Age - MCQ 2


20 Questions MCQ Test Quantitative Aptitude for Competitive Examinations | Age - MCQ 2


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QUESTION: 1

This year, the sum of ages of the Suresh’s family members is 78. Currently, there are 4 members namely viz Suresh, Rita, Deepa and Tarun. Suresh is 4 years older than Rita. Deepa is two years older than Tarun. If Suresh is 7 times as old as Tarun, how old is Deepa?

Solution:

S+R+D+T = 78 – (i)
S = R + 4 – (ii)
D = T + 2 – (iii)
S = 7T – (iv)
7T = R+4
R= 7T – 4
h+w+d+s = 78 – (i)
7T+7T-4+T+2+T = 78
16T = 80
T = 5
d = T+2 = 5+2 = 7

QUESTION: 2

Three years from now, Deepa will be three times as old as Emma and Emma will be six years younger than Femina. If Deepa’s age is three years less than twice Femina’s age, how old is Femina?

Solution:

D+3=3(E+3)
E+3=(F+3)-6
D=2F-3.
E+3=F-3 or E=F-6.
D+3=3(F-6+3)
D+3=3F-9
D=3F-12.
now, we have D=2F-3.
2F-3=3F-12
9=F

QUESTION: 3

Ravi is now 4 years older than Emma and half of that amount older than Ishu. If in 2 years, Ravi will be twice as old as Emma, then in 2 years what would be Ravi’s age multiplied by Ishu’s age?

Solution:

Ravi – x + 4
Emma – x
Ishu – x + 2
(Ravi 4 years older than Emma & 2 years older than Ishu)
Ages after 2 yrs
Ravi – x + 6
Emma – x + 2
Ishu – x + 4
x+6 = 2(x + 2)
x = 2
Ravi * Ishu = 8 * 6 = 48

QUESTION: 4

Jack is now twice as old as Femina, who is two years older than Suresh. Four years ago, Jack was four times as old as Suresh. How old is Jack now?

Solution:

j = 2f – (i)
f = s+2 – (ii)
four years ago
j - 4 = 4(s-4)
putting (ii) in (iii)
we get j - 4 = 4(f-2-4)
sub (i) in above equation we get
j - 4 = 4(j/2-6) ⇒ j = 20

QUESTION: 5

James is now 12 years younger than Mahesh. If 9 years from now Mahesh will be twice as old as James, how old will James be in 4 years?

Solution:

James = x years, Mahesh = x+12 years
9 from now,
2(x+9)=x+21
2x+18=x+21
x=3
x+4 =7 years

QUESTION: 6

Ravi has three children: two daughters and one son. All were born on the same date in different years. The sum of the ages of the two daughters today is smaller than the age of the son today, but a year from now the sum of the ages of the daughters will equal the age of the son. Three years from today, the difference between the age of the son and the combined ages of the daughters will be

Solution:

one year from now, D+d=s
two years after that, D+2+d+2=s+2
D+d-s=2-4=-2

QUESTION: 7

Leena’s age is 5 years more than twice Gayathri’s age. Vijay’s age is 13 years less than 10 times Gayathri’s age. If Vijay is 3 times as old as Leena, how old is Leena? 

Solution:

given l = 2g+5
v = 10g-13
v = 3l
f rom v = 3l ⇒ 10g-13 = 6g+15
g = 7
l = 2g+5 = 19

QUESTION: 8

8 years ago, Geeta was half as old as Saina. Saina is now 20 years older than Geeta. How old will Geeta be in 10 years? 

Solution:

Saina is now 20 years older than Geeta → S=G+20 → Saina 8 years ago was G+20-8=G+12 years old.
(G-8)*2=G+12 → G=28. 28+10=38 years.

QUESTION: 9

Lilly is 5 years older than Martin, and Catherine is twice as old as Lilly. If the sum of their ages is 95, how old is Lilly?

Solution:

L=5+M and C=2L and C+M+L=95 ⇒ 4L=100 ⇒ L=25

QUESTION: 10

Arun will be half as old as Lilly in 3 years. Arun will also be one-third as old as James in 5 years. If James is 15 years older than Lilly, how old is Arun?

Solution:

let age of Arun =x, Lilly =y James = z
(x+3) =1/2 *(y+3) so we have 2x-y =-3 -(1)
(x+5) =1/3 * (z+5) ; ⇒ 3x-z=-10 -(2)
From (1)&(2) we get, x+y-z =-7 -(3)
we have z =15+y – (4)
from equation 3 and 4 we get x=8

QUESTION: 11

The average runs scored by a cricketer is 42 innings, is 40. The difference between his maximum and minimum scores in an inning is 100. If these two innings are not taken into consideration, then the average score of remaining 40 inning is 37. Calculate the maximum runs scored by him in an innings?

Solution:

Let the minimum score = x
∴ Maximum score = x + 100
∴ x + (x + 100) = 42 x 40 – 40 x 37
⇒ 2x + 100 = 1680 – 1480 = 200
⇒ 2x = 200 – 100 = 100
⇒ x = 50
Maximum score = x + 100 = 50 + 100 = 150

QUESTION: 12

Silvia was married 8 year ago. Today her age is 9/7 times to that time of marriage. At present his son’s age is 1/6th of her age. What was her son’s age 3 year ago?

Solution:

Silvia’s age 8 year ago = x
Present age = x + 8
x + 8 = 9/7 x
7(x + 8)= 9x
x = 28; 28 + 8 = 36
Son’s age = 1/6 * 36 = 6
Son’s age 3 year ago = 6-3 =3

QUESTION: 13

Eight years ago, Pavi’s age was equal to the sum of the present ages of her one son and one daughter. Five years hence, the respective ratio between the ages of her daughter and her son that time will be 7:6. If Pavi’s husband is 7 years elder to her and his present age is three times the present age of their son, what is the present age of the daughter?

Solution:

P – 8 = S + D —(1)
6D + 30 = 7S + 35 —(2)
H = 7 + P
H = 3S
3S = 7 + P —(3)
Solving eqn (1),(2) and (3) D = 23

QUESTION: 14

Ravi’s present age is three times his son’s present age and 4/5th of his father’s present age. The average of the present ages of all of them is 62 years. What is the difference between the Ravi’s son’s present age and Ravi’s father’s present age?

Solution:

Present age of Ravi is = 4/5x
Present age of Ravi’s father is = 4/15x
Ratio = 15: 12 : 4
Difference between the Ravi’s son’s present age and Ravi’s father’s present age = 62/31 * 3(15 – 4).
= 2*3*11 = 66 year.

QUESTION: 15

The respective ratio between the present age of Mary and Deepa is x : 42. Mary is 8 years younger than Prema. Prema’s age after 8 years will be 33 years. The difference between Deepa’s and Mary’s age is same as the present age of Prema. What is the value of x?

Solution:

Prema’s age after 8 years = 33 years
Prema’s present age = 33 – 8= 25 years
Mary’s present age = 25 – 8 = 17 years
Deepa’s present age = 17 + 25 = 42 years
Ratio between Mary and Deepa = 17 : 42
X = 17

QUESTION: 16

Mr. Sharma has three sons namely Ram, Amit and Karan. Ram is the eldest son of Mr. Sharma while Karan is the youngest one. The present ages of all three of them are square numbers. The sum of their ages after 5 years is 44. What is the age of Ram after three years?

Solution:

Square numbers – x, y, z
(x + 5) + (y + 5)+ (z + 5) = 44
x + y + z = 44 – 15 = 29
Possible values of x, y, z = 4, 9, 16 [Out of 1, 4, 9, 16, 25] Ram’s present age = 16; after three years = 19

QUESTION: 17

One year before, Reshma was six times as old as her daughter. Six years hence, Reshma’s age will exceed her daughter’s age by 20 years. The ratio of the present ages of Reshma and her daughter is?

Solution:

Ages of Reshma and her daughter = 6x, x
[6x + 1 + 6] – [x + 1 + 6] = 20
5x = 20; x = 4
Ratio = 6x + 1 : x + 1 = 25 : 5 = 5:1

QUESTION: 18

The ratio of present ages of Ramya and Karan’s age is 4:5 and the ratio of present ages of Karan and Priya is 2:5 respectively. If Ramya is two-fifth of Priya’s age, What is Ramya’s age?

Solution:

Data Insufficient – Can not be determined.

QUESTION: 19

Ravi is as much younger than Surya as he is older than Suresh. If the sum of the ages of Surya and Suresh is 50 years, what is definitely the difference between Surya and Ravi’s age?

Solution:

Surya’s age – Ravi’s age = Ravi’s age – Surya’s age
Suresh’s age + Surya’s age = 2 Ravi’s age
Surya’s age + Suresh’s age = 50
Ravi’s age = 25; We can not find the difference between Surya and Ravi’s age

QUESTION: 20

Amit’s present age is 1.5 times of Hari’s present age. If after 4 years, Amit’s age will be twice of Hari’s age 4 years ago. What is the sum of the present ages of Amit and Hari?

Solution:

A = 1.5 * H
A + 4 = 2 * (H – 4)
A + 4 = 2H – 8
A + 12 = 2H
1.5H + 12 = 2H
H = 24
A = 24 * 1.5 = 36
Sum = 36 + 24 = 60

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