In this article, as mentioned we are sharing the memory based Quantitative Aptitude questions of IBPS PO Mains 2018 with Unique Solutions. Here we are going to discuss the questions from the topics like data sufficiency, number series and probability. Talking about the level of questions asked in the exam, this year’s quantitative aptitude section in IBPS PO Mains 2018 Exam came with a surprise and became the most difficult section among all. Let’s see what can be the unique method to solve these memory based questions.
Q.) Two different number series are given below, but the pattern in both of them follows the same. So considering this fact of same pattern in both the series, find the value of ‘?’.
1 3 9 31 129
5 _ _ _ ?
Solution : If we look carefully at the first series given above, the numbers are increasing rapidly which means that some kind of multiplication is involved in this series. So the pattern of the series can be as follows.
1 × 1 + 2 = 3
3 × 2 + 3 = 9
9 × 3 + 4 = 31
31 × 4 + 5 = 129
On following the same pattern we can find out the value of ‘?’ and the second number series will proceed like this as follows.
5 × 1 + 2 = 7
7 × 2 + 3 = 17
17 × 3 + 4 = 55
55 × 4 + 5 = 225 (=?)
Hence the value of ‘?’ will be 225, which is the answer in our case.
Q.) Observe the series given below carefully and answer the following question.
4 2 2 3 6 15 ……. 4835
If 4835 is the nth term then find the value of n.
(a) 12
(b) 9
(c) 10
(d) 7
(e) 15
Solution : Inorder to solve this question, we first need to find the pattern which is being followed in this number series. And on observing the series carefully we can understand that some kind of multiplication is involved in the series.
So the pattern can be as follows.
(i) 4 × 0.5 = 2
(ii) 2 × 1 = 2
(iii) 2 × 1.5 = 3
(iv) 3 × 2 = 6
(v) 6 × 2.5 = 15
Since now we have understood the pattern, next step which we need to do is to find out that following the same pattern at which turn we arrive at 4835 number as given in the question.
Hence,
(vi) 15 × 3 = 75
(vii) 75 × 3.5 = 262.5
(viii) 262.5 × 4 = 1050 [Tip : To multiple 262.5 by 4 smartly, one should double the number two times]
(ix) 1050 × 4.5 = 4835
One more smart job can be that instead of multiplying the number 1050 by 4.5 completely one can assume that while multiplying 1050 by 4, we can arrive at the number close to 4835 which can assure us that it can only be the last term to arrive at the given number.
Therefore the number of terms required to find the nth term as given in the question will be 9 (=n). And hence option (b) will be our answer.
Q.) A’s investment is half of the initial investment of ‘B’. But ‘B’ withdraw some money after 4 month. ‘C’ joins the business after ‘B’ left, but not in the same month. ‘C’ joins with amount X. The initial investment of ‘B’ was ₹2400. If in the annual profit the share of ‘A’ and ‘C’ are same, then which is the possible value of X ?
(a) 1800
(b) 7200
(c) 3600
(d) 1200
(e) 5400
Find which of the above given options are sufficient to answer the question ?
(i) Only (a) and (b)
(ii) Only (b) and (c)
(iii) Only (a), (b) and (c)
(iv) Only (b), (d) and (e)
(v) Only (a), (b), (c) and (e)
Solution : This questionof data sufficiency is based on the concept of partnership. Talking about the basic formula of Partnership, we all know that,
Money Invested × Time Period = Profit
So in this case the ratio of annual profits for A, B and C will be
A : B : C = (1200 × 12) : (2400 × 4) : (X × t)
• Here ‘t’ is the time period for which C has invested the money. The value of ‘t’ may lie from 1 to 8 as ‘C’ joins the business after ‘B’ left, but not in the same month. So the maximum value of ‘t’ can be (12-4 = 8) and the minimum value of ‘t’ can be 1.
• As the investment of ‘A’ is equal to half the initial investment of ‘B’. So the money invested by ‘A’ will be 1200 ₹ because the money invested by ‘B’ is given as 2400 ₹.
• Since the time period for which ‘A’ invested the money is not given in the question directly we can assume it to be 12 months which is equivalent to 1 year.
Now according to the question the share in annual profit of ‘A’ is equal to the share of ‘B’.
Hence, 1200 × 12 = X × ‘t’
Case (I) : If ‘t’ = 8
then X = (1200 × 12) / 8 = 1800.
And X = 1800
Case (II) : If ‘t’ < 8 then ultimately the value of X will be greater than 1800 as per the rule of division when the value of denominator decreases the value of answer increases
Means X > 1800
Therefore the possible value of X will be greater than or equal to 1800. Means (X ≥ 1800) and thus Option (v) – Only (a), (b), (c) and (e) are sufficient to answer this question.
Q.) In a bag there are total 15 balls of three colours Red, Blue and Green. The number of different coloured balls are not the same. The difference between Red and Green colour balls is equal to difference between Green and Blue colour balls. Then what can be the number of Blue coloured balls, if possibility to choose a Blue colour ball is always greater than 0.2 ?
(a) 4
(b) 7
(c) 5
(d) 8
(e) more than one options are correct.
Solution : This question is based on the concept of probability.
Let the number of Red, Green and Blue balls be R, G and B.
As given, total number of balls = 15
(i) The number of balls of different colours are not same means R ≠ G ≠ B.
(ii) The difference between Red and Green colour balls is equal to difference between Green and Blue colour balls, means |R – G| = |G – B|
(iii) The probability to choose a blue colour ball is greater than 0.2 means the value of B is greater than 20% of 15 or the value of B is greater than 3.
To calculate 20% of 15 we know that,
Since the value of B is greater than 3, the possible values of B will be 4 or greater than 4 which means that (B ≥ 4).
Now the best way to solve this question is to go by options.
Option (a) B = 4,
Remaining number of balls = 15-4 = 11
Now divide 11 between R and G in such a way that both (i) and (ii) condition follows.
So the values can be, R = 6 and G = 5 and hence option (a) can be our answer.
Let’s check another option,
Option (b) B = 7
Remaining number of balls = 15-7 = 8
So if we divide 8 in such a way that R = 3 and G = 5, then both (i) and (ii) condition will follow and hence option (b) can also be our answer.
Now here since on checking two values we came to know that only option (e) more than one options are correct, can be our answer as it’s the only choice left, we mark it and move on. There’s no need to check the other options.
But for understanding we can also check the remaining two options.
Option (c) B = 5
Remaining number of balls = 15-5= 10
Now we can take any value such as R=4 and G=6 or R=3 and G=7 but in both the cases none of the condition follows. Hence option (c) cannot be the answer.
Option (d) B = 8
Remaining number of balls = 15-8= 7
Now we can take the values as R = 2 and G = 5, which completely satisfies both (i) and (ii) condition. And hence option (d) can also be our answer.
At last after checking all the options we came to a conclusion that option (e) – more than one options are correct is the right answer to this question.
Note : Try to do all the calculations mentally, as described above and cut short the sentences written above, they are just to make you understand. Although Explanations may look difficult or tedious, but once understood it will be easy and less time consuming to deal with.
That’s all in this article, hope this can be a great learning for our readers. Also if there is any feedback or suggestion then feel free to comment below. And don’t forget to share this with your near and dear ones as said Sharing is Caring. Happy Learning.
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