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 [#1695]
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Q1. 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
Q1. 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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Related MCQ Quizzes
Q1. If 20% of a is b, then b% of 20 is the same as :
Q1. 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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Q2. What is the term for a number that can be divided by 2?
Q2. What is the term for a number that can be divided by 2?
(A) Prime number
(A) Prime number
(A) Prime number
(B) Odd number
(B) Odd number
(B) Odd number
(C) Even number
(C) Even number
(C) Even number
(D) Fraction
(D) Fraction
(D) Fraction
Answer: (C) Even number
Answer: (C) Even number
Answer: (C) Even number
An even number is a whole number that is divisible by 2, such as 4, 6, or 8.
An even number is a whole number that is divisible by 2, such as 4, 6, or 8.
An even number is a whole number that is divisible by 2, such as 4, 6, or 8.
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Q3. If 2x = 32, then the value of x is
Q3. If 2x = 32, then the value of x is
(A) 4
(A) 4
(A) 4
(B) 5
(B) 5
(B) 5
(C) 6
(C) 6
(C) 6
(D) 7
(D) 7
(D) 7
Answer: (B) 5
Answer: (B) 5
Answer: (B) 5
25 = 2*2*2*2*2 = 32
25 = 2*2*2*2*2 = 32
25 = 2*2*2*2*2 = 32
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Q4. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is
Q4. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is
(A) 1:6
(A) 1:6
(A) 1:6
(B) 2:9
(B) 2:9
(B) 2:9
(C) 1:9
(C) 1:9
(C) 1:9
(D) 6:9
(D) 6:9
(D) 6:9
Answer: (C) 1:9
Answer: (C) 1:9
Answer: (C) 1:9
Area = πr2
π12 : π32
= π1 : π9
= 1:9
Area = πr2 π12 : π32 = π1 : π9 = 1:9
Area = πr2 π12 : π32 = π1 : π9 = 1:9
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Q5. 4/5 of a number is 64. Then half of the number is
Q5. 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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Q6. The remainder when –76 is divided by 3, is
Q6. 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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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. Two numbers are in the ratio 2 : 3 and the product of their HCF and LCM is 96. The sum of the two numbers is
Q8. Two numbers are in the ratio 2 : 3 and the product of their HCF and LCM is 96. The sum of the two numbers is
(A) 18
(A) 18
(A) 18
(B) 20
(B) 20
(B) 20
(C) 22
(C) 22
(C) 22
(D) 24
(D) 24
(D) 24
Answer: (B) 20
Answer: (B) 20
Answer: (B) 20
2:3 = 8:12
N1*N2 = HCF*LCM
8*12 = 96
N1+N2 = 8+12 = 20
2:3 = 8:12 N1*N2 = HCF*LCM 8*12 = 96 N1+N2 = 8+12 = 20
2:3 = 8:12 N1*N2 = HCF*LCM 8*12 = 96 N1+N2 = 8+12 = 20
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Q9. The sum of two numbers is 15 and the sum of their squares is 113. The numbers are
Q9. 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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Q10. ‘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 :
Q10. ‘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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Related Questions
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