#$&* course mth151 008. Arithmetic Sequences*********************************************
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Given Solution: Pairing 4 with 900, 8 with 896, etc., we get pairs adding up to 904. The difference between 4 and 900 is 896. The numbers 'jump' by 4, so there are 896 / 4 = 224 'jumps'. None of these 'jumps' ends at the first number so there are 224 + 1 = 225 numbers. Thus we have 225 / 2 = 112.5 pairs each adding up to 904, and our total is 112.5 * 904. ********************************************* Question: `q007. What expression would stand for the sum 1 + 2 + 3 + ... + n, where n is some whole number? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: n/2 * (n+1) confidence rating #$&*: very confident ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ====================================================================== 008. `Query 8 ********************************************* Question: `q1.3.6 9 and 11 yr old hosses; sum of ages 122. How many 9- and 11-year-old horses are there? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5 9 year olds and 7 11 year olds confidence rating #$&*: very confident ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `qQuery 1.3.10 divide clock into segments each with same total YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1+2+3+4+5+6+7+8+9+10+11+12 = 78 if you create 3 horizontal sections each section adds up to 26 confidence rating #$&*: very confident ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `qQuery 1.3.18 M-F 32 acorns each day, half of all acorns eaten, 35 acorns left after Friday YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: they started out with 4 acrons on Monday. M +32-16=16 T 16+32-24=24 W 24+32-28=28 TH 28+32-30=30 F 30+32-31=31 S 31 +4 = 35 confidence rating #$&*: did not get this one at all ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `qQuery 1.3.30 Frog in well, 4 ft jump, 3 ft back. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: by the 17th day he reaches the 20 foot mark for the first time. since he is at the top of the well, you can assume that he does not slip backward 3 feet on that day. confidence rating #$&*: very confident; but i did make the common error the first time out. when i worked it on paper, i could see that he would indeed get to the top for the first time on day 17 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `qQuery 1.3.48 How many ways to pay 15 cents? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 6 ways: 15 pennnies 1 dime 5 pennies 1 dime 1 nickel 1 nickel 10 pennies 2 nicles 5 pennies 3 nickels confidence rating #$&*: very confident. i listed them on paper ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `qQuery 1.3.52 Given 8 coins, how do you find the unbalanced one in 3 weighings YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3 weighings: divide the coins into 2 stacks of 4 coins each and weigh each stack. whichever stack is lighter contains the bad coin. now take the lighter stack and divide it into two stacks of two coins each. the lighter stack again contains the bad coin. now weigh the two remaining coins separately. the lighter coin is the bad coin. 2 weighings: divide the coins into 2 stacks of 4 coins each and weigh each stack. whichever stack is lighter contains the bad coin. dividing the weight of the good stack by 4 tells you how much each good coin should weigh. then drop the remaining 4 coins into a vending machine in turn. the bad coin will not register. then you can weigh it and cofirm that it actually is lighter (could not figure this one out) confidence rating #$&*: very confident about 3 weighings, not confident about two weighings ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^