HCF & LCM

PrepBank · পাতা ১ / ৩ · ১–১০০ / ২২৭

১.

The sum of the L.C.M. and H.C.F. of two numbers is 1260, and the L.C.M. is 900 more than the H.C.F. What is the product of these two numbers?

ক
194800
খ
194000
গ
149400
ঘ
194400

ব্যাখ্যা

Question: The sum of the L.C.M. and H.C.F. of two numbers is 1260, and the L.C.M. is 900 more than the H.C.F. What is the product of these two numbers?

Solution:
Let the HCF be x

LCM = HCF + 900
LCM = x + 900 ...............(1)

And, LCM + HCF = 1260
LCM + x = 1260 .................(2)

From (1) and (2) equation,
(x + 900) + x = 1260
⇒ 2x + 900 = 1260
⇒ 2x = 1260 - 900
⇒ 2x = 360
⇒ x = 360/2
∴ x = 180

∴ HCF = 180
And, from equation (1),
LCM = HCF + 900
LCM = 180 + 900
∴ LCM = 1080

By formula, the product of the numbers is equal to the product of their HCF and LCM.

Product of numbers = HCF × LCM = 180 × 1080
∴ Product = 194400

২.

HCF and LCM of two fractions is 1/35 and 15/1, if one fraction is 3/5, then the second fraction is:

ক
3/7
খ
7/3
গ
5/7
ঘ
7/5

ব্যাখ্যা

Question: HCF and LCM of two fractions is 1/35 and 15/1, if one fraction is 3/5, then the second fraction is:

Solution:
Given,
HCF of two fractions = 1/35
LCM of two fractions = 15/1
One fraction = 3/5

HCF of fractions = HCF of numerators/LCM of denominators
LCM of fractions = LCM of numerators/HCF of denominators
LCM × HCF = Product of two fractions

Let second fraction be a/b,
then
a/b × 3/5 = 15/1 × 1/35
⇒ a/b = 15/1 × 1/35 × 5/3
∴ a/b = 5/7

৩.

What is the greatest 4-digit number which is divisible by 15, 25, 40, and 75?

ক
9000
খ
9400
গ
9600
ঘ
9900

ব্যাখ্যা

Question: What is the greatest 4-digit number which is divisible by 15, 25, 40, and 75?

Solution:
Greatest number of 4-digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
∴ Required number (9999 - 399) = 9600.

৪.

The HCF and LCM of the two numbers are 21 and 84 respectively. If the ratio of the two numbers is 1 : 4 then the larger of the two numbers is =?

ক
72
খ
84
গ
96
ঘ
100

ব্যাখ্যা

Question: The HCF and LCM of the two numbers are 21 and 84 respectively. If the ratio of the two numbers is 1 : 4 then the larger of the two numbers is =?

Solution:
We know,
LCM × HCF = 1st number × 2nd number

Let, 1st number = P
2nd number = 4P

ATQ,
P × 4P = 21 × 84
⇒ 4P² = 21 × 84
⇒ P² = 21 × 21
∴ P = 21

Then, the 1st number = 21
2nd Number = 4 × 21 = 84

So, the larger number = 84

৫.

The product of two numbers is 2028 and their HCF is 13. The number of such pairs is =?

ক
4
খ
3
গ
2
ঘ
1

ব্যাখ্যা

Question: The product of two numbers is 2028 and their HCF is 13. The number of such pairs is =?

Solution:
Let the two numbers be x and y respectively.

It is given that the product of the two numbers is 2028, therefore,
xy = 2028

Also, 13 is their HCF, thus both numbers must be divisible by 13.

So, let x = 13a and y = 13b, then,
13a × 13b = 2028
⇒ 169ab = 2028
⇒ ab = 2028
∴ ab = 12

Therefore, the required possible pair of values of x and y which are prime to each other are (1, 12) and (3, 4).
Thus, the required numbers are (12, 156) and (39, 52).

Hence, the number of possible pairs is 2.

৬.

Find the LCM of the fractions 2/3, 5/9, 1/3.

ক
10/3
খ
10/9
গ
1/3
ঘ
1/9

ব্যাখ্যা

Question: Find the LCM of the fractions 2/3, 5/9, 1/3.

Solution:
Given,
the fractions are 2/3, 5/9 and 1/3

We know,
LCM of the fraction = LCM of numerator/HCF of denominator

LCM of numerators = LCM of (2, 5, 1) = 2 × 5 × 1 = 10
HCF of denominators = HCF of ( 3, 9, 3) = 3

∴ LCM of fraction = 10/3

৭.

H.C.F. of 513, 1134 and 1215 is-

ক
18
খ
27
গ
33
ঘ
36

ব্যাখ্যা

Question: H.C.F. of 513, 1134 and 1215 is-

Solution:

H.C.F. of 513, 1134 and 1215 = 3 × 3 × 3 = 27

৮.

The LCM and HCF of two numbers are 495 and 5. If the sum of the numbers is 100, what is the difference of the numbers?

ক
10
খ
15
গ
20
ঘ
5

ব্যাখ্যা

Question: The LCM and HCF of two numbers are 495 and 5. If the sum of the numbers is 100, what is the difference of the numbers?

Solution:
let,
one number is x
∴ another number is (100 - x)

ATQ,
x(100 - x) = 5 × 495
⇒ 100x - x² = 2475
⇒ x² - 100x + 2475 = 0
⇒ x² - 55x - 45x + 2475 = 0
⇒ x(x - 55) - 45(x - 55) = 0
⇒ (x - 55)(x - 45) = 0
∴ x = 45, 55

∴ the numbers are 45 and 55. difference = 55 - 45 = 10

৯.

Four bells ringing together and ring at an interval of 12 sec, 15 sec, 20 sec, and 30 sec respectively. How many times will they ring together in 8 hours?

ক
481
খ
480
গ
482
ঘ
483

ব্যাখ্যা

Question: Four bells ringing together and ring at an interval of 12 sec, 15 sec, 20 sec, and 30 sec respectively. How many times will they ring together in 8 hours?

Solution:
Four bells ringing timing is 12 sec, 15 sec, 20 sec,30 sec
Now we have to take LCM of time interval ⇒ LCM of (12, 15, 20, 30) = 60
Total seconds in 8 hours = 8 × 3600 = 28800

Number of times bell rings = 28800/60
⇒ Number of times bell rings = 480

If four bells ring together in starting ⇒ 480 + 1
∴ The bell ringing 481 times in 8 hours.

Mistake Points: The bells start tolling together, the first toll also needs to be counted, that is the number of times of tolling since the first time.

১০.

Find the greatest number that exactly divides each of the numbers 36, 60, and 84.

ক
6
খ
8
গ
12
ঘ
18

ব্যাখ্যা

Question: Find the greatest number that exactly divides each of the numbers 36, 60, and 84.

Solution:
We know,
The HCF (Highest Common Factor) of two or more numbers is the greatest number that divides each of them exactly.

Now,
Prime factorization of 36 = 2 × 2 × 3 × 3

Prime factorization of 60 = 2 × 2 × 3 × 5

Prime factorization of 84 = 2 × 2 × 3 × 7

∴ HCF of 36, 60, and 84 = 2 × 2 × 3 = 12

Therefore, the greatest number is 12.

১১.

The least number by which 108 must be multiplied to make it a perfect square is-

ক
2
খ
5
গ
3
ঘ
4

ব্যাখ্যা

Question: The least number by which 108 must be multiplied to make it a perfect square is-

Solution:
একটি সংখ্যা পূর্ণবর্গ সংখ্যা হতে হলে তার মৌলিক গুণনীয়কগুলোকে অবশ্যই জোড় সংখ্যায় (even power) থাকতে হবে।

108 = 2 × 2 × 3 × 3 × 3
= 2² × 3³
জোড়া গঠন করে পাই,(2 × 2) × (3 × 3) × 3
এখানে জোড়া বিহীন সংখ্যা 3

∴ 108 কে 3 দ্বারা গুণ করলে এটি পূর্ণবর্গ সংখ্যা হবে।

১২.

If the ratio of two numbers is 5 : 8, and their Least Common Multiple is 200, what are the two numbers?

ক
20, 32
খ
15, 24
গ
25, 40
ঘ
30, 48

ব্যাখ্যা

Question: If the ratio of two numbers is 5 : 8, and their Least Common Multiple is 200, what are the two numbers?

Solution:
ধরি,
সংখ্যা দুইটি যথাক্রমে 5x, 8x
5x, 8x এর লসাগু = 40x

প্রশ্নমতে,
40x = 200
⇒ x = 200/40
∴ x = 5

∴ সংখ্যা দুইটি যথাক্রমে = 5 × 5 = 25, 8 × 5 = 40

১৩.

The HCF and LCM of the two numbers are 21 and 84 respectively. If the ratio of the two numbers is 1 : 4 then the larger of the two numbers is =?

ক
24
খ
48
গ
84
ঘ
96

ব্যাখ্যা

Question: The HCF and LCM of the two numbers are 21 and 84 respectively. If the ratio of the two numbers is 1 : 4 then the larger of the two numbers is =?

Solution:
We know,
LCM × HCF = 1^st number × 2^nd number

Let 1^st number = P
2^nd number = 4P

P × 4P = 21 × 84
⇒ 4P² = 21 × 84
⇒ P² = 21 × 21
∴ P = 21

Then, the numbers = 21, 84
So, the larger number = 84

১৪.

HCF and LCM of two numbers are 7 and 140 respectively. If the numbers are between 20 and 45, the sum of the numbers is:

ক
52
খ
56
গ
60
ঘ
63
ঙ
66

ব্যাখ্যা

Question: HCF and LCM of two numbers are 7 and 140 respectively. If the numbers are between 20 and 45, the sum of the numbers is:

Solution:
Let, the numbers be 7x and 7y
where x and y are co-prime.

Now, LCM of = 7x
and 7y = 7xy

∴ 7xy = 140
⇒ xy = 140/7
⇒ xy = 20

Now, required values of x and y whose product is 50 and are coprime, will be 4 and 5.
∴ Numbers are 28 and 35 which lie between 20 and 45.

∴ Required sum = (28 + 35)
= 63

১৫.

The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

ক
286
খ
300
গ
322
ঘ
346

ব্যাখ্যা

Question: The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

Solution:
Let, the two numbers be 23 x and 23 y.
Since the other two factors of LCM are 13 and 14, the values of x and y are 13 and 14, respectively.
So, the largest number is 23y,
23y = 23 × 14 = 322

১৬.

Find the smallest number of five digits exactly divisible by 16, 24, 36, and 54.

ক
9632
খ
10362
গ
10638
ঘ
10368

ব্যাখ্যা

Question: Find the smallest number of five digits exactly divisible by 16, 24, 36, and 54.

Solution:
The smallest number of five digits is = 10000
The number must be divisible by the LCM of 16, 24, 36, and 54 = 432
On dividing 10000 by 432, we get 64 as the remainder.

So, required number is = 10000 + (432 - 64)
= 10000 + 368
= 10368

১৭.

Sum of two numbers is 384. H.C.F. of the numbers is 48. The difference of the numbers is-

ক
336
খ
288
গ
192
ঘ
100

ব্যাখ্যা

Question: Sum of two numbers is 384. H.C.F. of the numbers is 48. The difference of the numbers is-

Solution:
Let, the numbers be 48x and 48y
where x and y are co-primes.

ATQ,
48x + 48y = 384
⇒ 48 ( x + y) = 384
⇒ x + y = 384/48
⇒ x + y = 8 ........... (i)

Possible and acceptable pairs of x and y satisfying this condition are: (1, 7) and (3, 5).

∴ Numbers are = 48 × 1 = 48
and 48 × 7 = 336
Again 48 × 3 = 144
and 48 × 5 = 240

∴ Required difference = 336 - 48
= 288

১৮.

A craft store has ribbons of lengths 36 cm, 48 cm, and 60 cm. What is the longest possible length that can evenly cut all ribbons without waste?

ক
15 cm
খ
18 cm
গ
9 cm
ঘ
12 cm

ব্যাখ্যা

Question: A craft store has ribbons of lengths 36 cm, 48 cm, and 60 cm. What is the longest possible length that can evenly cut all ribbons without waste?

Solution:
Prime factorization of
36 = 2 × 2 × 3 × 3
48 = 2 × 2 × 2 × 2 × 3
60 = 2 × 2 × 3 × 5

∴ HCF = 2 × 2 × 3 = 12

So, the longest ribbon length that can evenly cut all three without waste is 12 cm.

১৯.

Two alarm clocks ring their alarms at regular intervals of 40 seconds and 60 seconds. If they first beep together at 10:30 AM, when will they next beep together for the first time?

ক
10 : 32 : 24 AM
খ
10 : 31 AM
গ
10 : 31 : 24 AM
ঘ
10 : 32 AM

ব্যাখ্যা

Question: Two alarm clocks ring their alarms at regular intervals of 40 seconds and 60 seconds. If they first beep together at 10:30 AM, when will they next beep together for the first time?

Solution:
They will ring together after the LCM of 40 and 60 seconds.
40 = 2 × 2 × 2 × 5
60 = 2 × 2 × 3 × 5
∴ LCM = 2 × 2 × 2 × 3 × 5 = 120 seconds
= 120 ÷ 60 = 2 minutes [ 1 min = 60 sec. ]

∴ They will beep together again at 10:30 AM + 2 minutes = 10:32 AM

২০.

What is the largest four-digit number that is exactly divisible by 15, 20, 25, and 30?

ক
9900
খ
9936
গ
9945
ঘ
9972

ব্যাখ্যা

Question: What is the largest four-digit number that is exactly divisible by 15, 20, 25, and 30?

Solution:
Greatest 4-digit number is 9999.
L.C.M. of 15, 20, 25 and 30 = 300.

On dividing 9999 by 300 the remainder is 99
(because 300 × 33 = 9900 and 9999 - 9900 = 99).

∴ Required number = 9999 - 99 = 9900.

২১.

What is the smallest number divisible by 3, 5 and 9 with remainder 2?

ক
135
খ
17
গ
45
ঘ
47

ব্যাখ্যা

Question: What is the smallest number divisible by 3, 5 and 9 with remainder 2?

Solution:
প্রদত্ত সংখ্যাগুলো দ্বারা বিভাজ্য ক্ষুদ্রতম সংখ্যা হবে সংখ্যাগুলোর ল. সা. গু।
তাহলে, ৩, ৫, ৯ এর ল. সা. গু এর সাথে ২ যোগ করলে ক্ষুদ্রতম সংখ্যাটি পাওয়া যাবে।

এখন,
৩ = ৩ × ১
৫ = ৫ × ১
৯ = ৩ × ৩

∴ ৩, ৫, ৯ এর ল. সা. গু = ৩ × ৩ × ৫ = ৪৫

নির্ণেয় ক্ষুদ্রতম সংখ্যা = (৪৫ + ২) = ৪৭

২২.

The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is-

ক
276
খ
299
গ
322
ঘ
345

ব্যাখ্যা

Question: The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is-

Solution:
Clearly, the numbers are (23 × 13) and (23 × 14).

∴ Larger number = (23 × 14) = 322.

২৩.

Six bells commence tolling together and toll at intervals of 3, 5, 6, 9, 10, and 15 seconds respectively. In 45 minutes, how many times do they toll together?

ক
29
খ
31
গ
33
ঘ
39

ব্যাখ্যা

Question: Six bells commence tolling together and toll at intervals of 3, 5, 6, 9, 10, and 15 seconds respectively. In 45 minutes, how many times do they toll together?

Solution:
3 = 3¹
5 = 5¹
6 = 2 × 3
9 = 3²
10 = 2 × 5
15 = 3 × 5

∴ ল.সা.গু. = 2¹ × 3² × 5¹
= 2 × 9 × 5
= 90

সুতরাং, ঘণ্টাগুলো প্রতি 90 সেকেন্ড পর পর একসাথে বাজবে।

এখন,
45 মিনিট = 45 × 60 = 2700 সেকেন্ড

মোট 2700 সেকেন্ডে ঘণ্টাগুলো যতবার একসাথে বাজবে = 2700/90 = 30 বার

যেহেতু ঘণ্টাগুলো প্রথমে একবার একসাথে বাজা শুরু করেছিল, তাই মোট সংখ্যাটি হবে 30 এর সাথে সেই প্রথমবারটি যোগ করে।
∴ মোট সংখ্যা = 30 + 1 = 31 বার।

সুতরাং, 45 মিনিটে ঘণ্টাগুলো মোট 31 বার একসাথে বাজবে।

২৪.

Find L.C.M. of 2/3, 8/9, 64/81, 10/27.

ক
250/9
খ
160/3
গ
128/9
ঘ
320/3

ব্যাখ্যা

Question: Find L.C.M. of 2/3, 8/9, 64/81, 10/27.

Solution:
L.C.M. = L.C.M. of Numerator/H.C.F. of Denominator

L.C.M. of numerators (2, 8, 64, 10)
2 = 2¹
8 = 2³
64 = 2⁶
10 = 2 × 5

L.C.M of 2, 8, 64, 10 = 2⁶ × 5 = 320

H.C.F. of denominators = 3, 9, 81, 27
3 = 3¹
9 = 3²
81 = 3⁴
27 = 3³

H.C.F. of 3, 9, 81, 27 = 3

∴ L.C.M. of 2/3, 8/9, 64/81, 10/27 = 320/3

২৫.

The sum of L.C.M. and H.C.F. of two numbers is 1260. If their L.C.M. is 900 more than their H.C.F., find the product of two numbers.

ক
203400
খ
194400
গ
198400
ঘ
205400

ব্যাখ্যা

Question: The sum of L.C.M. and H.C.F. of two numbers is 1260. If their L.C.M. is 900 more than their H.C.F., find the product of two numbers.

Solution:
Let the HCF be x

LCM = HCF + 900
LCM = x + 900 ...............(1)

And,
LCM + HCF = 1260
LCM + x = 1260 .................(2)

From (1) and (2) equation,
(x + 900) + x = 1260
⇒ 2x + 900 = 1260
⇒ 2x = 1260 - 900
⇒ 2x = 360
⇒ x = 360/2
⇒ x = 180

∴ HCF = 180

And, from equation (1),
LCM = HCF + 900
LCM = 180 + 900
∴ LCM = 1080

By formula, the product of the numbers is equal to the product of their HCF and LCM.

Product of numbers = HCF × LCM
= 180 × 1080
∴ Product = 194400

২৬.

Two numbers have a product of 2028 and a highest common factor (HCF) of 13. How many such pairs of numbers exist?

ক
1
খ
2
গ
3
ঘ
4

ব্যাখ্যা

Question: Two numbers have a product of 2028 and a highest common factor (HCF) of 13. How many such pairs of numbers exist?

Solution:
Let the two numbers be x and y respectively.
It is given that the product of the two numbers is 2028, therefore, xy = 2028

Also, 13 is their HCF, thus both numbers must be divisible by 13.

So, let x = 13a and y = 13b,

ATQ,
13a × 13b = 2028
⇒ 169ab = 2028
⇒ ab = 2028
∴ ab = 12

Therefore, the required possible pair of values of x and y which are prime to each other are (1, 12) and (3, 4).
Thus, the required numbers are (12, 156) and (39, 52).
Hence, the number of possible pairs is 2.

২৭.

The LCM of two numbers is 495 and their HCF is 5. If the sum of the numbers is 100, then their difference is = ?

ক
90
খ
10
গ
70
ঘ
46

ব্যাখ্যা

Question: The LCM of two numbers is 495 and their HCF is 5. If the sum of the numbers is 100, then their difference is = ?

Solution:
Given that,
LCM = 495, HCF = 5
And,
The sum of the numbers is, a + b = 100
The product of two numbers is equal to the product of their LCM and HCF.
Product(a × b) = LCM × HCF = 495 × 5 = 2475

We know,
(a - b)² = (a + b)² - 4ab
⇒ (a - b)² = 100² - 4 × 2475 ; [where, a + b = 100 and ab = 2475 ]
⇒ (a - b)² = 10000 - 9900
⇒ (a - b)² = 100
⇒ a - b = √100 = 10
∴ a - b = 10

So the difference between the numbers is 10.

২৮.

A person has to completely put each of three liquids: 403 litres of petrol, 465 litres of diesel and 496 litres of Mobil Oil in bottles of equal size without mixing any of the above three types of liquids such that each bottle is completely filled. What is the least possible number of bottles required?

ক
41
খ
44
গ
47
ঘ
52

ব্যাখ্যা

Question: A person has to completely put each of three liquids: 403 litres of petrol, 465 litres of diesel and 496 litres of Mobil Oil in bottles of equal size without mixing any of the above three types of liquids such that each bottle is completely filled. What is the least possible number of bottles required?

Solution: For the least number of bottles, the capacity of each bottle must be maximum.
The capacity of each bottle = HCF of 403 liters, 465 liters, and 496 liters
= 31 liters.

For petrol, required number of bottles = 403/31 = 13
For diesel, required number of bottles = 465/31 = 15
For mobil, required number of bottles = 496/31 = 16

Hence,
required number of bottles = 13 + 15 + 16 = 44

২৯.

Find the HCF of 1/3 , 8/7, 9/11.

ক
1/231
খ
231
গ
1/72
ঘ
72
ঙ
None of these

ব্যাখ্যা

Question: Find the HCF of 1/3 , 8/7, 9/11.

Solution:
HCF of given Numbers : HCF(1, 8, 9)/LCM(3, 7, 11) = 1/231.

৩০.

A student has three iron rods of lengths 44 cm, 22 cm, and 55 cm. He needs to cut them into rods of the largest possible length such that no iron is wasted. What is the longest rod length he can achieve?

ক
17 cm
খ
15 cm
গ
13 cm
ঘ
11 cm

ব্যাখ্যা

Question: A student has three iron rods of lengths 44 cm, 22 cm, and 55 cm. He needs to cut them into rods of the largest possible length such that no iron is wasted. What is the longest rod length he can achieve?
(একজন ছাত্রকে ৩টি বিভিন্ন দৈর্ঘ্যের লোহার টুকরা দেওয়া হয়েছে – যথাক্রমে ৪৪ সেমি, ২২ সেমি ও ৫৫ সেমি। তাকে এমন একটি সর্বোচ্চ দৈর্ঘ্যের রড তৈরি করতে হবে যাতে কোনো লোহার অপচয় না হয়। রডটির সর্বাধিক দৈর্ঘ্য নির্ণয় করো।)

Solution:
H.C.F. = গ.সা.গু
এ ধরনের রডের সর্বাধিক সম্ভাব্য দৈর্ঘ্য = (44, 22, 55 এর গ.সা.গু) cm = 11 cm.

৩১.

The traffic lights at three different road crossings change after every 24 sec., 36 sec. and 72 sec. respectively. If they all change simultaneously at 8 : 20 : 00 hrs; then they will again change simultaneously at

ক
8 : 21 : 12 hrs
খ
8 : 27 : 48 hrs
গ
8 : 28 : 22 hrs
ঘ
8 : 27 : 36 hrs
ঙ
None of these

ব্যাখ্যা

Question: The traffic lights at three different road crossings change after every 24 sec., 36 sec. and 72 sec. respectively. If they all change simultaneously at 8 : 20 : 00 hrs; then they will again change simultaneously at

Solution:
Interval of change = (L.C.M. of 24, 36, 72) sec. = 72 sec. So, the lights will change after every 72 seconds, i.e. 1 min. 12 sec.
So, the next simultaneous change will take place at 8 : 21 : 12 hrs.

৩২.

The H.C.F. of two numbers is 12 and their difference is 12. Which of the following can be the numbers?

ক
66, 77
খ
70, 84
গ
94, 108
ঘ
66, 106
ঙ
84, 96

ব্যাখ্যা

Question: The H.C.F. of two numbers is 12 and their difference is 12. Which of the following can be the numbers?

Solution:
The difference of requisite numbers must be 12 and each should be divisible by 12. Checking the options given, only the fifth option satisfies.

84, 96
96 - 84 = 12
HCF of 84 and 96 is 12

৩৩.

Find the LCM of the fractions 5/8, 3/4, 1/2.

ক
15
খ
40
গ
15/2
ঘ
15/8

ব্যাখ্যা

Question: Find the LCM of the fractions 5/8, 3/4, 1/2.

Solution:
The fractions are 5/8, 3/4 and 1/2
LCM of the fraction = LCM of numerator/HCF of denominator

LCM of numerators = LCM of (5, 3 , 1 = 5 × 3 × 1 = 15
HCF of denominators = HCF of ( 8, 4, 2) = 2

∴ LCM of fraction = 15/2

৩৪.

Which of the following is the least number which will leave the remainder 5, when divided by 8, 12, 16, and 20?

ক
235
খ
245
গ
255
ঘ
265
ঙ
None of the above

ব্যাখ্যা

Question: Which of the following is the least number which will leave the remainder 5, when divided by 8, 12, 16, and 20?

Solution:
First we need to find the least number, so we have to find out the LCM of 8, 12, 16, and 20.
8 = 2 x 2 x 2
12 = 2 x 2 x 3
16 = 2 x 2 x 2 x 2
20 = 2 x 2 x 5

LCM = 2 x 2 x 2 x 2 x 3 x 5 = 240

240 is the least number that is exactly divisible by 8, 12, 16, and 20.

So, the required number that will leave remainder 5 is -
240 + 5 = 245

৩৫.

The product of two co-prime numbers is 342. Then their LCM is =?

ক
272
খ
342
গ
552
ঘ
756

ব্যাখ্যা

Question: The product of two co-prime numbers is 342. Then their LCM is =?

Solution:
We know,
If the HCF of two or more numbers is 1, then two or more numbers are co-prime numbers.
HCF of co-prime number is always = 1

Now,
Product of number = LCM × HCF
⇒ LCM × 1 = 342
⇒ LCM = 342

৩৬.

The L.C.M of 1/3, 5/6, 2/9, 4/27 is?

ক
10/27
খ
1/54
গ
24/7
ঘ
20/3

ব্যাখ্যা

Question: The L.C.M of 1/3, 5/6, 2/9, 4/27 is?

Solution:
L.C.M of 1, 5, 2, 4 is 20
H.C.F of 3, 6, 9, 27 is 3

We know,
Required L.C.M = L.C.M of 1, 5, 2, 4/H.C.F of 3, 6, 9, 27
= 20/3

৩৭.

What is the H.C.F. of the following fractions?
3/6, 6/9, 9/12

ক
1/6
খ
1/2
গ
2/15
ঘ
1/12

ব্যাখ্যা

Question: What is the H.C.F. of the following fractions?
3/6, 6/9, 9/12

Solution:
আমরা জানি,
ভগ্নাংশের গসাগু = (লবের গসাগু)/(হরের লসাগু)

এখানে লব = 3, 6 এবং 9
3 = 3 × 1
6 = 3 × 2
9 = 3 × 3
∴ লবের গসাগু (H.C.F.) = 3

হর = 6, 9 এবং 12
6 = 2 × 3
9 = 3²
12 = 2² × 3
∴ হরের লসাগু (L.C.M.) = 2² × 3²
= 4 × 9 = 36

ভগ্নাংশের গসাগু = লবের গসাগু/হরের লসাগু
= 3/36
= 1/12

৩৮.

Find the greatest common divisor (GCD) of the numbers 0.9, 0.36, and 1.08

ক
0.108
খ
0.06
গ
1.8
ঘ
0.18

ব্যাখ্যা

Question: Find the greatest common divisor (GCD) of the numbers 0.9, 0.36, and 1.08

Solution:
প্রদত্ত সংখ্যাগুলো হলো 0.9, 0.36, 1.08
0.90 × 100 = 90
0.36 × 100 = 36
1.08 × 100 = 108

90, 36, 108, 36, এর গ.সা.গু = 18
∴ 0.9, 0.36, 1.08 এর গ.সা.গু = 0.18

৩৯.

The ratio of two numbers is 5:7 and their H.C.F is 6. What is their L.C.M?

ক
180
খ
210
গ
240
ঘ
252

ব্যাখ্যা

Question: The ratio of two numbers is 5:7 and their H.C.F is 6. What is their L.C.M?

Solution:
Let the numbers be 5x and 7x.
∴ HCF = x = 6

We know,
HCF × LCM = Product of two numbers
⇒ 6 × LCM = (5x) × (7x)
⇒ 6 × LCM = (5 × 6) × (7 × 6)
⇒ LCM = (30 × 42)/6
⇒ LCM = 1260/6
∴ LCM = 210

৪০.

A rectangular block measuring 8 cm by 12 cm by 16 cm is cut into an exact number of cubes of equal size. What will be the least possible number of cubes?

ক
6
খ
11
গ
15
ঘ
24
ঙ
None

ব্যাখ্যা

Question: A rectangular block measuring 8 cm by 12 cm by 16 cm is cut into an exact number of cubes of equal size. What will be the least possible number of cubes?

Solution:
Given that,
Sides of the rectangular block = 8 cm, 12 cm, 16 cm

Let n be the number of cubes.
HCF( 8, 12, 16) = 4 cm
And side of the square = 4 cm

We know,
Volume of the cuboid = n × Volume of the cube
⇒ Length × breadth × height = n × side³⇒ 8 × 12 × 16 = n × 4 × 4 × 4
⇒ 64n = 8 × 12 × 16
⇒ n = (8 × 12 × 16)/64
∴ n = 24

So the number of cubes is 24.

৪১.

Find the greatest number that will divide 964, 1238 and 1400 leaving remainders 41, 31 and 51 respectively.

ক
71
খ
75
গ
79
ঘ
81

ব্যাখ্যা

Question: Find the greatest number that will divide 964, 1238 and 1400 leaving remainders 41, 31 and 51 respectively.

Solution:
If a number divides the given numbers with the mentioned remainders, then
964 - 41 = 923
1238 - 31 = 1207
1400 - 51 = 1349

Required number:
= HCF of (964 - 41), (1238 - 31) and (1400 - 51)
HCF of 923, 1207 and 1349 = 71

৪২.

The traffic lights at three different road crossings change after every 24 sec, 36 sec, and 72 sec respectively. If they all change simultaneously at 8 : 20 : 00 hrs, then they will again change simultaneously at:

ক
8 : 20 : 48 hrs
খ
8 : 23 : 12 hrs
গ
8 : 32 : 10 hrs
ঘ
8 : 27 : 16 hrs
ঙ
8 : 21 : 12 hrs

ব্যাখ্যা

Question: The traffic lights at three different road crossings change after every 24 sec, 36 sec, and 72 sec respectively. If they all change simultaneously at 8 : 20 : 00 hrs, then they will again change simultaneously at:

Solution:

Time interval for simultaneous change = L.C.M. of 24, 36, 72 = 72 sec

72 seconds = 1 min 12 sec

Next simultaneous change = 8 : 20 : 00 + 1 min 12 sec = 8 : 21 : 12 hrs

৪৩.

The H.C.F. of two numbers, each having three digits , is 17 and their L.C.M. is 714. The sum of the numbers will be:

ক
219
খ
221
গ
233
ঘ
239

ব্যাখ্যা

Question: The H.C.F. of two numbers, each having three digits , is 17 and their L.C.M. is 714. The sum of the numbers will be:

Solution:
Let, the numbers be 17x and 17y
where x and y are co-prime.

∴ LCM of 17x and 17y = 17 xy

ATQ,
17xy = 714
⇒ xy = 714/17
⇒ xy = 42
⇒ xy = 6 × 7

⇒ x = 6 and y = 7
∴ x = 7 and y =6

First number = 17x = 17 × 6 = 102
Second number = 17y = 17 × 7 = 119

∴ Sum of the numbers = (102 + 119) = 221

৪৪.

Three bells ring at intervals of 12 sec, 18 sec, and 30 sec. If they start together, how many times will all three ring together in 2 hours?

ক
101
খ
76
গ
61
ঘ
41
ঙ
31

ব্যাখ্যা

Question: Three bells ring at intervals of 12 sec, 18 sec, and 30 sec. If they start together, how many times will all three ring together in 2 hours?

Solution:
Prime factorization:
12 = 2² × 3
18 = 2 × 3²
30 = 2 × 3 × 5

LCM = 2² × 3² × 5
= 4 × 9 × 5
= 180 sec
So, all three bells ring together every 180 seconds.

2 hours = 2 × 60 × 60
= 7200 sec

Number of times = 7200/180 + 1 [As they are starting together, they will ring at 0 sec. So, +1]
= 40 + 1 = 41

৪৫.

Two alarm clocks ring their alarms at regular intervals of 50 seconds and 48 seconds. If they first beep together at 2.30 PM, at what time will they beep again for first time?

ক
2 : 50 PM
খ
3 : 10 PM
গ
2 : 55 PM
ঘ
3 : 00 PM

ব্যাখ্যা

Question: Two alarm clocks ring their alarms at regular intervals of 50 seconds and 48 seconds. If they first beep together at 2 : 30 PM, at what time will they beep again for first time?

Solution:
They will ring together after,
LCM of 48 and 50 secs.
48 = 2 × 2 × 2 × 2 × 3
50 = 2 × 5 × 5

∴ LCM = 2 × 2 × 2 × 2 × 3 × 5 × 5 = 1200 secs
= 1200/60 = 20 min. [1 min = 60 sec]

∴ They will beep together at 2 : 30 + 20 = 2 : 50 PM

৪৬.

What is the H.C.F. of 4/9, 8/12, and 16/18?

ক
2/75
খ
1/9
গ
84
ঘ
3/52
ঙ
4/9

ব্যাখ্যা

Question: What is the H.C.F. of 4/9, 8/12, and 16/18?

Solution:
We know,
H.C.F. of fractions = (H.C.F. of numerators)/(L.C.M. of denominators)

H.C.F. of numerators:
H.C.F.(4, 8, 16) = 4

L.C.M. of denominators:
9 = 3²
12 = 2² × 3
18 = 2 × 3²∴ L.C.M. = 2² × 3² = 4 × 9 = 36

∴ Required H.C.F. = 4/36 = 1/9

৪৭.

The HCF and LCM of two numbers are 8 and 48 respectively. If one of the numbers is 16, then the other number is = ?

ক
24
খ
12
গ
36
ঘ
54

ব্যাখ্যা

Question: The HCF and LCM of two numbers are 8 and 48 respectively. If one of the numbers is 16, then the other number is = ?

Solution:
Here,
HCF = 8
LCM = 48

One number = 16

Let the other number be = p
∴ 16p = 48 × 8
⇒ p = 24

Hence, the other number is = 24

৪৮.

The LCM and HCF of two numbers are 495 and 5. If the sum of the numbers is 100, what is the difference of the numbers?

ক
5
খ
15
গ
20
ঘ
10

ব্যাখ্যা

Question: The LCM and HCF of two numbers are 495 and 5. If the sum of the numbers is 100, what is the difference of the numbers?

Solution:
let,
one number is x
∴ another number is (100 - x)

∴ x(100 - x) = 5 × 495
or, 100x - x² = 2475
or, x² - 100x + 2475 = 0
or, x² - 55x - 45x + 2475 = 0
or, x(x - 55) - 45(x - 55) = 0
or, (x - 55)(x - 45) = 0
∴ x = 45, 55
∴ the numbers are 45 and 55.
difference = 55 - 45 = 10

৪৯.

The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 308, then the other is:

ক
242
খ
275
গ
228
ঘ
280
ঙ
295

ব্যাখ্যা

Question: The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 308, then the other is:

Solution:
We know that,
L.C.M × H.C.F. = Product of two numbers
⇒ 7700 × 11 = 308 × other number
⇒ Other number = (7700 × 11)/308
∴ Other number = 275

৫০.

The maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is-

ক
101
খ
91
গ
1001
ঘ
910

ব্যাখ্যা

Question: The maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is-

Solution:
We use prime factorization,
1001 = 7 × 11 × 13
910 = 2 × 5 × 7 × 13
∴ Common factors are 7 and 13

∴ HCF = 7 × 13 = 91

∴ Maximum number of students = HCF of 1001 and 910 = 91

৫১.

The ratio of two numbers is 5 : 6 and their H.C.F is 7. Their L.C.M is -

ক
210
খ
252
গ
294
ঘ
320

ব্যাখ্যা

Question: The ratio of two numbers is 5 : 6 and their H.C.F is 7. Their L.C.M is -

Solution:
ধরি, সংখ্যা দুটি হলো 5x এবং 6x
∴গসাগু (H.C.F) = x = 7

∴ সংখ্যা দুটি: হলো 5 × 7 = 35 এবং 6 × 7 = 42

∴ সংখ্যাদ্বয়ের গুণফল = 35 × 42 = 1470
এবং H.C.F = 7

আমরা জানি,
L.C.M = (Product of two numbers)/H.C.F
= 1470/7
= 210

∴ সংখ্যা দুটির লসাগু (L.C.M) = 210

৫২.

What is the L.C.M. of 25, 30, 35 and 40?

ক
3800
খ
4200
গ
4400
ঘ
3200

ব্যাখ্যা

Question: What is the L.C.M. of 25, 30, 35 and 40?

Solution:
25 = 5 × 5,
30 = 5 × 3 × 2,
35 = 5 × 7,
40 = 2 × 2 × 2 × 5

Required LCM = 2 × 2 × 2 × 5 × 5 × 3 × 7 = 4200

৫৩.

An officer was appointed on maximum daily wages on contract money of Tk. 6720. But on being absent for some days, he was paid Tk. 5600. For how many days was he absent?

ক
5 days
খ
4 days
গ
3 days
ঘ
2 days
ঙ
1 day

ব্যাখ্যা

Question: An officer was appointed on maximum daily wages on contract money of Tk. 6720. But on being absent for some days, he was paid Tk. 5600. For how many days was he absent?

Solution:
Maximum daily wages of the officers = H.C.F of Tk. 6720 and Tk. 5600

H.C.F of 6720 & 5600 = 1120

Maximum daily wage = Tk. 1120
Total days appointed = 6720 ÷ 1120 = 6 days

Absent days = 6 − 5 = 1 day

৫৪.

The product of two numbers is 2028 and their HCF is 13. The number of such pair is -

ক
1
খ
2
গ
4
ঘ
3

ব্যাখ্যা

Question: The product of two numbers is 2028 and their HCF is 13. The number of such pair is -

Solution:
let,
the numbers are 13x and 13y
then,
13x × 13y = 2028
or, xy = 2024/169
∴ xy = 12

number pair that have the product of 12 is (1, 12) and (3, 4)

∴ number-pairs are = (13 × 1, 13 × 12) and (13 × 3, 13 × 4)
= (13, 156), (39, 52)

৫৫.

The HCF and LCM of two numbers are 8 and 48 respectively. If one of the numbers is 16, then the other number is = ?

ক
22
খ
28
গ
16
ঘ
24

ব্যাখ্যা

Question: The HCF and LCM of two numbers are 8 and 48 respectively. If one of the numbers is 16, then the other number is = ?

Solution:
Here,
HCF = 8
LCM = 48

One number = 16

Let the other number be = p
∴ 16p = 48 × 8
⇒ p = 24

Hence, the other number is = 24

৫৬.

What is the H.C.F. of the following fractions?
6/10, 9/15, 12/20.

ক
1/10
খ
12
গ
1/15
ঘ
1/20

ব্যাখ্যা

Question: What is the H.C.F. of the following fractions?
6/10, 9/15, 12/20.

Solution:
আমরা জানি,
ভগ্নাংশের গসাগু = লবের গসাগু/হরের লসাগু

এখানে,
লব = 6, 9, 12
6 = 2 × 3
9 = 3 × 3
12 = 2 × 2 × 3

∴ গসাগু (HCF) = 3

হর = 10, 15, 20
10 = 2 × 5
15 = 3 × 5
20 = 2 × 2 × 5

∴ লসাগু (LCM) = 2 × 2 × 3 × 5
= 60

ভগ্নাংশের গসাগু = লবের গসাগু/হরের লসাগু
= 3/60
=1/20

৫৭.

84 Maths books, 90 Physics books, and 120 Chemistry books have to be stacked topicwise. How many books will be there in each stack so that each stack will have the same height too?

ক
6
খ
12
গ
4
ঘ
None of these

ব্যাখ্যা

Question: 84 Maths books, 90 Physics books, and 120 Chemistry books have to be stacked topicwise. How many books will be there in each stack so that each stack will have the same height too?

Solution:
As the height of each stack is the same, the required number of books in each stack
= HCF of 84, 90 and 120

84 = 2 × 2 × 3 × 7
90 = 2 × 3 × 3 × 5
120 = 2 × 2 × 2 × 3 × 5

∴ HCF = 2 × 3 = 6

Hence, The required number of books in each stack is 6.

৫৮.

What is the maximum side length of a square slab that can be used to tile the floor of a room with dimensions 5 meters 44 cm by 3 meters 74 cm?

ক
26 cm
খ
30 cm
গ
32 cm
ঘ
34 cm

ব্যাখ্যা

Question: What is the maximum side length of a square slab that can be used to tile the floor of a room with dimensions 5 meters 44 cm by 3 meters 74 cm?

Solution:
length = 5 meters 44 cm
= 500 + 44 cm
= 544 cm

breadth = 3 meters 74 cm
= 300 + 74
= 374 cm

∴ The side of the square slab is the H.C.F. of 544 and 374 cm i.e. 34.

৫৯.

Let x be a number. GCF of 2/5, 3/5 and 1/4 of that number is 5. find the number.

ক
100
খ
200
গ
120
ঘ
90

ব্যাখ্যা

Question: Let x be a number. GCF of 2/5, 3/5 and 1/4 of that number is 5. find the number.

Solution:
the numbers are 2x/5, 3x/5, x/4.
GCF of 2x, 3x, and x is x
and LCM of 5, 5 and 4 is 20
∴ the GCF of 2x/5, 3x/5, x/4 is x/20

ATQ,
x/20 = 5
x = 100

৬০.

Four people are running around a circular ground from a point on the circumference at 9.00 am. For one round, these four persons take respectively 40, 50, 60 and 30 minutes. At what time will they meet together again?

ক
5.00 pm
খ
6.00 pm
গ
7.00 pm
ঘ
9.00 pm

ব্যাখ্যা

Question: Four people are running around a circular ground from a point on the circumference at 9.00 am. For one round, these four persons take respectively 40, 50, 60 and 30 minutes. At what time will they meet together again?

Solution:
L.C.M. of 40, 50, 60 and 30
= 600 minutes
= 10 hours
So, they meet again 10 hours after they start.
They meet together again = 9.00 am + 10 hours
= 7.00 pm

৬১.

If the least common multiple of two numbers is twelve times their highest common factor, and their sum (HCF + LCM) equals 403, then what is the other number when one number is 93?

ক
124
খ
126
গ
128
ঘ
132

ব্যাখ্যা

Question: If the least common multiple of two numbers is twelve times their highest common factor, and their sum (HCF + LCM) equals 403, then what is the other number when one number is 93?

Solution:
Let HCF be h and LCM be l
Then l = 12h and
l + h = 403

∴12h + h = 403
⇒ h = 31

So, l = (403 − 31) = 372

Hence, the other number = (31 × 372)/93 = 124

৬২.

Three bells ring at intervals of 9, 12, and 15 minutes respectively. If they all ring together at 3:00 PM, when will they ring together again?

ক
4 : 20 PM
খ
5 : 30 PM
গ
6 : 00 PM
ঘ
6 : 10 PM

ব্যাখ্যা

Question: Three bells ring at intervals of 9, 12, and 15 minutes respectively. If they all ring together at 3:00 PM, when will they ring together again?

Solution:
তিনটি ঘণ্টা বিকাল ৩ টায় একত্রে বাজলে 9, 12, 15 এর ল.সা.গুর সমান সময়ের পর ঘণ্টাগুলো পুনরায় একত্রে বাজবে।

সংখ্যা গুলোর মৌলিক উৎপাদক:
9 = 3 × 3 = 3²
12 = 2 × 2 × 3 = 2² × 3
15 = 3 × 5

9, 12, 15 এর ল.সা.গু. = 2² × 3² × 5
= 4 × 9 × 5
= 180 অর্থাৎ 180 মিনিট

সুতরাং, ঘণ্টা গুলো একবার বিকাল 3 টায় বাজার পর পুনরায় বাজবে = 3 টা + 180 মিনিটে
= 3 টা + (60 + 60 + 60) মিনিটে
= 3 টা + 3 ঘণ্টা
= 6 টা
∴ তারা আবার 6:00 PM-এ একত্রে বাজবে।

৬৩.

What is the LCM of 12, 15, and 20?

ক
60
খ
120
গ
180
ঘ
240

ব্যাখ্যা

Question: What is the LCM of 12, 15, and 20?

Solution:
12 = 2 × 2 × 3
15 = 3 × 5
20 = 2 × 2 × 5

∴ LCM of 12, 15, and 20 is = 2 × 2 × 3 × 5 = 60

৬৪.

The LCM of two numbers is 1920 and their HCF is 16. If one of the number is 128, find the other number.

ক
216
খ
224
গ
230
ঘ
240
ঙ
260

ব্যাখ্যা

Question: The LCM of two numbers is 1920 and their HCF is 16. If one of the number is 128, find the other number.

Solution:
1st number × 2nd number = L.C. M. × H.C.F

We have,
First number × second number = LCM × HCF
∴ Second number = (1920 × 16)/128
= 240

৬৫.

The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then the sum of the numbers is:

ক
64
খ
32
গ
40
ঘ
28

ব্যাখ্যা

Question: The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then the sum of the numbers is:

Solution:

Let the numbers be 2x and 3x

Then, their L.C.M. = 6x

ATQ,

6x = 48

or x = 8

∴ The numbers are 16 and 24

Hence, required sum = (16 + 24) = 40

৬৬.

Find the least number that leaves a remainder of 5 when divided by 6, 9, 15, and 20.

ক
170
খ
159
গ
185
ঘ
190

ব্যাখ্যা

Question: Find the least number that leaves a remainder of 5 when divided by 6, 9, 15, and 20.

Solution:
We have to find the least number,
therefore, we find out the LCM of 6, 9, 15, and 20.

6 = 3 × 2
9 = 3 × 3
15 = 3 × 5
20 = 2 × 2 × 5

∴ LCM = 2 × 2 × 3 × 3 × 5
= 180

This is the least number which is exactly divisible by 6, 9, 15, and 20.
So, required number leaves remainder of 5 is = 180 + 5 = 185

৬৭.

If HCF = 3, LCM = 18, and sum of reciprocals = 7/18, find the numbers.

ক
19, 36
খ
14, 3
গ
18, 3
ঘ
13, 36
ঙ
5, 13

ব্যাখ্যা

Question: If HCF = 3, LCM = 18, and sum of reciprocals = 7/18, find the numbers.

Solution:
Given that,
If HCF = 3, LCM = 18, and sum of reciprocals = 7/18
Let the numbers be a and b

Then, (1/a) + (1/b) = 7/18 and ab = 3 × 18 = 54 (product of the numbers = (HCF × LCM)

∴ (1/a) + (1/b) = 7/18
⇒ (a + b)/ab = 7/18
∴ a + b = 21
From Quadratic Equation
We know:
x² - 21x + 54 = 0
⇒ x² - 18x - 3x + 54 = 0
⇒ (x -18) (x -3) = 0
∴ x = 18, 3

৬৮.

John, Smith and Kate start at same time, same point and in same direction to run around a circular ground. John completes a round in 250 seconds, Smith in 300 seconds and Kate in 150 seconds. Find after what time will they meet again at the starting point?

ক
30 min
খ
25 min
গ
20 min
ঘ
15 min

ব্যাখ্যা

Question: John, Smith and Kate start at same time, same point and in same direction to run around a circular ground. John completes a round in 250 seconds, Smith in 300 seconds and Kate in 150 seconds. Find after what time will they meet again at the starting point?

Solution:
L.C.M. of 250, 300 and 150 = 1500 sec
Dividing 1500 by 60 we get 25, which mean 25 minutes.
John, Smith and Kate meet after 25 minutes.

৬৯.

Three numbers which are co-prime to one another are such that product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is?

ক
69
খ
75
গ
83
ঘ
85

ব্যাখ্যা

Question: Three numbers which are co-prime to one another are such that product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is?

Solution:
Since the numbers are co-prime, their HCF = 1

Product of first two numbers = 551
Product of last two numbers = 1073

The middle number is common in both of these products.
Hence, if we take HCF of 551 and 1073, we get the common middle number.

HCF of 551 and 1073 = 29
So, Middle Number = 29
First Number = 551/29 = 19
Last Number = 1073/29 = 37

∴ Sum of the three numbers = (19 + 29 + 37) = 85.

৭০.

The least number which when divided by 4, 6, 8, 12 and 16 leaves remainder of 2 in each case is = ?

ক
46
খ
48
গ
56
ঘ
50

ব্যাখ্যা

Question: The least number which when divided by 4, 6, 8, 12 and 16 leaves remainder of 2 in each case is = ?

Solution:
LCM of (4, 6, 8, 12, 16)
⇒ 16 × 3 = 48

∴ The number when divided by (4, 6, 8, 12, 16) leaves remainder 2 is
= 48 + 2
= 50

৭১.

What is the LCM of 6, 9, and 15?

ক
90
খ
60
গ
120
ঘ
45

ব্যাখ্যা

Question: What is the LCM of 6, 9, and 15?

Solution:
Prime factorization:
6 = 2 × 3
9 = 3 × 3 = 3²
15 = 3 × 5

LCM = take the highest power of each prime:
2¹ × 3² × 5¹ = 2 × 9 × 5 = 90

৭২.

The sum of two numbers is 45. Their difference is 1/9 of their sum. Their LCM is -

ক
50
খ
100
গ
75
ঘ
150

ব্যাখ্যা

Question: The sum of two numbers is 45. Their difference is 1/9 of their sum. Their LCM is -

Solution:
let the numbers are p and q.
then,
p + q = 45..........(i)
p - q = 45/9 = 5...........(ii)

adding (i) and (ii).
2p = 50
p = 25

putting p on (i) we get,
25 + q = 45
q = 20

the LCM of 25, 20 is 100.

৭৩.

Find the sum of the reciprocals of two numbers whose total is 36, HCF is 3, and LCM is 105.

ক
4/35
খ
1/21
গ
2/33
ঘ
6/37
ঙ
3/35

ব্যাখ্যা

Question: Find the sum of the reciprocals of two numbers whose total is 36, HCF is 3, and LCM is 105.

Solution:
Let, the numbers be a and b.

Then, a + b = 36 and ab = 3 × 105 = 315 [∵ Product of the numbers = HCF×LCM]

∴ sum of their reciprocals
= (1/a) + (1/b)
= (a + b)/ab
= 36/315
= 4/35

৭৪.

The product of two co-prime numbers is 820. Then their LCM is =?

ক
410
খ
1640
গ
820
ঘ
1230

ব্যাখ্যা

Question: The product of two co-prime numbers is 820. Then their LCM is =?

Solution:
If hcf of two or more numbers is 1, then two or more numbers are co-prime numbers.
HCF of co-prime number is always 1

∴ Product of number = LCM × HCF
⇒ LCM × 1 = 820
⇒ LCM = 820

৭৫.

Find HCF of 84, 126, and 210.

ক
24
খ
38
গ
40
ঘ
42

ব্যাখ্যা

Question: Find HCF of 84, 126, and 210.

Solution:
Factorize each number into prime factors:
84 = 2² × 3 × 7
126 = 2 × 3² × 7
210 = 2 × 3 × 5 × 7

HCF is the product of common prime factors with smallest powers:
Common primes: 2, 3, 7
Smallest powers: 2¹, 3¹, 7¹
HCF = 2 × 3 × 7 = 42

৭৬.

Four bells ring simultaneously at the start and then at intervals of 6 seconds, 12 seconds, 15 seconds, and 20 seconds respectively. How many times do they ring together in 2 hours?

ক
120 times
খ
119 times
গ
122 times
ঘ
121 times

ব্যাখ্যা

Question: Four bells ring simultaneously at the start and then at intervals of 6 seconds, 12 seconds, 15 seconds, and 20 seconds respectively. How many times do they ring together in 2 hours?

Solution:
Given that,
Four bells ring simultaneously at starting and an interval of 6 sec, 12 sec, 15 sec and 20 sec respectively.

∴ LCM of (6, 12, 15, 20) = 60
All 4 bells ring together again after every 60 seconds

Now, In 2 Hours, they ring together = [(2 × 60 × 60)/60] times + 1 (at the starting)
= (120 + 1) times
= 121 times

∴ In 2 hours they ring together for 121 times

৭৭.

The product of two numbers is 2028 and their HCF is 13. The number of such pairs is =?

ক
2
খ
1
গ
4
ঘ
3

ব্যাখ্যা

৭৮.

The smallest number exactly divisible by 104, 78 and 260 is p. If p + 40 = q², then what is the positive value of q?

ক
40
খ
64
গ
55
ঘ
80
ঙ
30

ব্যাখ্যা

Question: The smallest number exactly divisible by 104, 78 and 260 is p. If p + 40 = q², then what is the positive value of q?

Solution:
The least number exactly divisible by 104, 78, and 260 is the LCM of these three numbers.
So the LCM of 104, 78, and 260 is 1560.

Therefore, p = 1560
The problem states that p + 40 = q²
Substituting p into this equation gives then we get,
⇒ 1560 + 40 = q²,
⇒ q² = 1600
⇒ q = √1600
∴ q = 40

Therefore, the positive value of q is 40.

৭৯.

A student is provided with three iron pieces of different lengths—44 cm, 22 cm, and 55 cm—and must form rods of the greatest possible length with no leftover iron. What is the maximum length of such rods?

ক
9 cm
খ
17 cm
গ
11 cm
ঘ
15 cm

ব্যাখ্যা

Question: A student is provided with three iron pieces of different lengths—44 cm, 22 cm, and 55 cm—and must form rods of the greatest possible length with no leftover iron. What is the maximum length of such rods?

Solution:
Maximum possible length of such rod = (H.C.F. of 44, 22, 55) cm = 11 cm.

৮০.

What is the least number which when doubled is exactly divisible by 8, 14, 18, and 24?

ক
240
খ
252
গ
264
ঘ
270

ব্যাখ্যা

Question: What is the least number which when doubled is exactly divisible by 8, 14, 18, and 24?

Solution:
Let the number be x.
Double the number is 2x.

8 = 2 × 2 × 2
14 = 2 × 7
18 = 2 × 3 × 3
24 = 2 × 2 × 2 × 3

∴ LCM = 2 × 2 × 2 × 3 × 3 ×7
= 504

∴ x = 504/2 = 252

৮১.

The number of number-pairs lying between 40 and 100 with their HCF as 15 is -

ক
4
খ
6
গ
10
ঘ
8

ব্যাখ্যা

Question: The number of number-pairs lying between 40 and 100 with their HCF as 15 is -

Solution:
গ.সা.গু ১৫ এমন প্রতিটি সংখ্যার উৎপাদক ১৫।
৪০ থেকে ১০০ এর মধ্যে ১৫ এর গুণিতক ৪৫, ৬০, ৭৫, ৯০

সংখ্যা জোড়: (৪৫, ৬০), (৪৫, ৭৫), (৪৫, ৯০), (৬০, ৭৫), (৬০, ৯০), (৭৫, ৯০)
কিন্তু এখানে (৪৫, ৯০) এর গ.সা.গু হলো ৪৫ এবং (৬০, ৯০) এর গ.সা.গু হলো ৩০।

তাই ৪ টি সংখ্যাজোড় আছে যাদের গ.সা.গু ১৫।

৮২.

Find the greatest number that will divide 39, 87, and 179 and leave the same remainder.

ক
7
খ
9
গ
4
ঘ
12

ব্যাখ্যা

Question: Find the greatest number that will divide 39, 87, and 179 and leave the same remainder.

Solution:
the number is the H.C.F of (87 - 39), (179 - 87) and (179 - 39)
= H.C.F of 48, 92 and 140
= 4

৮৩.

The highest common factor (HCF) of x² - 1, x⁴ - 1 and x⁴ - x³ + x - 1 is

ক
x⁶ - 1
খ
x³ + 1
গ
x² - 1
ঘ
x³ - 1

ব্যাখ্যা

Question: The highest common factor (HCF) of x² - 1, x⁴ - 1 and x⁴ - x³ + x - 1 is-

Solution:
১ম রাশি = x² - 1 = (x + 1)(x - 1)

২য় রাশি = x⁴ - 1 = (x² - 1)(x² + 1) = (x² + 1) (x + 1)(x - 1)

৩য় রাশি = x⁴ - x³ + x - 1 = (x⁴ - x³) + (x - 1)
= x³(x - 1) + 1(x - 1)
= (x³ + 1)(x - 1)
= (x + 1)(x² - x + 1)(x - 1)
= (x - 1)(x + 1)(x² - x + 1)

∴ নির্ণয়ে গ.সা.গু = (x + 1)(x - 1) = x² - 1

৮৪.

The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is-

ক
654
খ
548
গ
580
ঘ
476

ব্যাখ্যা

Question: The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is-

Solution:
The number leaves a remainder 8 when divided by 12, 15, 20 and 54.
So the required number = LCM(12, 15, 20, 54) + 8

Now,
12 = 2 × 2 × 3
15 = 3 × 5
20 = 2 × 2 × 5
54 = 2 × 3 × 3 × 3

∴ LCM(12, 15, 20, 54) = 540

∴ Required Number = 540 + 8 = 548

৮৫.

Find the LCM of 2.5, 0.5 and 0.175.

ক
7.5
খ
17.5
গ
2.5
ঘ
0.75

ব্যাখ্যা

Question: Find the LCM of 2.5, 0.5 and 0.175.

Solution:
Given that,
2.5 = 25/10
0.5 = 5/10
0.175 = 175/1000

Now,
LCM of Numerators is 25, 5, 175 = 175
and HCF of Denominators is 10, 10, 1000 = 10

We know,
LCM of two or more fractions is given by,
LCM = LCM of Numerators/HCF of Denominators
= 175/10
= 17.5

৮৬.

With bell chimes occurring every 12, 15, 20, and 30 seconds, how many times will the four bells toll in perfect synchrony within 8 hours?

ক
480
খ
481
গ
380
ঘ
381

ব্যাখ্যা

Question: With bell chimes occurring every 12, 15, 20, and 30 seconds, how many times will the four bells toll in perfect synchrony within 8 hours?

Solution:
Four bells ringing timing is 12 sec, 15 sec, 20 sec,30 sec
Now we have to take LCM of time interval ⇒ LCM of (12, 15, 20, 30) = 60
Total seconds in 8 hours = 8 × 3600 = 28800

Number of times bell rings = 28800/60
⇒ Number of times bell rings = 480

If four bells ring together in starting ⇒ 480 + 1
∴ The bell ringing 481 times in 8 hours.

Mistake Points: The bells start tolling together, the first toll also needs to be counted, that is the number of times of tolling since the first time.

৮৭.

The smallest number, which when increased by 5 is divisible by each of 24, 32, 36 and 64 is =?

ক
383
খ
467
গ
571
ঘ
212

ব্যাখ্যা

Question: The smallest number, which when increased by 5 is divisible by each of 24, 32, 36 and 64 is =?

Solution:
To find the least common multiple (LCM) of the given numbers (24, 32, 36, 64), we can first find the prime factorization of each number:

24 = 2³ × 3
32 = 2⁵
36 = 2² × 3²
64 = 2⁶

Then, we take the highest power of each prime factor that appears in any of the numbers:
The highest power of 2 is 2⁶
The highest power of 3 is 3²

So, the LCM is 2⁶ × 3² = 64 × 9 = 576

∴ The required number is 576 − 5 = 571

৮৮.

The product and LCM of two numbers are 54 and 18. Find their HCF-

ক
3
খ
2
গ
7
ঘ
6

ব্যাখ্যা

Question: The product and LCM of two numbers are 54 and 18. Find their HCF-

Solution:
আমরা জানি,
দুইটি সংখ্যার গুণফল = ল.সা.গু × গ.সা.গু।
গ.সা.গু = দুইটি সংখ্যার গুণফল/ল.সা.গু
গ.সা.গু = ৫৪/১৮
গ.সা.গু = ৩

৮৯.

Six bells commence tolling together and toll at intervals of 3, 5, 6, 9, 10, and 15 seconds respectively. In 45 minutes, how many times do they toll together?

ক
29
খ
30
গ
31
ঘ
34

ব্যাখ্যা

Question: Six bells commence tolling together and toll at intervals of 3, 5, 6, 9, 10, and 15 seconds respectively. In 45 minutes, how many times do they toll together?

Solution:
3 = 3¹
5 = 5¹
6 = 2 × 3
9 = 3²10 = 2 × 5
15 = 3 × 5

∴ ল.সা.গু. = 2¹ × 3² × 5¹ = 2 × 9 × 5 = 90।

সুতরাং, ঘণ্টাগুলো প্রতি 90 সেকেন্ড পর পর একসাথে বাজবে।

এখন, 45 মিনিট = 45 × 60 = 2700 সেকেন্ড।

মোট 2700 সেকেন্ডে ঘণ্টাগুলো যতবার একসাথে বাজবে তার সংখ্যা হলো = 2700/90 = 30 বার।

সুতরাং, 45 মিনিটে ঘণ্টাগুলো মোট 31 বার একসাথে বাজবে।

৯০.

The product of two co-prime numbers is 540. Then their LCM is =?

ক
540
খ
1640
গ
820
ঘ
1230

ব্যাখ্যা

Question: The product of two co-prime numbers is 540. Then their LCM is =?

Solution:
If HCF of two or more numbers is 1, then two or more numbers are co-prime numbers.
HCF of co-prime number is always 1

∴ Product of number = LCM × HCF
⇒ LCM × 1 = 540
⇒ LCM = 540

৯১.

What is the H.C.F. of 4/9, 10/21 and 20/63?

ক
4/189
খ
2/63
গ
6/63
ঘ
20/21

ব্যাখ্যা

Question: What is the H.C.F. of 4/9, 10/21 and 20/63?

Solution:
H.C.F of 4/9, 10/21 and 20/63 = (H.C.F of 4,10 and 20)/(L.C.M of 9,21 and 63)

H.C.F of 4, 10 and 20 = 2
L.C.M. of 9, 21 and 63 = 63.

∴ Required H.C.F. = 2/63

৯২.

Find the least multiple of 23. which when divided by 18, 21 and 24 leaves remainders 7, 10 and 13 respectively.

ক
3002
খ
3013
গ
3024
ঘ
3036

ব্যাখ্যা

Question: Find the least multiple of 23. which when divided by 18, 21 and 24 leaves remainders 7, 10 and 13 respectively.

Solution:
Here, (18 - 7) = 11, (21 - 10) = 11 and (24 - 13) = 11
L.C.M of 18, 21 and 24 is 504.

Let, the required number be 504x - 11.
Least value of x for which (504x - 11) is divisible by 23 is x = 6

Required number = (504 × 6) - 11 = 3024 - 11 = 3013.

৯৩.

Find the HCF of 3/4, 5/6 and 6/7 = ?

ক
1/48
খ
1/60
গ
1/84
ঘ
None of these

ব্যাখ্যা

Question: Find the HCF of 3/4, 5/6 and 6/7 = ?

Solution:
For the HCF of fractions, it has to be taken the HCF of numerators and LCM denominators.

HCF of 3, 5, 6 = 1
LCM of 4, 6, 7 = 84

HCF of numerators/LCM of denominators = 1/84

Hence, the HCF of 3/4, 5/6 and 6/7 = 1/84

৯৪.

Find the HCF of 210, 385, and 735.

ক
45
খ
35
গ
55
ঘ
27

ব্যাখ্যা

Question: Find the HCF of 210, 385, and 735.

Solution:
HCF of 210, 385, and 735.

Factor of 210 = 2 × 3 × 5 × 7
Factor of 385 = 5 × 7 × 11
Factor of 735 = 3 × 5 × 7 × 7
∴ HCF of (210, 385 and 735) = 35

৯৫.

What is the HCF of 408 and 1032?

ক
8
খ
12
গ
18
ঘ
24

ব্যাখ্যা

Question: What is the HCF of 408 and 1032?

Solution:
The prime factorisation of 408 is 2 × 2 × 2 × 3 × 17

The prime factorisation of 1032 is 2 × 2 × 2 × 3 × 43

Thus, the product of the prime factors of 2 × 2 × 2 × 3 is 24

Hence, the HCF of 408 and 1032 is 24.

৯৬.

The ratio of two numbers is 3 : 4 and their H.C.F is 5. Their L.C.M is -

ক
20
খ
45
গ
60
ঘ
90

ব্যাখ্যা

Question: The ratio of two numbers is 3 : 4 and their H.C.F is 5. Their L.C.M is -

Solution:
ধরি, সংখ্যা দুটি হলো 3x এবং 4x
∴ গসাগু (H.C.F) = x = 5

∴ সংখ্যা দুটি হলো: 3 × 5 = 15 এবং 4 × 5 = 20
∴ সংখ্যাদ্বয়ের গুণফল = 15 × 20 = 300 এবং H.C.F = 5

আমরা জানি, L.C.M = (Product of two numbers)/H.C.F
= 300/5 = 60

∴ সংখ্যা দুটির লসাগু (L.C.M) = 60

৯৭.

Four metal rods of lengths 78 cm, 104 cm, 117 cm, and 169 cm are to be cut into parts of equal length. Each part must be as long as possible. What is the maximum number of pieces that can be cut?

ক
27
খ
36
গ
40
ঘ
48

ব্যাখ্যা

Question: Four metal rods of lengths 78 cm, 104 cm, 117 cm, and 169 cm are to be cut into parts of equal length. Each part must be as long as possible. What is the maximum number of pieces that can be cut?

Solution:
Maximum length of each part
= H.C.F. of 78 cm, 104 cm, 117 cm, and 169 cm
= 13 cm.

That means, if we take the H.C.F the length we get can be used to cut all 4 rods equally, so that each piece will have the same length.

∴ Number of pieces = (78 + 104 + 117 + 169)/13
= 468/13
= 36

৯৮.

The number of students in a school leaves a remainder of 4 when divided by 5, 8, and 20. Find the total number of students.

ক
40
খ
44
গ
36
ঘ
60

ব্যাখ্যা

Question: The number of students in a school leaves a remainder of 4 when divided by 5, 8, and 20. Find the total number of students.

Solution:
Given,
When the number of students is divided by 5, 8, and 20, the remainder is always 4.
So, the number of students is 4 more than the LCM of 5, 8, and 20.

Now,
Prime factors of 5 = 5 × 1
Prime factors of 8 = 2 × 2 × 2
Prime factors of 20 = 2 × 2 × 5

LCM of 5, 8, and 20 = 2 × 2 × 2 × 5 = 40

Total number of students = (40 + 4) = 44

৯৯.

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15, and 18 is:

ক
364
খ
328
গ
264
ঘ
228

ব্যাখ্যা

Question: The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15, and 18 is:

Solution:
L.C.M. of 6, 9, 15 and 18 is 90.
Let the required number be 90k + 4, which is a multiple of 7

The least value of k for which (90k + 4) is divisible by 7 is k = 4

Hence, the equired number = (90 × 4) + 4 = 364

১০০.

A merchant has three different types of milk: 324 litres, 351 litres, and 459 litres. Find the minimum number of casks of equal size that can store the milk without mixing.

ক
21
খ
28
গ
36
ঘ
42
ঙ
64

ব্যাখ্যা

Question: A merchant has three different types of milk: 324 litres, 351 litres, and 459 litres. Find the minimum number of casks of equal size that can store the milk without mixing.

Solution:
The size of each cask should be the greatest possible that divides all three quantities, i.e., H.C.F of 324, 351, and 459.

H.C.F(324, 351, 459) = 27 litres

Total milk = 324 + 351 + 459 = 1134 litres

Number of casks required = 1134 ÷ 27 = 42