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 [#1087]
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Q1. 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
Q1. 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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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. If A : B = 3, then 144A : 36B is
Q2. If A : B = 3, then 144A : 36B is
(A) 4
(A) 4
(A) 4
(B) 12
(B) 12
(B) 12
(C) 3
(C) 3
(C) 3
(D) 16
(D) 16
(D) 16
Answer: (B) 12
Answer: (B) 12
Answer: (B) 12
A : B = 3
=> = 3
=
= x 3
= 4 x 3
= 12
A : B = 3 => = 3 = = x 3 = 4 x 3 = 12
A : B = 3 => = 3 = = x 3 = 4 x 3 = 12
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Q3. If the decimal number 34p5 is divisible by 9, then the value of p is
Q3. 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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Q4. Three numbers are in the ratio 3:4:6 and their product is 1944. The largest of the numbers is.
Q4. Three numbers are in the ratio 3:4:6 and their product is 1944. The largest of the numbers is.
(A) 21
(A) 21
(A) 21
(B) 12
(B) 12
(B) 12
(C) 18
(C) 18
(C) 18
(D) 9
(D) 9
(D) 9
Answer: (C) 18
Answer: (C) 18
Answer: (C) 18
18
18
18
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Q5. A 100 metre long train moving in a uniform speed of 20 m/sec crosses a bridge of length 1 km. The time taken by the train to cross the bridge is
Q5. A 100 metre long train moving in a uniform speed of 20 m/sec crosses a bridge of length 1 km. The time taken by the train to cross the bridge is
(A) 45 seconds
(A) 45 seconds
(A) 45 seconds
(B) 50 seconds
(B) 50 seconds
(B) 50 seconds
(C) 55 seconds
(C) 55 seconds
(C) 55 seconds
(D) 60 seconds
(D) 60 seconds
(D) 60 seconds
Answer: (C) 55 seconds
Answer: (C) 55 seconds
Answer: (C) 55 seconds
(1000+100)m / (20m/s) = 55 seconds
(1000+100)m / (20m/s) = 55 seconds
(1000+100)m / (20m/s) = 55 seconds
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Q6. 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
Q6. 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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Q7. A number was increased by 40% and thereafter decreased by 40%. The net change in the number in percentage is
Q7. A number was increased by 40% and thereafter decreased by 40%. The net change in the number in percentage is
(A) 16% increase
(A) 16% increase
(A) 16% increase
(B) 16% decrease
(B) 16% decrease
(B) 16% decrease
(C) 32% decrease
(C) 32% decrease
(C) 32% decrease
(D) No change
(D) No change
(D) No change
Answer: (B) 16% decrease
Answer: (B) 16% decrease
Answer: (B) 16% decrease
16% decrease
= 40-40-%
= -%
= -16%
16% decrease = 40-40-% = -% = -16%
16% decrease = 40-40-% = -% = -16%
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Q8. The count of prime numbers between 80 and 100 is
Q8. 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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Q9. Which is the smallest Whole Number?
Q9. Which is the smallest Whole Number?
(A) -1
(A) -1
(A) -1
(B) 0
(B) 0
(B) 0
(C) 1
(C) 1
(C) 1
(D) 2
(D) 2
(D) 2
Answer: (B) 0
Answer: (B) 0
Answer: (B) 0
Zero
Zero
Zero
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Q10. 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 :
Q10. 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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