Download Top-20 IBPS RRB PO Number System Questions PDF. Number System questions based on asked questions in previous year exam papers very important for the IBPS RRB PO (Officer Scale-I, II & III) exam.<\/p>\n
Download Number System Question For IBPS RRB PO<\/a><\/p>\n 35 IBPS RRB PO Mocks @ Rs. 149. Enroll Now<\/a><\/p>\n 70 IBPS RRB (PO + Clerk) Mocks @ Rs. 199<\/a><\/p>\n Take a free mock test for IBPS RRB PO<\/a><\/p>\n Download IBPS RRB PO 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 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 Free Mock Test for IBPS RRB PO<\/a><\/p>\n IBPS RRB Clerk Previous Papers<\/a><\/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 Instructions<\/b><\/p>\n Read the following passage and answer the questions that follow:<\/strong><\/p>\n Observe in the opening pages of the great novel of \u201cMiddlemarch\u201d how soon we pass from the outside dress to the inside reasons for it, from the costume to the motives which control it and color it. It was \u201conly to close observers that Celia\u2019s dress differed from her sister\u2019s,\u201d and had \u201ca shade of coquetry in its arrangements.\u201d Dorothea\u2019s \u201cplain dressing was due to mixed conditions, in most of which her sister shared.\u201d They were both influenced by \u201cthe pride of being ladies,\u201d of belonging to a stock not exactly aristocratic, but unquestionably \u201cgood.\u201d The very quotation of the word good is significant and suggestive. There were \u201cno parcel-tying forefathers\u201d in the Brooke pedigree. A Puritan forefather, \u201cwho served under Cromwell, but afterward conformed and managed to come out of all political troubles as the proprietor of a respectable family estate,\u201d had a hand in Dorothea\u2019s \u201cplain\u201d wardrobe. \u201cShe could not reconcile the anxieties of a spiritual life involving eternal consequences with a keen interest in gimp and artificial protrusions of drapery,\u201d but Celia \u201chad that common-sense which is able to accept momentous doctrines without any eccentric agitation.\u201d Both were examples of \u201creversion.\u201d Then, as an instance of heredity working itself out in character \u201cin Mr. Brooke, the hereditary strain of Puritan energy was clearly in abeyance, but in his niece Dorothea it glowed alike through faults and virtues.\u201d<\/p>\n Could anything be more natural than for a woman with this passion for, and skill in, \u201cunravelling certain human lots,\u201d to lay herself out upon the human lot of woman, with all her \u201cpassionate patience of genius?\u201d One would say this was inevitable. And, for a delineation of what that lot of woman really is, as made for her, there is nothing in all literature equal to what we find in \u201cMiddlemarch,\u201d \u201cRomola,\u201d \u201cDaniel Deronda,\u201d and \u201cJanet\u2019s Repentance.\u201d \u201cShe was a woman, and could not make her own lot.\u201d Never before, indeed, was so much got out of the word \u201clot.\u201d Never was that little word so hard worked, or well worked. \u201cWe women,\u201d says Gwendolen Harleth, \u201cmust stay where we grow, or where the gardeners like to transplant us. We are brought up like the flowers, to look as pretty as we can, and be dull without complaining. That is my notion about the plants, and that is the reason why some of them have got poisonous.\u201d To appreciate the work that George Eliot has done you must read her with the determination of finding out the reason why Gwendolen Harleth \u201cbecame poisonous,\u201d and Dorothea, with all her brains and \u201cplans,\u201d a failure; why \u201cthe many Theresas find for themselves no epic life, only a life of mistakes, the offspring of a certain spiritual grandeur ill-matched with the meanness of opportunity.\u201d You must search these marvellous studies in motives for the key to the blunders of \u201cthe blundering lives\u201d of woman which \u201csome have felt are due to the inconvenient indefiniteness with which the Supreme power has fashioned the natures of women.\u201d But as there is not \u201cone level of feminine incompetence as strict as the ability to count three and no more, the social lot of woman cannot be treated with scientific certitude.\u201d It is treated with a dissective delineation in the women of George Eliot unequalled in the pages of fiction.<\/p>\n Question 8:\u00a0<\/b>What does the author mean by the line \u201cone level of feminine incompetence as strict as the ability to count three and no more, the social lot of woman cannot be treated with scientific certitude\u201d?<\/p>\n a)\u00a0The feminine incompetence prevents the social lot of woman to be treated with scientific certitude.<\/p>\n b)\u00a0Since women cannot be treated with scientific certitude, a clear definition of feminine incompetence cannot be reached.<\/p>\n c)\u00a0Since there is no specific way to define scientific certitude, the feminine incompetence cannot be measured or defined.<\/p>\n d)\u00a0Since women have different levels of competence, we cannot study their social lot with scientific certitude.<\/p>\n Instructions<\/b><\/p>\n Read the following passage and answer the questions that follow.<\/strong><\/p>\n The human desire to create ever bigger and more impressive structures is insatiable. The pyramids of Ancient Egypt, the Great Wall of China and the Burj Khalifa in Dubai – now the tallest edifice in the world at over 828 metres (2,722 ft) – are a consequence of pushing engineering to its limits. But huge buildings aren\u2019t just monuments to human ambition: they might also hold the key to humanity\u2019s progress in the space-faring age.<\/p>\n Proposals are now circulating for a free-standing tower or \u2018space elevator\u2019 that could reach up into the geosynchronous orbit around the Earth. Such a tower would be an alternative to rocket-based transport, and drastically reduce the amount of energy it takes to get into space. Beyond that, we can imagine space-based megastructures many kilometres in size, powered by solar energy, perhaps encompassing whole planets or even stars.<\/p>\n In recent years, engineers have been able to build on grander scales thanks to the strength and reliability of substances such as novel steel alloys. But as we enter the realm of megastructures – those of 1,000 km or more in dimension – maintaining safety and structural integrity has become a fiendish challenge.<\/p>\n It turns out that biological design, equipped with around 3.8 billion years of experience, might help solve this puzzle. Before the age of materials science, engineers had<\/em> to look to nature for creative tricks to help them overcome the restrictions of their materials.<\/p>\n This led to a sub-discipline known as \u2018reliability engineering\u2019. Designers started to make structures that were much stronger than the maximum possible load they needed to bear – which meant the stress on the materials stayed within a range where the probability of breakage was very low. Once structures turn into megastructures, though, calculations show that this risk-averse approach places a cap on their size. Megastructures necessarily push materials to their limits, and remove the luxury of weathering comfortable levels of stress.<\/p>\n However, neither the bones nor tendons in our bodies enjoy this luxury. In fact, they\u2019re often compressed and stretched well beyond the point at which their underlying substances might be expected to break. Yet these components of human bodies are still much more \u2018reliable\u2019 than their sheer material strength would suggest. For example, merely running can push the Achilles\u2019 tendon to over 75 per cent of its ultimate tensile strength, whereas weightlifters can experience stresses of over 90 per cent of the strength of their lumbar spines, when they are hefting hundreds of kilogrammes.<\/p>\n Sean Sun & Dan Popescu<\/p>\n This article was originally published at Aeon<\/a> and has been republished under Creative Commons.<\/p>\n Question 9:\u00a0<\/b>What is the \u2018puzzle\u2019 as mentioned in the fourth paragraph?<\/p>\n a)\u00a0The increasing size of structures poses difficulties in keeping the building upright.<\/p>\n b)\u00a0As the height of a building increases, its strength decreases.<\/p>\n c)\u00a0Creative tricks from nature do not work anymore in constructing large structures.<\/p>\n d)\u00a0The dilemma over whether we should build megastructures extending into space or not.<\/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>If the product of two successive natural numbers is 5402, then find out the sum of the integers.<\/p>\n a)\u00a0139<\/p>\n b)\u00a0143<\/p>\n c)\u00a0135<\/p>\n d)\u00a0147<\/p>\n e)\u00a0None of the above<\/p>\n Question 14:\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 15:\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 Quantitative Aptitude formulas PDF<\/a><\/p>\n 520 Banking Mocks – Just Rs. 499<\/a><\/p>\n Question 16:\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 Question 17:\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 18:\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 19:\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 20:\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 18000 Free Solved Questions – Banking Study Material<\/a><\/p>\n 35 IBPS RRB PO Mocks @ Rs. 149. Enroll Now<\/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(D)<\/strong><\/p>\n Through the given line, the author states that we cannot treat the female lot with scientific certitude since we cannot generalize the competence of women. The author does not intend to convey that females are incompetent. Options A,\u00a0B and C talk about feminine incompetence, which can be termed as a misinterpretation of the given line. Therefore, we can eliminate these options. Only option D captures the fact that the different levels of competence prevents us from taking a scientific approach. Hence, option D is the right answer.<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n From the last line of the third paragraph, we can infer that maintaining safety and structural integrity is the puzzle. 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 \u2018$a$\u2019 be the smaller number. Then, 14)\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 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n We know that for any two numbers \u2018a\u2019 and \u2018b\u2019. 16)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Let \u2018a\u2019 and \u2018b\u2019 be the two numbers where (a > b). It is given that 17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let \u2018a\u2019 and \u2018b\u2019 be the two numbers where (a > b). It is given that 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let us assume that the original two digit number is \u2018ab\u2019. 19)\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. 20)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Number of days in March = 31
\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
\nIt is not mentioned in the passage that the strength of building decreases with increasing height.
\nOptions C and D are out of the scope of the passage.
\nHence, option A 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
\n$\\Rightarrow$ $a(a+1) = 5402$
\n$\\Rightarrow$ $a^2 + a = 5402$
\n$\\Rightarrow$ $a = 73, -74$
\n\u2018$a$\u2019 can\u2019t be -74 as \u2018$a$\u2019 is a natural number. Hence, two successive natural numbers are 73 and 74.
\nTherefore, sum of the numbers = 74 + 73 = 147.
\nHence, option D 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