When the numerator of a fraction is multiplied by 4 and the denominator by 9, the fraction reverses. The fraction is [#1701]
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Q1. When the numerator of a fraction is multiplied by 4 and the denominator by 9, the fraction reverses. The fraction is
Q1. When the numerator of a fraction is multiplied by 4 and the denominator by 9, the fraction reverses. The fraction is
(A)
(A)
(A)
(B)
(B)
(B)
(C)
(C)
(C)
(D)
(D)
(D)
Answer: (D)
Answer: (D)
Answer: (D)
>
=
=
> = =
> = =
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Related MCQ Quizzes
Q1. A wheel can cover a distance of 22 km in 1000 rounds. The radius of the wheel is
Q1. 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
(A) 4.5 m
(B) 2.1 m
(B) 2.1 m
(B) 2.1 m
(C) 2.8 m
(C) 2.8 m
(C) 2.8 m
(D) 3.5 m
(D) 3.5 m
(D) 3.5 m
Answer: (D) 3.5 m
Answer: (D) 3.5 m
Answer: (D) 3.5 m
3.5 m
3.5 m
3.5 m
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Q2. When the numerator of a fraction increases by 3, the fraction increases by its three-fourth. The numerator of the fraction is
Q2. When the numerator of a fraction increases by 3, the fraction increases by its three-fourth. The numerator of the fraction is
(A) 3
(A) 3
(A) 3
(B) 4
(B) 4
(B) 4
(C) 5
(C) 5
(C) 5
(D) 6
(D) 6
(D) 6
Answer: (B) 4
Answer: (B) 4
Answer: (B) 4
4
4
4
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Q3. A certain amount invested in a firm would become double at the end of one month, but it deducts an amount of ₹120 on every doubling. A person invests an amount of ₹105 and continues for 3 months without investing any additional amount. At the end of 3 months, his net income is
Q3. A certain amount invested in a firm would become double at the end of one month, but it deducts an amount of ₹120 on every doubling. A person invests an amount of ₹105 and continues for 3 months without investing any additional amount. At the end of 3 months, his net income is
(A) 55
(A) 55
(A) 55
(B) 0
(B) 0
(B) 0
(C) 270
(C) 270
(C) 270
(D) 45
(D) 45
(D) 45
Answer: (B) 0
Answer: (B) 0
Answer: (B) 0
1st Month -> (105 * 2) - 120 = 210 - 120 = 90
2nd Month -> (90 * 2) - 120 = 180 - 120 = 60
3rd Month -> (60 * 2) - 120 = 120 - 120 = 0
1st Month -> (105 * 2) - 120 = 210 - 120 = 90 2nd Month -> (90 * 2) - 120 = 180 - 120 = 60 3rd Month -> (60 * 2) - 120 = 120 - 120 = 0
1st Month -> (105 * 2) - 120 = 210 - 120 = 90 2nd Month -> (90 * 2) - 120 = 180 - 120 = 60 3rd Month -> (60 * 2) - 120 = 120 - 120 = 0
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Q4. If is equal to
Q4. 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
Answer: (A) 2
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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Q5. The difference of two numbers is 5 and the difference of their square is 135. Then the sum of the numbers is
Q5. The difference of two numbers is 5 and the difference of their square is 135. Then the sum of the numbers is
(A) 25
(A) 25
(A) 25
(B) 45
(B) 45
(B) 45
(C) 32
(C) 32
(C) 32
(D) 27
(D) 27
(D) 27
Answer: (D) 27
Answer: (D) 27
Answer: (D) 27
27
27
27
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Q6. The missing term in the sequence 7, 11, 19, 31, ______, 67 is
Q6. The missing term in the sequence 7, 11, 19, 31, ______, 67 is
(A) 43
(A) 43
(A) 43
(B) 47
(B) 47
(B) 47
(C) 51
(C) 51
(C) 51
(D) 45
(D) 45
(D) 45
Answer: (B) 47
Answer: (B) 47
Answer: (B) 47
47
47
47
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Q7. The product of two consecutive odd numbers is 19043. Which is the smaller number?
Q7. 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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Q8. What is the value of x in the equation 2x + 5 = 11?
Q8. What is the value of x in the equation 2x + 5 = 11?
(A) 2
(A) 2
(A) 2
(B) 3
(B) 3
(B) 3
(C) 4
(C) 4
(C) 4
(D) 5
(D) 5
(D) 5
Answer: (B) 3
Answer: (B) 3
Answer: (B) 3
To solve for x, subtract 5 from both sides of the equation, resulting in 2x = 6. Then, divide both sides by 2, giving x = 3.
To solve for x, subtract 5 from both sides of the equation, resulting in 2x = 6. Then, divide both sides by 2, giving x = 3.
To solve for x, subtract 5 from both sides of the equation, resulting in 2x = 6. Then, divide both sides by 2, giving x = 3.
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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. Find the median of the data set.
Q10. Find the median of the data set.
1, 2, 4, 1, 5, 12, 7, 8, 5, 1, 16, 17, 12
1, 2, 4, 1, 5, 12, 7, 8, 5, 1, 16, 17, 12
1, 2, 4, 1, 5, 12, 7, 8, 5, 1, 16, 17, 12
(A) 8
(A) 8
(A) 8
(B) 7
(B) 7
(B) 7
(C) 5
(C) 5
(C) 5
(D) 4
(D) 4
(D) 4
Answer: (C) 5
Answer: (C) 5
Answer: (C) 5
1, 2, 4, 1, 5, 12, 7, 8, 5, 1, 16, 17, 12
= 1, 1, 1, 2, 4, 5, 5, 7, 8, 12, 12, 16, 17
Total numbers of values N = 13
Hence median value = (N + 1)/2 = (13 + 1)/2 = 7th
7th Number = 5
1, 2, 4, 1, 5, 12, 7, 8, 5, 1, 16, 17, 12 = 1, 1, 1, 2, 4, 5, 5, 7, 8, 12, 12, 16, 17 Total numbers of values N = 13 Hence median value = (N + 1)/2 = (13 + 1)/2 = 7th 7th Number = 5
1, 2, 4, 1, 5, 12, 7, 8, 5, 1, 16, 17, 12 = 1, 1, 1, 2, 4, 5, 5, 7, 8, 12, 12, 16, 17 Total numbers of values N = 13 Hence median value = (N + 1)/2 = (13 + 1)/2 = 7th 7th Number = 5
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