The sum of 5 consecutive odd numbers is found to be 95. The largest of the numbers is [#1706]
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Q1. The sum of 5 consecutive odd numbers is found to be 95. The largest of the numbers is
Q1. 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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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. 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
Q2. 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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Q3. What is the name of the mathematical concept that describes a value that never changes, like the ratio of a circle's circumference to its diameter?
Q3. What is the name of the mathematical concept that describes a value that never changes, like the ratio of a circle's circumference to its diameter?
(A) Variable
(A) Variable
(A) Variable
(B) Constant
(B) Constant
(B) Constant
(C) Fraction
(C) Fraction
(C) Fraction
(D) Decimal
(D) Decimal
(D) Decimal
Answer: (B) Constant
Answer: (B) Constant
Answer: (B) Constant
A constant is a mathematical concept that represents a value that remains unchanged, like pi (π), which is approximately 3.14 and never changes.
A constant is a mathematical concept that represents a value that remains unchanged, like pi (π), which is approximately 3.14 and never changes.
A constant is a mathematical concept that represents a value that remains unchanged, like pi (π), which is approximately 3.14 and never changes.
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Q4. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is
Q4. 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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Q5. A 20 m long ladder is leaning on a vertical wall. It makes an angle of 30° with the ground. The height of the point the ladder touches wall is
Q5. A 20 m long ladder is leaning on a vertical wall. It makes an angle of 30° with the ground. The height of the point the ladder touches wall is
(A) 10 m
(A) 10 m
(A) 10 m
(B) 17.32 m
(B) 17.32 m
(B) 17.32 m
(C) 8.16 m
(C) 8.16 m
(C) 8.16 m
(D) 13 m
(D) 13 m
(D) 13 m
Answer: (A) 10 m
Answer: (A) 10 m
Answer: (A) 10 m
10 m
10 m
10 m
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Q6. What is the term for a triangle with three equal sides?
Q6. What is the term for a triangle with three equal sides?
(A) Isosceles triangle
(A) Isosceles triangle
(A) Isosceles triangle
(B) Equilateral triangle
(B) Equilateral triangle
(B) Equilateral triangle
(C) Scalene triangle
(C) Scalene triangle
(C) Scalene triangle
(D) Right triangle
(D) Right triangle
(D) Right triangle
Answer: (B) Equilateral triangle
Answer: (B) Equilateral triangle
Answer: (B) Equilateral triangle
An equilateral triangle is a triangle with three equal sides and angles, a regular polygon with three sides.
An equilateral triangle is a triangle with three equal sides and angles, a regular polygon with three sides.
An equilateral triangle is a triangle with three equal sides and angles, a regular polygon with three sides.
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Q7. ‘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 :
Q7. ‘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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Q8. A ladder of length 13 m is leaning against a vertical wall with the upper end at the height of 5 m. The horizontal distance between the foot of the wall and the lower end of the ladder is
Q8. A ladder of length 13 m is leaning against a vertical wall with the upper end at the height of 5 m. The horizontal distance between the foot of the wall and the lower end of the ladder is
(A) 9m
(A) 9m
(A) 9m
(B) 5m
(B) 5m
(B) 5m
(C) 11m
(C) 11m
(C) 11m
(D) 12m
(D) 12m
(D) 12m
Answer: (D) 12m
Answer: (D) 12m
Answer: (D) 12m
52 + X2 = 132
=> 25 + X2 = 169
=> X2 = 196 - 25
=> X2 = 144
=> X = 12
52 + X2 = 132 => 25 + X2 = 169 => X2 = 196 - 25 => X2 = 144 => X = 12
52 + X2 = 132 => 25 + X2 = 169 => X2 = 196 - 25 => X2 = 144 => X = 12
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Q9. The count of prime numbers between 80 and 100 is
Q9. The count of prime numbers between 80 and 100 is
(A) 2
(A) 2
(A) 2
(B) 5
(B) 5
(B) 5
(C) 3
(C) 3
(C) 3
(D) 4
(D) 4
(D) 4
Answer: (C) 3
Answer: (C) 3
Answer: (C) 3
Prime numbers are numbers that are only divisible by 1 and themselves. Between 80 and 100, the prime numbers are 83, 89, and 97. Therefore, there are 3 prime numbers in this range.
Prime numbers are numbers that are only divisible by 1 and themselves. Between 80 and 100, the prime numbers are 83, 89, and 97. Therefore, there are 3 prime numbers in this range.
Prime numbers are numbers that are only divisible by 1 and themselves. Between 80 and 100, the prime numbers are 83, 89, and 97. Therefore, there are 3 prime numbers in this range.
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Q10. Two numbers are in the ratio 2 : 3 and the product of their HCF and LCM is 96. The sum of the two numbers is
Q10. Two numbers are in the ratio 2 : 3 and the product of their HCF and LCM is 96. The sum of the two numbers is
(A) 18
(A) 18
(A) 18
(B) 20
(B) 20
(B) 20
(C) 22
(C) 22
(C) 22
(D) 24
(D) 24
(D) 24
Answer: (B) 20
Answer: (B) 20
Answer: (B) 20
2:3 = 8:12
N1*N2 = HCF*LCM
8*12 = 96
N1+N2 = 8+12 = 20
2:3 = 8:12 N1*N2 = HCF*LCM 8*12 = 96 N1+N2 = 8+12 = 20
2:3 = 8:12 N1*N2 = HCF*LCM 8*12 = 96 N1+N2 = 8+12 = 20
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