Three numbers are in the ratio 3:4:5. The sum of the largest and the smallest equals the sum of the third and 52. The smallest number is [#899]
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Q1. Three numbers are in the ratio 3:4:5. The sum of the largest and the smallest equals the sum of the third and 52. The smallest number is
Q1. Three numbers are in the ratio 3:4:5. The sum of the largest and the smallest equals the sum of the third and 52. The smallest number is
(A) 48
(A) 48
(A) 48
(B) 36
(B) 36
(B) 36
(C) 39
(C) 39
(C) 39
(D) 30
(D) 30
(D) 30
Answer: (C) 39
Answer: (C) 39
Answer: (C) 39
39
39
39
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Related MCQ Quizzes
Q1. The product of two consecutive odd numbers is 19043. Which is the smaller number?
Q1. The product of two consecutive odd numbers is 19043. Which is the smaller number?
(A) 133
(A) 133
(A) 133
(B) 131
(B) 131
(B) 131
(C) 137
(C) 137
(C) 137
(D) 129
(D) 129
(D) 129
Answer: (C) 137
Answer: (C) 137
Answer: (C) 137
137
137
137
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Q2. At what speed should a car travel on a highway to reach a destination 12 km away in 15 minutes?
Q2. 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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Q3. A person moves along a path such that he is always away from a given point by 7 m. After moving for some time, he again reaches his starting point. The approximate distance the person moved during the time is
Q3. A person moves along a path such that he is always away from a given point by 7 m. After moving for some time, he again reaches his starting point. The approximate distance the person moved during the time is
(A) 22 m
(A) 22 m
(A) 22 m
(B) 44 m
(B) 44 m
(B) 44 m
(C) 122 m
(C) 122 m
(C) 122 m
(D) 144 m
(D) 144 m
(D) 144 m
Answer: (B) 44 m
Answer: (B) 44 m
Answer: (B) 44 m
2 * * 7
= 2 * 22
= 44
2 * * 7 = 2 * 22 = 44
2 * * 7 = 2 * 22 = 44
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Q4. 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?
Q4. 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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Q5. Three fourth of a number is more than two third of the number by 5. Then the number is
Q5. Three fourth of a number is more than two third of the number by 5. Then the number is
(A) 72
(A) 72
(A) 72
(B) 60
(B) 60
(B) 60
(C) 80
(C) 80
(C) 80
(D) 48
(D) 48
(D) 48
Answer: (B) 60
Answer: (B) 60
Answer: (B) 60
60
60
60
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Q6. If the average age of A, B and C is 22 years and the average age of B and C is 25 years, then find the age of A after 9 years from now.
Q6. If the average age of A, B and C is 22 years and the average age of B and C is 25 years, then find the age of A after 9 years from now.
(A) 25 years
(A) 25 years
(A) 25 years
(B) 35 years
(B) 35 years
(B) 35 years
(C) 50 years
(C) 50 years
(C) 50 years
(D) 45 years
(D) 45 years
(D) 45 years
Answer: (A) 25 years
Answer: (A) 25 years
Answer: (A) 25 years
25 years
=> A+B+C = 22*3
=> A+(B+C) = 66
=> A+(25*2) = 66
=> A = 66-50
=> A = 16
After 9 years A = 16+9 = 25 years
25 years => A+B+C = 22*3 => A+(B+C) = 66 => A+(25*2) = 66 => A = 66-50 => A = 16 After 9 years A = 16+9 = 25 years
25 years => A+B+C = 22*3 => A+(B+C) = 66 => A+(25*2) = 66 => A = 66-50 => A = 16 After 9 years A = 16+9 = 25 years
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Q7. ‘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 :
Q7. ‘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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Q8. Which of the following is a prime number?
Q8. Which of the following is a prime number?
81,33,71,93
81,33,71,93
81,33,71,93
(A) 93
(A) 93
(A) 93
(B) 33
(B) 33
(B) 33
(C) 71
(C) 71
(C) 71
(D) 81
(D) 81
(D) 81
Answer: (C) 71
Answer: (C) 71
Answer: (C) 71
71
71
71
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Q9. 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
Q9. 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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Q10. The simple interest earned by 4,000 in 18 months at 12% per annum is
Q10. The simple interest earned by 4,000 in 18 months at 12% per annum is
(A) 216
(A) 216
(A) 216
(B) 720
(B) 720
(B) 720
(C) 360
(C) 360
(C) 360
(D) 960
(D) 960
(D) 960
Answer: (B) 720
Answer: (B) 720
Answer: (B) 720
720
720
720
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Related Questions
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