A person moves along a path such that he is always away from a given point by 7 m. After moving for some time, he again reaches his starting point. The approximate distance the person moved during the time is [#1713]
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Q1. A person moves along a path such that he is always away from a given point by 7 m. After moving for some time, he again reaches his starting point. The approximate distance the person moved during the time is
Q1. A person moves along a path such that he is always away from a given point by 7 m. After moving for some time, he again reaches his starting point. The approximate distance the person moved during the time is
(A) 22 m
(A) 22 m
(A) 22 m
(B) 44 m
(B) 44 m
(B) 44 m
(C) 122 m
(C) 122 m
(C) 122 m
(D) 144 m
(D) 144 m
(D) 144 m
Answer: (B) 44 m
Answer: (B) 44 m
Answer: (B) 44 m
2 * * 7
= 2 * 22
= 44
2 * * 7 = 2 * 22 = 44
2 * * 7 = 2 * 22 = 44
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Related MCQ Quizzes
Q1. The 8th term of the sequence 2, 6, 18, 54, ...... is
Q1. The 8th term of the sequence 2, 6, 18, 54, ...... is
(A) 4370
(A) 4370
(A) 4370
(B) 4374
(B) 4374
(B) 4374
(C) 7443
(C) 7443
(C) 7443
(D) 7434
(D) 7434
(D) 7434
Answer: (B) 4374
Answer: (B) 4374
Answer: (B) 4374
4374
4374
4374
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Q2. If 20% of a is b, then b% of 20 is the same as :
Q2. If 20% of a is b, then b% of 20 is the same as :
(A) 4% of a
(A) 4% of a
(A) 4% of a
(B) 8% of a
(B) 8% of a
(B) 8% of a
(C) 6% of a
(C) 6% of a
(C) 6% of a
(D) 10% of a
(D) 10% of a
(D) 10% of a
Answer: (A) 4% of a
Answer: (A) 4% of a
Answer: (A) 4% of a
20% of a is b
Hence b = 20% of a = 20a/100
b% of 20
= 20% of b
= (20/100) * 20a/100
= (20*20*a)/(100*100)
= 4a/100
= (4/100) * a
= 4% of a
20% of a is b Hence b = 20% of a = 20a/100 b% of 20 = 20% of b = (20/100) * 20a/100 = (20*20*a)/(100*100) = 4a/100 = (4/100) * a = 4% of a
20% of a is b Hence b = 20% of a = 20a/100 b% of 20 = 20% of b = (20/100) * 20a/100 = (20*20*a)/(100*100) = 4a/100 = (4/100) * a = 4% of a
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Q3. 100% of 100 when added to 200% of 200 would result
Q3. 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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Q4. A wheel can cover a distance of 22 km in 1000 rounds. The radius of the wheel is
Q4. 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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Q5. Two numbers are in the ratio of 2 : 3 and the product of their LCM and HCF is 96. The sum of the numbers is
Q5. Two numbers are in the ratio of 2 : 3 and the product of their LCM and HCF is 96. The sum of the numbers is
(A) 8
(A) 8
(A) 8
(B) 12
(B) 12
(B) 12
(C) 20
(C) 20
(C) 20
(D) 36
(D) 36
(D) 36
Answer: (C) 20
Answer: (C) 20
Answer: (C) 20
=> 2x * 3x = 96
=> 6x2 = 96
=> x2 = 96/6
=> x2 = 16
=> x = 4
2x + 3x = 5x = 5 * 4 = 20
=> 2x * 3x = 96 => 6x2 = 96 => x2 = 96/6 => x2 = 16 => x = 4 2x + 3x = 5x = 5 * 4 = 20
=> 2x * 3x = 96 => 6x2 = 96 => x2 = 96/6 => x2 = 16 => x = 4 2x + 3x = 5x = 5 * 4 = 20
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Q6. 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 :
Q6. 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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Q7. If the sum of five consecutive numbers is 190, then the lowest number amongst them is
Q7. If the sum of five consecutive numbers is 190, then the lowest number amongst them is
(A) 40
(A) 40
(A) 40
(B) 19
(B) 19
(B) 19
(C) 38
(C) 38
(C) 38
(D) 36
(D) 36
(D) 36
Answer: (D) 36
Answer: (D) 36
Answer: (D) 36
36
36
36
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Q8. In a range of consecutive numbers starting with 1, all the even numbers are removed. From the remaining, consider the first 7 numbers. The sum of these 7 numbers is
Q8. In a range of consecutive numbers starting with 1, all the even numbers are removed. From the remaining, consider the first 7 numbers. The sum of these 7 numbers is
(A) 35
(A) 35
(A) 35
(B) 42
(B) 42
(B) 42
(C) 49
(C) 49
(C) 49
(D) 56
(D) 56
(D) 56
Answer: (C) 49
Answer: (C) 49
Answer: (C) 49
1+3+5+7+9+11+13 = 49
1+3+5+7+9+11+13 = 49
1+3+5+7+9+11+13 = 49
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Q9. The count of prime numbers between 80 and 100 is
Q9. The count of prime numbers between 80 and 100 is
(A) 2
(A) 2
(A) 2
(B) 5
(B) 5
(B) 5
(C) 3
(C) 3
(C) 3
(D) 4
(D) 4
(D) 4
Answer: (C) 3
Answer: (C) 3
Answer: (C) 3
Prime numbers are numbers that are only divisible by 1 and themselves. Between 80 and 100, the prime numbers are 83, 89, and 97. Therefore, there are 3 prime numbers in this range.
Prime numbers are numbers that are only divisible by 1 and themselves. Between 80 and 100, the prime numbers are 83, 89, and 97. Therefore, there are 3 prime numbers in this range.
Prime numbers are numbers that are only divisible by 1 and themselves. Between 80 and 100, the prime numbers are 83, 89, and 97. Therefore, there are 3 prime numbers in this range.
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Q10. The conversion of in the fractional form is
Q10. The conversion of in the fractional form is
(A)
(A)
(A)
(B)
(B)
(B)
(C)
(C)
(C)
(D)
(D)
(D)
Answer: (C)
Answer: (C)
Answer: (C)
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