Solutions To Some Age Problems Post
Question no 1
2d - 3 = m............. eq 1
m = 63 - d .............eq 2....... > m + d
= 63 given
put in eq 1
2d - 3 = 63 - d
3d = 66
d = 22
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Question no 2
f - 10 = 7(s-10)
f - 10 = 7s - 70
f = 7s - 70 + 10
f = 7s - 60
:
2(f + 2) = 5(s+2)
2f + 4 = 5s + 10
2f = 5s + 10 - 4
2f = 5s + 6
Replace f with (7s-60)
2(7s - 60) = 5s + 6
14s - 120 = 5s + 6
14s = 5s + 6 + 120
14s - 5s = 126
9s = 126
s =
s = 14 yrs, son's present age
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Question no 3
f = 3s
2s + 15 = f
2s + 15 = 3s
s = 15
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Question no 4
THE SUM OF THE AGES OF
SADIA AND MOTHER IS 70 Y
sadia = s
mothrt = m
s + m = 70 ............eq 1
10 Y EARLER, HER AGE WAS 1/4 OF HER MOTHER'S
AGE
s - 10 = 1/4 ( m - 10 ) ......eq 2
eq 2 multiply by 4 both sides
we get
4s - 40 = m - 10 ......eq 3
from eq 1
s = 70 - m ...........eq 4
put the value of s in eq 3 we get
4 ( 70 - m ) - 40 = m - 10
280 - 30 = 5m
250 = 5m
m = 50
Question no 5
Question no 6
{ 44 - ( x - 8 ) }/(x - 8 ) = 6/5
{44 - x + 8 }/ (x - 8 ) = 6/5
5( 52 - x ) = 6x - 48
260 - 5x = 6x - 48
x = 28
Question 6 another method
A + B = 44
A /B = 6/5
A = B 6/5
6/5 B + B = 44
11B = 220
B = 20
after 8 years
B = 20 + 8 = 28
Question no 7
f = 15s
f + 18 = 3 (s +18 )
f + 18 = 3f/15 + 18
5f + 90 = f + 270
4f = 180
f = 45
Question no 8
f = 54
f - 4 = 5( s - 4 )
54 - 4 = 5s - 20
50 + 20 = 5s
s = 14
after 5 years
s = 19
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Question 9
x/3 + 20 = x
x + 60 = 3x
2x = 60
x= 30
Question no 10
a - b = 2
a^2 - b^2 = 16
a = 2 + b
(2 + b )^2 - b^2 = 16
4 + 4b + b^2 - b^2 = 16
4b = 12
b = 3
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