = 15 × 10 × 6 = 900 ft3 (b) To find the capacity of the box inner length () = 15 ft – 2 × 0.125 = 14.75 ft inner breadth (b) = 10 ft = 2 × 0.125 = 9.75 ft inner height (h) = 6 ft = 2 × 0.125 = 5.75 ft Now, V = × b × h = 14.75 × 9.75 × 5.75 = 826.9219 ft3 (c) Volume of wood = 900 ft3 – 826.9219 ft3 = 73.0781 Now, height of 1 ft3 wood = 6 kg height of 73.08 ft3 wood = 6 × 73.08 kg = 438.468 kg Hence, the weight of the box = 438.468 kg (d) The cost of 1 ft3 wood = Rs. 1225 The cost of 73.08 ft3 wood = Rs. 73.08 × 1225 = Rs. 89,520.67 (ii) Here, length of cistern () = 1.35 m broad of cistern (b) = 1.08 m deep of cistern (h) = 90 cm = 0.9 m Thickness of iron = 2.5 cm = 0.025 m (a) The outer volume of the cistern (V) = × b × h = 1.35 × 1.08 × 0.9 = 1.3122 m3 (b) For the capacity of the cistern, = 1.35 m – 2 × 0.025 = 1.3 m b = 1.08 m – 2 × 0.025 = 1.03 m h = 0.9 m – 2 × 0.025 = 0.85 m The capacity of the cistern (V) = 1.3 m × 1.03 m × 0.85 m = 1.13815 m3 = 1.13815 × 100 × 100 × 100 cm3 = 11,38,150 cm3

144 (c) Volume of the iron used = 1.3122 m3 – 1.13815 m3 = 0.17405 m3 = 174050 cm3 (d) For the cost of iron, 1 cubic cm iron = 50 gm 174050 cm3 iron = 174050 × 50 gm = 8702500 gm = 8702500 1000 kg = 8702.50 kg Now, the cost of 1 kg iron = Rs. 105 the cost of 8702.50 kg iron = 8702.50 × 105 = 913,762.50 13. (i) A box is made of iron 0.05 cm thick, its internal dimensions are 50cm × 20cm × 20cm. (a) Find the weight of the box if 1 cm3 weighs 7 gm. (b) Find the cost of the box if the iron costs Rs. 5.50/cm3 . (ii) The inner dimensions of a closed wooden box are 20 cm, 13 cm and 8 cm. Thickness of the wood is 1 cm. (a) Find the cost of wood required to make the box if 1 cm3 of wood costs Rs. 1.50. (b) Find the weight of the box if 1 cm3 wood weighs 5 gm. Solution: (i) Here, internal dimensions of iron box = 50 cm × 20 cm × 20 cm Volume of iron box (V1) = 20,000 cm3 Thickness of iron = 0.05 cm length of iron box with iron () = 50 + 0.5 + 0.05 = 50.1 cm b = 20.1 cm h = 20.1 cm Now, volume of box with iron (V2) = 50.1 × 20.1 × 20.1 = 20,240.901 cm3 Now, volume of iron only = V2 – V1 = 20,240.901 – 20,000 = 240.901 cm3 (a) By question weight of 1 cm3 iron = 7 gm weight of 240.901 cm3 iron = 7 × 240.901 gm = 1686.307 gm = 1.686 kg = 1 kg 686 gm (b) Cost of 1 cm3 iron = Rs. 5.50 Cost of 240.901 cm3 iron = Rs. 240.901 × 5.50 = Rs. 1324.96. (ii) Inner dimension of a closed wooden = 20 cm × 13 m × 8 m = 2080 cm3 Volume of box (inner) (V1) = 2080 cm3 Thickness of wood = 1 cm Now, volume of box with wooden (V2) = (20 + 2) × (13 + 2) × (8 + 2) = 22 × 15 × 10 = 3300 Volume of wooden in box = V2 – V1 = 3300 – 2080 = 1220 cm3 (a) Cost of 1 cm3 wood = Rs. 1.50 Cost of 1220 cm3 wood = Rs. 1220 × 1.50 = Rs. 1830 (b) Again, weight of 1 cm3 wood = 5 gm weight of 1220 cm3 wood = 1220 × 5 = 6100 gm = 6.1 kg

6.2 Rea Keep 1. ( ( ( ( Soluti ( ( ( ( 2. F Soluti ( ( ( 2 Surface Ar ad / Underst ping Skill Sha (a) What is cyl (b) Write the fo (c) What is the r cm? (d) What is the ion: (a) A circular b (b) Curved surfa (c) Total surfac (d) Volume of t Find the curved ion: (a) Here, radius height CSA = Now, CSA = (b) Here, radius height CSA = Now, CSA = (c) Here, diame radius height CSA = Now, CSA = rea and Vol tand / Think arp inder ? Define it ormula to calcul e total surface e volume of a rig ased prism is call face area of the cy e area of the cyli the cylinder with surface area of s (r) = 7 cm of cylinder (h) = = ? = 2rh = 2 × 22 7 × s (r) = 9 in (h) = 20 in = ? = 2rh = 2 × 22 7 × eter (d) = 10 cm (r) = d 2 = 10 2 = 5 c (h) = 12 cm = ? = 2rh = 2 × 22 7 × 145 lume of Cyl PRACTICE / Do t. late the curved s area of a cylind ght cylinder havi led a cylinder. ylinder with radiu nder with radius radius r and heig the given cylind 40 cm × 7 × 40 cm2 = 17 × 9 in × 20 in = 1 cm × 5 cm × 12 cm = linder E 6.2 surface area of t der having heig ing height h ft an us r and height h r and height h (T ght h (V) = r 2 h ders : 760 cm2 131.43 sq. in. = 377.14 cm2 the cylinder. ht h cm and ra nd radius r cm? (CSA) = 2rh TSA) = 2r(r + h) adius of its base ? ) e

( 3. C Soluti ( ( ( ( (d) Here, diame radius height CSA = Now, CSA = Calculate the vol ion: (a) Here, Area o height volume Now, V = A (b) Here, diame radius height volume Now, V = r (c) Here, height circum volume Now, C = 44 or, 2r = 4 or, 2 × 22 7 r = 44 Volum (d) Here, height CSA = Now, CSA = or, 2rh = or, 2 × 22 7 r = 21 2 × Volum eter (d) = 13 cm (r) = d 2 = 13 2 = 6.5 (h) = 21 cm = ? = 2rh = 2 × 22 7 × lume of the give of base (A) = 12 (h) = 30 cm e (V) = ? A × h = 12 cm2 × eter (d) = 21 cm (r) = d 2 = 21 2 = 10 (h) = 24 cm e (V) = ? r 2 h = 22 7 × (10.5) t (h) = 20 in mference (c) = 44 e (V) = ? 4 in 44 in × r = 44 in in × 7 44 = 7 in me (V) = r 2 h = 22 7 t (h) = 24 cm = 2112 cm2 = 2112 = 2112 × r × 24 = 2112 112 × 7 22 × 24 = 14 cm me (V) = r 2 h = 22 7 146 5 cm × 6.5 cm × 21 cm en cylinders : cm2 30 cm = 360 cm3 0.5 cm 2 × 24 = 8316 cm in 2 7 × 72 × 20 = 308 m 2 7 × 14 × 24 = 147 m = 858 cm2 3 m3 80 cu. in. 784 cm3