If A : B = 3, then 144A : 36B is [#1697]
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Q1. If A : B = 3, then 144A : 36B is
Q1. If A : B = 3, then 144A : 36B is
(A) 4
(A) 4
(A) 4
(B) 12
(B) 12
(B) 12
(C) 3
(C) 3
(C) 3
(D) 16
(D) 16
(D) 16
Answer: (B) 12
Answer: (B) 12
Answer: (B) 12
A : B = 3
=> = 3
=
= x 3
= 4 x 3
= 12
A : B = 3 => = 3 = = x 3 = 4 x 3 = 12
A : B = 3 => = 3 = = x 3 = 4 x 3 = 12
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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. 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
Q2. 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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Q3. The median of the sequence of numbers 2, – 1, 3, 1, – 2, 5, 6 is
Q3. The median of the sequence of numbers 2, – 1, 3, 1, – 2, 5, 6 is
(A) 1
(A) 1
(A) 1
(B) -1
(B) -1
(B) -1
(C) 2
(C) 2
(C) 2
(D) 3
(D) 3
(D) 3
Answer: (C) 2
Answer: (C) 2
Answer: (C) 2
In ascending order = -2, -1, 1, 2, 3, 5, 6
median = th term
= 4th term
= 2
In ascending order = -2, -1, 1, 2, 3, 5, 6 median = th term = 4th term = 2
In ascending order = -2, -1, 1, 2, 3, 5, 6 median = th term = 4th term = 2
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Q4. What is the value of x in the equation 2x + 5 = 11?
Q4. What is the value of x in the equation 2x + 5 = 11?
(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: (B) 3
Answer: (B) 3
Answer: (B) 3
To solve for x, subtract 5 from both sides of the equation, resulting in 2x = 6. Then, divide both sides by 2, giving x = 3.
To solve for x, subtract 5 from both sides of the equation, resulting in 2x = 6. Then, divide both sides by 2, giving x = 3.
To solve for x, subtract 5 from both sides of the equation, resulting in 2x = 6. Then, divide both sides by 2, giving x = 3.
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Q5. The smallest among is
Q5. The smallest among is
(A)
(A)
(A)
(B)
(B)
(B)
(C)
(C)
(C)
(D)
(D)
(D)
Answer: (C)
Answer: (C)
Answer: (C)
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Q6. If 16℅ of 40℅ of a number is 8, then the number is
Q6. 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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Q7. When three times of a given number is subtracted from the square of the number, the result is the number itself. The number is
Q7. When three times of a given number is subtracted from the square of the number, the result is the number itself. The number is
(A) 4
(A) 4
(A) 4
(B) 2
(B) 2
(B) 2
(C) -2
(C) -2
(C) -2
(D) -4
(D) -4
(D) -4
Answer: (A) 4
Answer: (A) 4
Answer: (A) 4
42 - (4*3)
= 16 - 12
= 4
42 - (4*3) = 16 - 12 = 4
42 - (4*3) = 16 - 12 = 4
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Q8. 98 toffees were distributed to some boys in a group. Each boy in the group got twice as many of the toffees as the number of boys. The number of boys in the group was
Q8. 98 toffees were distributed to some boys in a group. Each boy in the group got twice as many of the toffees as the number of boys. The number of boys in the group was
(A) 5
(A) 5
(A) 5
(B) 7
(B) 7
(B) 7
(C) 10
(C) 10
(C) 10
(D) 14
(D) 14
(D) 14
Answer: (B) 7
Answer: (B) 7
Answer: (B) 7
7
=> x * 2x = 98
=> x2 =
=> x2 = 49
=> x = 7
7 => x * 2x = 98 => x2 = => x2 = 49 => x = 7
7 => x * 2x = 98 => x2 = => x2 = 49 => x = 7
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Q9. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is
Q9. 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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Q10. 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
Q10. 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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