A man is 3 years older than his wife and four times as old as his son. If the son becomes 15 years old after 3 years, the present age of his wife is : [#1127]
Q1. A man is 3 years older than his wife and four times as old as his son. If the son becomes 15 years old after 3 years, the present age of his wife is : Q1. A man is 3 years older than his wife and four times as old as his son. If the son becomes 15 years old after 3 years, the present age of his wife is :
(A) 60 years (A) 60 years
(B) 51 years (B) 51 years
(C) 48 years (C) 48 years
(D) 45 years (D) 45 years
Answer: (D) 45 years Answer: (D) 45 years
The son becomes 15 years old after 3 years.
Hence son's present age = 15 years - 3 years = 12 years
Man is four times as old as his son.
Hence Man's present age = 12 years * 4 = 48 years
Man is 3 years older than his wife.
Hence present age of his wife = 48 years - 3 years = 45 years.The son becomes 15 years old after 3 years.
Hence son's present age = 15 years - 3 years = 12 years
Man is four times as old as his son.
Hence Man's present age = 12 years * 4 = 48 years
Man is 3 years older than his wife.
Hence present age of his wife = 48 years - 3 years = 45 years.
Q1. At what percentage of simple interest does an amount of money double in 12 years? Q1. At what percentage of simple interest does an amount of money double in 12 years?
(A) (A)
(B) (B)
(C) (C)
(D) (D)
Answer: (C) Answer: (C)
P * * 12
= P * * 12
= P *
= P * 100%P * * 12
= P * * 12
= P *
= P * 100%
Q2. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is Q2. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is
Q5. The price of an article is first decreased by 40% and then increased by 20%. The net change in the price will be : Q5. The price of an article is first decreased by 40% and then increased by 20%. The net change in the price will be :
Q7. The LCM of two numbers is 40 and their HCF is 4. If the difference between the two numbers is 12, then the sum of the numbers is Q7. The LCM of two numbers is 40 and their HCF is 4. If the difference between the two numbers is 12, then the sum of the numbers is
(A) 20 (A) 20
(B) 24 (B) 24
(C) 28 (C) 28
(D) 32 (D) 32
Answer: (C) 28 Answer: (C) 28
X * (X-12) = 40 * 4
X = 20
X + (X-12) = 20 + 20 - 12 = 28X * (X-12) = 40 * 4
X = 20
X + (X-12) = 20 + 20 - 12 = 28
Q8. The smallest positive integer that is simultaneously divisible by 6, 8 and 12 is Q8. The smallest positive integer that is simultaneously divisible by 6, 8 and 12 is
Q9. Three fourth of a number is more than two third of the number by 5. Then the number is Q9. Three fourth of a number is more than two third of the number by 5. Then the number is