Q1.1. 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
Answers
Explanation:
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
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:
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