Number system is important topic in quantitative aptitude section of SSC CHSL and CGL exams. You can download the SSC CHSL number system questions and answers PDF.<\/p>\n
Number System Questions for SSC CHSL Exam:<\/strong><\/span><\/p>\n 20 SSC CHSL (latest pattern mocks) – Rs. 117<\/a><\/p>\n You can practice SSC CHSL English question papers\u00a0<\/a><\/p>\n Download SSC CHSL 2018 Syllabus PDF<\/a><\/p>\n Download All SSC CHSL Questions Topic wise PDF<\/a><\/p>\n Take Free SSC CHSL Mock test 2018<\/a><\/p>\n Question 1:<\/strong> What smallest number should be added to 2957 so that the sum is completely divisible by 17 ?<\/p>\n a) 9 Question 2:<\/strong> Which of the following numbers is not a prime number?<\/p>\n a) 731 Question 3:<\/strong> The largest 5 digit number exactly divisible by 95 is<\/p>\n a) 99936 Question 4:<\/strong> If the number 583_437 is completely divisible by 9, then the smallest whole number in the place of the blank digit will be<\/p>\n a) 4 Question 5:<\/strong> The sum of all prime numbers between 30 and 42 is<\/p>\n a) 103 Take SSC CHSL Free Mock Test<\/a><\/p>\n Question 6:<\/strong> What smallest number should be added to 2401 so that the sum is completely divisible by 14 ?<\/p>\n a) 8 Question 7:<\/strong> Which of the following numbers is completely divisible by 99?<\/p>\n a) 51579 Question 8:<\/strong> If in a two digit number, the digit at unit place is z and the digit at tens place is 8, then the number is<\/p>\n a) 80z + z Question 9:<\/strong> What least number must be added to 1039, so that the sum obtained is completely divisible by 29?<\/p>\n a) 4 Question 10:<\/strong> Which of the following numbers is not a prime number?<\/p>\n a) 197 SSC CHSL 2018 Syllabus PDF<\/a><\/p>\n Solutions for\u00a0Number System Questions for SSC CHSL:<\/strong><\/p>\n Solutions:<\/strong><\/p>\n 1) Answer (D)<\/strong><\/p>\n If 2957 is divisible 17, => 2957 = 17 \\times<\/span>\u00d7<\/span><\/span><\/span><\/span> 173 + 16 2) Answer (A)<\/strong><\/p>\n Prime Factorization of 731 = 17\u00a0\\times<\/span>\u00d7<\/span><\/span><\/span><\/span>\u00a043 3) Answer (C)<\/strong><\/p>\n Largest 5 digit number = 99999 4) Answer (D)<\/strong><\/p>\n Number :\u00a0583_437 5) Answer (B)<\/strong><\/p>\n Sum of prime numbers between 30 and 42 : SSC CHSL Daily practice set<\/a><\/p>\n 6) Answer (B)<\/strong><\/p>\n Factorizing 2401 = 14 \\times<\/span>\u00d7<\/span><\/span><\/span><\/span> 171 + 7 7) Answer (A)<\/strong><\/p>\n For a number to be divisible by 99, it must be divisible by 9 and 11 8) Answer (B)<\/strong><\/p>\n Digit at unit’s place = z 9) Answer (B)<\/strong><\/p>\n Factorizing 1039 = 29 \\times<\/span>\u00d7<\/span><\/span><\/span><\/span> 35 + 24 10) Answer (D)<\/strong><\/p>\n Among the given numbers, there are no factors of 197, 313, 439 other than 1 and the numbers themselves. Number system questions and answers for SSC CHSL exam PDF are given here, you can solve more SSC questions<\/a><\/p>\n
\nb) 2
\nc) 3
\nd) 1<\/p>\n
\nb) 227
\nc) 347
\nd) 461<\/p>\n
\nb) 99935
\nc) 99940
\nd) 99933<\/p>\n
\nb) 5
\nc) 3
\nd) 6<\/p>\n
\nb) 109
\nc) 105
\nd) 104<\/p>\n
\nb) 7
\nc) 4
\nd) 5<\/p>\n
\nb) 51557
\nc) 55036
\nd) 49984<\/p>\n
\nb) 80 + z
\nc) 8z + 8
\nd) 80z + 1<\/p>\n
\nb) 5
\nc) 8
\nd) 6<\/p>\n
\nb) 313
\nc) 439
\nd) 391<\/p>\n
\nQuotient is 173 and remainder is 16
\nThus, smallest number that should be added to 2957\u00a0so that the sum is completely divisible by 17 = (17 – 16) = 1
\n=> Ans – (D)<\/p>\n
\nFor all the remaining numbers, the only factors are 1 and the number themselves.
\nThus, only 731 is not a prime number.
\n=> Ans – (A)<\/p>\n
\nIf 99999 is divided by 95, => 99999 = 95\u00a0\\times<\/span>\u00d7<\/span><\/span><\/span><\/span> 1052 + 59
\nQuotient is 1052 and remainder is 59
\nThus, largest 5 digit number which is divisible by 95 = 99999 – 59
\n= 99940
\n=> Ans – (C)<\/p>\n
\nLet the missing digit = $$x$$
\n=> Sum of digits = 5 + 8 + 3 + $$x$$ + 4 + 3 + 7 = 30 + $$x$$
\nClearly, sum of digits is between 30 and 39.\u00a0The only number divisible by 9 and which lies between 30 and 39 is 36
\n=> $$30 + x = 36$$
\n=> $$x = 36 – 30 = 6$$
\n=> Ans – (D)<\/p>\n
\n= 31 + 37 + 41
\n= 109
\n=> Ans – (B)<\/p>\n
\nThus, on dividing 2401 by 14, the remainder is 7
\n\u2234<\/span><\/span><\/span><\/span> The number that\u00a0must be added to 2401 so that the sum obtained is completely divisible by 14
\n= 14 – 7 = 7
\n=> Ans – (B)<\/p>\n
\n(A) : 51579 = 5 + 1 + 5 + 7 + 9 = 27 which is divisible by 9 and also by 11
\n(B) : 51557 = 5 + 1 + 5 + 5 + 7 = 23 which is not divisible by 9
\n(C) : 55036 = 5 + 5 + 0 + 3 + 6 = 19\u00a0which is not divisible by 9
\n(D) : 49984 = 4 + 9 + 9 + 8 + 4 = 34\u00a0which is not divisible by 9
\nThus, only option (A) is divisible by 99<\/p>\n
\nDigit at ten’s place = 8
\n=> 2-digit number = $$(10 \\times 8) + (1 \\times z)$$
\n= 80 + z
\n=> Ans – (B)<\/p>\n
\nThus, on dividing 1039 by 29, the remainder is 24
\n\u2234<\/span><\/span><\/span><\/span> The number that\u00a0must be added to 1039 so that the sum obtained is completely divisible by 29
\n= 29 – 24 = 5
\n=> Ans – (B)<\/p>\n
\n391 = 23 \\times<\/span>\u00d7<\/span><\/span><\/span><\/span> 17
\n=> 391 is not a prime number.
\n=> Ans – (D)<\/p>\n