100% of 100 when added to 200% of 200 would result [#1704]
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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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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. If the average age of A, B and C is 22 years and the average age of B and C is 25 years, then find the age of A after 9 years from now.
Q2. If the average age of A, B and C is 22 years and the average age of B and C is 25 years, then find the age of A after 9 years from now.
(A) 25 years
(A) 25 years
(A) 25 years
(B) 35 years
(B) 35 years
(B) 35 years
(C) 50 years
(C) 50 years
(C) 50 years
(D) 45 years
(D) 45 years
(D) 45 years
Answer: (A) 25 years
Answer: (A) 25 years
Answer: (A) 25 years
25 years
=> A+B+C = 22*3
=> A+(B+C) = 66
=> A+(25*2) = 66
=> A = 66-50
=> A = 16
After 9 years A = 16+9 = 25 years
25 years => A+B+C = 22*3 => A+(B+C) = 66 => A+(25*2) = 66 => A = 66-50 => A = 16 After 9 years A = 16+9 = 25 years
25 years => A+B+C = 22*3 => A+(B+C) = 66 => A+(25*2) = 66 => A = 66-50 => A = 16 After 9 years A = 16+9 = 25 years
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Q3. The difference between the smallest 3-digit even natural number and the largest 2-digit even natural number is
Q3. 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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Q4. When three times of a given number is subtracted from the square of the number, the result is the number itself. The number is
Q4. 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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Q5. 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
Q5. 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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Q6. What is the term for a triangle with three equal sides?
Q6. What is the term for a triangle with three equal sides?
(A) Isosceles triangle
(A) Isosceles triangle
(A) Isosceles triangle
(B) Equilateral triangle
(B) Equilateral triangle
(B) Equilateral triangle
(C) Scalene triangle
(C) Scalene triangle
(C) Scalene triangle
(D) Right triangle
(D) Right triangle
(D) Right triangle
Answer: (B) Equilateral triangle
Answer: (B) Equilateral triangle
Answer: (B) Equilateral triangle
An equilateral triangle is a triangle with three equal sides and angles, a regular polygon with three sides.
An equilateral triangle is a triangle with three equal sides and angles, a regular polygon with three sides.
An equilateral triangle is a triangle with three equal sides and angles, a regular polygon with three sides.
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Q7. The 8th term of the sequence 2, 6, 18, 54, ...... is
Q7. The 8th term of the sequence 2, 6, 18, 54, ...... is
(A) 4370
(A) 4370
(A) 4370
(B) 4374
(B) 4374
(B) 4374
(C) 7443
(C) 7443
(C) 7443
(D) 7434
(D) 7434
(D) 7434
Answer: (B) 4374
Answer: (B) 4374
Answer: (B) 4374
4374
4374
4374
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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. The value of is :
Q9. The value of is :
(A) 1.73
(A) 1.73
(A) 1.73
(B) 2.03
(B) 2.03
(B) 2.03
(C) 2.73
(C) 2.73
(C) 2.73
(D) 1.03
(D) 1.03
(D) 1.03
Answer: (C) 2.73
Answer: (C) 2.73
Answer: (C) 2.73
=
=
= 0.2 + 1.2 + 1.3 + 0.03
= 2.73
= = = 0.2 + 1.2 + 1.3 + 0.03 = 2.73
= = = 0.2 + 1.2 + 1.3 + 0.03 = 2.73
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Q10. 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?
Q10. 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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