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. ‘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 :
Q1. ‘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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Q2. Find the odd number - 15, 17, 6, 12.
Q2. Find the odd number - 15, 17, 6, 12.
(A) 6
(A) 6
(A) 6
(B) 12
(B) 12
(B) 12
(C) 15
(C) 15
(C) 15
(D) 17
(D) 17
(D) 17
Answer: (D) 17
Answer: (D) 17
Answer: (D) 17
17 which is a Prime number.
17 which is a Prime number.
17 which is a Prime number.
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Q3. The sum of 5 consecutive odd numbers is found to be 95. The largest of the numbers is
Q3. The sum of 5 consecutive odd numbers is found to be 95. The largest of the numbers is
(A) 17
(A) 17
(A) 17
(B) 21
(B) 21
(B) 21
(C) 23
(C) 23
(C) 23
(D) 19
(D) 19
(D) 19
Answer: (C) 23
Answer: (C) 23
Answer: (C) 23
X + (X+2) + (X+4) + (X+6) + (X+8) = 95
=> 5X + 20 = 95
=> 5X = 95-20
=> 5X = 75
=> X = 15
Hence (X+8) = 15+8 = 23
X + (X+2) + (X+4) + (X+6) + (X+8) = 95 => 5X + 20 = 95 => 5X = 95-20 => 5X = 75 => X = 15 Hence (X+8) = 15+8 = 23
X + (X+2) + (X+4) + (X+6) + (X+8) = 95 => 5X + 20 = 95 => 5X = 95-20 => 5X = 75 => X = 15 Hence (X+8) = 15+8 = 23
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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. 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 angles of a quadrilateral are in the ratio of 1 : 3 : 4 : 7. The difference between the largest and the smallest angle is
Q6. The angles of a quadrilateral are in the ratio of 1 : 3 : 4 : 7. The difference between the largest and the smallest angle is
(A) 120°
(A) 120°
(A) 120°
(B) 140°
(B) 140°
(B) 140°
(C) 144°
(C) 144°
(C) 144°
(D) 145°
(D) 145°
(D) 145°
Answer: (C) 144°
Answer: (C) 144°
Answer: (C) 144°
(360°/15) * (7-1) = 144°
(360°/15) * (7-1) = 144°
(360°/15) * (7-1) = 144°
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Q7. A person moves along a path such that he is always away from a given point by 7 m. After moving for some time, he again reaches his starting point. The approximate distance the person moved during the time is
Q7. A person moves along a path such that he is always away from a given point by 7 m. After moving for some time, he again reaches his starting point. The approximate distance the person moved during the time is
(A) 22 m
(A) 22 m
(A) 22 m
(B) 44 m
(B) 44 m
(B) 44 m
(C) 122 m
(C) 122 m
(C) 122 m
(D) 144 m
(D) 144 m
(D) 144 m
Answer: (B) 44 m
Answer: (B) 44 m
Answer: (B) 44 m
2 * * 7
= 2 * 22
= 44
2 * * 7 = 2 * 22 = 44
2 * * 7 = 2 * 22 = 44
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Q8. The difference between the smallest 3-digit even natural number and the largest 2-digit even natural number is
Q8. The difference between the smallest 3-digit even natural number and the largest 2-digit even natural number is
(A) 1
(A) 1
(A) 1
(B) 2
(B) 2
(B) 2
(C) 3
(C) 3
(C) 3
(D) 4
(D) 4
(D) 4
Answer: (B) 2
Answer: (B) 2
Answer: (B) 2
100 - 98 = 2
100 - 98 = 2
100 - 98 = 2
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Q9. The conversion of in the fractional form is
Q9. The conversion of in the fractional form is
(A)
(A)
(A)
(B)
(B)
(B)
(C)
(C)
(C)
(D)
(D)
(D)
Answer: (C)
Answer: (C)
Answer: (C)
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Q10. In a school with 10 teachers, one retires and immediately a new teacher of age 25 years joins. As a result, the average age of the teacher reduces by 3. The age of the retired teacher is
Q10. In a school with 10 teachers, one retires and immediately a new teacher of age 25 years joins. As a result, the average age of the teacher reduces by 3. The age of the retired teacher is
(A) 55 years
(A) 55 years
(A) 55 years
(B) 65 years
(B) 65 years
(B) 65 years
(C) 58 years
(C) 58 years
(C) 58 years
(D) 60 years
(D) 60 years
(D) 60 years
Answer: (A) 55 years
Answer: (A) 55 years
Answer: (A) 55 years
25 years + (3*10) years = 55 years
25 years + (3*10) years = 55 years
25 years + (3*10) years = 55 years
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