A number is as much greater than 22 as is less than 72. Then the number is [#898]
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Q1. A number is as much greater than 22 as is less than 72. Then the number is
Q1. A number is as much greater than 22 as is less than 72. Then the number is
(A) 44
(A) 44
(A) 44
(B) 50
(B) 50
(B) 50
(C) 52
(C) 52
(C) 52
(D) 47
(D) 47
(D) 47
Answer: (D) 47
Answer: (D) 47
Answer: (D) 47
47
47
47
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Related MCQ Quizzes
Q1. The sum of the four consecutive alternate numbers is 64. The smallest number amongst them is
Q1. The sum of the four consecutive alternate numbers is 64. The smallest number amongst them is
(A) 13
(A) 13
(A) 13
(B) 11
(B) 11
(B) 11
(C) 16
(C) 16
(C) 16
(D) 9
(D) 9
(D) 9
Answer: (A) 13
Answer: (A) 13
Answer: (A) 13
The smallest number amongst them is 13
=> X + (X+2) + (X+4) + (X+6) = 64
=> 4X + 12 = 64
=> 4X = 64 - 12
=> 4X = 52
=> X = 13
The smallest number amongst them is 13 => X + (X+2) + (X+4) + (X+6) = 64 => 4X + 12 = 64 => 4X = 64 - 12 => 4X = 52 => X = 13
The smallest number amongst them is 13 => X + (X+2) + (X+4) + (X+6) = 64 => 4X + 12 = 64 => 4X = 64 - 12 => 4X = 52 => X = 13
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Q2. If the sum of five consecutive numbers is 190, then the lowest number amongst them is
Q2. If the sum of five consecutive numbers is 190, then the lowest number amongst them is
(A) 40
(A) 40
(A) 40
(B) 19
(B) 19
(B) 19
(C) 38
(C) 38
(C) 38
(D) 36
(D) 36
(D) 36
Answer: (D) 36
Answer: (D) 36
Answer: (D) 36
36
36
36
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Q3. The sum of two numbers is 15 and the sum of their squares is 113. The numbers are
Q3. The sum of two numbers is 15 and the sum of their squares is 113. The numbers are
(A) 4 and 10
(A) 4 and 10
(A) 4 and 10
(B) 6 and 9
(B) 6 and 9
(B) 6 and 9
(C) 5 and 10
(C) 5 and 10
(C) 5 and 10
(D) 7 and 8
(D) 7 and 8
(D) 7 and 8
Answer: (D) 7 and 8
Answer: (D) 7 and 8
Answer: (D) 7 and 8
7 and 8
7 and 8
7 and 8
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Q4. Find the odd number - 15, 17, 6, 12.
Q4. 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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Q5. 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
Q5. 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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Q6. 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 :
Q6. 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
(A) 60 years
(B) 51 years
(B) 51 years
(B) 51 years
(C) 48 years
(C) 48 years
(C) 48 years
(D) 45 years
(D) 45 years
(D) 45 years
Answer: (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.
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.
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Q7. Three numbers are in the ratio 3:4:6 and their product is 1944. The largest of the numbers is.
Q7. 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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Q8. The simple interest earned by 4,000 in 18 months at 12% per annum is
Q8. 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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Q9. The sum of first 6 prime numbers is
Q9. The sum of first 6 prime numbers is
(A) 40
(A) 40
(A) 40
(B) 43
(B) 43
(B) 43
(C) 41
(C) 41
(C) 41
(D) 42
(D) 42
(D) 42
Answer: (C) 41
Answer: (C) 41
Answer: (C) 41
41
41
41
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Q10. 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?
Q10. 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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