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 [#1711]
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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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Related MCQ Quizzes
Q1. At what speed should a car travel on a highway to reach a destination 12 km away in 15 minutes?
Q1. At what speed should a car travel on a highway to reach a destination 12 km away in 15 minutes?
(A) 64 kmph
(A) 64 kmph
(A) 64 kmph
(B) 84 kmph
(B) 84 kmph
(B) 84 kmph
(C) 48 kmph
(C) 48 kmph
(C) 48 kmph
(D) 36 kmph
(D) 36 kmph
(D) 36 kmph
Answer: (C) 48 kmph
Answer: (C) 48 kmph
Answer: (C) 48 kmph
12
---- x 60 = 48 KMPH
15
12 ---- x 60 = 48 KMPH 15
12 ---- x 60 = 48 KMPH 15
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Q2. 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
Q2. 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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Q3. A 100 metre long train moving in a uniform speed of 20 m/sec crosses a bridge of length 1 km. The time taken by the train to cross the bridge is
Q3. A 100 metre long train moving in a uniform speed of 20 m/sec crosses a bridge of length 1 km. The time taken by the train to cross the bridge is
(A) 45 seconds
(A) 45 seconds
(A) 45 seconds
(B) 50 seconds
(B) 50 seconds
(B) 50 seconds
(C) 55 seconds
(C) 55 seconds
(C) 55 seconds
(D) 60 seconds
(D) 60 seconds
(D) 60 seconds
Answer: (C) 55 seconds
Answer: (C) 55 seconds
Answer: (C) 55 seconds
(1000+100)m / (20m/s) = 55 seconds
(1000+100)m / (20m/s) = 55 seconds
(1000+100)m / (20m/s) = 55 seconds
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Q4. ‘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 :
Q4. ‘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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Q5. The angles of a quadrilateral are in the ratio of 1 : 3 : 4 : 7. The difference between the largest and the smallest angle is
Q5. The angles of a quadrilateral are in the ratio of 1 : 3 : 4 : 7. The difference between the largest and the smallest angle is
(A) 120°
(A) 120°
(A) 120°
(B) 140°
(B) 140°
(B) 140°
(C) 144°
(C) 144°
(C) 144°
(D) 145°
(D) 145°
(D) 145°
Answer: (C) 144°
Answer: (C) 144°
Answer: (C) 144°
(360°/15) * (7-1) = 144°
(360°/15) * (7-1) = 144°
(360°/15) * (7-1) = 144°
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Q6. 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
Q6. 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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Q7. Find the odd number - 15, 17, 6, 12.
Q7. Find the odd number - 15, 17, 6, 12.
(A) 6
(A) 6
(A) 6
(B) 12
(B) 12
(B) 12
(C) 15
(C) 15
(C) 15
(D) 17
(D) 17
(D) 17
Answer: (D) 17
Answer: (D) 17
Answer: (D) 17
17 which is a Prime number.
17 which is a Prime number.
17 which is a Prime number.
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Q8. The remainder when –76 is divided by 3, is
Q8. The remainder when –76 is divided by 3, is
(A) -1
(A) -1
(A) -1
(B) 1
(B) 1
(B) 1
(C) 2
(C) 2
(C) 2
(D) -2
(D) -2
(D) -2
Answer: (C) 2
Answer: (C) 2
Answer: (C) 2
2
Because, Reminder must be positive and it should be less then the divisor
2 Because, Reminder must be positive and it should be less then the divisor
2 Because, Reminder must be positive and it should be less then the divisor
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Q9. Three numbers are in the ratio 3:4:6 and their product is 1944. The largest of the numbers is.
Q9. Three numbers are in the ratio 3:4:6 and their product is 1944. The largest of the numbers is.
(A) 21
(A) 21
(A) 21
(B) 12
(B) 12
(B) 12
(C) 18
(C) 18
(C) 18
(D) 9
(D) 9
(D) 9
Answer: (C) 18
Answer: (C) 18
Answer: (C) 18
18
18
18
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Q10. 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
Q10. 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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