‘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 : [#1122]
Q1. ‘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 : Q1. ‘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
Q1. The sum of 5 consecutive odd numbers is found to be 95. The largest of the numbers is Q1. The sum of 5 consecutive odd numbers is found to be 95. The largest of the numbers 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.
An equilateral triangle is a triangle with three equal sides and angles, a regular polygon with three sides.An equilateral triangle is a triangle with three equal sides and angles, a regular polygon with three sides.
Q4. If 20% of a is b, then b% of 20 is the same as : Q4. If 20% of a is b, then b% of 20 is the same as :
(A) 4% of a (A) 4% of a
(B) 8% of a (B) 8% of a
(C) 6% of a (C) 6% of a
(D) 10% of a (D) 10% 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 a20% 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
Q5. 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 Q5. 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
Q6. 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 Q6. 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
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
