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Name: Problems on Ages: Free Aptitude Mock Test – 4
Subject: Mathematics / Numerical Aptitude
Topic: Problems on Ages
Questions: 11 Objective Type Questions
Time Allowed: 25 Minutes
Language: English
Important for: Police Exam, SSC ( CGL, CHSL, GD etc), State PCS, UPSC CSE CSAT, Railway, IBPS Clerk, IBPS RRB, SBI Clerk, Engineering Entrance exam, CTET and State TET, समूह ग आदि 
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 Question 1 of 11
1. Question
1 pointsIn a family of 5 members, the average age at present is 33 years. The youngest member is 9 years old. The average age of the family just before the birth of the youngest member was
CorrectSum of Present age of all member of family = \(33 \times 5\) = 165
According to question, sum of their ages = \(165 – 9 \times 5\) = 120 Years
Required Average age = \(\frac{120}{4}\)IncorrectSum of Present age of all member of family = \(33 \times 5\) = 165
According to question, sum of their ages = \(165 – 9 \times 5\) = 120 Years
Required Average age = \(\frac{120}{4}\)UnattemptedSum of Present age of all member of family = \(33 \times 5\) = 165
According to question, sum of their ages = \(165 – 9 \times 5\) = 120 Years
Required Average age = \(\frac{120}{4}\)  Question 2 of 11
2. Question
1 pointsThe average age of 14 girls and their teacher’s age is 15 Years. If the teacher’s age is excluded the average reduces by 1. What is the teacher’s age?
CorrectTeacher’s age
= \(15\times15 – 14\times14\) = 225
225 – 196 = 29IncorrectTeacher’s age
= \(15\times15 – 14\times14\) = 225
225 – 196 = 29UnattemptedTeacher’s age
= \(15\times15 – 14\times14\) = 225
225 – 196 = 29  Question 3 of 11
3. Question
1 pointsThe respectively ratio between the present age of Aarti and Savita is 5:x. Aarti is 9 yr younger than Jahnavi’s age after 9 yr will be 33 yr. The difference between Savita’s and Aarti’s age is same as the present age of Jahnavi. What will come in place of x?
CorrectJahnavi’s present age = 33 – 9 = 24 yr
Aarti’s present age = 24 – 9 = 15 yr
now, Aarti : Savita = 5 : x = 15 : 3x
Savita’s present age = 3x yr
then, 3x – 15 = 24
3x = 24 + 15 = 39
\(x\frac{39}{3}=13\)IncorrectJahnavi’s present age = 33 – 9 = 24 yr
Aarti’s present age = 24 – 9 = 15 yr
now, Aarti : Savita = 5 : x = 15 : 3x
Savita’s present age = 3x yr
then, 3x – 15 = 24
3x = 24 + 15 = 39
\(x\frac{39}{3}=13\)UnattemptedJahnavi’s present age = 33 – 9 = 24 yr
Aarti’s present age = 24 – 9 = 15 yr
now, Aarti : Savita = 5 : x = 15 : 3x
Savita’s present age = 3x yr
then, 3x – 15 = 24
3x = 24 + 15 = 39
\(x\frac{39}{3}=13\)  Question 4 of 11
4. Question
1 pointsThe average of age of 8 family members is 35 years. If the age of the youngest member is 7 years, find the average age of the family at the birth of the youngest member.
CorrectAverage age of family at the birth of the youngest member
= \(\frac{8\times358\times7}{7}\)
= \(\frac{224}{7}\) = 32 YearsIncorrectAverage age of family at the birth of the youngest member
= \(\frac{8\times358\times7}{7}\)
= \(\frac{224}{7}\) = 32 YearsUnattemptedAverage age of family at the birth of the youngest member
= \(\frac{8\times358\times7}{7}\)
= \(\frac{224}{7}\) = 32 Years  Question 5 of 11
5. Question
1 pointsNiharika’s age is onefifth of her father’s age. Niharika’s father’s age will be thrice of Vihan’s age after 6 years. If Vihan’s sixth birthday was celebrated three years ago, then, what is Niharika’s age?
CorrectVihan’s present age = 6 + 3 = 9 Years.
Now, let the Niharika’s father age be F.
then, F + 6 = \(3\times(8+9)\)
F = 45 Years
Niharika’s Present age = \(1\frac{1}{5}\times45\) = 9 YearsIncorrectVihan’s present age = 6 + 3 = 9 Years.
Now, let the Niharika’s father age be F.
then, F + 6 = \(3\times(8+9)\)
F = 45 Years
Niharika’s Present age = \(1\frac{1}{5}\times45\) = 9 YearsUnattemptedVihan’s present age = 6 + 3 = 9 Years.
Now, let the Niharika’s father age be F.
then, F + 6 = \(3\times(8+9)\)
F = 45 Years
Niharika’s Present age = \(1\frac{1}{5}\times45\) = 9 Years  Question 6 of 11
6. Question
1 pointsThe ratio of the ages of Sonal and Arti is 5: 8. After 7 Years their ages will be in there ratio of 2:3. Find the sum of their present ages.
CorrectLet the age of Sonal and Arti be 5x and 8x Years,
\(\frac{5x+7}{8x+7}\)= \(\frac{2}{3}\)
15x + 21 = 16x + 14
x= 7
Sum of their ages = 5x + 8x = \(13\times7\) = 91 Years.IncorrectLet the age of Sonal and Arti be 5x and 8x Years,
\(\frac{5x+7}{8x+7}\)= \(\frac{2}{3}\)
15x + 21 = 16x + 14
x= 7
Sum of their ages = 5x + 8x = \(13\times7\) = 91 Years.UnattemptedLet the age of Sonal and Arti be 5x and 8x Years,
\(\frac{5x+7}{8x+7}\)= \(\frac{2}{3}\)
15x + 21 = 16x + 14
x= 7
Sum of their ages = 5x + 8x = \(13\times7\) = 91 Years.  Question 7 of 11
7. Question
1 pointsThe ratio of A’s and B’s age is 6:7. The products of their ages is 378 years. Find the age of ‘A’ after four years.
Correct\(6x\times 378 \) = x = 3
required answer = \(6\times3+4\) = 22 YearsIncorrect\(6x\times 378 \) = x = 3
required answer = \(6\times3+4\) = 22 YearsUnattempted\(6x\times 378 \) = x = 3
required answer = \(6\times3+4\) = 22 Years  Question 8 of 11
8. Question
1 pointsThe ratio of the ages of Ram and Rahim 10 years ago was 1:3. The ratio of the their ages five years hence will be 2 : 3. Then the ratio of the their present ages is
CorrectLet age of Ram and Rahim after 10 Years be x and 3x.
after 5 Years, \(\frac{20+x}{100+x}\) = \(\frac{3}{11}\)
6x + 30 = 3x + 45
3x = 45 – 30 = 15
x = 5
Ratio of their present age = (x+10) : (3x+10)
= 15 : 25 = 3 : 5IncorrectLet age of Ram and Rahim after 10 Years be x and 3x.
after 5 Years, \(\frac{20+x}{100+x}\) = \(\frac{3}{11}\)
6x + 30 = 3x + 45
3x = 45 – 30 = 15
x = 5
Ratio of their present age = (x+10) : (3x+10)
= 15 : 25 = 3 : 5UnattemptedLet age of Ram and Rahim after 10 Years be x and 3x.
after 5 Years, \(\frac{20+x}{100+x}\) = \(\frac{3}{11}\)
6x + 30 = 3x + 45
3x = 45 – 30 = 15
x = 5
Ratio of their present age = (x+10) : (3x+10)
= 15 : 25 = 3 : 5  Question 9 of 11
9. Question
1 pointsI am three times as old as my son. 15 years hence, I will be twice as old as my son. The sum of our ages is
CorrectLet age of the son = x Year
Present age of father = 3x year
After 15 year
3x + 15 = 2 (x + 15)
3x + 15 = 2x + 30
x = 30 – 15 = 15
Sum of the present age of both
= x + 3x = 4x
= \(4\times15\) = 60 Years.IncorrectLet age of the son = x Year
Present age of father = 3x year
After 15 year
3x + 15 = 2 (x + 15)
3x + 15 = 2x + 30
x = 30 – 15 = 15
Sum of the present age of both
= x + 3x = 4x
= \(4\times15\) = 60 Years.UnattemptedLet age of the son = x Year
Present age of father = 3x year
After 15 year
3x + 15 = 2 (x + 15)
3x + 15 = 2x + 30
x = 30 – 15 = 15
Sum of the present age of both
= x + 3x = 4x
= \(4\times15\) = 60 Years.  Question 10 of 11
10. Question
1 pointsAverage age of 38 student is 14 years. If the age of the teacher is included, the average becomes 14 years and 4 months. Then teacher’s age is
CorrectAge of teacher = \(\frac{39\times4}{12} + 14\)
= 13 + 14
= 27 yearsIncorrectAge of teacher = \(\frac{39\times4}{12} + 14\)
= 13 + 14
= 27 yearsUnattemptedAge of teacher = \(\frac{39\times4}{12} + 14\)
= 13 + 14
= 27 years  Question 11 of 11
11. Question
1 pointsThe ratio between the present ages of Soni and her husband is 4 : 7 respectively. Also, her husband is 18 year older than her. What is the present age of Soni’s brother who is half the present age of Soni’s husband?
CorrectLet the present age of Soni = 4x
and the present age of her husband = 7x
7x = 4x + 18
3x = 18
x = 6
The age of Soni’s husband = \(7\times x = 7\times6\) = 42 Years
Then, Age of Soni’s brother = \(\frac{42}{2}\)=21 yearsIncorrectLet the present age of Soni = 4x
and the present age of her husband = 7x
7x = 4x + 18
3x = 18
x = 6
The age of Soni’s husband = \(7\times x = 7\times6\) = 42 Years
Then, Age of Soni’s brother = \(\frac{42}{2}\)=21 yearsUnattemptedLet the present age of Soni = 4x
and the present age of her husband = 7x
7x = 4x + 18
3x = 18
x = 6
The age of Soni’s husband = \(7\times x = 7\times6\) = 42 Years
Then, Age of Soni’s brother = \(\frac{42}{2}\)=21 years