The age of father is 4 times the age of son. If 5 years ago father's age was 7 times the age of his son at that time, what is father's present age?
If the age of A, t years ago was \[n_1\] times the age of B and at present A's age is \[n_2\] times that of B, then
A's present age \[= \left(\frac{n_1-1}{n_1-n_2}\right)n_2t\]
\[\therefore \text{Required age}\]
\[=\left(\frac{7-1}{7-4}\right)4\times 5\]
\[=\frac{6\times 4\times 5}{3}\]
\[=40\] years.
The age of father is 4 times the age of son. If 5 years ago father's age was 7 times the age of his son at that time, what is son's present age?
If the age of A, t years ago was \[n_1\] times the age of B and at present A's age is \[n_2\] times that of B, then
B's present age \[= \left(\frac{n_1-1}{n_1-n_2}\right)t\]
\[\therefore \text{Required age}\]
\[=\left(\frac{7-1}{7-4}\right)4\]
\[=\frac{6\times 4}{3}\]
\[=\frac{24}{3}\]
\[=8\] years.
The age of Mr. Gupta is six times the age of his son. After 10 years, the age of Mr. Gupta will be four times the age of his son. Find the present age of Mr. Gupta.
The present age of A is \[n_1\] times the present age of B. If t years hence, the age of A would be \[n_2\] times that of B, then
A's present age \[= \left(\frac{n_2-1}{n_1-n_2}\right)t\]
\[\therefore \text{Required age}\]
\[\left(\frac{n_2-1}{n_1-n_2}\right)n_2t\]
\[=\left(\frac{4-1}{6-4}\right)4\times 10\]
\[=\frac{3}{2}\times 40\]
\[=60\] years.
The age of Mr. Gupta is six times the age of his son. After 10 years, the age of Mr. Gupta will be four times the age of his son. Find the present age of Mr. Gupta's son.
The present age of A is \[n_1\] times the present age of B. If t years hence, the age of A would be \[n_2\] times that of B, then
B's present age \[= \left(\frac{n_2-1}{n_1-n_2}\right)t\]
\[\therefore \text{Required age}\]
\[\left(\frac{n_2-1}{n_1-n_2}\right)t\]
\[=\left(\frac{4-1}{6-4}\right)10\]
\[=\frac{3}{2}\times 10\]
\[=15\] years.
10 years ago Anu's mother was 4 times older than her daughter. After 10 years, the mother will be twice older than the daughter. Find the present age of Anu's mother.
The age of A, \[t_1\] years ago, was \[n_1\] times the age of B. If \[t_2\] years hence A's age would be \[n_2\] times that of B, then
A's present age \[= \frac{n_1\left(t_1+t_2\right)\left(n_2-1\right)}{\left(n_1-n_2\right)}+t_1\]
\[\therefore \text{Required age}\]
\[=\frac{4\left(10+10\right)\left(2-1\right)}{\left(4-2\right)}+10\]
\[=\frac{4\left(20\right)}{2}+10\]
\[=\left(4\times 10\right)+10\]
\[=40+10\]
\[=50\] years.
10 years ago Anu's mother was 4 times older than her daughter. After 10 years, the mother will be twice older than the daughter. Find the present age of Anu.
The age of A, \[t_1\] years ago, was \[n_1\] times the age of B. If \[t_2\] years hence A's age would be \[n_2\] times that of B, then
B's present age \[= \frac{t_2\left(n_2-1\right)+t_1\left(n_1-1\right)}{n_1-n_2}\]
\[\therefore \text{Required age}\]
\[=\frac{10\left(2-1\right)+10\left(4-1\right)}{4-2}\]
\[=\frac{10+\left(10\times 3\right)}{2}\]
\[=\frac{10+30}{2}\]
\[=\frac{40}{2}\]
\[=20\] years.
The sum of the ages of A and B is 42 years. 3 years back, the age of A was 5 times the age of B. Find the present age of A.
The sum of the ages of A and B is S years. If t years back, the age of A was n times the age of B, then
A's present age
\[=\frac{Sn-t\left(n-1\right)}{n+1}\]
\[\therefore \text{Required age}\]
\[=\frac{42 \times 5 - 3\left(5-1\right)}{5+1}\]
\[=\frac{210 - 12}{6}\]
\[=\frac{198}{6}\]
\[=33\] years.
The sum of the ages of A and B is 42 years. 3 years back, the age of A was 5 times the age of B. Find the present age of B.
The sum of the ages of A and B is S years. If t years back, the age of A was n times the age of B, then
B's present age
\[=\frac{S + t\left(n-1\right)}{n+1}\]
\[\therefore \text{Required age}\]
\[=\frac{42 + 3 \left(5-1\right)}{5+1}\]
\[=\frac{42 + 12}{6}\]
\[=\frac{54}{6}\]
\[=9\] years.
The sum of a son and father is 56 years. After four years, the age of the father will be three times that of the son. Find the age of father.
The sum of present ages of A and B is S years. If, t years hence, the ages of A would be n times the age of B, then
Present age of A
\[=\frac{Sn+t\left(n-1\right)}{n+1}\]
\[\therefore \text{Required age}\]
\[=\frac{56\times 3+4\left(3-1\right)}{3+1}\]
\[=\frac{168+8}{4}\]
\[=\frac{176}{4}\]
\[=44\] years.
The ratio of the ages of father and son at present is 6 : 1. after 5 years the ratio will become 7 : 2. Find the present age of the father.
If the ratio of the present ages of A and B is a : b and t years hence, it will be c : d, then
Present age of A
\[=\frac{at(c-d)}{ad-bc}\]
\[\therefore \text{Required age}\]
\[=\frac{6\times 5\left(7-2\right)}{6\times 2-1\times 7}\]
\[=\frac{6\times 25}{12-7}\]
\[=\frac{150}{5}\]
\[=30\] years.
The ratio of the ages of father and son at present is 6 : 1. after 5 years the ratio will become 7 : 2. Find the present age of the son.
If the ratio of the present ages of A and B is a : b and t years hence, it will be c : d, then
Present age of B
\[=\frac{bt(c-d)}{ad-bc}\]
\[\therefore \text{Required age}\]
\[=\frac{1\times 5\left(7-2\right)}{6\times 2-1\times 7}\]
\[=\frac{1\times 25}{12-7}\]
\[=\frac{25}{5}\]
\[=5\] years.
Six years ago the ratio of the ages of Mahesh and Suresh was 2 : 1. The ratio of their present ages is 9 : 5 respectively. what is the present age of Mahesh?
t years ago the ratio of the ages of A and B was a : b. If the ratio of their present ages is c : d respectively, then
Present age of A
\[=\frac{-at(c-d)}{ad-bc}\]
\[\therefore \text{Required age}\]
\[=\frac{-9\times 6\left(2-1\right)}{1\times 9-5\times 2}\]
\[=\frac{-9\times 6}{9-10}\]
\[=\frac{-54}{-1}\]
\[=54\] years.
Six years ago the ratio of the ages of Mahesh and Suresh was 2 : 1. The ratio of their present ages is 9 : 5 respectively. what is the present age of Suresh?
t years ago the ratio of the ages of A and B was a : b. If the ratio of their present ages is c : d respectively, then
Present age of B
\[=\frac{-bt(c-d)}{ad-bc}\]
\[\therefore \text{Required age}\]
\[=\frac{-5\times 6\left(2-1\right)}{1\times 9-5\times 2}\]
\[=\frac{-5\times 6}{9-10}\]
\[=\frac{-30}{-1}\]
\[=30\] years.
A's age is one-sixth of B's age. B's age will be twice of C's age after 10 years. If C's eighth birthday was celebrated two years ago, then the present age of A must be
\[\therefore B=30\]
\[A=5\] years.
Sachin was twice as old as Ajay 10 years back. How old is Ajay today if Sachin will be 40 years old 10 years hence?
Sachin's age today = 30 years
Sachin's age 10 years back = 10 years
Ajay's age 10 years back = 10 years
Ajay's age today = 20 years
A father is twice as old as his son. 20 years back, he was twelve times as old as the son. What are their present ages?
Let the age of
\[F = 2S\]
\[\implies F - 20 = 12(S - 20)\]
\[\implies 2S - 20 = 12(S - 20)\]
\[\implies 2S - 20 = 12S - 240\]
\[\implies S = 22\]
\[\therefore F = 44\]
The average age of the mother and her 6 children is 12 years which is reduced by 5 years if the age of the mother is excluded by 5 years. How old is the mother?
Total age of the mother and six children
\[= 12 \times 7 = 84\] years
Total age of six children
\[= 7 \times 6 = 42\] years
\[\therefore\] Mother is 42 years old.