1In a two digit number, the tens digit greater than the units digit by 1.If the sum
of the square of the two digits is smaller than the number itself by 4,find the
number.
ANS is 65
2.Jack cycled from one place A to another place B at constant speed. If he had
increased his speed by 3 km/h, he would have arrived at B half an hour earlier.
Given that the distance between A and B is 30 km, find jack’s original cycling
Speed.
ANS is 12km/h
3.Joan’s mother is 3 years younger than Joan’s father, and Joan’s age is
exactly the square root of her mother’s age. If the sum of the ages of Joan and her father is 45, find the age of Joan’s father.
ANS is 39
THX!!!
MATHS...simultaneous equation
2010-07-12 12:54 am
回答 (2)
2010-07-12 4:47 am
2010-07-12 4:50 am
1. Let the tens digit be X , the units digit be X-1.
X^2 + ( X - 1 )^2 = ( 10X + ( X-1 ) ) - 4
2X^2 - 2X + 1 = 11X - 5
2X^2 - 13X + 6 = 0
( 2X - 1 ) ( X - 6 ) = 0
X = 1/2 ( rejected ) or 6
So the tens digit is 6. i.e. the unit digit is 5
So the number is 65.
2. Let x km/h be the jack’s original cycling speed. h hours be the time taken for Jack to cycled from one place A to another place B .
( x ) ( h ) = 30
h = 30 / x ---> ( i )
( x + 3 ) ( h - 0.5 ) = 30
xh - 0.5x + 3h - 1.5 = 30
xh - 0.5x + 3h = 31.5 ---> ( ii )
Sub ( i ) into ( ii ) ,
( x ) ( 30 / x ) - 0.5x + 3 ( 30/x ) = 31.5
30 - 0.5x + 90/x = 31.5
0.5x^2 + 1.5x - 90 = 0
x^2 + 3x - 180 = 0
( x+15 ) ( x-12 ) = 0
x = -15 ( rejected ) or 12
So the original speed is 12km/h
3. Let x be the age of joan.
( x^2 + 3 ) + ( x ) = 45 [ ( x^2 + 3 ) be the age of joan's father ]
x^2 + x - 42 = 0
( x-6 ) ( x+7 ) = 0
x = 6 or -7 ( rejected )
So the age of joan's father :
6^2 + 3
= 36 + 3
= 39
If you have any questions to the above solution , you can ask me and I will explain it to you in details ^^
X^2 + ( X - 1 )^2 = ( 10X + ( X-1 ) ) - 4
2X^2 - 2X + 1 = 11X - 5
2X^2 - 13X + 6 = 0
( 2X - 1 ) ( X - 6 ) = 0
X = 1/2 ( rejected ) or 6
So the tens digit is 6. i.e. the unit digit is 5
So the number is 65.
2. Let x km/h be the jack’s original cycling speed. h hours be the time taken for Jack to cycled from one place A to another place B .
( x ) ( h ) = 30
h = 30 / x ---> ( i )
( x + 3 ) ( h - 0.5 ) = 30
xh - 0.5x + 3h - 1.5 = 30
xh - 0.5x + 3h = 31.5 ---> ( ii )
Sub ( i ) into ( ii ) ,
( x ) ( 30 / x ) - 0.5x + 3 ( 30/x ) = 31.5
30 - 0.5x + 90/x = 31.5
0.5x^2 + 1.5x - 90 = 0
x^2 + 3x - 180 = 0
( x+15 ) ( x-12 ) = 0
x = -15 ( rejected ) or 12
So the original speed is 12km/h
3. Let x be the age of joan.
( x^2 + 3 ) + ( x ) = 45 [ ( x^2 + 3 ) be the age of joan's father ]
x^2 + x - 42 = 0
( x-6 ) ( x+7 ) = 0
x = 6 or -7 ( rejected )
So the age of joan's father :
6^2 + 3
= 36 + 3
= 39
If you have any questions to the above solution , you can ask me and I will explain it to you in details ^^
收錄日期: 2021-04-22 00:48:39
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