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ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ]

পরীক্ষাব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ]তারিখতারিখ অনির্ধারিতসময়17 minutes
মোট প্রশ্ন১৫
সিলেবাস
Exam - 2 Daily Quiz: Math: Topic:Problems on Number, HCF & LCM
ঘনত্ব
উত্তর
উত্তরিতবর্তমানপুনরায় দেখুনঅসম্পূর্ণ

ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ]

ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ] · তারিখ অনির্ধারিত · ১৫ প্রশ্ন

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How many such positive integers exist so that if the unit digit of the original integer is removed, the ratio of the new number to the original one becomes 1/14?
  1. 4
  2. 1
  3. 2
  4. 3
ব্যাখ্যা
Question 1: How many such positive integers exist so that if the unit digit of the original integer is removed, the ratio of the new number to the original one becomes 1/14?

Solution: 
Let,
X = tens digit
Y = units digit
X/(10X + Y) = 1/14
4X = Y
Y must be a one-digit multiple of 4, either 4 or 8
Then, X must be either 1 or 2

∴ 1/(10 × 1 + 4) = 1/14
And 2/(10 × 2 + 8) = 1/14

So the numbers are 14 and 28
.
What number should be added to 102255 to make it exactly divisible by 7?
  1. 1
  2. 2
  3. 5
  4. 6
ব্যাখ্যা
Question: What number should be added to 102255 to make it exactly divisible by 7?

Solution: 
if we divide 102255 by 7 the remainder is 6
to make it divisible by 7 we need to add 1 with that number.

adding 1 we get = 102255 + 1 = 102256
dividing 102256 by 7 we get 102256/7 = 14608
.
n is an integer between 40 and 90, then any of the following could be n + 8 except - 
  1. 52
  2. 65
  3. 75
  4. 99
ব্যাখ্যা
Question: n is an integer between 40 and 90, then any of the following could be n + 8 except - 

Solution: 
if n is a positive integer, then 40 < n < 90
from option (4)
n + 8 = 99
n = 91

its not possible
.
Three numbers are in a ratio of 3 : 4 : 5 and their L.C.M is 2400. Their H.C.F is - 
  1. 20
  2. 30
  3. 40
  4. 60
ব্যাখ্যা
Question: Three numbers are in a ratio of 3 : 4 : 5 and their L.C.M is 2400. Their H.C.F is - 

Solution: 
ধরি,
সংখ্যাগুলো যথাক্রমে ৩ক, ৪ক এবং ৫ক
∴ তাদের ল.সা.গু = ৬০ক
∴ ৬০ক = ২৪০০
বা, ক = ২৪০০/৬০ = ৪০
∴ সংখ্যাগুলো যথাক্রমে 
(৩ × ৪০) = ১২০
(৪ × ৪০) = ১৬০
(৫ × ৪০) = ২০০

তাহলে, গ.সা.গু = ৪০
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If p, q, r are the digits of a number beginning from the left, the number is - 
  1. 100q + 10p + r
  2. 100p + 10q + r
  3. 100r + 10q + p
  4. rqp
ব্যাখ্যা
Question: If p, q, r are the digits of a number beginning from the left, the number is - 

Solution: 
as the sequence is from the left,
the value of p is = 100p
the value of q is = 10q
the value of r = r

∴ the number is = 100p + 10q + r
.
The H.C.F and L.C.M of two numbers are 12 and 336 respectively. If one of the numbers is 84, the other one is - 
  1. 28
  2. 32
  3. 64
  4. 48
ব্যাখ্যা
Question: The H.C.F and L.C.M of two numbers are 12 and 336 respectively. If one of the numbers is 84, the other one is - 

Solution: 
দুইটি সংখ্যার ল.সা.গু ও গ.সা.গু এর গুণফল সংখ্যা ২টির গুণফলের সমান।
ধরি,
অপর সংখ্যাটি = ক

তাহলে,
ক × ৮৪ = ১২ × ৩৩৬
বা, ক = (১২ × ৩৩৬)/৮৪
∴ ক = ৪৮
.
If x is an odd integer, which of the following is true?
  1. 5x - 2 is even
  2. 5x2 + 2 is odd
  3. 5x3 + 3 is odd
  4. None of these
ব্যাখ্যা
Question: If x is an odd integer, which of the following is true?

Solution:
let, x = 1
putting x to each option we get,
5x - 2 = (5 × 1) - 2 = 3; here 3 is an odd integer. so, option 1 is not true.
5x2 + 2 = (5 × 1) + 2 = 7; here 7 is an odd integer. so, option 2 is true.
5x2 + 3 = (5 × 1) + 3 = 8; here 8 is an even integer. so, option 3 is not true.

so only option 2 is true.
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If the sum of two numbers is 36 and their H.C.F and L.C.M are 3 and 105, respectively, the sum of the reciprocals of the two number is - 
  1. 4/35
  2. 4/39
  3. 2/17
  4. 4/33
ব্যাখ্যা
Question: If the sum of two numbers is 36 and their H.C.F and L.C.M are 3 and 105, respectively, the sum of the reciprocals of the two number is - 

Solution: 
let,
the numbers are x and y 
∴ x + y = 36 
and xy = (3 × 105) = 315

sum of their reciprocals = 1/x + 1/y 
= (x + y)/xy
= 36/315
= 4/35
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The sum of three prime numbers is 100. If one of them exceeds another by 36, then one of the numbers is-
  1. 3
  2. 67
  3. 29
  4. 41
ব্যাখ্যা
Question: The sum of three prime numbers is 100. If one of them exceeds another by 36, then one of the numbers is-

Solution:
ধরি,
তিনটি মৌলিক সংখ্যা যথাক্রমে x, y এবং y + 36.
∴ x + y + y + 36 = 100
or, x + 2y = 100 - 36
∴ x + 2y = 64
যেহেতু, তিনটি সংখ্যার যোগফল জোড় সংখ্যা তাই,
- তিনটি সংখ্যাই বিজোড় হতে পারে না।
- ২ টি বিজোড় ও একটি জোড় সংখ্যার সমষ্টিই কেবল ১০০ হতে পারে।
- মৌলিক সংখ্যাগুলোর মধ্যে একমাত্র ২ হলো জোড় সংখ্যা। বাকি সব বিজোড় সংখ্যা। তাই এদের মধ্যে একটি সংখ্যা অবশ্যই ২ হবে।
ধরি,
x = 2
তাহলে,
2 + 2y = 64
2y = 62
y = 31

∴ অপর সংখ্যাটি = 31 + 36 = 67

সংখ্যা তিনটি হলো 2, 31, 67
১০.
What is the greatest number of three digits, which, when divided by 6, 9, and 12, leaves a remainder of 3 in each case?
  1. 991
  2. 981
  3. 975
  4. 973
ব্যাখ্যা
Question: What is the greatest number of three digits, which, when divided by 6, 9, and 12, leaves a remainder of 3 in each case?

Solution: 
the largest three-digit number is = 999
the L.C.M of 6, 9, 12 is = 36
dividing 999 by 36 we get the remainder of 27

so, the number is = (999 -27) + 3 = 975
১১.
7 is added to a certain number; the sum is multiplied by 5; the product is divided by 9 and 3 is subtracted from the quotient. Thus, if the remainder left is 12, what was the original number?
  1. 20
  2. 37
  3. 30
  4. 27
ব্যাখ্যা
Question: 7 is added to a certain number; the sum is multiplied by 5; the product is divided by 9 and 3 is subtracted from the quotient. Thus, if the remainder left is 12, what was the original number?

Solution: 
let the number be x.

ATQ,
[{5(x + 7)}/9] - 3 = 12
or, 5(x + 7) = 15 × 9
or, x + 7 = 135/5
or, x + 7 = 27
or, x = 27 - 7
∴ x = 20
১২.
Find the greatest number that will divide 43, 91, and 183 and leave the same remainder.
  1. 12
  2. 8
  3. 3
  4. 4
ব্যাখ্যা
Question: Find the greatest number that will divide 43, 91, and 183 and leave the same remainder.

Solution: 
the number is the H.C.F of (91 - 43), (183 - 91) and (183 - 43)
= H.C.F of 48, 92 and 140
= 4
১৩.
The difference between a number consisting of two digits and a number formed by interchanging the digits is always divisible by - 
  1. 11
  2. 9
  3. 7
  4. 2
ব্যাখ্যা
Question: The difference between a number consisting of two digits and a number formed by interchanging the digits is always divisible by - 

Solution: 
ধরি,
একক স্থানীয় অংকটি y 
দশক স্থানীয় অংকটি x

তাহলে সংখ্যাটি = 10x + y
ডিজিটগুলো জায়গা পরিবর্তন করলে নতুন সংখ্যাটি = 10y + x

পার্থক্য = 10x + y - 10y - x
= 9x - 9y
= 9(x - y)

এখানে 9(x - y) সংখ্যাটি অবশ্যই 9 দ্বারা বিভাজ্য হবে।
১৪.
Three sets of English, Mathematics, and Science books containing 336, 240, and 96 books, respectively, have to be stacked in such a way that all the books are stored subject-wise and the height of each stack is the same. The total number of stacks will be - 
  1. 18
  2. 14
  3. 22
  4. 24
ব্যাখ্যা
Question: Three sets of English, Mathematics, and Science books containing 336, 240, and 96 books, respectively, have to be stacked in such a way that all the books are stored subject-wise and the height of each stack is the same. The total number of stacks will be - 

Solution: 
প্রত্যেক তাকে বইয়ের সংখ্যা হবে ৩৩৬, ২৪০ এবং ৯৬ এর গ.সা.গু. এর সমান।
৩৩৬, ২৪০ এবং ৯৬ এর গ.সা.গু = ৪৮

মোট তাক হবে = (৩৩৬ + ২৪০ + ৯৬)/৪৮
= ১৪টি
১৫.
A number when divided by the sum of 55 and 45 gives two times their difference as quotient and 11 as the remainder. The number is - 
  1. 20011
  2. 2111
  3. 2011
  4. 3011
ব্যাখ্যা
Question: A number when divided by the sum of 55 and 45 gives two times their difference as quotient and 11 as the remainder. The number is - 

Solution: 
the sum of 55 and 45 is = (55 + 45) = 100
the difference of 55 and 45 is = (55 - 45) = 10

so, the number is = {(2 × 10) × 100} + 11
= 2000 + 11
= 2011