When a number is divided by 893 the remainder is 193. If the same number is divided by 47, the remainder will be [#1020]
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Q1. When a number is divided by 893 the remainder is 193. If the same number is divided by 47, the remainder will be
Q1. When a number is divided by 893 the remainder is 193. If the same number is divided by 47, the remainder will be
(A) 3
(A) 3
(A) 3
(B) 25
(B) 25
(B) 25
(C) 5
(C) 5
(C) 5
(D) 33
(D) 33
(D) 33
Answer: (C) 5
Answer: (C) 5
Answer: (C) 5
5
5
5
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Related MCQ Quizzes
Q1. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is
Q1. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is
(A) 1:6
(A) 1:6
(A) 1:6
(B) 2:9
(B) 2:9
(B) 2:9
(C) 1:9
(C) 1:9
(C) 1:9
(D) 6:9
(D) 6:9
(D) 6:9
Answer: (C) 1:9
Answer: (C) 1:9
Answer: (C) 1:9
Area = πr2
π12 : π32
= π1 : π9
= 1:9
Area = πr2 π12 : π32 = π1 : π9 = 1:9
Area = πr2 π12 : π32 = π1 : π9 = 1:9
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Q2. What is the term for a number that has no decimal places or fractional part?
Q2. What is the term for a number that has no decimal places or fractional part?
(A) Integer
(A) Integer
(A) Integer
(B) Fraction
(B) Fraction
(B) Fraction
(C) Decimal
(C) Decimal
(C) Decimal
(D) Percentage
(D) Percentage
(D) Percentage
Answer: (A) Integer
Answer: (A) Integer
Answer: (A) Integer
An integer is a whole number, either positive, negative, or zero, without a decimal or fractional part, such as 5, -3, or 0.
An integer is a whole number, either positive, negative, or zero, without a decimal or fractional part, such as 5, -3, or 0.
An integer is a whole number, either positive, negative, or zero, without a decimal or fractional part, such as 5, -3, or 0.
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Q3. The sum of first 6 prime numbers is
Q3. 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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Q4. The largest of 26, 35, 44 and 53 is
Q4. The largest of 26, 35, 44 and 53 is
(A) 26
(A) 26
(A) 26
(B) 35
(B) 35
(B) 35
(C) 44
(C) 44
(C) 44
(D) 53
(D) 53
(D) 53
Answer: (C) 44
Answer: (C) 44
Answer: (C) 44
26 = 64
35 = 243
44 = 256
53 = 125
26 = 64 35 = 243 44 = 256 53 = 125
26 = 64 35 = 243 44 = 256 53 = 125
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Q5. The sum of two numbers is 15 and the sum of their squares is 113. The numbers are
Q5. 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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Q6. The remainder when –76 is divided by 3, is
Q6. The remainder when –76 is divided by 3, is
(A) -1
(A) -1
(A) -1
(B) 1
(B) 1
(B) 1
(C) 2
(C) 2
(C) 2
(D) -2
(D) -2
(D) -2
Answer: (C) 2
Answer: (C) 2
Answer: (C) 2
2
Because, Reminder must be positive and it should be less then the divisor
2 Because, Reminder must be positive and it should be less then the divisor
2 Because, Reminder must be positive and it should be less then the divisor
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Q7. ‘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 :
Q7. ‘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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Q8. 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?
Q8. 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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Q9. When three times of a given number is subtracted from the square of the number, the result is the number itself. The number is
Q9. When three times of a given number is subtracted from the square of the number, the result is the number itself. The number is
(A) 4
(A) 4
(A) 4
(B) 2
(B) 2
(B) 2
(C) -2
(C) -2
(C) -2
(D) -4
(D) -4
(D) -4
Answer: (A) 4
Answer: (A) 4
Answer: (A) 4
42 - (4*3)
= 16 - 12
= 4
42 - (4*3) = 16 - 12 = 4
42 - (4*3) = 16 - 12 = 4
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Q10. Three numbers are in the ratio 3:4:6 and their product is 1944. The largest of the numbers is.
Q10. 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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