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. If the price of sugar increases by 20%, by what percentage should a household reduce its consumption of sugar so that the budget remains the same?
Q1. If the price of sugar increases by 20%, by what percentage should a household reduce its consumption of sugar so that the budget remains the same?
(A)
(A)
(A)
(B)
(B)
(B)
(C)
(C)
(C)
(D)
(D)
(D)
Answer: (D)
Answer: (D)
Answer: (D)
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Q2. A 100 metre long train moving in a uniform speed of 20 m/sec crosses a bridge of length 1 km. The time taken by the train to cross the bridge is
Q2. A 100 metre long train moving in a uniform speed of 20 m/sec crosses a bridge of length 1 km. The time taken by the train to cross the bridge is
(A) 45 seconds
(A) 45 seconds
(A) 45 seconds
(B) 50 seconds
(B) 50 seconds
(B) 50 seconds
(C) 55 seconds
(C) 55 seconds
(C) 55 seconds
(D) 60 seconds
(D) 60 seconds
(D) 60 seconds
Answer: (C) 55 seconds
Answer: (C) 55 seconds
Answer: (C) 55 seconds
(1000+100)m / (20m/s) = 55 seconds
(1000+100)m / (20m/s) = 55 seconds
(1000+100)m / (20m/s) = 55 seconds
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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. 45% of a number is same as 30% of another number. The ratio of the first number to the second number is :
Q4. 45% of a number is same as 30% of another number. The ratio of the first number to the second number is :
(A) 2 : 3
(A) 2 : 3
(A) 2 : 3
(B) 3 : 5
(B) 3 : 5
(B) 3 : 5
(C) 3 : 2
(C) 3 : 2
(C) 3 : 2
(D) 2 : 5
(D) 2 : 5
(D) 2 : 5
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Answer: (A) 2 : 3
Answer: (A) 2 : 3
=
=
= X:Y = 2:3
= = = X:Y = 2:3
= = = X:Y = 2:3
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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. What is the sum of the interior angles of a triangle?
Q6. What is the sum of the interior angles of a triangle?
(A) 180 degrees
(A) 180 degrees
(A) 180 degrees
(B) 270 degrees
(B) 270 degrees
(B) 270 degrees
(C) 360 degrees
(C) 360 degrees
(C) 360 degrees
(D) 450 degrees
(D) 450 degrees
(D) 450 degrees
Answer: (A) 180 degrees
Answer: (A) 180 degrees
Answer: (A) 180 degrees
The sum of the interior angles of a triangle is always 180 degrees, a fundamental principle in geometry.
The sum of the interior angles of a triangle is always 180 degrees, a fundamental principle in geometry.
The sum of the interior angles of a triangle is always 180 degrees, a fundamental principle in geometry.
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Q7. Two numbers are in the ratio of 2 : 3 and the product of their LCM and HCF is 96. The sum of the numbers is
Q7. Two numbers are in the ratio of 2 : 3 and the product of their LCM and HCF is 96. The sum of the numbers is
(A) 8
(A) 8
(A) 8
(B) 12
(B) 12
(B) 12
(C) 20
(C) 20
(C) 20
(D) 36
(D) 36
(D) 36
Answer: (C) 20
Answer: (C) 20
Answer: (C) 20
=> 2x * 3x = 96
=> 6x2 = 96
=> x2 = 96/6
=> x2 = 16
=> x = 4
2x + 3x = 5x = 5 * 4 = 20
=> 2x * 3x = 96 => 6x2 = 96 => x2 = 96/6 => x2 = 16 => x = 4 2x + 3x = 5x = 5 * 4 = 20
=> 2x * 3x = 96 => 6x2 = 96 => x2 = 96/6 => x2 = 16 => x = 4 2x + 3x = 5x = 5 * 4 = 20
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Q8. When the numerator of a fraction increases by 3, the fraction increases by its three-fourth. The numerator of the fraction is
Q8. 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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Q9. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is
Q9. 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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Q10. ‘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 :
Q10. ‘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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