If A : B = 3, then 144A : 36B is [#1697]
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Q1. If A : B = 3, then 144A : 36B is
Q1. 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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Related MCQ Quizzes
Q1. 4/5 of a number is 64. Then half of the number is
Q1. 4/5 of a number is 64. Then half of the number is
(A) 16
(A) 16
(A) 16
(B) 80
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(C) 32
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(D) 40
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(D) 40
Answer: (D) 40
Answer: (D) 40
Answer: (D) 40
40
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40
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Q2. A wheel can cover a distance of 22 km in 1000 rounds. The radius of the wheel is
Q2. A wheel can cover a distance of 22 km in 1000 rounds. The radius of the wheel is
(A) 4.5 m
(A) 4.5 m
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Answer: (D) 3.5 m
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Answer: (D) 3.5 m
3.5 m
3.5 m
3.5 m
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Q3. If is equal to
Q3. If is equal to
(A) 2
(A) 2
(A) 2
(B) 3
(B) 3
(B) 3
(C) 4
(C) 4
(C) 4
(D) 5
(D) 5
(D) 5
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Answer: (A) 2
Answer: (A) 2
Hence, X:Y:Z = 3:4:7
=
=
= 2
Hence, X:Y:Z = 3:4:7 = = = 2
Hence, X:Y:Z = 3:4:7 = = = 2
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Q4. Find the value of the following expression
Q4. Find the value of the following expression
48÷12+4×25÷5
48÷12+4×25÷5
48÷12+4×25÷5
(A) 15
(A) 15
(A) 15
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(B) 24
(B) 24
(C) 40
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(C) 40
(D) 25
(D) 25
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Answer: (B) 24
Answer: (B) 24
Answer: (B) 24
24
24
24
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Q5. 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.
Q5. 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
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(B) 35 years
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(D) 45 years
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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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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. In a range of consecutive numbers starting with 1, all the even numbers are removed. From the remaining, consider the first 7 numbers. The sum of these 7 numbers is
Q7. In a range of consecutive numbers starting with 1, all the even numbers are removed. From the remaining, consider the first 7 numbers. The sum of these 7 numbers is
(A) 35
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(B) 42
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(D) 56
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Answer: (C) 49
Answer: (C) 49
Answer: (C) 49
1+3+5+7+9+11+13 = 49
1+3+5+7+9+11+13 = 49
1+3+5+7+9+11+13 = 49
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Q8. 100% of 100 when added to 200% of 200 would result
Q8. 100% of 100 when added to 200% of 200 would result
(A) 300
(A) 300
(A) 300
(B) 400
(B) 400
(B) 400
(C) 500
(C) 500
(C) 500
(D) 600
(D) 600
(D) 600
Answer: (C) 500
Answer: (C) 500
Answer: (C) 500
100 * 100% + 200 * 200%
=
=
= 100 + 400
= 500
100 * 100% + 200 * 200% = = = 100 + 400 = 500
100 * 100% + 200 * 200% = = = 100 + 400 = 500
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Q9. 3/4 of a number is 19 less than the original number. The number is
Q9. 3/4 of a number is 19 less than the original number. The number is
(A) 62
(A) 62
(A) 62
(B) 64
(B) 64
(B) 64
(C) 79
(C) 79
(C) 79
(D) 76
(D) 76
(D) 76
Answer: (D) 76
Answer: (D) 76
Answer: (D) 76
76
76
76
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Q10. 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 :
Q10. 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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