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 : [#1119]
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Q1. 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 :
Q1. 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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Related MCQ Quizzes
Q1. In a school with 10 teachers, one retires and immediately a new teacher of age 25 years joins. As a result, the average age of the teacher reduces by 3. The age of the retired teacher is
Q1. In a school with 10 teachers, one retires and immediately a new teacher of age 25 years joins. As a result, the average age of the teacher reduces by 3. The age of the retired teacher is
(A) 55 years
(A) 55 years
(A) 55 years
(B) 65 years
(B) 65 years
(B) 65 years
(C) 58 years
(C) 58 years
(C) 58 years
(D) 60 years
(D) 60 years
(D) 60 years
Answer: (A) 55 years
Answer: (A) 55 years
Answer: (A) 55 years
25 years + (3*10) years = 55 years
25 years + (3*10) years = 55 years
25 years + (3*10) years = 55 years
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Q2. The sum of 5 consecutive odd numbers is found to be 95. The largest of the numbers is
Q2. 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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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. If 16℅ of 40℅ of a number is 8, then the number is
Q4. If 16℅ of 40℅ of a number is 8, then the number is
(A) 250
(A) 250
(A) 250
(B) 125
(B) 125
(B) 125
(C) 200
(C) 200
(C) 200
(D) 100
(D) 100
(D) 100
Answer: (B) 125
Answer: (B) 125
Answer: (B) 125
125
125
125
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Q5. The price of an article is first decreased by 40% and then increased by 20%. The net change in the price will be :
Q5. The price of an article is first decreased by 40% and then increased by 20%. The net change in the price will be :
(A) 20%
(A) 20%
(A) 20%
(B) 40%
(B) 40%
(B) 40%
(C) 24%
(C) 24%
(C) 24%
(D) 28%
(D) 28%
(D) 28%
Answer: (D) 28%
Answer: (D) 28%
Answer: (D) 28%
Net change (-40 + 20 + )%
= (-20 + )%
= (-20 - 8)%
= -28%
= 28%
Net change (-40 + 20 + )% = (-20 + )% = (-20 - 8)% = -28% = 28%
Net change (-40 + 20 + )% = (-20 + )% = (-20 - 8)% = -28% = 28%
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Q6. When a number is divided by 893 the remainder is 193. If the same number is divided by 47, the remainder will be
Q6. When a number is divided by 893 the remainder is 193. If the same number is divided by 47, the remainder will be
(A) 3
(A) 3
(A) 3
(B) 25
(B) 25
(B) 25
(C) 5
(C) 5
(C) 5
(D) 33
(D) 33
(D) 33
Answer: (C) 5
Answer: (C) 5
Answer: (C) 5
5
5
5
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Q7. 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
Q7. 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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Q8. When three times of a given number is subtracted from the square of the number, the result is the number itself. The number is
Q8. 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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Q9. 5 : 27 :: 9 : ________
Q9. 5 : 27 :: 9 : ________
Fill the blank.
Fill the blank.
Fill the blank.
(A) 83
(A) 83
(A) 83
(B) 81
(B) 81
(B) 81
(C) 36
(C) 36
(C) 36
(D) 18
(D) 18
(D) 18
Answer: (A) 83
Answer: (A) 83
Answer: (A) 83
5 : (52 + 2) = 5 : 27
Hence
9 : (92 + 2) = 9 : 83
5 : (52 + 2) = 5 : 27 Hence 9 : (92 + 2) = 9 : 83
5 : (52 + 2) = 5 : 27 Hence 9 : (92 + 2) = 9 : 83
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Q10. 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 :
Q10. 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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