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 [#1026]
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Q1. 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
Q1. 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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Related MCQ Quizzes
Q1. The difference of two numbers is 5 and the difference of their square is 135. Then the sum of the numbers is
Q1. The difference of two numbers is 5 and the difference of their square is 135. Then the sum of the numbers is
(A) 25
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(A) 25
(B) 45
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(C) 32
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Q2. Which is the smallest Whole Number?
Q2. Which is the smallest Whole Number?
(A) -1
(A) -1
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(B) 0
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Zero
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Q3. Which of the following is a prime number?
Q3. Which of the following is a prime number?
81,33,71,93
81,33,71,93
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Answer: (C) 71
71
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Q4. If the decimal number 34p5 is divisible by 9, then the value of p is
Q4. If the decimal number 34p5 is divisible by 9, then the value of p is
(A) 8
(A) 8
(A) 8
(B) 7
(B) 7
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Answer: (D) 6
Answer: (D) 6
34p5 -> 3+4+p+5 = 12+p
12+p = 9k
12+6 = 18 = 9k
p = 6
34p5 -> 3+4+p+5 = 12+p 12+p = 9k 12+6 = 18 = 9k p = 6
34p5 -> 3+4+p+5 = 12+p 12+p = 9k 12+6 = 18 = 9k p = 6
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Q5. Find the least number by which 1250 must be multiplied to make it a perfect square.
Q5. Find the least number by which 1250 must be multiplied to make it a perfect square.
(A) 4
(A) 4
(A) 4
(B) 3
(B) 3
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1250 = 2*5*5*5*5
1250*2 = 2*2*5*5*5*5 = 2500
= 2*5*5 = 50
1250 = 2*5*5*5*5 1250*2 = 2*2*5*5*5*5 = 2500 = 2*5*5 = 50
1250 = 2*5*5*5*5 1250*2 = 2*2*5*5*5*5 = 2500 = 2*5*5 = 50
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Q6. What is the name of the mathematical concept that describes a value that never changes, like the ratio of a circle's circumference to its diameter?
Q6. What is the name of the mathematical concept that describes a value that never changes, like the ratio of a circle's circumference to its diameter?
(A) Variable
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(B) Constant
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(C) Fraction
(C) Fraction
(D) Decimal
(D) Decimal
(D) Decimal
Answer: (B) Constant
Answer: (B) Constant
Answer: (B) Constant
A constant is a mathematical concept that represents a value that remains unchanged, like pi (π), which is approximately 3.14 and never changes.
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A constant is a mathematical concept that represents a value that remains unchanged, like pi (π), which is approximately 3.14 and never changes.
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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. 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. When a number is divided by 893 the remainder is 193. If the same number is divided by 47, the remainder will be
Q9. 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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Q10. The median of the sequence of numbers 2, – 1, 3, 1, – 2, 5, 6 is
Q10. The median of the sequence of numbers 2, – 1, 3, 1, – 2, 5, 6 is
(A) 1
(A) 1
(A) 1
(B) -1
(B) -1
(B) -1
(C) 2
(C) 2
(C) 2
(D) 3
(D) 3
(D) 3
Answer: (C) 2
Answer: (C) 2
Answer: (C) 2
In ascending order = -2, -1, 1, 2, 3, 5, 6
median = th term
= 4th term
= 2
In ascending order = -2, -1, 1, 2, 3, 5, 6 median = th term = 4th term = 2
In ascending order = -2, -1, 1, 2, 3, 5, 6 median = th term = 4th term = 2
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