Select the number pair in which the two numbers are related in the same way as 35 : 6. [#1130]
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Q1. Select the number pair in which the two numbers are related in the same way as 35 : 6.
Q1. 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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Related MCQ Quizzes
Q1. Which of the following is a prime number?
Q1. Which of the following is a prime number?
81,33,71,93
81,33,71,93
81,33,71,93
(A) 93
(A) 93
(A) 93
(B) 33
(B) 33
(B) 33
(C) 71
(C) 71
(C) 71
(D) 81
(D) 81
(D) 81
Answer: (C) 71
Answer: (C) 71
Answer: (C) 71
71
71
71
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Q2. When the numerator of a fraction increases by 3, the fraction increases by its three-fourth. The numerator of the fraction is
Q2. When the numerator of a fraction increases by 3, the fraction increases by its three-fourth. The numerator of the fraction is
(A) 3
(A) 3
(A) 3
(B) 4
(B) 4
(B) 4
(C) 5
(C) 5
(C) 5
(D) 6
(D) 6
(D) 6
Answer: (B) 4
Answer: (B) 4
Answer: (B) 4
4
4
4
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Q3. If HONESTY is written as 5132468 and POVERTY is as 7192068, then in the same code, HORSE is written as :
Q3. If HONESTY is written as 5132468 and POVERTY is as 7192068, then in the same code, HORSE is written as :
(A) 50124
(A) 50124
(A) 50124
(B) 51042
(B) 51042
(B) 51042
(C) 51024
(C) 51024
(C) 51024
(D) 52014
(D) 52014
(D) 52014
Answer: (B) 51042
Answer: (B) 51042
Answer: (B) 51042
H = 5
O = 1
R = 0
S = 4
E = 2
HORSE = 51042
H = 5 O = 1 R = 0 S = 4 E = 2 HORSE = 51042
H = 5 O = 1 R = 0 S = 4 E = 2 HORSE = 51042
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Q4. 98 toffees were distributed to some boys in a group. Each boy in the group got twice as many of the toffees as the number of boys. The number of boys in the group was
Q4. 98 toffees were distributed to some boys in a group. Each boy in the group got twice as many of the toffees as the number of boys. The number of boys in the group was
(A) 5
(A) 5
(A) 5
(B) 7
(B) 7
(B) 7
(C) 10
(C) 10
(C) 10
(D) 14
(D) 14
(D) 14
Answer: (B) 7
Answer: (B) 7
Answer: (B) 7
7
=> x * 2x = 98
=> x2 =
=> x2 = 49
=> x = 7
7 => x * 2x = 98 => x2 = => x2 = 49 => x = 7
7 => x * 2x = 98 => x2 = => x2 = 49 => x = 7
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Q5. The number when divided by 2, leaves remainder 1; when divided by 3, leaves remainder 2 and when divided by 4, leaves remainder 3, is
Q5. The number when divided by 2, leaves remainder 1; when divided by 3, leaves remainder 2 and when divided by 4, leaves remainder 3, is
(A) 25
(A) 25
(A) 25
(B) 41
(B) 41
(B) 41
(C) 13
(C) 13
(C) 13
(D) 11
(D) 11
(D) 11
Answer: (D) 11
Answer: (D) 11
Answer: (D) 11
11
11
11
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Q6. A man is 3 years older than his wife and four times as old as his son. If the son becomes 15 years old after 3 years, the present age of his wife is :
Q6. A man is 3 years older than his wife and four times as old as his son. If the son becomes 15 years old after 3 years, the present age of his wife is :
(A) 60 years
(A) 60 years
(A) 60 years
(B) 51 years
(B) 51 years
(B) 51 years
(C) 48 years
(C) 48 years
(C) 48 years
(D) 45 years
(D) 45 years
(D) 45 years
Answer: (D) 45 years
Answer: (D) 45 years
Answer: (D) 45 years
The son becomes 15 years old after 3 years.
Hence son's present age = 15 years - 3 years = 12 years
Man is four times as old as his son.
Hence Man's present age = 12 years * 4 = 48 years
Man is 3 years older than his wife.
Hence present age of his wife = 48 years - 3 years = 45 years.
The son becomes 15 years old after 3 years. Hence son's present age = 15 years - 3 years = 12 years Man is four times as old as his son. Hence Man's present age = 12 years * 4 = 48 years Man is 3 years older than his wife. Hence present age of his wife = 48 years - 3 years = 45 years.
The son becomes 15 years old after 3 years. Hence son's present age = 15 years - 3 years = 12 years Man is four times as old as his son. Hence Man's present age = 12 years * 4 = 48 years Man is 3 years older than his wife. Hence present age of his wife = 48 years - 3 years = 45 years.
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Q7. The LCM of two numbers is 40 and their HCF is 4. If the difference between the two numbers is 12, then the sum of the numbers is
Q7. The LCM of two numbers is 40 and their HCF is 4. If the difference between the two numbers is 12, then the sum of the numbers is
(A) 20
(A) 20
(A) 20
(B) 24
(B) 24
(B) 24
(C) 28
(C) 28
(C) 28
(D) 32
(D) 32
(D) 32
Answer: (C) 28
Answer: (C) 28
Answer: (C) 28
X * (X-12) = 40 * 4
X = 20
X + (X-12) = 20 + 20 - 12 = 28
X * (X-12) = 40 * 4 X = 20 X + (X-12) = 20 + 20 - 12 = 28
X * (X-12) = 40 * 4 X = 20 X + (X-12) = 20 + 20 - 12 = 28
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Q8. If the price of sugar increases by 20%, by what percentage should a household reduce its consumption of sugar so that the budget remains the same?
Q8. If the price of sugar increases by 20%, by what percentage should a household reduce its consumption of sugar so that the budget remains the same?
(A)
(A)
(A)
(B)
(B)
(B)
(C)
(C)
(C)
(D)
(D)
(D)
Answer: (D)
Answer: (D)
Answer: (D)
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Q9. ‘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 :
Q9. ‘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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Q10. The least number by which 2450 must be multiplied to make it a perfect square, is
Q10. 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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