The median of the sequence of numbers 2, – 1, 3, 1, – 2, 5, 6 is [#1712]
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Q1. The median of the sequence of numbers 2, – 1, 3, 1, – 2, 5, 6 is
Q1. 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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Related MCQ Quizzes
Q1. Which is the smallest prime number?
Q1. Which is the smallest prime number?
(A) 1
(A) 1
(A) 1
(B) 2
(B) 2
(B) 2
(C) 3
(C) 3
(C) 3
(D) 5
(D) 5
(D) 5
Answer: (B) 2
Answer: (B) 2
Answer: (B) 2
2
2
2
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Q2. ‘A’ starts his journey at 1:00 p.m. from a location P with a speed of 1 m/sec. ‘B’ starts his journey from the same location P and along the same direction at 1:10 p.m. with a speed of 2 m/sec. If ‘B’ meets ‘A’ at the location Q, then the distance PQ is :
Q2. ‘A’ starts his journey at 1:00 p.m. from a location P with a speed of 1 m/sec. ‘B’ starts his journey from the same location P and along the same direction at 1:10 p.m. with a speed of 2 m/sec. If ‘B’ meets ‘A’ at the location Q, then the distance PQ is :
(A) 1.5 km
(A) 1.5 km
(A) 1.5 km
(B) 1.75 km
(B) 1.75 km
(B) 1.75 km
(C) 1.2 km
(C) 1.2 km
(C) 1.2 km
(D) 1.25 km
(D) 1.25 km
(D) 1.25 km
Answer: (C) 1.2 km
Answer: (C) 1.2 km
Answer: (C) 1.2 km
A starts journey before B at a speed of 1 m/s. Hence A will be ahead of B
(10*60)s * 1m/s = 600m
Let A covers a distance of X after starting of B,
Then X + 600m = 2X
=> 2X - X = 600m
=> X = 600m
Hence B will cover a distance of 2X = 2 * 600m = 1200m = 1.2 km
A starts journey before B at a speed of 1 m/s. Hence A will be ahead of B (10*60)s * 1m/s = 600m Let A covers a distance of X after starting of B, Then X + 600m = 2X => 2X - X = 600m => X = 600m Hence B will cover a distance of 2X = 2 * 600m = 1200m = 1.2 km
A starts journey before B at a speed of 1 m/s. Hence A will be ahead of B (10*60)s * 1m/s = 600m Let A covers a distance of X after starting of B, Then X + 600m = 2X => 2X - X = 600m => X = 600m Hence B will cover a distance of 2X = 2 * 600m = 1200m = 1.2 km
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Q3. The sum of two numbers is 15 and the sum of their squares is 113. The numbers are
Q3. The sum of two numbers is 15 and the sum of their squares is 113. The numbers are
(A) 4 and 10
(A) 4 and 10
(A) 4 and 10
(B) 6 and 9
(B) 6 and 9
(B) 6 and 9
(C) 5 and 10
(C) 5 and 10
(C) 5 and 10
(D) 7 and 8
(D) 7 and 8
(D) 7 and 8
Answer: (D) 7 and 8
Answer: (D) 7 and 8
Answer: (D) 7 and 8
7 and 8
7 and 8
7 and 8
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Q4. If the price of sugar increases by 20%, by what percentage should a household reduce its consumption of sugar so that the budget remains the same?
Q4. If the price of sugar increases by 20%, by what percentage should a household reduce its consumption of sugar so that the budget remains the same?
(A)
(A)
(A)
(B)
(B)
(B)
(C)
(C)
(C)
(D)
(D)
(D)
Answer: (D)
Answer: (D)
Answer: (D)
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Q5. The price of an article is first decreased by 40% and then increased by 20%. The net change in the price will be :
Q5. 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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Q6. The missing term in the sequence 7, 11, 19, 31, ______, 67 is
Q6. The missing term in the sequence 7, 11, 19, 31, ______, 67 is
(A) 43
(A) 43
(A) 43
(B) 47
(B) 47
(B) 47
(C) 51
(C) 51
(C) 51
(D) 45
(D) 45
(D) 45
Answer: (B) 47
Answer: (B) 47
Answer: (B) 47
47
47
47
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Q7. The remainder when –76 is divided by 3, is
Q7. The remainder when –76 is divided by 3, is
(A) -1
(A) -1
(A) -1
(B) 1
(B) 1
(B) 1
(C) 2
(C) 2
(C) 2
(D) -2
(D) -2
(D) -2
Answer: (C) 2
Answer: (C) 2
Answer: (C) 2
2
Because, Reminder must be positive and it should be less then the divisor
2 Because, Reminder must be positive and it should be less then the divisor
2 Because, Reminder must be positive and it should be less then the divisor
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Q8. A number when divided by 6 is diminished by 40. The number is
Q8. 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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Q9. 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
Q9. 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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Q10. 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
Q10. 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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