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. When three times of a given number is subtracted from the square of the number, the result is the number itself. The number is
Q1. When three times of a given number is subtracted from the square of the number, the result is the number itself. The number is
(A) 4
(A) 4
(A) 4
(B) 2
(B) 2
(B) 2
(C) -2
(C) -2
(C) -2
(D) -4
(D) -4
(D) -4
Answer: (A) 4
Answer: (A) 4
Answer: (A) 4
42 - (4*3)
= 16 - 12
= 4
42 - (4*3) = 16 - 12 = 4
42 - (4*3) = 16 - 12 = 4
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Q2. 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
Q2. 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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Q3. Three fourth of a number is more than two third of the number by 5. Then the number is
Q3. Three fourth of a number is more than two third of the number by 5. Then the number is
(A) 72
(A) 72
(A) 72
(B) 60
(B) 60
(B) 60
(C) 80
(C) 80
(C) 80
(D) 48
(D) 48
(D) 48
Answer: (B) 60
Answer: (B) 60
Answer: (B) 60
60
60
60
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Q4. 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 :
Q4. 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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Q5. The smallest positive integer that is simultaneously divisible by 6, 8 and 12 is
Q5. The smallest positive integer that is simultaneously divisible by 6, 8 and 12 is
(A) 24
(A) 24
(A) 24
(B) 18
(B) 18
(B) 18
(C) 48
(C) 48
(C) 48
(D) 36
(D) 36
(D) 36
Answer: (A) 24
Answer: (A) 24
Answer: (A) 24
6*4 = 24
8*3 = 24
12*2 = 24
6*4 = 24 8*3 = 24 12*2 = 24
6*4 = 24 8*3 = 24 12*2 = 24
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Q6. If the decimal number 34p5 is divisible by 9, then the value of p is
Q6. 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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Q7. The conversion of in the fractional form is
Q7. The conversion of in the fractional form is
(A)
(A)
(A)
(B)
(B)
(B)
(C)
(C)
(C)
(D)
(D)
(D)
Answer: (C)
Answer: (C)
Answer: (C)
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Q8. 100% of 100 when added to 200% of 200 would result
Q8. 100% of 100 when added to 200% of 200 would result
(A) 300
(A) 300
(A) 300
(B) 400
(B) 400
(B) 400
(C) 500
(C) 500
(C) 500
(D) 600
(D) 600
(D) 600
Answer: (C) 500
Answer: (C) 500
Answer: (C) 500
100 * 100% + 200 * 200%
=
=
= 100 + 400
= 500
100 * 100% + 200 * 200% = = = 100 + 400 = 500
100 * 100% + 200 * 200% = = = 100 + 400 = 500
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Q9. If is equal to
Q9. 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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Q10. Find the value of
Q10. Find the value of
(A)
(A)
(A)
(B)
(B)
(B)
(C)
(C)
(C)
(D)
(D)
(D)
Answer: (D)
Answer: (D)
Answer: (D)
=
=
=
=
=
= = = = =
= = = = =
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