Question

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

Answer 1

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

Answer 2

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