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Thursday, April 17, 2014

POSTAL ASSISTANT EXAM MATERIAL- QUANTITATIVE APTITUDE-AGE RELATED PROBLEMS

Disclaimer:- All the Information provided in this post are prepared & compiled by A. Praveen Kumar, SPM, Papannapet SO-502303, Telangana State  for in good faith of Postal Assistant Exam Aspirants. Author of blog does not accepts any responsibility in relation to the accuracy, completeness, usefulness or otherwise, of contents.



                                     AGES

Shortcut Methods for Age related Problems

1.    t 1 years earlier the fathers age was x times that of his son. At present the fathers age is y times that of his son. What are the present ages of his son and father

Sons age = t1 (x-1) / x-y
2.    The present age of the father is y times the age of his son. t 2 years hence, the fathers age becomes z times the age of his son. What are the present ages of the fathesr and son
Sons age = (z-1) t2 / y-z
3.    T1 years earlier the age of the father was x times the age of his son. t2 years hence, the age of the father becomes z times the age of his son. the present ages of the son and father
Sons age =  (z-1)t2 + t1(x-1)
                   ----------------------
                            (x-z)

4.    n years ago age = Total + Number of Years (Times-1)
                              --------------------------------------------------
                                             Times+1
5.    N years hence age = Total - Number of Years (Times-1)
                              --------------------------------------------------
                                             Times+1
6.    Present age = x:y, After T years  = a:b
Sons age =
                                       y * T(a-b)/ difference of cross products
Fathers age=
x * T(a-b)/ difference of cross products
7.    n years before the ages of Father and Son were in the ratio x:y. In another n years from now the ratio will be a:b. what are their Present ages
2n(a-b) / Difference of cross products
8.    If a is n years older than b. After x years y times of a equal to z times b. then their ages now
a age = zn/ z-y,  b age = yn/y-z                                          
  
                                                          Examples

1 The age of the father 3 years ago was 7 times the age of his son. At present the father's age is 5 times that of his son. What are the present ages of the father and the son?

Sol: Let the present age of son = x years

Then the present age of father = 5x yrs

3 years ago, 7(x-3) =5x - 3

2x = 18, x= 9 yrs, so father's age = 45 yrs

2. The ratio of the father's age to that of son's age is 4:1 the product of their ages is 196.What will be ratio of their ages after 5 yrs?

Sol:  let the ratio of proportionality be x yrs

4x*x = 196 or 4x2 = 196 or x= 7

Thus father's age = 28 yrs, Son's age = 7 yrs

After 5 yrs, father's age = 33 yrs, son's age =12 yrs

Ratio = 33:12= 11:4

3. The ratio of Vimals age and Arunas age is 3: 5 and the sum of their ages is 80 years. The ratio of their ages after 10 years will be
Sol: Let their ages be 3x and 5x years. 
Then,3x + 5x = 8O so x = 1O. 
Ratio of their ages after 10 years = (3x+ 10) : (5x+ 10) 
40 : 60 = 2 : 3.

4. One year ago, Mrs Promila was four times as old as her daughter Swati. Six years hence, Mrs Promilas age will exceed her daugther’s age by 9 years. The ratio of the present ages of Promila and her daughter is:
Sol: Let Swatis age 1 year ago = x. 
Then, Promilas age 1 year ago = 4x. 
(4x+6) - (x+6) = 9 or x = 3. 
Present age of Promila = (12 + 1) years = 13 years. 
Present age of Swati = (3+ 1) years =4 years. 
Ratio of their ages = 13 : 4.

5. The age of father 10 years ago was thrice the age of his son. Ten years hence, the father’s age will be twice that of his son. The ratio of their present ages is:
Sol: Let sons age 10 years ago = x. 
Then, fathers age 10 years ago = 3x. 
2(x+ 10+ 10) = (3x + 10 + 10)so x = 20. 
Ratio of their present ages = (3x + 10) : Cx + 10) 70: 30 = 7 :3.

6. Snehs age is1/6 th of her fathers age. Snehs fathers age will be twice of Vimals age after 10 years. If Vimals eighth birthday was celebrated two years before, then what is Snehs present age?.
Sol: Vimals age after 10 years = (8 + 2 + 10) years = 20 years. 
Snehs father’s age after 10 years = 40 years. 
Snehs father’s present age = 30 years. 
Snehs age= 1/6 * 30 years = 5 years.

7. Ten years ago, Chandrawati s mother was four times older than her daughter. After ten years, the mother will be twice older than daughter. The present age of Chandrawati is
Sol: Let Chandrawatis age 10 years ago is x years. 
Her mother’s age 10 years ago 4x. 
(4x+10)+10=2(x+10+10) or 2x=20 or x=10. 
Present age of Chandrawati (x + 10) = 20 years.

8. Ten years ago A was half of B in age. If the ratio of their present ages is 3: 4, what will be the total of their present ages
Sol: Let As age 10 years ago = x. 
Then, Bs age 10 years ago = 2x.
(x + 10) / (2x+ lO) = 3/4 hence x = 5. Total of their present ages =(x + 10 + 2x + 10) 
(3x + 20) = 35 years.

9. Kamla got married 6 years ago. Today her age is 1.25 times her age at the time of marriage. Her son’s age is 1/10 times her age. Her sons age is:
Sol: Let Kamlas age 6 years ago = x years. Kamlas present age = (x + 6) years. 
x+6= (5/4)*x or x=24. 
Kamla’s present age= 30 years. 
Sons present age = (1/10)*30 years = 3 years.

10.Pushpa is twice as old as Rita was 2 years ago. If the difference of their ages be 2 years, how old is Pushpa today
Sol: Let Ritas age 2 years ago = x years. 
Pushpas age now = 2x years.
2x - (x + 2) = 2 so x = 4. 
Pushpas age now = 8 years.

11. Sushil was thrice as old as Snehal 6 years back. Sushil will be times as old as Snehal 6 years hence. How old is Snehal today?
Sol: Let Snehals age 6 years back = x. 
Then, Sushils age 6 years back = 3x. 
(5/3) * (X + 6 + 6) = (3X + 6 + 6) hence 5(x+ 12) = 3(3x+ 12) so x=6. 
Snehal today = (x+ 6) years = 12 years.

12. The sum of the ages of a father and son is 45 years. Five years ago, the product of their ages was four times the father’s age at that time. The present age of father and son, respectively are: 
Sol: Let sons age = x years. Then fathers age = (45 - x)years. 
(x—5)(45—x—5) = 4(45- x - 5) hence (x—5) = 4 so x = 9 
their ages are 36 years and 9 years.

13. Two years ago a man was 6 times as old as his son. After 18 years, he will be twice as old as his son. Their present ages (in years) are
Sol: Let sons age 2 years ago be x. 
Mans age 2 years ago = 6x. 
2(x + 2 + 18) = (6x + 2 + 18) hece 4x=20 so x = 5. 
Their present ages are (6x +2) and (x +2) i.e. 32 years and 7 years.

14. A father is twice as old as his son. 20 years ago, the age of the father was 12 times the age of the son. The present age of the father (in years) is:
Sol: Let sons age = x. Then, fathers age = 2x. 
12(x—2O) = (2x - 20) so x = 22 
Fathers present age = 44 years.

15. The ages of two persons differ by 20 years. If 5 years ago, the elder one be 5 times as old as the younger one, their present ages (in years) are respectively:
Sol: Let their ages be x and (x + 20) years. 
5 (x - 5) = (x + 20 - 5) or 4x = 40 or x = 10. 
Their present ages are 30 years and 10 years.

16. The ages of Ram and Mukta are in the ratio of 3: 5. After 9 years, the ratio of their ages will becomes 3:4. The present age of Mukta (in years) is.
Sol: Let Rams age = 3x & Muktas age = 5x. 
3x + 9 / 5x + 9 = 3/ 4 so 4*(3x+ 9) = 3 (5x+ 9) hence x= 3. 
Muktas age = 15 years.

17. If 6 years are subtracted from the present age of Gulzar and the remainder is divided by 18, then the present age of his grandson Anup is obtained. If Anup is 2 years younger to Mahesh whose age is 5 years, then what is the present age of Guizar
Sol: Anups age =(5—2)years = 3 years.
Let Guizar’s age x years 
Then x - 16/18 = 3 so x - 6 = 54 so x 60

18. The average age of an adult class is 40 years. Twelve new students with an average age of 32 years join the class, thereby decreasing the average of the class by 4 years. The original strength of the class was
Sol: Let, original strength = x. 
40x + 12*32 / (x + 12) = 36 or x = 12

19. 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
Sol: Sachins age today = 30 years. 
Sachins age 10 years back = 20 years. 
Ajays age 10 years back = 10 years. 
Ajays age today =20 years.

20. Five years ago, the total of the ages of father and his son was 40 years. The ratio of their present ages is 4 : 1. What is the present age of the father?
Sol: Let sons age = x. Then, fathers age = 4x. 
(x - 5)+(4x - 5) = 40 
 x = 10 
Present age of father = 40 years.

21.The ratio of the ages of Meena and Meera is 4 : 3. The sum of their ages is 2 years. The ratio Of their ages after 8 years will be
Sol: Let Meenas age = 4x & Meeras age = 3x. Then, 4x+3x=28 hence x=4. 
Meenas age = 16 years and Meeras age = 12 years. 
Ratio of their ages after 8 years = (16 + 8): (12 + 8) = 24 : 20 = 6 : 5.

22. Ratio of Ashoks age to Sandeeps age is 4 : 3. Ashok will be 26 years old after 6 years. How old is Sandeep now?
Sol: Let their ages be 4x and 3x years. 
Now 4x + 6 = 26 hence x = 5 
. so Age of Sandeep = 3x = 15 years.

23. In 10 years, A will be twice as old5as B was 10 years ago. If A is now 9 years older than B, the present age of B is:
Sol: Let Bs age = x years. Then, As age = (x+ 9) years. 
(x+9+10)=2(x—10) hence x=39. 
Present age of B = 39 years.

24. The average age of 12 students is 20 years1f the age of one more student is included, the average decreases by 1. What is the age of the new student?
Sol: New average = (20 - 1) = 19. 
Age of new student = (13*19 - 12*20) years = 7 years

25. Average age of A and B is 24 years and average age of B, C and D is 22 years. The sum of the ages of A, B, C and D is:
Sol: A+B = (24X2) = 48 and B + C + D = (22*3) = 66 
With this data, we cannot find A + B + C + D.
So, the data is inadequate,

26. The ratio of the ages of Swati and Varun is 2: 5. After 8 years, their ages will be in the ratio of 1: 2. The difference in their present ages (in years) is:
Sol: Let Swati age = 2x & Varuns age = 5x. 
2x + 8 / 5x + 8 = 1/2 hence 2(2x+8) = (5x+8) so x=8. 
Swatis age = 16 years & Varun’s age =40 years. 
Difference of their ages = 24 years.

27. The Average age of a class of 22 subjects in 21 years. The average increased by 1 when the teacher’s age also included. What is the age of the teacher?
Sol: Avg x Number = Total




 21 years x 22 nos = 462 years
... (1)


 22 years x 23 nos = 506 years
... (2)



Teacher’s age = (2) - (1) = 506 – 462 = 44 Years.


28. Ten years ago, Kumar was thrice as old as Selva was but 10 years hence, he will be only twice as old. Find Kumar’s present age
Sol: Let Kumar’s present age be X years and Selva’s present age be Y years.

Then, according to the first condition,

X - 10 = 3(y - 10)

or, X - 3Y = -20
..(1)


Now. Kumar's age after 10 years


=(X + 10) years


Selva's age after 10 years = (Y + 10)



(X+10) = 2(Y+10)

or, X - 2Y = 10
..(2)


Solving (1) and (2), we get


X = 70 and Y = 30




Kumar's age = 70 years and Selva's age = 30 years.

29. A father is four times as old as his son today. After 20 years, he would be just twice as old. At the time of birth of his son, how much old must the father be?

Sol: Let the ages of son and father today by 'X' and '4x' years respectively.




4X + 20 = 2 (X+20) ---> As per information




X = 10 years ==> 4X = 40 years.



At the time of birth of his son, father must be 40 - 10 = 30 old


Sushil was thrice as old as Snehal 6 years back. Sushil will be
5
 times as old as Snehal
3
6 years hence. How old is Snehal today?
Let Snehal's age 6 years back = X Then, Sushil's age 6 years back = 32. 
Then
5
 (X + 6 + 6) = (3X + 6 + 6)


3





5(X + 12) = 3 (3X + 12)






4X = 24

 X = 6



 Snehal's age today = (X + 6) years = 12 years.





30. The ages of Kamaraj and Deenan differ by 16 years, Six years ago, Mohan’s age was thrice as that of Kamaraj’s find their present ages.
Sol:  
Let Kamaraj's age = X years

So, Mohan's age = (X + 16) years

Also, 3(X - 6) = X + 16 - 6 or, X = 14



 Kamaraj's age = 14 years

and, Mohan's age = 14 + 16 = 30 years.


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