Preperation of Age Problem



Question No 1

In 1975, A was thrice as old as B but in 1979, A was only twice as old as B was. How old was A in 1985?

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Solution!
Let A's age in 1985 was X.! !Given that!X-10 = 3(B-10)!X-10 = 3B-30!3B-X = 20 ...(1)! !and!X-6 = 2(B-6)!X-6 = 2B-12!2B-X = 6 ...(2)! !By simultaneously solving equation(1) and (2) we have!X = 22! !Hence A's age in 1985 was 22 years..

Question No 2

The ratio of the ages of two friends A and B is 3/2. However if the age of A is decreased by 10 and that of B is increased by 10 then the original ratio is reversed. Age of A is

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Solution!
Given that!A/B = 3/2!2A = 3B!2A-3B = 0 ...(1)! !and also given that!(A-10)/(B+10) = 2/3!3A-30 = 2B+20!3A-2B = 50 ...(2)! !By simultaneously solving equation(1) and equation(2)!A = 30! !Hence age of A is 30 years..

Question No 3

The sum of the ages of a son and a father is 64. After 4 years the age of father will be three times that of his son's. The age of father is

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Solution!
Let present age of son is x and present age of father is y.! !Given that!x+y = 64 ...(1)! !and!y+4 = 3(x+4)!y+4 = 3x + 12!y-3x = 8 ...(2)! !Put value of x from equation(1) in equation(2)! !y-3(64-y) = 8!y-192+3y = 8!4y = 200!y = 50! !Hence present age of the father is 50 years..

Question No 4

The total age of A and B is 12 years more than the total age of B and C. C is how many year younger than A?

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Solution!
Given that A+B = 12 + B + C! !A - C = 12 + B - B = 12! !C is younger than A by 12 years..

Question No 5

The age of father 10 years ago was thrice the age of his son. Ten years hence, father's age will be twice that of his son. What is the ratio of their present ages?

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Solution!
Let the age of the son before 10 years = x!and age of the father before 10 years = 3x! !Now we can write as (3x + 20) = 2(x + 20)!x = 20! !Age of Father the son at present = x + 10!= 20 + 10 = 30! !Age of the father at present = 3x + 10!= 3 * 20 + 10 = 70! !Required ratio = 70/30 = 7/3.

Question No 6

A man is 24 years older than his son. In two years, his age will be twice the age of his son. What is the present age of his son?

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Solution!
Let the present age of the son = x years! !Then present age the man = (x+24) years! !Given that in 2 years, man's age will be twice the age of his son! !(x+24) +2 = 2(x+2)!x = 22.

Question No 7

A is as much younger than B and he is older than C. If the sum of the ages of B and C is 50 years, what is definitely the difference between B and A's age?

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Solution!
No answer description available for this question..

Question No 8

John is 35 years older than his new born cousin. How old will the cousin be, when his age is 1/6 that of John's?

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Solution!
We can simply check it from options. If John's cousins age is 7 then John's age is 42!and!1/6 0f 42 = 7!Hence 7 fulfills the required condition..

Question No 9

If 1/2 x year ago A was 12 and 1/2 x years from now he will be 2x years old, how old will he be 3x years from now?

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Solution!
Given that!A-(1/2)x = 12!2A-x = 24 ...(1)!and!A+(1/2)x = 2x!A+(1/2)x-2x = 0!A+(x-4x)/2 = 0!A-(3/2)x = 0!2A-3x = 0 ...(2)!By simultaneously solving equation(1) and (2) we have!x = 12!and!A = 18! !Now!3x = 3*12 = 36! !After 3x years A will be!18+36 = 54 years old.

Question No 10

The present age of A is 45 yeras. 5 years ago the age of A was 5 times the age of his son B. How old is B now?

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Solution!
Given that!A = 45! !and!A-5 = 5(B-5)!40 = 5B-25!65 = 5B!B = 13! !Hence B is 13 years old now..

Question No 11

A father said to his son, "I was as old as you are at the present at the time of your birth". If the father's age is 38 years now, what was the son's age five years back?

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Solution!
Let the son's present age be x years.! !Then! (38 - x) = x! !2x = 38!x = 38/2 = 19! !Son's age 5 years back = 19-5 = 14 years.

Question No 12

A is as much younger than B and he is older than C. If the sum of the ages of B and C is 50 years, what is definitely the difference between B and A's age?

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Solution!
Age of C < Age of A < Age of B! !Given that sum of the ages of B and C is 50 years.! !Let's take B's age = x and C's age = 50-x! !Now we need to find out B's age - A's age.!But we cannot find out this with the given data..

Question No 13

Sachin's age after 15 years will be 5 times his age 5 years back. Find out the present age of Sachin?

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Solution!
Let the present age of Sachin = x! !Then (x+15) = 5(x-5)!4x = 40!x = 10.

Question No 14

The sum of ages of 5 children born at the intervals of 3 years each is 50 years. Find out the age of the youngest child?

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Solution!
Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.! !Then,!x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50!5x = 20!x = 4! !Hence, Age of the youngest child = x = 4 years.

Question No 15

3 years ago, the average age of C and D was 18 years. With E joining them, the average becomes 22 years. How old is E now?

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Solution!
Given that![(C-3)+(D-3)]/2 = 18!C-3+D-3 = 36!C+D = 42 ...(1)! !and also given that!(C+D+E)/3 = 22!C+D+E = 66! !Put value of C+D from equation(1)!42+E = 66!E = 66-42!E = 24! !Hence E is 24 years old now..

Question No 16

X is 4 times as old as y. 4 years ago, X was 10 times as old as Y. What will be the ratio of their ages 6 years hence?

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Solution!
Given that!X = 4Y ...(1)! !and!X-4 = 10(Y-4)!X-4 = 10Y-40!X-10Y = -36!10Y-X = 36 ...(2)! !Put value of X from equation(1) in equation(2)!10Y-4Y = 36!6Y = 36!Y = 6! !Put value of Y in equation(1)!X = 4*6!X = 24! !After 6 years!Y = 24+6!Y = 30!and!X = 6+6!X = 12! !Ratio between X and Y is!30/12 = 5/2! !Hence after 6 years ratio between ages of X and Y will be 5/2..

Question No 17

A person's present age is two-fifth of the age of his mother. After 8 years, he will be one-half of the age of his mother. What is the present age of the mother?

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Solution!
Let the present age of the person = x! !Then present age of the mother = 5x/2! !Given that , after 8 years, the person will be one-half of the age of his mother.! !(x + 8) = (1/2)(5x/2 + 8)!2x + 16 = 5x/2 + 8!x/2 = 8!x = 16! !Present age of the mother = 5x/2!= 5 * 16/2 = 40.

Question No 18

Kiran is younger than Bineesh by 7 years and their ages are in the respective ratio of 7 / 9, how old is Kiran?

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Solution!
Let the ages of Kiran and Bineesh are 7x and 9x respectively! !7x = 9x-7!x = 7/2 = 3.5! !Kiran's age = 7x!= 7 * 3.5 = 24.5.

Question No 19

The sum of the ages of two children is 12 and their product is 32. The age of elder one is

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Solution!
Let age of younger child is x and age of elder child is y.! !Given that!x+y = 12 ...(1)! !and!xy = 32!x = 32/y! !Put value of x in equation(1)!(32/y)+y = 12!32+(y^2) = 12y!(y^2)-12y+32 = 0!(y^2)-8y-4y+32 = 0!y(y-8)+4(y-8) = 0!(y-8)(y+4) = 0!y = -4 (impossible)!or!y = 8! !Hence age of elder child is 8 years..

Question No 20

Ages of A and B are in ratio 8/7. 27 years ago their ages were in the ratio 5/4. The age of B is

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Solution!
Given that!A/B = 8/7 ...(1)! !and!(A-27)/(B-27) = 5/4!5B-135 = 4A-108 ...(2)! !From equation(1) we have!A = (8/7)B! !Put value of A in equation(2)!5B-135 = 4*[(8/7)B]-108!5B-27 = (32/7)B!5B-(32/7)B = 27!(35B-32B)/7 = 27!3B = 189!B = 63! !Hence the age of B is 63 years..

Question No 21

The Average age of a class of 22 students is 21 years.The average increased by 1 when the teacher's age also included.What is the age of the teacher?

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Solution!
No answer description available for this question..

Question No 22

Kamal was 4 times as old as his son 8 years ago. After 8 years, Kamal will be twice as old as his son. Find out the present age of Kamal.

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Solution!
Let the age of the son before 8 years = x! !Then age of Kamal before 8 years ago = 4x! !After 8 years, Kamal will be twice as old as his son! !4x + 16 = 2(x + 16)!x = 8! !Present age of Kamal = 4x + 8 = (4*8) + 8 = 40.

Question No 23

A mother's age after 5 years, will be twice the age of her daughter. The difference of their ages is 19. What is the age of mother?

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Solution!
Let present age of daughter is x and present age of mother is y.! !Given that!y+5 = 2(x+5)!y+5 = 2x+10!5-10 = 2x-y!y-2x = 5 ...(1)! !and also given that!y-x = 19 ...(2)! !By simultaneously solving equation(1) and equation(2) we can find!y = 33! !Hence mothers present age is 33 years..

Question No 24

The sum of ages of 5 children born at the intervals of 3 years each is 50 years. Find out the age of the youngest child?

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Solution!
Let the age of the youngest child = x! !Then the ages of 5 children can be written as x, (x+3), (x+6),(x+9) and (x+12).! !Then,!X + (x+3) + (x+6) + (x+9) + (x+12) = 50!5x + 30 =50!5x = 20!x = 20/5 = 4.

Question No 25

C's mother was four times as old as C ten years ago. After 10 years she will be twice as old as C is now. How old C is now?

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Solution!
Let C's present age is x and his mother's age is y.! !Given that!y = 4(x-10)!y = 4x-40!4x-y = 40 ...(1)! !and!y+10 = 2x!2x-y = 10 ...(2).! !By simultaneously solving equation(1) and equation(2) we find! x = 15! !Hence C is 15 years old..

Question No 26

Sandeep's age after six years will be three-seventh of his father's age. Ten years ago the ratio of their ages was 1/5. What is Sandeep's father's age at present?

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Solution!
Let the age of Sandeep and his father before 10 years be x and 5x respectively.! !Given that Sandeep's age after six years will be three-seventh of his father's age! !x + 16 = (3/7)(5x + 16)!7x + 112 = 15x + 48!8x = 64!x = 8! !Sandeep's father's present age = 5x + 10!= 5*8 + 10 = 50.

Question No 27

The ages of C and D are in 6/11 and E is 6 years younger than D. If the sum of ages is 134, then what is the age of D?

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Solution!
Given that!C/D = 6/11!C = (6/11)D ...(1)! !and!C+D+E = 134! !As E is 6 years younger from D so we can put D-6 instead of D!C+D+D-6 = 134!C+2D = 140!Put value of C from equation(1)!(6/11)D+2D = 140!6D+22D = 140!28D = 1540!D = 55! !Hence D is 55 years old..

Question No 28

Ten years ago Akram was thrice as old as Aslam was, but 10 years hence, he will be only twice as old. What is Akram's present age?

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Solution!
Let Aslam's present age is x and Akram's present age is y.! !Given that!y-10 = 3(x-10)!y-10 = 3x-30!y-3x = -20!3x-y = 20 ...(1)! !and also given that!y+10 = 2(x+10)!y+10 = 2x+20!y-2x = 10 ...(2)! !By simultaneously solving equation(1) and equation(2)!y=70! !Hence Akram's present age is 70 years..

Question No 29

The difference of the ages of two brothers is 2 and the difference of the square of their ages is 16. The age of smaller one is?

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Solution!
Let present age of elder brother is y and present age of younger brother is x.! !Given that!y-x = 2 ...(1)! !and!(y^2)-(x^2) = 16!(y+x)(y-x) = 16! !Put value of (y-x) from equation(1)!2(y+x) = 16!y+x = 8 ...(2)! !By simultaneously solving equation(1) and equation(2) we have!y = 5!and!x = 3! !Hence present age of younger brother is 3 years..

Question No 30

Present ages of Sameer and Anand are in the ratio of 5/4 respectively. Three years hence, the ratio of their ages will become 11/9 respectively. What is Anand's present age in years?

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Solution!
Let the present ages of Sameer and Anand be 5x years and 4x years respectively.! !Then,!(5x+3)/(4x+3) = 11/9!9(5x + 3) = 11(4x + 3)!45x + 27 = 44x + 33!45x - 44x = 33 - 27!x = 6! !nand's present age = 4x = 24 years.

Question No 31

A is two years older than B who is twice as old as C. The total of the ages of A, B and C is 27.How old is B?

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Solution!
Let C's age be x years. Then, B's age = 2x years. A's age = (2x + 2) years.! !So, (2x + 2) + 2x + x = 27!5x = 25!x = 5! !Hence, B's age = 2x = 10 years.

Question No 32

The sum of the present ages of a mother and daughter is 63. Three years back if the age of the mother was double that of the daughter, then the present age of the daughter is

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Solution!
Let mother's present age is x and daughters present age is y! !Given that!x+y = 63 ...(1)!and! !x-3 = 2(y-3)!x-3 = 2y-6!x-2y = -6+3!x-2y = -3 ...(2)! !By simultaneously solving equation(1) and equation(2) We'll find!y = 22! !Hence present age of the daughter is 22 years..

Question No 33

A is two years older than B who is twice as old as C. The total of the ages of A, B and C is 27. How old is B?

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Solution!
Let the age of C = x.! !Then! !Age of B = 2x! !Age of A = 2 + 2x! !The total age of A,B and C =27! !(2+2x) + 2x + x = 27!5x = 25!B's age = 2x = 2 * 5 = 10.

Question No 34

The product of the ages of two class fellows is 24. If the greater fellow is 3/2 times the smaller one, then the age of the smaller fellow is

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Solution!
Let present age of the greater fellow is x and present age of smaller fellow is y.! !Given that xy = 24 ...(1)!and! !x = (3/2)y!Put value of x in equation(1).! !(3/2)y^2 = 24!y^2 = 24*(2/3)!y^2 = 16!y = 4! !Hence age of smaller fellow is 4 years..

Question No 35

A person's present age is two-fifth of the age of his mother. After 8 years, he will be one-half of the age of his mother. How old is the mother at present?

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Solution!
Let the mother's present age be x years.! !Then,! the person's present age = (2/5)x years.! !(2/5)x + 8 = (1/2)(x + 8)!2(2x + 40) = 5(x + 8)!x = 40!.

Question No 36

Ten years ago, P was half of Q in age. If the ratio of their present ages is 3/4, what will be the total of their present ages?

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Solution!
Let the present age of P and Q be 3x and 4x respectively.! !Ten years ago, P was half of Q in age.! !(3x - 10) = 1/2 * (4x - 10)!6x - 20 = 4x - 10!2x = 10!x = 5! !total of their present ages!= 3x + 4x = 7x = 7 * 5!= 35.

Question No 37

Bill's age is 1/3 of Mike's age, who is 60% of James' age. Tom is twice as old as Mike. What is the ratio of Bill's age to Tom's age?

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Solution!
Let Bill's age is Z, Mike's age is Y and Tom's age is X.! !Given that!Z = (1/3)Y!and!Y = (60/100)X!Y = (6/10)X! !Tom's Age is double of Mike's age mean his age is 2y.!Now Bill's age is (1/3)Y and Tom's age is 2Y and ratio between them is![(1/3)Y] / 2Y!(1/3)*(1/2) = 1/6! !Hence ratio between Bill's and Tom's age is 1/6..

Question No 38

The sum of the present ages of a father and his son is 60 years. Six years ago, father's age was five times the age of the son. After 6 years, son's age will be

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Solution!
Let the present ages of son and father be x and (60 -x) years respectively.! !Then,!(60 - x) - 6 = 5(x - 6)!54 - x = 5x - 30!6x = 84!x = 14! !Hence, Son's age after 6 years = (x+ 6)!= 20 years.

Question No 39

Kamal's age is 1/6 of his father's age. If kamal's 8th birthday was celebrated 2 years ago, what is father's present age?

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Solution!
Let Kamal's present age is x and his father's present age is y.! !Given that!y = 6x ...(1)! !and!x-2 = 8!x = 10! !Put value of x in equation(1)!y = 6*10!y = 60! !Hence father's present age is 60 years..

Question No 40

20 years ago my age was 1/3 of what it is now. What is my present age?

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Solution!
Let my present age is x.! !Given that!x-20 = (1/3)x!3x-60 = x!2x = 60!x = 30! !Hence my present age is 30 years..

Question No 41

Six years ago, the ratio of the ages of Kunal and Sagar was 6/5. Four years hence, the ratio of their ages will be 11/10. What is Sagar's age at present?

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Solution!
Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.! !Then,![(6x + 6) + 4]/[(5x + 6) + 4] = 11/10!10(6x + 10) = 11(5x + 10)!5x = 10!x = 2! !Hence, Sagar's present age = (5x + 6) = 16 years..

Question No 42

One year ago, the ratio of Sooraj's and Vimal's age was 6/7 respectively. Four years hence, this ratio would become 7/8. How old is Vimal?

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Solution!
Let take the age of Sooraj and Vimal, 1 year ago as 6x and 7x respectively.! !Given that, four years hence, this ratio would become 7/8.! !(6x + 5)/(7x + 5) = 7/8!48x + 40 = 49x + 35!x = 5! !Vimal's present age = 7x + 1 = 7*5 + 1 = 36.

Question No 43

A father is 15 times as old as his son. however, 18 years later, he will be only thrice as old as the son. What is the father's present age?

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Solution!
Let present age of the son is x and present age of Father is y.! !Given that!y = 15x!15x-y = 0 ...(1)! !and also given that!y+18 = 3(x+18)!y+18 = 3x+54!y-3x = 36 ...(2)! !By simultaneously solving equation(1) and equation(2)!y = 45! !Hence father's present age is 45 years..

Question No 44

A's age is 3/4 of that of B's. Two years ago A's age was 1/2 of what B will be in 4 years. How old is A now?

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Solution!
Given that!A = (3/4)B ...(1)! !and!A-2 = (1/2)(B+4)!2A-4 = B+4!2A-B = 8! !Put value of B from equation(1)!2A-(4/3)A = 8!6A-4A = 24!2A = 24!A = 12! !Hence A is 12 years old now..

Question No 45

Meena is thrice as old as Sheraz. Fatima will be twice as old as Meena 6 years hence. Six years ago Sheraz was five years old. What is Fatima's present age?

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Solution!
Let present age of Sheraz is x and present age of Fatima is y.! !So,!Meena's age = 3x! !Given that!y+6 = 2(3x+6) ...(1)!and!x-6 = 5!x = 11! !Put value of x in equation(1)! !y+6 = 2(33+6)!y+6 = 78!y = 72! !Hence Fatima's present age is 72 years..

Question No 46

At present, the ratio between the ages of Arun and Deepak is 4/3. After 6 years, Arun's age will be 26 years. What is the age of Deepak at present ?

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Solution!
Let the present ages of Arun and Deepak be 4x years and 3x years respectively.! !Then,!4x + 6 = 26!4x = 20!x = 5! !Hence, Deepak's age = 3x = 15 years..

Question No 47

Tina is twice as old as David. 15 years ago, Tina's age was 5 times the age of David. What is the present age of Tina?

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Solution!
Let David's age is x and Tina's age is y.! !Given that!y = 2x!y-2x = 0 ...(1)! !and also given that!y-15 = 5(x-15)!y-15 = 5x-75!y-5x = -60 ...(2)! !By simultaneously solving equation(1) and equation(2)!y = 40! !Hence Tina's present age is 40 years..

Question No 48

The age of father 10 years ago was thrice the age of his son. Ten years hence, father's age will be twice that of his son. The ratio of their present ages is

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Solution!
Let the ages of father and son 10 years ago be 3x and x years respectively.! !Then,!(3x + 10) + 10 = 2[(x + 10) + 10]!3x + 20 = 2x + 40!x = 20!So, 3x + 10 = 70!and x + 10 = 30! !Hence, ratio is 70/30 = 7/3..

Question No 49

A father said to his son, "I was as old as you are at the present at the time of your birth". If the father's age is 38 years now, the son's age five years back was?

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Solution!
Let the son's present age be x years. Then, (38 - x) = x! !2x = 38!x = 19! !Hence, Son's age 5 years back (19 - 5) = 14 years..

Question No 50

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?

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Solution!
Total age of all students = 22 * 21 = 462! !Total age of all students + age of the teacher = 23 * 22 = 506! !Age of the teacher = 506 - 462 = 44.

Question No 51

The average age of A and B is 20. If C were to replace A, the average would be 19, and if B was replaced by C, the average would be 21. The age of C is

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Solution!
Given that!(A+B)/2 = 20!A+B = 40 ...(1)! !and!(C+B)/2 = 19!C+B = 38 ...(2)! !and!(A+C)/2 = 21!A+C = 42 ...(3)! !Put value of A from equation(1) in equation(3)!(40-B)+C = 42!C-B = 2 ...(4)! !By simultaneously solving equation(2) and equation(4)!C = 20! !Hence C is 20 years old..

Question No 52

The ratio of A's age and B's age is 3/7, and the difference of their ages is 16 years. Find the age of B.

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Solution!
Given that!A/B = 3/7 ...(1)! !and!B-A = 16 ...(2)!Put value of A from equation(1) in equation(2)!B-(3/7)B = 16!(7B-3B)/7 = 16!4B = 112!B = 28! !Hence Age of B is 28 years..

Question No 53

A man's age is three times his son's. In 15 years it will be double that of his son's. How old is the son now?

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Solution!
Let Son's age is x!Mans Age = 3x! !Given that!3x+15 = 2(x+15)!3x+15 = 2x+30!3x-2x = 30-15! !x = 15! !Hence son is 15 years old..

Question No 54

Sachin is younger than Rahul by 7 years. If their ages are in the respective ratio of 7/9, how old is Sachin?

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Solution!
Let Rahul's age be x years.! !Then,!Sachin's age = (x - 7) years.! !(x-7)/x = 7/9!9x - 63 = 7x!2x = 63!x = 31.5! !Hence, Sachin's age =(x - 7)= 24.5 years.

Question No 55

Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit's age. After further 8 years, how many times would he be of Ronit's age?

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Solution!
Let Ronit's present age be x years. Then, father's present age =(x + 3x) years!= 4x years.! !(4x + 8) = 5/2 (x+8)!8x + 16 = 5x + 40!3x = 24!x = 8! !Hence, required ratio = (4x+16)/(x+160)!= 48/24 = 2.

Question No 56

Q is as much younger than R as he is older than T. If the sum of the ages of R and T is 50 years, what is definitely the difference between R and Q's age?

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Solution!
Given that! !(i)The difference of age b/w R and Q = The difference of age b/w Q and T!(ii) (R + T) = 50! !So,R - Q = Q - T! !(R + T) = 2Q! !Now given that, (R + T) = 50! !So, 50 = 2Q and therefore Q = 25! !Question is (R - Q) = ?! !Here we know the value(age) of Q (25), but we don't know the age of R.! !Therefore, (R-Q) cannot be determined..

Question No 57

A father is 3 years older than 3 times the age of his son. If the sum of their ages is 31 years, how old was father 3 years ago?

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Solution!
Let present age of son is x and present age of father is y.! !Given that!y = 3x+3 ...(1)! !and!x+y = 31!x = 31-y! !Put value of x in equation(1)!y = 3(31-y)+3!y = 93-3y+3!4y = 96!y = 24! !Three years ago!y = 24-3!y = 21! !Hence father was 21 years old 3 years ago..

Question No 58

The ages of two persons differ by 16 years. 6 years ago, the elder one was 3 times as old as the younger one. What are their present ages of the elder person?

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Solution!
Let's take the present age of the elder person = x! !and the present age of the younger person = x - 16! !(x - 6) = 3 (x-16-6)!x - 6 = 3x - 66!2x = 60!x = 60/2 = 30.

Question No 59

The ages of X and Y are in the ratio 7/4. If the sum of their present ages is 55, then the age of X is

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Solution!
Given that!x/y = 7/4 ...(1)! !and!x+y = 55 ...(2)! !From equation(1)!4x = 7y!y = (4/7)x! !Put value of y in equation(2)!x + (4/7)x = 55!(7x+4x)/7 = 55!11x = 385!x = 35! !Hence the age of x is 35 years..

Question No 60

Ten years ago a father was seven times as old as his son, two years hence, twice his age will be equal to five times his son's. What is the present age of son?

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Solution!
Let present age of son is x and present age of father is y.! !Given that!y-10 = 7(x-10)!y-10 = 7x-70!y-7x = -70+10!7x-y = 60 ...(1)! ! and also given that!2(y+2) = 5(x+2)!2y+4 = 5x+10!2y-5x = 6 ...(2)By simultaneously soving equation(1) and equation(2)!x = 14! !Hence present age of son is 14 years..
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