A 20 m long ladder is leaning on a vertical wall. It makes an angle of 30° with the ground. The height of the point the ladder touches wall is [#1022]
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Q1. A 20 m long ladder is leaning on a vertical wall. It makes an angle of 30° with the ground. The height of the point the ladder touches wall is
Q1. A 20 m long ladder is leaning on a vertical wall. It makes an angle of 30° with the ground. The height of the point the ladder touches wall is
(A) 10 m
(A) 10 m
(A) 10 m
(B) 17.32 m
(B) 17.32 m
(B) 17.32 m
(C) 8.16 m
(C) 8.16 m
(C) 8.16 m
(D) 13 m
(D) 13 m
(D) 13 m
Answer: (A) 10 m
Answer: (A) 10 m
Answer: (A) 10 m
10 m
10 m
10 m
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Related MCQ Quizzes
Q1. 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
Q1. 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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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 term for all positive and negative numbers as a whole including zero?
Q3. What is the term for all positive and negative numbers as a whole including zero?
(A) Real Numbers
(A) Real Numbers
(A) Real Numbers
(B) Natural Numbers
(B) Natural Numbers
(B) Natural Numbers
(C) Whole Numbers
(C) Whole Numbers
(C) Whole Numbers
(D) Integer Numbers
(D) Integer Numbers
(D) Integer Numbers
Answer: (D) Integer Numbers
Answer: (D) Integer Numbers
Answer: (D) Integer Numbers
Integers
Integers
Integers
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Q4. A lady deposits money in her savings bank in such a way that every next day her deposit amount is ₹ 12 more than her previous day deposit. If she starts her deposit with ₹ 12 on the first day, the total amount deposited by Liza at the end of 30 days will be :
Q4. A lady deposits money in her savings bank in such a way that every next day her deposit amount is ₹ 12 more than her previous day deposit. If she starts her deposit with ₹ 12 on the first day, the total amount deposited by Liza at the end of 30 days will be :
(A) 5,420
(A) 5,420
(A) 5,420
(B) 5,580
(B) 5,580
(B) 5,580
(C) 5,620
(C) 5,620
(C) 5,620
(D) 5,780
(D) 5,780
(D) 5,780
Answer: (B) 5,580
Answer: (B) 5,580
Answer: (B) 5,580
The formula for the sum of the first n terms of an arithmetic progression is
= 15*(12+360)
= 15 * 372
= 5580
The formula for the sum of the first n terms of an arithmetic progression is = 15*(12+360) = 15 * 372 = 5580
The formula for the sum of the first n terms of an arithmetic progression is = 15*(12+360) = 15 * 372 = 5580
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Q5. 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
Q5. 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
(A) 5
(B) 7
(B) 7
(B) 7
(C) 10
(C) 10
(C) 10
(D) 14
(D) 14
(D) 14
Answer: (B) 7
Answer: (B) 7
Answer: (B) 7
7
=> x * 2x = 98
=> x2 =
=> x2 = 49
=> x = 7
7 => x * 2x = 98 => x2 = => x2 = 49 => x = 7
7 => x * 2x = 98 => x2 = => x2 = 49 => x = 7
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Q6. A number was increased by 40% and thereafter decreased by 40%. The net change in the number in percentage is
Q6. A number was increased by 40% and thereafter decreased by 40%. The net change in the number in percentage is
(A) 16% increase
(A) 16% increase
(A) 16% increase
(B) 16% decrease
(B) 16% decrease
(B) 16% decrease
(C) 32% decrease
(C) 32% decrease
(C) 32% decrease
(D) No change
(D) No change
(D) No change
Answer: (B) 16% decrease
Answer: (B) 16% decrease
Answer: (B) 16% decrease
16% decrease
= 40-40-%
= -%
= -16%
16% decrease = 40-40-% = -% = -16%
16% decrease = 40-40-% = -% = -16%
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Q7. The LCM of two numbers is 40 and their HCF is 4. If the difference between the two numbers is 12, then the sum of the numbers is
Q7. The LCM of two numbers is 40 and their HCF is 4. If the difference between the two numbers is 12, then the sum of the numbers is
(A) 20
(A) 20
(A) 20
(B) 24
(B) 24
(B) 24
(C) 28
(C) 28
(C) 28
(D) 32
(D) 32
(D) 32
Answer: (C) 28
Answer: (C) 28
Answer: (C) 28
X * (X-12) = 40 * 4
X = 20
X + (X-12) = 20 + 20 - 12 = 28
X * (X-12) = 40 * 4 X = 20 X + (X-12) = 20 + 20 - 12 = 28
X * (X-12) = 40 * 4 X = 20 X + (X-12) = 20 + 20 - 12 = 28
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Q8. The largest of 26, 35, 44 and 53 is
Q8. The largest of 26, 35, 44 and 53 is
(A) 26
(A) 26
(A) 26
(B) 35
(B) 35
(B) 35
(C) 44
(C) 44
(C) 44
(D) 53
(D) 53
(D) 53
Answer: (C) 44
Answer: (C) 44
Answer: (C) 44
26 = 64
35 = 243
44 = 256
53 = 125
26 = 64 35 = 243 44 = 256 53 = 125
26 = 64 35 = 243 44 = 256 53 = 125
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Q9. The conversion of in the fractional form is
Q9. The conversion of in the fractional form is
(A)
(A)
(A)
(B)
(B)
(B)
(C)
(C)
(C)
(D)
(D)
(D)
Answer: (C)
Answer: (C)
Answer: (C)
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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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