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Number System Question for SSC CGL Mains Asked by TCS

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Number System Question for SSC CGL Mains Asked by TCS

SSC CGL Tier 2 2019 SSC CGL Tier 1 2019 SSC CHSL 2019 SSC CPO 2020 SSC MTS 2019
SSC CGL Tier 2 2018 SSC CGL Tier 1 2018 SSC CHSL 2018 SSC CPO 2019 SSC MTS 2017

Q1. If a nine-digit number 389x6378y is divisible by 72, then the value of

will be :

यदि नौ अंकों की एक संख्या 389x6378y, 72 से विभाजित है, तो 6x+7y का मान होगा : 

SSC CGL Tier-II- 11 September 2019 

(a) 6

(b) 13     

(c) 46

(d) 8

Q2. When 12, 16, 18, 20, and 25 divide the least number x, the remainder in each case is 4 but x is divisible by 7. What is the digit at the thousands’ place in x?

जब 12, 16, 18, 20 और 25 न्यूनतम संख्या x को विभाजित करते हैं, तो हर मामले में शेषफल 4 आता है लेकिन x, 7 से विभाजित है | x के हजारवें स्थान पर कौन सा अंक है ?

SSC CGL Tier-II 11 September 2019 

(a) 5

(b) 8      

(c) 4

(d) 3

Q3. When 7897, 8110, and 8536 are divided by the greatest number x, then the remainder in each case is the same. The sum of the digits of x is :

जब 7897, 8110 और 8536 को सबसे बड़ी संख्या x से विभाजित किया जाता  है, तो प्रत्येक मामले में शेषफल समान आता है | x के अंकों का जोड़ है :

SSC CGL Tier-II 11 September 2019 

(a) 14

(b) 5                 

(c) 9

(d) 6

Q4. One of the factors of  (82k+52k), where k is an odd number, is :

 (82k+52k) का एक गुणक ज्ञात करें, जहाँ k एक विषम संख्या है | 

SSC CGL Tier-II- 11 September 2019 

(a) 86

(b) 88    

(c) 84

(d) 89

Q5. Let x = (633)24-(277)38+ +(266)54. what is the unit digit of x?

मान लीजिये कि  x =(633)24- -(277)38+(266)54 है | x का इकाई अंक क्या है ? 

SSC CGL Tier II- 11 September 2019 

(a) 7

(b) 6      

(c) 4

(d) 8

Q6. The sum of the digits of a two-digit number is 1⁄7of the number. The units digit is 4 less than the tens digit. If the number obtained on reversing its digit is divided by 7, the remainder will be :

दो अंकों की एक संख्या के अंकों का जोड़ संख्या का 1/7 है | इकाई अंक दहाईं के अंक से 4 कम है | यदि इसके अंकों को पलटने से बनी संख्या को 7 से भाग दिया जाए, तो शेषफल होगा :

SSC CGL Tier-II 11 September 2019 

(a) 4

(b) 5   

(c) 1

(d) 6

Q7. If the 11-digit number 5678x43267y is divisible by 72, then the value of  is :

यदि 11-अंकों की संख्या 5678x43267y , 72 से विभाजित है, तो  का मान क्या होगा ?

SSC CGL Tier-II 12 September 2019

(a) 6

(b) 4   

(c) 7

(d) 8 

Q8. The number of factors of 3600 is : 3600 के गुणकों की संख्या है

SSC CGL Tier-II 12 September 2019 

(a) 45

(b) 44     

(c) 43

(d) 42

Q9. If x = (164)169+(333)337-(727)726, then what is the unit digit of x?

यदि x =  (164)169+(333)337-(727)726 है, तो x का इकाई अंक क्या है ?

SSC CGL Tier-II 12 September 2019

(a) 5

(b) 7      

(c) 8

(d) 9

Q10. Let x be the least number which when divided by 15, 18, 20, and 27, the remainder in each case is 10 and x is a multiple of 31. What least number should be added to x to make it a perfect square?

मान लीजिये कि x वह सबसे छोटी संख्या है जिसे 15, 18, 20 और 27 से विभाजित करने पर प्रत्येक स्थिति में शेषफल 10 आता है और x, 31 का एक गुणज है | x में न्यूनतम कौन सी संख्या जोड़ी जानी चाहिए ताकि यह पूर्ण वर्ग बन जाए ?

SSC CGL Tier-II 12 September 2019 

(a) 39

(b) 37                

(c) 43

(d) 36

Here are the solutions to the above questions in short methods. It will help you to save your precious time in the exam.

Sol 1.

Ans. (d)

Since 389x6378y is divisible by 72 it must be divisible by 9 and 8 (coprime factors of 72) and y must be an even number. So sum of digits of this number must be divisible by 9 and last three digits by 8. 3+8+9+x+6+3+7+8+y = 44+x+y

x+y must be 1 or 10 as after 45 nearest multiple of 9 is 45 and 54.

For x+y =1 x must be 1 as y can’t be odd number. And 780 is not divisible by 8 so it will get neglected.

Pairs for 10 = (1,9)(2,8), (3,7), (4,6), (5,5), (6,4)(7,3)(8,2)(9,1)

Only pair which satisfies these conditions is (6,4). So the required value is= 8

Sol 2.

Ans. (b)

12 = 2x2x3

16 = 2x2x2x2

18 = 2x3x3

20=2x2x5

25 = 5x5

LCM of 12,16,18,20 and 25 = 2x2x2x2x3x3x5x5 = 3600

x must be = 3600k+4

Where 3600k+4 is multiple of 7

The condition gets satisfied when k=5

Required number = 3600(5)+4 = 18004

digit at the thousands’ place in x =8 

Sol 3.

Ans. (d)

let the number be n 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

Sol 4. (d)

Put any odd value of k. For example k=1 

(82k+52k) ⇒ (82(1)+52(1)) = 89

Clearly, 89 will be the factor.

Sol 5. (d)

Given, x = (633)24-(277)38+(266)54

Unit digit of (633)24= 34= 1

Unit digit of (277)38= 72= 9

Unit digit of (266)54= 62

Unit digit of x = 1-9+6 = -2

But unit digit can’t be negative so, required unit digit = 10 + (-2) = 8

Sol 6.(d)

Let the number = 10x+y

According to the question 

(x+y) = 1/7(10x+y)  ………..(1)

And 

y = x-4

Put this value in equation (1)

(x+x-4) = 1/7(10x+x-4)

⇒ 14x-28 = 11x-4

⇒ x=8and y = 8-4 = 4

the number obtained on reversing the digit = 10y+x = 10(4)+8 = 48

Required remainder = 48/7= 6

Sol 7. (a) Since, 5678x43267y is divisible by 72 it must be divisible by 9 and 8 (coprime factors of 72) and y must be an even number. So sum of digits of this number must be divisible by 9 and last three digits by 8. 5+6+7+8+x+4+3+2+6+7+y = 48+x+y

x+y must be 6 or 15 as after 48 nearest multiples of 9 are 54 and 63.

Pairs for 6 = (1,5)(2,4), (3,3), (4,2), (5,1), (6,0)

y can’t be an odd number. Only pair that satisfies all the condition is (4,2)

Pairs for 15 = (6,9), (7,8), (8,7), (9,6)

None of the pairs of 15 satisfies the given conditions.

So the required value is

Sol 8. (a)

3600 = 24 × 32 × 52

Number of factors of 3600= (4+1) (2+1) (2+1) = 45

Sol 9. (c)

Given, 

x = (164)169+(333)337-(727)726

Unit digit of (164)169= 41= 4

Unit digit of (333)337= 31= 3

Unit digit of (727)726= 72= 9

Unit digit of x = 4+3-9 = -2

But unit digit can’t be negative so, required unit digit = 10 + (-2) = 8

Sol 10. (a)

15 = 3x5

18 = 2x3x3

20 = 2x2x5

27 = 3x3x3

LCM of 15,18,20 and 27 = 2x2x3x3x3x5 = 540

⇒ x must be = 540k+10

Where 540k+10 is multiple of 31

The condition gets satisfied when k=4

Required number = 540 (4)+10 = 2170

Nearest square to 2170 = 2209

Required number = 2209-2170 = 39

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