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All questions of LCM and HCF for Civil Engineering (CE) Exam
Find the least number exactly divisible by 10, 15, 18 and 30.- a)85
- b)90
- c)88
- d)93
- e)105
Correct answer is option 'B'. Can you explain this answer?
Find the least number exactly divisible by 10, 15, 18 and 30.
a)
85
b)
90
c)
88
d)
93
e)
105
|
Nikita Singh answered |
It will be the LCM of these numbers.
Find the 4-digit smallest number which when divided by 12, 15, 25, 30 leaves no remainder?- a)1020
- b)1120
- c)1200
- d)1800
- e)None of these
Correct answer is option 'C'. Can you explain this answer?
Find the 4-digit smallest number which when divided by 12, 15, 25, 30 leaves no remainder?
a)
1020
b)
1120
c)
1200
d)
1800
e)
None of these
|
Kavya Saxena answered |
LCM of 12, 15, 25 and 30 is 300
least number of 4-digit divided by 300 is 1200
least number of 4-digit divided by 300 is 1200
The HCF and LCM of two numbers is 84 and 840 respectively. If the first number is 168, find the second one.- a)210
- b)420
- c)225
- d)326
- e)None of these
Correct answer is option 'B'. Can you explain this answer?
The HCF and LCM of two numbers is 84 and 840 respectively. If the first number is 168, find the second one.
a)
210
b)
420
c)
225
d)
326
e)
None of these
|
Bank Exams India answered |
Product of two numbers = HCF * LCM
So 2nd number = 84*840/168
So 2nd number = 84*840/168
HCF of 4/3, 8/6, 36/63 and 20/42- a)4/126
- b)4/8
- c)4/36
- d)4/42
- e)None of these
Correct answer is option 'A'. Can you explain this answer?
HCF of 4/3, 8/6, 36/63 and 20/42
a)
4/126
b)
4/8
c)
4/36
d)
4/42
e)
None of these
|
Preeti Khanna answered |
HCF of numerator(4,8,36,20) = 4
LCM of denominator(3,6,63,42) = 126
LCM of denominator(3,6,63,42) = 126
If the product of two numbers is 2496 and HCF is 8,then the ratio of HCF and LCM is- a)1:32
- b)39:1
- c)1:39
- d)4:63
- e)None of these
Correct answer is option 'C'. Can you explain this answer?
If the product of two numbers is 2496 and HCF is 8,then the ratio of HCF and LCM is
a)
1:32
b)
39:1
c)
1:39
d)
4:63
e)
None of these
|
|
Preeti Khanna answered |
HCF = 8
LCM = Product/HCF Product of the 2 number = 2496
HCF:LCM = 8 : (2496/8) ⇒ 8 : 312 ⇒ 1: 39
LCM = Product/HCF Product of the 2 number = 2496
HCF:LCM = 8 : (2496/8) ⇒ 8 : 312 ⇒ 1: 39
In a seminar the number of participants in Technology, Economics and Science are 150, 90 and 180 respectively. Find the minimum number of rooms required, where in each room the same number of participants are to be seated and all of them being in the same subject.
- a)27
- b)32
- c)30
- d)14
- e)None of these
Correct answer is option 'D'. Can you explain this answer?
In a seminar the number of participants in Technology, Economics and Science are 150, 90 and 180 respectively. Find the minimum number of rooms required, where in each room the same number of participants are to be seated and all of them being in the same subject.
a)
27
b)
32
c)
30
d)
14
e)
None of these
|
|
Bank Exams India answered |
HCF of 150, 90 and 180 = 30
No of participants can be seated in each room = 30
No of participants can be seated in each room = 30
There are total 420 participants.
As the HCF is 30 i.e. mininum number of participants in each room.
so total rooms required will be 420/30 = 14
Sum of 2 numbers is 128 and their HCF is 8. How many numbers of pairs of numbers will satisfy this condition?- a)2
- b)5
- c)3
- d)6
- e)4
Correct answer is option 'E'. Can you explain this answer?
Sum of 2 numbers is 128 and their HCF is 8. How many numbers of pairs of numbers will satisfy this condition?
a)
2
b)
5
c)
3
d)
6
e)
4
|
|
Preeti Khanna answered |
Since HCF = 8 is highest common factor among those numbers
So, let first no = 8x, 2nd number = 8y
So 8x + 8y = 128
x + y = 16
the co-prime numbers which sum up 16 are (1,15), (3,13), (5,11), and (7,9) *co-prime numbers are those which do no have any factor in common. These pairs are taken here because in HCF highest common factor is taken, so the remaining multiples x and y must have no common factors.
Since there are 4 pairs of co-prime numbers here, there will be 4 pairs of numbers satisfying given conditions. These are (8*1, 8*15), (8*3, 8*13) , (8*5, 8*11) , (8*7, 8*9)
So, let first no = 8x, 2nd number = 8y
So 8x + 8y = 128
x + y = 16
the co-prime numbers which sum up 16 are (1,15), (3,13), (5,11), and (7,9) *co-prime numbers are those which do no have any factor in common. These pairs are taken here because in HCF highest common factor is taken, so the remaining multiples x and y must have no common factors.
Since there are 4 pairs of co-prime numbers here, there will be 4 pairs of numbers satisfying given conditions. These are (8*1, 8*15), (8*3, 8*13) , (8*5, 8*11) , (8*7, 8*9)
A gardener had a number of shrubs to plant in rows. At first he tried to plant 8,then 12 and then 16 in a row but he had always 3 shrubs left with him. On trying 7 he had none left. Find the total number of shrubs.- a)154
- b)147
- c)137
- d)150
- e)None of these
Correct answer is option 'B'. Can you explain this answer?
A gardener had a number of shrubs to plant in rows. At first he tried to plant 8,then 12 and then 16 in a row but he had always 3 shrubs left with him. On trying 7 he had none left. Find the total number of shrubs.
a)
154
b)
147
c)
137
d)
150
e)
None of these
|
Cstoppers Instructors answered |
Find the LCM of 3/8, 9/32, 33/48, 18/72- a)3/8
- b)8/33
- c)198/8
- d)8/3
- e)None of these
Correct answer is option 'C'. Can you explain this answer?
Find the LCM of 3/8, 9/32, 33/48, 18/72
a)
3/8
b)
8/33
c)
198/8
d)
8/3
e)
None of these
|
Yash Patel answered |
LCM of numerator(3,9,33,18) = 198
HCF of denominator(8,32,48,72) = 8
HCF of denominator(8,32,48,72) = 8
Find the smallest number of 4 digits which is exactly divisible by 12, 15, 21 and 30.- a)1580
- b)1420
- c)1260
- d)1056
- e)None of these
Correct answer is option 'C'. Can you explain this answer?
Find the smallest number of 4 digits which is exactly divisible by 12, 15, 21 and 30.
a)
1580
b)
1420
c)
1260
d)
1056
e)
None of these
|
|
Preeti Khanna answered |
Least 4 digit number = 1000
LCM of (12, 15, 21, 30) = 420
On dividing 420 by 1000, leaves remainder = 160
So answer = 1000 + (420 – 160)
LCM of (12, 15, 21, 30) = 420
On dividing 420 by 1000, leaves remainder = 160
So answer = 1000 + (420 – 160)
Find the least number which when divided by 2, 5, 9 and 12 leaves a remainder 3 but leaves no remainder when same number is divided by 11.- a)281
- b)357
- c)360
- d)363
- e)224
Correct answer is option 'D'. Can you explain this answer?
Find the least number which when divided by 2, 5, 9 and 12 leaves a remainder 3 but leaves no remainder when same number is divided by 11.
a)
281
b)
357
c)
360
d)
363
e)
224
|
Logith S answered |
(USE THIS METHOD ONLY IF ANY ONE OF THE OPTIONS REMAINDERS AS 0 and don't do this method if all the options got the remainder as 0).
So the question was by dividing 11 there will be no remainder. I divided the numbers in the option with 11 and I got remainders as
a) remainder is 6
b)remainder is 5
c)remainder is 8
d) remainder is 0
e) remainder is different
so the option d leaves a remainder is 0 the answer is 363.
I solved this problem by using this method.
So the question was by dividing 11 there will be no remainder. I divided the numbers in the option with 11 and I got remainders as
a) remainder is 6
b)remainder is 5
c)remainder is 8
d) remainder is 0
e) remainder is different
so the option d leaves a remainder is 0 the answer is 363.
I solved this problem by using this method.
The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:- a)1
- b)2
- c)3
- d)5
- e)None of these
Correct answer is option 'B'. Can you explain this answer?
The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:
a)
1
b)
2
c)
3
d)
5
e)
None of these
|
Faizan Khan answered |
Let the numbers 13a and 13b.
Then, 13a x 13b = 2028
⇒ ab = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).
Clearly, there are 2 such pairs.
Then, 13a x 13b = 2028
⇒ ab = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).
Clearly, there are 2 such pairs.
The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is- a) 1677
- b) 1683
- c) 2523
- d) 3363
Correct answer is option 'B'. Can you explain this answer?
The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is
a)
1677
b)
1683
c)
2523
d)
3363
|
Sagar Sharma answered |
Understanding the Problem:
The problem states that the least number when divided by 5, 6, 7, and 8 leaves a remainder of 3, but when divided by 9 leaves no remainder.
Solution:
To find the least number that satisfies these conditions, we need to consider the LCM (Least Common Multiple) of 5, 6, 7, and 8 first.
Finding LCM:
- LCM of 5, 6, 7, and 8 = 840
Finding Number:
- Let x be the number we are looking for
- According to the problem, x leaves a remainder of 3 when divided by 5, 6, 7, and 8
- This can be represented as x = 840k + 3, where k is an integer
- Now, x should leave no remainder when divided by 9
- This implies that x should be a multiple of 9
- Substitute x = 840k + 3 into x being a multiple of 9, we get 840k + 3 ≡ 0 (mod 9)
- Solving this congruence, we find k = 2
- Substitute k = 2 back into x = 840k + 3, we get x = 1683
Therefore, the least number that satisfies the given conditions is 1683, which is option B.
The problem states that the least number when divided by 5, 6, 7, and 8 leaves a remainder of 3, but when divided by 9 leaves no remainder.
Solution:
To find the least number that satisfies these conditions, we need to consider the LCM (Least Common Multiple) of 5, 6, 7, and 8 first.
Finding LCM:
- LCM of 5, 6, 7, and 8 = 840
Finding Number:
- Let x be the number we are looking for
- According to the problem, x leaves a remainder of 3 when divided by 5, 6, 7, and 8
- This can be represented as x = 840k + 3, where k is an integer
- Now, x should leave no remainder when divided by 9
- This implies that x should be a multiple of 9
- Substitute x = 840k + 3 into x being a multiple of 9, we get 840k + 3 ≡ 0 (mod 9)
- Solving this congruence, we find k = 2
- Substitute k = 2 back into x = 840k + 3, we get x = 1683
Therefore, the least number that satisfies the given conditions is 1683, which is option B.
The HCF of 3 different no is 17, Which of the following cannot be their LCM ?- a)540
- b)289
- c)340
- d)425
- e)None of these
Correct answer is option 'A'. Can you explain this answer?
The HCF of 3 different no is 17, Which of the following cannot be their LCM ?
a)
540
b)
289
c)
340
d)
425
e)
None of these
|
|
Nikita Singh answered |
To check whether the given number is it's LCM or not just divide the LCM by HCF .
1) 540 ÷ 17 = 31.76
2) 289 ÷ 17 = 17
3) 340 ÷ 17 = 20
4) 425 ÷ 17 = 25 .
540 is not divisible by 17.
Hence : 540 is not the LCM .
1) 540 ÷ 17 = 31.76
2) 289 ÷ 17 = 17
3) 340 ÷ 17 = 20
4) 425 ÷ 17 = 25 .
540 is not divisible by 17.
Hence : 540 is not the LCM .
The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:- a)534
- b)486
- c)544
- d)548
- e)None of these
Correct answer is option 'D'. Can you explain this answer?
The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:
a)
534
b)
486
c)
544
d)
548
e)
None of these
|
Anu Reddy answered |
Required number = (L.C.M. of 12, 15, 20, 54) + 8
= 540 + 8
= 548.
= 540 + 8
= 548.
Three buckets contains balloons filled with water. First bucket contains 243 balloons. Second contains 304 balloons and last bucket contains 127 balloons. Find the largest number of balloons that can be given equally to the children such that 3, 4 and 7 balloons are left in first, second and third bucket respectively?- a)20
- b)30
- c)40
- d)60
- e)None of these
Correct answer is option 'D'. Can you explain this answer?
Three buckets contains balloons filled with water. First bucket contains 243 balloons. Second contains 304 balloons and last bucket contains 127 balloons. Find the largest number of balloons that can be given equally to the children such that 3, 4 and 7 balloons are left in first, second and third bucket respectively?
a)
20
b)
30
c)
40
d)
60
e)
None of these
|
Chirag Makkar answered |
HCF (240, 300, 120) = 60
The greatest possible length which can be used to measure exactly the lengths 1m 92cm,3m 84cm ,23m 4cm- a)23
- b)32
- c)36
- d)34
- e)None of these
Correct answer is option 'B'. Can you explain this answer?
The greatest possible length which can be used to measure exactly the lengths 1m 92cm,3m 84cm ,23m 4cm
a)
23
b)
32
c)
36
d)
34
e)
None of these
|
Alok Verma answered |
192 = 42×22×3
384 = 42×22×6
2304 = 42×2×62
HCF = 42×2 = 16×2 = 32
384 = 42×22×6
2304 = 42×2×62
HCF = 42×2 = 16×2 = 32
HCF and LCM of two numbers are 11 and 385 .If one number lies between 75 and 125 then that number is- a)123
- b)73
- c)77
- d)154
- e)None of these
Correct answer is option 'C'. Can you explain this answer?
HCF and LCM of two numbers are 11 and 385 .If one number lies between 75 and 125 then that number is
a)
123
b)
73
c)
77
d)
154
e)
None of these
|
|
Sagar Sharma answered |
HCF and LCM are important concepts in number theory. HCF stands for Highest Common Factor, which is the largest number that divides two numbers evenly. LCM stands for Lowest Common Multiple, which is the smallest number that is a multiple of two numbers.
Given that the HCF is 11 and the LCM is 385, we can use these values to find the two numbers.
To find the two numbers, we can use the formula:
HCF x LCM = Product of the two numbers
So, 11 x 385 = Product of the two numbers
The product of the two numbers is 4235.
Now, we need to find the two numbers that have a product of 4235 and one of them lies between 75 and 125.
Let's consider the factors of 4235:
1 x 4235 = 4235
5 x 847 = 4235
11 x 385 = 4235
35 x 121 = 4235
55 x 77 = 4235
Out of these pairs, only one number lies between 75 and 125, which is 77. Therefore, the other number must be 55.
So, the two numbers are 77 and 55.
Now, let's verify if the HCF and LCM of these two numbers are indeed 11 and 385.
The HCF of 77 and 55 is 11.
To find the LCM, we can use the formula:
LCM = (Product of the two numbers) / HCF
LCM = (77 x 55) / 11
LCM = 4235 / 11
LCM = 385
Therefore, the HCF and LCM of 77 and 55 are indeed 11 and 385, respectively.
Hence, the number that lies between 75 and 125 is 77, which is option C.
Given that the HCF is 11 and the LCM is 385, we can use these values to find the two numbers.
To find the two numbers, we can use the formula:
HCF x LCM = Product of the two numbers
So, 11 x 385 = Product of the two numbers
The product of the two numbers is 4235.
Now, we need to find the two numbers that have a product of 4235 and one of them lies between 75 and 125.
Let's consider the factors of 4235:
1 x 4235 = 4235
5 x 847 = 4235
11 x 385 = 4235
35 x 121 = 4235
55 x 77 = 4235
Out of these pairs, only one number lies between 75 and 125, which is 77. Therefore, the other number must be 55.
So, the two numbers are 77 and 55.
Now, let's verify if the HCF and LCM of these two numbers are indeed 11 and 385.
The HCF of 77 and 55 is 11.
To find the LCM, we can use the formula:
LCM = (Product of the two numbers) / HCF
LCM = (77 x 55) / 11
LCM = 4235 / 11
LCM = 385
Therefore, the HCF and LCM of 77 and 55 are indeed 11 and 385, respectively.
Hence, the number that lies between 75 and 125 is 77, which is option C.
The HCF and LCM of two numbers is 70 and 1050 respectively. If the first number is 210, find the second one.- a)450
- b)350
- c)510
- d)260
- e)None of these
Correct answer is option 'B'. Can you explain this answer?
The HCF and LCM of two numbers is 70 and 1050 respectively. If the first number is 210, find the second one.
a)
450
b)
350
c)
510
d)
260
e)
None of these
|
|
Sagar Sharma answered |
Given:
The highest common factor (HCF) of two numbers is 70.
The least common multiple (LCM) of two numbers is 1050.
The first number is 210.
To find:
The second number.
Solution:
We know that the product of two numbers is equal to the product of their HCF and LCM.
So, we can write the equation as:
First number * Second number = HCF * LCM
Substituting the given values, we have:
210 * Second number = 70 * 1050
Simplifying the equation:
Second number = (70 * 1050) / 210
Second number = 350
Therefore, the second number is 350.
Answer:
The second number is 350.
The highest common factor (HCF) of two numbers is 70.
The least common multiple (LCM) of two numbers is 1050.
The first number is 210.
To find:
The second number.
Solution:
We know that the product of two numbers is equal to the product of their HCF and LCM.
So, we can write the equation as:
First number * Second number = HCF * LCM
Substituting the given values, we have:
210 * Second number = 70 * 1050
Simplifying the equation:
Second number = (70 * 1050) / 210
Second number = 350
Therefore, the second number is 350.
Answer:
The second number is 350.
HCF and LCM of two numbers is 5 and 275 respectively and the sum of these two numbers is 80. Find the sum of the reciprocals of these numbers- a)16/125
- b)32/275
- c)32/125
- d)16/275
- e)None of these
Correct answer is option 'D'. Can you explain this answer?
HCF and LCM of two numbers is 5 and 275 respectively and the sum of these two numbers is 80. Find the sum of the reciprocals of these numbers
a)
16/125
b)
32/275
c)
32/125
d)
16/275
e)
None of these
|
Pallabi Deshpande answered |
Let the 2 number be x and y
Given:
x + y = 80,
xy = 5*275 = 1375
sum of reciprocals = 1/x + 1/y = (x+y)/xy = 80/1375 = 16/275
A rectangular courtyard 140cm long 525cm wide is to be paved exactly with square tiles, all of the same size. what is the largest size of the tile which could be used for the purpose?- a)64cm
- b)35cm
- c)21cm
- d)28cm
- e)None of these
Correct answer is option 'B'. Can you explain this answer?
A rectangular courtyard 140cm long 525cm wide is to be paved exactly with square tiles, all of the same size. what is the largest size of the tile which could be used for the purpose?
a)
64cm
b)
35cm
c)
21cm
d)
28cm
e)
None of these
|
Gowri Chakraborty answered |
30 cm = 4 x 5 x 7
525 cm = 5 x 5 x 3 x 7
Hence common factors are 5 and 7
Hence LCM = 5 x 7 = 35
Find the least number which when divided by15, 21, 24 and 32 leaves the same remainder 2 in each case.- a)3362
- b)3360
- c)3456
- d)3266
- e)3364
Correct answer is option 'A'. Can you explain this answer?
Find the least number which when divided by15, 21, 24 and 32 leaves the same remainder 2 in each case.
a)
3362
b)
3360
c)
3456
d)
3266
e)
3364
|
|
Sagar Sharma answered |
Step 1: Find the LCM of 15, 21, 24, and 32
To find the least number that leaves a remainder of 2 when divided by 15, 21, 24, and 32, we need to find their least common multiple (LCM).
The prime factors of the numbers are:
- 15 = 3 * 5
- 21 = 3 * 7
- 24 = 2 * 2 * 2 * 3
- 32 = 2 * 2 * 2 * 2 * 2
Now, find the LCM by taking the highest power of each prime factor:
LCM = 2^5 * 3^1 * 5^1 * 7^1 = 3360
Step 2: Add the common remainder to the LCM
Since the number leaves a remainder of 2 when divided by each of the given numbers, we need to add this common remainder to the LCM calculated in step 1.
3360 + 2 = 3362
Therefore, the least number that leaves a remainder of 2 when divided by 15, 21, 24, and 32 is 3362, which corresponds to option A.
To find the least number that leaves a remainder of 2 when divided by 15, 21, 24, and 32, we need to find their least common multiple (LCM).
The prime factors of the numbers are:
- 15 = 3 * 5
- 21 = 3 * 7
- 24 = 2 * 2 * 2 * 3
- 32 = 2 * 2 * 2 * 2 * 2
Now, find the LCM by taking the highest power of each prime factor:
LCM = 2^5 * 3^1 * 5^1 * 7^1 = 3360
Step 2: Add the common remainder to the LCM
Since the number leaves a remainder of 2 when divided by each of the given numbers, we need to add this common remainder to the LCM calculated in step 1.
3360 + 2 = 3362
Therefore, the least number that leaves a remainder of 2 when divided by 15, 21, 24, and 32 is 3362, which corresponds to option A.
Find the least number which when divided by 12, 27 and 35 leaves 6 as a remainder?- a)3774
- b)3780
- c)3786
- d)4786
- e)None of these
Correct answer is option 'C'. Can you explain this answer?
Find the least number which when divided by 12, 27 and 35 leaves 6 as a remainder?
a)
3774
b)
3780
c)
3786
d)
4786
e)
None of these
|
Rajeev Kumar answered |
Number = LCM (12, 17, 35) + 6 = 3780 + 6 = 3786
A person has 3 bars whose length are 12,16,24m respectively. He want to cuts the longest possible pieces, all of the same length from each of the 3 bars, what is the length of the each piece, if he is cut without any wastage- a)12m
- b)20m
- c)6m
- d)4m
- e)None of these
Correct answer is option 'D'. Can you explain this answer?
A person has 3 bars whose length are 12,16,24m respectively. He want to cuts the longest possible pieces, all of the same length from each of the 3 bars, what is the length of the each piece, if he is cut without any wastage
a)
12m
b)
20m
c)
6m
d)
4m
e)
None of these
|
|
Aarav Sharma answered |
Solution:
To find the length of the longest possible pieces that can be cut from the three bars, we need to find the greatest common divisor (GCD) of the three lengths.
Step 1: Find the factors of the lengths of the bars
The factors of 12 are 1, 2, 3, 4, 6, and 12.
The factors of 16 are 1, 2, 4, 8, and 16.
The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.
Step 2: Find the common factors
The common factors of the three lengths are 1, 2, 4, and 6.
Step 3: Find the largest common factor
The largest common factor is 6. Therefore, the longest possible pieces that can be cut from each of the three bars will be 6 meters long.
Answer: Option D (6m)
To find the length of the longest possible pieces that can be cut from the three bars, we need to find the greatest common divisor (GCD) of the three lengths.
Step 1: Find the factors of the lengths of the bars
The factors of 12 are 1, 2, 3, 4, 6, and 12.
The factors of 16 are 1, 2, 4, 8, and 16.
The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.
Step 2: Find the common factors
The common factors of the three lengths are 1, 2, 4, and 6.
Step 3: Find the largest common factor
The largest common factor is 6. Therefore, the longest possible pieces that can be cut from each of the three bars will be 6 meters long.
Answer: Option D (6m)
In a college, all the students can stand in a row, so that each row has 9, 7 and 12 students. Find the least no of students in the school ?- a)145
- b)265
- c)186
- d)252
- e)None of these
Correct answer is option 'D'. Can you explain this answer?
In a college, all the students can stand in a row, so that each row has 9, 7 and 12 students. Find the least no of students in the school ?
a)
145
b)
265
c)
186
d)
252
e)
None of these
|
|
Aarav Sharma answered |
Approach:
- The least number of students in the school will be the LCM of 9, 7, and 12.
- We can find the LCM by prime factorizing the numbers and taking the highest power of each prime factor.
- Once we have the LCM, we can check if it satisfies the given conditions.
Solution:
- Prime factorizing 9: 3^2
- Prime factorizing 7: 7
- Prime factorizing 12: 2^2 * 3
- Taking the highest power of each prime factor: 2^2 * 3^2 * 7 = 252
- So the least number of students in the school is 252.
- We can check if 252 satisfies the given conditions:
- 252/9 = 28 (rows of 9 students)
- 252/7 = 36 (rows of 7 students)
- 252/12 = 21 (rows of 12 students)
- So 252 satisfies the given conditions.
- Therefore, the correct answer is option D) 252.
- The least number of students in the school will be the LCM of 9, 7, and 12.
- We can find the LCM by prime factorizing the numbers and taking the highest power of each prime factor.
- Once we have the LCM, we can check if it satisfies the given conditions.
Solution:
- Prime factorizing 9: 3^2
- Prime factorizing 7: 7
- Prime factorizing 12: 2^2 * 3
- Taking the highest power of each prime factor: 2^2 * 3^2 * 7 = 252
- So the least number of students in the school is 252.
- We can check if 252 satisfies the given conditions:
- 252/9 = 28 (rows of 9 students)
- 252/7 = 36 (rows of 7 students)
- 252/12 = 21 (rows of 12 students)
- So 252 satisfies the given conditions.
- Therefore, the correct answer is option D) 252.
The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:- a)30
- b)22
- c)40
- d)60
- e)None of these
Correct answer is option 'C'. Can you explain this answer?
The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:
a)
30
b)
22
c)
40
d)
60
e)
None of these
|
|
Anaya Patel answered |
Let the numbers be 2x and 3x.
Then, their L.C.M. = 6x.
So, 6x = 48 or x = 8.
The numbers are 16 and 24.
Hence, required sum = (16 + 24) = 40.
Then, their L.C.M. = 6x.
So, 6x = 48 or x = 8.
The numbers are 16 and 24.
Hence, required sum = (16 + 24) = 40.
The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:- a)124
- b)100
- c)111
- d)175
- e)None of these
Correct answer is option 'C'. Can you explain this answer?
The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
a)
124
b)
100
c)
111
d)
175
e)
None of these
|
|
Aarav Sharma answered |
The given question is related to finding the greater number out of two numbers whose product is given and the highest common factor (HCF) is also given. Let's solve it step by step.
Given:
Product of two numbers = 4107
HCF of the two numbers = 37
Step 1: Prime Factorization of 4107
To find the prime factorization of 4107, we divide it by prime numbers starting from 2 until we cannot divide any further. The prime factors of 4107 are:
4107 ÷ 3 = 1369
1369 ÷ 37 = 37
So, the prime factorization of 4107 is 3 × 37 × 37.
Step 2: Determining the Numbers
Now that we have the prime factorization, we can determine the two numbers whose product is 4107 and whose HCF is 37. Since the HCF is 37, one of the numbers should be a multiple of 37. Let's assign the factors as follows:
Number 1 = 37 × a
Number 2 = 37 × b
Multiplying the two numbers, we get:
Number 1 × Number 2 = (37 × a) × (37 × b) = 4107
Expanding the equation:
(37 × 37) × (a × b) = 4107
Simplifying:
1369 × (a × b) = 4107
Dividing by 1369 on both sides:
a × b = 3
Step 3: Finding the Greater Number
We need to find the greater number out of the two. To do this, we need to consider the values of a and b. Since a and b are positive integers and their product is 3, the possible values for a and b are:
a = 1, b = 3
a = 3, b = 1
Substituting the values back into the equation, we get the two numbers:
Number 1 = 37 × a = 37 × 1 = 37
Number 2 = 37 × b = 37 × 3 = 111
Comparing the two numbers, we can see that the greater number is 111.
Therefore, the correct answer is option C) 111.
Given:
Product of two numbers = 4107
HCF of the two numbers = 37
Step 1: Prime Factorization of 4107
To find the prime factorization of 4107, we divide it by prime numbers starting from 2 until we cannot divide any further. The prime factors of 4107 are:
4107 ÷ 3 = 1369
1369 ÷ 37 = 37
So, the prime factorization of 4107 is 3 × 37 × 37.
Step 2: Determining the Numbers
Now that we have the prime factorization, we can determine the two numbers whose product is 4107 and whose HCF is 37. Since the HCF is 37, one of the numbers should be a multiple of 37. Let's assign the factors as follows:
Number 1 = 37 × a
Number 2 = 37 × b
Multiplying the two numbers, we get:
Number 1 × Number 2 = (37 × a) × (37 × b) = 4107
Expanding the equation:
(37 × 37) × (a × b) = 4107
Simplifying:
1369 × (a × b) = 4107
Dividing by 1369 on both sides:
a × b = 3
Step 3: Finding the Greater Number
We need to find the greater number out of the two. To do this, we need to consider the values of a and b. Since a and b are positive integers and their product is 3, the possible values for a and b are:
a = 1, b = 3
a = 3, b = 1
Substituting the values back into the equation, we get the two numbers:
Number 1 = 37 × a = 37 × 1 = 37
Number 2 = 37 × b = 37 × 3 = 111
Comparing the two numbers, we can see that the greater number is 111.
Therefore, the correct answer is option C) 111.
Four bells commence tolling together and toll at the intervals of 3,9,12,15 seconds resp. In 60 minutes how many times they will toll together.- a)20
- b)21
- c)24
- d)30
- e)None of these
Correct answer is option 'B'. Can you explain this answer?
Four bells commence tolling together and toll at the intervals of 3,9,12,15 seconds resp. In 60 minutes how many times they will toll together.
a)
20
b)
21
c)
24
d)
30
e)
None of these
|
Ishani Rane answered |
LCM of 3, 9, 12, 15 = 180s = 3m
In 60m = 60/3 = 20=> 20+1 = 21
Three boxes of lengths 60m,30m and 45m are to be cut into pieces of equal length. What is the greatest possible length of each pieces ?- a)15m
- b)18m
- c)5m
- d)3m
- e)None of these
Correct answer is option 'A'. Can you explain this answer?
Three boxes of lengths 60m,30m and 45m are to be cut into pieces of equal length. What is the greatest possible length of each pieces ?
a)
15m
b)
18m
c)
5m
d)
3m
e)
None of these
|
|
Sagar Sharma answered |
Explanation:
- To find the greatest possible length of each piece, we need to find the greatest common divisor (GCD) of the lengths of the boxes.
- The lengths of the boxes are 60m, 30m, and 45m.
- We can find the GCD of these lengths by finding the GCD of the pairwise combinations of the lengths.
- GCD(60,30) = 30
- GCD(30,45) = 15
- GCD(60,45) = 15
- Therefore, the greatest common divisor of 60, 30, and 45 is 15m.
- So, the greatest possible length of each piece that can be cut from the boxes is 15m.
Therefore, the correct answer is option A) 15m.
- To find the greatest possible length of each piece, we need to find the greatest common divisor (GCD) of the lengths of the boxes.
- The lengths of the boxes are 60m, 30m, and 45m.
- We can find the GCD of these lengths by finding the GCD of the pairwise combinations of the lengths.
- GCD(60,30) = 30
- GCD(30,45) = 15
- GCD(60,45) = 15
- Therefore, the greatest common divisor of 60, 30, and 45 is 15m.
- So, the greatest possible length of each piece that can be cut from the boxes is 15m.
Therefore, the correct answer is option A) 15m.
The HCF of 2511 and 3402 is- a)31
- b)42
- c)76
- d)81
- e)None of these
Correct answer is option 'D'. Can you explain this answer?
The HCF of 2511 and 3402 is
a)
31
b)
42
c)
76
d)
81
e)
None of these
|
|
Alok Verma answered |
2511 = 81×31
3402 = 81×42
Hence HCF is 81
3402 = 81×42
Hence HCF is 81
Two containers contain 50 and 125 litres of water respectively. Find the maximum capacity of a container which can measure the water in each container an exact number of times(in litres)- a)25
- b)11
- c)12
- d)15
- e)None of these
Correct answer is option 'A'. Can you explain this answer?
Two containers contain 50 and 125 litres of water respectively. Find the maximum capacity of a container which can measure the water in each container an exact number of times(in litres)
a)
25
b)
11
c)
12
d)
15
e)
None of these
|
Arya Roy answered |
The maximum capacity of a container which can measure the water in each container an exact number of times will be the HCF of the capacity of the two containers:
HCF of 50 and 125 = 25
A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds, all starting at the same point. After what time will they meet again at the starting point?- a)15 minutes 15 seconds
- b)42 minutes 30 seconds
- c)42 minutes
- d)46 minutes 12 seconds
- e)None of these
Correct answer is option 'D'. Can you explain this answer?
A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds, all starting at the same point. After what time will they meet again at the starting point?
a)
15 minutes 15 seconds
b)
42 minutes 30 seconds
c)
42 minutes
d)
46 minutes 12 seconds
e)
None of these
|
|
Nikita Singh answered |
L.C.M. of 252, 308 and 198 = 2772.So, A, B and C will again meet at the starting point in 2772 see i.e., 46 min. 12 sec
The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 50, 62, 98 and 130 as remainders respectively, is:- a)11115
- b)15110
- c)15130
- d)15310
- e)None of these
Correct answer is option 'B'. Can you explain this answer?
The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 50, 62, 98 and 130 as remainders respectively, is:
a)
11115
b)
15110
c)
15130
d)
15310
e)
None of these
|
|
Varun Rane answered |
Here (48 – 38) = 10, (60 – 50) = 10, (72 – 62) = 10, (108 – 98) = 10 & (140 – 130) = 10.
Required number = (L.C.M. of 48, 60, 72, 108, 140) – 10
= 15120 – 10 = 15110
Required number = (L.C.M. of 48, 60, 72, 108, 140) – 10
= 15120 – 10 = 15110
HCF and LCM of two numbers is 5 and 275 respectively and the sum of these two numbers is 80. Find the sum of the reciprocals of these numbers.- a)15/242
- b)16/275
- c)32/221
- d)20/268
- e)None of these
Correct answer is option 'B'. Can you explain this answer?
HCF and LCM of two numbers is 5 and 275 respectively and the sum of these two numbers is 80. Find the sum of the reciprocals of these numbers.
a)
15/242
b)
16/275
c)
32/221
d)
20/268
e)
None of these
|
|
Sagar Sharma answered |
Understanding the Problem
To solve for the two numbers given their HCF, LCM, and sum, we can use the relationship between HCF, LCM, and the product of the two numbers.
Key Relationships
- HCF (Highest Common Factor) = 5
- LCM (Least Common Multiple) = 275
- Sum of the two numbers = 80
Finding the Product of the Numbers
Using the relationship:
- HCF × LCM = Product of the two numbers
Therefore, we calculate:
- 5 × 275 = 1375
Let the two numbers be x and y. Hence, we have:
- x + y = 80
- xy = 1375
Forming the Quadratic Equation
From the equations, we can form:
- t^2 - (x+y)t + xy = 0
- t^2 - 80t + 1375 = 0
Calculating the Roots
Using the quadratic formula:
- t = [80 ± sqrt((80)^2 - 4 × 1 × 1375)] / 2
Calculating the discriminant:
- (80)^2 - 4 × 1375 = 6400 - 5500 = 900
Thus, we have:
- t = [80 ± 30] / 2
This gives us:
- t = 55 and t = 25
So, the two numbers are 55 and 25.
Finding the Sum of the Reciprocals
The sum of the reciprocals of the numbers is given by:
- (1/x) + (1/y) = (y + x) / (xy)
Substituting the values:
- (55 + 25) / (55 × 25) = 80 / 1375 = 16 / 275
Conclusion
The sum of the reciprocals of the two numbers is:
- 16/275
Thus, the correct answer is option 'B'.
To solve for the two numbers given their HCF, LCM, and sum, we can use the relationship between HCF, LCM, and the product of the two numbers.
Key Relationships
- HCF (Highest Common Factor) = 5
- LCM (Least Common Multiple) = 275
- Sum of the two numbers = 80
Finding the Product of the Numbers
Using the relationship:
- HCF × LCM = Product of the two numbers
Therefore, we calculate:
- 5 × 275 = 1375
Let the two numbers be x and y. Hence, we have:
- x + y = 80
- xy = 1375
Forming the Quadratic Equation
From the equations, we can form:
- t^2 - (x+y)t + xy = 0
- t^2 - 80t + 1375 = 0
Calculating the Roots
Using the quadratic formula:
- t = [80 ± sqrt((80)^2 - 4 × 1 × 1375)] / 2
Calculating the discriminant:
- (80)^2 - 4 × 1375 = 6400 - 5500 = 900
Thus, we have:
- t = [80 ± 30] / 2
This gives us:
- t = 55 and t = 25
So, the two numbers are 55 and 25.
Finding the Sum of the Reciprocals
The sum of the reciprocals of the numbers is given by:
- (1/x) + (1/y) = (y + x) / (xy)
Substituting the values:
- (55 + 25) / (55 × 25) = 80 / 1375 = 16 / 275
Conclusion
The sum of the reciprocals of the two numbers is:
- 16/275
Thus, the correct answer is option 'B'.
Naren, Suraj and Praveen start running around a circular stadium and complete one round in 12s, 9s and 15s respectively. In how much time will they meet again at the starting point ?- a)2m
- b)2.30m
- c)3m
- d)3.30m
- e)None of these
Correct answer is option 'C'. Can you explain this answer?
Naren, Suraj and Praveen start running around a circular stadium and complete one round in 12s, 9s and 15s respectively. In how much time will they meet again at the starting point ?
a)
2m
b)
2.30m
c)
3m
d)
3.30m
e)
None of these
|
Akash Pandey answered |
LCM of 12, 9, 15 =180s = 3m
Two baskets contain 183 and 242 Apples respectively, which are distributed in equal number among children. Find the largest number of apples that can be given, so that 3 apples are left over from the first basket and 2 from the second.- a)45
- b)40
- c)60
- d)56
- e)None of these
Correct answer is option 'C'. Can you explain this answer?
Two baskets contain 183 and 242 Apples respectively, which are distributed in equal number among children. Find the largest number of apples that can be given, so that 3 apples are left over from the first basket and 2 from the second.
a)
45
b)
40
c)
60
d)
56
e)
None of these
|
Sameer Rane answered |
183-3 = 180
142-2 = 240
HCF of 240, 180 = 60
Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together?- a)8
- b)11
- c)13
- d)16
- e)None of these
Correct answer is option 'D'. Can you explain this answer?
Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together?
a)
8
b)
11
c)
13
d)
16
e)
None of these
|
|
Pallabi Deshpande answered |
LCM of 2, 4, 6, 8 10 and 12 is 120.
So, after each 120 seconds, they would toll together.
Hence, in 30 minutes, they would toll 30*60 seconds / 120 seconds = 15 times
But then the question says they commence tolling together. So, they basically also toll at the "beginning" ("0" second).
So, total tolls together = 15+1 = 16
What is the HCF of 3/5, 7/10, 2/15, 6/21?- a)5/123
- b)6/121
- c)2/105
- d)1/210
- e)None of these
Correct answer is option 'D'. Can you explain this answer?
What is the HCF of 3/5, 7/10, 2/15, 6/21?
a)
5/123
b)
6/121
c)
2/105
d)
1/210
e)
None of these
|
|
Nikita Singh answered |
HCF will be HCF of numerators/LCM of denominators
So HCF = HCF of 3,7,2,6)/(LCM of 5,10, 15, 21)
= 1/210
So HCF = HCF of 3,7,2,6)/(LCM of 5,10, 15, 21)
= 1/210
Find the least number which when divided by 10, 15, 18 and 30 leaves remainders 6, 11, 14 and 26 respectively.- a)72
- b)86
- c)85
- d)90
- e)94
Correct answer is option 'B'. Can you explain this answer?
Find the least number which when divided by 10, 15, 18 and 30 leaves remainders 6, 11, 14 and 26 respectively.
a)
72
b)
86
c)
85
d)
90
e)
94
|
|
Aisha Gupta answered |
Since 10-6 = 4, 15-11 = 4, 18-14 = 4, 30-26 = 4
So answer will be the LCM of these numbers – 4
So 90 – 4
So answer will be the LCM of these numbers – 4
So 90 – 4
In a college all the students are made to stand in four rows. 4 rows contains 12, 8, 22, 30 students respectively. Find the least number of students in the college?- a)3360
- b)3630
- c)3960
- d)3990
- e)None of these
Correct answer is option 'C'. Can you explain this answer?
In a college all the students are made to stand in four rows. 4 rows contains 12, 8, 22, 30 students respectively. Find the least number of students in the college?
a)
3360
b)
3630
c)
3960
d)
3990
e)
None of these
|
|
Alok Verma answered |
LCM (12, 8, 22, 30) = 3960
What will be the least number which when doubled will be exactly divisible by 12,18,21 and 30?- a)630
- b)360
- c)603
- d)306
- e)None of these
Correct answer is option 'A'. Can you explain this answer?
What will be the least number which when doubled will be exactly divisible by 12,18,21 and 30?
a)
630
b)
360
c)
603
d)
306
e)
None of these
|
|
Rajeev Kumar answered |
LCM of 12,18,21 and 30 = 1260
Doubled (divide by 2) = 1260/2 = 630
Doubled (divide by 2) = 1260/2 = 630
If the L.C.M of x and y is z, their H.C.F is- a)xy/z
- b)xyz
- c)(x + y)/z
- d)z/xy
- e)None of these
Correct answer is option 'A'. Can you explain this answer?
If the L.C.M of x and y is z, their H.C.F is
a)
xy/z
b)
xyz
c)
(x + y)/z
d)
z/xy
e)
None of these
|
|
Yash Patel answered |
HCF = Product of the 2 number/LCM = XY/Z
The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:- a)68
- b)98
- c)180
- d)364
- e)None of these
Correct answer is option 'D'. Can you explain this answer?
The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
a)
68
b)
98
c)
180
d)
364
e)
None of these
|
|
Chirag Nair answered |
L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
Required number = (90 x 4) + 4 = 364.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
Required number = (90 x 4) + 4 = 364.
The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:- a)10
- b)14
- c)23
- d)30
- e)None of these
Correct answer is option 'C'. Can you explain this answer?
The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:
a)
10
b)
14
c)
23
d)
30
e)
None of these
|
|
Rajesh Khatri answered |
L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
L.C.M. of 5, 6, 4 and 3 = 60. Number to be added = (60 - 37) = 23.
Riya, Anil and Rishi start running around a circular stadium and complete one round in 15s, 12s and 21s respectively. In how much time will they meet again at the starting point?- a)6min
- b)7min
- c)8min
- d)9min
- e)None of these
Correct answer is option 'B'. Can you explain this answer?
Riya, Anil and Rishi start running around a circular stadium and complete one round in 15s, 12s and 21s respectively. In how much time will they meet again at the starting point?
a)
6min
b)
7min
c)
8min
d)
9min
e)
None of these
|
|
Akash Pandey answered |
LCM (15, 12, 21) = 420 second = 7 minutes
Chapter doubts & questions for LCM and HCF - RRB JE Mock Test Series for Civil Engineering (CE) 2025 2025 is part of Civil Engineering (CE) exam preparation. The chapters have been prepared according to the Civil Engineering (CE) exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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