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. Two trains of 200 m and 100 m long are running parallel at the rate of 20 m/sec. and 22 m/sec. respectively. How much time will they take to cross each other, if the trains are running along the same direction?
Q1. Two trains of 200 m and 100 m long are running parallel at the rate of 20 m/sec. and 22 m/sec. respectively. How much time will they take to cross each other, if the trains are running along the same direction?
(A) 120 sec
(A) 120 sec
(A) 120 sec
(B) 100 sec
(B) 100 sec
(B) 100 sec
(C) 60 sec
(C) 60 sec
(C) 60 sec
(D) 50 sec
(D) 50 sec
(D) 50 sec
Answer: (D) 50 sec
Answer: (D) 50 sec
Answer: (D) 50 sec
The difference of speed = 22m/s - 20m/s = 2m/s
As the train with speed 22m/s will cross the other, hence this train have to go the smallest length of the trains distance ahead of the other to cross completely.
Hence This train will be ahead of 100m with a speed of 2m/s.
Hence The time to cross each other will take = 100m / (2m/s)
= 50 sec
The difference of speed = 22m/s - 20m/s = 2m/s As the train with speed 22m/s will cross the other, hence this train have to go the smallest length of the trains distance ahead of the other to cross completely. Hence This train will be ahead of 100m with a speed of 2m/s. Hence The time to cross each other will take = 100m / (2m/s) = 50 sec
The difference of speed = 22m/s - 20m/s = 2m/s As the train with speed 22m/s will cross the other, hence this train have to go the smallest length of the trains distance ahead of the other to cross completely. Hence This train will be ahead of 100m with a speed of 2m/s. Hence The time to cross each other will take = 100m / (2m/s) = 50 sec
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Q2. 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 :
Q2. 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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Q3. The smallest among is
Q3. 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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Q4. A ladder of length 13 m is leaning against a vertical wall with the upper end at the height of 5 m. The horizontal distance between the foot of the wall and the lower end of the ladder is
Q4. A ladder of length 13 m is leaning against a vertical wall with the upper end at the height of 5 m. The horizontal distance between the foot of the wall and the lower end of the ladder is
(A) 9m
(A) 9m
(A) 9m
(B) 5m
(B) 5m
(B) 5m
(C) 11m
(C) 11m
(C) 11m
(D) 12m
(D) 12m
(D) 12m
Answer: (D) 12m
Answer: (D) 12m
Answer: (D) 12m
52 + X2 = 132
=> 25 + X2 = 169
=> X2 = 196 - 25
=> X2 = 144
=> X = 12
52 + X2 = 132 => 25 + X2 = 169 => X2 = 196 - 25 => X2 = 144 => X = 12
52 + X2 = 132 => 25 + X2 = 169 => X2 = 196 - 25 => X2 = 144 => X = 12
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Q5. The largest of 26, 35, 44 and 53 is
Q5. The largest of 26, 35, 44 and 53 is
(A) 26
(A) 26
(A) 26
(B) 35
(B) 35
(B) 35
(C) 44
(C) 44
(C) 44
(D) 53
(D) 53
(D) 53
Answer: (C) 44
Answer: (C) 44
Answer: (C) 44
26 = 64
35 = 243
44 = 256
53 = 125
26 = 64 35 = 243 44 = 256 53 = 125
26 = 64 35 = 243 44 = 256 53 = 125
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Q6. If the salary of A is 25% more than that of B, how much percentage is the salary of B lower than that of A?
Q6. If the salary of A is 25% more than that of B, how much percentage is the salary of B lower than that of A?
(A) 20%
(A) 20%
(A) 20%
(B) 25%
(B) 25%
(B) 25%
(C) 12%
(C) 12%
(C) 12%
(D) 16%
(D) 16%
(D) 16%
Answer: (A) 20%
Answer: (A) 20%
Answer: (A) 20%
20%
= %
%
= 20%
20% = % % = 20%
20% = % % = 20%
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Q7. 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
Q7. 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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Q8. At what percentage of simple interest does an amount of money double in 12 years?
Q8. At what percentage of simple interest does an amount of money double in 12 years?
(A)
(A)
(A)
(B)
(B)
(B)
(C)
(C)
(C)
(D)
(D)
(D)
Answer: (C)
Answer: (C)
Answer: (C)
P * * 12
= P * * 12
= P *
= P * 100%
P * * 12 = P * * 12 = P * = P * 100%
P * * 12 = P * * 12 = P * = P * 100%
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Q9. The price of an article is first decreased by 40% and then increased by 20%. The net change in the price will be :
Q9. The price of an article is first decreased by 40% and then increased by 20%. The net change in the price will be :
(A) 20%
(A) 20%
(A) 20%
(B) 40%
(B) 40%
(B) 40%
(C) 24%
(C) 24%
(C) 24%
(D) 28%
(D) 28%
(D) 28%
Answer: (D) 28%
Answer: (D) 28%
Answer: (D) 28%
Net change (-40 + 20 + )%
= (-20 + )%
= (-20 - 8)%
= -28%
= 28%
Net change (-40 + 20 + )% = (-20 + )% = (-20 - 8)% = -28% = 28%
Net change (-40 + 20 + )% = (-20 + )% = (-20 - 8)% = -28% = 28%
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Q10. 100% of 100 when added to 200% of 200 would result
Q10. 100% of 100 when added to 200% of 200 would result
(A) 300
(A) 300
(A) 300
(B) 400
(B) 400
(B) 400
(C) 500
(C) 500
(C) 500
(D) 600
(D) 600
(D) 600
Answer: (C) 500
Answer: (C) 500
Answer: (C) 500
100 * 100% + 200 * 200%
=
=
= 100 + 400
= 500
100 * 100% + 200 * 200% = = = 100 + 400 = 500
100 * 100% + 200 * 200% = = = 100 + 400 = 500
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