138 Verbal reasoning problems on ages

Here are some reasoning problems related to ages, designed to test your logical thinking and arithmetic skills:

Example 1: Age Difference

Question:
The age of Alice is 6 years more than the age of Bob. The sum of their ages is 48 years. What are their ages?

Answer:
Let Alice’s age be AAA and Bob’s age be BBB.
From the problem, we have:

A=B+6A = B + 6A=B+6
A+B=48A + B = 48A+B=48

Substitute A=B+6A = B + 6A=B+6 into the second equation:
(B+6)+B=48(B + 6) + B = 48(B+6)+B=48
2B+6=482B + 6 = 482B+6=48
2B=422B = 422B=42
B=21B = 21B=21

So, Bob’s age is 21, and Alice’s age is 21+6=2721 + 6 = 2721+6=27.

Example 2: Age in Future

Question:
John is 5 years older than Sarah. In 10 years, the sum of their ages will be 70 years. How old are John and Sarah now?

Answer:
Let John’s current age be JJJ and Sarah’s current age be SSS.
From the problem, we know:

J=S+5J = S + 5J=S+5
(J+10)+(S+10)=70(J + 10) + (S + 10) = 70(J+10)+(S+10)=70

Substitute J=S+5J = S + 5J=S+5 into the second equation:
(S+5+10)+(S+10)=70(S + 5 + 10) + (S + 10) = 70(S+5+10)+(S+10)=70
S+15+S+10=70S + 15 + S + 10 = 70S+15+S+10=70
2S+25=702S + 25 = 702S+25=70
2S=452S = 452S=45
S=22.5S = 22.5S=22.5

Since Sarah’s age is 22.5, John’s age is 22.5+5=27.522.5 + 5 = 27.522.5+5=27.5. So, Sarah is 22.5 years old, and John is 27.5 years old.

Example 3: Sum of Ages

Question:
The sum of the ages of Alice and Bob is 50 years. Ten years ago, Alice was three times as old as Bob. How old are Alice and Bob now?

Answer:
Let Alice’s age be AAA and Bob’s age be BBB.
From the problem, we have two equations:

A+B=50A + B = 50A+B=50
A−10=3(B−10)A – 10 = 3(B – 10)A−10=3(B−10)

Substitute the second equation into the first:
A=50−BA = 50 – BA=50−B

Now substitute A=50−BA = 50 – BA=50−B into A−10=3(B−10)A – 10 = 3(B – 10)A−10=3(B−10):
(50−B)−10=3(B−10)(50 – B) – 10 = 3(B – 10)(50−B)−10=3(B−10)
40−B=3B−3040 – B = 3B – 3040−B=3B−30
40+30=3B+B40 + 30 = 3B + B40+30=3B+B
70=4B70 = 4B70=4B
B=17.5B = 17.5B=17.5

So, Bob is 17.5 years old, and Alice is 50−17.5=32.550 – 17.5 = 32.550−17.5=32.5.

Example 4: Relative Age

Question:
The sum of the ages of A and B is 36 years. Ten years ago, A was twice as old as B. How old are A and B now?

Answer:
Let A’s current age be AAA and B’s current age be BBB.
From the problem, we have two equations:

A+B=36A + B = 36A+B=36
A−10=2(B−10)A – 10 = 2(B – 10)A−10=2(B−10)

Substitute A=36−BA = 36 – BA=36−B into the second equation:
(36−B)−10=2(B−10)(36 – B) – 10 = 2(B – 10)(36−B)−10=2(B−10)
26−B=2B−2026 – B = 2B – 2026−B=2B−20
26+20=2B+B26 + 20 = 2B + B26+20=2B+B
46=3B46 = 3B46=3B
B=15.33B = 15.33B=15.33

So, B’s age is approximately 15.33 years, and A’s age is 36−15.33=20.6736 – 15.33 = 20.6736−15.33=20.67.

Example 5: Father’s Age

Question:
The father’s age is four times the son’s age. In 5 years, the father will be three times as old as the son. What are their current ages?

Answer:
Let the son’s age be SSS and the father’s age be FFF.
From the problem, we know:

F=4SF = 4SF=4S
F+5=3(S+5)F + 5 = 3(S + 5)F+5=3(S+5)

Substitute F=4SF = 4SF=4S into the second equation:
4S+5=3(S+5)4S + 5 = 3(S + 5)4S+5=3(S+5)
4S+5=3S+154S + 5 = 3S + 154S+5=3S+15
4S−3S=15−54S – 3S = 15 – 54S−3S=15−5
S=10S = 10S=10

The son is 10 years old, and the father’s age is 4S=4×10=404S = 4 \times 10 = 404S=4×10=40. So, the father is 40 years old.

Example 6: Three Siblings

Question:
The sum of the ages of three siblings is 75 years. The first sibling is 5 years older than the second, and the second sibling is 3 years older than the third. How old is each sibling?

Answer:
Let the third sibling’s age be XXX.
Then the second sibling’s age is X+3X + 3X+3, and the first sibling’s age is (X+3)+5=X+8(X + 3) + 5 = X + 8(X+3)+5=X+8.
The sum of their ages is:
X+(X+3)+(X+8)=75X + (X + 3) + (X + 8) = 75X+(X+3)+(X+8)=75
3X+11=753X + 11 = 753X+11=75
3X=643X = 643X=64
X=21.33X = 21.33X=21.33

Thus, the third sibling is about 21.33 years old, the second sibling is 21.33+3=24.3321.33 + 3 = 24.3321.33+3=24.33, and the first sibling is 24.33+5=29.3324.33 + 5 = 29.3324.33+5=29.33.

Example 7: Difference in Ages

Question:
A mother is 24 years older than her son. In 8 years, the mother will be twice as old as her son. How old are they now?

Answer:
Let the son’s age be SSS and the mother’s age be MMM.
From the problem, we know:

M=S+24M = S + 24M=S+24
M+8=2(S+8)M + 8 = 2(S + 8)M+8=2(S+8)

Substitute M=S+24M = S + 24M=S+24 into the second equation:
(S+24)+8=2(S+8)(S + 24) + 8 = 2(S + 8)(S+24)+8=2(S+8)
S+32=2S+16S + 32 = 2S + 16S+32=2S+16
32−16=2S−S32 – 16 = 2S – S32−16=2S−S
16=S16 = S16=S

So, the son is 16 years old, and the mother’s age is 16+24=4016 + 24 = 4016+24=40.

Example 8: In Five Years

Question:
The sum of the ages of Lisa and her brother is 50. In five years, Lisa will be twice as old as her brother. How old are they now?

Answer:
Let Lisa’s age be LLL and her brother’s age be BBB.
From the problem, we know:

L+B=50L + B = 50L+B=50
L+5=2(B+5)L + 5 = 2(B + 5)L+5=2(B+5)

Substitute L=50−BL = 50 – BL=50−B into the second equation:
(50−B)+5=2(B+5)(50 – B) + 5 = 2(B + 5)(50−B)+5=2(B+5)
55−B=2B+1055 – B = 2B + 1055−B=2B+10
55−10=2B+B55 – 10 = 2B + B55−10=2B+B
45=3B45 = 3B45=3B
B=15B = 15B=15

So, Lisa’s brother is 15 years old, and Lisa’s age is 50−15=3550 – 15 = 3550−15=35.

Example 9: Age Relationship

Question:
The sum of the ages of Mark and John is 60 years. Mark is 5 years older than John. How old are Mark and John?

Answer:
Let Mark’s age be MMM and John’s age be JJJ.
From the problem, we have the following relationships:

M+J=60M + J = 60M+J=60
M=J+5M = J + 5M=J+5

Substitute M=J+5M = J + 5M=J+5 into the first equation:
(J+5)+J=60(J + 5) + J = 60(J+5)+J=60
2J+5=602J + 5 = 602J+5=60
2J=552J = 552J=55
J=27.5J = 27.5J=27.5

So, John is 27.5 years old, and Mark is 27.5+5=32.527.5 + 5 = 32.527.5+5=32.5.

Example 10: Father and Daughter

Question:
The father is four times as old as his daughter. Five years ago, he was five times as old as his daughter. How old are they now?

Answer:
Let the daughter’s age be DDD and the father’s age be FFF.
From the problem, we have:

F=4DF = 4DF=4D
F−5=5(D−5)F – 5 = 5(D – 5)F−5=5(D−5)

Substitute F=4DF = 4DF=4D into the second equation:
4D−5=5(D−5)4D – 5 = 5(D – 5)4D−5=5(D−5)
4D−5=5D−254D – 5 = 5D – 254D−5=5D−25
4D−5D=−25+54D – 5D = -25 + 54D−5D=−25+5
−D=−20-D = -20−D=−20
D=20D = 20D=20

So, the daughter is 20 years old, and the father is 4×20=804 \times 20 = 804×20=80.

Example 11: Brothers’ Ages

Question:
The sum of the ages of two brothers is 48 years. The elder brother is 4 years older than the younger brother. How old are they?

Answer:
Let the younger brother’s age be YYY and the elder brother’s age be EEE.
From the problem, we have:

E+Y=48E + Y = 48E+Y=48
E=Y+4E = Y + 4E=Y+4

Substitute E=Y+4E = Y + 4E=Y+4 into the first equation:
(Y+4)+Y=48(Y + 4) + Y = 48(Y+4)+Y=48
2Y+4=482Y + 4 = 482Y+4=48
2Y=442Y = 442Y=44
Y=22Y = 22Y=22

So, the younger brother is 22 years old, and the elder brother is 22+4=2622 + 4 = 2622+4=26.

Example 12: Age Difference

Question:
The sum of the ages of mother and daughter is 55 years. The mother is 30 years older than the daughter. What are their ages?

Answer:
Let the daughter’s age be DDD and the mother’s age be MMM.
From the problem, we have:

M+D=55M + D = 55M+D=55
M=D+30M = D + 30M=D+30

Substitute M=D+30M = D + 30M=D+30 into the first equation:
(D+30)+D=55(D + 30) + D = 55(D+30)+D=55
2D+30=552D + 30 = 552D+30=55
2D=252D = 252D=25
D=12.5D = 12.5D=12.5

So, the daughter is 12.5 years old, and the mother is 12.5+30=42.512.5 + 30 = 42.512.5+30=42.5.

Example 13: Ages of Siblings

Question:
The sum of the ages of two siblings is 48 years. The elder sibling is 6 years older than the younger one. What are their ages?

Answer:
Let the younger sibling’s age be YYY and the elder sibling’s age be EEE.
From the problem, we have:

E+Y=48E + Y = 48E+Y=48
E=Y+6E = Y + 6E=Y+6

Substitute E=Y+6E = Y + 6E=Y+6 into the first equation:
(Y+6)+Y=48(Y + 6) + Y = 48(Y+6)+Y=48
2Y+6=482Y + 6 = 482Y+6=48
2Y=422Y = 422Y=42
Y=21Y = 21Y=21

So, the younger sibling is 21 years old, and the elder sibling is 21+6=2721 + 6 = 2721+6=27.

Example 14: Ages of Two Friends

Question:
The sum of the ages of Alice and Bob is 46 years. In 5 years, Alice will be twice as old as Bob. What are their current ages?

Answer:
Let Alice’s age be AAA and Bob’s age be BBB.
From the problem, we have:

A+B=46A + B = 46A+B=46
A+5=2(B+5)A + 5 = 2(B + 5)A+5=2(B+5)

Substitute A=46−BA = 46 – BA=46−B into the second equation:
(46−B)+5=2(B+5)(46 – B) + 5 = 2(B + 5)(46−B)+5=2(B+5)
51−B=2B+1051 – B = 2B + 1051−B=2B+10
51−10=2B+B51 – 10 = 2B + B51−10=2B+B
41=3B41 = 3B41=3B
B=13.67B = 13.67B=13.67

So, Bob is approximately 13.67 years old, and Alice’s age is 46−13.67=32.3346 – 13.67 = 32.3346−13.67=32.33.

Example 15: Age Relation in Years

Question:
The difference between the father’s age and his son’s age is 30 years. In 6 years, the father will be twice as old as his son. How old are the father and son?

Answer:
Let the son’s age be SSS and the father’s age be FFF.
From the problem, we know:

F−S=30F – S = 30F−S=30
F+6=2(S+6)F + 6 = 2(S + 6)F+6=2(S+6)

Substitute F=S+30F = S + 30F=S+30 into the second equation:
(S+30)+6=2(S+6)(S + 30) + 6 = 2(S + 6)(S+30)+6=2(S+6)
S+36=2S+12S + 36 = 2S + 12S+36=2S+12
36−12=2S−S36 – 12 = 2S – S36−12=2S−S
24=S24 = S24=S

So, the son is 24 years old, and the father’s age is 24+30=5424 + 30 = 5424+30=54.

Example 16: Father and Son’s Age

Question:
The father is 40 years old, and the son is 10 years old. In how many years will the father be three times as old as his son?

Answer:
Let the number of years from now be xxx.
In xxx years, the father’s age will be 40+x40 + x40+x and the son’s age will be 10+x10 + x10+x.
We are told that the father will be three times as old as the son in xxx years, so:
40+x=3(10+x)40 + x = 3(10 + x)40+x=3(10+x)

Simplify the equation:
40+x=30+3×40 + x = 30 + 3×40+x=30+3x
40−30=3x−x40 – 30 = 3x – x40−30=3x−x
10=2×10 = 2×10=2x
x=5x = 5x=5

So, in 5 years, the father will be three times as old as the son.

Example 17: Age Difference of Three

Question:
The sum of the ages of three friends, Jack, Jill, and John, is 70 years. Jack is 10 years older than Jill, and John is 5 years younger than Jack. How old are they?

Answer:
Let Jill’s age be JJJ, Jack’s age be KKK, and John’s age be NNN.
From the problem, we have:

K=J+10K = J + 10K=J+10
N=K−5=J+10−5=J+5N = K – 5 = J + 10 – 5 = J + 5N=K−5=J+10−5=J+5
J+K+N=70J + K + N = 70J+K+N=70

Substitute K=J+10K = J + 10K=J+10 and N=J+5N = J + 5N=J+5 into the third equation:
J+(J+10)+(J+5)=70J + (J + 10) + (J + 5) = 70J+(J+10)+(J+5)=70
3J+15=703J + 15 = 703J+15=70
3J=553J = 553J=55
J=18.33J = 18.33J=18.33

So, Jill is 18.33 years old, Jack is 18.33+10=28.3318.33 + 10 = 28.3318.33+10=28.33, and John is 18.33+5=23.3318.33 + 5 = 23.3318.33+5=23.33.

Example 18: Age Difference in 10 Years

Question:
The sum of the ages of two people is 60 years. Ten years ago, one person was three times as old as the other. What are their current ages?

Answer:
Let the current ages be AAA and BBB.
From the problem, we have:

A+B=60A + B = 60A+B=60
A−10=3(B−10)A – 10 = 3(B – 10)A−10=3(B−10)

Substitute the second equation into the first:
A=60−BA = 60 – BA=60−B

Now substitute into A−10=3(B−10)A – 10 = 3(B – 10)A−10=3(B−10):
(60−B)−10=3(B−10)(60 – B) – 10 = 3(B – 10)(60−B)−10=3(B−10)
50−B=3B−3050 – B = 3B – 3050−B=3B−30
50+30=3B+B50 + 30 = 3B + B50+30=3B+B
80=4B80 = 4B80=4B
B=20B = 20B=20

So, B=20B = 20B=20, and A=60−20=40A = 60 – 20 = 40A=60−20=40.

Thus, one person is 40 years old and the other is 20 years old.

Example 19: The Grandfather’s Age

Question:
The age of a grandfather is 20 years more than his grandson. In 5 years, the grandfather will be three times as old as his grandson. What are their current ages?

Answer:
Let the grandson’s age be GGG and the grandfather’s age be FFF.
From the problem, we have:

F=G+20F = G + 20F=G+20
F+5=3(G+5)F + 5 = 3(G + 5)F+5=3(G+5)

Substitute F=G+20F = G + 20F=G+20 into the second equation:
(G+20)+5=3(G+5)(G + 20) + 5 = 3(G + 5)(G+20)+5=3(G+5)
G+25=3G+15G + 25 = 3G + 15G+25=3G+15
25−15=3G−G25 – 15 = 3G – G25−15=3G−G
10=2G10 = 2G10=2G
G=5G = 5G=5

So, the grandson is 5 years old, and the grandfather’s age is 5+20=255 + 20 = 255+20=25.

Example 20: Age of a Sister

Question:
Maria is 6 years older than her sister. The sum of their ages is 48 years. How old is each sister?

Answer:
Let Maria’s age be MMM and her sister’s age be SSS.
From the problem, we know:

M=S+6M = S + 6M=S+6
M+S=48M + S = 48M+S=48

Substitute M=S+6M = S + 6M=S+6 into the second equation:
(S+6)+S=48(S + 6) + S = 48(S+6)+S=48
2S+6=482S + 6 = 482S+6=48
2S=422S = 422S=42
S=21S = 21S=21

So, Maria’s sister is 21 years old, and Maria is 21+6=2721 + 6 = 2721+6=27.

Example 21: The Age of a Father and Son

Question:
The father’s age is 5 times the son’s age. In 10 years, the father will be three times as old as the son. How old are the father and the son?

Answer:
Let the son’s age be SSS and the father’s age be FFF.
From the problem, we have:

F=5SF = 5SF=5S
F+10=3(S+10)F + 10 = 3(S + 10)F+10=3(S+10)

Substitute F=5SF = 5SF=5S into the second equation:
5S+10=3(S+10)5S + 10 = 3(S + 10)5S+10=3(S+10)
5S+10=3S+305S + 10 = 3S + 305S+10=3S+30
5S−3S=30−105S – 3S = 30 – 105S−3S=30−10
2S=202S = 202S=20
S=10S = 10S=10

So, the son is 10 years old, and the father’s age is 5×10=505 \times 10 = 505×10=50.

Example 22: The Sum of Ages

Question:
The sum of the ages of A and B is 70 years. Ten years ago, A was twice as old as B. How old are they now?

Answer:
Let A’s age be AAA and B’s age be BBB.
From the problem, we have:

A+B=70A + B = 70A+B=70
A−10=2(B−10)A – 10 = 2(B – 10)A−10=2(B−10)

Substitute A=70−BA = 70 – BA=70−B into the second equation:
(70−B)−10=2(B−10)(70 – B) – 10 = 2(B – 10)(70−B)−10=2(B−10)
60−B=2B−2060 – B = 2B – 2060−B=2B−20
60+20=2B+B60 + 20 = 2B + B60+20=2B+B
80=3B80 = 3B80=3B
B=26.67B = 26.67B=26.67

So, B is approximately 26.67 years old, and A’s age is 70−26.67=43.3370 – 26.67 = 43.3370−26.67=43.33.

Example 23: Two People’s Ages

Question:
The sum of the ages of two friends is 90 years. One friend is 10 years older than the other. What are their ages?

Answer:
Let the younger friend’s age be YYY and the older friend’s age be OOO.
From the problem, we know:

O=Y+10O = Y + 10O=Y+10
O+Y=90O + Y = 90O+Y=90

Substitute O=Y+10O = Y + 10O=Y+10 into the second equation:
(Y+10)+Y=90(Y + 10) + Y = 90(Y+10)+Y=90
2Y+10=902Y + 10 = 902Y+10=90
2Y=802Y = 802Y=80
Y=40Y = 40Y=40

So, the younger friend is 40 years old, and the older friend is 40+10=5040 + 10 = 5040+10=50.

Example 24: Age of a Couple

Question:
A man is 8 years older than his wife. In 12 years, the man will be twice as old as his wife. How old are they now?

Answer:
Let the wife’s age be WWW and the man’s age be MMM.
From the problem, we have:

M=W+8M = W + 8M=W+8
M+12=2(W+12)M + 12 = 2(W + 12)M+12=2(W+12)

Substitute M=W+8M = W + 8M=W+8 into the second equation:
(W+8)+12=2(W+12)(W + 8) + 12 = 2(W + 12)(W+8)+12=2(W+12)
W+20=2W+24W + 20 = 2W + 24W+20=2W+24
20−24=2W−W20 – 24 = 2W – W20−24=2W−W
−4=W-4 = W−4=W

This seems like an unrealistic answer, as a person cannot have a negative age, meaning there may be an issue with the constraints or the set up of the problem. In these cases, it’s important to recheck the problem for errors or unrealistic assumptions.

Example 25: Mother and Daughter

Question:
The mother’s age is 12 years more than twice her daughter’s age. The sum of their ages is 54 years. What are their ages?

Answer:
Let the daughter’s age be DDD and the mother’s age be MMM.
From the problem, we have:

M=2D+12M = 2D + 12M=2D+12
M+D=54M + D = 54M+D=54

Substitute M=2D+12M = 2D + 12M=2D+12 into the second equation:
(2D+12)+D=54(2D + 12) + D = 54(2D+12)+D=54
3D+12=543D + 12 = 543D+12=54
3D=423D = 423D=42
D=14D = 14D=14

So, the daughter is 14 years old, and the mother’s age is 2×14+12=402 \times 14 + 12 = 402×14+12=40.

Example 26: The Sum of Ages of Two Sisters

Question:
The sum of the ages of two sisters is 48 years. The elder sister is 6 years older than the younger sister. What are their ages?

Answer:
Let the younger sister’s age be YYY and the elder sister’s age be EEE.
From the problem, we have:

E=Y+6E = Y + 6E=Y+6
E+Y=48E + Y = 48E+Y=48

Substitute E=Y+6E = Y + 6E=Y+6 into the second equation:
(Y+6)+Y=48(Y + 6) + Y = 48(Y+6)+Y=48
2Y+6=482Y + 6 = 482Y+6=48
2Y=422Y = 422Y=42
Y=21Y = 21Y=21

So, the younger sister is 21 years old, and the elder sister is 21+6=2721 + 6 = 2721+6=27.

Example 27: Father and Son’s Ages

Question:
The age of a father is 4 times the age of his son. Five years ago, the father was 5 times as old as his son. What are their current ages?

Answer:
Let the son’s age be SSS and the father’s age be FFF.
From the problem, we have:

F=4SF = 4SF=4S
F−5=5(S−5)F – 5 = 5(S – 5)F−5=5(S−5)

Substitute F=4SF = 4SF=4S into the second equation:
4S−5=5(S−5)4S – 5 = 5(S – 5)4S−5=5(S−5)
4S−5=5S−254S – 5 = 5S – 254S−5=5S−25
−5+25=5S−4S-5 + 25 = 5S – 4S−5+25=5S−4S
20=S20 = S20=S

So, the son is 20 years old, and the father’s age is 4×20=804 \times 20 = 804×20=80.

Example 28: Two People’s Age Sum

Question:
The sum of the ages of Alice and Bob is 72 years. Alice is 8 years older than Bob. How old are Alice and Bob?

Answer:
Let Bob’s age be BBB and Alice’s age be AAA.
From the problem, we have:

A+B=72A + B = 72A+B=72
A=B+8A = B + 8A=B+8

Substitute A=B+8A = B + 8A=B+8 into the first equation:
(B+8)+B=72(B + 8) + B = 72(B+8)+B=72
2B+8=722B + 8 = 722B+8=72
2B=642B = 642B=64
B=32B = 32B=32

So, Bob is 32 years old, and Alice is 32+8=4032 + 8 = 4032+8=40.

Example 29: Age Relationship in Five Years

Question:
In 5 years, a father will be 4 times as old as his son. Ten years ago, he was 6 times as old as his son. What are their current ages?

Answer:
Let the son’s age be SSS and the father’s age be FFF.
From the problem, we have:

F+5=4(S+5)F + 5 = 4(S + 5)F+5=4(S+5)
F−10=6(S−10)F – 10 = 6(S – 10)F−10=6(S−10)

From the first equation:
F+5=4(S+5)F + 5 = 4(S + 5)F+5=4(S+5)
F=4S+15F = 4S + 15F=4S+15

Substitute F=4S+15F = 4S + 15F=4S+15 into the second equation:
(4S+15)−10=6(S−10)(4S + 15) – 10 = 6(S – 10)(4S+15)−10=6(S−10)
4S+5=6S−604S + 5 = 6S – 604S+5=6S−60
5+60=6S−4S5 + 60 = 6S – 4S5+60=6S−4S
65=2S65 = 2S65=2S
S=32.5S = 32.5S=32.5

So, the son is approximately 32.5 years old, and the father’s age is 4×32.5+15=1404 \times 32.5 + 15 = 1404×32.5+15=140.

Example 30: Age of a Grandfather and Grandson

Question:
A grandfather is 25 years older than his grandson. In 10 years, the grandfather will be twice as old as the grandson. What are their current ages?

Answer:
Let the grandson’s age be GGG and the grandfather’s age be FFF.
From the problem, we have:

F=G+25F = G + 25F=G+25
F+10=2(G+10)F + 10 = 2(G + 10)F+10=2(G+10)

Substitute F=G+25F = G + 25F=G+25 into the second equation:
(G+25)+10=2(G+10)(G + 25) + 10 = 2(G + 10)(G+25)+10=2(G+10)
G+35=2G+20G + 35 = 2G + 20G+35=2G+20
35−20=2G−G35 – 20 = 2G – G35−20=2G−G
15=G15 = G15=G

So, the grandson is 15 years old, and the grandfather’s age is 15+25=4015 + 25 = 4015+25=40.

Example 31: Ages of Two Friends

Question:
The sum of the ages of two friends, Sam and Tom, is 80 years. Sam is 12 years older than Tom. What are their current ages?

Answer:
Let Sam’s age be SSS and Tom’s age be TTT.
From the problem, we have:

S+T=80S + T = 80S+T=80
S=T+12S = T + 12S=T+12

Substitute S=T+12S = T + 12S=T+12 into the first equation:
(T+12)+T=80(T + 12) + T = 80(T+12)+T=80
2T+12=802T + 12 = 802T+12=80
2T=682T = 682T=68
T=34T = 34T=34

So, Tom is 34 years old, and Sam is 34+12=4634 + 12 = 4634+12=46.

Example 32: Age of a Mother and Son

Question:
The mother’s age is 3 times the age of her son. In 10 years, the mother will be twice as old as her son. How old are the mother and son?

Answer:
Let the son’s age be SSS and the mother’s age be MMM.
From the problem, we have:

M=3SM = 3SM=3S
M+10=2(S+10)M + 10 = 2(S + 10)M+10=2(S+10)

Substitute M=3SM = 3SM=3S into the second equation:
3S+10=2(S+10)3S + 10 = 2(S + 10)3S+10=2(S+10)
3S+10=2S+203S + 10 = 2S + 203S+10=2S+20
3S−2S=20−103S – 2S = 20 – 103S−2S=20−10
S=10S = 10S=10

So, the son is 10 years old, and the mother’s age is 3×10=303 \times 10 = 303×10=30.

Example 33: Difference Between Ages of Two People

Question:
The difference in ages between two people is 18 years. Five years ago, the elder was 3 times as old as the younger. What are their current ages?

Answer:
Let the younger person’s age be YYY and the elder person’s age be EEE.
From the problem, we have:

E−Y=18E – Y = 18E−Y=18
E−5=3(Y−5)E – 5 = 3(Y – 5)E−5=3(Y−5)

Substitute E=Y+18E = Y + 18E=Y+18 into the second equation:
(Y+18)−5=3(Y−5)(Y + 18) – 5 = 3(Y – 5)(Y+18)−5=3(Y−5)
Y+13=3Y−15Y + 13 = 3Y – 15Y+13=3Y−15
13+15=3Y−Y13 + 15 = 3Y – Y13+15=3Y−Y
28=2Y28 = 2Y28=2Y
Y=14Y = 14Y=14

So, the younger person is 14 years old, and the elder person is 14+18=3214 + 18 = 3214+18=32.

Example 34: Father’s Age After Some Years

Question:
A father is 7 years older than his daughter. After 6 years, the father will be twice as old as his daughter. How old are they now?

Answer:
Let the daughter’s age be DDD and the father’s age be FFF.
From the problem, we have:

F=D+7F = D + 7F=D+7
F+6=2(D+6)F + 6 = 2(D + 6)F+6=2(D+6)

Substitute F=D+7F = D + 7F=D+7 into the second equation:
(D+7)+6=2(D+6)(D + 7) + 6 = 2(D + 6)(D+7)+6=2(D+6)
D+13=2D+12D + 13 = 2D + 12D+13=2D+12
13−12=2D−D13 – 12 = 2D – D13−12=2D−D
1=D1 = D1=D

So, the daughter is 1 year old, and the father’s age is 1+7=81 + 7 = 81+7=8.

Example 35: Age Difference Between Two Sisters

Question:
The sum of the ages of two sisters is 56 years. The elder sister is 10 years older than the younger one. What are their current ages?

Answer:
Let the younger sister’s age be YYY and the elder sister’s age be EEE.
From the problem, we have:

E=Y+10E = Y + 10E=Y+10
E+Y=56E + Y = 56E+Y=56

Substitute E=Y+10E = Y + 10E=Y+10 into the second equation:
(Y+10)+Y=56(Y + 10) + Y = 56(Y+10)+Y=56
2Y+10=562Y + 10 = 562Y+10=56
2Y=462Y = 462Y=46
Y=23Y = 23Y=23

So, the younger sister is 23 years old, and the elder sister is 23+10=3323 + 10 = 3323+10=33.

Example 36: Mother and Daughter’s Ages

Question:
The sum of the ages of a mother and her daughter is 50 years. The mother is 4 times as old as her daughter. How old is each?

Answer:
Let the daughter’s age be DDD and the mother’s age be MMM.
From the problem, we have:

M+D=50M + D = 50M+D=50
M=4DM = 4DM=4D

Substitute M=4DM = 4DM=4D into the first equation:
4D+D=504D + D = 504D+D=50
5D=505D = 505D=50
D=10D = 10D=10

So, the daughter is 10 years old, and the mother’s age is 4×10=404 \times 10 = 404×10=40.

Example 37: Difference in Ages of Father and Son

Question:
The father is 8 times as old as his son. In 4 years, the father will be 5 times as old as his son. What are their current ages?

Answer:
Let the son’s age be SSS and the father’s age be FFF.
From the problem, we have:

F=8SF = 8SF=8S
F+4=5(S+4)F + 4 = 5(S + 4)F+4=5(S+4)

Substitute F=8SF = 8SF=8S into the second equation:
8S+4=5(S+4)8S + 4 = 5(S + 4)8S+4=5(S+4)
8S+4=5S+208S + 4 = 5S + 208S+4=5S+20
8S−5S=20−48S – 5S = 20 – 48S−5S=20−4
3S=163S = 163S=16
S=5.33S = 5.33S=5.33

This result doesn’t provide an integer value for age, suggesting there may be an error in the problem’s conditions or assumptions.

Example 38: The Age of Two People

Question:
The sum of the ages of two people is 72 years. One is 12 years older than the other. What are their ages?

Answer:
Let the younger person’s age be YYY and the older person’s age be OOO.
From the problem, we have:

O+Y=72O + Y = 72O+Y=72
O=Y+12O = Y + 12O=Y+12

Substitute O=Y+12O = Y + 12O=Y+12 into the first equation:
(Y+12)+Y=72(Y + 12) + Y = 72(Y+12)+Y=72
2Y+12=722Y + 12 = 722Y+12=72
2Y=602Y = 602Y=60
Y=30Y = 30Y=30

So, the younger person is 30 years old, and the older person is 30+12=4230 + 12 = 4230+12=42.

Example 39: The Father and Daughter Age Problem

Question:
A father is three times as old as his daughter. The sum of their ages is 48 years. What are their ages?

Answer:
Let the daughter’s age be DDD and the father’s age be FFF.
From the problem, we have:

F=3DF = 3DF=3D
F+D=48F + D = 48F+D=48

Substitute F=3DF = 3DF=3D into the second equation:
3D+D=483D + D = 483D+D=48
4D=484D = 484D=48
D=12D = 12D=12

So, the daughter is 12 years old, and the father’s age is 3×12=363 \times 12 = 363×12=36.

Example 40: Ages of Two Brothers

Question:
The sum of the ages of two brothers is 50 years. Five years ago, one of them was 5 times the age of the other. What are their current ages?

Answer:
Let the younger brother’s age be YYY and the older brother’s age be OOO.
From the problem, we have:

O+Y=50O + Y = 50O+Y=50
O−5=5(Y−5)O – 5 = 5(Y – 5)O−5=5(Y−5)

Substitute O=50−YO = 50 – YO=50−Y into the second equation:
(50−Y)−5=5(Y−5)(50 – Y) – 5 = 5(Y – 5)(50−Y)−5=5(Y−5)
45−Y=5Y−2545 – Y = 5Y – 2545−Y=5Y−25
45+25=5Y+Y45 + 25 = 5Y + Y45+25=5Y+Y
70=6Y70 = 6Y70=6Y
Y=11.67Y = 11.67Y=11.67

This result isn’t an integer, so there may be an issue with the problem’s given conditions.

Example 41: The Age of a Grandfather and Grandchild

Question:
The age of a grandfather is 48 years more than his grandchild. In 12 years, the grandfather will be twice as old as his grandchild. What are their ages?

Answer:
Let the grandchild’s age be GGG and the grandfather’s age be FFF.
From the problem, we have:

F=G+48F = G + 48F=G+48
F+12=2(G+12)F + 12 = 2(G + 12)F+12=2(G+12)

Substitute F=G+48F = G + 48F=G+48 into the second equation:
(G+48)+12=2(G+12)(G + 48) + 12 = 2(G + 12)(G+48)+12=2(G+12)
G+60=2G+24G + 60 = 2G + 24G+60=2G+24
60−24=2G−G60 – 24 = 2G – G60−24=2G−G
36=G36 = G36=G

So, the grandchild is 36 years old, and the grandfather’s age is 36+48=8436 + 48 = 8436+48=84.

Example 42: Age of a Brother

Question:
A brother is 4 years older than his sister. The sum of their ages is 32 years. How old are they?

Answer:
Let the sister’s age be SSS and the brother’s age be BBB.
From the problem, we have:

B=S+4B = S + 4B=S+4
B+S=32B + S = 32B+S=32

Substitute B=S+4B = S + 4B=S+4 into the second equation:
(S+4)+S=32(S + 4) + S = 32(S+4)+S=32
2S+4=322S + 4 = 322S+4=32
2S=282S = 282S=28
S=14S = 14S=14

So, the sister is 14 years old, and the brother is 14+4=1814 + 4 = 1814+4=18.

Example 43: A Father and Son’s Ages

Question:
The sum of the father and son’s ages is 72 years. The father is 36 years older than his son. What are their ages?

Answer:
Let the son’s age be SSS and the father’s age be FFF.
From the problem, we have:

F+S=72F + S = 72F+S=72
F=S+36F = S + 36F=S+36

Substitute F=S+36F = S + 36F=S+36 into the first equation:
(S+36)+S=72(S + 36) + S = 72(S+36)+S=72
2S+36=722S + 36 = 722S+36=72
2S=362S = 362S=36
S=18S = 18S=18

So, the son is 18 years old, and the father is 18+36=5418 + 36 = 5418+36=54.

Example 44: Age of a Brother and Sister

Question:
The sum of the ages of two siblings is 46 years. The brother is 6 years older than his sister. How old are they?

Answer:
Let the sister’s age be SSS and the brother’s age be BBB.
From the problem, we have:

B=S+6B = S + 6B=S+6
B+S=46B + S = 46B+S=46

Substitute B=S+6B = S + 6B=S+6 into the second equation:
(S+6)+S=46(S + 6) + S = 46(S+6)+S=46
2S+6=462S + 6 = 462S+6=46
2S=402S = 402S=40
S=20S = 20S=20

So, the sister is 20 years old, and the brother is 20+6=2620 + 6 = 2620+6=26.

Example 45: Age of a Father and Daughter

Question:
The sum of the ages of a father and his daughter is 60 years. The father is 4 times as old as his daughter. What are their ages?

Answer:
Let the daughter’s age be DDD and the father’s age be FFF.
From the problem, we have:

F+D=60F + D = 60F+D=60
F=4DF = 4DF=4D

Substitute F=4DF = 4DF=4D into the first equation:
4D+D=604D + D = 604D+D=60
5D=605D = 605D=60
D=12D = 12D=12

So, the daughter is 12 years old, and the father’s age is 4×12=484 \times 12 = 484×12=48.

Example 46: Mother and Son’s Ages

Question:
A mother is 5 times as old as her son. In 10 years, she will be 3 times as old as her son. What are their current ages?

Answer:
Let the son’s age be SSS and the mother’s age be MMM.
From the problem, we have:

M=5SM = 5SM=5S
M+10=3(S+10)M + 10 = 3(S + 10)M+10=3(S+10)

Substitute M=5SM = 5SM=5S into the second equation:
5S+10=3(S+10)5S + 10 = 3(S + 10)5S+10=3(S+10)
5S+10=3S+305S + 10 = 3S + 305S+10=3S+30
5S−3S=30−105S – 3S = 30 – 105S−3S=30−10
2S=202S = 202S=20
S=10S = 10S=10

So, the son is 10 years old, and the mother’s age is 5×10=505 \times 10 = 505×10=50.

Example 47: The Age of Two Friends

Question:
The sum of the ages of two friends is 52 years. One of them is 6 years older than the other. What are their ages?

Answer:
Let the younger person’s age be YYY and the older person’s age be OOO.
From the problem, we have:

O+Y=52O + Y = 52O+Y=52
O=Y+6O = Y + 6O=Y+6

Substitute O=Y+6O = Y + 6O=Y+6 into the first equation:
(Y+6)+Y=52(Y + 6) + Y = 52(Y+6)+Y=52
2Y+6=522Y + 6 = 522Y+6=52
2Y=462Y = 462Y=46
Y=23Y = 23Y=23

So, the younger person is 23 years old, and the older person is 23+6=2923 + 6 = 2923+6=29.

Example 48: The Difference in Ages of Two Sisters

Question:
The age of the elder sister is 4 times that of her younger sister. 10 years ago, the elder sister was 6 times as old as the younger sister. What are their current ages?

Answer:
Let the younger sister’s age be YYY and the elder sister’s age be EEE.
From the problem, we have:

E=4YE = 4YE=4Y
E−10=6(Y−10)E – 10 = 6(Y – 10)E−10=6(Y−10)

Substitute E=4YE = 4YE=4Y into the second equation:
4Y−10=6(Y−10)4Y – 10 = 6(Y – 10)4Y−10=6(Y−10)
4Y−10=6Y−604Y – 10 = 6Y – 604Y−10=6Y−60
−10+60=6Y−4Y-10 + 60 = 6Y – 4Y−10+60=6Y−4Y
50=2Y50 = 2Y50=2Y
Y=25Y = 25Y=25

So, the younger sister is 25 years old, and the elder sister is 4×25=1004 \times 25 = 1004×25=100.

Example 49: Father’s Age Relative to Son’s

Question:
A father is twice as old as his son. 5 years ago, he was 3 times as old as his son. What are their current ages?

Answer:
Let the son’s age be SSS and the father’s age be FFF.
From the problem, we have:

F=2SF = 2SF=2S
F−5=3(S−5)F – 5 = 3(S – 5)F−5=3(S−5)

Substitute F=2SF = 2SF=2S into the second equation:
2S−5=3(S−5)2S – 5 = 3(S – 5)2S−5=3(S−5)
2S−5=3S−152S – 5 = 3S – 152S−5=3S−15
−5+15=3S−2S-5 + 15 = 3S – 2S−5+15=3S−2S
10=S10 = S10=S

So, the son is 10 years old, and the father’s age is 2×10=202 \times 10 = 202×10=20.

Example 50: Father and Daughter’s Age Relationship

Question:
The father is 4 times as old as his daughter. In 8 years, the father will be twice as old as his daughter. What are their current ages?

Answer:
Let the daughter’s age be DDD and the father’s age be FFF.
From the problem, we have:

F=4DF = 4DF=4D
F+8=2(D+8)F + 8 = 2(D + 8)F+8=2(D+8)

Substitute F=4DF = 4DF=4D into the second equation:
4D+8=2(D+8)4D + 8 = 2(D + 8)4D+8=2(D+8)
4D+8=2D+164D + 8 = 2D + 164D+8=2D+16
4D−2D=16−84D – 2D = 16 – 84D−2D=16−8
2D=82D = 82D=8
D=4D = 4D=4

So, the daughter is 4 years old, and the father’s age is 4×4=164 \times 4 = 164×4=16.

Example 51: The Sum of Ages of Two People

Question:
The sum of the ages of two persons is 80 years. One person is 10 years older than the other. What are their ages?

Answer:
Let the younger person’s age be YYY and the older person’s age be OOO.
From the problem, we have:

O+Y=80O + Y = 80O+Y=80
O=Y+10O = Y + 10O=Y+10

Substitute O=Y+10O = Y + 10O=Y+10 into the first equation:
(Y+10)+Y=80(Y + 10) + Y = 80(Y+10)+Y=80
2Y+10=802Y + 10 = 802Y+10=80
2Y=702Y = 702Y=70
Y=35Y = 35Y=35

So, the younger person is 35 years old, and the older person is 35+10=4535 + 10 = 4535+10=45.

Example 52: The Age of Two Friends

Question:
Two friends have ages that differ by 6 years. In 3 years, the sum of their ages will be 54 years. What are their current ages?

Answer:
Let the younger friend’s age be YYY and the older friend’s age be OOO.
From the problem, we have:

O=Y+6O = Y + 6O=Y+6
O+3+Y+3=54O + 3 + Y + 3 = 54O+3+Y+3=54

Substitute O=Y+6O = Y + 6O=Y+6 into the second equation:
(Y+6+3)+(Y+3)=54(Y + 6 + 3) + (Y + 3) = 54(Y+6+3)+(Y+3)=54
2Y+12=542Y + 12 = 542Y+12=54
2Y=422Y = 422Y=42
Y=21Y = 21Y=21

So, the younger friend is 21 years old, and the older friend is 21+6=2721 + 6 = 2721+6=27.

Example 53: Sum of Ages of a Mother and Daughter

Question:
The sum of the ages of a mother and her daughter is 72 years. The mother is 6 times as old as her daughter. What are their ages?

Answer:
Let the daughter’s age be DDD and the mother’s age be MMM.
From the problem, we have:

M+D=72M + D = 72M+D=72
M=6DM = 6DM=6D

Substitute M=6DM = 6DM=6D into the first equation:
6D+D=726D + D = 726D+D=72
7D=727D = 727D=72
D=10.29D = 10.29D=10.29

The result isn’t a whole number, suggesting there might be an issue with the problem’s conditions or assumptions.

Example 54: A Father’s Age and His Son

Question:
The father is 5 times as old as his son. In 10 years, the father will be 3 times as old as his son. What are their current ages?

Answer:
Let the son’s age be SSS and the father’s age be FFF.
From the problem, we have:

F=5SF = 5SF=5S
F+10=3(S+10)F + 10 = 3(S + 10)F+10=3(S+10)

So, the son is 10 years old, and the father’s age is 5×10=505 \times 10 = 505×10=50.

Example 55: Age of a Father and Daughter

Question:
The father is 5 times as old as his daughter. In 10 years, he will be 3 times as old as his daughter. What are their current ages?

Answer:
Let the daughter’s age be DDD and the father’s age be FFF.
From the problem, we have:

F=5DF = 5DF=5D
F+10=3(D+10)F + 10 = 3(D + 10)F+10=3(D+10)

Substitute F=5DF = 5DF=5D into the second equation:
5D+10=3(D+10)5D + 10 = 3(D + 10)5D+10=3(D+10)
5D+10=3D+305D + 10 = 3D + 305D+10=3D+30
5D−3D=30−105D – 3D = 30 – 105D−3D=30−10
2D=202D = 202D=20
D=10D = 10D=10

So, the daughter is 10 years old, and the father’s age is 5×10=505 \times 10 = 505×10=50.

Example 56: Age Difference Problem

Question:
The difference in age between a father and his daughter is 24 years. In 4 years, the father will be 4 times as old as his daughter. What are their current ages?

Answer:
Let the daughter’s age be DDD and the father’s age be FFF.
From the problem, we have:

F=D+24F = D + 24F=D+24
F+4=4(D+4)F + 4 = 4(D + 4)F+4=4(D+4)

Substitute F=D+24F = D + 24F=D+24 into the second equation:
(D+24)+4=4(D+4)(D + 24) + 4 = 4(D + 4)(D+24)+4=4(D+4)
D+28=4D+16D + 28 = 4D + 16D+28=4D+16
28−16=4D−D28 – 16 = 4D – D28−16=4D−D
12=3D12 = 3D12=3D
D=4D = 4D=4

So, the daughter is 4 years old, and the father’s age is 4+24=284 + 24 = 284+24=28.

Example 57: A Father and Son’s Age

Question:
The father is 20 years older than his son. In 10 years, the father will be twice as old as his son. What are their ages now?

Answer:
Let the son’s age be SSS and the father’s age be FFF.
From the problem, we have:

F=S+20F = S + 20F=S+20
F+10=2(S+10)F + 10 = 2(S + 10)F+10=2(S+10)

Substitute F=S+20F = S + 20F=S+20 into the second equation:
(S+20)+10=2(S+10)(S + 20) + 10 = 2(S + 10)(S+20)+10=2(S+10)
S+30=2S+20S + 30 = 2S + 20S+30=2S+20
30−20=2S−S30 – 20 = 2S – S30−20=2S−S
10=S10 = S10=S

So, the son is 10 years old, and the father’s age is 10+20=3010 + 20 = 3010+20=30.

Example 58: Age of a Mother and Child

Question:
A mother is 30 years older than her child. In 5 years, she will be twice as old as her child. What are their current ages?

Answer:
Let the child’s age be CCC and the mother’s age be MMM.
From the problem, we have:

M=C+30M = C + 30M=C+30
M+5=2(C+5)M + 5 = 2(C + 5)M+5=2(C+5)

Substitute M=C+30M = C + 30M=C+30 into the second equation:
(C+30)+5=2(C+5)(C + 30) + 5 = 2(C + 5)(C+30)+5=2(C+5)
C+35=2C+10C + 35 = 2C + 10C+35=2C+10
35−10=2C−C35 – 10 = 2C – C35−10=2C−C
25=C25 = C25=C

So, the child is 25 years old, and the mother’s age is 25+30=5525 + 30 = 5525+30=55.

Example 59: A Brother and Sister’s Ages

Question:
A brother is 6 years older than his sister. The sum of their ages is 36 years. What are their ages?

Answer:
Let the sister’s age be SSS and the brother’s age be BBB.
From the problem, we have:

B=S+6B = S + 6B=S+6
B+S=36B + S = 36B+S=36

Substitute B=S+6B = S + 6B=S+6 into the second equation:
(S+6)+S=36(S + 6) + S = 36(S+6)+S=36
2S+6=362S + 6 = 362S+6=36
2S=302S = 302S=30
S=15S = 15S=15

So, the sister is 15 years old, and the brother is 15+6=2115 + 6 = 2115+6=21.

Example 60: Age Problem Involving a Mother and Daughter

Question:
A mother is 10 times as old as her daughter. In 20 years, the mother will be 4 times as old as her daughter. What are their current ages?

Answer:
Let the daughter’s age be DDD and the mother’s age be MMM.
From the problem, we have:

M=10DM = 10DM=10D
M+20=4(D+20)M + 20 = 4(D + 20)M+20=4(D+20)

Substitute M=10DM = 10DM=10D into the second equation:
10D+20=4(D+20)10D + 20 = 4(D + 20)10D+20=4(D+20)
10D+20=4D+8010D + 20 = 4D + 8010D+20=4D+80
10D−4D=80−2010D – 4D = 80 – 2010D−4D=80−20
6D=606D = 606D=60
D=10D = 10D=10

So, the daughter is 10 years old, and the mother’s age is 10×10=10010 \times 10 = 10010×10=100.

Example 61: Mother and Son’s Age

Question:
A mother is 30 years older than her son. The sum of their ages is 90 years. What are their current ages?

Answer:
Let the son’s age be SSS and the mother’s age be MMM.
From the problem, we have:

M=S+30M = S + 30M=S+30
M+S=90M + S = 90M+S=90

Substitute M=S+30M = S + 30M=S+30 into the second equation:
(S+30)+S=90(S + 30) + S = 90(S+30)+S=90
2S+30=902S + 30 = 902S+30=90
2S=602S = 602S=60
S=30S = 30S=30

So, the son is 30 years old, and the mother’s age is 30+30=6030 + 30 = 6030+30=60.

Example 62: Age Difference Between Two People

Question:
The difference in age between two friends is 9 years. Five years ago, one friend was 4 times as old as the other. What are their current ages?

Answer:
Let the younger friend’s age be YYY and the older friend’s age be OOO.
From the problem, we have:

O=Y+9O = Y + 9O=Y+9
O−5=4(Y−5)O – 5 = 4(Y – 5)O−5=4(Y−5)

Substitute O=Y+9O = Y + 9O=Y+9 into the second equation:
(Y+9)−5=4(Y−5)(Y + 9) – 5 = 4(Y – 5)(Y+9)−5=4(Y−5)
Y+4=4Y−20Y + 4 = 4Y – 20Y+4=4Y−20
4+20=4Y−Y4 + 20 = 4Y – Y4+20=4Y−Y
24=3Y24 = 3Y24=3Y
Y=8Y = 8Y=8

So, the younger friend is 8 years old, and the older friend is 8+9=178 + 9 = 178+9=17.

Example 63: Mother and Daughter’s Age Problem

Question:
A mother is 7 times as old as her daughter. In 6 years, the mother will be 4 times as old as her daughter. What are their current ages?

Answer:
Let the daughter’s age be DDD and the mother’s age be MMM.
From the problem, we have:

M=7DM = 7DM=7D
M+6=4(D+6)M + 6 = 4(D + 6)M+6=4(D+6)

Substitute M=7DM = 7DM=7D into the second equation:
7D+6=4(D+6)7D + 6 = 4(D + 6)7D+6=4(D+6)
7D+6=4D+247D + 6 = 4D + 247D+6=4D+24
7D−4D=24−67D – 4D = 24 – 67D−4D=24−6
3D=183D = 183D=18
D=6D = 6D=6

So, the daughter is 6 years old, and the mother’s age is 7×6=427 \times 6 = 427×6=42.

Example 64: A Man and His Wife

Question:
A man is 6 years older than his wife. In 8 years, the man will be twice as old as his wife. What are their current ages?

Answer:
Let the wife’s age be WWW and the man’s age be MMM.
From the problem, we have:

M=W+6M = W + 6M=W+6
M+8=2(W+8)M + 8 = 2(W + 8)M+8=2(W+8)

Substitute M=W+6M = W + 6M=W+6 into the second equation:
(W+6)+8=2(W+8)(W + 6) + 8 = 2(W + 8)(W+6)+8=2(W+8)
W+14=2W+16W + 14 = 2W + 16W+14=2W+16
14−16=2W−W14 – 16 = 2W – W14−16=2W−W
−2=W-2 = W−2=W

This solution does not make sense since ages cannot be negative, indicating the problem may not have been set up properly.

Example 65: The Ages of Two Friends

Question:
Two friends have a combined age of 60 years. One friend is 10 years older than the other. What are their ages?

Answer:
Let the younger friend’s age be YYY and the older friend’s age be OOO.
From the problem, we have:

O=Y+10O = Y + 10O=Y+10
O+Y=60O + Y = 60O+Y=60

Substitute O=Y+10O = Y + 10O=Y+10 into the second equation:
(Y+10)+Y=60(Y + 10) + Y = 60(Y+10)+Y=60
2Y+10=602Y + 10 = 602Y+10=60
2Y=502Y = 502Y=50
Y=25Y = 25Y=25

So, the younger friend is 25 years old, and the older friend is 25+10=3525 + 10 = 3525+10=35.

Example 66: Grandfather and Grandson’s Age

Question:
The grandfather is 42 years older than his grandson. In 6 years, the grandfather will be 3 times as old as his grandson. What are their current ages?

Answer:
Let the grandson’s age be GGG and the grandfather’s age be FFF.
From the problem, we have:

F=G+42F = G + 42F=G+42
F+6=3(G+6)F + 6 = 3(G + 6)F+6=3(G+6)

Substitute F=G+42F = G + 42F=G+42 into the second equation:
(G+42)+6=3(G+6)(G + 42) + 6 = 3(G + 6)(G+42)+6=3(G+6)
G+48=3G+18G + 48 = 3G + 18G+48=3G+18
48−18=3G−G48 – 18 = 3G – G48−18=3G−G
30=2G30 = 2G30=2G
G=15G = 15G=15

So, the grandson is 15 years old, and the grandfather’s age is 15+42=5715 + 42 = 5715+42=57.

Example 67: Age Problem Involving a Sister

Question:
A sister is 4 years younger than her brother. In 10 years, the brother will be 3 times as old as the sister. What are their ages?

Answer:
Let the sister’s age be SSS and the brother’s age be BBB.
From the problem, we have:

S=B−4S = B – 4S=B−4
B+10=3(S+10)B + 10 = 3(S + 10)B+10=3(S+10)

Substitute S=B−4S = B – 4S=B−4 into the second equation:
B+10=3((B−4)+10)B + 10 = 3((B – 4) + 10)B+10=3((B−4)+10)
B+10=3(B+6)B + 10 = 3(B + 6)B+10=3(B+6)
B+10=3B+18B + 10 = 3B + 18B+10=3B+18
10−18=3B−B10 – 18 = 3B – B10−18=3B−B
−8=2B-8 = 2B−8=2B
B=−4B = -4B=−4

This solution does not make sense, as ages cannot be negative. The problem likely contains an error or is unsolvable.

Example 68: A Woman and Her Son

Question:
A woman is 8 times as old as her son. In 4 years, the woman will be 6 times as old as her son. What are their current ages?

Answer:
Let the son’s age be SSS and the woman’s age be WWW.
From the problem, we have:

W=8SW = 8SW=8S
W+4=6(S+4)W + 4 = 6(S + 4)W+4=6(S+4)

Substitute W=8SW = 8SW=8S into the second equation:
8S+4=6(S+4)8S + 4 = 6(S + 4)8S+4=6(S+4)
8S+4=6S+248S + 4 = 6S + 248S+4=6S+24
8S−6S=24−48S – 6S = 24 – 48S−6S=24−4
2S=202S = 202S=20
S=10S = 10S=10

So, the son is 10 years old, and the woman’s age is 8×10=808 \times 10 = 808×10=80.

Example 69: Father and Daughter’s Age

Question:
A father is 30 years older than his daughter. In 10 years, the father will be 3 times as old as his daughter. What are their current ages?

Answer:
Let the daughter’s age be DDD and the father’s age be FFF.
From the problem, we have:

F=D+30F = D + 30F=D+30
F+10=3(D+10)F + 10 = 3(D + 10)F+10=3(D+10)

Substitute F=D+30F = D + 30F=D+30 into the second equation:
(D+30)+10=3(D+10)(D + 30) + 10 = 3(D + 10)(D+30)+10=3(D+10)
D+40=3D+30D + 40 = 3D + 30D+40=3D+30
40−30=3D−D40 – 30 = 3D – D40−30=3D−D
10=2D10 = 2D10=2D
D=5D = 5D=5

So, the daughter is 5 years old, and the father’s age is 5+30=355 + 30 = 355+30=35.

Example 70: Mother and Daughter

Question:
A mother is 4 times as old as her daughter. In 8 years, she will be 5 times as old as her daughter. What are their current ages?

Answer:
Let the daughter’s age be DDD and the mother’s age be MMM.
From the problem, we have:

M=4DM = 4DM=4D
M+8=5(D+8)M + 8 = 5(D + 8)M+8=5(D+8)

Substitute M=4DM = 4DM=4D into the second equation:
4D+8=5(D+8)4D + 8 = 5(D + 8)4D+8=5(D+8)
4D+8=5D+404D + 8 = 5D + 404D+8=5D+40
8−40=5D−4D8 – 40 = 5D – 4D8−40=5D−4D
−32=D-32 = D−32=D

This result indicates a contradiction, meaning the problem might be impossible as phrased.

Also read : Verbal reasoning question on Insert missing character

Example 71: Brother and Sister’s Age

Question:
A brother is 3 times as old as his sister. The sum of their ages is 40 years. What are their current ages?

Answer:
Let the sister’s age be SSS and the brother’s age be BBB.
From the problem, we have:

B=3SB = 3SB=3S
B+S=40B + S = 40B+S=40

Substitute B=3SB = 3SB=3S into the second equation:
3S+S=403S + S = 403S+S=40
4S=404S = 404S=40
S=10S = 10S=10

So, the sister is 10 years old, and the brother is 3×10=303 \times 10 = 303×10=30.

Example 72: Father and Son Age Problem

Question:
A father is 6 times as old as his son. 20 years ago, the father was 8 times as old as his son. What are their current ages?

Answer:
Let the son’s age be SSS and the father’s age be FFF.
From the problem, we have:

F=6SF = 6SF=6S
F−20=8(S−20)F – 20 = 8(S – 20)F−20=8(S−20)

Substitute F=6SF = 6SF=6S into the second equation:
6S−20=8(S−20)6S – 20 = 8(S – 20)6S−20=8(S−20)
6S−20=8S−1606S – 20 = 8S – 1606S−20=8S−160
−20+160=8S−6S-20 + 160 = 8S – 6S−20+160=8S−6S
140=2S140 = 2S140=2S
S=70S = 70S=70

So, the son is 70 years old, and the father’s age is 6×70=4206 \times 70 = 4206×70=420, which clearly doesn’t make sense in the real-world context, suggesting a flaw in the problem or its assumptions.

Example 73: Age of a Grandfather

Question:
A grandfather is 45 years older than his granddaughter. In 5 years, he will be 5 times as old as his granddaughter. What are their current ages?

Answer:
Let the granddaughter’s age be GGG and the grandfather’s age be FFF.
From the problem, we have:

F=G+45F = G + 45F=G+45
F+5=5(G+5)F + 5 = 5(G + 5)F+5=5(G+5)

Substitute F=G+45F = G + 45F=G+45 into the second equation:
(G+45)+5=5(G+5)(G + 45) + 5 = 5(G + 5)(G+45)+5=5(G+5)
G+50=5G+25G + 50 = 5G + 25G+50=5G+25
50−25=5G−G50 – 25 = 5G – G50−25=5G−G
25=4G25 = 4G25=4G
G=6.25G = 6.25G=6.25

This is not a valid solution because ages must be whole numbers, indicating that the problem setup may need to be reconsidered.

Example 74: A Man and His Daughter

Question:
A man is 20 years older than his daughter. In 4 years, he will be twice as old as his daughter. What are their current ages?

Answer:
Let the daughter’s age be DDD and the man’s age be MMM.
From the problem, we have:

M=D+20M = D + 20M=D+20
M+4=2(D+4)M + 4 = 2(D + 4)M+4=2(D+4)

Substitute M=D+20M = D + 20M=D+20 into the second equation:
(D+20)+4=2(D+4)(D + 20) + 4 = 2(D + 4)(D+20)+4=2(D+4)
D+24=2D+8D + 24 = 2D + 8D+24=2D+8
24−8=2D−D24 – 8 = 2D – D24−8=2D−D
16=D16 = D16=D

So, the daughter is 16 years old, and the man’s age is 16+20=3616 + 20 = 3616+20=36.

Example 75: Age of a Sister

Question:
A sister is 4 years younger than her brother. In 5 years, she will be half as old as her brother. What are their current ages?

Answer:
Let the sister’s age be SSS and the brother’s age be BBB.
From the problem, we have:

S=B−4S = B – 4S=B−4
S+5=12(B+5)S + 5 = \frac{1}{2}(B + 5)S+5=21(B+5)

Substitute S=B−4S = B – 4S=B−4 into the second equation:
(B−4)+5=12(B+5)(B – 4) + 5 = \frac{1}{2}(B + 5)(B−4)+5=21(B+5)
B+1=12(B+5)B + 1 = \frac{1}{2}(B + 5)B+1=21(B+5)
2(B+1)=B+52(B + 1) = B + 52(B+1)=B+5
2B+2=B+52B + 2 = B + 52B+2=B+5
2B−B=5−22B – B = 5 – 22B−B=5−2
B=3B = 3B=3

So, the brother is 3 years old, and the sister is 3−4=−13 – 4 = -13−4=−1, which doesn’t make sense. This again suggests the problem setup might not be valid.

Example 76: Age Problem Involving a Family

Question:
The sum of the ages of a mother and her daughter is 50 years. The mother is 10 years older than twice the daughter’s age. What are their current ages?

Answer:
Let the daughter’s age be DDD and the mother’s age be MMM.
From the problem, we have:

M+D=50M + D = 50M+D=50
M=2D+10M = 2D + 10M=2D+10

Substitute M=2D+10M = 2D + 10M=2D+10 into the first equation:
(2D+10)+D=50(2D + 10) + D = 50(2D+10)+D=50
3D+10=503D + 10 = 503D+10=50
3D=403D = 403D=40
D=403≈13.33D = \frac{40}{3} \approx 13.33D=340≈13.33

So, the daughter is about 13.33 years old, and the mother’s age is 2×13.33+10=36.662 \times 13.33 + 10 = 36.662×13.33+10=36.66, which doesn’t yield an integer solution, suggesting an issue with the problem.

Example 77: A Father’s Age

Question:
A father is 4 times as old as his son. In 12 years, the father will be twice as old as his son. What are their current ages?

Answer:
Let the son’s age be SSS and the father’s age be FFF.
From the problem, we have:

F=4SF = 4SF=4S
F+12=2(S+12)F + 12 = 2(S + 12)F+12=2(S+12)

Substitute F=4SF = 4SF=4S into the second equation:
4S+12=2(S+12)4S + 12 = 2(S + 12)4S+12=2(S+12)
4S+12=2S+244S + 12 = 2S + 244S+12=2S+24
4S−2S=24−124S – 2S = 24 – 124S−2S=24−12
2S=122S = 122S=12
S=6S = 6S=6

So, the son is 6 years old, and the father’s age is 4×6=244 \times 6 = 244×6=24.

Example 78: Mother and Daughter’s Age

Question:
A mother is 3 times as old as her daughter. In 10 years, she will be twice as old as her daughter. What are their current ages?

Answer:
Let the daughter’s age be DDD and the mother’s age be MMM.
From the problem, we have:

M=3DM = 3DM=3D
M+10=2(D+10)M + 10 = 2(D + 10)M+10=2(D+10)

Substitute M=3DM = 3DM=3D into the second equation:
3D+10=2(D+10)3D + 10 = 2(D + 10)3D+10=2(D+10)
3D+10=2D+203D + 10 = 2D + 203D+10=2D+20
3D−2D=20−103D – 2D = 20 – 103D−2D=20−10
D=10D = 10D=10

So, the daughter is 10 years old, and the mother’s age is 3×10=303 \times 10 = 303×10=30.

Example 79: Brother’s and Sister’s Age

Question:
A brother is 2 years older than his sister. In 4 years, he will be 3 times as old as his sister. What are their current ages?

Answer:
Let the sister’s age be SSS and the brother’s age be BBB.
From the problem, we have:

B=S+2B = S + 2B=S+2
B+4=3(S+4)B + 4 = 3(S + 4)B+4=3(S+4)

Substitute B=S+2B = S + 2B=S+2 into the second equation:
(S+2)+4=3(S+4)(S + 2) + 4 = 3(S + 4)(S+2)+4=3(S+4)
S+6=3S+12S + 6 = 3S + 12S+6=3S+12
6−12=3S−S6 – 12 = 3S – S6−12=3S−S
−6=2S-6 = 2S−6=2S
S=−3S = -3S=−3

This result is not possible, which indicates that the problem setup is not correct or the data is flawed.

Example 80: A Father and His Twin Sons

Question:
A father is 30 years older than his twin sons. In 5 years, he will be 4 times as old as one of his sons. What are their current ages?

Answer:
Let the sons’ age be SSS and the father’s age be FFF.
From the problem, we have:

F=S+30F = S + 30F=S+30
F+5=4(S+5)F + 5 = 4(S + 5)F+5=4(S+5)

Substitute F=S+30F = S + 30F=S+30 into the second equation:
(S+30)+5=4(S+5)(S + 30) + 5 = 4(S + 5)(S+30)+5=4(S+5)
S+35=4S+20S + 35 = 4S + 20S+35=4S+20
35−20=4S−S35 – 20 = 4S – S35−20=4S−S
15=3S15 = 3S15=3S
S=5S = 5S=5

So, the sons are each 5 years old, and the father is 5+30=355 + 30 = 355+30=35.

Example 81: A Man’s Age

Question:
A man is 12 years older than his son. In 4 years, he will be twice as old as his son. What are their current ages?

Answer:
Let the son’s age be SSS and the man’s age be MMM.
From the problem, we have:

M=S+12M = S + 12M=S+12
M+4=2(S+4)M + 4 = 2(S + 4)M+4=2(S+4)

Substitute M=S+12M = S + 12M=S+12 into the second equation:
(S+12)+4=2(S+4)(S + 12) + 4 = 2(S + 4)(S+12)+4=2(S+4)
S+16=2S+8S + 16 = 2S + 8S+16=2S+8
16−8=2S−S16 – 8 = 2S – S16−8=2S−S
8=S8 = S8=S

So, the son is 8 years old, and the man’s age is 8+12=208 + 12 = 208+12=20.

Example 82: A Father’s Age

Question:
A father is 5 times as old as his son. In 15 years, he will be 3 times as old as his son. What are their current ages?

Answer:
Let the son’s age be SSS and the father’s age be FFF.
From the problem, we have:

F=5SF = 5SF=5S
F+15=3(S+15)F + 15 = 3(S + 15)F+15=3(S+15)

Substitute F=5SF = 5SF=5S into the second equation:
5S+15=3(S+15)5S + 15 = 3(S + 15)5S+15=3(S+15)
5S+15=3S+455S + 15 = 3S + 455S+15=3S+45
5S−3S=45−155S – 3S = 45 – 155S−3S=45−15
2S=302S = 302S=30
S=15S = 15S=15

So, the son is 15 years old, and the father’s age is 5×15=755 \times 15 = 755×15=75.

Example 83: Sister and Brother’s Age

Question:
A brother is 6 years older than his sister. In 3 years, he will be 3 times as old as his sister. What are their current ages?

Answer:
Let the sister’s age be SSS and the brother’s age be BBB.
From the problem, we have:

B=S+6B = S + 6B=S+6
B+3=3(S+3)B + 3 = 3(S + 3)B+3=3(S+3)

Substitute B=S+6B = S + 6B=S+6 into the second equation:
(S+6)+3=3(S+3)(S + 6) + 3 = 3(S + 3)(S+6)+3=3(S+3)
S+9=3S+9S + 9 = 3S + 9S+9=3S+9
9−9=3S−S9 – 9 = 3S – S9−9=3S−S
0=2S0 = 2S0=2S
S=0S = 0S=0

This result is not possible because the sister cannot be 0 years old, suggesting the problem setup may be flawed.

Example 84: Age Problem with a Teacher and Student

Question:
A teacher is 7 years older than his student. In 6 years, the teacher will be twice as old as the student. What are their current ages?

Answer:
Let the student’s age be SSS and the teacher’s age be TTT.
From the problem, we have:

T=S+7T = S + 7T=S+7
T+6=2(S+6)T + 6 = 2(S + 6)T+6=2(S+6)

Substitute T=S+7T = S + 7T=S+7 into the second equation:
(S+7)+6=2(S+6)(S + 7) + 6 = 2(S + 6)(S+7)+6=2(S+6)
S+13=2S+12S + 13 = 2S + 12S+13=2S+12
13−12=2S−S13 – 12 = 2S – S13−12=2S−S
1=S1 = S1=S

So, the student is 1 year old, and the teacher is 1+7=81 + 7 = 81+7=8.

Example 85: Father and Son’s Age

Question:
A father is 10 years older than his son. In 10 years, he will be twice as old as his son. What are their current ages?

Answer:
Let the son’s age be SSS and the father’s age be FFF.
From the problem, we have:

F=S+10F = S + 10F=S+10
F+10=2(S+10)F + 10 = 2(S + 10)F+10=2(S+10)

Substitute F=S+10F = S + 10F=S+10 into the second equation:
(S+10)+10=2(S+10)(S + 10) + 10 = 2(S + 10)(S+10)+10=2(S+10)
S+20=2S+20S + 20 = 2S + 20S+20=2S+20
20−20=2S−S20 – 20 = 2S – S20−20=2S−S
0=S0 = S0=S

So, the son is 0 years old, and the father’s age is 0+10=100 + 10 = 100+10=10, which means the problem setup is invalid in real-world terms.

Example 86: Mother and Daughter’s Age

Question:
A mother is 4 times as old as her daughter. In 5 years, the mother will be 3 times as old as her daughter. What are their current ages?

Answer:
Let the daughter’s age be DDD and the mother’s age be MMM.
From the problem, we have:

M=4DM = 4DM=4D
M+5=3(D+5)M + 5 = 3(D + 5)M+5=3(D+5)

Substitute M=4DM = 4DM=4D into the second equation:
4D+5=3(D+5)4D + 5 = 3(D + 5)4D+5=3(D+5)
4D+5=3D+154D + 5 = 3D + 154D+5=3D+15
4D−3D=15−54D – 3D = 15 – 54D−3D=15−5
D=10D = 10D=10

So, the daughter is 10 years old, and the mother’s age is 4×10=404 \times 10 = 404×10=40.

Example 87: Grandfather and Grandson

Question:
A grandfather is 60 years older than his grandson. In 15 years, he will be 4 times as old as his grandson. What are their current ages?

Answer:
Let the grandson’s age be GGG and the grandfather’s age be FFF.
From the problem, we have:

F=G+60F = G + 60F=G+60
F+15=4(G+15)F + 15 = 4(G + 15)F+15=4(G+15)

Substitute F=G+60F = G + 60F=G+60 into the second equation:
(G+60)+15=4(G+15)(G + 60) + 15 = 4(G + 15)(G+60)+15=4(G+15)
G+75=4G+60G + 75 = 4G + 60G+75=4G+60
75−60=4G−G75 – 60 = 4G – G75−60=4G−G
15=3G15 = 3G15=3G
G=5G = 5G=5

So, the grandson is 5 years old, and the grandfather’s age is 5+60=655 + 60 = 655+60=65.

Example 88: A Father’s Age

Question:
A father is 25 years older than his son. In 5 years, the father will be twice as old as the son. What are their current ages?

Answer:
Let the son’s age be SSS and the father’s age be FFF.
From the problem, we have:

F=S+25F = S + 25F=S+25
F+5=2(S+5)F + 5 = 2(S + 5)F+5=2(S+5)

Substitute F=S+25F = S + 25F=S+25 into the second equation:
(S+25)+5=2(S+5)(S + 25) + 5 = 2(S + 5)(S+25)+5=2(S+5)
S+30=2S+10S + 30 = 2S + 10S+30=2S+10
30−10=2S−S30 – 10 = 2S – S30−10=2S−S
20=S20 = S20=S

So, the son is 20 years old, and the father’s age is 20+25=4520 + 25 = 4520+25=45.

Example 89: Age of a Man and His Son

Question:
A man is 8 times as old as his son. In 12 years, the man will be 5 times as old as his son. What are their current ages?

Answer:
Let the son’s age be SSS and the man’s age be MMM.
From the problem, we have:

M=8SM = 8SM=8S
M+12=5(S+12)M + 12 = 5(S + 12)M+12=5(S+12)

Substitute M=8SM = 8SM=8S into the second equation:
8S+12=5(S+12)8S + 12 = 5(S + 12)8S+12=5(S+12)
8S+12=5S+608S + 12 = 5S + 608S+12=5S+60
8S−5S=60−128S – 5S = 60 – 128S−5S=60−12
3S=483S = 483S=48
S=16S = 16S=16

So, the son is 16 years old, and the man’s age is 8×16=1288 \times 16 = 1288×16=128.

Example 90: Teacher and Student

Question:
A teacher is 18 years older than her student. In 6 years, the teacher will be 3 times as old as the student. What are their current ages?

Answer:
Let the student’s age be SSS and the teacher’s age be TTT.
From the problem, we have:

T=S+18T = S + 18T=S+18
T+6=3(S+6)T + 6 = 3(S + 6)T+6=3(S+6)

Substitute T=S+18T = S + 18T=S+18 into the second equation:
(S+18)+6=3(S+6)(S + 18) + 6 = 3(S + 6)(S+18)+6=3(S+6)
S+24=3S+18S + 24 = 3S + 18S+24=3S+18
24−18=3S−S24 – 18 = 3S – S24−18=3S−S
6=2S6 = 2S6=2S
S=3S = 3S=3

So, the student is 3 years old, and the teacher’s age is 3+18=213 + 18 = 213+18=21.

Example 91: Grandfather and Granddaughter

Question:
A grandfather is 48 years older than his granddaughter. In 4 years, the grandfather will be 3 times as old as his granddaughter. What are their current ages?

Answer:
Let the granddaughter’s age be GGG and the grandfather’s age be FFF.
From the problem, we have:

F=G+48F = G + 48F=G+48
F+4=3(G+4)F + 4 = 3(G + 4)F+4=3(G+4)

Substitute F=G+48F = G + 48F=G+48 into the second equation:
(G+48)+4=3(G+4)(G + 48) + 4 = 3(G + 4)(G+48)+4=3(G+4)
G+52=3G+12G + 52 = 3G + 12G+52=3G+12
52−12=3G−G52 – 12 = 3G – G52−12=3G−G
40=2G40 = 2G40=2G
G=20G = 20G=20

So, the granddaughter is 20 years old, and the grandfather’s age is 20+48=6820 + 48 = 6820+48=68.

Example 92: A Father and His Two Sons

Question:
A father is 4 times as old as his older son. In 10 years, the father will be 3 times as old as his older son. The younger son is half as old as the older son. What are their current ages?

Answer:
Let the older son’s age be SSS, the younger son’s age be YYY, and the father’s age be FFF.
From the problem, we have:

F=4SF = 4SF=4S
F+10=3(S+10)F + 10 = 3(S + 10)F+10=3(S+10)
Y=12SY = \frac{1}{2}SY=21S

Substitute F=4SF = 4SF=4S into the second equation:
4S+10=3(S+10)4S + 10 = 3(S + 10)4S+10=3(S+10)
4S+10=3S+304S + 10 = 3S + 304S+10=3S+30
4S−3S=30−104S – 3S = 30 – 104S−3S=30−10
S=20S = 20S=20

So, the older son is 20 years old. From equation (3), the younger son is Y=12×20=10Y = \frac{1}{2} \times 20 = 10Y=21×20=10 years old, and the father’s age is F=4×20=80F = 4 \times 20 = 80F=4×20=80.

Example 93: A Woman’s Age

Question:
A woman is 6 years older than her daughter. In 12 years, she will be twice as old as her daughter. What are their current ages?

Answer:
Let the daughter’s age be DDD and the woman’s age be WWW.
From the problem, we have:

W=D+6W = D + 6W=D+6
W+12=2(D+12)W + 12 = 2(D + 12)W+12=2(D+12)

Substitute W=D+6W = D + 6W=D+6 into the second equation:
(D+6)+12=2(D+12)(D + 6) + 12 = 2(D + 12)(D+6)+12=2(D+12)
D+18=2D+24D + 18 = 2D + 24D+18=2D+24
18−24=2D−D18 – 24 = 2D – D18−24=2D−D
−6=D-6 = D−6=D

This result is not possible because the daughter cannot have a negative age. Thus, the setup is invalid, and we can conclude that either the problem contains a mistake or the provided information is inconsistent.

Example 94: A Brother’s Age

Question:
A brother is 3 times as old as his sister. In 6 years, he will be 2 years older than twice his sister’s age. What are their current ages?

Answer:
Let the sister’s age be SSS and the brother’s age be BBB.
From the problem, we have:

B=3SB = 3SB=3S
B+6=2(S+6)+2B + 6 = 2(S + 6) + 2B+6=2(S+6)+2

Substitute B=3SB = 3SB=3S into the second equation:
(3S)+6=2(S+6)+2(3S) + 6 = 2(S + 6) + 2(3S)+6=2(S+6)+2
3S+6=2S+12+23S + 6 = 2S + 12 + 23S+6=2S+12+2
3S+6=2S+143S + 6 = 2S + 143S+6=2S+14
3S−2S=14−63S – 2S = 14 – 63S−2S=14−6
S=8S = 8S=8

So, the sister is 8 years old, and the brother’s age is 3×8=243 \times 8 = 243×8=24.

Example 95: A Teacher and a Student

Question:
A teacher is 15 years older than her student. In 5 years, she will be 2 times as old as her student. What are their current ages?

Answer:
Let the student’s age be SSS and the teacher’s age be TTT.
From the problem, we have:

T=S+15T = S + 15T=S+15
T+5=2(S+5)T + 5 = 2(S + 5)T+5=2(S+5)

Substitute T=S+15T = S + 15T=S+15 into the second equation:
(S+15)+5=2(S+5)(S + 15) + 5 = 2(S + 5)(S+15)+5=2(S+5)
S+20=2S+10S + 20 = 2S + 10S+20=2S+10
20−10=2S−S20 – 10 = 2S – S20−10=2S−S
10=S10 = S10=S

So, the student is 10 years old, and the teacher’s age is 10+15=2510 + 15 = 2510+15=25.

Example 96: Age Difference Between Two People

Question:
A man is 20 years older than his daughter. In 20 years, he will be 3 times as old as she is. What are their current ages?

Answer:
Let the daughter’s age be DDD and the man’s age be MMM.
From the problem, we have:

M=D+20M = D + 20M=D+20
M+20=3(D+20)M + 20 = 3(D + 20)M+20=3(D+20)

Substitute M=D+20M = D + 20M=D+20 into the second equation:
(D+20)+20=3(D+20)(D + 20) + 20 = 3(D + 20)(D+20)+20=3(D+20)
D+40=3D+60D + 40 = 3D + 60D+40=3D+60
40−60=3D−D40 – 60 = 3D – D40−60=3D−D
−20=2D-20 = 2D−20=2D
D=10D = 10D=10

So, the daughter is 10 years old, and the man’s age is 10+20=3010 + 20 = 3010+20=30.

Example 97: Age of a Teacher

Question:
A teacher is 10 years older than her student. In 4 years, she will be twice as old as the student. What are their current ages?

Answer:
Let the student’s age be SSS and the teacher’s age be TTT.
From the problem, we have:

T=S+10T = S + 10T=S+10
T+4=2(S+4)T + 4 = 2(S + 4)T+4=2(S+4)

Substitute T=S+10T = S + 10T=S+10 into the second equation:
(S+10)+4=2(S+4)(S + 10) + 4 = 2(S + 4)(S+10)+4=2(S+4)
S+14=2S+8S + 14 = 2S + 8S+14=2S+8
14−8=2S−S14 – 8 = 2S – S14−8=2S−S
6=S6 = S6=S

So, the student is 6 years old, and the teacher’s age is 6+10=166 + 10 = 166+10=16.

Example 98: Age of a Man and His Son

Question:
A man is 3 times as old as his son. In 5 years, he will be twice as old as his son. What are their current ages?

Answer:
Let the son’s age be SSS and the man’s age be MMM.
From the problem, we have:

M=3SM = 3SM=3S
M+5=2(S+5)M + 5 = 2(S + 5)M+5=2(S+5)

Substitute M=3SM = 3SM=3S into the second equation:
(3S)+5=2(S+5)(3S) + 5 = 2(S + 5)(3S)+5=2(S+5)
3S+5=2S+103S + 5 = 2S + 103S+5=2S+10
3S−2S=10−53S – 2S = 10 – 53S−2S=10−5
S=5S = 5S=5

So, the son is 5 years old, and the man’s age is 3×5=153 \times 5 = 153×5=15.

Example 99: Brother’s Age

Question:
A brother is 5 years older than his sister. In 10 years, the brother will be twice as old as his sister. What are their current ages?

Answer:
Let the sister’s age be SSS and the brother’s age be BBB.
From the problem, we have:

B=S+5B = S + 5B=S+5
B+10=2(S+10)B + 10 = 2(S + 10)B+10=2(S+10)

Substitute B=S+5B = S + 5B=S+5 into the second equation:
(S+5)+10=2(S+10)(S + 5) + 10 = 2(S + 10)(S+5)+10=2(S+10)
S+15=2S+20S + 15 = 2S + 20S+15=2S+20
15−20=2S−S15 – 20 = 2S – S15−20=2S−S
−5=S-5 = S−5=S

This result is not valid, as age cannot be negative, indicating a flaw in the problem setup.

Example 100: A Man and His Child

Question:
A man is 30 years older than his child. In 10 years, he will be 3 times as old as his child. What are their current ages?

Answer:
Let the child’s age be CCC and the man’s age be MMM.
From the problem, we have:

M=C+30M = C + 30M=C+30
M+10=3(C+10)M + 10 = 3(C + 10)M+10=3(C+10)

Substitute M=C+30M = C + 30M=C+30 into the second equation:
(C+30)+10=3(C+10)(C + 30) + 10 = 3(C + 10)(C+30)+10=3(C+10)
C+40=3C+30C + 40 = 3C + 30C+40=3C+30
40−30=3C−C40 – 30 = 3C – C40−30=3C−C
10=2C10 = 2C10=2C
C=5C = 5C=5

So, the child is 5 years old, and the man’s age is 5+30=355 + 30 = 355+30=35.

Example 101: Age of a Teacher and Student

Question:
A teacher is 12 years older than her student. In 4 years, she will be 3 times as old as her student. What are their current ages?

Answer:
Let the student’s age be SSS and the teacher’s age be TTT.
From the problem, we have:

T=S+12T = S + 12T=S+12
T+4=3(S+4)T + 4 = 3(S + 4)T+4=3(S+4)

Substitute T=S+12T = S + 12T=S+12 into the second equation:
(S+12)+4=3(S+4)(S + 12) + 4 = 3(S + 4)(S+12)+4=3(S+4)
S+16=3S+12S + 16 = 3S + 12S+16=3S+12
16−12=3S−S16 – 12 = 3S – S16−12=3S−S
4=2S4 = 2S4=2S
S=2S = 2S=2

So, the student is 2 years old, and the teacher’s age is 2+12=142 + 12 = 142+12=14.

Example 102: A Father and His Daughter

Question:
A father is 30 years older than his daughter. In 10 years, he will be 3 times as old as she is. What are their current ages?

Answer:
Let the daughter’s age be DDD and the father’s age be FFF.
From the problem, we have:

F=D+30F = D + 30F=D+30
F+10=3(D+10)F + 10 = 3(D + 10)F+10=3(D+10)

So, the daughter is 5 years old, and the father’s age is 5+30=355 + 30 = 355+30=35.

Example 103: A Teacher and Her Student

Question:
A teacher is 20 years older than her student. In 10 years, the teacher will be twice as old as the student. What are their current ages?

Answer:
Let the student’s age be SSS and the teacher’s age be TTT.
From the problem, we have:

T=S+20T = S + 20T=S+20
T+10=2(S+10)T + 10 = 2(S + 10)T+10=2(S+10)

Substitute T=S+20T = S + 20T=S+20 into the second equation:
(S+20)+10=2(S+10)(S + 20) + 10 = 2(S + 10)(S+20)+10=2(S+10)
S+30=2S+20S + 30 = 2S + 20S+30=2S+20
30−20=2S−S30 – 20 = 2S – S30−20=2S−S
10=S10 = S10=S

So, the student is 10 years old, and the teacher’s age is 10+20=3010 + 20 = 3010+20=30.

Example 104: A Father and His Son’s Age

Question:
A father is 5 times as old as his son. In 20 years, he will be 3 times as old as his son. What are their current ages?

Answer:
Let the son’s age be SSS and the father’s age be FFF.
From the problem, we have:

F=5SF = 5SF=5S
F+20=3(S+20)F + 20 = 3(S + 20)F+20=3(S+20)

Substitute F=5SF = 5SF=5S into the second equation:
(5S)+20=3(S+20)(5S) + 20 = 3(S + 20)(5S)+20=3(S+20)
5S+20=3S+605S + 20 = 3S + 605S+20=3S+60
5S−3S=60−205S – 3S = 60 – 205S−3S=60−20
2S=402S = 402S=40
S=20S = 20S=20

So, the son is 20 years old, and the father’s age is 5×20=1005 \times 20 = 1005×20=100.

Example 105: A Teacher and Student’s Age Difference

Question:
A teacher is 15 years older than her student. In 10 years, she will be twice as old as the student. What are their current ages?

Answer:
Let the student’s age be SSS and the teacher’s age be TTT.
From the problem, we have:

T=S+15T = S + 15T=S+15
T+10=2(S+10)T + 10 = 2(S + 10)T+10=2(S+10)

Substitute T=S+15T = S + 15T=S+15 into the second equation:
(S+15)+10=2(S+10)(S + 15) + 10 = 2(S + 10)(S+15)+10=2(S+10)
S+25=2S+20S + 25 = 2S + 20S+25=2S+20
25−20=2S−S25 – 20 = 2S – S25−20=2S−S
5=S5 = S5=S

So, the student is 5 years old, and the teacher’s age is 5+15=205 + 15 = 205+15=20.

Example 106: Age Difference Between Siblings

Question:
A brother is 4 years older than his sister. In 6 years, the brother will be twice as old as his sister. What are their current ages?

Answer:
Let the sister’s age be SSS and the brother’s age be BBB.
From the problem, we have:

B=S+4B = S + 4B=S+4
B+6=2(S+6)B + 6 = 2(S + 6)B+6=2(S+6)

Substitute B=S+4B = S + 4B=S+4 into the second equation:
(S+4)+6=2(S+6)(S + 4) + 6 = 2(S + 6)(S+4)+6=2(S+6)
S+10=2S+12S + 10 = 2S + 12S+10=2S+12
10−12=2S−S10 – 12 = 2S – S10−12=2S−S
−2=S-2 = S−2=S

This result is not valid, as age cannot be negative, indicating an issue with the problem setup.

Example 107: Parent and Child Age Relationship

Question:
A parent is 40 years older than their child. In 8 years, the parent will be 5 times as old as the child. What are their current ages?

Answer:
Let the child’s age be CCC and the parent’s age be PPP.
From the problem, we have:

P=C+40P = C + 40P=C+40
P+8=5(C+8)P + 8 = 5(C + 8)P+8=5(C+8)

Substitute P=C+40P = C + 40P=C+40 into the second equation:
(C+40)+8=5(C+8)(C + 40) + 8 = 5(C + 8)(C+40)+8=5(C+8)
C+48=5C+40C + 48 = 5C + 40C+48=5C+40
48−40=5C−C48 – 40 = 5C – C48−40=5C−C
8=4C8 = 4C8=4C
C=2C = 2C=2

So, the child is 2 years old, and the parent’s age is 2+40=422 + 40 = 422+40=42.

Example 108: A Mother and Daughter

Question:
A mother is 24 years older than her daughter. In 6 years, the mother will be 3 times as old as her daughter. What are their current ages?

Answer:
Let the daughter’s age be DDD and the mother’s age be MMM.
From the problem, we have:

M=D+24M = D + 24M=D+24
M+6=3(D+6)M + 6 = 3(D + 6)M+6=3(D+6)

Substitute M=D+24M = D + 24M=D+24 into the second equation:
(D+24)+6=3(D+6)(D + 24) + 6 = 3(D + 6)(D+24)+6=3(D+6)
D+30=3D+18D + 30 = 3D + 18D+30=3D+18
30−18=3D−D30 – 18 = 3D – D30−18=3D−D
12=2D12 = 2D12=2D
D=6D = 6D=6

So, the daughter is 6 years old, and the mother’s age is 6+24=306 + 24 = 306+24=30.

Example 109: A Father and Son’s Age

Question:
A father is 35 years older than his son. In 5 years, the father will be twice as old as the son. What are their current ages?

Answer:
Let the son’s age be SSS and the father’s age be FFF.
From the problem, we have:

F=S+35F = S + 35F=S+35
F+5=2(S+5)F + 5 = 2(S + 5)F+5=2(S+5)

Substitute F=S+35F = S + 35F=S+35 into the second equation:
(S+35)+5=2(S+5)(S + 35) + 5 = 2(S + 5)(S+35)+5=2(S+5)
S+40=2S+10S + 40 = 2S + 10S+40=2S+10
40−10=2S−S40 – 10 = 2S – S40−10=2S−S
30=S30 = S30=S

So, the son is 30 years old, and the father’s age is 30+35=6530 + 35 = 6530+35=65.

Example 110: A Teacher and Her Student

Question:
A teacher is 10 years older than her student. In 4 years, she will be twice as old as her student. What are their current ages?

Answer:
Let the student’s age be SSS and the teacher’s age be TTT.
From the problem, we have:

T=S+10T = S + 10T=S+10
T+4=2(S+4)T + 4 = 2(S + 4)T+4=2(S+4)

So, the student is 6 years old, and the teacher’s age is 6+10=166 + 10 = 166+10=16.

Example 111: A Sister and Brother

Question:
A brother is 8 years older than his sister. In 4 years, he will be twice as old as she is. What are their current ages?

Answer:
Let the sister’s age be SSS and the brother’s age be BBB.
From the problem, we have:

B=S+8B = S + 8B=S+8
B+4=2(S+4)B + 4 = 2(S + 4)B+4=2(S+4)

Substitute B=S+8B = S + 8B=S+8 into the second equation:
(S+8)+4=2(S+4)(S + 8) + 4 = 2(S + 4)(S+8)+4=2(S+4)
S+12=2S+8S + 12 = 2S + 8S+12=2S+8
12−8=2S−S12 – 8 = 2S – S12−8=2S−S
4=S4 = S4=S

So, the sister is 4 years old, and the brother’s age is 4+8=124 + 8 = 124+8=12.

Example 112: A Father and Son Age Problem

Question:
A father is 40 years older than his son. In 20 years, the father will be twice as old as his son. What are their current ages?

Answer:
Let the son’s age be SSS and the father’s age be FFF.
From the problem, we have:

F=S+40F = S + 40F=S+40
F+20=2(S+20)F + 20 = 2(S + 20)F+20=2(S+20)

Substitute F=S+40F = S + 40F=S+40 into the second equation:
(S+40)+20=2(S+20)(S + 40) + 20 = 2(S + 20)(S+40)+20=2(S+20)
S+60=2S+40S + 60 = 2S + 40S+60=2S+40
60−40=2S−S60 – 40 = 2S – S60−40=2S−S
20=S20 = S20=S

So, the son is 20 years old, and the father’s age is 20+40=6020 + 40 = 6020+40=60.

Example 113: A Father, Son, and Daughter

Question:
A father is 36 years older than his son. In 12 years, the father will be 3 times as old as his son. What are their current ages?

Answer:
Let the son’s age be SSS and the father’s age be FFF.
From the problem, we have:

F=S+36F = S + 36F=S+36
F+12=3(S+12)F + 12 = 3(S + 12)F+12=3(S+12)

Substitute F=S+36F = S + 36F=S+36 into the second equation:
(S+36)+12=3(S+12)(S + 36) + 12 = 3(S + 12)(S+36)+12=3(S+12)
S+48=3S+36S + 48 = 3S + 36S+48=3S+36
48−36=3S−S48 – 36 = 3S – S48−36=3S−S
12=2S12 = 2S12=2S
S=6S = 6S=6

So, the son is 6 years old, and the father’s age is 6+36=426 + 36 = 426+36=42.

Example 114: A Teacher and Her Student

Question:
A teacher is 18 years older than her student. In 4 years, the teacher will be twice as old as her student. What are their current ages?

Answer:
Let the student’s age be SSS and the teacher’s age be TTT.
From the problem, we have:

T=S+18T = S + 18T=S+18
T+4=2(S+4)T + 4 = 2(S + 4)T+4=2(S+4)

Substitute T=S+18T = S + 18T=S+18 into the second equation:
(S+18)+4=2(S+4)(S + 18) + 4 = 2(S + 4)(S+18)+4=2(S+4)
S+22=2S+8S + 22 = 2S + 8S+22=2S+8
22−8=2S−S22 – 8 = 2S – S22−8=2S−S
14=S14 = S14=S

So, the student is 14 years old, and the teacher’s age is 14+18=3214 + 18 = 3214+18=32.

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