Two numbers are in the ratio of 2 : 3 and the product of their LCM and HCF is 96. The sum of the numbers is [#1110]
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Q1. Two numbers are in the ratio of 2 : 3 and the product of their LCM and HCF is 96. The sum of the numbers is
Q1. Two numbers are in the ratio of 2 : 3 and the product of their LCM and HCF is 96. The sum of the numbers is
(A) 8
(A) 8
(A) 8
(B) 12
(B) 12
(B) 12
(C) 20
(C) 20
(C) 20
(D) 36
(D) 36
(D) 36
Answer: (C) 20
Answer: (C) 20
Answer: (C) 20
=> 2x * 3x = 96
=> 6x2 = 96
=> x2 = 96/6
=> x2 = 16
=> x = 4
2x + 3x = 5x = 5 * 4 = 20
=> 2x * 3x = 96 => 6x2 = 96 => x2 = 96/6 => x2 = 16 => x = 4 2x + 3x = 5x = 5 * 4 = 20
=> 2x * 3x = 96 => 6x2 = 96 => x2 = 96/6 => x2 = 16 => x = 4 2x + 3x = 5x = 5 * 4 = 20
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Related MCQ Quizzes
Q1. Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is
Q1. Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is
(A) 2 : 5
(A) 2 : 5
(A) 2 : 5
(B) 3 : 5
(B) 3 : 5
(B) 3 : 5
(C) 4 : 5
(C) 4 : 5
(C) 4 : 5
(D) 7 : 5
(D) 7 : 5
(D) 7 : 5
Answer: (C) 4 : 5
Answer: (C) 4 : 5
Answer: (C) 4 : 5
4 : 5
4 : 5
4 : 5
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Q2. 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
Q2. 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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Q3. The smallest positive integer that is simultaneously divisible by 6, 8 and 12 is
Q3. The smallest positive integer that is simultaneously divisible by 6, 8 and 12 is
(A) 24
(A) 24
(A) 24
(B) 18
(B) 18
(B) 18
(C) 48
(C) 48
(C) 48
(D) 36
(D) 36
(D) 36
Answer: (A) 24
Answer: (A) 24
Answer: (A) 24
6*4 = 24
8*3 = 24
12*2 = 24
6*4 = 24 8*3 = 24 12*2 = 24
6*4 = 24 8*3 = 24 12*2 = 24
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Q4. 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
Q4. 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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Q5. If 2x = 32, then the value of x is
Q5. If 2x = 32, then the value of x is
(A) 4
(A) 4
(A) 4
(B) 5
(B) 5
(B) 5
(C) 6
(C) 6
(C) 6
(D) 7
(D) 7
(D) 7
Answer: (B) 5
Answer: (B) 5
Answer: (B) 5
25 = 2*2*2*2*2 = 32
25 = 2*2*2*2*2 = 32
25 = 2*2*2*2*2 = 32
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Q6. Two numbers are in the ratio of 2 : 3 and the product of their LCM and HCF is 96. The sum of the numbers is
Q6. Two numbers are in the ratio of 2 : 3 and the product of their LCM and HCF is 96. The sum of the numbers is
(A) 8
(A) 8
(A) 8
(B) 12
(B) 12
(B) 12
(C) 20
(C) 20
(C) 20
(D) 36
(D) 36
(D) 36
Answer: (C) 20
Answer: (C) 20
Answer: (C) 20
=> 2x * 3x = 96
=> 6x2 = 96
=> x2 = 96/6
=> x2 = 16
=> x = 4
2x + 3x = 5x = 5 * 4 = 20
=> 2x * 3x = 96 => 6x2 = 96 => x2 = 96/6 => x2 = 16 => x = 4 2x + 3x = 5x = 5 * 4 = 20
=> 2x * 3x = 96 => 6x2 = 96 => x2 = 96/6 => x2 = 16 => x = 4 2x + 3x = 5x = 5 * 4 = 20
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Q7. Find the odd number - 15, 17, 6, 12.
Q7. 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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Q8. 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 :
Q8. 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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Q9. If is equal to
Q9. If is equal to
(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
Hence, X:Y:Z = 3:4:7
=
=
= 2
Hence, X:Y:Z = 3:4:7 = = = 2
Hence, X:Y:Z = 3:4:7 = = = 2
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Q10. In a range of consecutive numbers starting with 1, all the even numbers are removed. From the remaining, consider the first 7 numbers. The sum of these 7 numbers is
Q10. In a range of consecutive numbers starting with 1, all the even numbers are removed. From the remaining, consider the first 7 numbers. The sum of these 7 numbers is
(A) 35
(A) 35
(A) 35
(B) 42
(B) 42
(B) 42
(C) 49
(C) 49
(C) 49
(D) 56
(D) 56
(D) 56
Answer: (C) 49
Answer: (C) 49
Answer: (C) 49
1+3+5+7+9+11+13 = 49
1+3+5+7+9+11+13 = 49
1+3+5+7+9+11+13 = 49
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Related Questions
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