১,০০১.
What is 3% of 0.07?
ব্যাখ্যা
Question: What is 3% of 0.07?
Solution:
0.07 এর 3%
= 0.07 এর 3/100
= 0.21/100
= 0.0021
Solution:
0.07 এর 3%
= 0.07 এর 3/100
= 0.21/100
= 0.0021
PrepBank · বিষয়ভিত্তিক প্রশ্ন
Number System, Problems on Number
Number System, Problems on Number
PrepBank · পাতা ১১ / ১৮ · ১,০০১–১,১০০ / ১,৭৩৬
১,০০২.
ব্যাখ্যা
Question:
Solution:
Solution:
১,০০৩.
0.1 is how many times greater than 0.001?
ব্যাখ্যা
প্রশ্ন: 0.1 is how many times greater than 0.001?
সমাধান:
0.1/0.001
= 0.1/(1/1000)
= 0.1 × 1000
= 100
সমাধান:
0.1/0.001
= 0.1/(1/1000)
= 0.1 × 1000
= 100
১,০০৪.
Which of the following number is divisible by 9?
ব্যাখ্যা
Question: Which of the following number is divisible by 9?
Solution:
A number is divisible by 9 if the sum of its digits is divisible by 9.
The Sum of the digits of the given numbers are;
5+6+7+8+5= 31
4+5+6+7+8= 30
6+5+8+8+9= 36
6+7+5+7+8= 33
The sum of the digits of the number 65889 is 36, which is divisible by 9, so the correct answer is 65889.
Solution:
A number is divisible by 9 if the sum of its digits is divisible by 9.
The Sum of the digits of the given numbers are;
5+6+7+8+5= 31
4+5+6+7+8= 30
6+5+8+8+9= 36
6+7+5+7+8= 33
The sum of the digits of the number 65889 is 36, which is divisible by 9, so the correct answer is 65889.
১,০০৫.
What is the sum of all positive integers up to 1000, which are divisible by 5 and are not divisible by 2?
ব্যাখ্যা
The positive integers, which are divisible by 5 are 5, 10, 15, ....., 1000
Out of these 10, 20, 30, ......, 1000 are divisible by 2
Thus, we have to find the sum of the positive integers 5, 15, 25, ......, 995
If n is the number of terms in it the sequence then
995 = 5 + 10(n - 1)
⇒ 1000 = 10n
∴ n = 100
Thus the sum of the series = (1st term + last term) × Number of term ÷ 2
= (5 + 995) × 100 ÷ 2 = 50000
Out of these 10, 20, 30, ......, 1000 are divisible by 2
Thus, we have to find the sum of the positive integers 5, 15, 25, ......, 995
If n is the number of terms in it the sequence then
995 = 5 + 10(n - 1)
⇒ 1000 = 10n
∴ n = 100
Thus the sum of the series = (1st term + last term) × Number of term ÷ 2
= (5 + 995) × 100 ÷ 2 = 50000
১,০০৬.
a, b, c, d and e are five consecutive integers in increasing order of size. Which one of the following expressions is not odd?
ব্যাখ্যা
Best way to solve this question is assuming the value of a, b, c, d and e and test the option.
১,০০৭.
The sum of twice a number and three times of 42 is 238. What is the sum thrice the number and two times of 42?
ব্যাখ্যা
Question: The sum of twice a number and three times of 42 is 238. What is the sum thrice the number and two times of 42?
Solution:
মনে করি,
সংখ্যাটি = x
১ম শর্তমতে,
2x + (3 × 42) = 238
বা, 2x + 126 = 238
বা, 2x = 112
বা, x = 56
∴ 3x + (2 × 42)
= (3 × 56) + (2 × 42)
= 168 + 84
= 252
Solution:
মনে করি,
সংখ্যাটি = x
১ম শর্তমতে,
2x + (3 × 42) = 238
বা, 2x + 126 = 238
বা, 2x = 112
বা, x = 56
∴ 3x + (2 × 42)
= (3 × 56) + (2 × 42)
= 168 + 84
= 252
১,০০৮.
How many times does the digit '4' come to write numbers from 10 to 100?
ব্যাখ্যা
14 থেকে 34 পর্যন্ত = 3 টি
40 থেকে 49 পর্যন্ত = 11 টি
54 থেকে 94 পর্যন্ত = 5 টি
_____________________________
মোট = 19 টি
১,০০৯.
The difference between two numbers is 12. If 1 added to the greater number, it becomes twice the smaller number. Calculate the two numbers?
ব্যাখ্যা
Question: The difference between two numbers is 12. If 1 added to the greater number, it becomes twice the smaller number. Calculate the two numbers?
Solution:
ধরি,
বড় সংখ্যাটি = x
ছোট সংখ্যাটি = y
প্রশ্নমতে,
x - y = 12 ................... (1)
x + 1 = 2y
বা, x - 2y = - 1 .................... (2)
(1) নং থেকে (2) নং বিয়োগ করে পাই,
x - y = 12
x - 2y = - 1
y = 13
y এর মান (1) নং বসিয়ে পাই,
x - 13 = 12
বা, x = 12 + 13
∴ x = 25
Solution:
ধরি,
বড় সংখ্যাটি = x
ছোট সংখ্যাটি = y
প্রশ্নমতে,
x - y = 12 ................... (1)
x + 1 = 2y
বা, x - 2y = - 1 .................... (2)
(1) নং থেকে (2) নং বিয়োগ করে পাই,
x - y = 12
x - 2y = - 1
y = 13
y এর মান (1) নং বসিয়ে পাই,
x - 13 = 12
বা, x = 12 + 13
∴ x = 25
১,০১০.
The difference between a number and its one-fourth is 60. What is the number?
ব্যাখ্যা
Question: The difference between a number and its one-fourth is 60. What is the number?
Solution:
Let the number be x
ATQ,
x - (x/4) = 60
⇒ (4x - x)/4 = 60
⇒ 3x/4 = 60
⇒ 3x = 240
⇒ x = 80
Solution:
Let the number be x
ATQ,
x - (x/4) = 60
⇒ (4x - x)/4 = 60
⇒ 3x/4 = 60
⇒ 3x = 240
⇒ x = 80
১,০১১.
The square root of (6 + 5√2)(6 - 5√2) is:
ব্যাখ্যা
Question: The square root of (6 + 5√2)(6 - 5√2) is:
Solution:
√{(6 + 5√2)(6 - 5√2)}
= √{62 - (5√2)2} [∵ (a + b)(a - b) = a2 - b2]
= √{36 - (25 × 2)}
= √{36 - 50}
= √(-14)
= √{14(- 1)}
= √14 × √(- 1)
= i√14 [যেখানে i2 = - 1]
১,০১২.
The product P of two prime numbers is between 9 and 55. If one of the prime numbers is greater than 2 but less than 6 and the other is greater than 13 but less than 25, then P =
ব্যাখ্যা
one of the prime numbers is greater than 2 but less than 6 = 3 or 5
the other is greater than 13 but less than 25 = 17, 19 or 23
17 × 3, 19 × 3 or 23 × 3 ⇒ 51, 57 or 69
17 × 5, 19 × 5 or 23 × 5 ⇒ 85, 95 or 115
Option 51 is among 51, 57, 69, 85, 95 or 115
the other is greater than 13 but less than 25 = 17, 19 or 23
17 × 3, 19 × 3 or 23 × 3 ⇒ 51, 57 or 69
17 × 5, 19 × 5 or 23 × 5 ⇒ 85, 95 or 115
Option 51 is among 51, 57, 69, 85, 95 or 115
১,০১৩.
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 -
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
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
১,০১৪.
A prime number N, in the range of 10 to 40, remains unchanged when its digits are reversed. The square of such a number is:
ব্যাখ্যা
Question: A prime number N, in the range of 10 to 40, remains unchanged when its digits are reversed. The square of such a number is:
Solution:
মৌলিক সংখ্যা ১১ এর একক ও দশক স্থানীয় অঙ্ক একই।
১১২
= ১২১
Solution:
মৌলিক সংখ্যা ১১ এর একক ও দশক স্থানীয় অঙ্ক একই।
১১২
= ১২১
১,০১৫.
A number is decreased by 10% and then increased by 10%. The number of obtained is 10 less than the original number. What was the original number?
ব্যাখ্যা
ধরি,
মূল সংখ্যা = 100x
10% কমলে দাঁড়ায় = 100x - 100x এর 10%
= 90x
90x সংখ্যাটি 10% বাড়ালে নতুন সংখ্যা = 90x + 90x এর 10%
= 99x
প্রশ্নমতে,
100x - 99x = 10
x = 10
নির্ণেয় সংখ্যা = 100x
=100 × 10 = 1000
মূল সংখ্যা = 100x
10% কমলে দাঁড়ায় = 100x - 100x এর 10%
= 90x
90x সংখ্যাটি 10% বাড়ালে নতুন সংখ্যা = 90x + 90x এর 10%
= 99x
প্রশ্নমতে,
100x - 99x = 10
x = 10
নির্ণেয় সংখ্যা = 100x
=100 × 10 = 1000
১,০১৬.
If n is negative, all but which of the following must also be negative?
ব্যাখ্যা
Since option C has a positive power so that it will be positive.
১,০১৭.
A number when divided by 195 leaves a remainder 47. If the same number is divided by 15, the remainder will be?
ব্যাখ্যা
Question: A number when divided by 195 leaves a remainder 47. If the same number is divided by 15, the remainder will be?
Solution:
Let the number be x and the quotient is q.
Then,
x = 195q + 47
= (15 × 13q) + (15 × 3) + 2
= 15 (13q + 3) + 2
So, the given number when divided by 15 gives 2 as remainder.
Solution:
Let the number be x and the quotient is q.
Then,
x = 195q + 47
= (15 × 13q) + (15 × 3) + 2
= 15 (13q + 3) + 2
So, the given number when divided by 15 gives 2 as remainder.
১,০১৮.
There are two numbers such that the sum of twice the first and thrice the second is 39, while the sum of thrice the first, and twice the second is 36. The larger of the two is -
ব্যাখ্যা
Let the numbers be x and y.
Then, 2x + 3y = 39 .......(i) and
3x + 2y = 30 .........(ii)
On solving (i) and (ii), we get : x = 6 and y = 9.
larger number = 9
১,০১৯.
Three numbers are in the ratio of 3 : 4 : 5 and their H.C.F. is 20. Their L.C.M. is -
ব্যাখ্যা
Question: Three numbers are in the ratio of 3 : 4 : 5 and their H.C.F. is 20. Their L.C.M. is -
Solution:
সংখ্যা তিনটির গ. সা. গু ২০
সংখ্যা তিনটি হচ্ছে (৩ × ২০) বা ৬০, (৪ × ২০) বা ৮০, (৫ × ২০) বা ১০০
৬০ = ৩ × ২ × ২ × ৫
৮০ = ২ × ২ × ২ × ২ × ৫
১০০ = ২ × ২ × ৫ ×৫
ল. সা. গু = ২ × ২ × ২ × ২ × ৩ × ৫ × ৫
= ১২০০
Solution:
সংখ্যা তিনটির গ. সা. গু ২০
সংখ্যা তিনটি হচ্ছে (৩ × ২০) বা ৬০, (৪ × ২০) বা ৮০, (৫ × ২০) বা ১০০
৬০ = ৩ × ২ × ২ × ৫
৮০ = ২ × ২ × ২ × ২ × ৫
১০০ = ২ × ২ × ৫ ×৫
ল. সা. গু = ২ × ২ × ২ × ২ × ৩ × ৫ × ৫
= ১২০০
১,০২০.
If 382A is divisible by 9, find the value of the smallest natural number A?
ব্যাখ্যা
Question: If 382A is divisible by 9, find the value of the smallest natural number A?
Solution:
একটি সংখ্যা ৯ দ্বারা বিভাজ্য হবে, যদি সংখ্যাটির অংকগুলোর যোগফল ৯ দ্বারা বিভাজ্য হয়।
৩ + ৮ + ২ = ১৩। ১৩ এর সাথে ৫ যোগ করলে ১৮ হয় যা ৯ দ্বারা বিভাজ্য।
অতএব, A অঙ্কটি হবে ৫।
Solution:
একটি সংখ্যা ৯ দ্বারা বিভাজ্য হবে, যদি সংখ্যাটির অংকগুলোর যোগফল ৯ দ্বারা বিভাজ্য হয়।
৩ + ৮ + ২ = ১৩। ১৩ এর সাথে ৫ যোগ করলে ১৮ হয় যা ৯ দ্বারা বিভাজ্য।
অতএব, A অঙ্কটি হবে ৫।
১,০২১.
What is the greatest prime factor of (42)2 - 1?
ব্যাখ্যা
এখানে
(42)2 - 1 = (42)2 - 12
= (42 + 1)(42 - 1)
= (16 + 1)(16 - 1)
= 17 × 15
= 17 × 3 × 5
সুতরাং, বৃহত্তম মৌলিক উৎপাদক হলো 17
(42)2 - 1 = (42)2 - 12
= (42 + 1)(42 - 1)
= (16 + 1)(16 - 1)
= 17 × 15
= 17 × 3 × 5
সুতরাং, বৃহত্তম মৌলিক উৎপাদক হলো 17
১,০২২.
100 × 10 - 100 + 2000 ÷ 100 = ?
ব্যাখ্যা
Question: 100 × 10 - 100 + 2000 ÷ 100 = ?
Solution:
100 × 10 - 100 + 2000 ÷ 100
=100 × 10 - 100 + 20
= 1000 - 100 + 20
= 1020 - 100
= 920
Solution:
100 × 10 - 100 + 2000 ÷ 100
=100 × 10 - 100 + 20
= 1000 - 100 + 20
= 1020 - 100
= 920
১,০২৩.
The sum of 3 consecutive integers is less than 75. What is the greatest possible value of the smallest one?
ব্যাখ্যা
প্রশ্ন:The sum of 3 consecutive integers is less than 75. What is the greatest possible value of the smallest one?
সমাধান:
ধরি,
সংখ্যা তিনটি যথাক্রমে x, x + 1, x + 2
প্রশ্নমতে,
x + 2 + x + x + 1 < 75
3x + 3 < 75
3x + 3 - 3 < 75 - 3
3x < 72
x/3 < 72/3
x < 24
ছোট সংখ্যাটি = ( 24 - 1 ) = 23
১,০২৪.
If the sum of (3, 7 and x) is 18, then the average of (3, 7 and x) is
ব্যাখ্যা
Question: If the sum of (3, 7 and x) is 18, then the average of (3, 7 and x) is
Solution:
3 + 7 + x = 18
⇒ x = 18 - 10
∴ x = 8
∴ the average of (3, 7 and 8) is = the sum of (3, 7 and 8)/3 = 18/3 = 6
Solution:
3 + 7 + x = 18
⇒ x = 18 - 10
∴ x = 8
∴ the average of (3, 7 and 8) is = the sum of (3, 7 and 8)/3 = 18/3 = 6
১,০২৫.
The difference between the two numbers is 1365. When the large number is divided by the smaller one, the quotient is 6 and the remainder is 15. The smaller number is :
ব্যাখ্যা
Let the numbers be x and (x + 1365)
Then,
⇒ x + 1365 = 6x + 15
⇒ 5x = 1350
⇒ x = 270
Then,
⇒ x + 1365 = 6x + 15
⇒ 5x = 1350
⇒ x = 270
১,০২৬.
The ratio of two numbers is 3 : 4 and their sum is 630. The smaller one of the two numbers is
ব্যাখ্যা
প্রশ্ন: The ratio of two numbers is 3 : 4 and their sum is 630. The smaller one of the two numbers is
সমাধান:
দেয়া আছে ,
দুটি সংখ্যার অনুপাত = 3 : 4
ধরি,
ক্ষুদ্রতম সংখ্যাটি = 3x
বৃহত্তম সংখ্যাটি = 4x
প্রশ্নমতে,
3x + 4x = 630
7x = 630
x = 90
ক্ষুদ্রতম সংখ্যাটি = 3× 90 = 270
সমাধান:
দেয়া আছে ,
দুটি সংখ্যার অনুপাত = 3 : 4
ধরি,
ক্ষুদ্রতম সংখ্যাটি = 3x
বৃহত্তম সংখ্যাটি = 4x
প্রশ্নমতে,
3x + 4x = 630
7x = 630
x = 90
ক্ষুদ্রতম সংখ্যাটি = 3× 90 = 270
১,০২৭.
A ferry can carry 24 buses or 36 cars at a time. If there are 18 buses on the ferry, how many cars can be loaded onto it?
ব্যাখ্যা
এখানে 24টি বাস = 36টি গাড়ি
1 টি বাস = 36/24 = 1.5টি গাড়ি
যেহেতু ফেরিটি 18 টি বাস বহন করে ফেলছে, সুতরাং আরো বহন করতে পারবে = 24 - 18 = 6 টি বাস = (6×1.5) টি গাড়ি অর্থাৎ 9 টি গাড়ি।
১,০২৮.
By how many of the following numbers is 212 - 1 divisible?
2, 3, 5, 7, 10, 11, 13, 14
2, 3, 5, 7, 10, 11, 13, 14
ব্যাখ্যা
Question: By how many of the following numbers is 212 - 1 divisible?
2, 3, 5, 7, 10, 11, 13, 14
Solution:
(212 - 1) = (4096 - 1) = 4095, which is clearly divisible by 3, 5, 7 and 13
∴ 4 numbers in all.
2, 3, 5, 7, 10, 11, 13, 14
Solution:
(212 - 1) = (4096 - 1) = 4095, which is clearly divisible by 3, 5, 7 and 13
∴ 4 numbers in all.
১,০২৯.
Students of a class stand in a queue. Suppose Rakib is 13th from the start and 7th from the end. How many students are there?
ব্যাখ্যা
Question: Students of a class stand in a queue. Suppose Rakib is 13th from the start and 7th from the end. How many students are there?
Solution:
যেহেতু রাকিব শুরু থেকে ১৩তম অবস্থানে আছে, তাই শুরু থেকে রাকিব সহ মোট ছাত্র আছে ১৩ জন।
আবার যেহেতু রাকিব শেষ থেকে ৭ম অবস্থানে আছে, তাই শেষ থেকে রাকিব ছাড়া ছাত্র আছে ৬ জন।
∴ মোট ছাত্র = ১৩ + ৬ = ১৯ জন।
Solution:
যেহেতু রাকিব শুরু থেকে ১৩তম অবস্থানে আছে, তাই শুরু থেকে রাকিব সহ মোট ছাত্র আছে ১৩ জন।
আবার যেহেতু রাকিব শেষ থেকে ৭ম অবস্থানে আছে, তাই শেষ থেকে রাকিব ছাড়া ছাত্র আছে ৬ জন।
∴ মোট ছাত্র = ১৩ + ৬ = ১৯ জন।
১,০৩০.
The H.C.F and L.C.M of two numbers are 12 and 288 respectively. If one of the numbers is 96. Find the other.
ব্যাখ্যা
আমরা জানি,
গ.সা.গু x ল.সা.গু = সংখ্যাদ্বয়ের গুণফল
⇒ 12 x 288 = 96 x অপর সংখ্যা
⇒ অপর সংখ্যা = (12×288)/96
∴ অপর সংখ্যা = 36
১,০৩১.
Max goes to the gym every fourth day, Ellen's exercise routine is every third day. Today is Monday and both Max and Ellen are at the gym. What will the day of the week be the next time they are both at the gym?
ব্যাখ্যা
LCM of 4 and 3 is 12. So they meet after every 12 day.
12 day after Monday is Saturday.
So, they will meet on Saturday again after the meeting on Monday.
১,০৩২.
Two numbers are in the ratio 5 : 6. If the difference of their squares is 99, then the smallest number is -
ব্যাখ্যা
Question: Two numbers are in the ratio 5 : 6. If the difference of their squares is 99, then the smallest number is -
Solution:
Let
the numbers are 5x, 6x.
Now,
(6x)2 - (5x)2 = 99
⇒ 36x2 - 25x2 = 99
⇒ 11x2 = 99
⇒ x2 = 9
∴ x = 3
The smallest number is = 5 × 3 = 15
Solution:
Let
the numbers are 5x, 6x.
Now,
(6x)2 - (5x)2 = 99
⇒ 36x2 - 25x2 = 99
⇒ 11x2 = 99
⇒ x2 = 9
∴ x = 3
The smallest number is = 5 × 3 = 15
১,০৩৩.
A natural number when increased by 5, equals 126 times reciprocal. The number is-
ব্যাখ্যা
Question: A natural number when increased by 5, equals 126 times reciprocal. The number is-
Solution:
Let,
the number be x.
Then,
x + 5 = 126 × (1/x)
⇒ x2 + 5x - 126 = 0
⇒ x2 + 14x - 9x - 126 = 0
⇒ x(x + 14) - 9(x + 14) = 0
⇒ (x + 14)(x - 9) = 0
∴ x= - 14, 9
Therefore, the required number is 9.
Solution:
Let,
the number be x.
Then,
x + 5 = 126 × (1/x)
⇒ x2 + 5x - 126 = 0
⇒ x2 + 14x - 9x - 126 = 0
⇒ x(x + 14) - 9(x + 14) = 0
⇒ (x + 14)(x - 9) = 0
∴ x= - 14, 9
Therefore, the required number is 9.
১,০৩৪.
The sum of twice a number and three times of 42 is 238. What is the sum of thrice the number and two times of 42?
ব্যাখ্যা
Question: The sum of twice a number and three times of 42 is 238. What is the sum of thrice the number and two times of 42?
Solution:
Let the number be 'p'
According to the question,
⇒ 2p + 3 × 42 = 238
⇒ 2p + 126 = 238
⇒ 2p = 112
⇒ p = 56
∴ Required sum:
= 3p + 2 × 42
= 3 × 56 + 2 × 42
=168 + 84
= 252
Solution:
Let the number be 'p'
According to the question,
⇒ 2p + 3 × 42 = 238
⇒ 2p + 126 = 238
⇒ 2p = 112
⇒ p = 56
∴ Required sum:
= 3p + 2 × 42
= 3 × 56 + 2 × 42
=168 + 84
= 252
১,০৩৫.
What is the square root of 0.16?
ব্যাখ্যা
Question: What is the square root of 0.16?
Solution:
√0.16 = 0.4
Solution:
√0.16 = 0.4
১,০৩৬.
একটি বিজোড় পূর্ণ সংখ্যার সাত গুণের সাথে পরবর্তী বিজোড় পূর্ণ সংখ্যার পাঁচ গুণ যোগ করলে ৯৪ হয়। প্রথম বিজোড় সংখ্যাটি কত?
ব্যাখ্যা
প্রশ্ন: একটি বিজোড় পূর্ণ সংখ্যার সাত গুণের সাথে পরবর্তী বিজোড় পূর্ণ সংখ্যার পাঁচ গুণ যোগ করলে ৯৪ হয়। প্রথম বিজোড় সংখ্যাটি কত?
সমাধান:
ধরি,
প্রথম বিজোড় পূর্ণ সংখ্যাটি = ক
তাহলে, পরবর্তী বিজোড় পূর্ণ সংখ্যা = (ক + ২)
প্রশ্নমতে,
৭ক + ৫(ক + ২) = ৯৪
⇒ ৭ক + ৫ক + ১০ = ৯৪
⇒ ১২ক = ৯৪ - ১০
⇒ ১২ক = ৮৪
∴ ক = ৭
সমাধান:
ধরি,
প্রথম বিজোড় পূর্ণ সংখ্যাটি = ক
তাহলে, পরবর্তী বিজোড় পূর্ণ সংখ্যা = (ক + ২)
প্রশ্নমতে,
৭ক + ৫(ক + ২) = ৯৪
⇒ ৭ক + ৫ক + ১০ = ৯৪
⇒ ১২ক = ৯৪ - ১০
⇒ ১২ক = ৮৪
∴ ক = ৭
১,০৩৭.
If the sum of five consecutive odd integers is 255, what is the largest number?
ব্যাখ্যা
Question: If the sum of five consecutive odd integers is 255, what is the largest number?
Solution:
ধরি, মাঝের সংখ্যাটি = x
সুতরাং, 5টি ক্রমিক বিজোড় সংখ্যা হবে যথাক্রমে: x - 4, x - 2, x, x + 2 এবং x + 4
প্রশ্নমতে,
(x - 4) + (x - 2) + x + (x + 2) + (x + 4) = 255
⇒ 5x = 255
⇒ x = 51
∴মাঝের সংখ্যা, x = 51
∴ সবচেয়ে বড় সংখ্যা = x + 4 = 51 + 4 = 55
১,০৩৮.
There are three numbers, these are co-prime to each other such that the product of the first two is 551 and that of the last two is 1073. What will be the sum of three numbers :
ব্যাখ্যা
As given the questions these numbers are co-primes, so there is only 1 as their common factor.
It is also given that two products have a middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29;
So the first number is: 551/29 = 19
Third number = 1073/29 = 37
So sum of these numbers is = (19 + 29 + 37) = 85
১,০৩৯.
Find the next number in the following pattern: 320, 160, 80, 40,…
ব্যাখ্যা
320, 160, 80, 40 ..... in this series every number is half of the preceding number.
So, the next number is 20
So, the next number is 20
১,০৪০.
The least number by which 150 must be multiplied to make it a perfect square is:
ব্যাখ্যা
Question: The least number by which 150 must be multiplied to make it a perfect square is:
Solution:
Prime factorization of 150:
150 = 2 × 3 × 5 × 5
= 21 × 31 × 52
Here, the powers of 2 and 3 are odd.
∴ To make it a perfect square, we need to multiply by 2 × 3 = 6.
১,০৪১.
A scored 30% marks and failed by 15 marks. B scored 40% marks and obtained 35 marks more than those required to pass. The pass percentage is-
ব্যাখ্যা
Question: A scored 30% marks and failed by 15 marks. B scored 40% marks and obtained 35 marks more than those required to pass. The pass percentage is-
Solution:
Let, total mark is x
Now
0.3x + 15 = 0.4x - 35
⇒ 0.4x - 0.3x = 35 + 15
⇒ 0.1x = 50
⇒ x = 50/0.1
∴ x = 500
Pass mark = 0.3 × 500 + 15
= 150 + 15
= 165
Pass percentage = (165/500) × 100%
= 33%
Solution:
Let, total mark is x
Now
0.3x + 15 = 0.4x - 35
⇒ 0.4x - 0.3x = 35 + 15
⇒ 0.1x = 50
⇒ x = 50/0.1
∴ x = 500
Pass mark = 0.3 × 500 + 15
= 150 + 15
= 165
Pass percentage = (165/500) × 100%
= 33%
১,০৪২.
Three numbers which are co-prime to each other are such that the product of the first two is 119 and that of the last two is 391. What is the sum of the three numbers?
ব্যাখ্যা
Since the numbers are co-prime, their HCF = 1
Product of first two numbers = 119
Product of last two numbers = 391
The middle number is common in both of these products. Hence,
if we take HCF of 119 and 391, we get the common middle number.
HCF of 119 and 391 = 17
⇒ Middle Number = 17
First Number = 119/17 = 7
Last Number = 391/17 = 23
Sum of the three numbers = 7 + 17 + 23 = 47.
১,০৪৩.
Two numbers when divided by 13, leaves remainder 11 and 9 respectively. If the sum of those two numbers is divided by 13, the remainder will be?
ব্যাখ্যা
Question: Two numbers when divided by 13, leaves remainder 11 and 9 respectively. If the sum of those two numbers is divided by 13, the remainder will be?
Solution:
Here,
Dividend = divisor × quotient + remainder
First number = (13 × n) + 11
And, Second number = (13 × n) + 9
Let,
n = 1
∴ first number = (13 × 1) + 11 = 24
and, second number = (13 ×) + 9 = 22
Now,
after adding these two the reminder is = (24 + 22)/13
∴ reminder = 7
Solution:
Here,
Dividend = divisor × quotient + remainder
First number = (13 × n) + 11
And, Second number = (13 × n) + 9
Let,
n = 1
∴ first number = (13 × 1) + 11 = 24
and, second number = (13 ×) + 9 = 22
Now,
after adding these two the reminder is = (24 + 22)/13
∴ reminder = 7
১,০৪৪.
A family has several children. Each boy in the family has as many sisters as brothers and each girl has twice as many brothers as sisters. How many total brothers and sisters are there?
ব্যাখ্যা
Question: A family has several children. Each boy in the family has as many sisters as brothers and each girl has twice as many brothers as sisters. How many total brothers and sisters are there? [একটি পরিবারে কয়েকটি শিশু আছে। ঐ পরিবারের প্রতিটি ছেলের যতজন ভাই আছে, তার সমান বোন রয়েছে এবং প্রতিটি মেয়ের যতজন বোন আছে তার দ্বিগুণ ভাই আছে। পরিবারে মোট কত জন ভাই বোন আছে?]
Solution:
Let the number of boys in the family be = x
Then, each boy has (x - 1) brothers and (x - 1) sisters. [He himself will be excluded during the counting of brothers]
As many sisters as a boy has, they are the girls of that family, so
So, the number of girls = (x - 1)
Now, each girl has x brothers and (x - 1 - 1) = (x - 2) sisters [She himself will be excluded during the counting of sisters]
According to the question,
x = 2(x - 2)
⇒ x = 2x - 4
⇒ x = 4
So, there are 4 brothers and (4 - 1) = 3 sisters.
∴ There are total brothers and sisters = (4 + 3) = 7
Solution:
Let the number of boys in the family be = x
Then, each boy has (x - 1) brothers and (x - 1) sisters. [He himself will be excluded during the counting of brothers]
As many sisters as a boy has, they are the girls of that family, so
So, the number of girls = (x - 1)
Now, each girl has x brothers and (x - 1 - 1) = (x - 2) sisters [She himself will be excluded during the counting of sisters]
According to the question,
x = 2(x - 2)
⇒ x = 2x - 4
⇒ x = 4
So, there are 4 brothers and (4 - 1) = 3 sisters.
∴ There are total brothers and sisters = (4 + 3) = 7
১,০৪৫.
When a two-digit number is reversed and added to itself we get 143. The product of the digits of that number is 36. What is the number?
ব্যাখ্যা
Let, the number be a and b.
When it is reversed and added to itself we get (10a + b) + (10b + a)
= 11a + 11b
= 11(a + b)
We are given,
143 = 11(a + b)
a + b = 143/11
a + b = 13
so the digits are a and 13 - a.
We are given their products as a(13 - a) = 36, which is a quadratic expression.
a(13 - a) = 36
13a - a2 = 36
-a2 + 13a - 36 = 0
a2 - 13a + 36 = 0
a2 - 9a - 4a + 36 = 0
a(a - 9) -4(a - 9) = 0
a - 9)(a - 4) = 0
a = 9 or a = 4
So, the number could be 49 or 94.
Hence the option is 49 then the answer will be 49.
১,০৪৬.
Find the number which when multiplied by 15 is increased by 196.
ব্যাখ্যা
Question: Find the number which when multiplied by 15 is increased by 196.
Solution:
Let,
the number be x.
Then,
15x - x = 196
⇒ 14x = 196
⇒ x = 14
Solution:
Let,
the number be x.
Then,
15x - x = 196
⇒ 14x = 196
⇒ x = 14
১,০৪৭.
If 0 ≤ x ≤ 4 and y < 6, which of the following cannot be the value of xy?
ব্যাখ্যা
Question: If 0 ≤ x ≤ 4 and y < 6, which of the following cannot be the value of xy?
Solution:
y < 6 হলে y এর মান 5, 4, 3, 2, 1, 0, - 1,..............
0 ≤ x ≤ 4 হলে x এর মান 0, 1, 2, 3, 4
এখন
x = 0, y = 1 হলে xy = 0 × 1 = 0
x = 2, y = - 1 হলে xy = 2 × (- 1) = - 2
x = 3, y = 2 হলে xy = 3 × 2 = 6
সঠিক উত্তর: None of these
১,০৪৮.
Sixty-five percent of a number is 21 less than; four-fifth of that number. What is the number?
ব্যাখ্যা
Let the number be x
As given 65% of x = 4/5 of x -21
So, solving above equation:
65 X x/100= 4 X x/5 - 21
⇒ 65 X x /100 = ( 4x - 105)/5
⇒ (65x) X 5 = (4x- 105) X 100
⇒ 65 x = (4x -105) X 20
⇒ 65 x = 80 x - 2100
⇒ 15x = 2100
∴ x = 140
So, the number is 140.
১,০৪৯.
A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. The fraction obtained is -
ব্যাখ্যা
Question: A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. The fraction obtained is -
Solution:
Let, the fraction is a/b
Now, (a - 1)/b = 1/3
⇒ b = 3a - 3
Again, a/(b + 8) = 1/4
⇒ b + 8 = 4a
⇒ b = 4a - 8
∴ 3a - 3 = 4a - 8
⇒ 4a - 3a = 8 - 3
⇒ a = 5
And, b = (3 × 5) - 3
= 15 - 3
= 12
∴ The fraction = a/b = 5/12
১,০৫০.
A number, when decreased by 175 and then increased by 130, equals 297. Find the number.
ব্যাখ্যা
Question: A number, when decreased by 175 and then increased by 130, equals 297. Find the number.
Solution:
Let the number be = x
According to the question,
(x - 175) + 130 = 297
⇒ x - 45 = 297
⇒ x = 297 + 45
⇒ x = 342
So, the number is 342
Solution:
Let the number be = x
According to the question,
(x - 175) + 130 = 297
⇒ x - 45 = 297
⇒ x = 297 + 45
⇒ x = 342
So, the number is 342
১,০৫১.
The difference between a two-digit number and the number obtained by interchanging the digits is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1: 2?
ব্যাখ্যা
Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit.
Lets, ten's and unit's be 2x and x respectively
Then, (10 times; 2x + x) - (10x + 2x) = 36
⇒ 9x = 36
⇒ x = 4
∴ Required difference (2x + x) - (2x - x ) = 2x = 4 × 2 = 8
Answer : 8
১,০৫২.
If n is an odd integer, which of the following must be an odd integer?
ব্যাখ্যা
Let, n = 3
n - 1 = 2
n + 1 = 4
4n + 1 = 13
3n + 1 = 10
১,০৫৩.
A and B are two positive integers such that AB = 60. Which of the following cannot be the value of A + B?
ব্যাখ্যা
Question: A and B are two positive integers such that AB = 60. Which of the following cannot be the value of A + B?
Solution:
Factor pairs of 60:
(1, 60) → A + B = 61
(2, 30) → A + B = 32
(3, 20) → A + B = 23
(4, 15) → A + B = 19
(5, 12) → A + B = 17
(6, 10) → A + B = 16
So, possible values of A + B are: 61, 32, 23, 19, 17, 16.
Among the options, 18 is not possible.
১,০৫৪.
The sum of three numbers is 144. The ratio of the first number to the second number is 2 : 3 and between the second and third numbers, this ratio is 3 : 4, find the second number.
ব্যাখ্যা
Question: The sum of three numbers is 144. The ratio of the first number to the second number is 2 : 3 and between the second and third numbers, this ratio is 3 : 4, find the second number.
Solution:
Given that,
The sum of three numbers is = 144
The ratio of first number to second number is = 2 : 3
The ratio of second number of third number is = 3 : 4
The ratio of first, second and third number = 2 : 3 : 4
Let the numbers are 2x, 3x , 4x
According to the question
2x + 3x + 4x = 144
⇒ 9x = 144
⇒ x = 144/9
⇒ x = 16
∴ The second number is
= 3 × 16
= 48
Solution:
Given that,
The sum of three numbers is = 144
The ratio of first number to second number is = 2 : 3
The ratio of second number of third number is = 3 : 4
The ratio of first, second and third number = 2 : 3 : 4
Let the numbers are 2x, 3x , 4x
According to the question
2x + 3x + 4x = 144
⇒ 9x = 144
⇒ x = 144/9
⇒ x = 16
∴ The second number is
= 3 × 16
= 48
১,০৫৫.
If n and p are both odd numbers, which of the following is an even number?
ব্যাখ্যা
Option 1: When n and p is odd number then product of n and p may be odd.
Option 2: When n and p is odd number then n + p is always even number
Option 3: When n and p is odd number then product of n and p may be odd.
⇒ np is odd
⇒ np + 2is odd
Option 4:
When n and p is odd number then n + p is always even number
⇒ n + p is even ⇒ n + p + 1 is become odd
∴ n + p will give even number
Option 2: When n and p is odd number then n + p is always even number
Option 3: When n and p is odd number then product of n and p may be odd.
⇒ np is odd
⇒ np + 2is odd
Option 4:
When n and p is odd number then n + p is always even number
⇒ n + p is even ⇒ n + p + 1 is become odd
∴ n + p will give even number
১,০৫৬.
The sum of three consecutive odd integers is 40 more than the first of the numbers. What is the middle number?
ব্যাখ্যা
Question: The sum of three consecutive odd integers is 40 more than the first of the numbers. What is the middle number?
Solution:
Let's denote the three consecutive odd integers as x, x + 2, and x + 4
x + (x + 2) + (x + 4) = x + 40
⇒ 3x + 6 = x + 40
⇒ 2x + 6 = 40
⇒ 2x = 34
∴ x = 17
The three consecutive odd integers are 17, 19, and 21
∴ The middle number is 19
Solution:
Let's denote the three consecutive odd integers as x, x + 2, and x + 4
x + (x + 2) + (x + 4) = x + 40
⇒ 3x + 6 = x + 40
⇒ 2x + 6 = 40
⇒ 2x = 34
∴ x = 17
The three consecutive odd integers are 17, 19, and 21
∴ The middle number is 19
১,০৫৭.
In first 1000 natural numbers, how many integers exist such that they leave a remainder 4 when divided by 7 and a remainder 9 when divided by 11?
ব্যাখ্যা
When Divided by 7,
A = 7x + 4
So, numbers can be: 4, 11, 18, 25, 32, 39, 46, 53…….
Again,
when divided by 11,
A = 11y + 9
So, numbers can be: 9, 20, 31, 42, 53…….
Here, 53 is common.
Now, LCM of 7 & 11 is 77.
So, the numbers pattern is : 77x + 53
Then,
77x ≤ 1000
Or, x ≤ 12.2
x can be 0, 1, 2 ...... 12 = total 13 integers
It can't be x = 13, then the number will be bigger than 1000
১,০৫৮.
The sum of the digits of two - digit number is 10, while when the digits are reversed, the number decrease by 54. Find the changed number.
ব্যাখ্যা
Question: The sum of the digits of two - digit number is 10, while when the digits are reversed, the number decrease by 54. Find the changed number.
Solution:
Let number be (10x + y)
ATQ
(10x + y) - (10y + x) = 54
⇒ 10x - 10y + y - x = 54
⇒ 9x - 9y = 54
⇒ x - y = 6 ................ (1)
Sum of digits,
x + y = 10 .................(2)
(1) + (2)
x - y + x + y = 6 + 10
⇒ 2x = 16
∴ x = 8
Put the value of x in (2)
We get,
x + y = 10
⇒ y = 10 - 8
∴ y = 2
The required number is = (10x + y)
= (10 × 8) + 2
= 82
Changed number = 28
১,০৫৯.
The sum of two numbers is 528 and their H.C.F. is 33. The number of pairs of numbers satisfying the above conditions is:
ব্যাখ্যা
Let the required number numbers be 33a and 33b
Then, 33a + 33b = 528
⇒a + b = 16
Now, co - primes with sum 16 are (1,15), (3,13), (5,11) and (7,9)
∴ Required numbers are (33 × 1, 33 × 15), (33 × 3, 33 × 13), ( 33 × 5, 33 × 11), (33 × 7, 33 ×9)
The numbers of such pairs are 4
১,০৬০.
Find the least number which when divided by 4, 5, 6, and 7 leaves remainder 2 in each case.
ব্যাখ্যা
Question: Find the least number which when divided by 4, 5, 6, and 7 leaves remainder 2 in each case.
Solution:
LCM of 4, 5, 6, and 7 = 420
So, the number is = 420 + 2 = 422
Solution:
LCM of 4, 5, 6, and 7 = 420
So, the number is = 420 + 2 = 422
১,০৬১.
The product of two consecutive even numbers is 168. What is the smaller number?
ব্যাখ্যা
Question: The product of two consecutive even numbers is 168. What is the smaller number?
Solution:
Let the smaller number be x,
so the next even number is x + 2
ATQ,
⇒ x(x + 2) = 168
⇒ x2 + 2x = 168
⇒ x2 + 2x - 168 = 0
⇒ x + 14x - 12x - 168 = 0
⇒ x(x + 14) - 12(x + 14) = 0
⇒ (x + 14) (x - 12) = 0
∴ x= -14 and x = 12
So, the smaller number is 12.
Solution:
Let the smaller number be x,
so the next even number is x + 2
ATQ,
⇒ x(x + 2) = 168
⇒ x2 + 2x = 168
⇒ x2 + 2x - 168 = 0
⇒ x + 14x - 12x - 168 = 0
⇒ x(x + 14) - 12(x + 14) = 0
⇒ (x + 14) (x - 12) = 0
∴ x= -14 and x = 12
So, the smaller number is 12.
১,০৬২.
5/9 of a number equals twenty-five percent of the second number. Second number equals 1/4 of third number. The value of third number is 2960. What is 30% of first number?
ব্যাখ্যা
Question: 5/9 of a number equals twenty-five percent of the second number. Second number equals 1/4 of third number. The value of third number is 2960. What is 30% of first number?
Solution:
second number = 1/4 of third number = 2960/4
= 740
(5/9)first number = 25% of second number
first number = (1/4) × 740 × (9/5)
= 333
30% of first number = (3/10) × 333
= 99.9
১,০৬৩.
If you write down all the numbers from 1 to 100, then how many times do you write 5?
ব্যাখ্যা
Question: If you write down all the numbers from 1 to 100, then how many times do you write 5?
Solution:
১ - ১০ পর্যন্ত ৫ আছে = ১ টি
১১ - ২০ পর্যন্ত ৫ আছে = ১ টি
২১ - ৩০ পর্যন্ত ৫ আছে = ১ টি
৩১ - ৪০ পর্যন্ত ৫ আছে = ১ টি
৪১ - ৫০ পর্যন্ত ৫ আছে = ২ টি
৫১ - ৬০ পর্যন্ত ৫ আছে = ১০ টি
৬১ - ৭০ পর্যন্ত ৫ আছে = ১ টি
৭১ - ৮০ পর্যন্ত ৫ আছে = ১ টি
৮১ - ৯০ পর্যন্ত ৫ আছে = ১ টি
৯১ - ১০০ পর্যন্ত ৫ আছে = ১ টি
∴ মোট ৫ রয়েছে = ২০ টি
Solution:
১ - ১০ পর্যন্ত ৫ আছে = ১ টি
১১ - ২০ পর্যন্ত ৫ আছে = ১ টি
২১ - ৩০ পর্যন্ত ৫ আছে = ১ টি
৩১ - ৪০ পর্যন্ত ৫ আছে = ১ টি
৪১ - ৫০ পর্যন্ত ৫ আছে = ২ টি
৫১ - ৬০ পর্যন্ত ৫ আছে = ১০ টি
৬১ - ৭০ পর্যন্ত ৫ আছে = ১ টি
৭১ - ৮০ পর্যন্ত ৫ আছে = ১ টি
৮১ - ৯০ পর্যন্ত ৫ আছে = ১ টি
৯১ - ১০০ পর্যন্ত ৫ আছে = ১ টি
∴ মোট ৫ রয়েছে = ২০ টি
১,০৬৪.
(0.75 × 4.4 × 2.4)/0.6 =?
ব্যাখ্যা
Question: (0.75 × 4.4 × 2.4)/0.6 =?
Solution:
(0.75 × 4.4 × 2.4)/0.6
= 7.92/0.6
= 13.2
Solution:
(0.75 × 4.4 × 2.4)/0.6
= 7.92/0.6
= 13.2
১,০৬৫.
Which one of the following is not a prime number?
ব্যাখ্যা
Question: Which one of the following is not a prime number?
Solution:
69 = 3 × 23 which is not a prime number.
29, 79, 89 are prime numbers.
Solution:
69 = 3 × 23 which is not a prime number.
29, 79, 89 are prime numbers.
১,০৬৬.
On dividing an individual number by 342, we get 47 as remainder. If the same number is divided by 18, what will be the remainder?
ব্যাখ্যা
Consider the number be 342 + 47 = 389.
Therefore 389 divided by 18 leave remainder 11
Therefore 389 divided by 18 leave remainder 11
১,০৬৭.
If one-third of one-fourth of a number is 15, then three-tenth of that number is-
ব্যাখ্যা
Question: If one-third of one-fourth of a number is 15, then three-tenth of that number is-
Solution:
Let the number is x Then,
ATQ,
⇒ (1/3) of (1/4) of x = 15
⇒ x/12
⇒ x = 15 × 12 = 180
So,required number is = (3/10) × 180 = 54
Solution:
Let the number is x Then,
ATQ,
⇒ (1/3) of (1/4) of x = 15
⇒ x/12
⇒ x = 15 × 12 = 180
So,required number is = (3/10) × 180 = 54
১,০৬৮.
If (x/y) > 0, which of the following must be true?
ব্যাখ্যা
Question: If (x/y) > 0, which of the following must be true?
Solution:
যেহেতু (x/y) > 0 তাই
x ও y এর দুইটি একই সাথে ধণাত্মক বা ঋণাত্বক হবে।
আমরা জানি
দুইটি একই সাথে ধণাত্মক বা ঋণাত্বক সংখ্যার গুণফল ধণাত্মক হয়।
( - x) ( - y) = xy
x × y = xy
i) xy > 0 অবশ্যই ধণাত্মক
১,০৬৯.
A number is tripled, then 7 is subtracted from it. If the result is then doubled, it becomes 58. What is the number?
ব্যাখ্যা
Question: A number is tripled, then 7 is subtracted from it. If the result is then doubled, it becomes 58. What is the number?
Solution:
Let,
the number be x
ATQ,
2(3x - 7) = 58
⇒ 6x - 14 = 58
⇒ 6x = 58 + 14
⇒ 6x = 72
⇒ x = 72/6
∴ x = 12
So the number is 12.
Solution:
Let,
the number be x
ATQ,
2(3x - 7) = 58
⇒ 6x - 14 = 58
⇒ 6x = 58 + 14
⇒ 6x = 72
⇒ x = 72/6
∴ x = 12
So the number is 12.
১,০৭০.
The average of a non-zero number and its square is 11 times the number. The number is-
ব্যাখ্যা
Question: The average of a non-zero number and its square is 11 times the number. The number is-
Solution:
Let the number be x (x ≠ 0).
According to the question,
The average of the number and its square is 11 times the number.
⇒ (x + x2)/2 = 11x
⇒ x + x2 = 22x
⇒ x2 + x - 22x = 0
⇒ x2 - 21x = 0
⇒ x(x - 21) = 0
So, x = 0 or x = 21
But the number is non-zero, so we discard x = 0.
Therefore, the number is 21.
১,০৭১.
Which among 21/2, 31/3, 41/4, 61/6, and 121/12 is the largest?
ব্যাখ্যা
Question: Which among 21/2, 31/3, 41/4, 61/6, and 121/12 is the largest?
Solution:
21/2, 31/3, 41/4, 61/6, and 121/12
= 2(1/2) × 12, 3(1/3) × 12, 4(1/4) × 12, 6(1/6) × 12, and 12(1/12) × 12
= 26, 34, 43, 62 and 121
= 64, 81, 64, 36, and 12
Here, 31/3 is the largest number.
Solution:
21/2, 31/3, 41/4, 61/6, and 121/12
= 2(1/2) × 12, 3(1/3) × 12, 4(1/4) × 12, 6(1/6) × 12, and 12(1/12) × 12
= 26, 34, 43, 62 and 121
= 64, 81, 64, 36, and 12
Here, 31/3 is the largest number.
১,০৭২.
The least perfect square number divisible by 3, 4, 5, 6 and 8 is = ?
ব্যাখ্যা
L.C.M. of 3, 4, 5, 6, 8 is 120
Now 120 = 2 × 2 × 2 × 3 × 5
To make it a perfect square, it must be multiplied by 2 × 3 × 5
So, required number
=22×22×32×52
=3600
১,০৭৩.
What is the value of n if (8-3) (8 - n) = 40?
ব্যাখ্যা
(8-3) (8 - n) = 40
=> 5(8 - n) = 40
=> 8 - n = 8
=> n = 0
=> 5(8 - n) = 40
=> 8 - n = 8
=> n = 0
১,০৭৪.
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. The remainder left is 12. Find the number.
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. The remainder left is 12. Find the number.
Solution:
Let the original number be x
ATQ,
{5(x + 7)/9} - 3 = 12
⇒ {5(x + 7) - 27}/9 = 12
⇒ 5(x + 7) - 27 = 108
⇒ 5x + 35 - 27 = 108
⇒ 5x + 8 = 108
⇒ 5x = 100
∴ x = 20
Solution:
Let the original number be x
ATQ,
{5(x + 7)/9} - 3 = 12
⇒ {5(x + 7) - 27}/9 = 12
⇒ 5(x + 7) - 27 = 108
⇒ 5x + 35 - 27 = 108
⇒ 5x + 8 = 108
⇒ 5x = 100
∴ x = 20
১,০৭৫.
What is the difference between the place value and the face value of 6 in the numeral 2962?
ব্যাখ্যা
Place value of 6 = 60
Face value of 6 = 6
Difference = 60 - 6
= 54
Face value of 6 = 6
Difference = 60 - 6
= 54
১,০৭৬.
Add 2√2 + 5√3 and √2 - 3√3.
ব্যাখ্যা
Question: Add 2√2 + 5√3 and √2 - 3√3.
Solution:
(2√2 + 5√3) + (√2 - 3√3)
= 2√2 + √2 + 5√3 - 3√3
= (2 + 1)√2 + (5 - 3)√3
= 3√2 + 2√3
Solution:
(2√2 + 5√3) + (√2 - 3√3)
= 2√2 + √2 + 5√3 - 3√3
= (2 + 1)√2 + (5 - 3)√3
= 3√2 + 2√3
১,০৭৭.
Which number is to be added to the numerator and denominator of 7/17 to form 3/5?
ব্যাখ্যা
Question: Which number is to be added to the numerator and denominator of 7/17 to form 3/5?
Solution:
Let, the number is x.
ATQ,
(7 + x)/(17 + x) = 3/5
Or, 5x + 35 = 51 + 3x
Or, 5x - 3x = 51 - 35
Or, 2x = 16
∴ x = 8
∴ The number is 8.
Solution:
Let, the number is x.
ATQ,
(7 + x)/(17 + x) = 3/5
Or, 5x + 35 = 51 + 3x
Or, 5x - 3x = 51 - 35
Or, 2x = 16
∴ x = 8
∴ The number is 8.
১,০৭৮.
Which of the following has the most number of divisors?
ব্যাখ্যা
Question: Which of the following has the most number of divisors?
Solution:
৯৮ = ২ × ৭ × ৭
১০১ = ১ × ১০১
১৭৬ = ২ × ২ × ২ × ২ × ১১
৩৫ = ৫ × ৭
অর্থাৎ ১৭৬ এর উৎপাদক সবচেয়ে বেশি।
Solution:
৯৮ = ২ × ৭ × ৭
১০১ = ১ × ১০১
১৭৬ = ২ × ২ × ২ × ২ × ১১
৩৫ = ৫ × ৭
অর্থাৎ ১৭৬ এর উৎপাদক সবচেয়ে বেশি।
১,০৭৯.
If the product of two numbers is 2,028, and their HCF is 13, what are the numbers?
ব্যাখ্যা
Let two numbers be 13a and 13b.
Their product is 2028.
So,13a x 13b=2028
169ab=2028
ab=2028/169
ab=12
Co primes for 12 are (1,12) and (3,4)
So the numbers are (13×1,13×12) or (13×3,13×4)
Two numbers are 13, 156 or 39, 52
===============================
মনে করি, সংখ্যা দুইটি ১৩ক ও ১৩খ
অতএব, ১৩ক × ১৩খ = ২০২৮
ক × খ = ২০২৮/(১৩ × ১৩) = ১২
১২ = ১ × ১২ = ৩ × ৪
অতএব, সংখ্যা গুলো হচ্ছে ১৩ × ১ ও ১৩ × ১২ অথবা ১৩ × ৩, ১৩ × ৪
সুতরাং, সংখ্যা দুইটি ১৩ ও ১৫৬ অথবা ৩৯ ও ৫২
------------------------------------------------------
Step 1: Suppose the two numbers be x and y.
So, as per given information x*y = 2028.
Step 2: Do the factor of this number:
2028 = 2*2*3*13*13
Step 3: Now, use the another given information, i.e H.C.F (x,y) = 13.
Therefore, factor 13 should be present in both x and y. So, from 2*2*3*13*13 factors 13*13 are settled and now we left with 2*2*3.
Since, we can not share factor ‘2’ in both x and y , otherwise the H.C.F will become 26, So, we have now two sub-cases.
i) Whole 2*2*3 will go to one of the number (x or y), suppose it goes to y, then numbers will be: x= 13, y = 2*2*3*13
ii) Factor ‘3’ goes to one of the number (x or y), suppose goes to x, then numbers will be: x=13*3 , y= 2*2*13.
Their product is 2028.
So,13a x 13b=2028
169ab=2028
ab=2028/169
ab=12
Co primes for 12 are (1,12) and (3,4)
So the numbers are (13×1,13×12) or (13×3,13×4)
Two numbers are 13, 156 or 39, 52
===============================
মনে করি, সংখ্যা দুইটি ১৩ক ও ১৩খ
অতএব, ১৩ক × ১৩খ = ২০২৮
ক × খ = ২০২৮/(১৩ × ১৩) = ১২
১২ = ১ × ১২ = ৩ × ৪
অতএব, সংখ্যা গুলো হচ্ছে ১৩ × ১ ও ১৩ × ১২ অথবা ১৩ × ৩, ১৩ × ৪
সুতরাং, সংখ্যা দুইটি ১৩ ও ১৫৬ অথবা ৩৯ ও ৫২
------------------------------------------------------
Step 1: Suppose the two numbers be x and y.
So, as per given information x*y = 2028.
Step 2: Do the factor of this number:
2028 = 2*2*3*13*13
Step 3: Now, use the another given information, i.e H.C.F (x,y) = 13.
Therefore, factor 13 should be present in both x and y. So, from 2*2*3*13*13 factors 13*13 are settled and now we left with 2*2*3.
Since, we can not share factor ‘2’ in both x and y , otherwise the H.C.F will become 26, So, we have now two sub-cases.
i) Whole 2*2*3 will go to one of the number (x or y), suppose it goes to y, then numbers will be: x= 13, y = 2*2*3*13
ii) Factor ‘3’ goes to one of the number (x or y), suppose goes to x, then numbers will be: x=13*3 , y= 2*2*13.
১,০৮০.
One-third of a number's quarter is 15. Find the value of three-tenths of the same number.
ব্যাখ্যা
Question: One-third of a number's quarter is 15. Find the value of three-tenths of the same number.
Solution:
Let,
The number be x
Then,
1/3 of 1/4 of x = 15
⇒ x = 15 × 12 = 180
So, required number = (3/10) × 180 = 54
Solution:
Let,
The number be x
Then,
1/3 of 1/4 of x = 15
⇒ x = 15 × 12 = 180
So, required number = (3/10) × 180 = 54
১,০৮১.
Two-fifth of one-fourth of three seventh of a number is 15. What is the half of the number?
ব্যাখ্যা
Question: Two-fifth of one-fourth of three seventh of a number is 15. What is the half of the number?
Solution:
let the number is P
ATQ,
P × (3/7) × (1/4) × (2/5) = 15
3P/70 = 15
P = 350
∴ half of the number is = 350/2 = 175
Solution:
let the number is P
ATQ,
P × (3/7) × (1/4) × (2/5) = 15
3P/70 = 15
P = 350
∴ half of the number is = 350/2 = 175
১,০৮২.
Three numbers are in the ratio 3 : 4 : 6 and their products is 1944. The smallest number is -
ব্যাখ্যা
Question: Three numbers are in the ratio 3 : 4 : 6 and their products is 1944. The smallest number is -
Solution:
Let the number be 3x, 4x, 6x.
ATQ,
3x × 4x × 6x = 1944
⇒ 72x3 = 1944
⇒ x3 = 1944/72
⇒ x3 = 27
∴ x = 3
So the smallest number = 3x = 3 × 3 = 9
Solution:
Let the number be 3x, 4x, 6x.
ATQ,
3x × 4x × 6x = 1944
⇒ 72x3 = 1944
⇒ x3 = 1944/72
⇒ x3 = 27
∴ x = 3
So the smallest number = 3x = 3 × 3 = 9
১,০৮৩.
Which of the following is irrational?
ব্যাখ্যা
Question: Which of the following is irrational?
Solution:
একটি সংখ্যা অমূলদ (irrational) হয় যদি এটি p/q আকারে প্রকাশ করা না যায়, যেখানে p এবং q পূর্ণসংখ্যা এবং q ≠ 0।
ক) 0.75 = 75/100 = 3/4 = এটি p/q আকারে প্রকাশ করা যায়, তাই এটি মূলদ সংখ্যা।
খ) √256 = 16 একটি পূর্ণবর্গ সংখ্যা (16² = 256), তাই √256 = 16 একটি মূলদ সংখ্যা।
গ) 2/5 = এটি ইতিমধ্যে p/q আকারে আছে, তাই এটি মূলদ সংখ্যা।
ঘ) √27 = √(9 × 3) = 3√3 একটি পূর্ণবর্গ সংখ্যা নয়, তাই √27 একটি অমূলদ সংখ্যা। এটি p/q আকারে প্রকাশ করা যায় না।
উত্তর: ঘ) √27 একটি অমূলদ (irrational) সংখ্যা।
১,০৮৪.
Find the number that exceeds 36 by the same amount it falls short of 86.
ব্যাখ্যা
Question: Find the number that exceeds 36 by the same amount it falls short of 86.
Solution:
Let the number be = x
According to the question,
x - 36 = 86 - x
⇒ x + x = 86 + 36
⇒ 2x = 122
⇒ x = 122/2
⇒ x = 61
Solution:
Let the number be = x
According to the question,
x - 36 = 86 - x
⇒ x + x = 86 + 36
⇒ 2x = 122
⇒ x = 122/2
⇒ x = 61
১,০৮৫.
Bowl S contains only marbles. If (1/4) of the marbles were removed, the bowl would be filled to (1/2) of its capacity. If 100 marbles were added, the bowl would be full. How many marbles are in bowl S?
ব্যাখ্যা
Question: Bowl S contains only marbles. If (1/4) of the marbles were removed, the bowl would be filled to (1/2) of its capacity. If 100 marbles were added, the bowl would be full. How many marbles are in bowl S?
Solution:
Let
there are x number of marbles and capacity of the bowl is y marbles
3x/4 = y/2
⇒ y = 3x/2
x + 100 = y
⇒ x + 100 = 3x/2
⇒ (3x/2) - x = 100
⇒ x/2 = 100
∴ x = 200
১,০৮৬.
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?
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
Days present = 5600 ÷ 1120 = 5 days
Absent days = 6 − 5 = 1
১,০৮৭.
The L.C.M of 3/4, 6/7, 9/8 is:
ব্যাখ্যা
Required L.C.M = (L.C.M of 3, 6, 9)/(H.C.F of 4, 7, 8)
=18/1 = 18
Answer : 18
১,০৮৮.
If the sum of seven consecutive odd integers is 385, what is the largest number?
ব্যাখ্যা
Question: If the sum of seven consecutive odd integers is 385, what is the largest number?
Solution:
ধরি, মাঝের সংখ্যাটি = x
সুতরাং, 7টি ক্রমিক বিজোড় সংখ্যা হবে যথাক্রমে: x - 6, x - 4, x - 2, x, x + 2, x + 4 এবং x + 6
প্রশ্নমতে,
(x - 6) + (x - 4) + (x - 2) + x + (x + 2) + (x + 4) + (x + 6) = 385
⇒ 7x = 385
⇒ x = 55
∴ মাঝের সংখ্যা, x = 55
∴ সবচেয়ে বড় সংখ্যা = x + 6 = 55 + 6 = 61
১,০৮৯.
The difference between the two numbers is 20% of the larger number. If the smaller number is 12, the larger one is:
ব্যাখ্যা
Question: The difference between the two numbers is 20% of the larger number. If the smaller number is 12, the larger one is:
Solution:
Let the number be x
Then,
x − 12 = 20% of x
⇒ x − 12 = x/5
⇒ x − x/5 = 12
⇒ 4x/5 = 12
⇒ x = (12 × 5)/4
∴ x = 15
Solution:
Let the number be x
Then,
x − 12 = 20% of x
⇒ x − 12 = x/5
⇒ x − x/5 = 12
⇒ 4x/5 = 12
⇒ x = (12 × 5)/4
∴ x = 15
১,০৯০.
The number n yields a remainder p when divided by 15 and a remainder q when divided by 5. If p = q + 10, then which one of the following could be the value of n?
ব্যাখ্যা
Question: The number n yields a remainder p when divided by 15 and a remainder q when divided by 5. If p = q+10, then which one of the following could be the value of n?
Solution:
প্রদত্ত শর্ত অনুযায়ী, আমাদের এমন একটি সংখ্যা n খুঁজে বের করতে হবে, যাকে 15 দ্বারা ভাগ করলে ভাগশেষ p এবং 5 দ্বারা ভাগ করলে ভাগশেষ q পাওয়া যায়, যেখানে p = q + 10।
• অপশন (ক):
ধরা যাক n = 24
24 ÷ 15 = 1 এবং ভাগশেষ p = 9।
24 ÷ 5 = 4 এবং ভাগশেষ q = 4।
শর্ত অনুযায়ী, p = q + 10। এখানে 9 ≠ 4 + 10 ; সুতরাং, এই অপশনটি সঠিক নয়।
• অপশন (খ):
ধরা যাক n = 32
32 ÷ 15 = 2 এবং ভাগশেষ p = 2।
32 ÷ 5 = 6 এবং ভাগশেষ q = 2।
শর্ত অনুযায়ী, p = q + 10। এখানে 2 ≠ 2 + 10 ; সুতরাং, এই অপশনটি সঠিক নয়।
• অপশন (গ):
ধরা যাক n = 43
43 ÷ 15 = 2 এবং ভাগশেষ p = 13।
43 ÷ 5 = 8 এবং ভাগশেষ q = 3।
শর্ত অনুযায়ী, p = q + 10। এখানে 13 = 3 + 10 = 13, যা সত্য। সুতরাং, এই অপশনটি সঠিক।
• অপশন (ঘ):
ধরা যাক n = 50
50 ÷ 15 = 3 এবং ভাগশেষ p = 5।
50 ÷ 5 = 10 এবং ভাগশেষ q = 0।
শর্ত অনুযায়ী p = q + 10। এখানে 5 ≠ 0 + 10 ; সুতরাং, এই অপশনটি সঠিক নয়।
১,০৯১.
- 1 হতে কত বিয়োগ করলে বিয়োগফল 0 (শূণ্য) হবে?
ব্যাখ্যা
প্রশ্ন: -1 হতে কত বিয়োগ করলে বিয়োগফল 0 (শূণ্য) হবে?
সমাধান:
- 1 - ( - 1)
= - 1 + 1
= 0
অর্থাৎ, -1 হতে -1 বিয়োগ করলে বিয়োগফল 0 (শূণ্য) হবে।
১,০৯২.
The sum of squares of three numbers is 138 and the sum of their products taken two at a time is 131. Find their sum.
ব্যাখ্যা
Question: The sum of squares of three numbers is 138 and the sum of their products taken two at a time is 131. Find their sum.
Solution:
Let the three numbers be x, y and z.
Sum of squares of three numbers is 138 and sum of their products taken two at a time is 131
Therefore,
x2 + y2 + z2 = 138
xy + yz + zx = 131
(x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)
(x + y + z )2 = 138 + 2(131)
(x + y + z )2 = 400
Hence, (x + y + z) = 20
Solution:
Let the three numbers be x, y and z.
Sum of squares of three numbers is 138 and sum of their products taken two at a time is 131
Therefore,
x2 + y2 + z2 = 138
xy + yz + zx = 131
(x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)
(x + y + z )2 = 138 + 2(131)
(x + y + z )2 = 400
Hence, (x + y + z) = 20
১,০৯৩.
The difference of two numbers is 12. If 1 is added to the greater number, it becomes twice the smaller number. Calculate the two numbers?
ব্যাখ্যা
Question: The difference of two numbers is 12. If 1 is added to the greater number, it becomes twice the smaller number. Calculate the two numbers?
Solution:
ধরি,
বড় সংখ্যাটি = x
ছোট সংখ্যাটি = y
১ম শর্তমতে,
x - y = 12 ............................. (1)
২য় শর্তমতে,
x + 1 = 2y ............................. (2)
(1) নং হতে (2) নং সমীকরণ বিয়োগ করে পাই,
x - y - x - 1 = 12 - 2y
বা, - y = - 13
∴ y = 13
(1) নং হতে পাই,
x - 13 = 12
বা, x = 12 + 13
∴ x = 25
∴ সংখ্যা দুটি 25 ও 13
Solution:
ধরি,
বড় সংখ্যাটি = x
ছোট সংখ্যাটি = y
১ম শর্তমতে,
x - y = 12 ............................. (1)
২য় শর্তমতে,
x + 1 = 2y ............................. (2)
(1) নং হতে (2) নং সমীকরণ বিয়োগ করে পাই,
x - y - x - 1 = 12 - 2y
বা, - y = - 13
∴ y = 13
(1) নং হতে পাই,
x - 13 = 12
বা, x = 12 + 13
∴ x = 25
∴ সংখ্যা দুটি 25 ও 13
১,০৯৪.
If a two-digit positive integer has its digits reversed, the resulting integer differs from the original by 27. By how much do the two digits differ?
ব্যাখ্যা
Given that, (10a + b) - (10b + a) = 27
⇒ 9a - 9b = 27
⇒ a - b = 3
১,০৯৫.
The difference between two positive numbers is 3 and the sum of their squares is 369. Then the sum of the numbers is
ব্যাখ্যা
Question: The difference between two positive numbers is 3 and the sum of their squares is 369. Then the sum of the numbers is-
Solution:
Let the numbers be x and (x + 3)
Then,
x2 + (x + 3)2=369
⇒ x2 + x2 + 9 + 6x=369
⇒ 2x2 + 6x - 360=0
⇒ x2 + 3x - 180=0
⇒ (x + 15)(x - 12)=0
⇒ x = 12
So, the numbers are 12 and 15
∴ Required sum = (12 + 15) = 27
Solution:
Let the numbers be x and (x + 3)
Then,
x2 + (x + 3)2=369
⇒ x2 + x2 + 9 + 6x=369
⇒ 2x2 + 6x - 360=0
⇒ x2 + 3x - 180=0
⇒ (x + 15)(x - 12)=0
⇒ x = 12
So, the numbers are 12 and 15
∴ Required sum = (12 + 15) = 27
১,০৯৬.
What is the total number of integers between 100 and 200 that are divisible by 3?
ব্যাখ্যা
First, identify the number that is a multiple of 3 more than 100.
That type of number is 102.
So, 102//3 = 34.
Second, we have to identify the number that is multiple of 3 but nearest less than 200.
That type of number is 198.
Now, 198/3 = 66.
Answer is (66 - 34) + 1 = 33
That type of number is 102.
So, 102//3 = 34.
Second, we have to identify the number that is multiple of 3 but nearest less than 200.
That type of number is 198.
Now, 198/3 = 66.
Answer is (66 - 34) + 1 = 33
১,০৯৭.
Soma was asked to multiply a certain number by 36. She multiplied it by 63 instead and got an answer 3834 more than the correct one. What was the number to be multiplied?
ব্যাখ্যা
Let,
the number be x
then, 63x - 36x = 3834
⇒ 27x = 3834
⇒ x = 3834/27
⇒ x = 142.
১,০৯৮.
A farmer had 17 hens. All but 9 died. How many live hens were left?
ব্যাখ্যা
Answer is given in the question. All but 9 died means - except 9 all other hens are died.
বাক্যটির বাংলা অর্থ - এক কৃষকের ১৭টি মুরগী ছিলো। নয়টি বাদে বাকি সবগুলো মারা গিয়েছিলো।
So there are 9 alive hens.
বাক্যটির বাংলা অর্থ - এক কৃষকের ১৭টি মুরগী ছিলো। নয়টি বাদে বাকি সবগুলো মারা গিয়েছিলো।
So there are 9 alive hens.
১,০৯৯.
The value of (489.1375 x 0.0483 x 1.956)/ (0.0873 x 92.581 x 99.749) is closest to:
ব্যাখ্যা
(489.1375 x 0.0483 x 1.956) / (0.0873 x 92.581 x 99.749)
≈ 489 x 0.05 x 2 / 0.09 x 93 x 100
= 489/ (9 x 93 x 10)
= (163/279)x (1/10)
= 0.58/ 10
= .058
≈ .06
১,১০০.
28√a + 1426 = three-fourths of 2984, then a =?
ব্যাখ্যা
Question: 28√a + 1426 = three-fourths of 2984, then a =?
Solution:
28√a + 1426 = (3/4) × 2984
⇒ 28√a + 1426 = 2238
⇒ 28√a = 2238 - 1426
⇒ 28√a = 812
⇒ √a = 812/28
⇒ √a = 29
⇒ (√a)2 = (29)2
∴ a = 841
Solution:
28√a + 1426 = (3/4) × 2984
⇒ 28√a + 1426 = 2238
⇒ 28√a = 2238 - 1426
⇒ 28√a = 812
⇒ √a = 812/28
⇒ √a = 29
⇒ (√a)2 = (29)2
∴ a = 841