Question

1. Manish's mother is one more than 5 times his present age.
4 years ago the product of their ages is 220. Find their present​

Answer 1

Given

Manish's Mother is one more than 5 times his present age.

4 years ago their product of their ages was 220.

To Find

We have to find their present ages

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

Let the present age of Manish be 'x'

Then ,his mother's present age be = 1+5x

4 years ago ,

➞Age of Manish 4 years ago = x-4

➞Age of Manish's mother 4 years ago = (1+5x)-4

According to the question:

Their product of their ages is 220 then,

➞(x-4)(1+5x)-4=220

➞(x-4)(5x+1)=220+4

➞x(5x+1)-4(5x+1)=224

➞5x²+x-20x-4=224

➞5x²-19x=228

➞5x²-19x-228=0

Factorise it by using quadratic formula

a=5,b= -19 & c= -228

x=-b±√b²-4ac ÷2a

➞x=-(-19)±√(-19)²-4(5)(-228) /2(5)

➞x=19±√361-(-4560) /10

➞x= 19±√4921/10

➞x=19±√4921/10

➞x=19/10±√4921/10

➞x= 8.91 or x= -5.11

Hence,age can't be negative so,we will take positive value of x

Hence,x = 8.91 so,we take 9 ( rounded off value)

Therefore, Present age of Manish is 9 years

Age of Manish's mother = 1+5x= 46 years

Answer 2

Answer:

Let the present age of Manish be 'x'

Then ,his mother's present age be = 1+5x

4 years ago ,

➞Age of Manish 4 years ago = x-4

➞Age of Manish's mother 4 years ago = (1+5x)-4

According to the question:

Their product of their ages is 220 then,

➞(x-4)(1+5x)-4=220

➞(x-4)(5x+1)=220+4

➞x(5x+1)-4(5x+1)=224

➞5x²+x-20x-4=224

➞5x²-19x=228

➞5x²-19x-228=0

Factorise it by using quadratic formula

a=5,b= -19 & c= -228

x=-b±√b²-4ac ÷2a

➞x=-(-19)±√(-19)²-4(5)(-228) /2(5)

➞x=19±√361-(-4560) /10

➞x= 19±√4921/10

➞x=19±√4921/10

➞x=19/10±√4921/10

➞x= 8.91 or x= -5.11

Hence,age can't be negative so,we will take positive value of x

Hence,x = 8.91 so,we take 9 ( rounded off value)

Therefore, Present age of Manish is 9 years

Age of Manish's mother = 1+5x= 46 years

Step-by-step explanation:

Hope it helps dear mark as brainliest please...