1.Chandrika's mother was 4 times her age 10 years ago.She will be twice as old as chandrika from now.find present age of candida's mother.
Pls answer this question with working
2.The smallest of three consecutive number is added to twice the largest.The result obtained is 7 less than 4 times the middle number.find the numbers. Answer them with working

2
are you sure 1st ques is complete... i'll solving 2nd one, review 1st ques one more time
1st question is incomplete
smthng needs to be filled in it
Its twice as old as chandrika 10 years from now...sry for the mistake
Guys pls answer 2nd one

Answers

The Brainliest Answer!
2014-12-30T19:10:02+05:30
1.
C is Chandrika
M is Chandrika's Mom

At present time,
age of C = x
age of M = y

Before 10 years,
age of C = (x - 10)
age of M = (y - 10)

After 10 years
age of C = (x + 10)
age of M = (y + 10)

now,
(y - 10) = 4 * (x - 10) => 4 * x - y = 30
(y + 10) = 2 * (x + 10) => 2 * x - y = -10

Solving above two equations, x = 20, y = 50

2. Three consecutive no. x - 1, x and x + 1
Now, (x - 1) + 2 * (x + 1) = 4 * x - 7
=> 3 * x + 1 = 4 * x - 7
=> x = 8
So, no. are 7, 8, 9
1 5 1
Thanks#PM using....It is great help
Thanks#AvmnuSng:-)
I solved 1st ques... you can check it
ya good
Thanks
2014-12-31T09:28:14+05:30
1. Let Chandrika's mothers age now = x,
               10 yrs ago her age= x-10
               10 yr hence her age= x+10
ATQ       10 yr ago  Chandrika's age = (x-10)/4 yr
ATQ      10 yr hence her age = (x+10)/2
Then we have (x-10)/4 + 20 = (x+10)/2
          Solving    x -10 + 80 = 2x +20
                                      x = 50
Present age of Chandrika's mother is 50 yrs.

2. Let the consecutive no.s be x-1, x and x+1
ATQ  (x-1) + 2(x+1) = 4x -7
     ⇒    x-1 + 2x +2 = 4x -7
     ⇒    x = 8
Then we have the numbers  8-1, 8, 8+1 i.e 7,8,9
0