If 2x = 32, then the value of x is [#1705]
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Q1. If 2x = 32, then the value of x is
Q1. 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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Related MCQ Quizzes
Q1. What is the term for all positive and negative numbers as a whole including zero?
Q1. What is the term for all positive and negative numbers as a whole including zero?
(A) Real Numbers
(A) Real Numbers
(A) Real Numbers
(B) Natural Numbers
(B) Natural Numbers
(B) Natural Numbers
(C) Whole Numbers
(C) Whole Numbers
(C) Whole Numbers
(D) Integer Numbers
(D) Integer Numbers
(D) Integer Numbers
Answer: (D) Integer Numbers
Answer: (D) Integer Numbers
Answer: (D) Integer Numbers
Integers
Integers
Integers
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Q2. When three times of a given number is subtracted from the square of the number, the result is the number itself. The number is
Q2. 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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Q3. 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
Q3. 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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Q4. 45% of a number is same as 30% of another number. The ratio of the first number to the second number is :
Q4. 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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Q5. Which is the smallest Natural Number?
Q5. Which is the smallest Natural Number?
(A) -1
(A) -1
(A) -1
(B) 0
(B) 0
(B) 0
(C) 1
(C) 1
(C) 1
(D) 2
(D) 2
(D) 2
Answer: (C) 1
Answer: (C) 1
Answer: (C) 1
1
1
1
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Q6. The least number by which 2450 must be multiplied to make it a perfect square, is
Q6. 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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Q7. What is the formula to calculate the area of a circle?
Q7. What is the formula to calculate the area of a circle?
(A) A = πr2
(A) A = πr2
(A) A = πr2
(B) A = 2πr
(B) A = 2πr
(B) A = 2πr
(C) A = πd
(C) A = πd
(C) A = πd
(D) A = 1/2πr2
(D) A = 1/2πr2
(D) A = 1/2πr2
Answer: (A) A = πr2
Answer: (A) A = πr2
Answer: (A) A = πr2
The formula to calculate the area of a circle is A = πr2, where A is the area and r is the radius of the circle.
The formula to calculate the area of a circle is A = πr2, where A is the area and r is the radius of the circle.
The formula to calculate the area of a circle is A = πr2, where A is the area and r is the radius of the circle.
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Q8. The sum of the four consecutive alternate numbers is 64. The smallest number amongst them is
Q8. The sum of the four consecutive alternate numbers is 64. The smallest number amongst them is
(A) 13
(A) 13
(A) 13
(B) 11
(B) 11
(B) 11
(C) 16
(C) 16
(C) 16
(D) 9
(D) 9
(D) 9
Answer: (A) 13
Answer: (A) 13
Answer: (A) 13
The smallest number amongst them is 13
=> X + (X+2) + (X+4) + (X+6) = 64
=> 4X + 12 = 64
=> 4X = 64 - 12
=> 4X = 52
=> X = 13
The smallest number amongst them is 13 => X + (X+2) + (X+4) + (X+6) = 64 => 4X + 12 = 64 => 4X = 64 - 12 => 4X = 52 => X = 13
The smallest number amongst them is 13 => X + (X+2) + (X+4) + (X+6) = 64 => 4X + 12 = 64 => 4X = 64 - 12 => 4X = 52 => X = 13
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Q9. 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
Q9. 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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Q10. The smallest positive integer that is simultaneously divisible by 6, 8 and 12 is
Q10. 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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