The simple interest earned by 4,000 in 18 months at 12% per annum is [#1019]
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Q1. The simple interest earned by 4,000 in 18 months at 12% per annum is
Q1. The simple interest earned by 4,000 in 18 months at 12% per annum is
(A) 216
(A) 216
(A) 216
(B) 720
(B) 720
(B) 720
(C) 360
(C) 360
(C) 360
(D) 960
(D) 960
(D) 960
Answer: (B) 720
Answer: (B) 720
Answer: (B) 720
720
720
720
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Related MCQ Quizzes
Q1. A hostel has 120 students and food supplies are for 45 days. If 30 more students joined the hostel, then how many days the hostel will run with the existing food?
Q1. A hostel has 120 students and food supplies are for 45 days. If 30 more students joined the hostel, then how many days the hostel will run with the existing food?
(A) 40 days
(A) 40 days
(A) 40 days
(B) 38 days
(B) 38 days
(B) 38 days
(C) 36 days
(C) 36 days
(C) 36 days
(D) 32 days
(D) 32 days
(D) 32 days
Answer: (C) 36 days
Answer: (C) 36 days
Answer: (C) 36 days
36 days
36 days
36 days
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Q2. What is the term for all positive and negative numbers as a whole including zero?
Q2. What is the term for all positive and negative numbers as a whole including zero?
(A) Real Numbers
(A) Real Numbers
(A) Real Numbers
(B) Natural Numbers
(B) Natural Numbers
(B) Natural Numbers
(C) Whole Numbers
(C) Whole Numbers
(C) Whole Numbers
(D) Integer Numbers
(D) Integer Numbers
(D) Integer Numbers
Answer: (D) Integer Numbers
Answer: (D) Integer Numbers
Answer: (D) Integer Numbers
Integers
Integers
Integers
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Q3. The missing term in the sequence 7, 11, 19, 31, ______, 67 is
Q3. 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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Q4. The sum of 5 consecutive odd numbers is found to be 95. The largest of the numbers is
Q4. 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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Q5. 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
Q5. 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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Q6. 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
Q6. 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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Q7. The product of two consecutive odd numbers is 19043. Which is the smaller number?
Q7. The product of two consecutive odd numbers is 19043. Which is the smaller number?
(A) 133
(A) 133
(A) 133
(B) 131
(B) 131
(B) 131
(C) 137
(C) 137
(C) 137
(D) 129
(D) 129
(D) 129
Answer: (C) 137
Answer: (C) 137
Answer: (C) 137
137
137
137
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Q8. 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 :
Q8. 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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Q9. 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
Q9. 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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Q10. 45% of a number is same as 30% of another number. The ratio of the first number to the second number is :
Q10. 45% of a number is same as 30% of another number. The ratio of the first number to the second number is :
(A) 2 : 3
(A) 2 : 3
(A) 2 : 3
(B) 3 : 5
(B) 3 : 5
(B) 3 : 5
(C) 3 : 2
(C) 3 : 2
(C) 3 : 2
(D) 2 : 5
(D) 2 : 5
(D) 2 : 5
Answer: (A) 2 : 3
Answer: (A) 2 : 3
Answer: (A) 2 : 3
=
=
= X:Y = 2:3
= = = X:Y = 2:3
= = = X:Y = 2:3
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