Comparing Quantities. = PxRxT TEXTBOOK QUESTIONS SOLVED. Learn and Remember. Exercise 8.1 (Page No. 157)

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$COMPARING QUANTITIES 27 Learn and Remember Comparing Quantities. To compare two quantities can be expressed in the form of ratio. 2. Two ratios can be compared by converting them to like fractions.$

1 COMPARING QUANTITIES 27 Learn and Remember Comparing Quantities. To compare two quantities can be expressed in the form of ratio. 2. Two ratios can be compared by converting them to like fractions.. Two fractions are equal if their ratios are equivalent. 4. Iftwo ratios are equivalent then the four quantities are said to be in proportion.. A way of comparing quantities is percentage. 6. Percentage is numerator offractions with denominator. 7. Per cent means per hundred. 8. S.P. means selling price and C.P. means cost price, (i) If S.P. > C.P. then there is profit, = S.P. - C.P. % = C.P. x (ii) If S.P. < C.P. then there is loss, Loss = C.P. - S.P. Loss Loss % = C.P. x 9. If P is principal, R is rate of interest per annum and T is the time in years Then simple interest (S.I.) is given by, SI = PxRxT.. 0. A = P + S.. where A is amount. TEXTBOOK QUESTIONS SOLVED Exercise 8. (Page No. 7) Q. Find the ratio of: (a) t to 0 paise (c) 9 m to 27 cm (d) kg to 20 g 0 days to 6 hours (a) t to 0 paise First we convert both quantities in same unit. t = x = 00 paise ('.: t = paise) (c) (d) Now ratio = = - = -,. 0 Thus, the required ratio = 0 : l. kg to 20 g Convert both weights in same unit. kg = x 0 = 000 g (-: kg = 0 g) Now ratio = = -- = -,. 2W 7 Thus, the required ratio = 00 : 7. 9 m to 27 cm Convert both lengths in same unit. 9 m = 9 x = 900 cm C' m = cm). 900 Now,ratIO=900:27= - =- 27 Thus, the required ratio = :. 0 days to 6 hours Convert both times in same unit. 0 days = 0 x 24 = 720 hours ('.: day = 24 hours) N. 720 " ow, ratio = : D) = 6 = T Thus, the required ratio = 20: l. Q2. In a computer lab, th- 're are computers for every 6 students. How many computers will be needed for 24 students? 6 students need = corn puters student needs = 6" computer Therefore, 24 students need = - x 24 computers 6 = x 4 = 2 computers Thus, 2 computers will be needed for 24 students. Q. Population of Rajasthan = 70 lakhs and population of U.P. = 660 lakhs. Area of Raj as than = lakh km2 and area of U.P. = 2 lakh km2?

2 28 (i) How many people are there per km2 in both these states? (ii) Which state is less populated?. Population (l) People are present per km2 = -=-.--- rea (ii) In Rajasthan = 70akhs" = 90 people per km2 lakhkm 660lakhs In U.P. = " = 80 people per km2 2lakhkm Rajasthan is less populated. Exercise 8.2 (Page No. 64-6) Q. Convert the given fractional numbers to per cents. (a) (a) "8 = "8 x % = 2" % = 2.%. (c) 40 - = - x % = x 2% = 2%. 4 4 (c) - = - x % = - x % = - % = 7.% (d) ;; (d) - = - x % = -% = 28-% Q2. Convert the given decimal fractions to per cents. (a) (c) 0.02 (d) (a) 0.6 = x % = 6% 2. = 0 x % = 20% '. (c) 0.02 = x % = 2% (d) 2.:::: x % = 2%. Q. Estimate what part of the figures is coloured and hence find the per cent which is coloured. Cj :... '.' c.,."....i'r -.'......'(\; :.' ";'\'_". ' t :'.. ' '... ', : '., ' :-' ;":,'-'t :" (i) (ii) (iii) COMPARING QUANTITIES. (l) Coloured part = '4 Q4. Find: Per cent of coloured part = '4 x % = 2%. (ii) Coloured part = ' Per cent of coloured part = ' x 0 = 60%. (iii) Coloured part = Per cent of coloured part = "8 x % = '2 x 2% = (. x 2)% = 7.%. (a) %of20 %ofhour (c) 20% of 200 (d) 7% of kg. (a) % of 20 = x 20 = x 2. = 7.. Q. % of hour We know that hour = 60 min. = 60 x 60 sec. Now, % of(60 x 60) sec. = - x (60 x 60) sec. = 6 x 6 sec. = 6 sec. 20 (c) 20% of 200 = x 200 = (20 x 2) = 00. (d) 7% of kg We know, kg = 0 g 7 Now, 7%of 0 g = x 0 g = 70 g or 0.70 kg. Find the whole quantity (a) % of it is % of it is 080. if (c) 40% of it is 00 km. (d) 70% of it is 4 minutes. (e) 8% of it is 40 litres. 29

0 Let the whole quantity (c) % of it is 600 % ofx = 600 x x = 600 be x in given problems: x = 600x = 2 000 ' Thus, x = 2,000, which is required quantity.

3 0 Let the whole quantity (c) % of it is 600 % ofx = 600 x x = 600 be x in given problems: x = 600x = ' Thus, x = 2,000, which is required quantity. Cb) 2%of it is 080 2%of x = xx= x x = 2 = 9,000 Thus, x = 9,000, which is required Cc) 40% of it is 00 km 40% ofx = xx=00 _ 00x -20 x- -, 40 Thus, x =,20 km, which is required Cd) 70% of it is 4 minutes 70% ofx = xx=4 quantity. quantity. x = 4x = Thus, x = 20 minutes, which is required quantity. (e) 8% of it is 40 litres 8% ofx = xx=40 x = 40x = 00 8 Thus, x = 00 litres, which is required quantity. COMPARING QUANTITIES Q6. Convert given per cents to decimal fractions and also to fractions in simplest forms: (a) 2% 0% (c) 20% (d) %. I S.No. I Per cents I Fractions I Simplest Decimal form form 2 Ca) I 2% I Cb) I 0% I - - I. 2 Cc) I 20% I (d) I % I I 0.2 I 0.0 Q7. In a city, 0% are females, 40% are males and remaining are children. What per cent are children? Given, percentage of females = 0% percentage of males = 40% Total percentage of females and males = 0% + 40% = 70% Percentage of children = Total percentage - Percentage offemales and males = (loo - 70)%= 0% Thus, the 0% are children. Q8. Out of,000 voters in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote? Total voter =,000 Percentage of voted candidates = 60% Percentage of not voted candidates = ( - 60)% = 40% Actual candidates who did not vote = 40% of, = x,000 = 6,000 Thus, 6,000 candidates did not vote. Q9. Meeta saves 400 from her salary. If this is 0% of her salary. What is her salary? Let Meeta's salary be x. Now, 0%of salary = 400

2 0%ofx =400 0 - xx=400 x = 400x = 4000 0 ' Thus, Meeta's salary is 4,000. Q.I0. A local cricket team played 20 matches in one season. It won 2% of them. How many matches did they win?

4 2 0%ofx = xx=400 x = 400x = ' Thus, Meeta's salary is 4,000. Q.I0. A local cricket team played 20 matches in one season. It won 2% of them. How many matches did they win? Given, number of matches played by cricket team = 20 Percentage of won matches = 2%. 2 Total matches won by them = 2% of 20 = x 20 = Thus, they won matches. Exercise 8. (Page No. 7-72) Q. Tell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent in each case. (a) Gardening shears bought for 20 and sold for 2. A refrigerator bought for 2,000 and sold at,00. (c) A cupboard bought for 2,00 and sold at,000. (d) A skirt bought for 20 and sold at 0. (a) Given, cost price of gardening shears = 20 Selling price of gardening shears = 2 On comparing the prices, the S.P. is greater so obtains profit, = S.P. - C.P. = (2-20) = 7 Now, per cent = C.P. x = 7 x = =0% Therefore, profit = 7, profit per cent = 0%. COMPARING QUANTITIES Given, cost price of refrigerator = 2,000 Selling price of refrigerator =,00 On comparing the prices, the S.P. is greater so obtained profit. = S.P. - C.P. = (,00-2,000) =,00 Now, per cent = C.P. x = 00 x 2000 = 2.% Therefore, profit =,00 and profit per cent = 2.%. (c) Given, cost price of cupboard = 2,00 Selling price of cupboard = 000 On comparing the price, S.P. is greater so obtained profit, = S.P. - C.P. = (,000-2,00) = 00 Now, per cent = -C-- x.p. = 00 x 200 =20% Therefore, profit = 00 and profit per cent = 20%. (d) Given, cost price of skirt = 20 Selling price of skirt = 0 On comparing the prices, S.P' is smaller so obtained loss. Loss = C.P. - S.P. = (20-0) Loss = Loss Now, Loss per cent = C.P. x = x 20 = 40% Therefore, loss = and loss per cent = 40%.

4 Q2. Convert each part of the ratio to percentage: (a) : 2: : (c) : 4 (d) : 2: Ca) : Total part = + = 4 Fractional of first part = "4 and second part = "4 Therefore, percentage of first part = "4 x

5 4 Q2. Convert each part of the ratio to percentage: (a) : 2: : (c) : 4 (d) : 2: Ca) : Total part = + = 4 Fractional of first part = "4 and second part = "4 Therefore, percentage of first part = "4 x = 7% Percentage of second part = "4 x = 2% Thus, : = 7% : 2%. Cb) 2: : Total part = = 0 Fractional of first part = 2 0' second part = 20 and third part = Therefore, Percentage of first part = 0 x = 20% Percentage of second part = 0 x = 0'0 Percentage of third part = 0x = 0% Thus, 2 : : = 20% : 0% : 0% Cc) : 4 Total part = + 4 = Fractional of first part = and second part = Percentage of first part = " x = 20% 4 Percentage of second part = " x = 80% Thus, : 4 = 20% : 80%. Cd) : 2 : Total part = = 8 part = - 8 Fractional of first part =, second part = and third COMPARING QUANTITIES Percentage of first part = 8" x = 2.% 2 Percentage of second part = 8" x = 2% Percentage of third part = % x = 62.% Thus, : 2 : = 2.% : 2% : 62.%. Q. The population of a city decreased from 2,000 to 24,00. Find the percentage decrease. The decreased population of a city from 2,000 to 24,00. Amount of change = 2,000-24,00 = 00 Amount of change Therefore, decreased percentage = O. al t x ngm amoun 00 =-- x=2% 2000 Thus, population decreased percentage is 2%. Q4. Arun bought a car for,0,000.the next year, the price went upto,70,000. What was the percentage of price increase? Increase in price of a car from,0,000 to,70,000. Amount of change = C,70,000 -,0,000) = 20,000 AmouIlt of change Therefore, increased percentage = Ori. al x ngm amount = x = % Thus, the percentage of price increased is "7 %. Q. I buy a T.V.for 0,000 and sell it at a profit of 20%.How much money do I get for it? Given, the cost price oft.v. = 0,000 per cent = 20% = % ofc.p. 20 = - x 0,000 = 2,000

6 Q6. Now, the selling price = C.P. + = (0,000 + 2,000) = 2,000 Thus, he gets 2,000 for selling his T.V. Juhi sells a washing machine for,00. She loses 20% in the bargain.

6 6 Q6. Now, the selling price = C.P. + = (0, ,000) = 2,000 Thus, he gets 2,000 for selling his T.V. Juhi sells a washing machine for,00. She loses 20% in the bargain. What was the price at which she bought it? Given, selling price of washing machine =,00 Loss per cent = 20% Let the cost price of washing machine be x. Loss = loss % of C.P. Therefore, 20 x Loss = 20% of x = - x x = - S.P. = C.P' - Loss x,00 =x-" 00 = 4x, 4x =,00 x,00x x= 4 or x = 6,87 Thus, the cost price ofthe washing machine is 6,87. Q7. (i) Chalk contains calcium, carbon and oxygen in the ratio 0 : : 2. Find the percentage of carbon (ii) in chalk. If in a stick of chalk, carbon is g, what is the weight of the chalk stick? (i) Given ratio = 0 : : 2 Total part = = 2 Now, part of carbon = - 2 Percentage ofcarbon part in chalk = x %= x 4% = 2% 2 Thus, the percentage of carbon in chalk is 2%. (ii) Given, quantity of carbon in chalk stick = g Let the weight of chalk be x g. COMPARING QUANTITIES 7 Q8. Q9. Then, 2% ofx = 2 x x = 2x = X= - or x=2 2 Thus, the weight of the chalk stick is 2 g. Amina buys a book for 27 and sells it at a loss of %. How much does she sell it for? Given, the cost price of a book = 27 Therefore, Loss per cent = % Loss = Loss % of C.P. = % of27 = ( X27) = 4,2 Loss = 4.2 S.P' = C.P. - Loss = (27-4.2) = 2.7 Thus, Amina sells a book for 2.7. Find the amount to be paid at the end of years in each case: (a) Principal =,200 at 2% p.a. Principal = 7,00 at % p.a. (a) Given, principal (P) =,200; rate (R) = 2% p.a.; time (T) = years S.. = P x R x T = 200 x 2 x = (2 x 2 x ) = 42 Amount = Principal + S.. = ( ) =,62 Given, principal (P) = 7,00; rate (R) = % p.a.; time (T) = years S.l. = P x R x T = 7,00 x x

7 8 = (7 x ) S.l. =,2 Amount = Principal + S.l. = (7,00 +,2) = 8,62. QI0. What rate gives 280 as interest on a sum of 6,000 in 2 years? Given, principal (P) = 6,000; simple interest (S..) = 280; time (T) = 2 years Let the rate of interest be R% p.a. PxRxT We know, S.l. = 280 = 6,000 x R x 2 R _ 280 x 2--.:8,---,00_0-6,000 x 2 -,2,000 or R=0.2% Thus, the rate on the sum is 0.2% p.a. QU. IfMeena gives an interest of 4 for one year at 9%rate p.a.. What is the sum she has borrowed? Given, Simple interest (S.L) = 4; rate (R) = 9% p.a.; time (T) = year Let the sum be P. Therefore, S.l. = PxRxT Px9x 4= ,00 = 9P P = 4,00 9 or P = 00 Thus, the sum, she has borrowed is 00. DD

8 COMPARING QUANTITIES

8 COMPARING QUANTITIES Exercise 8.1 Q.1. Find the ratio of : (a) Rs 5 to 50 paise (b) 15 kg to 210 gm (c) 9 m to 27 cm (d) 30 days to 36 hours Ans. (a) Ratio between Rs 5 to 50 paise Rs 1 paise Rs 5 500