In a range of consecutive numbers starting with 1, all the even numbers are removed. From the remaining, consider the first 7 numbers. The sum of these 7 numbers is [#1699]
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Q1. In a range of consecutive numbers starting with 1, all the even numbers are removed. From the remaining, consider the first 7 numbers. The sum of these 7 numbers is
Q1. In a range of consecutive numbers starting with 1, all the even numbers are removed. From the remaining, consider the first 7 numbers. The sum of these 7 numbers is
(A) 35
(A) 35
(A) 35
(B) 42
(B) 42
(B) 42
(C) 49
(C) 49
(C) 49
(D) 56
(D) 56
(D) 56
Answer: (C) 49
Answer: (C) 49
Answer: (C) 49
1+3+5+7+9+11+13 = 49
1+3+5+7+9+11+13 = 49
1+3+5+7+9+11+13 = 49
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Related MCQ Quizzes
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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Q2. What is the term for the result of multiplying a number by itself?
Q2. What is the term for the result of multiplying a number by itself?
(A) Factor
(A) Factor
(A) Factor
(B) Product
(B) Product
(B) Product
(C) Quotient
(C) Quotient
(C) Quotient
(D) Square
(D) Square
(D) Square
Answer: (D) Square
Answer: (D) Square
Answer: (D) Square
The result of multiplying a number by itself is called a square, such as 4 × 4 = 16, which is denoted as 42 (four squared).
The result of multiplying a number by itself is called a square, such as 4 × 4 = 16, which is denoted as 42 (four squared).
The result of multiplying a number by itself is called a square, such as 4 × 4 = 16, which is denoted as 42 (four squared).
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Q3. 4/5 of a number is 64. Then half of the number is
Q3. 4/5 of a number is 64. Then half of the number is
(A) 16
(A) 16
(A) 16
(B) 80
(B) 80
(B) 80
(C) 32
(C) 32
(C) 32
(D) 40
(D) 40
(D) 40
Answer: (D) 40
Answer: (D) 40
Answer: (D) 40
40
40
40
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Q4. If the sum of five consecutive numbers is 190, then the lowest number amongst them is
Q4. If the sum of five consecutive numbers is 190, then the lowest number amongst them is
(A) 40
(A) 40
(A) 40
(B) 19
(B) 19
(B) 19
(C) 38
(C) 38
(C) 38
(D) 36
(D) 36
(D) 36
Answer: (D) 36
Answer: (D) 36
Answer: (D) 36
36
36
36
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Q5. ‘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 :
Q5. ‘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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Q6. 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
Q6. 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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Q7. Which of the following is a prime number?
Q7. Which of the following is a prime number?
81,33,71,93
81,33,71,93
81,33,71,93
(A) 93
(A) 93
(A) 93
(B) 33
(B) 33
(B) 33
(C) 71
(C) 71
(C) 71
(D) 81
(D) 81
(D) 81
Answer: (C) 71
Answer: (C) 71
Answer: (C) 71
71
71
71
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Q8. A 20 m long ladder is leaning on a vertical wall. It makes an angle of 30° with the ground. The height of the point the ladder touches wall is
Q8. A 20 m long ladder is leaning on a vertical wall. It makes an angle of 30° with the ground. The height of the point the ladder touches wall is
(A) 10 m
(A) 10 m
(A) 10 m
(B) 17.32 m
(B) 17.32 m
(B) 17.32 m
(C) 8.16 m
(C) 8.16 m
(C) 8.16 m
(D) 13 m
(D) 13 m
(D) 13 m
Answer: (A) 10 m
Answer: (A) 10 m
Answer: (A) 10 m
10 m
10 m
10 m
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Q9. A wheel can cover a distance of 22 km in 1000 rounds. The radius of the wheel is
Q9. 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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Q10. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is
Q10. 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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