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 [#1087]
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Q1. 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
Q1. 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
(A) 20
(B) 24
(B) 24
(B) 24
(C) 28
(C) 28
(C) 28
(D) 32
(D) 32
(D) 32
Answer: (C) 28
Answer: (C) 28
Answer: (C) 28
X * (X-12) = 40 * 4
X = 20
X + (X-12) = 20 + 20 - 12 = 28
X * (X-12) = 40 * 4 X = 20 X + (X-12) = 20 + 20 - 12 = 28
X * (X-12) = 40 * 4 X = 20 X + (X-12) = 20 + 20 - 12 = 28
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Related MCQ Quizzes
Q1. A certain amount invested in a firm would become double at the end of one month, but it deducts an amount of ₹120 on every doubling. A person invests an amount of ₹105 and continues for 3 months without investing any additional amount. At the end of 3 months, his net income is
Q1. A certain amount invested in a firm would become double at the end of one month, but it deducts an amount of ₹120 on every doubling. A person invests an amount of ₹105 and continues for 3 months without investing any additional amount. At the end of 3 months, his net income is
(A) 55
(A) 55
(A) 55
(B) 0
(B) 0
(B) 0
(C) 270
(C) 270
(C) 270
(D) 45
(D) 45
(D) 45
Answer: (B) 0
Answer: (B) 0
Answer: (B) 0
1st Month -> (105 * 2) - 120 = 210 - 120 = 90
2nd Month -> (90 * 2) - 120 = 180 - 120 = 60
3rd Month -> (60 * 2) - 120 = 120 - 120 = 0
1st Month -> (105 * 2) - 120 = 210 - 120 = 90 2nd Month -> (90 * 2) - 120 = 180 - 120 = 60 3rd Month -> (60 * 2) - 120 = 120 - 120 = 0
1st Month -> (105 * 2) - 120 = 210 - 120 = 90 2nd Month -> (90 * 2) - 120 = 180 - 120 = 60 3rd Month -> (60 * 2) - 120 = 120 - 120 = 0
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Q2. What is the sum of the interior angles of a triangle?
Q2. What is the sum of the interior angles of a triangle?
(A) 180 degrees
(A) 180 degrees
(A) 180 degrees
(B) 270 degrees
(B) 270 degrees
(B) 270 degrees
(C) 360 degrees
(C) 360 degrees
(C) 360 degrees
(D) 450 degrees
(D) 450 degrees
(D) 450 degrees
Answer: (A) 180 degrees
Answer: (A) 180 degrees
Answer: (A) 180 degrees
The sum of the interior angles of a triangle is always 180 degrees, a fundamental principle in geometry.
The sum of the interior angles of a triangle is always 180 degrees, a fundamental principle in geometry.
The sum of the interior angles of a triangle is always 180 degrees, a fundamental principle in geometry.
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Q3. What is the term for a number that has no decimal places or fractional part?
Q3. What is the term for a number that has no decimal places or fractional part?
(A) Integer
(A) Integer
(A) Integer
(B) Fraction
(B) Fraction
(B) Fraction
(C) Decimal
(C) Decimal
(C) Decimal
(D) Percentage
(D) Percentage
(D) Percentage
Answer: (A) Integer
Answer: (A) Integer
Answer: (A) Integer
An integer is a whole number, either positive, negative, or zero, without a decimal or fractional part, such as 5, -3, or 0.
An integer is a whole number, either positive, negative, or zero, without a decimal or fractional part, such as 5, -3, or 0.
An integer is a whole number, either positive, negative, or zero, without a decimal or fractional part, such as 5, -3, or 0.
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Q4. The sum of 5 consecutive odd numbers is found to be 95. The largest of the numbers is
Q4. The sum of 5 consecutive odd numbers is found to be 95. The largest of the numbers is
(A) 17
(A) 17
(A) 17
(B) 21
(B) 21
(B) 21
(C) 23
(C) 23
(C) 23
(D) 19
(D) 19
(D) 19
Answer: (C) 23
Answer: (C) 23
Answer: (C) 23
X + (X+2) + (X+4) + (X+6) + (X+8) = 95
=> 5X + 20 = 95
=> 5X = 95-20
=> 5X = 75
=> X = 15
Hence (X+8) = 15+8 = 23
X + (X+2) + (X+4) + (X+6) + (X+8) = 95 => 5X + 20 = 95 => 5X = 95-20 => 5X = 75 => X = 15 Hence (X+8) = 15+8 = 23
X + (X+2) + (X+4) + (X+6) + (X+8) = 95 => 5X + 20 = 95 => 5X = 95-20 => 5X = 75 => X = 15 Hence (X+8) = 15+8 = 23
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Q5. A hostel has 120 students and food supplies are for 45 days. If 30 more students joined the hostel, then how many days the hostel will run with the existing food?
Q5. A hostel has 120 students and food supplies are for 45 days. If 30 more students joined the hostel, then how many days the hostel will run with the existing food?
(A) 40 days
(A) 40 days
(A) 40 days
(B) 38 days
(B) 38 days
(B) 38 days
(C) 36 days
(C) 36 days
(C) 36 days
(D) 32 days
(D) 32 days
(D) 32 days
Answer: (C) 36 days
Answer: (C) 36 days
Answer: (C) 36 days
36 days
36 days
36 days
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Q6. The count of prime numbers between 80 and 100 is
Q6. The count of prime numbers between 80 and 100 is
(A) 2
(A) 2
(A) 2
(B) 5
(B) 5
(B) 5
(C) 3
(C) 3
(C) 3
(D) 4
(D) 4
(D) 4
Answer: (C) 3
Answer: (C) 3
Answer: (C) 3
Prime numbers are numbers that are only divisible by 1 and themselves. Between 80 and 100, the prime numbers are 83, 89, and 97. Therefore, there are 3 prime numbers in this range.
Prime numbers are numbers that are only divisible by 1 and themselves. Between 80 and 100, the prime numbers are 83, 89, and 97. Therefore, there are 3 prime numbers in this range.
Prime numbers are numbers that are only divisible by 1 and themselves. Between 80 and 100, the prime numbers are 83, 89, and 97. Therefore, there are 3 prime numbers in this range.
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Q7. The least number by which 2450 must be multiplied to make it a perfect square, is
Q7. The least number by which 2450 must be multiplied to make it a perfect square, is
(A) 2
(A) 2
(A) 2
(B) 3
(B) 3
(B) 3
(C) 4
(C) 4
(C) 4
(D) 5
(D) 5
(D) 5
Answer: (A) 2
Answer: (A) 2
Answer: (A) 2
2
2
2
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Q8. At what percentage of simple interest does an amount of money double in 12 years?
Q8. At what percentage of simple interest does an amount of money double in 12 years?
(A)
(A)
(A)
(B)
(B)
(B)
(C)
(C)
(C)
(D)
(D)
(D)
Answer: (C)
Answer: (C)
Answer: (C)
P * * 12
= P * * 12
= P *
= P * 100%
P * * 12 = P * * 12 = P * = P * 100%
P * * 12 = P * * 12 = P * = P * 100%
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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. ‘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 :
Q10. ‘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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