Question

1)A's salary is same as 4 times B's salary. If together they earn Rs 3750 a month ,find the salary of each.

2)Seperate 178 into two parts so that the first part is 8 less than twice the second part.

3)The difference two no.s is 3 and the difference of their squares is 69.Find the numbers.

4)A man is 42 yrs old and his son is 12 yrs old. In how manu yrs will the age of the son is half the age of the man at that time?

5) The present age of a man is twice that of his son. 8 yrs hence their ages will be in ratio 7:4. Find their present ages.

Answer 1

1)
Let B's salary be x
then A's salary is 4x
By the problem , x +4x = 3750
5x=3750
x=3750/5 = 750
Therefore, A's salary is 4(750) = Rs.3000
B's salary =Rs.750

2)
Let the second part be 'x'
Then the first part is 2x-8
By the problem,x+2x-8=178
3x-8=178
3x=178+8=186
x=186/3 = 62
So the first part is 2x-8 = 2(62)-8 = 124-8=116
Second part is x = 62

3)
Let the numbers be x and y

The question says, their difference is 3

So, x - y = 3

x = 3 + y

Now we get the 2 numbers as y and (y+3)

Also, difference of their squares is 69

That is, x² - y² = 69

But x = y+3

So, (y+3)² - y² = 69

y² + 3² + 2(3)(y) - y² = 69

9 + 6y = 69

6y = 69 - 9 = 60

y = 60/6 = 10

x = y + 3 = 10 + 3 = 13

Therefore, the numbers are 10 and 13.

4)

Let us assume that it takes ‘x’ years for the son to be half of father’s age

So after ‘x’ years

Father’s age =42+x

Son’s age =12+x

Then as per given condition

Son’s age =1/2 of father’s age

That is 12+x=1/2(42+x)

2(12+x)=42+x

24+2x=42+x

2x-x=42–24

x=18

So there you go it takes 18years for the son to be half of his father’s age.

5)
Present age of son=x
present age of father=2x
after 8 years,

2x+8/x+8=7/4
8x+32=7x+56
x=24
father's age is 48 and son's age is 24