The smallest among is [#1691]
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Q1. The smallest among is
Q1. The smallest among 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. 4/5 of a number is 64. Then half of the number is
Q1. 4/5 of a number is 64. Then half of the number is
(A) 16
(A) 16
(A) 16
(B) 80
(B) 80
(B) 80
(C) 32
(C) 32
(C) 32
(D) 40
(D) 40
(D) 40
Answer: (D) 40
Answer: (D) 40
Answer: (D) 40
40
40
40
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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. 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. What is the value of x in the equation 2x + 5 = 11?
Q4. 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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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. Which is the smallest Whole Number?
Q6. 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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Q7. If is equal to
Q7. If is equal to
(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: (A) 2
Answer: (A) 2
Answer: (A) 2
Hence, X:Y:Z = 3:4:7
=
=
= 2
Hence, X:Y:Z = 3:4:7 = = = 2
Hence, X:Y:Z = 3:4:7 = = = 2
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Q8. The product of two consecutive odd numbers is 19043. Which is the smaller number?
Q8. The product of two consecutive odd numbers is 19043. Which is the smaller number?
(A) 133
(A) 133
(A) 133
(B) 131
(B) 131
(B) 131
(C) 137
(C) 137
(C) 137
(D) 129
(D) 129
(D) 129
Answer: (C) 137
Answer: (C) 137
Answer: (C) 137
137
137
137
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Q9. Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is
Q9. 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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Q10. 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 :
Q10. 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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Related Questions
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