The largest of 26, 35, 44 and 53 is [#1698]
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Q1. The largest of 26, 35, 44 and 53 is
Q1. The largest of 26, 35, 44 and 53 is
(A) 26
(A) 26
(A) 26
(B) 35
(B) 35
(B) 35
(C) 44
(C) 44
(C) 44
(D) 53
(D) 53
(D) 53
Answer: (C) 44
Answer: (C) 44
Answer: (C) 44
26 = 64
35 = 243
44 = 256
53 = 125
26 = 64 35 = 243 44 = 256 53 = 125
26 = 64 35 = 243 44 = 256 53 = 125
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Related MCQ Quizzes
Q1. What is the term for a number that has no decimal places or fractional part?
Q1. What is the term for a number that has no decimal places or fractional part?
(A) Integer
(A) Integer
(A) Integer
(B) Fraction
(B) Fraction
(B) Fraction
(C) Decimal
(C) Decimal
(C) Decimal
(D) Percentage
(D) Percentage
(D) Percentage
Answer: (A) Integer
Answer: (A) Integer
Answer: (A) Integer
An integer is a whole number, either positive, negative, or zero, without a decimal or fractional part, such as 5, -3, or 0.
An integer is a whole number, either positive, negative, or zero, without a decimal or fractional part, such as 5, -3, or 0.
An integer is a whole number, either positive, negative, or zero, without a decimal or fractional part, such as 5, -3, or 0.
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Q2. The angles of a quadrilateral are in the ratio of 1 : 3 : 4 : 7. The difference between the largest and the smallest angle is
Q2. The angles of a quadrilateral are in the ratio of 1 : 3 : 4 : 7. The difference between the largest and the smallest angle is
(A) 120°
(A) 120°
(A) 120°
(B) 140°
(B) 140°
(B) 140°
(C) 144°
(C) 144°
(C) 144°
(D) 145°
(D) 145°
(D) 145°
Answer: (C) 144°
Answer: (C) 144°
Answer: (C) 144°
(360°/15) * (7-1) = 144°
(360°/15) * (7-1) = 144°
(360°/15) * (7-1) = 144°
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Q3. 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
Q3. 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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Q4. If is equal to
Q4. 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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Q5. ‘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 :
Q5. ‘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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Q6. Select the number pair in which the two numbers are related in the same way as 35 : 6.
Q6. Select the number pair in which the two numbers are related in the same way as 35 : 6.
(A) 85 : 9
(A) 85 : 9
(A) 85 : 9
(B) 64 : 8
(B) 64 : 8
(B) 64 : 8
(C) 26 : 5
(C) 26 : 5
(C) 26 : 5
(D) 120 : 11
(D) 120 : 11
(D) 120 : 11
Answer: (D) 120 : 11
Answer: (D) 120 : 11
Answer: (D) 120 : 11
62-1 : 6 = 35 : 6
112-1 : 11 = 120 : 11
62-1 : 6 = 35 : 6 112-1 : 11 = 120 : 11
62-1 : 6 = 35 : 6 112-1 : 11 = 120 : 11
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Q7. The difference of two numbers is 5 and the difference of their square is 135. Then the sum of the numbers is
Q7. The difference of two numbers is 5 and the difference of their square is 135. Then the sum of the numbers is
(A) 25
(A) 25
(A) 25
(B) 45
(B) 45
(B) 45
(C) 32
(C) 32
(C) 32
(D) 27
(D) 27
(D) 27
Answer: (D) 27
Answer: (D) 27
Answer: (D) 27
27
27
27
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Q8. 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
Q8. 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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Q9. The least number by which 2450 must be multiplied to make it a perfect square, is
Q9. The least number by which 2450 must be multiplied to make it a perfect square, is
(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
2
2
2
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Q10. A number when divided by 6 is diminished by 40. The number is
Q10. A number when divided by 6 is diminished by 40. The number is
(A) 72
(A) 72
(A) 72
(B) 48
(B) 48
(B) 48
(C) 60
(C) 60
(C) 60
(D) 84
(D) 84
(D) 84
Answer: (B) 48
Answer: (B) 48
Answer: (B) 48
48
48
48
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Related Questions
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