solve blue questions for 50. points
Answers
2. let the numerator be x then the denominator would be x+3
fraction=x/(x+3)
A.T.Q.
x+1/x+3+1=2/3
x+1/x+4=2/3
(x+1)×3=(x+4)×2
3x+3=2x+8
3x-2x=11
x=11
3. Ratio of ages of Reema and Seema=7:2
let the ages be 7x and 2x respectively
8 year hence,
age of Reema=7x+8
age of Seema=2x+8
A.T.Q.
(7x+8)/(2x+8)=5/2
(7x+8)×2=(2x+8)×5
14x+16=10x+40
14x-10x=40-16
4x=24
x=6
present age of Reema=7x=42
present age of Seema=2x=12
4.let the unit's place digit be x
and tens place digit be y
So, the no.is 10y+x
A.T.Q.
x+y=7 .........(1)
also, A.T.Q.
10y+x+27=10x+y
9y-9x=-27
-9(x-y)=-27
x-y=3 .........(2)
On adding (1) and (2), we get,
x+y+x-y=7+3
2x=10
=>x=5
By putting x=5 in (1), we get,
5+y=7
=>y=2
So, the no. is 25
6. let the no. be x
A.T.Q.
5x-68=x
5x-x=68
4x=68
=>x=17
7. let age of John's sister be x
then,A.T.Q.
x/2+7=35
x/2=28
=>x=56
8.let the smaller no. be x
so, the larger no. be x+1
A.T.Q.
x+x+1=45
2x+1=45
2x=44
=>x=22
So, the smaller no. is 22 and larger no. is 23
2. let the numerator be x then the denominator would be x+3
fraction=x/(x+3)
A.T.Q.
x+1/x+3+1=2/3
x+1/x+4=2/3
(x+1)×3=(x+4)×2
3x+3=2x+8
3x-2x=11
x=11
3. Ratio of ages of Reema and Seema=7:2
let the ages be 7x and 2x respectively
8 year hence,
age of Reema=7x+8
age of Seema=2x+8
A.T.Q.
(7x+8)/(2x+8)=5/2
(7x+8)×2=(2x+8)×5
14x+16=10x+40
14x-10x=40-16
4x=24
x=6
present age of Reema=7x=42
present age of Seema=2x=12
4.let the unit's place digit be x
and tens place digit be y
So, the no.is 10y+x
A.T.Q.
x+y=7 .........(1)
also, A.T.Q.
10y+x+27=10x+y
9y-9x=-27
-9(x-y)=-27
x-y=3 .........(2)
On adding (1) and (2), we get,
x+y+x-y=7+3
2x=10
=>x=5
By putting x=5 in (1), we get,
5+y=7
=>y=2
So, the no. is 25
6. let the no. be x
A.T.Q.
5x-68=x
5x-x=68
4x=68
=>x=17
7. let age of John's sister be x
then,A.T.Q.
x/2+7=35
x/2=28
=>x=56
8.let the smaller no. be x
so, the larger no. be x+1
A.T.Q.
x+x+1=45
2x+1=45
2x=44
=>x=22
So, the smaller no. is 22 and larger no. is 23