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. ‘A’ starts his journey at 1:00 p.m. from a location P with a speed of 1 m/sec. ‘B’ starts his journey from the same location P and along the same direction at 1:10 p.m. with a speed of 2 m/sec. If ‘B’ meets ‘A’ at the location Q, then the distance PQ is :
Q1. ‘A’ starts his journey at 1:00 p.m. from a location P with a speed of 1 m/sec. ‘B’ starts his journey from the same location P and along the same direction at 1:10 p.m. with a speed of 2 m/sec. If ‘B’ meets ‘A’ at the location Q, then the distance PQ is :
(A) 1.5 km
(A) 1.5 km
(A) 1.5 km
(B) 1.75 km
(B) 1.75 km
(B) 1.75 km
(C) 1.2 km
(C) 1.2 km
(C) 1.2 km
(D) 1.25 km
(D) 1.25 km
(D) 1.25 km
Answer: (C) 1.2 km
Answer: (C) 1.2 km
Answer: (C) 1.2 km
A starts journey before B at a speed of 1 m/s. Hence A will be ahead of B
(10*60)s * 1m/s = 600m
Let A covers a distance of X after starting of B,
Then X + 600m = 2X
=> 2X - X = 600m
=> X = 600m
Hence B will cover a distance of 2X = 2 * 600m = 1200m = 1.2 km
A starts journey before B at a speed of 1 m/s. Hence A will be ahead of B (10*60)s * 1m/s = 600m Let A covers a distance of X after starting of B, Then X + 600m = 2X => 2X - X = 600m => X = 600m Hence B will cover a distance of 2X = 2 * 600m = 1200m = 1.2 km
A starts journey before B at a speed of 1 m/s. Hence A will be ahead of B (10*60)s * 1m/s = 600m Let A covers a distance of X after starting of B, Then X + 600m = 2X => 2X - X = 600m => X = 600m Hence B will cover a distance of 2X = 2 * 600m = 1200m = 1.2 km
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Q2. If 20% of a is b, then b% of 20 is the same as :
Q2. If 20% of a is b, then b% of 20 is the same as :
(A) 4% of a
(A) 4% of a
(A) 4% of a
(B) 8% of a
(B) 8% of a
(B) 8% of a
(C) 6% of a
(C) 6% of a
(C) 6% of a
(D) 10% of a
(D) 10% of a
(D) 10% of a
Answer: (A) 4% of a
Answer: (A) 4% of a
Answer: (A) 4% of a
20% of a is b
Hence b = 20% of a = 20a/100
b% of 20
= 20% of b
= (20/100) * 20a/100
= (20*20*a)/(100*100)
= 4a/100
= (4/100) * a
= 4% of a
20% of a is b Hence b = 20% of a = 20a/100 b% of 20 = 20% of b = (20/100) * 20a/100 = (20*20*a)/(100*100) = 4a/100 = (4/100) * a = 4% of a
20% of a is b Hence b = 20% of a = 20a/100 b% of 20 = 20% of b = (20/100) * 20a/100 = (20*20*a)/(100*100) = 4a/100 = (4/100) * a = 4% of a
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Q3. Consider the four numbers given numbers. If the digits of each number are arranged in ascending order, which new number would be the smallest?
Q3. Consider the four numbers given numbers. If the digits of each number are arranged in ascending order, which new number would be the smallest?
I. 376
II. 629
III. 921
IV. 397
I. 376 II. 629 III. 921 IV. 397
I. 376 II. 629 III. 921 IV. 397
(A) III
(A) III
(A) III
(B) II
(B) II
(B) II
(C) I
(C) I
(C) I
(D) IV
(D) IV
(D) IV
Answer: (A) III
Answer: (A) III
Answer: (A) III
376 -> 367
629 -> 269
921 -> 129 *
397 -> 379
376 -> 367 629 -> 269 921 -> 129 * 397 -> 379
376 -> 367 629 -> 269 921 -> 129 * 397 -> 379
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Q4. Find the odd number - 15, 17, 6, 12.
Q4. Find the odd number - 15, 17, 6, 12.
(A) 6
(A) 6
(A) 6
(B) 12
(B) 12
(B) 12
(C) 15
(C) 15
(C) 15
(D) 17
(D) 17
(D) 17
Answer: (D) 17
Answer: (D) 17
Answer: (D) 17
17 which is a Prime number.
17 which is a Prime number.
17 which is a Prime number.
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Q5. The simple interest earned by 4,000 in 18 months at 12% per annum is
Q5. The simple interest earned by 4,000 in 18 months at 12% per annum is
(A) 216
(A) 216
(A) 216
(B) 720
(B) 720
(B) 720
(C) 360
(C) 360
(C) 360
(D) 960
(D) 960
(D) 960
Answer: (B) 720
Answer: (B) 720
Answer: (B) 720
720
720
720
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Q6. The value of is :
Q6. The value of is :
(A) 1.73
(A) 1.73
(A) 1.73
(B) 2.03
(B) 2.03
(B) 2.03
(C) 2.73
(C) 2.73
(C) 2.73
(D) 1.03
(D) 1.03
(D) 1.03
Answer: (C) 2.73
Answer: (C) 2.73
Answer: (C) 2.73
=
=
= 0.2 + 1.2 + 1.3 + 0.03
= 2.73
= = = 0.2 + 1.2 + 1.3 + 0.03 = 2.73
= = = 0.2 + 1.2 + 1.3 + 0.03 = 2.73
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Q7. What is the term for a shape with five sides?
Q7. What is the term for a shape with five sides?
(A) Triangle
(A) Triangle
(A) Triangle
(B) Quadrilateral
(B) Quadrilateral
(B) Quadrilateral
(C) Pentagon
(C) Pentagon
(C) Pentagon
(D) Hexagon
(D) Hexagon
(D) Hexagon
Answer: (C) Pentagon
Answer: (C) Pentagon
Answer: (C) Pentagon
A pentagon is a polygon with five sides, like the shape of the Pentagon building in Washington, D.C.
A pentagon is a polygon with five sides, like the shape of the Pentagon building in Washington, D.C.
A pentagon is a polygon with five sides, like the shape of the Pentagon building in Washington, D.C.
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Q8. 45% of a number is same as 30% of another number. The ratio of the first number to the second number is :
Q8. 45% of a number is same as 30% of another number. The ratio of the first number to the second number is :
(A) 2 : 3
(A) 2 : 3
(A) 2 : 3
(B) 3 : 5
(B) 3 : 5
(B) 3 : 5
(C) 3 : 2
(C) 3 : 2
(C) 3 : 2
(D) 2 : 5
(D) 2 : 5
(D) 2 : 5
Answer: (A) 2 : 3
Answer: (A) 2 : 3
Answer: (A) 2 : 3
=
=
= X:Y = 2:3
= = = X:Y = 2:3
= = = X:Y = 2:3
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Q9. 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 :
Q9. 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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Q10. The product of two consecutive odd numbers is 19043. Which is the smaller number?
Q10. The product of two consecutive odd numbers is 19043. Which is the smaller number?
(A) 133
(A) 133
(A) 133
(B) 131
(B) 131
(B) 131
(C) 137
(C) 137
(C) 137
(D) 129
(D) 129
(D) 129
Answer: (C) 137
Answer: (C) 137
Answer: (C) 137
137
137
137
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Related Questions
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