Number System Question For IBPS RRB PO
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.
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Question 1: 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.
a) 30
b) 40
c) 15
d) 20
e) 25
Question 2: 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.
a) 3 : 8
b) 1 : 4
c) 2 : 7
d) 1 : 3
e) 1 : 5
Question 3: 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
a) 54
b) 18
c) 72
d) 36
e) 90
Question 4: 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 ‘N’.
a) 12
b) 22
c) 17
d) 14
e) 19
Question 5: 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 ‘N’.
a) 32
b) 31
c) 34
d) 28
e) 27
Question 6: Second term of an AP is twice the fifth term. What is the sum of the first 15 terms of that AP?
a) -12
b) 7
c) 11
d) 0
e) -3
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Question 7: The largest 4 digit number that leaves the same remainder when divided by 2, 3, 4 and 5 is
a) 9960
b) 9999
c) 9997
d) 9974
e) 9961
Instructions
Read the following passage and answer the questions that follow:
Observe in the opening pages of the great novel of “Middlemarch” 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 “only to close observers that Celia’s dress differed from her sister’s,” and had “a shade of coquetry in its arrangements.” Dorothea’s “plain dressing was due to mixed conditions, in most of which her sister shared.” They were both influenced by “the pride of being ladies,” of belonging to a stock not exactly aristocratic, but unquestionably “good.” The very quotation of the word good is significant and suggestive. There were “no parcel-tying forefathers” in the Brooke pedigree. A Puritan forefather, “who served under Cromwell, but afterward conformed and managed to come out of all political troubles as the proprietor of a respectable family estate,” had a hand in Dorothea’s “plain” wardrobe. “She could not reconcile the anxieties of a spiritual life involving eternal consequences with a keen interest in gimp and artificial protrusions of drapery,” but Celia “had that common-sense which is able to accept momentous doctrines without any eccentric agitation.” Both were examples of “reversion.” Then, as an instance of heredity working itself out in character “in Mr. Brooke, the hereditary strain of Puritan energy was clearly in abeyance, but in his niece Dorothea it glowed alike through faults and virtues.”
Could anything be more natural than for a woman with this passion for, and skill in, “unravelling certain human lots,” to lay herself out upon the human lot of woman, with all her “passionate patience of genius?” 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 “Middlemarch,” “Romola,” “Daniel Deronda,” and “Janet’s Repentance.” “She was a woman, and could not make her own lot.” Never before, indeed, was so much got out of the word “lot.” Never was that little word so hard worked, or well worked. “We women,” says Gwendolen Harleth, “must 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.” To appreciate the work that George Eliot has done you must read her with the determination of finding out the reason why Gwendolen Harleth “became poisonous,” and Dorothea, with all her brains and “plans,” a failure; why “the 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.” You must search these marvellous studies in motives for the key to the blunders of “the blundering lives” of woman which “some have felt are due to the inconvenient indefiniteness with which the Supreme power has fashioned the natures of women.” But as there is not “one 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.” It is treated with a dissective delineation in the women of George Eliot unequalled in the pages of fiction.
Question 8: What does the author mean by the line “one 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”?
a) The feminine incompetence prevents the social lot of woman to be treated with scientific certitude.
b) Since women cannot be treated with scientific certitude, a clear definition of feminine incompetence cannot be reached.
c) Since there is no specific way to define scientific certitude, the feminine incompetence cannot be measured or defined.
d) Since women have different levels of competence, we cannot study their social lot with scientific certitude.
Instructions
Read the following passage and answer the questions that follow.
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’t just monuments to human ambition: they might also hold the key to humanity’s progress in the space-faring age.
Proposals are now circulating for a free-standing tower or ‘space elevator’ 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.
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.
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 to look to nature for creative tricks to help them overcome the restrictions of their materials.
This led to a sub-discipline known as ‘reliability engineering’. 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.
However, neither the bones nor tendons in our bodies enjoy this luxury. In fact, they’re 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 ‘reliable’ than their sheer material strength would suggest. For example, merely running can push the Achilles’ 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.
Sean Sun & Dan Popescu
This article was originally published at Aeon and has been republished under Creative Commons.
Question 9: What is the ‘puzzle’ as mentioned in the fourth paragraph?
a) The increasing size of structures poses difficulties in keeping the building upright.
b) As the height of a building increases, its strength decreases.
c) Creative tricks from nature do not work anymore in constructing large structures.
d) The dilemma over whether we should build megastructures extending into space or not.
Question 10: 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?
a) 54
b) 60
c) 63
d) 57
e) 51
Question 11: A 2 digit number is such that the number is equal to 7 times the sum of its digits. The smallest such number is
a) 64
b) 42
c) 21
d) 12
e) 24
Question 12: A number leaves the same reminder on dividing 237 and 269. How many numbers satisfy this criterion?
a) 2
b) 5
c) 6
d) 8
e) 16
Question 13: If the product of two successive natural numbers is 5402, then find out the sum of the integers.
a) 139
b) 143
c) 135
d) 147
e) None of the above
Question 14: 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?
a) 49
b) 50
c) 24
d) 25
e) 99
Question 15: If the product of two prime numbers is 713, then find out the L.C.M. of these two numbers.
a) 713
b) 31
c) 23
d) 17
e) None of the above
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Question 16: The product of two natural numbers is 9222. If they differ by 19 then find out the sum of the numbers.
a) 205
b) 199
c) 197
d) 195
e) 193
Question 17: The product of two natural numbers is 4810. If they differ by 9 then find out the sum of the numbers.
a) 149
b) 139
c) 135
d) 145
e) 147
Question 18: 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?
a) 39
b) 48
c) 57
d) 93
e) None of the above
Question 19: 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?
a) 38
b) 28
c) 33
d) 25
e) 30
Question 20: Mohan starts saving Rs 3,6,9,……… 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)
a) 1100
b) 1224
c) 1222
d) 1348
e) 1456
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Answers & Solutions:
1) Answer (C)
Let ‘r’ be the common ratio of the given GP.
The ratio of 3rd and 6th term of a GP is 27.
$\dfrac{ar^2}{ar^5}$ = 27
$\Rightarrow$ $\dfrac{1}{r^3}$ = 27
$\Rightarrow$ $r = \dfrac{1}{3}$
Sum of infinite GP = $\dfrac{a}{1-r}$ = $\dfrac{10}{1-1/3}$ = 15
Hence, option C is the correct answer.
2) Answer (E)
Let ‘a’ and ‘d’ be the first term and common difference of the given AP.
$\Rightarrow$ $\dfrac{3}{2}(2a + 2d) = \dfrac{2}{9}[\dfrac{6}{2}(2a + 5d)]$
$\Rightarrow$ $3a+3d = \dfrac{2}{3}(2a + 5d)$
$\Rightarrow$ $9a+9d = 4a + 10d$
$\Rightarrow$ $5a = d$
$\Rightarrow$ $\dfrac{a}{d} = \dfrac{1}{5}$
Therefore, option E is the correct answer.
3) Answer (D)
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.
2599 – 2527 = 72
2599- 2419= 180
2599-2383= 216
2527- 2419= 108
2527- 2383= 144
2419- 2383= 36
As we can see, 36 is the smallest difference and hence, the largest number that satisfies the given condition is 36.
Therefore, option D is the right answer.
4) Answer (A)
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.
Therefore, the smallest number which is of this kind = (LCM of 5, 6, 7) – 3 = 210 – 3 = 207
Next such number = 207 + 210 = 333
We can say that
$\Rightarrow$ N $\geq$ 999
$\Rightarrow$ 207+(n – 1)210 $\geq$ 999
$\Rightarrow$ n > 4.77
Therefore, we can say that $n_{min}$ = 5.
Therefore, N = 207+4*210 = 1047
Hence, the sum of the digits of N = 1 + 0 + 4 + 7 = 12. (Option : A)
5) Answer (E)
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.
Therefore, the smallest number which is of this kind = (LCM of 6, 7, 8) – 3 = 168 – 3 = 165
Next such number = 165 + 168 = 333
We can say that
$\Rightarrow$ N $\leq$ 9999
$\Rightarrow$ 165+(n – 1)168 $\leq$ 9999
$\Rightarrow$ n < 59.53
Therefore, we can say that $n_{max}$ = 59.
Therefore, N = 165+58*168 = 9909
Hence, the sum of the digits of N = 9 + 9 + 0 + 9 = 27. (Option : E)
6) Answer (D)
let the first term be ‘a’ and the common difference be ‘d’.
As per the question, a + d = 2(a + 4d)
On solving, we get a + 7d = 0
or, Eighth term is 0.
Sum of the first 15 terms = $\frac{n}{2}[2a + (n – 1)d] = n(a + 8d)$ = 0
Hence, option D is the correct answer.
7) Answer (E)
We know that the number leaves the same remainder when divided by 2,3,4 and 5.
Therefore, the number should be of the form LCM(2,3,4,5) + k.
LCM of 2,3,4,5 is 60.
The largest multiple of 60 below 10,000 is 9960.
2 can leave a remainder of 0 or 1.
Therefore, the largest possible value of k is 1.
The largest 4 digit number that leaves the same remainder when divided by 2, 3, 4 and 5 is 9960 + 1 = 9961.
Therefore, option E is the right answer.
8) Answer (D)
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, B 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.
9) Answer (A)
From the last line of the third paragraph, we can infer that maintaining safety and structural integrity is the puzzle.
It is not mentioned in the passage that the strength of building decreases with increasing height.
Options C and D are out of the scope of the passage.
Hence, option A is the correct answer.
10) Answer (A)
Let the A.P be a, a+d, a+2d,……..
Given, 2(a+2d) = a+5d
i.e. 2a+4d = a+5d
Hence, a=d.
Also, a+3d+a+4d = 27
Hence, 9a = 27
Hence, a=d=3.
Therefore, 18th term = 18d = 18*3 = 54.
Hence, option A is the correct answer.
11) Answer (C)
Let the number be ab.
Given that 10a+b = 7(a+b)
i.e. 10a+b = 7a+7b
=> 3a = 6b => a =2b
Hence, smallest 2 digit number satisfying our condition is 21.
Hence, option C is the correct answer.
12) Answer (C)
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.
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.
13) Answer (D)
Let ‘$a$’ be the smaller number. Then,
$\Rightarrow$ $a(a+1) = 5402$
$\Rightarrow$ $a^2 + a = 5402$
$\Rightarrow$ $a = 73, -74$
‘$a$’ can’t be -74 as ‘$a$’ is a natural number. Hence, two successive natural numbers are 73 and 74.
Therefore, sum of the numbers = 74 + 73 = 147.
Hence, option D is the correct answer.
14) Answer (D)
Let the 2 natural numbers be a and b.
Since both of them are even numbers, we can write a = 2x and b = 2y.
It has been given that 2x + 2y = 100
=> x + y = 50
Also, we know that x and y are natural numbers.
Therefore, x can take all values from 1 to 49 and y will take up a value accordingly.
However, 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.
Therefore, 25 pairs of numbers will satisfy the condition and hence, option D is the right answer.
15) Answer (A)
We know that for any two numbers ‘a’ and ‘b’.
$\Rightarrow$ a*b = L.C.M. * H.C.F.
Since if is given that both the numbers are prime numbers hence H.C.F. = 1.
Therefore L.C.M. = a*b = 713
Hence option A is the correct answer.
16) Answer (E)
Let ‘a’ and ‘b’ be the two numbers where (a > b). It is given that
$\Rightarrow$ a*b = 9222 and a – b = 19
Solving for ‘a’ and ‘b’,
$\Rightarrow$ b(b+19) = 9222, b = 87 or -106
‘b’ can’t be -106 as ‘b’ is a natural number. Therefore, b = 87.
Hence, a = b + 19 = 87 + 19 = 106
Therefore, sum of the numbers = 106 + 87 = 193.
Option E is the correct answer.
17) Answer (B)
Let ‘a’ and ‘b’ be the two numbers where (a > b). It is given that
$\Rightarrow$ a*b = 4810 and a – b = 9
Solving for ‘a’ and ‘b’,
$\Rightarrow$ b(b+9) = 4810, b = 65 or -74
‘b’ can’t be -74 as ‘b’ is a natural number. Therefore, b = 65.
Hence, a = b + 9 = 65 + 9 = 74
Therefore, sum of the numbers = 65 + 74 = 139.
Option B is the correct answer.
18) Answer (C)
Let us assume that the original two digit number is ‘ab’.
Number obtained by interchanging its digits = ‘ba’
Given that (10b+a) – (10a+b) = 18
$\Rightarrow$ b – a = 2
Also sum of the digits, a + b = 12
Therefore, a = 5, b = 7
Hence, the original number = 57.
Option C is the correct answer.
19) Answer (E)
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,……… on day1, day 2 and so on. This forms an A.P. with a = 30 and d = 10.
Now we need to find smallest n such that $\frac{n}{2}$*(2a+(n-1)*d) > 5000.
i.e. $\frac{n}{2}$*(2*30+(n-1)*10)>5000$
i.e. $n(2*3+(n-1)*1)>1000$
i.e. $n(n+5)>1000$
i.e. $n^2+5n-1000>0$
Equation $n^2+5n-1000$ 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}$
= $ \frac{-5\pm\sqrt{25+4000}}{2}$
Now we know n cannot be -ve so we take n = $ \frac{-5+\sqrt{4025}}{2}$
Closest square root of 4025 is 63 therefore n will be slightly greater than $\frac{63-5}{2}$ = $29$.
Therefore, he will have money till 29th day, but on 30th day his money will get exhausted.
Therefore, our answer is option ‘e’.
20) Answer (C)
Number of days in March = 31
Number of days in April = 30
Number of days till 15th May = 15
Total number of days till his birthday = 31+30+15 = 76
He 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
Sum of the money he has on his birthday =
$\frac{n}{2}$*(2a+(n-1)*d) = $\frac{76}{2}$*(2*3+(76-1)*3)
= 8778
He needs 10000 Rupees. Therefore, he will need 10000-8778 = 1222 Rupees.
Therefore, our answer is option ‘c’.
