The conversion of in the fractional form is [#1032]
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Q1. The conversion of in the fractional form is
Q1. The conversion of in the fractional form is
(A)
(A)
(A)
(B)
(B)
(B)
(C)
(C)
(C)
(D)
(D)
(D)
Answer: (C)
Answer: (C)
Answer: (C)
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Related MCQ Quizzes
Q1. 3/4 of a number is 19 less than the original number. The number is
Q1. 3/4 of a number is 19 less than the original number. The number is
(A) 62
(A) 62
(A) 62
(B) 64
(B) 64
(B) 64
(C) 79
(C) 79
(C) 79
(D) 76
(D) 76
(D) 76
Answer: (D) 76
Answer: (D) 76
Answer: (D) 76
76
76
76
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Q2. If A : B = 3, then 144A : 36B is
Q2. If A : B = 3, then 144A : 36B is
(A) 4
(A) 4
(A) 4
(B) 12
(B) 12
(B) 12
(C) 3
(C) 3
(C) 3
(D) 16
(D) 16
(D) 16
Answer: (B) 12
Answer: (B) 12
Answer: (B) 12
A : B = 3
=> = 3
=
= x 3
= 4 x 3
= 12
A : B = 3 => = 3 = = x 3 = 4 x 3 = 12
A : B = 3 => = 3 = = x 3 = 4 x 3 = 12
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Q3. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is
Q3. 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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Q4. A wheel can cover a distance of 22 km in 1000 rounds. The radius of the wheel is
Q4. 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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Q5. 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
Q5. 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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Q6. Find the value of the following expression
Q6. 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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Q7. What is the value of x in the equation 2x + 5 = 11?
Q7. What is the value of x in the equation 2x + 5 = 11?
(A) 2
(A) 2
(A) 2
(B) 3
(B) 3
(B) 3
(C) 4
(C) 4
(C) 4
(D) 5
(D) 5
(D) 5
Answer: (B) 3
Answer: (B) 3
Answer: (B) 3
To solve for x, subtract 5 from both sides of the equation, resulting in 2x = 6. Then, divide both sides by 2, giving x = 3.
To solve for x, subtract 5 from both sides of the equation, resulting in 2x = 6. Then, divide both sides by 2, giving x = 3.
To solve for x, subtract 5 from both sides of the equation, resulting in 2x = 6. Then, divide both sides by 2, giving x = 3.
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Q8. If 20% of a is b, then b% of 20 is the same as :
Q8. 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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Q9. Find the least number by which 1250 must be multiplied to make it a perfect square.
Q9. 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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Q10. 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
Q10. 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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