The ratio of the radii of two circles is 1 : 3. The ratio of their areas is [#1703]
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Q1. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is
Q1. 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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Related MCQ Quizzes
Q1. The sum of the four consecutive alternate numbers is 64. The smallest number amongst them is
Q1. The sum of the four consecutive alternate numbers is 64. The smallest number amongst them is
(A) 13
(A) 13
(A) 13
(B) 11
(B) 11
(B) 11
(C) 16
(C) 16
(C) 16
(D) 9
(D) 9
(D) 9
Answer: (A) 13
Answer: (A) 13
Answer: (A) 13
The smallest number amongst them is 13
=> X + (X+2) + (X+4) + (X+6) = 64
=> 4X + 12 = 64
=> 4X = 64 - 12
=> 4X = 52
=> X = 13
The smallest number amongst them is 13 => X + (X+2) + (X+4) + (X+6) = 64 => 4X + 12 = 64 => 4X = 64 - 12 => 4X = 52 => X = 13
The smallest number amongst them is 13 => X + (X+2) + (X+4) + (X+6) = 64 => 4X + 12 = 64 => 4X = 64 - 12 => 4X = 52 => X = 13
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Q2. Find the median of the data set.
Q2. 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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Q3. p, q, r are three numbers such that the LCM of p and q is q and the LCM of q and r is r. The LCM of p, q and r will be
Q3. p, q, r are three numbers such that the LCM of p and q is q and the LCM of q and r is r. The LCM of p, q and r will be
(A) q
(A) q
(A) q
(B) r
(B) r
(B) r
(C) qr
(C) qr
(C) qr
(D) pqr
(D) pqr
(D) pqr
Answer: (B) r
Answer: (B) r
Answer: (B) r
LCM will be r.
px= q and qy = r, hence r = pxy.
LCM will be r. px= q and qy = r, hence r = pxy.
LCM will be r. px= q and qy = r, hence r = pxy.
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Q4. Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is
Q4. Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is
(A) 2 : 5
(A) 2 : 5
(A) 2 : 5
(B) 3 : 5
(B) 3 : 5
(B) 3 : 5
(C) 4 : 5
(C) 4 : 5
(C) 4 : 5
(D) 7 : 5
(D) 7 : 5
(D) 7 : 5
Answer: (C) 4 : 5
Answer: (C) 4 : 5
Answer: (C) 4 : 5
4 : 5
4 : 5
4 : 5
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Q5. The sum of 5 consecutive odd numbers is found to be 95. The largest of the numbers is
Q5. 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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Q6. If the sum of five consecutive numbers is 190, then the lowest number amongst them is
Q6. 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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Q7. The median of the sequence of numbers 2, – 1, 3, 1, – 2, 5, 6 is
Q7. 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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Q8. A lady deposits money in her savings bank in such a way that every next day her deposit amount is ₹ 12 more than her previous day deposit. If she starts her deposit with ₹ 12 on the first day, the total amount deposited by Liza at the end of 30 days will be :
Q8. A lady deposits money in her savings bank in such a way that every next day her deposit amount is ₹ 12 more than her previous day deposit. If she starts her deposit with ₹ 12 on the first day, the total amount deposited by Liza at the end of 30 days will be :
(A) 5,420
(A) 5,420
(A) 5,420
(B) 5,580
(B) 5,580
(B) 5,580
(C) 5,620
(C) 5,620
(C) 5,620
(D) 5,780
(D) 5,780
(D) 5,780
Answer: (B) 5,580
Answer: (B) 5,580
Answer: (B) 5,580
The formula for the sum of the first n terms of an arithmetic progression is
= 15*(12+360)
= 15 * 372
= 5580
The formula for the sum of the first n terms of an arithmetic progression is = 15*(12+360) = 15 * 372 = 5580
The formula for the sum of the first n terms of an arithmetic progression is = 15*(12+360) = 15 * 372 = 5580
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Q9. If 20% of a is b, then b% of 20 is the same as :
Q9. If 20% of a is b, then b% of 20 is the same as :
(A) 4% of a
(A) 4% of a
(A) 4% of a
(B) 8% of a
(B) 8% of a
(B) 8% of a
(C) 6% of a
(C) 6% of a
(C) 6% of a
(D) 10% of a
(D) 10% of a
(D) 10% of a
Answer: (A) 4% of a
Answer: (A) 4% of a
Answer: (A) 4% of a
20% of a is b
Hence b = 20% of a = 20a/100
b% of 20
= 20% of b
= (20/100) * 20a/100
= (20*20*a)/(100*100)
= 4a/100
= (4/100) * a
= 4% of a
20% of a is b Hence b = 20% of a = 20a/100 b% of 20 = 20% of b = (20/100) * 20a/100 = (20*20*a)/(100*100) = 4a/100 = (4/100) * a = 4% of a
20% of a is b Hence b = 20% of a = 20a/100 b% of 20 = 20% of b = (20/100) * 20a/100 = (20*20*a)/(100*100) = 4a/100 = (4/100) * a = 4% of a
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Q10. A hostel has 120 students and food supplies are for 45 days. If 30 more students joined the hostel, then how many days the hostel will run with the existing food?
Q10. A hostel has 120 students and food supplies are for 45 days. If 30 more students joined the hostel, then how many days the hostel will run with the existing food?
(A) 40 days
(A) 40 days
(A) 40 days
(B) 38 days
(B) 38 days
(B) 38 days
(C) 36 days
(C) 36 days
(C) 36 days
(D) 32 days
(D) 32 days
(D) 32 days
Answer: (C) 36 days
Answer: (C) 36 days
Answer: (C) 36 days
36 days
36 days
36 days
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