A number is as much greater than 22 as is less than 72. Then the number is [#898]
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Q1. A number is as much greater than 22 as is less than 72. Then the number is
Q1. A number is as much greater than 22 as is less than 72. Then the number is
(A) 44
(A) 44
(A) 44
(B) 50
(B) 50
(B) 50
(C) 52
(C) 52
(C) 52
(D) 47
(D) 47
(D) 47
Answer: (D) 47
Answer: (D) 47
Answer: (D) 47
47
47
47
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Related MCQ Quizzes
Q1. Which is the smallest Whole Number?
Q1. Which is the smallest Whole Number?
(A) -1
(A) -1
(A) -1
(B) 0
(B) 0
(B) 0
(C) 1
(C) 1
(C) 1
(D) 2
(D) 2
(D) 2
Answer: (B) 0
Answer: (B) 0
Answer: (B) 0
Zero
Zero
Zero
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Q2. The count of prime numbers between 80 and 100 is
Q2. The count of prime numbers between 80 and 100 is
(A) 2
(A) 2
(A) 2
(B) 5
(B) 5
(B) 5
(C) 3
(C) 3
(C) 3
(D) 4
(D) 4
(D) 4
Answer: (C) 3
Answer: (C) 3
Answer: (C) 3
Prime numbers are numbers that are only divisible by 1 and themselves. Between 80 and 100, the prime numbers are 83, 89, and 97. Therefore, there are 3 prime numbers in this range.
Prime numbers are numbers that are only divisible by 1 and themselves. Between 80 and 100, the prime numbers are 83, 89, and 97. Therefore, there are 3 prime numbers in this range.
Prime numbers are numbers that are only divisible by 1 and themselves. Between 80 and 100, the prime numbers are 83, 89, and 97. Therefore, there are 3 prime numbers in this range.
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Q3. When three times of a given number is subtracted from the square of the number, the result is the number itself. The number is
Q3. 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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Q4. 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
Q4. 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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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
(B) 3
(C) 5
(C) 5
(C) 5
(D) 2
(D) 2
(D) 2
Answer: (D) 2
Answer: (D) 2
Answer: (D) 2
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. 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
Q6. 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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Q7. The smallest positive integer that is simultaneously divisible by 6, 8 and 12 is
Q7. The smallest positive integer that is simultaneously divisible by 6, 8 and 12 is
(A) 24
(A) 24
(A) 24
(B) 18
(B) 18
(B) 18
(C) 48
(C) 48
(C) 48
(D) 36
(D) 36
(D) 36
Answer: (A) 24
Answer: (A) 24
Answer: (A) 24
6*4 = 24
8*3 = 24
12*2 = 24
6*4 = 24 8*3 = 24 12*2 = 24
6*4 = 24 8*3 = 24 12*2 = 24
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Q8. Find the value of the following expression
Q8. Find the value of the following expression
48÷12+4×25÷5
48÷12+4×25÷5
48÷12+4×25÷5
(A) 15
(A) 15
(A) 15
(B) 24
(B) 24
(B) 24
(C) 40
(C) 40
(C) 40
(D) 25
(D) 25
(D) 25
Answer: (B) 24
Answer: (B) 24
Answer: (B) 24
24
24
24
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Q9. A number was increased by 40% and thereafter decreased by 40%. The net change in the number in percentage is
Q9. A number was increased by 40% and thereafter decreased by 40%. The net change in the number in percentage is
(A) 16% increase
(A) 16% increase
(A) 16% increase
(B) 16% decrease
(B) 16% decrease
(B) 16% decrease
(C) 32% decrease
(C) 32% decrease
(C) 32% decrease
(D) No change
(D) No change
(D) No change
Answer: (B) 16% decrease
Answer: (B) 16% decrease
Answer: (B) 16% decrease
16% decrease
= 40-40-%
= -%
= -16%
16% decrease = 40-40-% = -% = -16%
16% decrease = 40-40-% = -% = -16%
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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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