Test: SSC CGL Previous Year Question: Number System and HCF & LCM (2023-2024) - 1 - SSC CGL MCQ

For a number to be divisible by 3, 7 and 11 it should be divisible by 231
And on dividing 67600 by 231 we get remainder as 148
So the next multiple is at the distance of 231 - 148 = 83
The next multiple is 67600 + 83 = 67683
So, x = 8 and y = 3
3x - 5y = 24 - 15 = 9

If a number is divisible by 3 then its sum should also be divisible by 3.
5 + 3 + 0 + 6 + P + 2 = 16 + P
Least value = P = 2 as 16 + 2 = 18 (divisible by 3)
Maximum value P = 8 as 16 + 8 = 24 (divisible by 3)
Difference in Square = 64 - 4 = 60

72 = 8 × 9
For divisibility of 8, last 3 digits should be divisible by 8
Possible value of y = 3 and 7
For divisibility of 9, the sum of digits should be divisible by 9.
9 + 4 + x + 2 + 9 + 3 + 6 = 9 or its multiple x = 3 (x = y)
Check with 7
9 + 4 + x + 2 + 9 + 7 + 6 = 9
or its multiple x = 8
So , 2x + 3y = 2(8) + 3(7) = 37

P = 2x , Q = 3y
P × Q = 6xy
6xy is divisible by 6
P × Q is also divisible by 6 but not 5
P + Q = 2x + 3y
2x + 3y is not divisible by 5 and 6
So, P + Q is not divisible by 5 and 6.

Let five consecutive odd natural numbers are
(x - 4), (x - 2), x, (x + 2), (x + 4)
(x - 4)² + (x - 2)² + x² + (x + 2)² + (x + 4)² = 233 x 5 ⇒ 5x² + 40 = 1165
5x² = 1125 ⇒ x = 15,
Largest number = 19
Smallest number = 11
Average = 15

LCM of 3, 7, 11 = 231
When 59399 is divided by 231, Reminder = 32
Required number = 59399 - 32 = 59367
Compare 59367 with 593ab
so, a = 6 , b = 7 , (a² - b² + ab) = 29

If a number is divisible by 7
11 and 13 then it must be divisible by 1001
And if the number is divisible by 1001 the its first three digits would be same as its next three digits (xyzxyz)
So, z = 4, x = 5 and y = 3 and x + y - z = 8 - 4 = 4.

4a067b
Sum of odd placed digits = 4 + 0 + 7
Sum of even placed digits = a + 6 + b
So, 11 - (6 + a + b) = 0 or multiple of 11
a + b = 5 or 16
So the sum all the possible values of (a + b) = 5 + 16 = 21

Let larger number = a
Smaller number = b
a - b = 1280 ....(1)
a = 7b + 50
a - 7b = 50 .....(2)
Multiplying eq (1) by 7
7a - 7b = 8960 ...(3)
Subtracting eq (2) from eq (3)
6a = 8910 ⇒ a = 1485

(4x - 2y + 3z)
= 4 × 17 − 24 × 2 + 27 × 3
(4x - 2y + 3z) = 101
When 101 is divided by 31 we get reminder = 8

LCM of 3, 7, 11 is = 231
Let the number be 23599
When 23599 is divided by 231 we get remainder as 37
So the number is = 23599 - 37 = 23562
x = 6 and y = 2
(3x - 4y) = 18 - 8 = 10

Let the number be x which divides 7897, 8110 and 8536 leaving a reminder r.
The required number then becomes H.C.F of (7897 - r), (8110 - r) and (8536 - r)
It could also be the H.C.F of (8536 - r) - (8110 - r) and (8110 - r) - (7897 - r) .i.e. 426 and 213
⇒ H.C.F of 426 and 213 = 213 the required sum = 2 + 1 + 3 = 6

Given, a + b + c = 1(11/12) ...(1) and (c/a) = 5/2
Let c = 5 unit and a = 2 unit
According to the question
b = (5/2) - (7/4) = 3/4
Put this value in eq (1)
a + c = (23/12) - (3/4) = 7/6
(5 + 2) unit = 7/6 ⇒ 1 unit = 1/6
⇒ 5 unit = 5/6 ⇒ 2 unit = 1/3
Required difference = 5/6 - 1/3 = 1/2.

Given,
x + y + z = 1(13/24) ...(1) and z/x = 9/16
Let z = 9 unit and x = 16 unit
According to the question
y = (9/16) - 0.0625
= (9/16) - (625/10000) = 1/2
Put this value in eq (1)
⇒ x + z = 

Since, 517X324 is divisible by 12 it must be divisible by 3 and 4(coprime factors of 12).
For a number to be divisible by 3, the sum of its digits must be divisible by 3.
So, 5 + 1 + 7 + x + 3 + 2 + 4 ⇒ 22 + x must be divisible by 3
Possible values of x = 2, 5, 8
Cleary the smallest value of x = 2

As, HCF of (13, a), (13, b) and (13, c) is 13 where a, b and c must be multiples of 13.
Only possible value for a, b and c such that a < b < c < 60 is 26, 39 and 52.
So, a = 26, b = 39 and c = 52.

L.C.M of 2, 3, 5, 6, 9 and 18 = 90
90 = 2 × 3 × 3 × 5
As 2 and 5 don’t have any pair
So, the required number = 90 × 2 × 5 = 900

Let the numbers be 10x and 10y
Then, 10x × 10y = 1500 ⇒ xy = 15
Possible value of x and y are
⇒ (1, 15) and (3, 5).

Let, the no. are 15a and 15b.
36x² - 81y² = 9(2x -3y)(2x + 3y)
= 9(2 x 15a - 3 x 15b)(2 x 15a + 3 x 15b)
= 9 x 15{(2a - 3b)(2a + 3b)}
And, 81x² - 9y² = 9(3x - y)(3x + y)
= 9 (3 x 15a − 15b)(3 x 15a + 15b)
= 9 x 15 {(3a − b)(3a + b)}
Their H.C.F.= 135

Least number which when double is divisible by 15, 18, 25 and 32

H.C.F. 

Let the number be 3x and 8x, then the L.C.M. will be 3 × 8 × x .
According to the question ,
120 = 24 x ⇒ x = 5
Now, sum of the number = 3x + 8x
= 11x = 11 × 5 = 55

Let the LCM and HCF of the given no’s be L and H respectively
Then, L = 56H
According to question,
L + H = 1710 ⇒ 56H + H = 1710
⇒ 57H = 1710 ⇒ H = 30
L = 56 × 30 = 1680
As we know that, First No. × Second No. = LCM × HCF
240 × Second number = 1680 × 30
Second number = (1680 x 30)/240 = 210.

LCM (8, 9, 12, 15) = 360
So, the number is 360k + 5, which is divisible by 7 ⇒ Putting k = 3, we have ;
Required no = 360 × 3 + 5 = 1080 + 5 = 1085

since, the given no’s are prime no’s. So, the HCF of (A, B) = 1 and
LCM of (A, B) = A × B = AB
According to the question,
A × B = 209 = 19 × 11
So, A = 19 and B = 11 (∵ A > B)
Hence, A² - B = 19² - 11 = 361 - 11 = 350

LCM (6 , 7 and 8) = 168 and, (6 – 4) = (7 – 5) = (8 – 6) = 2
When, the greatest four-digit number i.e.9999 is divided by 168 leaves remainder 87
Required number = (9999 - 87) - 2 = 9910.

According to the question,

Let three numbers be 6x, 8x and 9x
So, 9x – 6x = 33 ⇒ 3x = 33 ⇒ x = 11
HCF (6x, 8x, 9x) = x = 11

To find the required greatest number, we need to find the HCF of the difference between all the three given
numbers.
1134 - 1062 = 72, 1182 - 1134 = 48
HCF of 72 and 48 = 24
Now, When 1062 is divided by 24 leaves, remainder 6.
So, x - y = 24 - 6 = 18

To find the value of x, we need to find the HCF of the difference between all the three given numbers.
1027 - 955 = 72, 1075 - 1027 = 48
HCF of (72 and 48) = 24
Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
Hence, 16 is not the factor of x.

Let the two no’s be 12x and 12y
According to question,
12x - 12y = 60 ⇒ x - y = 5
And, 12 × xy = 204 × 12 ⇒ xy = 204
Now, (x + y) = 

So, the required sum = 12(x + y) = 12 × 29 = 348.