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# Recent problems solved by 'venugopalramana'

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 Triangles/23698: How can you tell by looking at the lengths of three line segments whether they can be used to form a triangle?1 solutions Answer 12443 by venugopalramana(3286)   on 2006-01-05 23:56:20 (Show Source): You can put this solution on YOUR website!THE SUM OF ANY TWO SIDES SHOULD BE MORE THAN THE THIRD TO FORM A TRIANGLE
 Quadratic_Equations/23670: Hi, I am really having a hard time analyzing this one, I hope someone could help me on this and I would really really appreciate it so much. Question 1 : Explain a sequence of steps that can be used to solve any quadratic equation in the QUICKEST WAY. Question 2 : Derive the quadratic formula used in solving quadratic equations. 1 solutions Answer 12439 by venugopalramana(3286)   on 2006-01-05 23:04:37 (Show Source): You can put this solution on YOUR website!LET THE QUADRATIC BE AX^2+BX+C=0...SINCE THIS IS A QUADRATIC ,A IS NOT EQUAL TO ZERO..SO DIVIDING WITH A WE GET X^2+BX/A+C/A=0 WRITE IT AS A PERFECT SQUARE USING X^2 AND X TERMS {X^2+2*X*(B/2A)+(B/2A)^2}-(B/2A)^2+C/A=0 {X+(B/2A)}^2=(B/2A)^2-C/A=B^2/4A^2-C/A=(B^2-4AC)/4A^2 TAKING SQUARE ROOT X+B/2A= OR
 Quadratic_Equations/23687: Please help me solve the following quadratic equation. Any help or guidance you offer is greatly appreciated because I am very confused with these types of problems. Find the vertex and intercepts for each quadratic function, and graph. y = -x^2 + 2x + 3 Thank you1 solutions Answer 12433 by venugopalramana(3286)   on 2006-01-05 22:37:12 (Show Source): You can put this solution on YOUR website!Find the vertex and intercepts for each quadratic function, and graph. y = -x^2 + 2x + 3 WRITE AS A SQUARE USING X^2 AND X TERMS Y=3-{(X^2-2*X*1+1^2)-1^2} Y=3-(X-1)^2+1=4-(X-1)^2 SINCE (X-1)^2 IS A PERFECT SQUARE IT IS ALWAYS POSITIVE OR MINIMUM ZERO.SO WE GET A MAXIMUM VALUE FOR Y AT X=1..VALUE IS Y=4..THIS IS CALLED VERTEX.YOU CAN SEE IT IN THE GRAPH. INTERCEPTS MEAN WHERE THE CURVE CUTS THE AXES PUT X=0 AND FIND Y TO GET Y INTERCEPTS..WE GET Y=3 HENCE 3 IS THE Y INTERCEPT. PUT Y=0 AND FIND X TO GET X INTERCEPTS...WE GET 0=4-(X-1)^2 (X-1)^2=4 X-1=2....OR...X-1=-2 X=3.....OR......-1... YOU CAN SEE IT FROM GRAPH GIVEN BELOW.
 Quadratic_Equations/23680: Please show me how to solve the inequality stating the solution set using intervla notation and graph the solution set. Any help and explanation will be appreciated. 4x^2 - 8x ≥ 01 solutions Answer 12427 by venugopalramana(3286)   on 2006-01-05 22:12:20 (Show Source): You can put this solution on YOUR website!4x^2-8x > = 0 4X(X-2) > =0 EITHER X=0 OR X=2...FOR 4x^2-8x = 0..SINCE A PRODUCT IS ZERO IF EITHER OF THE 2 FACTORS IS ZERO. FOR 4x^2-8x > 0....EITHER BOTH FACTORS SHOULD BE POSITIVE OR NEGATIVE. X > 0 AND X > 2....OR....X < 0 AND X < 2...THAT IS....... X > 2...............OR.......X < 0
 Polynomials-and-rational-expressions/23672: (2u^3-13u^2-8u+7)divided by (u-7) I have the quotient as 2u^2+u-15 with a remainder of 105 Useing long division This seems wrong because the reamainder was 0-105. Please help me out. Thank you1 solutions Answer 12417 by venugopalramana(3286)   on 2006-01-05 21:43:55 (Show Source): You can put this solution on YOUR website!(2u^3-13u^2-8u+7)divided by (u-7)...YOU CAN USE SHORT DIVISION FOR THIS ... LET U-7=0...U=7 7....|.......2........-13.......-8..........7 .....|.......0........14........7..........-7 ------------------------------------------------- .....|.......2.........1.........-1.........0 HENCE REMINDER IS ZERO QUOTIENT IS 2U^2+U-1.... IF YOU SHOW YOUR WORKING , WE CAN TELL WHERE YOU HAVE ERRED.
 Parallelograms/23675: Can not figure out how to solve the following. I am to find the value of x and y to ensure the quadrilateral is a parallelogram. there is a square with x from corner to corner. in top triangle right side says 7x in bottom triangle right side says y + 2 right triangle is empty. Left side triangle has 4x on top line and 2y on bottom... Please help!1 solutions Answer 12415 by venugopalramana(3286)   on 2006-01-05 21:37:27 (Show Source): You can put this solution on YOUR website!WE SHALL DEFINITELY HELP YOU.BUT YOUR PROBLEM IS NOT CLEAR.DRAW A SKETCH ,SHOWING WHAT YOU MEAN ,SCAN AND SEND ,WE SHALL ANSWER YOU IMMEDIATELY
 Quadratic_Equations/23585: I need your help The problem is 4x^2-8x greater than or equal to 0.1 solutions Answer 12371 by venugopalramana(3286)   on 2006-01-05 10:43:59 (Show Source): You can put this solution on YOUR website!4x^2-8x > = 0 4X(X-2) > =0 EITHER X=0 OR X=2...FOR 4x^2-8x = 0..SINCE A PRODUCT IS ZERO IF EITHER OF THE 2 FACTORS IS ZERO. FOR 4x^2-8x > 0....EITHER BOTH FACTORS SHOULD BE POSITIVE OR NEGATIVE. X > 0 AND X > 2....OR....X < 0 AND X < 2...THAT IS....... X > 2...............OR.......X < 0
 Quadratic_Equations/23586: 3x-x^2> 0 please can somone help me with these?1 solutions Answer 12370 by venugopalramana(3286)   on 2006-01-05 10:33:34 (Show Source): You can put this solution on YOUR website!3x-x^2> 0 X(3-X)>0 FOR A PRODUCT TO BE POSITIVE EITHER BOTH FACTORS SHOULD BE POSITIVE OR BOTH SHOULD BE POSITIVE. HENCE X > 0 AND 3 > X...THAT IS 0 < X < 3.......OR............... X < 0 AND 3 < X....THAT IS NOT POSSIBLE SO THE SOLUTION IS 0 < X < 3
 Quadratic_Equations/23587: x^2+25<10x I need some help on this one also.1 solutions Answer 12369 by venugopalramana(3286)   on 2006-01-05 10:26:44 (Show Source): You can put this solution on YOUR website!x^2+25<10x x^2+25-10x <0 (X)^2+(5)^2-2*(X)*(5)<0 (X-5)^2<0 A PERFECT SQUARE IS ALWAYS POSITIVE.ITS MINIMUM VALUE IS 0.IT CAN BE NEVER NEGATIVE OR LESS THAN ZERO..HENCE THIS PROBLEM HAS NO SOLUTION
 test/23550: 11)The units digit of a two-digit number is one-third the tens digit.When the digits are reversed, the new number is two more than four times the tens digit. Find the number.1 solutions Answer 12368 by venugopalramana(3286)   on 2006-01-05 10:21:21 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK ---------------------------------------------------------------------------- A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!! LET THE ORIGINAL NUMBER BE HT0 WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND 0 IN UNITS DIGIT SINCE THE NUMBER IS DIVISIBLE BY 10....ITS VALUE =100H+10T WE ARE GIVEN THAT T-H=1.................I SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1 52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T 104H+52=100H+10(H+1)=110H+10 110H-104H=52-10=42 6H=42 H=7 HENCE T=H+1=7+1=8 THE NUMBER IS 780 SEE THE FOLLOWING EXAMPLE. ----------------------------------------------------------------------------- LET THE ORIGINAL NUMBER BE HTU WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND U IN UNITS DIGIT....ITS VALUE =100H+10T+U the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.thank you!! TENS DIGIT EXCEEDS HUNDREDS DIGIT BY T-H UNITS DIGIT EXCEEDS TENS DIGIT BY U-T THESE ARE SAME .HENCE T-H=U-T...OR 2T=U+H..................................................I WHEN DIGITS ARE REVERSED TNE NEW NUMBER BECOMES UTH...ITS VALUE = 100U+10T+H NEW NUMBER EXCEEDS ORIGINAL NUMBER BY =100U+10T+H-100H-10T-U =99U-99H=99(U-H) THIS IS EQUAL TO 198 HENCE 99(U-H)=198 U-H=198/99=2.........................................................II U+H=2T..................FROM....I ADDING,WE GET 2U=2T+2.....DIVIDING BY 2,WE GET U=T+1...OR........T=U-1......................AND H=U-2...........FROM.......II SO THE ORIGINAL NUMBER IS HTU = (U-2)(U-1)U...NOW SINCE U IS NOT SPECIFIED WE CAN HAVE VARIOUS VALUES FOR U STARTING FROM 9 TO 2...SINCE U-2 CANNOT BE NEGATIVE..SO THE SOLUTIONS ARE U=9..........789.............987-789=198 U=8...........678............876-678=198...ETC U=7..........567 U=6..........456 U=5............345 U=4............234 U=3............123 U=2.............012...WE MAY DISCARD THIS ALSO AS IT CANOT BE CONSIDERED AS 3 DIGIT NUMBER,THOUGH THIS ALSO SATISFIES THE CONDITION THAT 210-12=198 --------------------------------------------------------------------------------
 test/23551: 1)The sum of the digits of a two-digit number is 12. The value of the number is two more than 11 times the tens digit. Find the number. PLEASE HELP!! T_T1 solutions Answer 12367 by venugopalramana(3286)   on 2006-01-05 10:20:25 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK ---------------------------------------------------------------------------- A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!! LET THE ORIGINAL NUMBER BE HT0 WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND 0 IN UNITS DIGIT SINCE THE NUMBER IS DIVISIBLE BY 10....ITS VALUE =100H+10T WE ARE GIVEN THAT T-H=1.................I SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1 52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T 104H+52=100H+10(H+1)=110H+10 110H-104H=52-10=42 6H=42 H=7 HENCE T=H+1=7+1=8 THE NUMBER IS 780 SEE THE FOLLOWING EXAMPLE. ----------------------------------------------------------------------------- LET THE ORIGINAL NUMBER BE HTU WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND U IN UNITS DIGIT....ITS VALUE =100H+10T+U the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.thank you!! TENS DIGIT EXCEEDS HUNDREDS DIGIT BY T-H UNITS DIGIT EXCEEDS TENS DIGIT BY U-T THESE ARE SAME .HENCE T-H=U-T...OR 2T=U+H..................................................I WHEN DIGITS ARE REVERSED TNE NEW NUMBER BECOMES UTH...ITS VALUE = 100U+10T+H NEW NUMBER EXCEEDS ORIGINAL NUMBER BY =100U+10T+H-100H-10T-U =99U-99H=99(U-H) THIS IS EQUAL TO 198 HENCE 99(U-H)=198 U-H=198/99=2.........................................................II U+H=2T..................FROM....I ADDING,WE GET 2U=2T+2.....DIVIDING BY 2,WE GET U=T+1...OR........T=U-1......................AND H=U-2...........FROM.......II SO THE ORIGINAL NUMBER IS HTU = (U-2)(U-1)U...NOW SINCE U IS NOT SPECIFIED WE CAN HAVE VARIOUS VALUES FOR U STARTING FROM 9 TO 2...SINCE U-2 CANNOT BE NEGATIVE..SO THE SOLUTIONS ARE U=9..........789.............987-789=198 U=8...........678............876-678=198...ETC U=7..........567 U=6..........456 U=5............345 U=4............234 U=3............123 U=2.............012...WE MAY DISCARD THIS ALSO AS IT CANOT BE CONSIDERED AS 3 DIGIT NUMBER,THOUGH THIS ALSO SATISFIES THE CONDITION THAT 210-12=198 --------------------------------------------------------------------------------
 test/23552: ..The units digit of a two-digit number exceeds thrice the tens digit by 1. the sum of the digits is 9. Find the number. I NEED HELP...1 solutions Answer 12366 by venugopalramana(3286)   on 2006-01-05 10:18:55 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK ---------------------------------------------------------------------------- A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!! LET THE ORIGINAL NUMBER BE HT0 WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND 0 IN UNITS DIGIT SINCE THE NUMBER IS DIVISIBLE BY 10....ITS VALUE =100H+10T WE ARE GIVEN THAT T-H=1.................I SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1 52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T 104H+52=100H+10(H+1)=110H+10 110H-104H=52-10=42 6H=42 H=7 HENCE T=H+1=7+1=8 THE NUMBER IS 780 SEE THE FOLLOWING EXAMPLE. ----------------------------------------------------------------------------- LET THE ORIGINAL NUMBER BE HTU WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND U IN UNITS DIGIT....ITS VALUE =100H+10T+U the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.thank you!! TENS DIGIT EXCEEDS HUNDREDS DIGIT BY T-H UNITS DIGIT EXCEEDS TENS DIGIT BY U-T THESE ARE SAME .HENCE T-H=U-T...OR 2T=U+H..................................................I WHEN DIGITS ARE REVERSED TNE NEW NUMBER BECOMES UTH...ITS VALUE = 100U+10T+H NEW NUMBER EXCEEDS ORIGINAL NUMBER BY =100U+10T+H-100H-10T-U =99U-99H=99(U-H) THIS IS EQUAL TO 198 HENCE 99(U-H)=198 U-H=198/99=2.........................................................II U+H=2T..................FROM....I ADDING,WE GET 2U=2T+2.....DIVIDING BY 2,WE GET U=T+1...OR........T=U-1......................AND H=U-2...........FROM.......II SO THE ORIGINAL NUMBER IS HTU = (U-2)(U-1)U...NOW SINCE U IS NOT SPECIFIED WE CAN HAVE VARIOUS VALUES FOR U STARTING FROM 9 TO 2...SINCE U-2 CANNOT BE NEGATIVE..SO THE SOLUTIONS ARE U=9..........789.............987-789=198 U=8...........678............876-678=198...ETC U=7..........567 U=6..........456 U=5............345 U=4............234 U=3............123 U=2.............012...WE MAY DISCARD THIS ALSO AS IT CANOT BE CONSIDERED AS 3 DIGIT NUMBER,THOUGH THIS ALSO SATISFIES THE CONDITION THAT 210-12=198 --------------------------------------------------------------------------------
 test/23549: The units digit of a two-digit number is one-third the tens digit. when the digits are reversed, the new number is two more than four times the tens digit. Find the original number.1 solutions Answer 12365 by venugopalramana(3286)   on 2006-01-05 10:17:45 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK ---------------------------------------------------------------------------- A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!! LET THE ORIGINAL NUMBER BE HT0 WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND 0 IN UNITS DIGIT SINCE THE NUMBER IS DIVISIBLE BY 10....ITS VALUE =100H+10T WE ARE GIVEN THAT T-H=1.................I SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1 52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T 104H+52=100H+10(H+1)=110H+10 110H-104H=52-10=42 6H=42 H=7 HENCE T=H+1=7+1=8 THE NUMBER IS 780 SEE THE FOLLOWING EXAMPLE. ----------------------------------------------------------------------------- LET THE ORIGINAL NUMBER BE HTU WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND U IN UNITS DIGIT....ITS VALUE =100H+10T+U the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.thank you!! TENS DIGIT EXCEEDS HUNDREDS DIGIT BY T-H UNITS DIGIT EXCEEDS TENS DIGIT BY U-T THESE ARE SAME .HENCE T-H=U-T...OR 2T=U+H..................................................I WHEN DIGITS ARE REVERSED TNE NEW NUMBER BECOMES UTH...ITS VALUE = 100U+10T+H NEW NUMBER EXCEEDS ORIGINAL NUMBER BY =100U+10T+H-100H-10T-U =99U-99H=99(U-H) THIS IS EQUAL TO 198 HENCE 99(U-H)=198 U-H=198/99=2.........................................................II U+H=2T..................FROM....I ADDING,WE GET 2U=2T+2.....DIVIDING BY 2,WE GET U=T+1...OR........T=U-1......................AND H=U-2...........FROM.......II SO THE ORIGINAL NUMBER IS HTU = (U-2)(U-1)U...NOW SINCE U IS NOT SPECIFIED WE CAN HAVE VARIOUS VALUES FOR U STARTING FROM 9 TO 2...SINCE U-2 CANNOT BE NEGATIVE..SO THE SOLUTIONS ARE U=9..........789.............987-789=198 U=8...........678............876-678=198...ETC U=7..........567 U=6..........456 U=5............345 U=4............234 U=3............123 U=2.............012...WE MAY DISCARD THIS ALSO AS IT CANOT BE CONSIDERED AS 3 DIGIT NUMBER,THOUGH THIS ALSO SATISFIES THE CONDITION THAT 210-12=198 --------------------------------------------------------------------------------
 Numbers_Word_Problems/23554: ...The sum of the digits of a two-digit number is one fourth of the number. The number is four times the sum of the digits. Find the number.-nid ur help pls-1 solutions Answer 12364 by venugopalramana(3286)   on 2006-01-05 10:17:00 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK ---------------------------------------------------------------------------- A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!! LET THE ORIGINAL NUMBER BE HT0 WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND 0 IN UNITS DIGIT SINCE THE NUMBER IS DIVISIBLE BY 10....ITS VALUE =100H+10T WE ARE GIVEN THAT T-H=1.................I SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1 52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T 104H+52=100H+10(H+1)=110H+10 110H-104H=52-10=42 6H=42 H=7 HENCE T=H+1=7+1=8 THE NUMBER IS 780 SEE THE FOLLOWING EXAMPLE. ----------------------------------------------------------------------------- LET THE ORIGINAL NUMBER BE HTU WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND U IN UNITS DIGIT....ITS VALUE =100H+10T+U the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.thank you!! TENS DIGIT EXCEEDS HUNDREDS DIGIT BY T-H UNITS DIGIT EXCEEDS TENS DIGIT BY U-T THESE ARE SAME .HENCE T-H=U-T...OR 2T=U+H..................................................I WHEN DIGITS ARE REVERSED TNE NEW NUMBER BECOMES UTH...ITS VALUE = 100U+10T+H NEW NUMBER EXCEEDS ORIGINAL NUMBER BY =100U+10T+H-100H-10T-U =99U-99H=99(U-H) THIS IS EQUAL TO 198 HENCE 99(U-H)=198 U-H=198/99=2.........................................................II U+H=2T..................FROM....I ADDING,WE GET 2U=2T+2.....DIVIDING BY 2,WE GET U=T+1...OR........T=U-1......................AND H=U-2...........FROM.......II SO THE ORIGINAL NUMBER IS HTU = (U-2)(U-1)U...NOW SINCE U IS NOT SPECIFIED WE CAN HAVE VARIOUS VALUES FOR U STARTING FROM 9 TO 2...SINCE U-2 CANNOT BE NEGATIVE..SO THE SOLUTIONS ARE U=9..........789.............987-789=198 U=8...........678............876-678=198...ETC U=7..........567 U=6..........456 U=5............345 U=4............234 U=3............123 U=2.............012...WE MAY DISCARD THIS ALSO AS IT CANOT BE CONSIDERED AS 3 DIGIT NUMBER,THOUGH THIS ALSO SATISFIES THE CONDITION THAT 210-12=198 --------------------------------------------------------------------------------
 Numbers_Word_Problems/23556: 17) The sum of the digits of a three-digit number is 19. The hundreds digit is three times the tens digit. The number is 198 more than the number reversed. Find the original number.1 solutions Answer 12363 by venugopalramana(3286)   on 2006-01-05 10:16:15 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK ---------------------------------------------------------------------------- A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!! LET THE ORIGINAL NUMBER BE HT0 WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND 0 IN UNITS DIGIT SINCE THE NUMBER IS DIVISIBLE BY 10....ITS VALUE =100H+10T WE ARE GIVEN THAT T-H=1.................I SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1 52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T 104H+52=100H+10(H+1)=110H+10 110H-104H=52-10=42 6H=42 H=7 HENCE T=H+1=7+1=8 THE NUMBER IS 780 SEE THE FOLLOWING EXAMPLE. ----------------------------------------------------------------------------- LET THE ORIGINAL NUMBER BE HTU WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND U IN UNITS DIGIT....ITS VALUE =100H+10T+U the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.thank you!! TENS DIGIT EXCEEDS HUNDREDS DIGIT BY T-H UNITS DIGIT EXCEEDS TENS DIGIT BY U-T THESE ARE SAME .HENCE T-H=U-T...OR 2T=U+H..................................................I WHEN DIGITS ARE REVERSED TNE NEW NUMBER BECOMES UTH...ITS VALUE = 100U+10T+H NEW NUMBER EXCEEDS ORIGINAL NUMBER BY =100U+10T+H-100H-10T-U =99U-99H=99(U-H) THIS IS EQUAL TO 198 HENCE 99(U-H)=198 U-H=198/99=2.........................................................II U+H=2T..................FROM....I ADDING,WE GET 2U=2T+2.....DIVIDING BY 2,WE GET U=T+1...OR........T=U-1......................AND H=U-2...........FROM.......II SO THE ORIGINAL NUMBER IS HTU = (U-2)(U-1)U...NOW SINCE U IS NOT SPECIFIED WE CAN HAVE VARIOUS VALUES FOR U STARTING FROM 9 TO 2...SINCE U-2 CANNOT BE NEGATIVE..SO THE SOLUTIONS ARE U=9..........789.............987-789=198 U=8...........678............876-678=198...ETC U=7..........567 U=6..........456 U=5............345 U=4............234 U=3............123 U=2.............012...WE MAY DISCARD THIS ALSO AS IT CANOT BE CONSIDERED AS 3 DIGIT NUMBER,THOUGH THIS ALSO SATISFIES THE CONDITION THAT 210-12=198 --------------------------------------------------------------------------------
 test/23548: 11)The units digit of a two-digit number is one-third the tens digit.When the higits are reversed, the new number is two more than four times the tens digit. Find the number.1 solutions Answer 12362 by venugopalramana(3286)   on 2006-01-05 10:04:58 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK ---------------------------------------------------------------------------- A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!! LET THE ORIGINAL NUMBER BE HT0 WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND 0 IN UNITS DIGIT SINCE THE NUMBER IS DIVISIBLE BY 10....ITS VALUE =100H+10T WE ARE GIVEN THAT T-H=1.................I SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1 52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T 104H+52=100H+10(H+1)=110H+10 110H-104H=52-10=42 6H=42 H=7 HENCE T=H+1=7+1=8 THE NUMBER IS 780 SEE THE FOLLOWING EXAMPLE. ----------------------------------------------------------------------------- LET THE ORIGINAL NUMBER BE HTU WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND U IN UNITS DIGIT....ITS VALUE =100H+10T+U the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.thank you!! TENS DIGIT EXCEEDS HUNDREDS DIGIT BY T-H UNITS DIGIT EXCEEDS TENS DIGIT BY U-T THESE ARE SAME .HENCE T-H=U-T...OR 2T=U+H..................................................I WHEN DIGITS ARE REVERSED TNE NEW NUMBER BECOMES UTH...ITS VALUE = 100U+10T+H NEW NUMBER EXCEEDS ORIGINAL NUMBER BY =100U+10T+H-100H-10T-U =99U-99H=99(U-H) THIS IS EQUAL TO 198 HENCE 99(U-H)=198 U-H=198/99=2.........................................................II U+H=2T..................FROM....I ADDING,WE GET 2U=2T+2.....DIVIDING BY 2,WE GET U=T+1...OR........T=U-1......................AND H=U-2...........FROM.......II SO THE ORIGINAL NUMBER IS HTU = (U-2)(U-1)U...NOW SINCE U IS NOT SPECIFIED WE CAN HAVE VARIOUS VALUES FOR U STARTING FROM 9 TO 2...SINCE U-2 CANNOT BE NEGATIVE..SO THE SOLUTIONS ARE U=9..........789.............987-789=198 U=8...........678............876-678=198...ETC U=7..........567 U=6..........456 U=5............345 U=4............234 U=3............123 U=2.............012...WE MAY DISCARD THIS ALSO AS IT CANOT BE CONSIDERED AS 3 DIGIT NUMBER,THOUGH THIS ALSO SATISFIES THE CONDITION THAT 210-12=198 --------------------------------------------------------------------------------
 Numbers_Word_Problems/23553: 13)The sum of the digits of a two-digit number is 15.When the digits are reversed, the new number obtained is 9 more than the original number. Find the original number.1 solutions Answer 12361 by venugopalramana(3286)   on 2006-01-05 10:04:11 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK ---------------------------------------------------------------------------- A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!! LET THE ORIGINAL NUMBER BE HT0 WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND 0 IN UNITS DIGIT SINCE THE NUMBER IS DIVISIBLE BY 10....ITS VALUE =100H+10T WE ARE GIVEN THAT T-H=1.................I SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1 52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T 104H+52=100H+10(H+1)=110H+10 110H-104H=52-10=42 6H=42 H=7 HENCE T=H+1=7+1=8 THE NUMBER IS 780 SEE THE FOLLOWING EXAMPLE. ----------------------------------------------------------------------------- LET THE ORIGINAL NUMBER BE HTU WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND U IN UNITS DIGIT....ITS VALUE =100H+10T+U the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.thank you!! TENS DIGIT EXCEEDS HUNDREDS DIGIT BY T-H UNITS DIGIT EXCEEDS TENS DIGIT BY U-T THESE ARE SAME .HENCE T-H=U-T...OR 2T=U+H..................................................I WHEN DIGITS ARE REVERSED TNE NEW NUMBER BECOMES UTH...ITS VALUE = 100U+10T+H NEW NUMBER EXCEEDS ORIGINAL NUMBER BY =100U+10T+H-100H-10T-U =99U-99H=99(U-H) THIS IS EQUAL TO 198 HENCE 99(U-H)=198 U-H=198/99=2.........................................................II U+H=2T..................FROM....I ADDING,WE GET 2U=2T+2.....DIVIDING BY 2,WE GET U=T+1...OR........T=U-1......................AND H=U-2...........FROM.......II SO THE ORIGINAL NUMBER IS HTU = (U-2)(U-1)U...NOW SINCE U IS NOT SPECIFIED WE CAN HAVE VARIOUS VALUES FOR U STARTING FROM 9 TO 2...SINCE U-2 CANNOT BE NEGATIVE..SO THE SOLUTIONS ARE U=9..........789.............987-789=198 U=8...........678............876-678=198...ETC U=7..........567 U=6..........456 U=5............345 U=4............234 U=3............123 U=2.............012...WE MAY DISCARD THIS ALSO AS IT CANOT BE CONSIDERED AS 3 DIGIT NUMBER,THOUGH THIS ALSO SATISFIES THE CONDITION THAT 210-12=198 --------------------------------------------------------------------------------
 Numbers_Word_Problems/23555: 16.The sum of the digits of a three-digit number number is 11. The tens digit is three times the hundreds digit and twice the units digit.Find the number.1 solutions Answer 12360 by venugopalramana(3286)   on 2006-01-05 10:03:15 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK ---------------------------------------------------------------------------- A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!! LET THE ORIGINAL NUMBER BE HT0 WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND 0 IN UNITS DIGIT SINCE THE NUMBER IS DIVISIBLE BY 10....ITS VALUE =100H+10T WE ARE GIVEN THAT T-H=1.................I SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1 52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T 104H+52=100H+10(H+1)=110H+10 110H-104H=52-10=42 6H=42 H=7 HENCE T=H+1=7+1=8 THE NUMBER IS 780 SEE THE FOLLOWING EXAMPLE. ----------------------------------------------------------------------------- LET THE ORIGINAL NUMBER BE HTU WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND U IN UNITS DIGIT....ITS VALUE =100H+10T+U the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.thank you!! TENS DIGIT EXCEEDS HUNDREDS DIGIT BY T-H UNITS DIGIT EXCEEDS TENS DIGIT BY U-T THESE ARE SAME .HENCE T-H=U-T...OR 2T=U+H..................................................I WHEN DIGITS ARE REVERSED TNE NEW NUMBER BECOMES UTH...ITS VALUE = 100U+10T+H NEW NUMBER EXCEEDS ORIGINAL NUMBER BY =100U+10T+H-100H-10T-U =99U-99H=99(U-H) THIS IS EQUAL TO 198 HENCE 99(U-H)=198 U-H=198/99=2.........................................................II U+H=2T..................FROM....I ADDING,WE GET 2U=2T+2.....DIVIDING BY 2,WE GET U=T+1...OR........T=U-1......................AND H=U-2...........FROM.......II SO THE ORIGINAL NUMBER IS HTU = (U-2)(U-1)U...NOW SINCE U IS NOT SPECIFIED WE CAN HAVE VARIOUS VALUES FOR U STARTING FROM 9 TO 2...SINCE U-2 CANNOT BE NEGATIVE..SO THE SOLUTIONS ARE U=9..........789.............987-789=198 U=8...........678............876-678=198...ETC U=7..........567 U=6..........456 U=5............345 U=4............234 U=3............123 U=2.............012...WE MAY DISCARD THIS ALSO AS IT CANOT BE CONSIDERED AS 3 DIGIT NUMBER,THOUGH THIS ALSO SATISFIES THE CONDITION THAT 210-12=198 --------------------------------------------------------------------------------
 Numbers_Word_Problems/23557: can u help me please? ....the tens digit of a two-digit number is 3 less than the units digit. The number is four times the sum of the digits. Find the number.1 solutions Answer 12359 by venugopalramana(3286)   on 2006-01-05 10:02:10 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK ---------------------------------------------------------------------------- A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!! LET THE ORIGINAL NUMBER BE HT0 WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND 0 IN UNITS DIGIT SINCE THE NUMBER IS DIVISIBLE BY 10....ITS VALUE =100H+10T WE ARE GIVEN THAT T-H=1.................I SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1 52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T 104H+52=100H+10(H+1)=110H+10 110H-104H=52-10=42 6H=42 H=7 HENCE T=H+1=7+1=8 THE NUMBER IS 780 SEE THE FOLLOWING EXAMPLE. ----------------------------------------------------------------------------- LET THE ORIGINAL NUMBER BE HTU WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND U IN UNITS DIGIT....ITS VALUE =100H+10T+U the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.thank you!! TENS DIGIT EXCEEDS HUNDREDS DIGIT BY T-H UNITS DIGIT EXCEEDS TENS DIGIT BY U-T THESE ARE SAME .HENCE T-H=U-T...OR 2T=U+H..................................................I WHEN DIGITS ARE REVERSED TNE NEW NUMBER BECOMES UTH...ITS VALUE = 100U+10T+H NEW NUMBER EXCEEDS ORIGINAL NUMBER BY =100U+10T+H-100H-10T-U =99U-99H=99(U-H) THIS IS EQUAL TO 198 HENCE 99(U-H)=198 U-H=198/99=2.........................................................II U+H=2T..................FROM....I ADDING,WE GET 2U=2T+2.....DIVIDING BY 2,WE GET U=T+1...OR........T=U-1......................AND H=U-2...........FROM.......II SO THE ORIGINAL NUMBER IS HTU = (U-2)(U-1)U...NOW SINCE U IS NOT SPECIFIED WE CAN HAVE VARIOUS VALUES FOR U STARTING FROM 9 TO 2...SINCE U-2 CANNOT BE NEGATIVE..SO THE SOLUTIONS ARE U=9..........789.............987-789=198 U=8...........678............876-678=198...ETC U=7..........567 U=6..........456 U=5............345 U=4............234 U=3............123 U=2.............012...WE MAY DISCARD THIS ALSO AS IT CANOT BE CONSIDERED AS 3 DIGIT NUMBER,THOUGH THIS ALSO SATISFIES THE CONDITION THAT 210-12=198 --------------------------------------------------------------------------------
 Numbers_Word_Problems/23559: 18)A three-digit number between 300 and 400 is 40 times the sum of its digits. The tens digit is six more than the units digit. Find the number.1 solutions Answer 12358 by venugopalramana(3286)   on 2006-01-05 10:01:23 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK ---------------------------------------------------------------------------- A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!! LET THE ORIGINAL NUMBER BE HT0 WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND 0 IN UNITS DIGIT SINCE THE NUMBER IS DIVISIBLE BY 10....ITS VALUE =100H+10T WE ARE GIVEN THAT T-H=1.................I SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1 52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T 104H+52=100H+10(H+1)=110H+10 110H-104H=52-10=42 6H=42 H=7 HENCE T=H+1=7+1=8 THE NUMBER IS 780 SEE THE FOLLOWING EXAMPLE. ----------------------------------------------------------------------------- LET THE ORIGINAL NUMBER BE HTU WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND U IN UNITS DIGIT....ITS VALUE =100H+10T+U the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.thank you!! TENS DIGIT EXCEEDS HUNDREDS DIGIT BY T-H UNITS DIGIT EXCEEDS TENS DIGIT BY U-T THESE ARE SAME .HENCE T-H=U-T...OR 2T=U+H..................................................I WHEN DIGITS ARE REVERSED TNE NEW NUMBER BECOMES UTH...ITS VALUE = 100U+10T+H NEW NUMBER EXCEEDS ORIGINAL NUMBER BY =100U+10T+H-100H-10T-U =99U-99H=99(U-H) THIS IS EQUAL TO 198 HENCE 99(U-H)=198 U-H=198/99=2.........................................................II U+H=2T..................FROM....I ADDING,WE GET 2U=2T+2.....DIVIDING BY 2,WE GET U=T+1...OR........T=U-1......................AND H=U-2...........FROM.......II SO THE ORIGINAL NUMBER IS HTU = (U-2)(U-1)U...NOW SINCE U IS NOT SPECIFIED WE CAN HAVE VARIOUS VALUES FOR U STARTING FROM 9 TO 2...SINCE U-2 CANNOT BE NEGATIVE..SO THE SOLUTIONS ARE U=9..........789.............987-789=198 U=8...........678............876-678=198...ETC U=7..........567 U=6..........456 U=5............345 U=4............234 U=3............123 U=2.............012...WE MAY DISCARD THIS ALSO AS IT CANOT BE CONSIDERED AS 3 DIGIT NUMBER,THOUGH THIS ALSO SATISFIES THE CONDITION THAT 210-12=198 --------------------------------------------------------------------------------
 Numbers_Word_Problems/23560: help please!! /The units digit of a two-digit number is 5 more than the tens digit. If the sum of the digits is one-thir of the number, find the number.1 solutions Answer 12357 by venugopalramana(3286)   on 2006-01-05 10:00:33 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK ---------------------------------------------------------------------------- A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!! LET THE ORIGINAL NUMBER BE HT0 WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND 0 IN UNITS DIGIT SINCE THE NUMBER IS DIVISIBLE BY 10....ITS VALUE =100H+10T WE ARE GIVEN THAT T-H=1.................I SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1 52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T 104H+52=100H+10(H+1)=110H+10 110H-104H=52-10=42 6H=42 H=7 HENCE T=H+1=7+1=8 THE NUMBER IS 780 SEE THE FOLLOWING EXAMPLE. ----------------------------------------------------------------------------- LET THE ORIGINAL NUMBER BE HTU WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND U IN UNITS DIGIT....ITS VALUE =100H+10T+U the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.thank you!! TENS DIGIT EXCEEDS HUNDREDS DIGIT BY T-H UNITS DIGIT EXCEEDS TENS DIGIT BY U-T THESE ARE SAME .HENCE T-H=U-T...OR 2T=U+H..................................................I WHEN DIGITS ARE REVERSED TNE NEW NUMBER BECOMES UTH...ITS VALUE = 100U+10T+H NEW NUMBER EXCEEDS ORIGINAL NUMBER BY =100U+10T+H-100H-10T-U =99U-99H=99(U-H) THIS IS EQUAL TO 198 HENCE 99(U-H)=198 U-H=198/99=2.........................................................II U+H=2T..................FROM....I ADDING,WE GET 2U=2T+2.....DIVIDING BY 2,WE GET U=T+1...OR........T=U-1......................AND H=U-2...........FROM.......II SO THE ORIGINAL NUMBER IS HTU = (U-2)(U-1)U...NOW SINCE U IS NOT SPECIFIED WE CAN HAVE VARIOUS VALUES FOR U STARTING FROM 9 TO 2...SINCE U-2 CANNOT BE NEGATIVE..SO THE SOLUTIONS ARE U=9..........789.............987-789=198 U=8...........678............876-678=198...ETC U=7..........567 U=6..........456 U=5............345 U=4............234 U=3............123 U=2.............012...WE MAY DISCARD THIS ALSO AS IT CANOT BE CONSIDERED AS 3 DIGIT NUMBER,THOUGH THIS ALSO SATISFIES THE CONDITION THAT 210-12=198 --------------------------------------------------------------------------------
 Numbers_Word_Problems/23562: /. the tens digit of a two-digit number is 5 more than the units digit. The number is 8 times the sum of the digits. Find the number. I really need ur help..please!!1 solutions Answer 12356 by venugopalramana(3286)   on 2006-01-05 09:59:42 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK ---------------------------------------------------------------------------- A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!! LET THE ORIGINAL NUMBER BE HT0 WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND 0 IN UNITS DIGIT SINCE THE NUMBER IS DIVISIBLE BY 10....ITS VALUE =100H+10T WE ARE GIVEN THAT T-H=1.................I SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1 52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T 104H+52=100H+10(H+1)=110H+10 110H-104H=52-10=42 6H=42 H=7 HENCE T=H+1=7+1=8 THE NUMBER IS 780 SEE THE FOLLOWING EXAMPLE. ----------------------------------------------------------------------------- LET THE ORIGINAL NUMBER BE HTU WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND U IN UNITS DIGIT....ITS VALUE =100H+10T+U the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.thank you!! TENS DIGIT EXCEEDS HUNDREDS DIGIT BY T-H UNITS DIGIT EXCEEDS TENS DIGIT BY U-T THESE ARE SAME .HENCE T-H=U-T...OR 2T=U+H..................................................I WHEN DIGITS ARE REVERSED TNE NEW NUMBER BECOMES UTH...ITS VALUE = 100U+10T+H NEW NUMBER EXCEEDS ORIGINAL NUMBER BY =100U+10T+H-100H-10T-U =99U-99H=99(U-H) THIS IS EQUAL TO 198 HENCE 99(U-H)=198 U-H=198/99=2.........................................................II U+H=2T..................FROM....I ADDING,WE GET 2U=2T+2.....DIVIDING BY 2,WE GET U=T+1...OR........T=U-1......................AND H=U-2...........FROM.......II SO THE ORIGINAL NUMBER IS HTU = (U-2)(U-1)U...NOW SINCE U IS NOT SPECIFIED WE CAN HAVE VARIOUS VALUES FOR U STARTING FROM 9 TO 2...SINCE U-2 CANNOT BE NEGATIVE..SO THE SOLUTIONS ARE U=9..........789.............987-789=198 U=8...........678............876-678=198...ETC U=7..........567 U=6..........456 U=5............345 U=4............234 U=3............123 U=2.............012...WE MAY DISCARD THIS ALSO AS IT CANOT BE CONSIDERED AS 3 DIGIT NUMBER,THOUGH THIS ALSO SATISFIES THE CONDITION THAT 210-12=198 --------------------------------------------------------------------------------
 Numbers_Word_Problems/23563: 20) the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.1 solutions Answer 12355 by venugopalramana(3286)   on 2006-01-05 09:58:42 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK ---------------------------------------------------------------------------- A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!! LET THE ORIGINAL NUMBER BE HT0 WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND 0 IN UNITS DIGIT SINCE THE NUMBER IS DIVISIBLE BY 10....ITS VALUE =100H+10T WE ARE GIVEN THAT T-H=1.................I SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1 52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T 104H+52=100H+10(H+1)=110H+10 110H-104H=52-10=42 6H=42 H=7 HENCE T=H+1=7+1=8 THE NUMBER IS 780 SEE THE FOLLOWING EXAMPLE. ----------------------------------------------------------------------------- LET THE ORIGINAL NUMBER BE HTU WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND U IN UNITS DIGIT....ITS VALUE =100H+10T+U the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.thank you!! TENS DIGIT EXCEEDS HUNDREDS DIGIT BY T-H UNITS DIGIT EXCEEDS TENS DIGIT BY U-T THESE ARE SAME .HENCE T-H=U-T...OR 2T=U+H..................................................I WHEN DIGITS ARE REVERSED TNE NEW NUMBER BECOMES UTH...ITS VALUE = 100U+10T+H NEW NUMBER EXCEEDS ORIGINAL NUMBER BY =100U+10T+H-100H-10T-U =99U-99H=99(U-H) THIS IS EQUAL TO 198 HENCE 99(U-H)=198 U-H=198/99=2.........................................................II U+H=2T..................FROM....I ADDING,WE GET 2U=2T+2.....DIVIDING BY 2,WE GET U=T+1...OR........T=U-1......................AND H=U-2...........FROM.......II SO THE ORIGINAL NUMBER IS HTU = (U-2)(U-1)U...NOW SINCE U IS NOT SPECIFIED WE CAN HAVE VARIOUS VALUES FOR U STARTING FROM 9 TO 2...SINCE U-2 CANNOT BE NEGATIVE..SO THE SOLUTIONS ARE U=9..........789.............987-789=198 U=8...........678............876-678=198...ETC U=7..........567 U=6..........456 U=5............345 U=4............234 U=3............123 U=2.............012...WE MAY DISCARD THIS ALSO AS IT CANOT BE CONSIDERED AS 3 DIGIT NUMBER,THOUGH THIS ALSO SATISFIES THE CONDITION THAT 210-12=198 --------------------------------------------------------------------------------
 Numbers_Word_Problems/23561: 19) A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.1 solutions Answer 12354 by venugopalramana(3286)   on 2006-01-05 09:17:47 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK ---------------------------------------------------------------------------- A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!! LET THE ORIGINAL NUMBER BE HT0 WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND 0 IN UNITS DIGIT SINCE THE NUMBER IS DIVISIBLE BY 10....ITS VALUE =100H+10T WE ARE GIVEN THAT T-H=1.................I SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1 52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T 104H+52=100H+10(H+1)=110H+10 110H-104H=52-10=42 6H=42 H=7 HENCE T=H+1=7+1=8 THE NUMBER IS 780 SEE THE FOLLOWING EXAMPLE. ----------------------------------------------------------------------------- LET THE ORIGINAL NUMBER BE HTU WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND U IN UNITS DIGIT....ITS VALUE =100H+10T+U the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.thank you!! TENS DIGIT EXCEEDS HUNDREDS DIGIT BY T-H UNITS DIGIT EXCEEDS TENS DIGIT BY U-T THESE ARE SAME .HENCE T-H=U-T...OR 2T=U+H..................................................I WHEN DIGITS ARE REVERSED TNE NEW NUMBER BECOMES UTH...ITS VALUE = 100U+10T+H NEW NUMBER EXCEEDS ORIGINAL NUMBER BY =100U+10T+H-100H-10T-U =99U-99H=99(U-H) THIS IS EQUAL TO 198 HENCE 99(U-H)=198 U-H=198/99=2.........................................................II U+H=2T..................FROM....I ADDING,WE GET 2U=2T+2.....DIVIDING BY 2,WE GET U=T+1...OR........T=U-1......................AND H=U-2...........FROM.......II SO THE ORIGINAL NUMBER IS HTU = (U-2)(U-1)U...NOW SINCE U IS NOT SPECIFIED WE CAN HAVE VARIOUS VALUES FOR U STARTING FROM 9 TO 2...SINCE U-2 CANNOT BE NEGATIVE..SO THE SOLUTIONS ARE U=9..........789.............987-789=198 U=8...........678............876-678=198...ETC U=7..........567 U=6..........456 U=5............345 U=4............234 U=3............123 U=2.............012...WE MAY DISCARD THIS ALSO AS IT CANOT BE CONSIDERED AS 3 DIGIT NUMBER,THOUGH THIS ALSO SATISFIES THE CONDITION THAT 210-12=198 --------------------------------------------------------------------------------
 Numbers_Word_Problems/23558: 18)A three-digit number between 300 and 400 is 40 times the sum of its digits. The tens digit is six more than the units digit. Find the number.1 solutions Answer 12353 by venugopalramana(3286)   on 2006-01-05 09:16:43 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK ---------------------------------------------------------------------------- A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!! LET THE ORIGINAL NUMBER BE HT0 WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND 0 IN UNITS DIGIT SINCE THE NUMBER IS DIVISIBLE BY 10....ITS VALUE =100H+10T WE ARE GIVEN THAT T-H=1.................I SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1 52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T 104H+52=100H+10(H+1)=110H+10 110H-104H=52-10=42 6H=42 H=7 HENCE T=H+1=7+1=8 THE NUMBER IS 780 SEE THE FOLLOWING EXAMPLE. ----------------------------------------------------------------------------- LET THE ORIGINAL NUMBER BE HTU WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND U IN UNITS DIGIT....ITS VALUE =100H+10T+U the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.thank you!! TENS DIGIT EXCEEDS HUNDREDS DIGIT BY T-H UNITS DIGIT EXCEEDS TENS DIGIT BY U-T THESE ARE SAME .HENCE T-H=U-T...OR 2T=U+H..................................................I WHEN DIGITS ARE REVERSED TNE NEW NUMBER BECOMES UTH...ITS VALUE = 100U+10T+H NEW NUMBER EXCEEDS ORIGINAL NUMBER BY =100U+10T+H-100H-10T-U =99U-99H=99(U-H) THIS IS EQUAL TO 198 HENCE 99(U-H)=198 U-H=198/99=2.........................................................II U+H=2T..................FROM....I ADDING,WE GET 2U=2T+2.....DIVIDING BY 2,WE GET U=T+1...OR........T=U-1......................AND H=U-2...........FROM.......II SO THE ORIGINAL NUMBER IS HTU = (U-2)(U-1)U...NOW SINCE U IS NOT SPECIFIED WE CAN HAVE VARIOUS VALUES FOR U STARTING FROM 9 TO 2...SINCE U-2 CANNOT BE NEGATIVE..SO THE SOLUTIONS ARE U=9..........789.............987-789=198 U=8...........678............876-678=198...ETC U=7..........567 U=6..........456 U=5............345 U=4............234 U=3............123 U=2.............012...WE MAY DISCARD THIS ALSO AS IT CANOT BE CONSIDERED AS 3 DIGIT NUMBER,THOUGH THIS ALSO SATISFIES THE CONDITION THAT 210-12=198 --------------------------------------------------------------------------------
 test/23547: The units digit of a two-digit number is one-third the tens digit. when the digits are reversed, the new number is two more than four times the tens digit. Find the original number.1 solutions Answer 12352 by venugopalramana(3286)   on 2006-01-05 09:08:53 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK ---------------------------------------------------------------------------- A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!! LET THE ORIGINAL NUMBER BE HT0 WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND 0 IN UNITS DIGIT SINCE THE NUMBER IS DIVISIBLE BY 10....ITS VALUE =100H+10T WE ARE GIVEN THAT T-H=1.................I SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1 52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T 104H+52=100H+10(H+1)=110H+10 110H-104H=52-10=42 6H=42 H=7 HENCE T=H+1=7+1=8 THE NUMBER IS 780 SEE THE FOLLOWING EXAMPLE. ----------------------------------------------------------------------------- LET THE ORIGINAL NUMBER BE HTU WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND U IN UNITS DIGIT....ITS VALUE =100H+10T+U the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.thank you!! TENS DIGIT EXCEEDS HUNDREDS DIGIT BY T-H UNITS DIGIT EXCEEDS TENS DIGIT BY U-T THESE ARE SAME .HENCE T-H=U-T...OR 2T=U+H..................................................I WHEN DIGITS ARE REVERSED TNE NEW NUMBER BECOMES UTH...ITS VALUE = 100U+10T+H NEW NUMBER EXCEEDS ORIGINAL NUMBER BY =100U+10T+H-100H-10T-U =99U-99H=99(U-H) THIS IS EQUAL TO 198 HENCE 99(U-H)=198 U-H=198/99=2.........................................................II U+H=2T..................FROM....I ADDING,WE GET 2U=2T+2.....DIVIDING BY 2,WE GET U=T+1...OR........T=U-1......................AND H=U-2...........FROM.......II SO THE ORIGINAL NUMBER IS HTU = (U-2)(U-1)U...NOW SINCE U IS NOT SPECIFIED WE CAN HAVE VARIOUS VALUES FOR U STARTING FROM 9 TO 2...SINCE U-2 CANNOT BE NEGATIVE..SO THE SOLUTIONS ARE U=9..........789.............987-789=198 U=8...........678............876-678=198...ETC U=7..........567 U=6..........456 U=5............345 U=4............234 U=3............123 U=2.............012...WE MAY DISCARD THIS ALSO AS IT CANOT BE CONSIDERED AS 3 DIGIT NUMBER,THOUGH THIS ALSO SATISFIES THE CONDITION THAT 210-12=198 --------------------------------------------------------------------------------
 Numbers_Word_Problems/23564: 20) the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.1 solutions Answer 12351 by venugopalramana(3286)   on 2006-01-05 09:04:38 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK ---------------------------------------------------------------------------- A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!! LET THE ORIGINAL NUMBER BE HT0 WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND 0 IN UNITS DIGIT SINCE THE NUMBER IS DIVISIBLE BY 10....ITS VALUE =100H+10T WE ARE GIVEN THAT T-H=1.................I SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1 52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T 104H+52=100H+10(H+1)=110H+10 110H-104H=52-10=42 6H=42 H=7 HENCE T=H+1=7+1=8 THE NUMBER IS 780 SEE THE FOLLOWING EXAMPLE. ----------------------------------------------------------------------------- LET THE ORIGINAL NUMBER BE HTU WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND U IN UNITS DIGIT....ITS VALUE =100H+10T+U the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.thank you!! TENS DIGIT EXCEEDS HUNDREDS DIGIT BY T-H UNITS DIGIT EXCEEDS TENS DIGIT BY U-T THESE ARE SAME .HENCE T-H=U-T...OR 2T=U+H..................................................I WHEN DIGITS ARE REVERSED TNE NEW NUMBER BECOMES UTH...ITS VALUE = 100U+10T+H NEW NUMBER EXCEEDS ORIGINAL NUMBER BY =100U+10T+H-100H-10T-U =99U-99H=99(U-H) THIS IS EQUAL TO 198 HENCE 99(U-H)=198 U-H=198/99=2.........................................................II U+H=2T..................FROM....I ADDING,WE GET 2U=2T+2.....DIVIDING BY 2,WE GET U=T+1...OR........T=U-1......................AND H=U-2...........FROM.......II SO THE ORIGINAL NUMBER IS HTU = (U-2)(U-1)U...NOW SINCE U IS NOT SPECIFIED WE CAN HAVE VARIOUS VALUES FOR U STARTING FROM 9 TO 2...SINCE U-2 CANNOT BE NEGATIVE..SO THE SOLUTIONS ARE U=9..........789.............987-789=198 U=8...........678............876-678=198...ETC U=7..........567 U=6..........456 U=5............345 U=4............234 U=3............123 U=2.............012...WE MAY DISCARD THIS ALSO AS IT CANOT BE CONSIDERED AS 3 DIGIT NUMBER,THOUGH THIS ALSO SATISFIES THE CONDITION THAT 210-12=198 --------------------------------------------------------------------------------
 Numbers_Word_Problems/23565: 20) the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.1 solutions Answer 12350 by venugopalramana(3286)   on 2006-01-05 09:03:13 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK ---------------------------------------------------------------------------- A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!! LET THE ORIGINAL NUMBER BE HT0 WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND 0 IN UNITS DIGIT SINCE THE NUMBER IS DIVISIBLE BY 10....ITS VALUE =100H+10T WE ARE GIVEN THAT T-H=1.................I SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1 52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T 104H+52=100H+10(H+1)=110H+10 110H-104H=52-10=42 6H=42 H=7 HENCE T=H+1=7+1=8 THE NUMBER IS 780 SEE THE FOLLOWING EXAMPLE. ----------------------------------------------------------------------------- LET THE ORIGINAL NUMBER BE HTU WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND U IN UNITS DIGIT....ITS VALUE =100H+10T+U the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.thank you!! TENS DIGIT EXCEEDS HUNDREDS DIGIT BY T-H UNITS DIGIT EXCEEDS TENS DIGIT BY U-T THESE ARE SAME .HENCE T-H=U-T...OR 2T=U+H..................................................I WHEN DIGITS ARE REVERSED TNE NEW NUMBER BECOMES UTH...ITS VALUE = 100U+10T+H NEW NUMBER EXCEEDS ORIGINAL NUMBER BY =100U+10T+H-100H-10T-U =99U-99H=99(U-H) THIS IS EQUAL TO 198 HENCE 99(U-H)=198 U-H=198/99=2.........................................................II U+H=2T..................FROM....I ADDING,WE GET 2U=2T+2.....DIVIDING BY 2,WE GET U=T+1...OR........T=U-1......................AND H=U-2...........FROM.......II SO THE ORIGINAL NUMBER IS HTU = (U-2)(U-1)U...NOW SINCE U IS NOT SPECIFIED WE CAN HAVE VARIOUS VALUES FOR U STARTING FROM 9 TO 2...SINCE U-2 CANNOT BE NEGATIVE..SO THE SOLUTIONS ARE U=9..........789.............987-789=198 U=8...........678............876-678=198...ETC U=7..........567 U=6..........456 U=5............345 U=4............234 U=3............123 U=2.............012...WE MAY DISCARD THIS ALSO AS IT CANOT BE CONSIDERED AS 3 DIGIT NUMBER,THOUGH THIS ALSO SATISFIES THE CONDITION THAT 210-12=198 --------------------------------------------------------------------------------
 Numbers_Word_Problems/23566: /.. I really can't understand this. please help.. The sum of the digits of a two-digit number is 13. The units digit exceeds twice the tens digit by 1. Find the number.1 solutions Answer 12349 by venugopalramana(3286)   on 2006-01-05 08:53:25 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK ---------------------------------------------------------------------------- A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!! LET THE ORIGINAL NUMBER BE HT0 WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND 0 IN UNITS DIGIT SINCE THE NUMBER IS DIVISIBLE BY 10....ITS VALUE =100H+10T WE ARE GIVEN THAT T-H=1.................I SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1 52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T 104H+52=100H+10(H+1)=110H+10 110H-104H=52-10=42 6H=42 H=7 HENCE T=H+1=7+1=8 THE NUMBER IS 780 SEE THE FOLLOWING EXAMPLE. ----------------------------------------------------------------------------- LET THE ORIGINAL NUMBER BE HTU WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND U IN UNITS DIGIT....ITS VALUE =100H+10T+U the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.thank you!! TENS DIGIT EXCEEDS HUNDREDS DIGIT BY T-H UNITS DIGIT EXCEEDS TENS DIGIT BY U-T THESE ARE SAME .HENCE T-H=U-T...OR 2T=U+H..................................................I WHEN DIGITS ARE REVERSED TNE NEW NUMBER BECOMES UTH...ITS VALUE = 100U+10T+H NEW NUMBER EXCEEDS ORIGINAL NUMBER BY =100U+10T+H-100H-10T-U =99U-99H=99(U-H) THIS IS EQUAL TO 198 HENCE 99(U-H)=198 U-H=198/99=2.........................................................II U+H=2T..................FROM....I ADDING,WE GET 2U=2T+2.....DIVIDING BY 2,WE GET U=T+1...OR........T=U-1......................AND H=U-2...........FROM.......II SO THE ORIGINAL NUMBER IS HTU = (U-2)(U-1)U...NOW SINCE U IS NOT SPECIFIED WE CAN HAVE VARIOUS VALUES FOR U STARTING FROM 9 TO 2...SINCE U-2 CANNOT BE NEGATIVE..SO THE SOLUTIONS ARE U=9..........789.............987-789=198 U=8...........678............876-678=198...ETC U=7..........567 U=6..........456 U=5............345 U=4............234 U=3............123 U=2.............012...WE MAY DISCARD THIS ALSO AS IT CANOT BE CONSIDERED AS 3 DIGIT NUMBER,THOUGH THIS ALSO SATISFIES THE CONDITION THAT 210-12=198 --------------------------------------------------------------------------------
 Numbers_Word_Problems/23567: 11) The units digit of a two-digit number is one-third the tens digit. When the digits are reversed, the new number is two more than four times the tens digit. Find the original number. thank you!!1 solutions Answer 12348 by venugopalramana(3286)   on 2006-01-05 08:52:18 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK ---------------------------------------------------------------------------- A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!! LET THE ORIGINAL NUMBER BE HT0 WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND 0 IN UNITS DIGIT SINCE THE NUMBER IS DIVISIBLE BY 10....ITS VALUE =100H+10T WE ARE GIVEN THAT T-H=1.................I SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1 52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T 104H+52=100H+10(H+1)=110H+10 110H-104H=52-10=42 6H=42 H=7 HENCE T=H+1=7+1=8 THE NUMBER IS 780 SEE THE FOLLOWING EXAMPLE. ----------------------------------------------------------------------------- LET THE ORIGINAL NUMBER BE HTU WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND U IN UNITS DIGIT....ITS VALUE =100H+10T+U the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.thank you!! TENS DIGIT EXCEEDS HUNDREDS DIGIT BY T-H UNITS DIGIT EXCEEDS TENS DIGIT BY U-T THESE ARE SAME .HENCE T-H=U-T...OR 2T=U+H..................................................I WHEN DIGITS ARE REVERSED TNE NEW NUMBER BECOMES UTH...ITS VALUE = 100U+10T+H NEW NUMBER EXCEEDS ORIGINAL NUMBER BY =100U+10T+H-100H-10T-U =99U-99H=99(U-H) THIS IS EQUAL TO 198 HENCE 99(U-H)=198 U-H=198/99=2.........................................................II U+H=2T..................FROM....I ADDING,WE GET 2U=2T+2.....DIVIDING BY 2,WE GET U=T+1...OR........T=U-1......................AND H=U-2...........FROM.......II SO THE ORIGINAL NUMBER IS HTU = (U-2)(U-1)U...NOW SINCE U IS NOT SPECIFIED WE CAN HAVE VARIOUS VALUES FOR U STARTING FROM 9 TO 2...SINCE U-2 CANNOT BE NEGATIVE..SO THE SOLUTIONS ARE U=9..........789.............987-789=198 U=8...........678............876-678=198...ETC U=7..........567 U=6..........456 U=5............345 U=4............234 U=3............123 U=2.............012...WE MAY DISCARD THIS ALSO AS IT CANOT BE CONSIDERED AS 3 DIGIT NUMBER,THOUGH THIS ALSO SATISFIES THE CONDITION THAT 210-12=198 --------------------------------------------------------------------------------
 Numbers_Word_Problems/23568: 11) The units digit of a two-digit number is one-third the tens digit. When the digits are reversed, the new number is two more than four times the tens digit. Find the original number. thank you!!1 solutions Answer 12347 by venugopalramana(3286)   on 2006-01-05 08:51:24 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING EXAMPLES AND TRY TO WORK OUT YOUR PROBLEM.IF YOU STILL HAVE DIFFICULTY COME BACK ---------------------------------------------------------------------------- A three-digit number divisible by ten has a hundreds digit that is one less than its tens digit. The number is 52 times the sum of the digits. find the number.Thank you!! LET THE ORIGINAL NUMBER BE HT0 WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND 0 IN UNITS DIGIT SINCE THE NUMBER IS DIVISIBLE BY 10....ITS VALUE =100H+10T WE ARE GIVEN THAT T-H=1.................I SUM OF DIGITS=H+T+0=H+T...BUT T=H+1 FROM I..SO SUM OF DIGITS =H+H+1=2H+1 52 TIMES THE ABOVE =52(2H+1)=VALUE OF NUMBER =100H+10T 104H+52=100H+10(H+1)=110H+10 110H-104H=52-10=42 6H=42 H=7 HENCE T=H+1=7+1=8 THE NUMBER IS 780 SEE THE FOLLOWING EXAMPLE. ----------------------------------------------------------------------------- LET THE ORIGINAL NUMBER BE HTU WITH H IN HUNDREDS DIGIT ,T IN TENS DIGIT AND U IN UNITS DIGIT....ITS VALUE =100H+10T+U the tens digit of a three-digit number exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. When the digits are reversed, the new number exceeds the original numer by 198. Find the number.thank you!! TENS DIGIT EXCEEDS HUNDREDS DIGIT BY T-H UNITS DIGIT EXCEEDS TENS DIGIT BY U-T THESE ARE SAME .HENCE T-H=U-T...OR 2T=U+H..................................................I WHEN DIGITS ARE REVERSED TNE NEW NUMBER BECOMES UTH...ITS VALUE = 100U+10T+H NEW NUMBER EXCEEDS ORIGINAL NUMBER BY =100U+10T+H-100H-10T-U =99U-99H=99(U-H) THIS IS EQUAL TO 198 HENCE 99(U-H)=198 U-H=198/99=2.........................................................II U+H=2T..................FROM....I ADDING,WE GET 2U=2T+2.....DIVIDING BY 2,WE GET U=T+1...OR........T=U-1......................AND H=U-2...........FROM.......II SO THE ORIGINAL NUMBER IS HTU = (U-2)(U-1)U...NOW SINCE U IS NOT SPECIFIED WE CAN HAVE VARIOUS VALUES FOR U STARTING FROM 9 TO 2...SINCE U-2 CANNOT BE NEGATIVE..SO THE SOLUTIONS ARE U=9..........789.............987-789=198 U=8...........678............876-678=198...ETC U=7..........567 U=6..........456 U=5............345 U=4............234 U=3............123 U=2.............012...WE MAY DISCARD THIS ALSO AS IT CANOT BE CONSIDERED AS 3 DIGIT NUMBER,THOUGH THIS ALSO SATISFIES THE CONDITION THAT 210-12=198 --------------------------------------------------------------------------------