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. The smallest positive integer that is simultaneously divisible by 6, 8 and 12 is Q1. The smallest positive integer that is simultaneously divisible by 6, 8 and 12 is
Q2. 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 : Q2. 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.
Q3. In a school with 10 teachers, one retires and immediately a new teacher of age 25 years joins. As a result, the average age of the teacher reduces by 3. The age of the retired teacher is Q3. In a school with 10 teachers, one retires and immediately a new teacher of age 25 years joins. As a result, the average age of the teacher reduces by 3. The age of the retired teacher is
(A) 55 years (A) 55 years
(B) 65 years (B) 65 years
(C) 58 years (C) 58 years
(D) 60 years (D) 60 years
Answer: (A) 55 years Answer: (A) 55 years
25 years + (3*10) years = 55 years25 years + (3*10) years = 55 years
Q4. What is the value of x in the equation 2x + 5 = 11? Q4. What is the value of x in the equation 2x + 5 = 11?
(A) 2 (A) 2
(B) 3 (B) 3
(C) 4 (C) 4
(D) 5 (D) 5
Answer: (B) 3 Answer: (B) 3
To solve for x, subtract 5 from both sides of the equation, resulting in 2x = 6. Then, divide both sides by 2, giving x = 3.To solve for x, subtract 5 from both sides of the equation, resulting in 2x = 6. Then, divide both sides by 2, giving x = 3.
Q5. A ladder of length 13 m is leaning against a vertical wall with the upper end at the height of 5 m. The horizontal distance between the foot of the wall and the lower end of the ladder is Q5. A ladder of length 13 m is leaning against a vertical wall with the upper end at the height of 5 m. The horizontal distance between the foot of the wall and the lower end of the ladder is
Q6. The sum of two numbers is 15 and the sum of their squares is 113. The numbers are Q6. The sum of two numbers is 15 and the sum of their squares is 113. The numbers are
Q7. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is Q7. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is
Q8. The least number by which 2450 must be multiplied to make it a perfect square, is Q8. The least number by which 2450 must be multiplied to make it a perfect square, is
