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. If '÷' means ‘addition’, ‘+’ means ‘subtraction’, ‘–’ means ‘multiplication’ and ‘×’ means ‘division’, then the value of 18 ÷ 12 × 4 – 5 is
Q1. If '÷' means ‘addition’, ‘+’ means ‘subtraction’, ‘–’ means ‘multiplication’ and ‘×’ means ‘division’, then the value of 18 ÷ 12 × 4 – 5 is
(A) 25
(A) 25
(A) 25
(B) 35
(B) 35
(B) 35
(C) 40
(C) 40
(C) 40
(D) 33
(D) 33
(D) 33
Answer: (D) 33
Answer: (D) 33
Answer: (D) 33
33
33
33
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Q2. A number was increased by 40% and thereafter decreased by 40%. The net change in the number in percentage is
Q2. A number was increased by 40% and thereafter decreased by 40%. The net change in the number in percentage is
(A) 16% increase
(A) 16% increase
(A) 16% increase
(B) 16% decrease
(B) 16% decrease
(B) 16% decrease
(C) 32% decrease
(C) 32% decrease
(C) 32% decrease
(D) No change
(D) No change
(D) No change
Answer: (B) 16% decrease
Answer: (B) 16% decrease
Answer: (B) 16% decrease
16% decrease
= 40-40-%
= -%
= -16%
16% decrease = 40-40-% = -% = -16%
16% decrease = 40-40-% = -% = -16%
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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. 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 :
Q4. 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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Q5. The sum of first 6 prime numbers is
Q5. The sum of first 6 prime numbers is
(A) 40
(A) 40
(A) 40
(B) 43
(B) 43
(B) 43
(C) 41
(C) 41
(C) 41
(D) 42
(D) 42
(D) 42
Answer: (C) 41
Answer: (C) 41
Answer: (C) 41
41
41
41
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Q6. The sum of 5 consecutive odd numbers is found to be 95. The largest of the numbers is
Q6. 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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Q7. If the decimal number 34p5 is divisible by 9, then the value of p is
Q7. If the decimal number 34p5 is divisible by 9, then the value of p is
(A) 8
(A) 8
(A) 8
(B) 7
(B) 7
(B) 7
(C) 4
(C) 4
(C) 4
(D) 6
(D) 6
(D) 6
Answer: (D) 6
Answer: (D) 6
Answer: (D) 6
34p5 -> 3+4+p+5 = 12+p
12+p = 9k
12+6 = 18 = 9k
p = 6
34p5 -> 3+4+p+5 = 12+p 12+p = 9k 12+6 = 18 = 9k p = 6
34p5 -> 3+4+p+5 = 12+p 12+p = 9k 12+6 = 18 = 9k p = 6
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Q8. 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
Q8. 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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Q9. Shyam stored Rs 35 in the form of 1 rupee coin and 50 paise coins in the ratio 2 : 3. The number of 50 paise coins are
Q9. Shyam stored Rs 35 in the form of 1 rupee coin and 50 paise coins in the ratio 2 : 3. The number of 50 paise coins are
(A) 20
(A) 20
(A) 20
(B) 25
(B) 25
(B) 25
(C) 30
(C) 30
(C) 30
(D) 35
(D) 35
(D) 35
Answer: (C) 30
Answer: (C) 30
Answer: (C) 30
1*20 + 0.5*30 = 20+15 = 35
1*20 + 0.5*30 = 20+15 = 35
1*20 + 0.5*30 = 20+15 = 35
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Q10. The count of prime numbers between 80 and 100 is
Q10. 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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