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how many kg of sugar can be bought for rf:69.if one kg costs rf:3.75

How many kg of sugar can be bought for RF:69.if one KG costs RF:3.71

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The Cost Calculator for MASS - Cleave Books


The Cost Calculator for MASS. Quantity: ... A tonne 't' is the metric tonne of 1000 kg. The troy ... the entry under 'Cost per unit' is the correct one for the ...
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Diabetes mellitus - Wikipedia


... the onset of diabetes can be triggered by one or more ... high blood sugar can be reversed by a variety of measures ... two hours after a 75 g oral glucose ...
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How much does sugar cost - Answers.com


How much does sugar cost? ... One pound of sugar costs $1.45 a pound. ... A bag of sugar typically costs around 1.69 or less.
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Pupils should be taught to: As outcomes, Year 4 pupils should ...


... As outcomes, Year 4 pupils should, ... 3/99 Y456 examples 75 Making decisions ... • Each letter from A to G is a code for one of these digits: 1, 3, 4, 5, 6, 8, ...
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Science Questions including "How is plexiglass made"


Science Questions including "How is plexiglass made" and "Is it ... Answer Over a period of time the sugar in Gatorade can ... If you have one kilogram of ...
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Amazon.com : Domino Sugar, Granulated, 10-Pound Bags : White ...


... Domino Sugar, Granulated, 10-Pound Bags : ... I bought this despite the negative ... The sugar isn't as fine as powdered sugar, but looks like it's one grind away ...

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If each CD costs $15.99, how many CDs can she ... If 2 cups of sugar equal one pound, how many pounds ... A third pump can drain the tank in 75 minutes. If all 3 ...
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Burning Desire for Efficiency | Do the Math


Burning Desire for Efficiency. ... is how much energy it takes to heat one kilogram ... One small pedantic note; the RF oscillator in a microwave oven is a magnetron,
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Development of Sugar Industry in Sri Lanka | SpringerLink


The sugar industry has faced many ... on the order of Rs 100/kg. In addition, sugar mills in ... problems of the sugar sector, one of the main ...
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Village milk processing - Food and Agriculture Organization


... and 3.75 litres of cream with a fat content of 400 ... The collection costs, therefore, for one litre of milk are: ... then the salary costs per kg of cheese ...
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Suggested Questions And Answer :


how many kg of sugar can be bought for rf:69.if one kg costs rf:3.75

69/3.75=18.4 ........
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word problem

Sean bought a bag of mixed candies. In the bag there were 25 Smarties an Reese's peanut Butter cups. Smarties cost $0.15 each an reese's are $0.75 each. If he paid $15.15 , how many Smarties an reese's did he buy? A. write an expression for the number of smarties . write an expression for the number of Reese's. B. write an algebraic equation using the expressions above. C. solve your equation (show all work) D. tell how many smarties an reese's were bought. please write : a,b,c,d after each answer S = 25 - R    R = 25 - S             (a)  These don't get us anywhere S + R = 25 and 0.15S + 0.75R = 15.15        (b)  One equation won't do   3 * (S + R) = 3 * 25     3S + 3R = 75    to equalize the R terms 4 * (0.15S + 0.75R) = 4 * 15.15 0.6S + 3R = 60.60 We have two equations we can subtract, leaving only the S term   3S    + 3R = 75 -(0.6S + 3R = 60.60) ----------------------------- 2.4S           = 14.40 2.4S = 14.40 S = 6 R = 25 - S = 25 - 6 = 19       (c) There were 6 Smarties and 19 Reese's    (d) 6 * 0.15 + 19 * 0.75 = 0.90 + 14.25 = 15.15
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How many pounds of the less expensive pears did Sally buy?

x+y=10  pounds, so y=10-x 2.5*x +3.5*y=27$ 2.5x +3.5*(10-x)=27 2.5x -3.5x +35=27 -x=27-35=-8 x=8 y=10-x=2
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Please answer soon! Algebra - show your work

Let n be the number of calculators so each one cost 240/n. She sold n-1 calculators for $300 so each calculator was sold for 300/(n-1). The profit=5=300/(n-1)-240/n. Multiply both sides of the equation by n(n-1): 5n(n-1)=300n-240n+240. So 5n^2-5n=60n+240. This is a quadratic equation: 5n^2-65n-240=0.  Divide through by 5: n^2-13n-48=0; (n-16)(n+3)=0. So n=16, because n=-3 is not applicable in this case. Let's see if 16 fits the facts. Each calculator costs 240/16=$15. She sold 15 for $300, making them $20 each. So she made a profit of 20-15=$5 on each calculator. That fits!
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how to solve this word problem

A retailer incurs a fixed cost of N$ 330when purchasing sugar for his stock. He pays N$15.00 per packet which he resells at N$18.00 per packet. How many packets should he purchase and sell in order to break even? Let us suppose that he buys X packets. Then his costs are: X packets at $15 per packet comes to $15X Fixed cost comes to $330 Total costs = $(330 + 15X)   Sales revenue He sells the X packets at $18 per packet Income is $18X Profit is P = Income - Costs P = 18X - (330 +15X) If he breaks even then there is no profit and P = 0, i.e. 18X - (330 + 15X) = 0 18X - 15X = 330 3X = 330 X = 110 Our retailer will break even if he sells 110 packets of sugar To make a profit, The retailer must buy, and sell, more than 110 packets
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word problem

word problem Sean bought a bag of mixed candies. In the bag there were 25 Smarties and Reese's peanut Butter cups. Smarties cost $0.15 each an reese's are $0.75 each. If he paid $15.15 , how many Smarties an reese's did he buy? A. write an expression for the number of smarties . write an expression for the number of Reese's.   B. write an algebraic equation using the expressions above. C. solve your equation (show all work) D. tell how many smarties an reese's were bought.   Smarties + Reese = 25 call this eq1 Reese = 25 - Smarties calll this eq2   Smarties = $0.15 Reese = $ 0.75 Total = $ 15.15   0.15 Smarties + 0.75 Reese = $ 15.15  call this eq3 substitute eq2 to eq3   0.15 Smarties + 0.75 (25 - Smaties) = $ 15.15 0.15 Smarties + 18.75 - 0.75Smaties) = $ 15.15  - 0.6 Smarties = $ 15.15 - 18.75 Smarties = - 3.6 / - 0.6   Smarties = 6 ◄ Ans sbstitute Smarties in eq1 Smarties + Reese = 25 call this eq1 6 + Reese = 25 Reese = 25 - 6 Reese = 19  ◄ Ans  
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Jane bought 7 mangoes and 5 oranges at atotal of ksh 48 each.If she had bouhgt 3 mangoes and 8 oranges she would have saved ksh 12. Find the price of a mango and an orange

A=cost of one mango, B=cost of one orange. 3A+8B=7A+5B-12, so 4A-3B=12. From this we can relate the cost of the fruits: B=4A/3-4. The question is puzzling because it says 7A+5B are bought "at a cost of ksh 48 each". The context for "each"  needs clarification. Assume it means 7A+5B=48 or 3A+8B=36. Then we have: 3A+8(4A/3-4)=36; 41A/3-32=36; 41A/3=68 so A=204/41=ksh 4.98 approx, so B=ksh 2.63 approx. The combined cost of a mango and an orange=ksh 7.61.
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using a system of linear equation involving cost

3x+y=32 or y=32-3x 2x+5y=49.50 2x+5(32-3x)=49.50 2x-15x+160=49.5 -13x=49.5-160 -13x=-110.5 x=110.5/13=8.50$ y=32-3x=32-3*8.5=32-25.5=6.5
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A movie theater in town posted these prices for tickets and snacks.

Sally and Anne both needed a ticket, so each paid $9 for the ticket. Then they each bought a drink costing $1.35. But they shared popcorn and a veggie cup costing $3.50+$1.74, so the cost to each woman was ½(3.50+1.74). We're looking for 9+1.35+½(3.50+1.74) in the answer. So answer 2 must be the right answer!
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Find how many cups of each blend you could drink to meet you minimum daily requirement with a minimum cost.

x   amount of A y  amount of B (25x +50y)/(75x +50y)=3/5 125x +250y = 225x +150y 100y =100x therefore, y=x. Since total number of cups is 8, one will have to drink 4 cups of each.
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