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]
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
Q1. 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 : Q1. 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
(B) 51 years (B) 51 years
(C) 48 years (C) 48 years
(D) 45 years (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.
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Q2. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is Q2. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is
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Q4. 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 Q4. 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
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Q5. Three numbers are in the ratio 3:4:6 and their product is 1944. The largest of the numbers is. Q5. Three numbers are in the ratio 3:4:6 and their product is 1944. The largest of the numbers is.
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Q6. At what percentage of simple interest does an amount of money double in 12 years? Q6. At what percentage of simple interest does an amount of money double in 12 years?
(A) (A)
(B) (B)
(C) (C)
(D) (D)
Answer: (C) Answer: (C)
P * * 12
= P * * 12
= P *
= P * 100%P * * 12
= P * * 12
= P *
= P * 100%
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Q7. 98 toffees were distributed to some boys in a group. Each boy in the group got twice as many of the toffees as the number of boys. The number of boys in the group was Q7. 98 toffees were distributed to some boys in a group. Each boy in the group got twice as many of the toffees as the number of boys. The number of boys in the group was
(A) 5 (A) 5
(B) 7 (B) 7
(C) 10 (C) 10
(D) 14 (D) 14
Answer: (B) 7 Answer: (B) 7
7
=> x * 2x = 98
=> x2 =
=> x2 = 49
=> x = 77
=> x * 2x = 98
=> x2 =
=> x2 = 49
=> x = 7
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Q8. ‘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 : Q8. ‘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
(B) 1.75 km (B) 1.75 km
(C) 1.2 km (C) 1.2 km
(D) 1.25 km (D) 1.25 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 kmA 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
