ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

The answer is divisible by 987.
So we can use the hit and trial method to find out the number divisible by 987 from the given choices.
553681/987 gives a remainder not equal to 0
555181/987 gives a remainder not equal to 0
556581/987 gives a remainder not equal to 0
But 555681/987 gives 0 as a remainder. Hence this is the answer

The sum of all prime numbers from 1 to 20
= (2 + 3 + 5 + 7 + 11 + 13 + 17 + 19)
= 77

a + b + c = 6
ab + bc + ca = 10
∴ (a + b + c)²= 36
⇒ a²+ b²+ c²+ 2ab + 2bc + 2ca = 36
⇒ a²+ b²+ c²+ 2(ab + bc + ca) = 36
⇒ a²+ b²+ c²+ 2 × 10 = 36
⇒ a²+ b²+ c²= 16
As we know
a³ + b³ + c³ - 3abc/(a² + b² + c² - ab - bc - ca) = a + b + c
⇒a³ + b³ + c³ - 3abc/16 - (ab + bc + ca) = 6
⇒a³ + b³ + c³ - 3abc/(16 - 10) = 6
⇒a³ + b³ + c³ - 3abc = 6× 6
⇒a³ + b³ + c³ - 3abc = 36.

Let A is to be added them 2x² + 3x - 5 + A = x² - x + 1
A = x² - x + 1 - (2x² + 3x - 5)
A = x² - x + 1 - 2x² - 3x + 5
A = -x² - 4x + 6.

Multiples of 7 between 11 and 200 are 14, 21, 28, 35, 42, ..........., 189, 196.
T_m = 196
14 + (m - 1) × 7 = 196
⇒ (m - 1) × 7 = 196 - 14
⇒ (m - 1) × 7 = 182
⇒ (m - 1) = 182/7
⇒ (m - 1) = 26
⇒ m = 27.
Multiples of 7 and 3 both, i.e that of 21 are 21, 42, 63, ........, 189
T_n = 189
21 + (n - 1) × 21 = 189
⇒ (n - 1) × 21 = 189 - 21
⇒ (n - 1) × 21 = 168
⇒ (n - 1) = 168/21
⇒ (n - 1) = 8
⇒ n = 9
∴ Required number of terms = 27 - 9
= 18.

Let the number be x and the quotient is q.
Then, x = 899q + 63 = (29 × 31q) + (29 × 2) + 5
= 29(31q + 2) + 5
So, the given number when divided by 29 gives 5 as remainder.

Let the smaller number be x.
Then, a large number
= x + 1365.
∴ x + 1365 = 6x + 15
5x = 1350
x = 270.

Let x = 13p + 11
and x = 17q + 9
Then,
13p + 11= 17q + 9
17q - 13p = 2
q = (2 + 13p)/17
The least value of p for which q = (2 + 13p)/17 is a whole number, is p = 26.
∴ x = (13 × 26 + 11)
= 338 + 11
= 349.

Greatest number of four digits = 9999
LCM of 15, 25, 40 and 75 = 600
9999/600 = 16,
remainder = 399
Hence, greatest number of four digits which is divisible by 15, 25, 40 and 75
= 9999 - 399
= 9600

LCM of 5, 6, 7 and 8 = 840
Hence the number can be written in the form (840k + 3) which is divisible by 9.
If k = 1, number = (840 × 1) + 3 = 843 which is not divisible by 9.
If k = 2, number = (840 × 2) + 3 = 1683 which is divisible by 9.
Hence 1683 is the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3,
but when divided by 9 leaves no remainder.

Product of two numbers = Product of their HCF and LCM.
Let one number = x
25 × x = 5 × 150
⇒ x = (5 × 150)/25
⇒ x = 30.

If the remainder is same in each case and remainder is not given,
HCF of the differences of the numbers is the required greatest number.
34 - 24 = 10
34 - 28 = 6
28 - 24 = 4
Hence, the greatest number which divides 24, 28 and 34 and gives the same remainder
= HCF of 10, 6, 4
= 2

LCM of 4, 8, 10, 12, 15 and 20 = 120
120 seconds = 2 minutes
Hence all the six bells will ring together in every 2 minutes
Hence, number of times they will ring together in 60 minutes = 1 + (60/2)
= (2+ 60)/2
= 31.

The HCF of a group of numbers will always be a factor of their LCM.
HCF is the product of all common prime factors using the least power of each common prime factor.
LCM is the product of the highest powers of all prime factors.

HCF of the two numbers = 23
Since HCF will always be a factor of LCM, 23 is a factor of the LCM.
Given that other two factors in the LCM are 13 and 14.
Hence factors of the LCM are 23, 13, 14
So, numbers can be taken as (23 × 13) and (23 × 14)
= 299 and 322
Hence, largest number = 322.

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.

LCM of 6, 9, 15 and 18 = 90
Required Number = (90k + 4) which is a multiple of 7
Put k = 1. We get numbers as (90 × 1) + 4 = 94. But this is not a multiple of 7
Put k = 2. We get numbers as (90 × 2) + 4 = 184. But this is not a multiple of 7
Put k = 3. We get numbers as (90 × 3) + 4 = 274. But this is not a multiple of 7
Put k = 4. We get numbers as (90 × 4) + 4 = 364. This is a multiple of 7
Hence 364 is the answer.

If the remainder is same in each case and remainder is not given,
HCF of the differences of the numbers is the required largest number.

9997 - 7654 = 2343
9997 - 8506 = 1491
8506 - 7654 = 852
Hence, the greatest number which divides 7654, 8506 and 9997 and leaves same remainder
= HCF of 2343, 1491, 852
= 213
Now we need to find out the common remainder.
Take any of the given numbers from 7654, 8506 and 9997, say 7654
7654/213 = 35,
remainder = 199.

Let the two parts be x and (50 - x).
Then,
1/x + 1/(50 - x) = 1/12
(50 - x + x)/x(50 - x) = 1/12
x2 - 50x + 600 = 0
⇒ (x - 30)(x - 20) = 0
⇒ x = 30 or x = 20.
So, the parts are 30 and 20.

Let, the number be z,
Then, 7z – 15 = 2z + 10
⇒ 5z = 25
⇒ z = 5.
Hence, the required number is 5.

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.

Let the four numbers be, A, B, C and D
Let A + 3 = B - 3 = 3C = D/3 = x
Then,
A = x - 3
B = x + 3
C = x/3
D = 3x
⇒ (A + B + C + D) = 64
⇒ (x - 3) + (x + 3) + x/3 + 3x) = 64
⇒ 5x + x/3 = 64
⇒ 16x = 192
⇒ x = 12.
Thus the numbers are 9, 15, 4 and 36.
Required difference
= 36 - 4
= 32.

Let the ten’s digit be x.
Then, unit’s digit = (x+3).
Sum of the digits = x + (x +3) = 2x + 3.
Number = 10x + (x+3) = 11x + 3.
(11x + 3)/(2x + 3) = 4/1
11x + 3 = 8x + 12
11x - 8x = 12 - 3
3x = 9
x = 3
Hence, Required number = 11x + 3 = 36.

Let the numbers be x and (184-x).
Then,
x/3 - (184 -x)/7 = 8
⇒ 7x – 3(184-x) = 168
⇒ 10x = 720
⇒ x = 72.
So, the numbers are 72 and 112.
Hence, a smaller number = 72.

Let the ten's digit be x.
Then, the unit's digit = 2x + 1.
[10x + (2x + 1)] - [{10 (2x + 1) + x} - {10x + (2x + 1)}] = 1
⇒ (12x + 1) - (9x + 9) = 1
⇒ 3x = 9, x = 3.
So, ten's digit = 3 and unit's digit = 7.
Hence, original number = 37.

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

Let the ten's digit be x.
Then, the unit's digit = x + 2.
Number = 10x + (x + 2) = 11x + 2
Sum of digits = x + (x + 2) = 2x + 2.
⇒ (11x + 2)(2x + 2) = 144
⇒ 22x² + 26x- 140 = 0
⇒ 11x² + 13x - 70 = 0
⇒ (x - 2)(11x + 35) = 0
⇒ x = 2
Hence, Required Number = 11x + 2 = 24