AGE Problems With Explanations

 Q) Andrew will be half as old as Larry in 3 years. Andrew will also be one-third as old as Jerome in 5 years. if Jerome is 15 years older than Larry, how old is Andrew?.

Sol: A+3 = 1/2(L+3)
=> 2A + 3 = L ----(1)

A + 5 = 1/3 (J+5)
=> 3A + 10 = J
Since J = L + 15

=> 3A - 5 = L ---(2)

Using (1) and (2)

A = 8

Or

Best way to do these problems is to write out every date.

Now || 3yrs from now || 5yrs from now
A || A+3 || A+5
J || J+3 || J+5
L || L+3 || L+5

Now make the equations for the appropriate years.

Since Andrew will be half as old as jerome in 3 years. you can rewrite it like this: 2(A+3)=L+3

Next equation: rewrite it just like above. Make the two sides equal to each other.

3(A+5)=J+5

Final one 15+L=J (since J is 15yrs older, adding 15 to L will equal J).

Now solve:
J=15+L--> 3(A+5)=J+5--> 3A+15=15+L+5 --> 3A-5=L

2(A+3)=L+3--> 2A+6=3A-5+3 -->2A+6=3A-2 ---> A=8

.

2.

In, 1950, Richard was 4 times as old as Robert. In 1955 , Richard was 3 times as old as Robert, in which year was Richard twice as old as Robert?

1960
1965
1970
1975
1980 

Sol:

trial and error method:

1950 (4x) 1955 (3x) 1970 (2x)
age 10 15 30
age 40 45 60

Or

Rich---X ,Rob---Y

1950----> X=4Y ,1955----> X+5=3(Y+5)-------->X= 3Y+15-5

3Y+10=4Y ,Y=10 ,X=40
1970
as X=60, Y=30

3.8 years ago, George was half as old as Sarah. Sarah is now 20 years older than George. How old will George be in 10 years? 

Sol: Sarah is now 20 years older than George --> S=G+20 --> Sarah 8 years ago was G+20-8=G+12 years old.

We also know that 8 years ago, George was half as old as Sarah --> (G-8)*2=G+12 --> G=28.

In 10 years George will be 28+10=38 years old.

Answer: 38.

4.The sum of the ages of 22 boys and 24 girls is 160.What is the sum of ages of one boy plus one girl, if all the boys are of the same age and all the girls are of the same age, and only full years are counted? 

Sol: 22B + 24G = 160
it is 11B + 12G = 80
 subst G=1 in the eq will not work (B should result in a integer since it is BOYS)
G=2 in the eq will not work
G=3 and B=4 works
so it is 7

Or

22B + 24G = 150. (B and G are integers) ==> 11B + 12G = 80. ==> 11(B+G) +G = 80
==> B+G = (80-G) / 11. Since B and G are intergers, then B + G must be an integer ==> 80 - G must be a multiple of 11. This implies G is the reaminder left when 80 is divided by 11. Ie G = 3.

Substituting G in B +G we get (80 - 3) / 11 = 7 which is the ans. 

Or

22X+24Y = 160
22(X+Y) + 2Y = 160

We know from the answers that Y must be less than or equal to 8.

So, 22(X+Y) must be greater than or equal to 144, and less that 160. Only answer possible is 7.
5. 8 years from now, the bottle of wine labeled 'Aged' will be 7 times older than the bottle of wine labe1ed 'Table.' I year ago, the bottle of wine labeled 'Table' was one-fourth as old as the bottle of wine labeled 'Vintage.' If the'Aged' bottle was 20 times older than the 'Vintage' bottle 2 years ago, then how old is each bottle now?

Sol: Let the current age of each wine be A - Aged, V - Vintage & T - Table.

You get three eq from the wordings:
A+8 = 7(t+8)
A-2 = 20(V-2)
T-1 = 0.25(V-1)

Solving them you get T = 2, V = 5, A = 62

or

a+8 = 7( t+8)
t-1 = ( v-1)/4
& a-2 =20( v-2)

i am just curious to know if first and third equations are translated as given in problem.
what i perceived is it should be
a+8 = (a+8) + 7( t+8)
t-1 = ( v-1)/4
& a-2 =(v-2)+20( v-2)
as question stem says 7 times older than.. and 20times older than..

6. This year, the sum of ages of the Perkins family members is 78. Currently, there are 4 members: husband, wife, daughter, and son. The husband is 4 years older than the wife. The daughter is two years older than the son. If the husband is 7 times older than the son, how old is the daughter?

(A) 5
(B) 7
(C) 10
(D) 13
(E) 14

Sol: h+w+d+s = 78 .............................i
h = w + 4 ...................................ii
d = s + 2 .......................................iii
h = 7s ........................................... iv

7s = w+4
w = 7s - 4

h+w+d+s = 78 .............................i
7s+7s-4+s+2+s = 78
16s = 80
s = 5
d = s+2 = 5+2 = 7

or

let the answer choice c) 10 yrs be daughter , son will be 8 yrs , husband is 56yrs , wife will be 52 yrs ....sum of their ages is 126 ( much higher than 78 )
so eliminate C , D , E

lets work with A) 5 yrs be daughter , son will be 3 yrs , husband is 21 yrs , wife will be 17 yrs ....sum of their ages is 46 ( lower than 78 )
so eliminate A

Answer is B

check: daughter 7 yrs , son 5 yrs , father 35 yrs , wife 31 yrs ...sum 78 yrs

or

Let H=Husband, W=Wife, D=Daughter, S=Son.

Express all of them in terms of S so that you have only one unknown.

S = S
D = S+2 [2 y. older than the son]
H = 7S [7 times older than the son]
W = 7S-4 [wife 4 years younger than the hubby]

S+(S+2)+7S+(7S-4) = 78
16S - 2 = 78
16S = 80
S=5

D=S+2
D=5+2=7

Or

Explanation:Assuming wife's age to be x,husband age is x+4.
Assuming son's age to be y,daughter's age is y+2.

Given,
x+(x+4)+y+(y+2) = 78---(1)
7y=x+4------------------(2)

Solving (1) and (2) gives
y=5
So,daughter's age is 7.

7.

If steve were half his age older, he would be twice the age of Eve. Twelve years ago, Eve was two-thirds Steve's age. How old is Steve?

A. 12
B. 24
C. 36
D. 48
E. 72 

Sol: S+S/2=2E equation 1
E-12=2/3*(S-12) equation 2
Solve them and find that Steve is 48 yrs

8.

Joe is older to Lloyd by five years. Ten years ago, John was 10 years older than Mary. What is Mary's age today?
(1) Mary's age today is three times the age of Joe.
(2) Lloyd today is 5 years old.

Sol:

. Joe=L+5
. Jh-10=M-10 -> Jh=M
-> what is M?

Statement 1: M=3Joe -> not sufficient
Statement 2: L=5 -> We can get Joe, but can't get Mary -> not sufficient
Together: from Statement 2 we got Joe; plug that into M=3Joe from statement 1 and we get Mary.

Answer is C.

9. five years ago,Beth's age was three times that of Amy.Ten years ago ,Beth's age was one half that of Chelsea.If "C" represents Chelsea's current age ,which of the following represents Amy's current age?



A)(c/6)+5
B)2c
C)(c-10)/3
D)3c-5
E)(5c/3)-10

Sol:starting  with Chelsey. Her age 10 years ago was C-10, so Beth's age 10 years ago was (c-10)/2. Next stage: Beth's age 5 years ago was (c-10)/2+5, and Amy's age 5 years ago was ((c-10)/2+5)/3. Now: Amy's age is ((c-10)/2+5)/3+5.
Now we should simplify the algebraic expression:
((c-10)/2+10/2)/3+5=c/6+5
Answer A

Or

How old is Beth today?...Let's call it "B"

Then, how old was Beth five years ago?..."B-5"

Similarly, how old was Amy five years ago?..."A-5"

At this point in time, Beth's age was three times that of Amy. So:

B-5 = 3*(A-5) (1)

B-10 = (C-10)/2 (2)

We want to solve for A in terms of C. So, we want to get rid of B. The easiest way is to equate the two equations. But the left-hand sides of each equation are different. So, just subtract 5 from both sides of (1) (OR add 5 to both sides of (2)):

B-5 -5 = 3*(A-5) - 5
B-10 = 3*(A-5) -5

Because B-10 = B-10, we have:

3*(A-5) - 5 = (C-10)/2

Solving for A, we have:

A = C/6 + 5

Choose A!

10. If the average age of three people is 21 years, is the youngest older than 13?

1) The oldest is 25.
2) One person is 24.

Sol: sum of the ages of three people is 63.

1st is SUFF. If the oldest is 25, then 38 is left for second oldest and the youngest. Youngest may be older than 13, may be 13, or may be younger than 13. 1st statement is true only when yougest is older than 13.

2nd is INSUFF. The statement gives just age, it can be any age.

Sol:  a 

Or

A.

Sum/#of terms = Avg.

Sum/3=21----> Sum of all ages is 63.

63-25= 38 (age of two younger people combined). I'm guessing that two of the people both can't be age 25. So the next oldest person has can be 24.999999 at the most. Well just call this 24.

38-24 = 14 could be 13.1111111111111. But this age is still greater than 13. Suff.

B. Insuff. Anyone could be older or younger.

Could be: 1,24,39 or could be 14,24,25

11.

  

Sol: The question is F+y=?

D+F=y
D=F+12
-> F=y/2 -6
->F+y= 3y/2 -6

D

12. Ann, Bob, Chris, Dana, and Eric are different ages such that their average age is 16. Bob is twice as old as Ann was three years ago. Eric is nine years younger than Bob. The sum of Chris and Ann's ages is equal to twice Dana's age. Dana is one year older than Ann. How old is Eric?

A) 13

B) 14

C) 15

D) 22

E) 23

Sol: -A-

A+B+C+D+E=80 -(1)

B=2(A-3)
E=B-9
C+A=2D
D=A+1
----------------
A=A
B=2A-6
C=A+2
D=A+1
E=2A-15

A+2A-6+A+2+A+1+2A-15=80
A=14
E=13

13. The average age of a group of 5 members is 20 years. Two years later, a new member joins the group. The average age of the group becomes 22 years. What is the age of the new member ?

A. 20 years
B. 21 years
C. 22 years
D. 23 years
E. 24 years

Sol: The average age of a group of 5 members is 20 years --> the sum of the ages is 5*20=100;
Two years later the sum of the ages of these 5 members would be 100+5*2=110;

Now, say the age of the new member is x years, so the sum of the ages of a new 6 member group is 110+x. Since given that the average age of this group of 6 members is 22 years then: 22*6=110+x --> x=22.

Answer: C.

or

The average age of the 5 members is 20 years
=> The sum of the ages of the 5 members is 20*5 = 100 years

Once the new member joins, the average age becomes 22 years.
Let the age of the new member be x.
Then (110+x)/6 = 22
=> x = 22 year

14. "A man's age is 4 times that of his son. In 20 years time the man's age will be double that of his son. How old is the man now?" 

Sol: m = man's age ; s = son's age

now: m = 4*s
20 years from now: m+20 = 2*(s+20)

m+20 = 2*((m/4) + 20)
solve for m and you will get m=40, and s=10 (now)
if had 20 to m and s, you will get m=60, s=30.

15. A Town T has 40,000 residents, 25 percent of whom are age 60 or older. What percent of the residents are female?

(1) The number of female residents of Town T age 60 or older is 30 percent of the number of female residents who are not 60 or older.

(2) 37 percent of the residents of Town T are males who are not 60 or older.
sol: You can solve this problem by  choosing double-set matrix. Given:

Numbers in black are given, numbers in red are calculated. We need to find the value of yellow box.

(1) The number of female residents of Town T age 60 or older is 30 percent of the number of female residents who are not 60 or older.

Not sufficient.

(2) 37 percent of the residents of Town T are males who are not 60 or older.

Not sufficient.

(1)+(2)

We know x=15200, hence we can get 1.3x. Sufficient.

Answer: C.