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. How many prime numbers exist, which are less than 40?
Q1. How many prime numbers exist, which are less than 40?
(A) 10
(A) 10
(A) 10
(B) 11
(B) 11
(B) 11
(C) 12
(C) 12
(C) 12
(D) 13
(D) 13
(D) 13
Answer: (C) 12
Answer: (C) 12
Answer: (C) 12
12
12
12
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Q2. A certain amount invested in a firm would become double at the end of one month, but it deducts an amount of ₹120 on every doubling. A person invests an amount of ₹105 and continues for 3 months without investing any additional amount. At the end of 3 months, his net income is
Q2. A certain amount invested in a firm would become double at the end of one month, but it deducts an amount of ₹120 on every doubling. A person invests an amount of ₹105 and continues for 3 months without investing any additional amount. At the end of 3 months, his net income is
(A) 55
(A) 55
(A) 55
(B) 0
(B) 0
(B) 0
(C) 270
(C) 270
(C) 270
(D) 45
(D) 45
(D) 45
Answer: (B) 0
Answer: (B) 0
Answer: (B) 0
1st Month -> (105 * 2) - 120 = 210 - 120 = 90
2nd Month -> (90 * 2) - 120 = 180 - 120 = 60
3rd Month -> (60 * 2) - 120 = 120 - 120 = 0
1st Month -> (105 * 2) - 120 = 210 - 120 = 90 2nd Month -> (90 * 2) - 120 = 180 - 120 = 60 3rd Month -> (60 * 2) - 120 = 120 - 120 = 0
1st Month -> (105 * 2) - 120 = 210 - 120 = 90 2nd Month -> (90 * 2) - 120 = 180 - 120 = 60 3rd Month -> (60 * 2) - 120 = 120 - 120 = 0
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Q3. The value of is :
Q3. The value of is :
(A) 1.73
(A) 1.73
(A) 1.73
(B) 2.03
(B) 2.03
(B) 2.03
(C) 2.73
(C) 2.73
(C) 2.73
(D) 1.03
(D) 1.03
(D) 1.03
Answer: (C) 2.73
Answer: (C) 2.73
Answer: (C) 2.73
=
=
= 0.2 + 1.2 + 1.3 + 0.03
= 2.73
= = = 0.2 + 1.2 + 1.3 + 0.03 = 2.73
= = = 0.2 + 1.2 + 1.3 + 0.03 = 2.73
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Q4. Consider the four numbers given numbers. If the digits of each number are arranged in ascending order, which new number would be the smallest?
Q4. Consider the four numbers given numbers. If the digits of each number are arranged in ascending order, which new number would be the smallest?
I. 376
II. 629
III. 921
IV. 397
I. 376 II. 629 III. 921 IV. 397
I. 376 II. 629 III. 921 IV. 397
(A) III
(A) III
(A) III
(B) II
(B) II
(B) II
(C) I
(C) I
(C) I
(D) IV
(D) IV
(D) IV
Answer: (A) III
Answer: (A) III
Answer: (A) III
376 -> 367
629 -> 269
921 -> 129 *
397 -> 379
376 -> 367 629 -> 269 921 -> 129 * 397 -> 379
376 -> 367 629 -> 269 921 -> 129 * 397 -> 379
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Q5. The ratio of the radii of two circles is 1 : 3. The ratio of their areas is
Q5. 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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Q6. 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
Q6. 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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Q7. Find the odd number - 15, 17, 6, 12.
Q7. Find the odd number - 15, 17, 6, 12.
(A) 6
(A) 6
(A) 6
(B) 12
(B) 12
(B) 12
(C) 15
(C) 15
(C) 15
(D) 17
(D) 17
(D) 17
Answer: (D) 17
Answer: (D) 17
Answer: (D) 17
17 which is a Prime number.
17 which is a Prime number.
17 which is a Prime number.
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Q8. What is the term for a triangle with three equal sides?
Q8. 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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Q9. The median of the sequence of numbers 2, – 1, 3, 1, – 2, 5, 6 is
Q9. The median of the sequence of numbers 2, – 1, 3, 1, – 2, 5, 6 is
(A) 1
(A) 1
(A) 1
(B) -1
(B) -1
(B) -1
(C) 2
(C) 2
(C) 2
(D) 3
(D) 3
(D) 3
Answer: (C) 2
Answer: (C) 2
Answer: (C) 2
In ascending order = -2, -1, 1, 2, 3, 5, 6
median = th term
= 4th term
= 2
In ascending order = -2, -1, 1, 2, 3, 5, 6 median = th term = 4th term = 2
In ascending order = -2, -1, 1, 2, 3, 5, 6 median = th term = 4th term = 2
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Q10. If the decimal number 34p5 is divisible by 9, then the value of p is
Q10. If the decimal number 34p5 is divisible by 9, then the value of p is
(A) 8
(A) 8
(A) 8
(B) 7
(B) 7
(B) 7
(C) 4
(C) 4
(C) 4
(D) 6
(D) 6
(D) 6
Answer: (D) 6
Answer: (D) 6
Answer: (D) 6
34p5 -> 3+4+p+5 = 12+p
12+p = 9k
12+6 = 18 = 9k
p = 6
34p5 -> 3+4+p+5 = 12+p 12+p = 9k 12+6 = 18 = 9k p = 6
34p5 -> 3+4+p+5 = 12+p 12+p = 9k 12+6 = 18 = 9k p = 6
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