Problems on ages

4 years back the age of husband and wife = \(25\times 2\)= 50 years

After 4 years , the sum will be increase by \(2\times4 \)=8

Sum of the present age of husband and wife = 50+8 =58 years

Sum of the present age of husband, wife and child = \(3\times 20 = 60 years\)

Child's present age = 60 - 58 = 2 years.

Sum of present ages = 68 years

Sum of ages 6 years ago = 68- (6+6) = 56 years

Ratio of ages of son and father 6 year ago = 1 : 2.5 = 2 : 5

Sum of ratios = 2+5= 7 ratio = 56 years

\( 1\: ratio = \frac{56}{7}=\:8\:years\\\\ Age\:of\:father\:6\:years\:ago=\:5\:ratio=\:5\times 8 =\:40\:years \)

Present age of father = 40+6= 46 years

\( A\::\:\frac{A+B}{2}=\:16\::\:19\\\\ \frac{A}{\frac{A+B}{2}}=\:\frac{16}{19}\\\\ \frac{2A}{A+B}=\:\frac{16}{19}\\\\ 38A=\:16A+16B\\\\ 22A=\:16B\\\\ \frac{A}{B}=\:\frac{16}{22}=\:\frac{8}{11}\\\\ A\::\:B=\:16\::22=\:8\::\:11\\\\ Ratio\:of\:present\:ages\:of\:A\:and\:B=\:8\::\:11\\\\ Ratio\:of\:ages\:after\:8\:years=\:10\::\:13\\\\ Ratio\:difference\:same=\:2\:ratio=\:8\:years\\\\ 1\:ratio=\:4\:years\\\\ Present\:age\:of\:B=\:11\:ratio=\:11\times 4=\:44\:years \)

Let the father's age is 5x years.

Sum of present ages of \( X, Y \) &\( Z \) = \( 126 \) years
10 years ago, Sum of the ages, = \( 96 \) years
Ratio =\( 5 : 6 : 1 \)
\( (5 + 6 + 1) p = 96 \)
\( 12p = 96 \)
\( p = 8 \) years
Present age of \( Y = (48 + 10 ) = 58 \) years
Present age of \( Z = (8 + 10 ) = 18 \) years
Sum of their ages = \( 76 \) years.

let the total number of members in the group be x .

Let B's age 6 years back = x.

Age of Rubi \(4\) years ago \(= 16x\: years\)
Age of Sri \(4\) years ago \(= 11x\: years\)
Then,
\(16x+4-11x-4=15\)
\(5x=15\: years\)
\(x=3\: years\)
Thus, Rubi’s age \(4\) years ago \(= 16x=16\times 3=48\: years\)
Then Rubi’s present age \(= 52\: years\)
And Mala’s present age \(=(\frac{1}{4})\times 52= 13\:years\)
Thus Mala’s age after \(23\: years=13+23=36 \:years\)

As per question,
\(\frac{Ram}{Sam} =\frac{6}{5}\)
\(\frac{Sam}{Venu} = \frac{5}{8}\)
Ratio of the ages of Ram, Sam and Venu\( = 6: 5: 8\)
Renu’s age\( = 6x \times \frac{150}{100} = 9x\)
\((6x + 5x + 8x + 9x) = 42 \times 4\)
\(28x = 168\)
\(x = 6\:years\)
Sam age \(= 5 \times 6 = 30 \:years\)
Vinu \(= 30 – 6 = 24\: years\)
Venu \(= 8 \times 6 = 48\: years\)
Renu\( = 9 \times 6 = 54\: years\)
Required average \(= \frac{(54 + 48 + 24)}{3} = 42 \:years\)

Let 'x' be the number of girls
Total age of girls = 14 x ............(1)
No : of boys = (750-x)
Total age of boys = 15\(\times\) (750-x).............(2)
Average age of total students = 14 years and 6 months = \(14 \tfrac{6}{12}\) = \(\frac{29}{2}\)
Total age of all students = 750\(\times\) \(\frac{29}{2}\) = 10875................(3)
From (1),(2) &(3)
14 x + (15\(\times\) (750-x) ) = 10875
14 x + 11250 - 15 x = 10875
x = 375
No : of boys = 750 - 375 = 375

Let the age of A = x

Age of B = x+8

Age of C = \( \frac{x+x+8}{2}=\:\frac{2x+8}{2}=\:x+4 \)

Average age of A, B and C = 40 years

\(\frac{x+x+8+x+4}{3}=\:40\\\\ \frac{3x+12}{3}=\:40\\\\ 3x+12=\:120\\\\ 3x=\:108\\\\ x=\:\frac{108}{3}=\:36\\\\\)

Age of A = 36

Age of B = 36+8= 44

Age of C = 36+4= 40

Ratio of ages after 6 years = 36+6 : 44+6 : 40+6

= 42 : 50 : 46

= 21 : 25 : 23

Let one year ago
Age of son = \( x \) , Age of Raju = \( 6x \)
Then Present ages,
Son =\( x + 1 \), Raju =\( 6x + 1 \)
After 8 years ages
Son = \( x + 1 + 8 = x + 9 \)
Raju = \( 6x + 1 + 8 = 6x + 9 \)
\( 6x + 9 = 2(x + 9) + 7 \)
\( 4x = 16 \)
\( x = 4 \)
\(\therefore\) Present age of son is\( x + 1 = \)5 years

Let the present age of the elder person = x

Present age of the younger person = y

x - y = 20

x = 20 + y

( 20 + y) - 10 = 5(y - 10)

10 + y = 5y - 50

4y = 60

y = 15

Age of the elder person = 15 + 20

= 35 years

Sum of ages of A and B = 47 years

Difference of ages = \( \frac{23\:\frac{19}{47}}{100}\times 47\\\\ =\:\frac{1100}{4700}\times 47\\\\ =\:11\:years \)

A+B = 47

A - B = 11

2A= 58

A = 29 years

B = 47-29 = 18 years

Required ratio = 29+6 : 18+3

= 35 : 21

= 5 : 3

Let 5x be Abi's present age then 7x will be Balu's age .

Given \( \frac{(5x - 5 )}{(7x + 5)} = \frac{1}{2} \)

On simplifying 10x - 7x = 10 + 5 = 15), we get x = 5

So Abi's present age is 25 and Balu's present age is 35 . 11 years after Abi's age will be 36 and Balu's ahe before 17 years was 18 , so ratio is 2 : 1 .

Let A be Anu , B be Brintha and R be Renu.

Average ages of A and B is

\(\frac{(A + B)}{2} – \frac{(B + R)}{2} = 12\)

\(A + B – B – R = 24\)

\(A – R = 24\)........(1)

Average age of A and R after 4 years,

\(A + R = 46 \times 2 – 8 = 84\)......(2)

solving eqn 1 and 2

2A = 108

A = 54

Present age of R=A- R

= 54 -24 = 30years

Let, Tanya age 16 years ago = \( x \)

Grandfather's age 16 years ago = \( 8x \)

8 years from now, \( 3(x+16+8) = (8x+16+8) => x = \( \frac{48}{58}\)

years ago ratio was: \( \frac{x+8}{8x+8}\)= \( \frac{ \frac{48}{5} +8 }{8 \times\frac{48}{5}}+8 \)

\( \frac{ 88}{424}=\frac{ 11}{53}\)

=11:53

Let the age of son = x years

The age of father = (45 - x)years

Five years before, son's age would be = x -5

Five years ago, father's age would be =(45- x )-5

From the question ,

\((x- 5)(45- x- 5) = 4(45- x - 5)\)

Hence (x—5) = 4

So x = 9

Their ages are 36 years and 9 years.

Let the ages be x and y
x + y = 12
xy = 35
Substitute the identity \( x^{2}\) + \( y^{2}\) = \( (x+y)^{2}\) - 2xy = 74

Let the average age of the whole team be x years.

let the age of the mother be M and that of the son is S.

Let the age of P and Q five years ago was 9x & 8x respectively.

P's present age = (9x + 5)

Q's present age = (8x + 5)

4 years hence,

\(\frac{(9x + 5) + 4}{(8x + 5) + 4} = \frac{12}{11}\)

99x + 99 = 96x + 108

x = 3

Q's present age = 24 + 5 = 29 years

OR

In case of age ratio problems, check the ratio values changes and number of years.

Means, the ratio values, 9 and 8 is changed to 12 and 11, which is incremented by 3.

Year gap is \(5+4=9\)

So we can see that 3 is used to represent the change of 9 years.

\(\therefore 3\) can be taken as the proportionality constant

Age of Q 4 years hence =\(11 \times 3 = 33\) <br> Present age = \(33-4=29\)

Ratio of present ages of A and B = 7 : 9

Sum of ages = 9+7 = 16 ratio

Difference of ages = 9-7 = 2 ratio

16-2= 14 ratio = 56 years

1 ratio = 4 years

Present age of A = 7 ratio = \(7\times 4\) = 28 years

Present age of B = 9 ratio = \(9\times 4\) = 36 years

Required ratio = \( 28+12 : 36+12 \)

= \( 40 : 48 \)

= \( 5 : 6 \)

Let the age of daughter is x

• Let the present ages of son and father be \(x\) and \((60 -x)\) years respectively.

  Then, \((60 - x) - 6 = 5(x - 6)\)

  \(54 + x = 5x - 30\)

  \(4x = 24\)

  \(x = 6\)

• Son's age after \(6\) years = \((x+ 6) = 12 \) years.

• Answer is 12.

Let the son's present age be x years. Then, \((38 - x) = x\)

\(2x = 38\).

\(x = 19\) .

Son's age \(5\) years back\( (19 - 5) = 14\) years.

Let Ronit's present age be x years. Then, father's present age =(x + 3x) years = 4x years.
\(4x+8= \frac{5}{2}(x+8)\)
\(8x+16=5x+40\)
\(x=8\)
Hence, required ratio = \(\frac{4x+16}{x+16} = \frac{48}{24}=2\)

• Total age of 10 students = 150 years

• Total age of 15 students = 240 years

• Total age of 5 new students = 240 - 150 = 90 years

• Average age of 5 new students = \(\frac{90}{5}\) =18 years

Let, Total of current ages of the 2 daughters is A years.
Then, father’s current age = 3A years.

\( (3A + 5) = 2 (A +10) \)

\( 3A + 5 = 2A + 20 \)

A = 15

Therefore, father’s current age = 45 years.

Let the present ages of the A & B be 3x & 2x respectively.

6 years back ,

\(\frac{(3x - 6)}{(2x - 6)} = \frac{3}{1}\)

3x - 6 = 6x - 18

x = 4

Age of A = 3x = 12

Age of B = 2x = 8

After, 12 years

A= 12+12 = 48

B = 8+12 = 20

The ratio of their ages is :

24 : 20 = 6 : 5

Let the present ages of A, B and C be 4k years, 5k years and 2k years respectively.

So, \(\frac{(4k + 6) + (5k + 6)}{2} = 3×2k - 3\)

9k + 12 = 12k - 6

k = 6

Average of their present ages \(= \frac{(4k + 5k + 2k)}{3} = 22\) years

Let the present age of Jothi and Viji be 13Xxand 11x respectively.

Given, Jothi's age 4 years hence and Viji's age 4 years ago in the ratio 15:9.

That is, \(\frac{13x+4}{11x-4}=\frac{15}{9}\)

9(13x + 4) = 15(11x - 4)

117x + 36 = 165x - 60

48x = 96

x = 2.

Now, required ratio is \(\frac{13x+4}{11x-4}\)

= \(\frac{13\times 2+4}{11\times 2-4}\)

=\(\frac{22}{26}\)=\(\frac{11}{13}\)= 11:13

Let the friend's age be x and that of the M is \(\frac{2}{5}x\)
after 15 years,

\(\frac{2}{5}x + 15 =\frac{1}{2}(x + 15) \)

solving for x, x = 75

sum of their ages after 10 years = \(75 + 10 + \frac{2}{5}\times75 + 10 = 125\)

Let present age of father be x and son be y.

The age of father 4 years ago,

x - 4 = 5 (y - 4) . . . . . ( 1 )

At present, the father’s age is 4 times that of his son, i.e,

x = 4y . . . . . ( 2 )

On solving equation ( 1 ) and ( 2) :

x = 64 and y = 16

Therefore,

Sum of their ages = 64 + 16 = 80

Let their ages one year ago be \( 4x \) and\( 3x \) years

\(\frac{4x+2}{5x+2} = \frac{5}{4}\)

\(4(4x+2)= 5(3x+2)\)

\( 16x+8=15x+10 \)

\( 16x-15x=10-2 \)

\(x=2\)
Sum of their present ages =\((4x +1+3x+1)\)

=\(4\times 2+1+3\times2+1\)

=\( 16 \) year

Let, the number of girls = G

The number of boys = B = (500 – G)

Average age of 500 students=

\(\frac{10G + 8B}{500} = 9.3 \)

10 G + 8 B = 4650

10G +8(500-G)= 4650

2G= 4650- 4000

2G= 650,

G= 325

Present age of A be 3x and Z be 5x \(\frac{3x-8}{5x+5}=\frac{2}{7}\)

21x-56=10x+10

x=6

Age of A=\(3x=3\times6=18\)

Age of Z=\(5x=5\times6=30\)

Ratio of A's age 12 years hence and Z's 12 years ago=(18+12):(30-12)

30:18=5:3

5 years ago average age of A,B,C,D = 45 years
5 years ago total age of A, B, C, D =\(45 \times 4 = 180\) years
Total present age of A, B, C, D = \(180 + 5\times 4 = 180+20=200\) years

If E's present age is x years

\( \frac{200+x}{5}=49 \)

\(200+x=245\)

\(x=45\).

four years ago

\(\frac{1}{2}A = 4B\)

\(\frac{\frac{1}{2}A}{4B} = \frac{5}{12}\)

\(\frac{A}{8B} = \frac{5}{12}\)

\(\frac{A}{B} = \frac{10}{3}\)

A : B = 10 : 3

A = 10x, B = 3x

8 years hence

A = 10x + 4 + 8 , B = 3x + 4 + 8

\(\frac{1}{2}(10x + 4 + 8) = 3x + 4 + 8\)

5x + 6 = 3x + 10

x = 2

present age of B = 3x + 4 = 3(2) + 4 = 10 years

Let R be Raveen's present age and M be Melvin's present age .

Given \( \frac{R + 11}{M - 6} = \frac{5}{7} \) ,

On simplifying \( 7R + 77 = 5M - 30 \)

And \( \frac{R + 8}{M + 9} = \frac{1}{3} \) ,

On simplifying \( 3R + 24 = M + 9 \)

Solving we get \( R = 4 , M = 27 \)

So ratio is \( 4 : 27 \) .

Let the son's present age be x years.

Let X & Y be the ages of A & B respectively.

Let the present ages of A & B be 9x and 5x respectively.

Let their present ages be 9x, 11x and 12x
\((9x-17)+(11x-17)+(12x-17)=109\)
32x-51=109
x=5
Age of Rajasree=9x=\(9\times5=45 years \)
Age of Rajashree=11x=\(11\times5=55 years \)
Age of Rajasri=12x=\(12\times5=60 years \)
Required difference=60-45=15 years

Let the present age of A = x

present age of B = 3x

3x - 5 = 4 ( x- 5 )

x = 15

Age of B = 15 x 3 = 45

After 5 years,

Age of B = 45 + 5 = 50 years

Let the present age of\( A, B \) & \( C \) are : \( 3x, 5x \) & \( 7x \)
\( (3x – 4) + (5x – 4) + (7x – 4) = 48 \)
\( 15x – 12 = 48 \)
\( 15x = 60 \)
\( x = 4 \)
The present ages are \( 12 \) years, \( 20 \) years and \( 28 \) years respectively.
Sum of the ages of \( A \) & \( C \) = \( 12 + 28 = 40 \) years

Total age of A and B = 12 + (Total age of B and C)

we can write it as,

Age of(A + B)= 12+ Age of (B+C)

\(\Rightarrow\)Age of A + Age of B = Age of B + Age of C

\(\Rightarrow\)Age of A= 12+Age of C

Therefore, C is 12 years younger than A.

Let the son's age = x years

Then mother's age = (50 - x)years

(x - 5)(50 - x - 5) = 3 (50 - x - 5 )

(x - 5) = 3

x = 8
mother's age = 50 - 8 =42 years

Their ages are 42 years and 8 years.

Let, Maala’s age = 4A and Kala’s age = 3A
Then \(4A + 3A = 28\)
A = 4
Maala’s age = 16 years and Kala’s age = 12 years
Proportion of their ages after 8 is =\( (16 + 8) : (12 + 8) = 24 : 20 = 6 : 5 \)

• Current ages of Yuvaraj = 4A and Ganguly = 3A

• So that \( 4A + 6 = 26 \) , \( A = 5 \) ,

  Current age of Ganguly = \( 3(A) => 3 \times 5 =15 \) years

• Suppose the ages of three persons are x, 2x, 3x years respectively .
• \(\frac{x+2x+3x}{3}=180\) \(6x=180\) \(x=30\)
• Hence, age of youngest person = x = 30 years.

Let their ages are \( 4x, 5x \) and \( 9x \) respectively.
Average of ages =\(\frac{4x + 5x + 9x}{3}\) \( = 24 \)
\( 6x = 24 \)
\( x = 4 \)
eldest one’s age is \( 9x = \) \(9 \times 4 = 36 \) years

• My brother was born 6 years after my sister was born and 3 years before I was born.
• Age of father during my brother’s birth = (30 + 6) = 36 years.
• Age of mother during my brother’s birth = (38 – 3) = 35 years.
• Hence, during the birth of my brother, my father was 36 years old and my mother was 35 years old.

• Let the son’s age 15 years ago be x years.

• Therefore, the father’s age 15 years ago was 3x years.

• The present age of father and son is (3x + 15) and (x + 15) respectively.

• 3 years hence their ages will be (3x + 18) and (x + 18) respectively. \((3x + 18) = 2 (x + 18) =3x + 18 = 2x + 36\)

• The sum of the present ages of father and son = (3x + 15) + (x + 15)

  = \(3 \times 18 + 15\)+ (18 + 15)

  = 102.

Let the number of students be P & Q for both groups respectively.

Average age of group 1 = 15 years

Total age of group 1 =15×P=15P

Average age of group 2 = 25 years

Total age of group 2 = 25×Q=25Q

Average age of third group (combined group) = 23 years

Total age of combined group will be = 23 times (P + Q)

15P + 25Q = 23(P + Q)

8P = 2Q

P : Q = 2 : 8