When the numerator of a fraction increases by 3, the fraction increases by its three-fourth. The numerator of the fraction is [#1089]
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Q1. When the numerator of a fraction increases by 3, the fraction increases by its three-fourth. The numerator of the fraction is
Q1. 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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Related MCQ Quizzes
Q1. 100% of 100 when added to 200% of 200 would result
Q1. 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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Q2. ‘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 :
Q2. ‘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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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. 5 : 27 :: 9 : ________
Q4. 5 : 27 :: 9 : ________
Fill the blank.
Fill the blank.
Fill the blank.
(A) 83
(A) 83
(A) 83
(B) 81
(B) 81
(B) 81
(C) 36
(C) 36
(C) 36
(D) 18
(D) 18
(D) 18
Answer: (A) 83
Answer: (A) 83
Answer: (A) 83
5 : (52 + 2) = 5 : 27
Hence
9 : (92 + 2) = 9 : 83
5 : (52 + 2) = 5 : 27 Hence 9 : (92 + 2) = 9 : 83
5 : (52 + 2) = 5 : 27 Hence 9 : (92 + 2) = 9 : 83
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Q5. Find the value of
Q5. 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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Q6. The least number by which 2450 must be multiplied to make it a perfect square, is
Q6. The least number by which 2450 must be multiplied to make it a perfect square, is
(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
2
2
2
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Q7. At what speed should a car travel on a highway to reach a destination 12 km away in 15 minutes?
Q7. 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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Q8. Find the median of the data set.
Q8. 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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Q9. Three numbers are in the ratio 3:4:6 and their product is 1944. The largest of the numbers is.
Q9. 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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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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