‘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]
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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
(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 MCQ Quizzes
Q1. What is the term for the point where two or more lines intersect?
Q1. What is the term for the point where two or more lines intersect?
(A) Vertex
(A) Vertex
(A) Vertex
(B) Edge
(B) Edge
(B) Edge
(C) Face
(C) Face
(C) Face
(D) Intersection
(D) Intersection
(D) Intersection
Answer: (A) Vertex
Answer: (A) Vertex
Answer: (A) Vertex
In geometry, a vertex (plural: vertices) is the point where two or more lines, rays, or edges meet, like the corner of a triangle or the point where two streets intersect.
In geometry, a vertex (plural: vertices) is the point where two or more lines, rays, or edges meet, like the corner of a triangle or the point where two streets intersect.
In geometry, a vertex (plural: vertices) is the point where two or more lines, rays, or edges meet, like the corner of a triangle or the point where two streets intersect.
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Q2. Two trains of 200 m and 100 m long are running parallel at the rate of 20 m/sec. and 22 m/sec. respectively. How much time will they take to cross each other, if the trains are running along the same direction?
Q2. Two trains of 200 m and 100 m long are running parallel at the rate of 20 m/sec. and 22 m/sec. respectively. How much time will they take to cross each other, if the trains are running along the same direction?
(A) 120 sec
(A) 120 sec
(A) 120 sec
(B) 100 sec
(B) 100 sec
(B) 100 sec
(C) 60 sec
(C) 60 sec
(C) 60 sec
(D) 50 sec
(D) 50 sec
(D) 50 sec
Answer: (D) 50 sec
Answer: (D) 50 sec
Answer: (D) 50 sec
The difference of speed = 22m/s - 20m/s = 2m/s
As the train with speed 22m/s will cross the other, hence this train have to go the smallest length of the trains distance ahead of the other to cross completely.
Hence This train will be ahead of 100m with a speed of 2m/s.
Hence The time to cross each other will take = 100m / (2m/s)
= 50 sec
The difference of speed = 22m/s - 20m/s = 2m/s As the train with speed 22m/s will cross the other, hence this train have to go the smallest length of the trains distance ahead of the other to cross completely. Hence This train will be ahead of 100m with a speed of 2m/s. Hence The time to cross each other will take = 100m / (2m/s) = 50 sec
The difference of speed = 22m/s - 20m/s = 2m/s As the train with speed 22m/s will cross the other, hence this train have to go the smallest length of the trains distance ahead of the other to cross completely. Hence This train will be ahead of 100m with a speed of 2m/s. Hence The time to cross each other will take = 100m / (2m/s) = 50 sec
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Q3. What is the name of the mathematical concept that describes a value that never changes, like the ratio of a circle's circumference to its diameter?
Q3. What is the name of the mathematical concept that describes a value that never changes, like the ratio of a circle's circumference to its diameter?
(A) Variable
(A) Variable
(A) Variable
(B) Constant
(B) Constant
(B) Constant
(C) Fraction
(C) Fraction
(C) Fraction
(D) Decimal
(D) Decimal
(D) Decimal
Answer: (B) Constant
Answer: (B) Constant
Answer: (B) Constant
A constant is a mathematical concept that represents a value that remains unchanged, like pi (π), which is approximately 3.14 and never changes.
A constant is a mathematical concept that represents a value that remains unchanged, like pi (π), which is approximately 3.14 and never changes.
A constant is a mathematical concept that represents a value that remains unchanged, like pi (π), which is approximately 3.14 and never changes.
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Q4. p, q, r are three numbers such that the LCM of p and q is q and the LCM of q and r is r. The LCM of p, q and r will be
Q4. p, q, r are three numbers such that the LCM of p and q is q and the LCM of q and r is r. The LCM of p, q and r will be
(A) q
(A) q
(A) q
(B) r
(B) r
(B) r
(C) qr
(C) qr
(C) qr
(D) pqr
(D) pqr
(D) pqr
Answer: (B) r
Answer: (B) r
Answer: (B) r
LCM will be r.
px= q and qy = r, hence r = pxy.
LCM will be r. px= q and qy = r, hence r = pxy.
LCM will be r. px= q and qy = r, hence r = pxy.
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Q5. 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
Q5. 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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Q6. The 8th term of the sequence 2, 6, 18, 54, ...... is
Q6. The 8th term of the sequence 2, 6, 18, 54, ...... is
(A) 4370
(A) 4370
(A) 4370
(B) 4374
(B) 4374
(B) 4374
(C) 7443
(C) 7443
(C) 7443
(D) 7434
(D) 7434
(D) 7434
Answer: (B) 4374
Answer: (B) 4374
Answer: (B) 4374
4374
4374
4374
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Q7. What is the term for a line that divides a shape into two equal parts?
Q7. What is the term for a line that divides a shape into two equal parts?
(A) Axis
(A) Axis
(A) Axis
(B) Median
(B) Median
(B) Median
(C) Vertex
(C) Vertex
(C) Vertex
(D) Bisector
(D) Bisector
(D) Bisector
Answer: (D) Bisector
Answer: (D) Bisector
Answer: (D) Bisector
A bisector is a line that divides a shape into two equal parts, like a line that cuts a triangle into two equal areas or a line that divides a circle into two equal parts (semi-circles).
A bisector is a line that divides a shape into two equal parts, like a line that cuts a triangle into two equal areas or a line that divides a circle into two equal parts (semi-circles).
A bisector is a line that divides a shape into two equal parts, like a line that cuts a triangle into two equal areas or a line that divides a circle into two equal parts (semi-circles).
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Q8. Find the value of
Q8. Find the value of
(A)
(A)
(A)
(B)
(B)
(B)
(C)
(C)
(C)
(D)
(D)
(D)
Answer: (D)
Answer: (D)
Answer: (D)
=
=
=
=
=
= = = = =
= = = = =
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Q9. ‘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 :
Q9. ‘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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Q10. 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
Q10. 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
(A) 9m
(A) 9m
(A) 9m
(B) 5m
(B) 5m
(B) 5m
(C) 11m
(C) 11m
(C) 11m
(D) 12m
(D) 12m
(D) 12m
Answer: (D) 12m
Answer: (D) 12m
Answer: (D) 12m
52 + X2 = 132
=> 25 + X2 = 169
=> X2 = 196 - 25
=> X2 = 144
=> X = 12
52 + X2 = 132 => 25 + X2 = 169 => X2 = 196 - 25 => X2 = 144 => X = 12
52 + X2 = 132 => 25 + X2 = 169 => X2 = 196 - 25 => X2 = 144 => X = 12
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Related Questions
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