Download SBI Clerk Number System Questions & Answers PDF for SBI Clerk Prelims and Mains exam. Very Important SBI Clerk Number System questions on with solutions.<\/p>\n
Download Number System Questions For SBI Clerk Exam PDF<\/a><\/p>\n 20 SBI Clerk mocks for Rs. 149. Enroll here<\/a><\/p>\n Take a free mock test for SBI Clerk<\/a><\/p>\n Download SBI Clerk previous papers PDF<\/a><\/p>\n Question 1:\u00a0<\/b>The ratio of 3rd and 6th term of a GP is 27. If the first term of the GP is 10, then find out the sum of infinite terms of this convergent GP.<\/p>\n a)\u00a030<\/p>\n b)\u00a040<\/p>\n c)\u00a015<\/p>\n d)\u00a020<\/p>\n e)\u00a025<\/p>\n Question 2:\u00a0<\/b>Sum of the first 3 terms of an AP is 2\/9 times the sum of first 6 terms of the same AP. Find out the ratio of first term to the common difference of the same AP.<\/p>\n a)\u00a03 : 8<\/p>\n b)\u00a01 : 4<\/p>\n c)\u00a02 : 7<\/p>\n d)\u00a01 : 3<\/p>\n e)\u00a01 : 5<\/p>\n Question 3:\u00a0<\/b>N is a number that leaves the same remainder on dividing 2527, 2419, 2383, and 2599. The largest value of N that satisfies the condition is<\/p>\n a)\u00a054<\/p>\n b)\u00a018<\/p>\n c)\u00a072<\/p>\n d)\u00a036<\/p>\n e)\u00a090<\/p>\n Question 4:\u00a0<\/b>N is the smallest four digit number which leaves remainder 2, 3 and 4 when it is divided by 5, 6 and 7 respectively. Find the sum of the digits of the number \u2018N\u2019.<\/p>\n a)\u00a012<\/p>\n b)\u00a022<\/p>\n c)\u00a017<\/p>\n d)\u00a014<\/p>\n e)\u00a019<\/p>\n Download SBI Clerk Previous Papers PDF<\/a><\/p>\n Take a free mock test for SBI Clerk<\/a><\/p>\n Question 5:\u00a0<\/b>N is the largest four digit number which leaves remainder 3, 4 and 5 when it is divided by 6, 7 and 8 respectively. Find the sum of the digits of the number \u2018N\u2019.<\/p>\n a)\u00a032<\/p>\n b)\u00a031<\/p>\n c)\u00a034<\/p>\n d)\u00a028<\/p>\n e)\u00a027<\/p>\n Question 6:\u00a0<\/b>Second term of an AP is twice the fifth term. What is the sum of the first 15 terms of that AP?<\/p>\n a)\u00a0-12<\/p>\n b)\u00a07<\/p>\n c)\u00a011<\/p>\n d)\u00a00<\/p>\n e)\u00a0-3<\/p>\n Question 7:\u00a0<\/b>The largest 4 digit number that leaves the same remainder when divided by 2, 3, 4 and 5 is<\/p>\n a)\u00a09960<\/p>\n b)\u00a09999<\/p>\n c)\u00a09997<\/p>\n d)\u00a09974<\/p>\n e)\u00a09961<\/p>\n Question 8:\u00a0<\/b>A number leaves the same remainder on dividing 1120, 1248, 992, and 1184. The largest number that satisfies this condition is<\/p>\n a)\u00a032<\/p>\n b)\u00a064<\/p>\n c)\u00a0128<\/p>\n d)\u00a0256<\/p>\n e)\u00a016<\/p>\n Question 9:\u00a0<\/b>Ramesh starts saving 3 rupees from 1st March and adds 4 rupees to the amount he saved the previous day. He saved till April 30th and stopped after that.What is the amount with Ramesh on May 1?<\/p>\n a)\u00a07602 rupees<\/p>\n b)\u00a07503 rupees<\/p>\n c)\u00a07413 rupees<\/p>\n d)\u00a07300 rupees<\/p>\n e)\u00a07366 rupees<\/p>\n Banking Study Material – 18000 Questions<\/a><\/p>\n Daily Free SBI Online Tests<\/a><\/p>\n Question 10:\u00a0<\/b>In an Arithmetic Progression the sum of 4th and 5th term is 27. Twice the 3rd term is equal to the 6th term. What is the 18th term of the series?<\/p>\n a)\u00a054<\/p>\n b)\u00a060<\/p>\n c)\u00a063<\/p>\n d)\u00a057<\/p>\n e)\u00a051<\/p>\n Question 11:\u00a0<\/b>A 2 digit number is such that the number is equal to 7 times the sum of its digits. The smallest such number is<\/p>\n a)\u00a064<\/p>\n b)\u00a042<\/p>\n c)\u00a021<\/p>\n d)\u00a012<\/p>\n e)\u00a024<\/p>\n Question 12:\u00a0<\/b>A number leaves the same reminder on dividing 237 and 269. How many numbers satisfy this criterion?<\/p>\n a)\u00a02<\/p>\n b)\u00a05<\/p>\n c)\u00a06<\/p>\n d)\u00a08<\/p>\n e)\u00a016<\/p>\n Question 13:\u00a0<\/b>Two natural numbers add up to 100. If it is known that both the numbers are even, how many pair of numbers satisfy this condition?<\/p>\n a)\u00a049<\/p>\n b)\u00a050<\/p>\n c)\u00a024<\/p>\n d)\u00a025<\/p>\n e)\u00a099<\/p>\n Question 14:\u00a0<\/b>If the product of two prime numbers is 713, then find out the L.C.M. of these two numbers.<\/p>\n a)\u00a0713<\/p>\n b)\u00a031<\/p>\n c)\u00a023<\/p>\n d)\u00a017<\/p>\n e)\u00a0None of the above<\/p>\n Question 15:\u00a0<\/b>The product of two natural numbers is 9222. If they differ by 19 then find out the sum of the numbers.<\/p>\n a)\u00a0205<\/p>\n b)\u00a0199<\/p>\n c)\u00a0197<\/p>\n d)\u00a0195<\/p>\n e)\u00a0193<\/p>\n 100+ Free GK Tests<\/a><\/p>\n Question 16:\u00a0<\/b>The product of two natural numbers is 4810. If they differ by 9 then find out the sum of the numbers.<\/p>\n a)\u00a0149<\/p>\n b)\u00a0139<\/p>\n c)\u00a0135<\/p>\n d)\u00a0145<\/p>\n e)\u00a0147<\/p>\n Question 17:\u00a0<\/b>Number obtained by interchanging the digits of a two digit number is 18 more than the original number. If the sum of the digits of the number is 12, then find out what is the original number?<\/p>\n a)\u00a039<\/p>\n b)\u00a048<\/p>\n c)\u00a057<\/p>\n d)\u00a093<\/p>\n e)\u00a0None of the above<\/p>\n Question 18:\u00a0<\/b>A man wins a lottery of 5000 Rs. He starts spending it starting with 30rs on day 1 and keeps on increasing the money he spends by 10rs every day. On which day will all of his lottery money get exhausted?<\/p>\n a)\u00a038<\/p>\n b)\u00a028<\/p>\n c)\u00a033<\/p>\n d)\u00a025<\/p>\n e)\u00a030<\/p>\n Question 19:\u00a0<\/b>Mohan starts saving Rs 3,6,9,\u2026\u2026\u2026 on each day starting from March 1st. He aims to buy a bicycle for himself costing Rs 10000. If his birthday is on 15th May, how much money will he have to take from his dad if he wishes to buy the bicycle on the day of his birthday itself?. (Assume that he will take the remaining amount from his dad)<\/p>\n a)\u00a01100<\/p>\n b)\u00a01224<\/p>\n c)\u00a01222<\/p>\n d)\u00a01348<\/p>\n e)\u00a01456<\/p>\n Question 20:\u00a0<\/b>There are 100 cards which are numbered from 1 to 100. Diya and Rahul are playing a game wherein Diya has to choose a certain number of cards, such that any number that Rahul guesses\u00a0(only from 1 to 100), she should be able to express it using the sum of cards she has with her. What is the minimum number of cards that she should choose?<\/p>\n a)\u00a05<\/p>\n b)\u00a050<\/p>\n c)\u00a011<\/p>\n d)\u00a07<\/p>\n e)\u00a0100<\/p>\n General Knowledge Questions & Answers PDF<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let \u2018r\u2019 be the common ratio of the given GP. 2)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Let \u2018a\u2019 and \u2018d\u2019 be the first term and common difference of the given AP. 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n N leaves the same remainder on dividing 2527, 2419, 2383, and 2599, the difference between any pair of these numbers should be divisible by the number. The smallest difference will be the limiting factor. 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n It is given that N leaves remainder 2, 3 and 4 when it is divided by 5, 6 and 7 respectively. We can also say that N leaves -3 as remainder when it is divided by 5, 6 and 7. 5)\u00a0Answer\u00a0(E)<\/strong><\/p>\n It is given that N leaves remainder 3, 4 and 5 when it is divided by 6, 7 and 8 respectively. We can also say that N leaves -3 as the remainder when it is divided by 6, 7 and 8. 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n let the first term be ‘a’ and the common difference be ‘d’. 7)\u00a0Answer\u00a0(E)<\/strong><\/p>\n We know that the number leaves the same remainder when divided by 2,3,4 and 5. Therefore, option E is the right answer.<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Since the number leaves the same remainder on dividing 1120, 1248, 992 and 1184, the difference between any pair of these numbers should be divisible by the number. The smallest difference will be the limiting factor.<\/p>\n 1248 – 1120 = 128 As we can see, 64 is the smallest difference and hence, the largest number that satisfies the given condition is 64. 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n This is clearly an Arithmetic Progression with a=3 and d=4. 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the A.P be a, a+d, a+2d,…….. 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let the number be ab. 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n It has been given that the number leaves the same remainder when dividing 237 and 269. Therefore, the difference between the 2 numbers (269 – 237 = 32) must be divisible by the number. Therefore, the number must be a factor of 32.<\/p>\n The factors of 32 are 1, 2, 4, 8, 16 and 32. The number can take 6 values. Therefore, option C is the right answer.<\/p>\n 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let the 2 natural numbers be a and b. Therefore, 25 pairs of numbers will satisfy the condition and hence, option D is the right answer.<\/p>\n 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n We know that for any two numbers \u2018a\u2019 and \u2018b\u2019. 15)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Let \u2018a\u2019 and \u2018b\u2019 be the two numbers where (a > b). It is given that 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let \u2018a\u2019 and \u2018b\u2019 be the two numbers where (a > b). It is given that 17)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let us assume that the original two digit number is \u2018ab\u2019. 18)\u00a0Answer\u00a0(E)<\/strong><\/p>\n The man starts spending 30rs on day 1 and keeps on increasing the money he spends by 10rs every day, i.e. he spends 30, 40, 50,\u2026\u2026\u2026 on day1, day 2 and so on. This forms an A.P. with a = 30 and d = 10. 19)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Number of days in March = 31 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Since 2 is the smallest number in the base system and all the numbers can be expressed in binary, the powers of 2 will only be needed to express the sum.<\/p>\n Sum of powers of 2 – 1, 2, 4, .. This should be greater than or equal to 100.<\/p>\n $2^n – 1 \\geq 100$<\/p>\n $2^7 = 128$<\/p>\n Hence, n = 7 is the answer.<\/p>\n Thus, if she has the cards which are numbered 1, 2, 4, 8, 16, 32 and 64, she can express all the numbers from 1 to 100 as sum of these cards.<\/p>\n
\nThe ratio of 3rd and 6th term of a GP is 27.
\n$\\dfrac{ar^2}{ar^5}$ = 27
\n$\\Rightarrow$ $\\dfrac{1}{r^3}$ = 27
\n$\\Rightarrow$ $r = \\dfrac{1}{3}$
\nSum of infinite GP = $\\dfrac{a}{1-r}$ = $\\dfrac{10}{1-1\/3}$ = 15
\nHence, option C is the correct answer.<\/p>\n
\n$\\Rightarrow$ $\\dfrac{3}{2}(2a + 2d) = \\dfrac{2}{9}[\\dfrac{6}{2}(2a + 5d)]$
\n$\\Rightarrow$ $3a+3d = \\dfrac{2}{3}(2a + 5d)$
\n$\\Rightarrow$ $9a+9d = 4a + 10d$
\n$\\Rightarrow$ $5a = d$
\n$\\Rightarrow$ $\\dfrac{a}{d} = \\dfrac{1}{5}$
\nTherefore, option E is the correct answer.<\/p>\n
\n2599 – 2527 = 72
\n2599- 2419= 180
\n2599-2383= 216
\n2527- 2419= 108
\n2527- 2383= 144
\n2419- 2383= 36
\nAs we can see, 36 is the smallest difference and hence, the largest number that satisfies the given condition is 36.
\nTherefore, option D is the right answer.<\/p>\n
\nTherefore, the smallest number which is of this kind = (LCM of 5, 6, 7) – 3 = 210 – 3 = 207
\nNext such number = 207 + 210 = 333
\nWe can say that
\n$\\Rightarrow$ N $\\geq$ 999
\n$\\Rightarrow$ 207+(n – 1)210 $\\geq$ 999
\n$\\Rightarrow$ n > 4.77
\nTherefore, we can say that $n_{min}$ = 5.
\nTherefore, N = 207+4*210 = 1047
\nHence, the sum of the digits of N = 1 + 0 + 4 + 7 = 12. (Option : A)<\/p>\n
\nTherefore, the smallest number which is of this kind = (LCM of 6, 7, 8) – 3 = 168 – 3 = 165
\nNext such number = 165 + 168 = 333
\nWe can say that
\n$\\Rightarrow$ N $\\leq$ 9999
\n$\\Rightarrow$ 165+(n – 1)168 $\\leq$ 9999
\n$\\Rightarrow$ n < 59.53
\nTherefore, we can say that $n_{max}$ = 59.
\nTherefore, N = 165+58*168 = 9909
\nHence, the sum of the digits of N = 9 + 9 + 0 + 9 = 27. (Option : E)<\/p>\n
\nAs per the question, a + d = 2(a + 4d)
\nOn solving, we get a + 7d = 0
\nor, Eighth term is 0.
\nSum of the first 15 terms = $\\frac{n}{2}[2a + (n – 1)d] = n(a + 8d)$ = 0
\nHence, option D is the correct answer.<\/p>\n
\nTherefore, the number should be of the form LCM(2,3,4,5) + k.
\nLCM of 2,3,4,5 is 60.
\nThe largest multiple of 60 below 10,000 is 9960.
\n2 can leave a remainder of 0 or 1.
\nTherefore, the largest possible value of k is 1.
\nThe largest 4 digit number that leaves the same remainder when divided by 2, 3, 4 and 5 is 9960 + 1 = 9961.<\/p>\n
\n1184 – 1120 = 64
\n1120-992 = 128
\n1248 – 992= 256
\n1184 – 992 = 192
\n1120 – 992 = 128<\/p>\n
\nTherefore, option B is the right answer.<\/p>\n
\nTill 1st May he has 61 days.
\nThus, till 1st May he will have = $\\dfrac{n}{2}*(2a+(n-1)d)$
\nHence, Amount of money with Ramesh on 1st May = $\\dfrac{61}{2}*(2*3+(61-1)4)$ = 7503 rupees.
\nHence, option B is the correct answer.<\/p>\n
\nGiven, 2(a+2d) = a+5d
\ni.e. 2a+4d = a+5d
\nHence, a=d.
\nAlso, a+3d+a+4d = 27
\nHence, 9a = 27
\nHence, a=d=3.
\nTherefore, 18th term = 18d = 18*3 = 54.
\nHence, option A is the correct answer.<\/p>\n
\nGiven that 10a+b = 7(a+b)
\ni.e. 10a+b = 7a+7b
\n=> 3a = 6b => a =2b
\nHence, smallest 2 digit number satisfying our condition is 21.
\nHence, option C is the correct answer.<\/p>\n
\nSince both of them are even numbers, we can write a = 2x and b = 2y.
\nIt has been given that 2x + 2y = 100
\n=> x + y = 50
\nAlso, we know that x and y are natural numbers.
\nTherefore, x can take all values from 1 to 49 and y will take up a value accordingly.
\nHowever, after x = 25, y = 25, the values will not be unique. The values of x and y will get interchanged. Since we have to find the unordered pairs, we can neglect these terms.<\/p>\n
\n$\\Rightarrow$ a*b = L.C.M. * H.C.F.
\nSince if is given that both the numbers are prime numbers hence H.C.F. = 1.
\nTherefore L.C.M. = a*b = 713
\nHence option A is the correct answer.<\/p>\n
\n$\\Rightarrow$ a*b = 9222 and a – b = 19
\nSolving for \u2018a\u2019 and \u2018b\u2019,
\n$\\Rightarrow$ b(b+19) = 9222, b = 87 or -106
\n\u2018b\u2019 can\u2019t be -106 as \u2018b\u2019 is a natural number. Therefore, b = 87.
\nHence, a = b + 19 = 87 + 19 = 106
\nTherefore, sum of the numbers = 106 + 87 = 193.
\nOption E is the correct answer.<\/p>\n
\n$\\Rightarrow$ a*b = 4810 and a – b = 9
\nSolving for \u2018a\u2019 and \u2018b\u2019,
\n$\\Rightarrow$ b(b+9) = 4810, b = 65 or -74
\n\u2018b\u2019 can\u2019t be -74 as \u2018b\u2019 is a natural number. Therefore, b = 65.
\nHence, a = b + 9 = 65 + 9 = 74
\nTherefore, sum of the numbers = 65 + 74 = 139.
\nOption B is the correct answer.<\/p>\n
\nNumber obtained by interchanging its digits = \u2018ba\u2019
\nGiven that (10b+a) – (10a+b) = 18
\n$\\Rightarrow$ b – a = 2
\nAlso sum of the digits, a + b = 12
\nTherefore, a = 5, b = 7
\nHence, the original number = 57.
\nOption C is the correct answer.<\/p>\n
\nNow we need to find smallest n such that $\\frac{n}{2}$*(2a+(n-1)*d) > 5000.
\ni.e. $\\frac{n}{2}$*(2*30+(n-1)*10)>5000$
\ni.e. $n(2*3+(n-1)*1)>1000$
\ni.e. $n(n+5)>1000$
\ni.e. $n^2+5n-1000>0$
\nEquation $n^2+5n-1000$ becomes equal to 0 for n = $\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}$ = $ \\frac{-5\\pm\\sqrt{5^2-4*1*(-1000)}}{2*1}$
\n= $ \\frac{-5\\pm\\sqrt{25+4000}}{2}$
\nNow we know n cannot be -ve so we take n = $ \\frac{-5+\\sqrt{4025}}{2}$
\nClosest square root of 4025 is 63 therefore n will be slightly greater than $\\frac{63-5}{2}$ = $29$.
\nTherefore, he will have money till 29th day, but on 30th day his money will get exhausted.
\nTherefore, our answer is option \u2018e\u2019.<\/p>\n
\nNumber of days in April = 30
\nNumber of days till 15th May = 15
\nTotal number of days till his birthday = 31+30+15 = 76
\nHe starts with Rs. 3 and keeps on increasing the amount he saves by 3 i.e. the series is an A.P. with a = 3, d = 3 and n = 76
\nSum of the money he has on his birthday =
\n$\\frac{n}{2}$*(2a+(n-1)*d) = $\\frac{76}{2}$*(2*3+(76-1)*3)
\n= 8778
\nHe needs 10000 Rupees. Therefore, he will need 10000-8778 = 1222 Rupees.
\nTherefore, our answer is option \u2018c\u2019.<\/p>\n
\n= $1(2^n – 1)$<\/p>\n