Two numbers are in the ratio 2 : 3 and the product of their HCF and LCM is 96. The sum of the two numbers is [#1693]
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Q1. Two numbers are in the ratio 2 : 3 and the product of their HCF and LCM is 96. The sum of the two numbers is
Q1. Two numbers are in the ratio 2 : 3 and the product of their HCF and LCM is 96. The sum of the two numbers is
(A) 18
(A) 18
(A) 18
(B) 20
(B) 20
(B) 20
(C) 22
(C) 22
(C) 22
(D) 24
(D) 24
(D) 24
Answer: (B) 20
Answer: (B) 20
Answer: (B) 20
2:3 = 8:12
N1*N2 = HCF*LCM
8*12 = 96
N1+N2 = 8+12 = 20
2:3 = 8:12 N1*N2 = HCF*LCM 8*12 = 96 N1+N2 = 8+12 = 20
2:3 = 8:12 N1*N2 = HCF*LCM 8*12 = 96 N1+N2 = 8+12 = 20
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Related MCQ Quizzes
Q1. The remainder when –76 is divided by 3, is
Q1. The remainder when –76 is divided by 3, is
(A) -1
(A) -1
(A) -1
(B) 1
(B) 1
(B) 1
(C) 2
(C) 2
(C) 2
(D) -2
(D) -2
(D) -2
Answer: (C) 2
Answer: (C) 2
Answer: (C) 2
2
Because, Reminder must be positive and it should be less then the divisor
2 Because, Reminder must be positive and it should be less then the divisor
2 Because, Reminder must be positive and it should be less then the divisor
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Q2. Two trains of 200 m and 100 m long are running parallel at the rate of 20 m/sec. and 22 m/sec. respectively. How much time will they take to cross each other, if the trains are running along the same direction?
Q2. Two trains of 200 m and 100 m long are running parallel at the rate of 20 m/sec. and 22 m/sec. respectively. How much time will they take to cross each other, if the trains are running along the same direction?
(A) 120 sec
(A) 120 sec
(A) 120 sec
(B) 100 sec
(B) 100 sec
(B) 100 sec
(C) 60 sec
(C) 60 sec
(C) 60 sec
(D) 50 sec
(D) 50 sec
(D) 50 sec
Answer: (D) 50 sec
Answer: (D) 50 sec
Answer: (D) 50 sec
The difference of speed = 22m/s - 20m/s = 2m/s
As the train with speed 22m/s will cross the other, hence this train have to go the smallest length of the trains distance ahead of the other to cross completely.
Hence This train will be ahead of 100m with a speed of 2m/s.
Hence The time to cross each other will take = 100m / (2m/s)
= 50 sec
The difference of speed = 22m/s - 20m/s = 2m/s As the train with speed 22m/s will cross the other, hence this train have to go the smallest length of the trains distance ahead of the other to cross completely. Hence This train will be ahead of 100m with a speed of 2m/s. Hence The time to cross each other will take = 100m / (2m/s) = 50 sec
The difference of speed = 22m/s - 20m/s = 2m/s As the train with speed 22m/s will cross the other, hence this train have to go the smallest length of the trains distance ahead of the other to cross completely. Hence This train will be ahead of 100m with a speed of 2m/s. Hence The time to cross each other will take = 100m / (2m/s) = 50 sec
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Q3. If 16℅ of 40℅ of a number is 8, then the number is
Q3. If 16℅ of 40℅ of a number is 8, then the number is
(A) 250
(A) 250
(A) 250
(B) 125
(B) 125
(B) 125
(C) 200
(C) 200
(C) 200
(D) 100
(D) 100
(D) 100
Answer: (B) 125
Answer: (B) 125
Answer: (B) 125
125
125
125
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Q4. If the average age of A, B and C is 22 years and the average age of B and C is 25 years, then find the age of A after 9 years from now.
Q4. If the average age of A, B and C is 22 years and the average age of B and C is 25 years, then find the age of A after 9 years from now.
(A) 25 years
(A) 25 years
(A) 25 years
(B) 35 years
(B) 35 years
(B) 35 years
(C) 50 years
(C) 50 years
(C) 50 years
(D) 45 years
(D) 45 years
(D) 45 years
Answer: (A) 25 years
Answer: (A) 25 years
Answer: (A) 25 years
25 years
=> A+B+C = 22*3
=> A+(B+C) = 66
=> A+(25*2) = 66
=> A = 66-50
=> A = 16
After 9 years A = 16+9 = 25 years
25 years => A+B+C = 22*3 => A+(B+C) = 66 => A+(25*2) = 66 => A = 66-50 => A = 16 After 9 years A = 16+9 = 25 years
25 years => A+B+C = 22*3 => A+(B+C) = 66 => A+(25*2) = 66 => A = 66-50 => A = 16 After 9 years A = 16+9 = 25 years
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Q5. 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 :
Q5. 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
(A) 60 years
(B) 51 years
(B) 51 years
(B) 51 years
(C) 48 years
(C) 48 years
(C) 48 years
(D) 45 years
(D) 45 years
(D) 45 years
Answer: (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.
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.
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Q6. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is
Q6. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is
(A) 1:6
(A) 1:6
(A) 1:6
(B) 2:9
(B) 2:9
(B) 2:9
(C) 1:9
(C) 1:9
(C) 1:9
(D) 6:9
(D) 6:9
(D) 6:9
Answer: (C) 1:9
Answer: (C) 1:9
Answer: (C) 1:9
Area = πr2
π12 : π32
= π1 : π9
= 1:9
Area = πr2 π12 : π32 = π1 : π9 = 1:9
Area = πr2 π12 : π32 = π1 : π9 = 1:9
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Q7. Find the value of
Q7. Find the value of
(A)
(A)
(A)
(B)
(B)
(B)
(C)
(C)
(C)
(D)
(D)
(D)
Answer: (D)
Answer: (D)
Answer: (D)
=
=
=
=
=
= = = = =
= = = = =
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Q8. The angles of a quadrilateral are in the ratio of 1 : 3 : 4 : 7. The difference between the largest and the smallest angle is
Q8. The angles of a quadrilateral are in the ratio of 1 : 3 : 4 : 7. The difference between the largest and the smallest angle is
(A) 120°
(A) 120°
(A) 120°
(B) 140°
(B) 140°
(B) 140°
(C) 144°
(C) 144°
(C) 144°
(D) 145°
(D) 145°
(D) 145°
Answer: (C) 144°
Answer: (C) 144°
Answer: (C) 144°
(360°/15) * (7-1) = 144°
(360°/15) * (7-1) = 144°
(360°/15) * (7-1) = 144°
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Q9. A person moves along a path such that he is always away from a given point by 7 m. After moving for some time, he again reaches his starting point. The approximate distance the person moved during the time is
Q9. A person moves along a path such that he is always away from a given point by 7 m. After moving for some time, he again reaches his starting point. The approximate distance the person moved during the time is
(A) 22 m
(A) 22 m
(A) 22 m
(B) 44 m
(B) 44 m
(B) 44 m
(C) 122 m
(C) 122 m
(C) 122 m
(D) 144 m
(D) 144 m
(D) 144 m
Answer: (B) 44 m
Answer: (B) 44 m
Answer: (B) 44 m
2 * * 7
= 2 * 22
= 44
2 * * 7 = 2 * 22 = 44
2 * * 7 = 2 * 22 = 44
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Q10. What is the term for all positive and negative numbers as a whole including zero?
Q10. What is the term for all positive and negative numbers as a whole including zero?
(A) Real Numbers
(A) Real Numbers
(A) Real Numbers
(B) Natural Numbers
(B) Natural Numbers
(B) Natural Numbers
(C) Whole Numbers
(C) Whole Numbers
(C) Whole Numbers
(D) Integer Numbers
(D) Integer Numbers
(D) Integer Numbers
Answer: (D) Integer Numbers
Answer: (D) Integer Numbers
Answer: (D) Integer Numbers
Integers
Integers
Integers
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Related Questions
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