100% of 100 when added to 200% of 200 would result [#1704]
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Q1. 100% of 100 when added to 200% of 200 would result
Q1. 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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Related MCQ Quizzes
Q1. At what speed should a car travel on a highway to reach a destination 12 km away in 15 minutes?
Q1. At what speed should a car travel on a highway to reach a destination 12 km away in 15 minutes?
(A) 64 kmph
(A) 64 kmph
(A) 64 kmph
(B) 84 kmph
(B) 84 kmph
(B) 84 kmph
(C) 48 kmph
(C) 48 kmph
(C) 48 kmph
(D) 36 kmph
(D) 36 kmph
(D) 36 kmph
Answer: (C) 48 kmph
Answer: (C) 48 kmph
Answer: (C) 48 kmph
12
---- x 60 = 48 KMPH
15
12 ---- x 60 = 48 KMPH 15
12 ---- x 60 = 48 KMPH 15
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Q2. The sum of first 6 prime numbers is
Q2. The sum of first 6 prime numbers is
(A) 40
(A) 40
(A) 40
(B) 43
(B) 43
(B) 43
(C) 41
(C) 41
(C) 41
(D) 42
(D) 42
(D) 42
Answer: (C) 41
Answer: (C) 41
Answer: (C) 41
41
41
41
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Q3. The sum of the four consecutive alternate numbers is 64. The smallest number amongst them is
Q3. 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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Q4. 100% of 100 when added to 200% of 200 would result
Q4. 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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Q5. 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
Q5. 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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Q6. 4/5 of a number is 64. Then half of the number is
Q6. 4/5 of a number is 64. Then half of the number is
(A) 16
(A) 16
(A) 16
(B) 80
(B) 80
(B) 80
(C) 32
(C) 32
(C) 32
(D) 40
(D) 40
(D) 40
Answer: (D) 40
Answer: (D) 40
Answer: (D) 40
40
40
40
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Q7. 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 :
Q7. 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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Q8. The number when divided by 2, leaves remainder 1; when divided by 3, leaves remainder 2 and when divided by 4, leaves remainder 3, is
Q8. The number when divided by 2, leaves remainder 1; when divided by 3, leaves remainder 2 and when divided by 4, leaves remainder 3, is
(A) 25
(A) 25
(A) 25
(B) 41
(B) 41
(B) 41
(C) 13
(C) 13
(C) 13
(D) 11
(D) 11
(D) 11
Answer: (D) 11
Answer: (D) 11
Answer: (D) 11
11
11
11
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Q9. Three numbers are in the ratio 3:4:5. The sum of the largest and the smallest equals the sum of the third and 52. The smallest number is
Q9. Three numbers are in the ratio 3:4:5. The sum of the largest and the smallest equals the sum of the third and 52. The smallest number is
(A) 48
(A) 48
(A) 48
(B) 36
(B) 36
(B) 36
(C) 39
(C) 39
(C) 39
(D) 30
(D) 30
(D) 30
Answer: (C) 39
Answer: (C) 39
Answer: (C) 39
39
39
39
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Q10. 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?
Q10. 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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