The smallest positive integer that is simultaneously divisible by 6, 8 and 12 is [#1714]
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Q1. The smallest positive integer that is simultaneously divisible by 6, 8 and 12 is
Q1. 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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Related MCQ Quizzes
Q1. 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
Q1. 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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Q2. A number when divided by 6 is diminished by 40. The number is
Q2. A number when divided by 6 is diminished by 40. The number is
(A) 72
(A) 72
(A) 72
(B) 48
(B) 48
(B) 48
(C) 60
(C) 60
(C) 60
(D) 84
(D) 84
(D) 84
Answer: (B) 48
Answer: (B) 48
Answer: (B) 48
48
48
48
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Q3. In a school with 10 teachers, one retires and immediately a new teacher of age 25 years joins. As a result, the average age of the teacher reduces by 3. The age of the retired teacher is
Q3. In a school with 10 teachers, one retires and immediately a new teacher of age 25 years joins. As a result, the average age of the teacher reduces by 3. The age of the retired teacher is
(A) 55 years
(A) 55 years
(A) 55 years
(B) 65 years
(B) 65 years
(B) 65 years
(C) 58 years
(C) 58 years
(C) 58 years
(D) 60 years
(D) 60 years
(D) 60 years
Answer: (A) 55 years
Answer: (A) 55 years
Answer: (A) 55 years
25 years + (3*10) years = 55 years
25 years + (3*10) years = 55 years
25 years + (3*10) years = 55 years
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Q4. The least number by which 2450 must be multiplied to make it a perfect square, is
Q4. 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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Q5. What is the term for a line that divides a shape into two equal parts?
Q5. What is the term for a line that divides a shape into two equal parts?
(A) Axis
(A) Axis
(A) Axis
(B) Median
(B) Median
(B) Median
(C) Vertex
(C) Vertex
(C) Vertex
(D) Bisector
(D) Bisector
(D) Bisector
Answer: (D) Bisector
Answer: (D) Bisector
Answer: (D) Bisector
A bisector is a line that divides a shape into two equal parts, like a line that cuts a triangle into two equal areas or a line that divides a circle into two equal parts (semi-circles).
A bisector is a line that divides a shape into two equal parts, like a line that cuts a triangle into two equal areas or a line that divides a circle into two equal parts (semi-circles).
A bisector is a line that divides a shape into two equal parts, like a line that cuts a triangle into two equal areas or a line that divides a circle into two equal parts (semi-circles).
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Q6. The median of the sequence of numbers 2, – 1, 3, 1, – 2, 5, 6 is
Q6. 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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Q7. The sum of 5 consecutive odd numbers is found to be 95. The largest of the numbers is
Q7. 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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Q8. A wheel can cover a distance of 22 km in 1000 rounds. The radius of the wheel is
Q8. A wheel can cover a distance of 22 km in 1000 rounds. The radius of the wheel is
(A) 4.5 m
(A) 4.5 m
(A) 4.5 m
(B) 2.1 m
(B) 2.1 m
(B) 2.1 m
(C) 2.8 m
(C) 2.8 m
(C) 2.8 m
(D) 3.5 m
(D) 3.5 m
(D) 3.5 m
Answer: (D) 3.5 m
Answer: (D) 3.5 m
Answer: (D) 3.5 m
3.5 m
3.5 m
3.5 m
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Q9. A number is as much greater than 31 as is less 55. Then the number is
Q9. A number is as much greater than 31 as is less 55. Then the number is
(A) 39
(A) 39
(A) 39
(B) 32
(B) 32
(B) 32
(C) 43
(C) 43
(C) 43
(D) 47
(D) 47
(D) 47
Answer: (C) 43
Answer: (C) 43
Answer: (C) 43
43
43
43
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Q10. Find the median of the data set.
Q10. Find the median of the data set.
1, 2, 4, 1, 5, 12, 7, 8, 5, 1, 16, 17, 12
1, 2, 4, 1, 5, 12, 7, 8, 5, 1, 16, 17, 12
1, 2, 4, 1, 5, 12, 7, 8, 5, 1, 16, 17, 12
(A) 8
(A) 8
(A) 8
(B) 7
(B) 7
(B) 7
(C) 5
(C) 5
(C) 5
(D) 4
(D) 4
(D) 4
Answer: (C) 5
Answer: (C) 5
Answer: (C) 5
1, 2, 4, 1, 5, 12, 7, 8, 5, 1, 16, 17, 12
= 1, 1, 1, 2, 4, 5, 5, 7, 8, 12, 12, 16, 17
Total numbers of values N = 13
Hence median value = (N + 1)/2 = (13 + 1)/2 = 7th
7th Number = 5
1, 2, 4, 1, 5, 12, 7, 8, 5, 1, 16, 17, 12 = 1, 1, 1, 2, 4, 5, 5, 7, 8, 12, 12, 16, 17 Total numbers of values N = 13 Hence median value = (N + 1)/2 = (13 + 1)/2 = 7th 7th Number = 5
1, 2, 4, 1, 5, 12, 7, 8, 5, 1, 16, 17, 12 = 1, 1, 1, 2, 4, 5, 5, 7, 8, 12, 12, 16, 17 Total numbers of values N = 13 Hence median value = (N + 1)/2 = (13 + 1)/2 = 7th 7th Number = 5
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