১.
Add 2√2 + 5√3 and √2 - 4√3
ব্যাখ্যা
(2√2 + 5√3) + (√2 - 4√3)
= (2 + 1)√2 + (5 - 4)√3
= 3√2 + √3
= (2 + 1)√2 + (5 - 4)√3
= 3√2 + √3
ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স
মোট প্রশ্ন২১
সিলেবাস
Math - 01: Number System, Problems on Number, HCF & LCM
ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স
ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স · তারিখ অনির্ধারিত · ২১ প্রশ্ন
২.
The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number?
ব্যাখ্যা
Let
the ten's digit be x
unit's digit be y.
Then,
(10x + y) - (10y + x) = 36
9(x - y) = 36
x - y = 4
the ten's digit be x
unit's digit be y.
Then,
(10x + y) - (10y + x) = 36
9(x - y) = 36
x - y = 4
৩.
Which of the following has the most number of divisors?
ব্যাখ্যা
Divisors of 99 are 1, 3, 9, 11, 33, .99
Divisors of 101 are 1 and 101
Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176
Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.
Hence, 176 has the most number of divisors.
Divisors of 101 are 1 and 101
Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176
Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.
Hence, 176 has the most number of divisors.
৪.
Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is:
ব্যাখ্যা
Let
The three integers be x, x + 2 and x + 4.
Then,
3x = 2(x + 4) + 3
3x = 2x + 8 + 3
3x - 2x = 11
x = 11.
Third integer = x + 4 = 11 + 4 = 15
The three integers be x, x + 2 and x + 4.
Then,
3x = 2(x + 4) + 3
3x = 2x + 8 + 3
3x - 2x = 11
x = 11.
Third integer = x + 4 = 11 + 4 = 15
৫.
Which of the following numbers should be added to 8567 to make it exactly divisible by 4?
ব্যাখ্যা
On dividing 8567 by 4, the remainder is 3.
To make it divisible by 4, we must add (4 - 3) =1
To make it divisible by 4, we must add (4 - 3) =1
৬.
The product of two numbers is 9375 and the quotient, when the larger one is divided by the smaller, is 15. The sum of the numbers is:
ব্যাখ্যা
Let
the larger numbers be x and
the smaller numbers be y.
Then, xy = 9375 and
x/y = 15
now
xy/(x/y) = 9375/15
y2 = 625
y2 = 252
y = 25
x = 25 × 15 = 375
x + y = 375 + 25 = 400
the larger numbers be x and
the smaller numbers be y.
Then, xy = 9375 and
x/y = 15
now
xy/(x/y) = 9375/15
y2 = 625
y2 = 252
y = 25
x = 25 × 15 = 375
x + y = 375 + 25 = 400
৭.
The HCF and LCM of two numbers are 24 and 168 and the numbers are in the ratio 1 ∶ 7. Find the greater of the two numbers.
ব্যাখ্যা
Given that
HCF = 24
LCM = 168
Ratio of numbers = 1 ∶ 7.
Let numbers be x and 7x.
We know
Product of numbers = LCM × HCF
x × 7x = 24 × 168
⇒ x2 = 24 × 24
⇒ x = 24
∴ Larger number = 7x
= 24 × 7
= 168
HCF = 24
LCM = 168
Ratio of numbers = 1 ∶ 7.
Let numbers be x and 7x.
We know
Product of numbers = LCM × HCF
x × 7x = 24 × 168
⇒ x2 = 24 × 24
⇒ x = 24
∴ Larger number = 7x
= 24 × 7
= 168
৮.
HCF of two number is 4 and the sum of those two numbers is 36. Find how many such pair of number is possible.
ব্যাখ্যা
HCF of two number is 4
Let, those numbers be 4x and 4y where x and y are prime to each other Accordingly,
4x + 4y = 36
⇒ 4(x + y) = 36
⇒ (x + y) = 9
Now,
9 = 8 + 1
9 = 7 + 2
9 = 6 + 3
9 = 5 + 4
In all these cases only (8, 1); (7, 2); and (5, 4) are prime to each other.
So, such three pair is possible
∴ Such three pair of number is possible.
Let, those numbers be 4x and 4y where x and y are prime to each other Accordingly,
4x + 4y = 36
⇒ 4(x + y) = 36
⇒ (x + y) = 9
Now,
9 = 8 + 1
9 = 7 + 2
9 = 6 + 3
9 = 5 + 4
In all these cases only (8, 1); (7, 2); and (5, 4) are prime to each other.
So, such three pair is possible
∴ Such three pair of number is possible.
To find the pairs we only can take those pair, which are prime to each other.
We can't take those pair, which are not prime to each other.
For example: In this problem, 9 can be represent as 6 + 3 = 9
But, 6 & 3 are not prime to each other.
So, it can't be a possible pair.
৯.
What is the sum of two consecutive even numbers, the difference of whose squares is 84?
ব্যাখ্যা
Let
the numbers be x and x + 2.
Then,
(x + 2)2 - x2 = 84
x2 + 2.x.2 + 22 - x2 =84
4x + 4 = 84
4x = 80
x = 20.
The required sum = x + (x + 2)
= 2x + 2
= (2 × 20) + 2
= 42
the numbers be x and x + 2.
Then,
(x + 2)2 - x2 = 84
x2 + 2.x.2 + 22 - x2 =84
4x + 4 = 84
4x = 80
x = 20.
The required sum = x + (x + 2)
= 2x + 2
= (2 × 20) + 2
= 42
১০.
Find the least number which when divided by 20, 25, 35, and 40 leaves remainders 14, 19, 29 and 34 respectively.
ব্যাখ্যা
Here
20 − 14 = 6
25 − 19 = 6
35 − 29 = 6
40 − 34= 6
Required number (L.C.M of 20, 25, 35, and 40) - 6
L.C.M of 20, 25, 35, and 40 = 1400
Required number= 1400 - 6 = 1394
20 − 14 = 6
25 − 19 = 6
35 − 29 = 6
40 − 34= 6
Required number (L.C.M of 20, 25, 35, and 40) - 6
L.C.M of 20, 25, 35, and 40 = 1400
Required number= 1400 - 6 = 1394
১১.
What is the least perfect square that is a multiple of 7, 11 and 12?
ব্যাখ্যা
Let us assume the least perfect square be X
⇒ 7 = 7 × 1
⇒ 11 = 11 × 1
⇒ 12 = 22 × 3 ⇒
The LCM of (7, 11, 12) = 22 × 3 × 11 × 7
⇒ The least perfect square = 22 × 32 × 112 × 72 = 213444
∴ The required result will be 213444.
⇒ 7 = 7 × 1
⇒ 11 = 11 × 1
⇒ 12 = 22 × 3 ⇒
The LCM of (7, 11, 12) = 22 × 3 × 11 × 7
⇒ The least perfect square = 22 × 32 × 112 × 72 = 213444
∴ The required result will be 213444.
১২.
The sum of two numbers is 40 and their difference is 4. The ratio of the number is
ব্যাখ্যা
Given:
The sum of two numbers is 40.
The difference of two number is 4.
Let
the larger number be x.
the smaller number be y.
According to the question,
(x + y)/(x - y) = 40/4
(x + y)/(x - y) =10/1
10(x - y) = x + y
10x - 10y = x + y
10x - x = 10y + y
9x = 11y
x/y = 11/9
x : y = 11 : 9
∴ The ratio of the number is 11 : 9
The sum of two numbers is 40.
The difference of two number is 4.
Let
the larger number be x.
the smaller number be y.
According to the question,
(x + y)/(x - y) = 40/4
(x + y)/(x - y) =10/1
10(x - y) = x + y
10x - 10y = x + y
10x - x = 10y + y
9x = 11y
x/y = 11/9
x : y = 11 : 9
∴ The ratio of the number is 11 : 9
১৩.
The number of times 99 is subtracted from 1111 so that the remainder is less than 99 is -
ব্যাখ্যা
On dividing 1111 by 99, the quotient is 11 and the remainder is 22.
Hence, the required number is 11
Hence, the required number is 11
১৪.
6*2 is three digit number with * as a missing digit. If the number is divisible by 6, the missing digit is
ব্যাখ্যা
Divisibility of 6 = Number should be multiple of 3 and 2
6*2 it is definitely divisible by 2
To be divisible by 3 Sum of all digits of the number be divisible by 3
We just try to put value in place of *
632 = 6 + 3 + 2 = 11(not divisible by 3)
662 = 6 + 6 + 2 = 14(not divisible by 3)
672 = 6 + 7 + 2 = 15( divisible by 3)
622 = 6 + 2 + 2 = 10(not divisible by 3)
∴ * is replaced by 7
6*2 it is definitely divisible by 2
To be divisible by 3 Sum of all digits of the number be divisible by 3
We just try to put value in place of *
632 = 6 + 3 + 2 = 11(not divisible by 3)
662 = 6 + 6 + 2 = 14(not divisible by 3)
672 = 6 + 7 + 2 = 15( divisible by 3)
622 = 6 + 2 + 2 = 10(not divisible by 3)
∴ * is replaced by 7
১৫.
The L.C.M. of (6x2 - x - 1), (3x2 + 7x + 2) and (2x2 + 3x - 2) is
ব্যাখ্যা
6x2 - x - 1 = 6x2 - 3x + 2x - 1
= 3x(2x - 1) + 1(2x - 1)
= (2x - 1) (3x + 1)
(3x2 + 7x + 2) = 3x2 + 6x + x + 2
= 3x(x + 2) + 1(x + 2)
= (x + 2)(3x + 1)
(2x2 + 3x - 2) = 2x2 + 4x - x - 2
= 2x(x + 2) - 1(x + 2)
= (x + 2) (2x - 1)
∴ Required L.C.M. is = (2x - 1) (3x + 1) (x + 2)
= 3x(2x - 1) + 1(2x - 1)
= (2x - 1) (3x + 1)
(3x2 + 7x + 2) = 3x2 + 6x + x + 2
= 3x(x + 2) + 1(x + 2)
= (x + 2)(3x + 1)
(2x2 + 3x - 2) = 2x2 + 4x - x - 2
= 2x(x + 2) - 1(x + 2)
= (x + 2) (2x - 1)
∴ Required L.C.M. is = (2x - 1) (3x + 1) (x + 2)
১৬.
Three numbers are in the ratio of 3 : 4 : 5 and their L.C.M is 2400. Their H.C.F is
ব্যাখ্যা
Let us consider the number be 3x,4x and 5x
The L.C.M of the three numbers be =60x
60x = 2400
x = 2400/60
x = 40
Then the number be
3x = 3 × 40 = 120
4x = 4 × 40 = 160
5x = 5 × 40 = 200
∴ The HCF of the three numbers be 40
The L.C.M of the three numbers be =60x
60x = 2400
x = 2400/60
x = 40
Then the number be
3x = 3 × 40 = 120
4x = 4 × 40 = 160
5x = 5 × 40 = 200
∴ The HCF of the three numbers be 40
১৭.
Out of six consecutive natural numbers if the sum of the first three is 30, what is the sum of the other three?
ব্যাখ্যা
Let
the first three consecutive numbers be x, x + 1 and x + 2.
Then,
x + x + 1 + x + 2 = 30
⇒ 3x + 3 = 30
⇒ 3x = 27
⇒ x = 9
∴ The first three numbers are 9, 10,11
⇒ The next three numbers are 12, 13, 14.
Hence, the required sum =12 + 13 + 14
= 39
the first three consecutive numbers be x, x + 1 and x + 2.
Then,
x + x + 1 + x + 2 = 30
⇒ 3x + 3 = 30
⇒ 3x = 27
⇒ x = 9
∴ The first three numbers are 9, 10,11
⇒ The next three numbers are 12, 13, 14.
Hence, the required sum =12 + 13 + 14
= 39
১৮.
If the numbers from 1 to 24, Which are divisible by 2 are arranged in descending order, which will be at the 7th place from the bottom?
ব্যাখ্যা
The descending order of the number from 1 to 24 which are divisible by 2 is:
24, 22, 20, 18, 16, 14, 12, 10, 8, 6, 4, 2
Now count the 7th place number from the right.
The number will be at the 8th place from the bottom is 14.
24, 22, 20, 18, 16, 14, 12, 10, 8, 6, 4, 2
Now count the 7th place number from the right.
The number will be at the 8th place from the bottom is 14.
১৯.
One-fifth of a number is equal to 5/8 of another number . If 35 is added to the first number, it becomes four times of the second number. The second number is-
ব্যাখ্যা
Let
the numbers be a and b.
According to the problem,
the first equation is:
a/5=5b/8
a/5=5b/8
25b = 8a
a = 25b/8
a + 35=4b
⟹a = 4b − 35
(25b/8) - 4b = - 35
(25b - 32b)/8 = - 35
- 7b/8 = -35
b = (35 × 8)/7
b = 40
the numbers be a and b.
According to the problem,
the first equation is:
a/5=5b/8
a/5=5b/8
25b = 8a
a = 25b/8
a + 35=4b
⟹a = 4b − 35
(25b/8) - 4b = - 35
(25b - 32b)/8 = - 35
- 7b/8 = -35
b = (35 × 8)/7
b = 40
২০.
A number whose fifth part increased by 4 is equal to its fourth part reduced by 10. Find the number.
ব্যাখ্যা
Let
the number be x
According to problems condition
(x/5) + 4 = (x/4) - 10
(x/5) - (x/4) = - 10 - 4
(4x - 5x)/20 = - 14
-x/20 = - 14
x = 280
the number be x
According to problems condition
(x/5) + 4 = (x/4) - 10
(x/5) - (x/4) = - 10 - 4
(4x - 5x)/20 = - 14
-x/20 = - 14
x = 280
২১.
If the three numbers are added in pairs , the sum is equal to 10, 16, 24. What is the difference between third and first number?
ব্যাখ্যা
Let,
the numbers are a, b and c
According to the question,
⇒ a + b = 10
⇒ b + c = 16
⇒ c + a = 24
From above question,
⇒ 2(a + b + c) = 50
⇒ (a + b + c) = 25
Value of a = 25 - (b + c) = 25 - 16 = 9
Value of c = 25 - (a + b) = 25 - 10 = 15
Required difference = (15 - 9) = 6
∴ Difference is 6
the numbers are a, b and c
According to the question,
⇒ a + b = 10
⇒ b + c = 16
⇒ c + a = 24
From above question,
⇒ 2(a + b + c) = 50
⇒ (a + b + c) = 25
Value of a = 25 - (b + c) = 25 - 16 = 9
Value of c = 25 - (a + b) = 25 - 10 = 15
Required difference = (15 - 9) = 6
∴ Difference is 6