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question:one alloy there is a certain percentage of chromium while in another alloy it is 8%. A certain amount of the first alloy was melted together with 35 kg of the second one to form a third alloy. The percentage of chromium in the new alloy is 10.1%. How much of the first alloy was used?
answer:Let's denote the amount of the first alloy used as ( x ) kg and the percentage of chromium in the first alloy as ( C_1 ) (in decimal form, so for example, 20% would be 0.20). The total weight of the new alloy is ( x + 35 ) kg. The total amount of chromium from the first alloy is ( C_1 times x ) kg, and from the second alloy is ( 0.08 times 35 ) kg. The new alloy has a chromium percentage of 10.1%, which means the total chromium in the new alloy is ( 0.101 times (x + 35) ) kg. The equation representing the chromium content before and after mixing is: [ C_1 times x + 0.08 times 35 = 0.101 times (x + 35) ] Now we can solve for ( x ): [ C_1 times x + 2.8 = 0.101x + 3.535 ] Since we don't know the value of ( boxed{C_1} ), we can't solve for ( x ) directly. We need the percentage of chromium in the first alloy to find the exact amount of the first alloy used. If you can provide that information, I can continue with the calculation.
question:There are eight balls numbered from 1 to 8, among which six balls are of the same weight, and the other two are 1 gram lighter. To find out these two lighter balls, a balance scale was used three times, and the results are as follows: the first time, balls 1 and 2 are heavier than balls 3 and 4; the second time, balls 5 and 6 are lighter than balls 7 and 8; the third time, the weight of balls 1, 3, and 5 combined is equal to the weight of balls 2, 4, and 8 combined. Therefore, the numbers of the two lighter balls are.
answer:Given that balls 1 and 2 are heavier than balls 3 and 4, it implies that at least one of the lighter balls is among balls 3 and 4. Since balls 5 and 6 are lighter than balls 7 and 8, it indicates that at least one of the lighter balls is among balls 5 and 6. Considering that the combined weight of balls 1, 3, and 5 is equal to the combined weight of balls 2, 4, and 8, we can deduce that the numbers of the two lighter balls are boxed{4 text{ and } 5}.
question:Olivia and Nigel are traveling in Asia. Olivia has 112 and Nigel has 139. If together they buy six one day passes to a tourist attraction for 28 per ticket, how much money do they have left?
answer:To solve this problem, we'll break it down into steps: 1. **Calculate the total amount of money Olivia and Nigel have together**: [112 + 139 = 251] 2. **Calculate the total cost of the six one-day passes**: [6 times 28 = 168] 3. **Determine how much money they have left after purchasing the passes**: [251 - 168 = 83] Therefore, after buying the six one-day passes to the tourist attraction, Olivia and Nigel have boxed{83} left.
question:Find the distance between the planes 2x 4y + 4z = 10 and 4x 8y + 8z = 20.
answer:First, simplify the equation of the second plane: [ 4x - 8y + 8z = 20 implies 2x - 4y + 4z = 10. ] Notice that these two equations are identical, indicating that they are not just parallel, but the same plane. Since they are the same plane, the distance between them is: [ boxed{0}. ]