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 number when divided by 2, leaves remainder 1; when divided by 3, leaves remainder 2 and when divided by 4, leaves remainder 3, is
Q1. The number when divided by 2, leaves remainder 1; when divided by 3, leaves remainder 2 and when divided by 4, leaves remainder 3, is
(A) 25
(A) 25
(A) 25
(B) 41
(B) 41
(B) 41
(C) 13
(C) 13
(C) 13
(D) 11
(D) 11
(D) 11
Answer: (D) 11
Answer: (D) 11
Answer: (D) 11
11
11
11
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Q2. The sum of the four consecutive alternate numbers is 64. The smallest number amongst them is
Q2. 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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Q3. Shyam stored Rs 35 in the form of 1 rupee coin and 50 paise coins in the ratio 2 : 3. The number of 50 paise coins are
Q3. Shyam stored Rs 35 in the form of 1 rupee coin and 50 paise coins in the ratio 2 : 3. The number of 50 paise coins are
(A) 20
(A) 20
(A) 20
(B) 25
(B) 25
(B) 25
(C) 30
(C) 30
(C) 30
(D) 35
(D) 35
(D) 35
Answer: (C) 30
Answer: (C) 30
Answer: (C) 30
1*20 + 0.5*30 = 20+15 = 35
1*20 + 0.5*30 = 20+15 = 35
1*20 + 0.5*30 = 20+15 = 35
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Q4. The count of prime numbers between 80 and 100 is
Q4. The count of prime numbers between 80 and 100 is
(A) 2
(A) 2
(A) 2
(B) 5
(B) 5
(B) 5
(C) 3
(C) 3
(C) 3
(D) 4
(D) 4
(D) 4
Answer: (C) 3
Answer: (C) 3
Answer: (C) 3
Prime numbers are numbers that are only divisible by 1 and themselves. Between 80 and 100, the prime numbers are 83, 89, and 97. Therefore, there are 3 prime numbers in this range.
Prime numbers are numbers that are only divisible by 1 and themselves. Between 80 and 100, the prime numbers are 83, 89, and 97. Therefore, there are 3 prime numbers in this range.
Prime numbers are numbers that are only divisible by 1 and themselves. Between 80 and 100, the prime numbers are 83, 89, and 97. Therefore, there are 3 prime numbers in this range.
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Q5. Which is the smallest Whole Number?
Q5. Which is the smallest Whole Number?
(A) -1
(A) -1
(A) -1
(B) 0
(B) 0
(B) 0
(C) 1
(C) 1
(C) 1
(D) 2
(D) 2
(D) 2
Answer: (B) 0
Answer: (B) 0
Answer: (B) 0
Zero
Zero
Zero
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Q6. Find the value of
Q6. 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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Q7. 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
Q7. 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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Q8. 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 :
Q8. 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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Q9. If A : B = 3, then 144A : 36B is
Q9. If A : B = 3, then 144A : 36B is
(A) 4
(A) 4
(A) 4
(B) 12
(B) 12
(B) 12
(C) 3
(C) 3
(C) 3
(D) 16
(D) 16
(D) 16
Answer: (B) 12
Answer: (B) 12
Answer: (B) 12
A : B = 3
=> = 3
=
= x 3
= 4 x 3
= 12
A : B = 3 => = 3 = = x 3 = 4 x 3 = 12
A : B = 3 => = 3 = = x 3 = 4 x 3 = 12
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Q10. The ratio of the number of boys and girls in a school of 720 students is 7 : 5. How many more girls should be admitted to make the ratio 1 : 1?
Q10. The ratio of the number of boys and girls in a school of 720 students is 7 : 5. How many more girls should be admitted to make the ratio 1 : 1?
(A) 60
(A) 60
(A) 60
(B) 120
(B) 120
(B) 120
(C) 180
(C) 180
(C) 180
(D) 240
(D) 240
(D) 240
Answer: (B) 120
Answer: (B) 120
Answer: (B) 120
120
120
120
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