Sol. Ans.(b) Working backwards from options is the right technique here.

Option (a) will give –

2 + 12 + 52 + 22 + 21 = 109 (which is not 122). So option (a) is wrong.

Option (b) gives us –

3 + 13 + 53 + 33 + 31 = 133. So option (b) is correct.

At the most, in any question, you will need to check three options, and if first three are wrong, then fourth has to be right (no need to check).

[You only need to score more than 33% or 66 marks for CSAT paper. So pick and choose questions carefully, without wasting too much time on any one. This one was an easy question based on basic mathematical symbols and operations, and should have been done.]

Sol. Ans.(b). When we have to measure the “minimum length”, it is the length common to both the given sticks. So it will be the HCF.

Correct answer is HCF of 7.5 and 3.25, which is 0.25.

(How to calculate? Find the HCF of 750 and 325, and then put decimal point appropriately)

[You only need to score more than 33% or 66 marks for CSAT paper. So pick and choose questions carefully, without wasting too much time on any one. This one was an easy question based on HCF and LCM, and should have been done.]

Sol. Ans.(c). The given sequence is: 14, 18, 20, 24, 30, 32, …. We are told that by applying an operation, we get prime numbers from this series.

We can see that by subtracting 1 from each term, we will get a sequence of primes.

14 – 1 = 13; 18 – 1 = 17; 20 – 1 = 19; 24 – 1 = 23; 30 -1 = 29; 32 – 1 = 31.

So the next term will be 38 – 1 = 37 which is also a prime number.

So the correct option is (c).

[An easy question, that must be attempted.]

Sol. Ans.(b).

We know that the sum of n consecutive natural numbers = n(n+1)/2.

Natural numbers are integers starting with 1.

Now according the information given in question our value of n(n+1)/2 should be greater than 195.

Why? Because one page is torn, so the sum is less than what it should be. Right?

If we take n = 19 we will get n(n+1)/2 = 19 × 20/2 = 190 which is less than 195.

So taking next number let’s assume n = 20.

In this case – n(n+1)/2 = 20×21/2 = 210. But two consecutive numbers were removed, so 210 – 195 = 15. It means that the sum of the two consecutive torn pages should be 15. Only option (b) matches.

[A question based on basic logic and numbers. Do not attempt this question. It is complex and you may invest a lot of time. Avoid it.]

Sol. Ans.(c). Observing the sequence and by putting various options you can find that the correct sequence is ababa bcbcb cdcdc deded. You have to find a repeated pattern. Only (c) gives that pattern.

By putting (a), we would get – ababa bbbcb cdcdc deddd – there is no pattern in it

By putting (b), we would get – ababa bbbcb ddcdc deded – there is no pattern in it

By putting (d), we would get – ababb bcbcb ddcdc deded – there is no pattern in it

[A simple series questions, based on simple reasoning. Must be attempted.]

Sol. Ans.(c).

Each letter represents different digits.

Unknown digits are 6 (A, B, C, D, E, F). Each digit is greater than 3 (possible digits are 4, 5, 6, 7, 8, 9). So each of these digits (4-9) will be used. We have to find A – D (or D – A).

Here, C + F is ending with 2. So, C and take two values 8 or 7. If C is 8 then F is 4 and if C is 7 then F is 5. In any case B + 2 + 1 (carry) is ending with 0 so B = 7. It means C can’t take value 7, so C = 8 and F = 4.

Now F + 3 + 1 (carry) is ending with 9 so E = 5. Hence there is no carry forward in the next term i.e. A + D = 15 and remaining digits are 9 and 6. Thus 9+ 6 = 6 + 9 = 15.

So the difference between A and D is 3. So correct option is (c).

[A question based on logic and number systems. A long and not easy question, to be avoided.]

Sol. Ans.(c). This is a Data Sufficiency question. It had vanished from CSAT but made a big entry in 2020!

Question – Are there more than 100 figures in that book?

Using S1 alone, number of figures on any page could be 0, 1 or 2. So minimum 0 (if every page will contain 0 figures) and maximum 102 figures (if every page will contain 2 figures) are possible. So S1 is not sufficient to answer the question uniquely.

Using S2 alone, number of figures on any single page could 1, 2, 3, … up to infinity. So it is also not sufficient to answer the question alone.

Even combining both we can have either 1 or 2 figures on any page. Which gives us minimum 51 (if every page will contain 1 figure) and maximum 102 figure (if every page will contain 2 figures). So it can’t be answered uniquely even by combining S1 and S2.

[Data Sufficiency questions made a comeback in this CSAT, and can be done if your are comfortable. Otherwise there are many more to be tried. You have to anyhow cross 33% net score.]

Sol. Ans.(a).

Let the number of boys = b and number of girls = g.

So by the condition given in question we have 9g + 8b = 8.8 (b + g). That gives us ratio b:g = 1:4. It means that if there are p boys number girls will be 4p, where p is any positive integer.

Again by the condition given in question 8×4p + 7×p = X (4p + p)

=> 32p + 7p = 5p × X => 39p = 5pX => 39 = 5X or X = 39/5 or X = 7.8.

[A question base on Averages. Not so easy. Avoid it (at least in the first round of solving).]

Sol. Ans.(b).

Given – 1. Two generations; 2. Six members P, Q, R, S, T and U. Three males, and three females. Two married couples (which means out of these four two will be male and two will be female). Two unmarried siblings.

Logical Conclusions - Out of two unmarried siblings one will be male and one will be female.

There are only two generations. Let the older generation be G1 and younger generation be G2.

P and Q will be part of G1 because of being a parent and mother-in-law.

U, R and T (unmarried male) will be part of G2.

Considering all the information we can get following order (circle representing male and triangle representing female –

[You should skip this question in actual exam, as this is a qualifying paper, and this may take up too much time. Find other easier ones to solve, first]

Sol. Ans.(c).

Let’s calculate what day of week will be on July 12 (Considering Jan 12 is Sunday and it is a non-leap year).

Total number of odd days between Jan 12 and July 12 = 19 (remaining day of Jan) + 28 (Feb) + 31 (March) + 30 (April) + 31 (May) + 30 (June) + 12 (July) = 181.

Now, 181 when divided by 7 gives remainder of 6. So final odd days = 6.

So If Jan 12 is Sunday July 12 will be Saturday (in a non-leap year).

In a leap year July 12 will be Sunday as one extra day will be added of Feb 29.

[A question based on calendars and dates. Not difficulty, but time consuming. Pick and choose carefully.]

Sol. Ans.(d). Here is the diagram representing the movement –

From the diagram given above, it is clear that he will be in North-West direction from the starting point.

[A simple Directions based question. Must be done.]

Sol. Ans.(a).

Given statements are – first, all numbers are divisible by 2 and second, all numbers are divisible by 3. If a number is divisible by 3 and 2 then it will surely be divisible by 6 (which is 2 x 3), but will not be necessarily divided by 4.

For example – 6, 18, 30 … are all such numbers. Hence only conclusion 1 is true (logically follows).

[Another logical question based on Statement-Conclusion format. Can be done.]

Sol. Ans.(c).

Neither of the conclusions will be true for the given syllogism argument. To understand let’s have look on the following Euler/Venn diagrams. The circle represents Cats, Dogs and Black (dotted circle).

Above argument can be drawn in the two ways. In both the cases you can see that in some cases all dogs can be blacks but then in some cases it is not. Hence conclusion I not follows. Similarly ‘every dog will be black’ if the dotted circle will be congruent to the circle representing dog. So in this case conclusion II does not follow. To be a valid conclusion it should be true in every case. Hence both the conclusions are not valid.

[Another logical question based on Statement-Conclusion format. Can be done.]

Sol. Ans.(b).

Given sequence is – 5 1 4 7 3 9 8 5 7 2 6 3 1 5 8 6 3 8 5 2 2 4 3 4 9 6.

It happens six times when an odd number is followed by another odd number. See the bold digits (51, 73, 39, 57, 31 and 15).

[Another simple number series identification question. Must be done.]

Sol. Ans.(b). Refer to the diagram given below.

A is 16th from left so 15 students will be there to A’s left. G is 11th to right from A and 3rd to right from V. So there will be 7 students between A and V and 2 students between V and G. Consequently there will 14 students to the right of G. So total number of Students will 15 + 7 + 2 + 14 + A(1) + V(1) + G (1) = 41.

[A slightly tricky question based on seating arrangements. Can be done but may take some extra time to do carefully. Think and decide!]

Sol. Ans.(d).

Question asked is ‘Who among A, B, C and D is the youngest?’

>Option (a): S1 alone is not sufficient. Because there is no clarity about A and B as to who will be younger.

Option (b): S1 and S2 together will also be not sufficient as again there will be not clarity about A and B as who will be younger.

Option (c): Combination S2 and S3 will not be sufficient as it does not provide any information about C.

Option (d): Combination S1 and S3 will give order D > C > A > B. So, the youngest person will be B.

[Logical reasoning question, of the Data Sufficiency variety. Not too tough, but not direct either. Choose carefully.]

Sol. Ans.(c).

Between 1 and 100 there are 19 such integers which have 4 as a digit – 4, 14, 24, 34, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 54, 64, 74, 84 and 94.

Out of these, only 7 integers – 4, 24, 40, 44, 48, 64 and 84 are divisible by 4.

So correct answer will be 19 – 7 = 12. A mechanical question but can be done.

[Numbers based direct question. Can be attempted in first round itself.]

Sol. Ans.(a).

Let the side of P = a, so the side of Q = 2a.

Volume of P (x) = a³ volume of Q (y) = 8a³

Also, It is given that, mass of P (m) = u × X = u × a³ ….. (1) and

Mass of Q (n) = v × y = v × 8a³ …..(2)

It is also given that Mass of Q (n) = 2 × mass of P (m).

So, by using equations 1 and 2 we have v × 8a³ = 2 × u × a³  4v = u.

So option (a) is correct.

[Avoid this algebra + mensuration based question, if you can do other questions first!]

Sol. Ans.(c).

Traditional Method Let the age of Teachers is T years and the three students is a years (Given age of all three student is same). Then according to the information given in question we have (T + 3a)/4 = 20 or T + 3a = 80 ….(1)

T – a = 20 ….(2) (assuming that teachers age will be greater than the students)

By solving equations 1 and 2 simultaneously we have a = 15 and T = 35. Option (c) is correct.

Working backward from options method:

Option (a) – If teacher’s age is 25 years, then students’ age would be 5 years (assuming that teachers age will be greater than the students). This will give average of teacher and students age to (25 + 3×5)/4 = 10. Hence it is not correct.

Option (b) – If teacher’s age is 30 years, then students age would be 10 years (assuming that teachers age will be greater than the students). This will give average of teacher and students age to (30 + 3×10)/4 = 15 .Hence it is not correct.

Option (c) – If teacher’s age is 35 years, then students age would be 15 years (assuming that teachers age will be greater than the students). This will give average of teacher and students age to (35 + 3×15)/4 = 20. This matches the information given in question. Hence it is the correct answer.

No need to check option (d).

[Averages based question. Not too easy. To be avoided in the first round.]

Sol. Ans.(d).

Traditional Method –

SP = Rs 3,00,000; Loss% = 20%.

CP = (SP × 100)/(100 – Loss%)

=>P = 3,00,000 × 100 / 80 =>CP = Rs. 3, 75,000. So, option D is correct.

Alternative method – Since loss is 20%, so 80% of CP = 3,00,000. => 0.8 CP = 300000 So CP = 3,75,000.

[A very direct question based on Profit and Loss. Must be attempted!]

Sol. Ans.(d). Theme is “Jobs in India”.

Clearly option (d) is the answer.

Option (a) is wrong as the passage does not talk about reducing labour movement out of rural areas.

Option (b) is wrong as that is not the core idea in the passage (“best implied”).

Option (c) is just opposite to what is given and is evident from ‘Services, where labour tends to be most productive, are not generating the additional jobs the country needs.’ It is given that ‘Only those sectors that drive domestic demand such as health and education (social sector) can comfortably fill other half’ so (d) is true.

Option (d) is slightly indirect but ultimately means what is being asked. Health and education sectors comprise the social sector, and large investments there can generate many jobs.

[An easy passage, if you have been reading newspapers regularly. Carefully scan through all options before marking.]

Sol. Ans.(a). Theme is “Rights of Citizens in 21st century”.

Statement 3 is clearly wrong, as it is not assumed anywhere in the passage. So options (b) and (c) are ruled out.

Look at the options – we need to just check if Statement 1 is right or not. It is indeed correct because “to actively ensure” is possible only through institutional mechanisms (not one-off or ad hoc responses). Hence only (a) seems correct.

The passage talks about ‘right to privacy’, then it provides insight on meaningfulness of right with an example on ‘right to free expression’. At the end of the passage there is mention of ‘role of state’ in ensuring the ‘right to free expression’ through acts of omission and commission.

[A not so easy passage. Pick it at your own risk!]

Sol. Ans.(c). Theme is “Water as a resource”.

Passage does not talk about ‘Ramsar Convention’ neither does it talk about ‘cost of its supply’, so options (a) and (d) are ruled out.

Passage talks about ‘strengthening the capacities of ecological systems’. For this you have to give more respect to the wetlands, from where the water is sourced, and that can be done by considering them more than just open sources of water. So option (c) is correct.

Option (b) is not the most logical inference as modernization of water systems does not necessarily mean respecting natural ecosystems.

[A not so easy passage. Needs some careful deliberation and elimination. But can be tried.]

Sol. Ans.(a). Theme is “Investment and Asset Allocation”.

The word ‘Asset allocation’ itself means the implementation of an investment strategy that attempts to balance risk versus reward by adjusting the percentage of each asset in an investment portfolio according to the investor's risk tolerance, goals and investment time frame.

Option (c) can be eliminated as that’s not mentioned anywhere.

Option (d) is worded rather loosely and can be eliminated.

Option (b) would be eliminated due to the words “to generate wealth”.

We are left with Option (a) now. At the end, the passage talks about diversifying the assets for lower risk [‘All these extreme forms of risks we know will not manifest at the same time’].

Passage is not talking about ‘risk-taking’ but to ‘diversify the risk’ and there is no mention of ‘asset management organisation’.

[A not so easy passage. Be careful with the options. Use elimination technique aggressively.]

Sol. Ans.(d). Theme is “Evolving nature of biotechnology”.

This passage has two questions and hence must be attempted.

Statement 1 is surely correct. It is evolving constantly. So option (c) is eliminated.

Between statements 2 and 3, st. 3 is much better (and worded as per passage). In fact, it includes what is written in statement 2. So 3 is surely correct, and 2 may or may not be correct.

Statement 4 is definitely correct, as without the political executive it is not possible.

So the only option with all of 1, 3 and 4 is option (d) which also has statement 2.

[A tough question. Handle with care!]

Sol. Ans.(c). Theme is “Evolving nature of biotechnology”.

We are now asked the “crux of the passage”.

The passage is talking mainly about what biotechnology is being used for today, and what is can be and must be used for in the coming years. Hence, option (c) best summarises the crux.

[This was an easy question. Since the passage had 2 questions, it must be solved.]

Sol. Ans.(a).

In a multiplication problem, 0 can be produced in two ways. First when a number is multiplied by 10 or second when 5 is multiplied by 2 or any even number.

So to count the number of zeroes in the given multiple let’s break the whole multiplication in to 10, 5 and 2.

We get 1 × 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 × 55 × 60

= 1 × 5 × 10 × (3 × 5) × (2×10) × (5×5) × (3×10) × (7×5) × (4×10) × (9×5) × (5×10) × (11×5) × (6×10)

= 1 × (10 × 10 × 10 × 10 × 10 × 10) × (5 × 5 × 5 × 5 × 5 × 5 × 5 × 5) × (2 × 4 × 6) × (3 × 3 × 7 × 9 × 11)

= 1 × (10 × 10 × 10 × 10 × 10 × 10) × (5 × 5 × 5 × 5 × 5 × 5 × 5 × 5) × {2 × (2 × 2) × (2×3)} × (3 × 3 × 7 × 9 × 11)

Now you can see in above multiplication we have six 10s, eight 5s and four 2s.

Six 10s will give us six 0s while four 2s when combined with four 5s (out of the eight available) will give us additional four 0s.

So in the multiplication six + four = ten 0s will be there.

[An easy question based on number systems, that required some factoring to reach the final answer. If you can it quickly, then attempt this in the first round, else wait till you have solved all easier ones first.]

Sol. Ans.(b).

XYZ is a 3-digit number not divisible by 3.

We know that expanded form of XYZ will be 100 X + 10Y + Z. (basic decimal number system)

So, XYZ + YZX + ZXY = (100 X + 10Y + Z) + (100 Y + 10Z + X) + (100 Z + 10X + Y)

= 111X + 111Y + 111Z = 111 (X + Y + Z)

(note 3 and 37 are prime numbers and X + Y + Z is not a multiple of 3 (given), so it will not be multiple of 9 too)

So, the correct option is (b).

[A tough question based on properties of numbers, and can be avoided.]

Sol. Ans.(c).

From the given relations, we have p – q = 4033; r – q = 4035 which clearly means r > p > q. Also r – s = 4037 which means r > s.

Thus, r is the greatest of all. So the correct option is (c).

[This is a simple question based on simple calculations, and must be done.]

Sol. Ans.(a).

This looks tough, but is not. A number is prime if it is divisible by none other than 1 and itself.

Now take the digits 1,2,3,4 and 5. Just add them up once. We get 15 as the sum. Recall the rule of divisibility by 3.

“If the sum of the digits of a number is divisible by 3 then number will be divisible by 3”.

So any number formed using these five digits will be divisible by 3. So it cannot be prime. Hence zero.

[A beautiful and logical question based on Prime numbers, which can be attempted if you remembered the rules of divisibility]

Sol. Ans.(b). Theme is “nature of private investment and employment”.

The passage is talking about the nature of private investment and its fickle tendency. It explains why employment generation cannot be left to foreign private investment, and even its positive effects may get diluted due to inherent problems in Indian human capital.

So, statement 1 is surely correct.It is definitely assumed in it. Why? When we are constantly worried about the fickle nature of FDI (and local private investment), and its negative relation with employment generation, how can we rely on it as an economic policy? If 1 is correct, then options (a) and (b) are possible answers but not (c) or (d).

Now between (a) and (b), we have to check 2 and 3.

Focus on 2 – India can do nothing to reduce volatility in FDI because it is the inherent nature of FDI to seek higher returns (hence fickle – moves from here to there). Note the phrase “a necessary consequence”. So 2 is wrong. But 3 may be right, and 5 is surely right (last part of the passage).

Why is 5 right? Education and health are bottlenecks (in absorbing new technology) so public investments will help.

Hence, correct answer is option (b). Also note that 4 is not explicitly talked about anywhere.

[A tough one, and must be tackled very carefully]

Sol. Ans.(a). Theme is “Water resources and utilization”.

Passage is talking about shortages and excesses of water due to unpredictability and nature of Monsoon in India. It also talks about ‘unutilized water which flows with the rivers’.

Statement 3 is not an implication we can get from the passage. Eliminate it. So we are left with options (a) and (b) only.

Now, how do we tackle the headache of water discharging into the oceans, unutilized? Statements 1 and 2 indicate possible solutions. Inter-linking of rivers will allow usage of excess water in deficit regions. And network of dams and canals will ensure timely harnessing of skewed monsoon flows.

Hence, (a) is best. Also, statement 3 is wrong, so (c) and (d) are ruled out.

[A tough question, but can be solved through elimination.]

Sol. Ans.(a). Theme is “Education and Taxation”.

The term “human capital” is not explicitly mentioned in the passage, though some may feel it is an automatic outcome of more education for more people. But we cannot be sure if it is indeed assumed. So 3 is ruled out, and hence options (c) and (d) are ruled out.

Statement 2 is not correct. Investment in children’s education (by parents) will be decided by factors like whether they will be able to use their education later, properly in an economic sense. It is not proper to derive the relation that “investment in education ensures economic freeom”.

So best answer is (a).

[A tough question. Handle with care!]

Sol. Ans.(a). Theme is “Cost of supplying water”.

Try this passage as it has two questions.

Option (c) is clearly wrong, as it says “…cannot be solved”.

Option (d) is wrong as it is nowhere mentioned in the passage.

Between (a) and (b), when we look at (b), it talks of “ULBs not being efficient enough”. But efficiency is not being discussed in the passage, rather financial viability is. So (b) is not entirely correct.

So (b) is not entirely correct. crucial message being given – “recovering costs through user charges” – which can be a financing mechanism to ensure sustainability of water supply.

[A tough question that needs closer inspection of options.]

Sol. Ans.(b). Theme is “Cost of supplying water”.

Statement 1 is not correct, as “rich cities could supply water for free” is mentioned. We are talking of sustainability in statement 1.

Statement 2 is correct as sustainable delivery includes many aspects other than just supplying water to households.

Hence option (b) is right.

[A tricky question, but can be solve with some thought.]

Sol. Ans.(b). Theme is “Agri prices and poverty reduction”.

An excellent topic with great historical background.

Statement 2 is definitely correct (an assumption made in the para) because good price incentives indeed trigger agri investments (mentioned clearly). So options (b) and (c) are only possibilities.

Statement 3 is not mentioned anywhere, even indirectly. So only option (b) is left.

Now statement 4 may look a bit silly when you think of India’s poverty reduction being contingent on higher global agri prices. But it is a fact that it does help directly.

[Not an easy one to solve.]

Sol. Ans.(d). Theme is “Agri prices and poverty reduction”.

The passage does not talk about ‘off-farm rural employment’, ‘farmer produce companies’ or the priority of private investment in agriculture over public. These are not mentioned anywhere at all, and it would be a pure assumption to consider them relevant.

So options (a), (b) and (c) are ruled out.

The passage clearly implies that the rate of poverty reduction was higher when the growth rate of agri-GDP was more. Hence option (d) is correct.

[This was not an easy one to crack!]

Sol. Ans.(c).

The question is ‘Among four numbers P, Q, R and S, which one is LARGEST?’

S1 alone is not sufficient to answer the question, as it doesn’t talk about S.

S2 alone can’t be true as it says that S is not the largest, but doesn’t give any hint about R, P and Q.

By combining S1 and S2 we get – ‘there are four numbers P, Q, R and S; S cannot be largest; So largest number will be among P, Q and R but ‘R is greater than P and Q’, so R is the largest.

Hence, answer is (c).

[An easy question based on Data Sufficiency logic. Must be tried.]

Sol. Ans.(d).

Neither of the S1 and S2 together or alone can give you the value of n.

Using S1 alone you have four prime numbers between 10 and 20 -> 11, 13, 17, and 19.

Using S2 alone there are two numbers 13 (= 4×3 + 1) and 17 (= 4×4 + 1) are possible.

Even after combining both you will get set of two numbers i.e., 13 and 17. So correct answer is option (d).

[A moderately tough question based on Data Sufficiency logic. May be tried.]

Sol. Ans.(c).

S1 alone gives you a × b = 21 (where a and b are two numbers) which have multiple possibilities like 3×7 = 21, 1.5×14 = 21 or 6 × 3.5 = 21. (it is not mentioned that numbers are integers)

So S1 alone is not sufficient.

S2 alone will also not sufficient as it will give you a + b = 10, and again multiple possibilities of a and b like (1, 9); (2, 8); (3, 7)… etc.

By combining bothwe have a + b = 10 or b = 10 – a …(1) and a × b = 21 …(2)

Putting the value of b in equation 2, we have a × (10 – a) = 21 =>a² – 10a +21 = 0

=> a² – 7a – 3a +21 = 0 => a(a – 7) – 3 (a – 7) = 0 =>a – 7) (a – 3) = 0

=> If a = 7, b = 3 and If a = 3, b = 7.

So both S1 and S2 are required to find the numbers. Option (c) is correct.

[A moderately tough question based on Data Sufficiency logic. May be tried.]

Sol. Ans.(d).

[A good question based on HCF/LCM that can be done directly using options.]

Sol. Ans.(d).

Growth rate = Birth rate – Death rate

Growth rate for 1911-1921 = 48.1 – 35.5 = 12.6

Growth rate for 1941-1951 = 39.9 – 27.4 = 12.5

Growth rate for 1961-1971 = 41.1 – 18.9 = 22.2

Growth rate for 1971-1981 = 37.1 – 14.8 = 22.3

We have to assume the simple formula for natural growth rate, as above, because nothing else is mentioned. And check only for the options given, not all seven years.

[A very simple question that must be attempted.]

Sol. Ans.(b).

Given that X = 1.272727…

So 100 x = 127.272727…

=>Subtract the first from second, to get – 99 X = 126

=>X = 126/99 = 14/11.

So the correct option is (c).

[If you use the direct method of converting recurring decimals to fractions, then you get 127/99 but that is not exactly 1.27 but closer to 1.28. So if you use 127272727/99999999, it gets closer and closer to 1.272727… So avoid that and go for the method shown, as option (b) is given as 14/11 which is exactly 1.272727…..]

Sol. Ans.(b).

Desired number = (Least multiple of four digit of LCM of 3, 4 and 5) + 2

So the LCM of 3, 4, 5 and 6 is 60.

The least 4-digit number divisible by 60 is 1020.

So with remainder 2 it will be 1020 + 2 = 1022. Answer is (b).

Alternative method: Work backward from the options and check all the options.

[A direct question based on basic arithmetic]

Sol. Ans.(c).

Traditional Method - Let total no. of men = X and Total population be 100.

Total women = (100 − X)

Married men = 40 X/100 …(i) and Married women = (100 − X) × 30/100 …(ii)

But, No. of married men = No. of married women. (as per conditions given)

So, from (i) and (ii) X = 300/7

So, married Men = 300/7 X 40/100 = 120/7 and Married women = 120/7.

Total married population = Men + Women = 240/7

Required percentage = 240/7 % = 342/7 %. So the correct option is (c).

Alternative method: No man marries more than one woman and no woman marries more than one man. It means the Number of married men = Number of married women = 120 (let). Married adult population = 240.

Number of men = 120×100/40 = 300 and Number of women 120 × 100/30 = 400. So, total population = 300 + 400 = 700.

Required percentage 240×100/700 = = 240/7 % = 342/7 %

(We took 120 due to ease of calculations with given details)

[A tough question that can be left.]

Sol. Ans.(a).

When a number is divided by 100, last two digits of that number will always be remainder. For example 312/100 will give 12 as remainder or 5759 will give 59 as remainder.

So our question is basically reduced to find last two digits of 51 × 27 × 35 × 62 × 75.

However, present question can be solved by looking at only unit digit of the above multiplication.

In the multiplication of two numbers, the unit’s place of the resultant number always be the product of the unit’s place digits of the number which are multiplying. For example, the unit’s digit of 51 × 27 will 1 × 7 = 7.

In above we will get unit digit as 1 × 7 × 5 × 2 × 5 = ..0. Only matching option is (a).

[A not so easy question based on basics of mathematical operations.]

Sol. Ans.(c).

Ratio 1/2 : 3/4 : 5/6 can be written as 6 : 9 : 10.

The total amount is Rs. 2500. Clearly the shares are Rs. 600, Rs. 900 and Rs. 1000 respectively.

The difference between the greatest and the smallest is Rs. 400. So the correct answer is (c).

[A very direct question based on shares and partnerships. Must be attempted.]

Sol. Ans.(b).

(10ⁿ + 1) will have sum of digits 2 for any whole number n. Whole numbers are integers starting with 0. Just put different value of n and observe yourself.

For n = 0, 10⁰ + 1 = 1 + 1. So sum is 2.

For n = 1, 10¹ + 1 = 11. So sum is 1 + 1 = 2.

For n = 2, 10² + 1 = 100 + 1 = 101. So sum is 1 + 1 = 2.

So the correct option is (b).

[A very easy question that can be solved by direct substitution.]

Sol. Ans.(c).

Class Average will remain same as nobody is coming in or going out of the class. So average weight equilibrium will be same as before shifting. A change can happen only if someone is added to the group(s) or someone leaves the group(s), not otherwise.

[A direct logical question, worth attempting!]

Sol. Ans.(a).

We have to choose at least three denominations out of five. So we can either choose 3 denominations out of 5 or choose 4 out of 5 or choose all 5.

Mathematically it will be written as –

Total number of ways = ⁵C₃ + ⁵C₄ + ⁵C₅ = 10 + 5 + 1 = 16 ways.

Don’t worry about ‘different sums’, they will be ‘different’ as each time we are taking a different set of denominations.

Manually, these combinations will be –

(a) Three denomination combinations –

50+100+200; 50+100+500; 50+100+2000; 50+200+500; 50+200+2000; 50+500+2000; 100+200+500; 100+200+2000; 100+500+2000; 200+500+2000; TOTAL 10

(b) Four denomination combinations –

50+100+200+500; 50+100+200+2000; 50+100+500+2000; 50+200+500+2000; 100+200+500+2000; TOTAL 5

(c) Five denomination combinations –

50+100+200+500+2000; TOTAL 1

So total = 10 + 5 + 1 = 16.

[You can either do it manually, or use the theory of permutations and combinations to crack it.]

Sol. Ans.(a).

Regular method:

X can complete 20% of the work in 8 days. So X can complete the entire work in 40 days. Thus, work done by X in 1 day = 1/40.

Y can complete 25% of the same work in 6 days. So work entire work in 24 days and work done by Y in 1 day = 1/24.

Work done by X and Y in 1 day = 1/40 + 1/24 = 8/120 = 1/15.

That means X and Y can together complete the entire work in 15 days.

So, time taken to finish 40% or 2/5th of the work = 15 × 2/5 = 6 days. So the correct option is (a).

Alternative method:

X can complete 20% of the work in 8 days so he will complete 40% of the work in 16 days or 48/3 days.

Y can complete 25% of the work in 6 days so he will complete 40% of the work in 6×40/25 = 48/5 days.

If both work together, the work will be completed in 1/(5/48+3/48)=1/(8/48)=1/(1/6)=6 days

[A standard question from Work and Time, but slightly tricky.]

Sol. Ans.(b).

Regular Method: Average Speed = Total distance ÷ Total time.

Let us assume that distance between X and Y be ‘a’. Then, total distance = a + a + a + a = 4a.

And total time = Distance/Speed = a/v + a/2v + a/3v +a/4v = 25a/12v

So, average speed for the entire journey = 4a/ (25a/12v) = 48v/25 = 1 .92v.

So the correct option is (b) i.e. between v and 2v.

Short-cut method: If same distance is travelled four times with different speeds say a kmph, b kmph, c kmph and d kmph, then Average Speed of total journey can be given by 4abcd/(abc+abd+acd+bcd).

By putting the value directly in formula we get

[A not so easy question that may be dropped]

Sol. Ans.(c).

As shown in figure, minimum numbers of points of intersection of a square and a circle is 2 and maximum number of points of intersection is 8.

So both statements are correct.

[A completely logical question from geometry. Must be tried!]

Sol. Ans.(d).

Regular method: Let the speed in still water = s kmph and the speed of current = c kmph. Let distance is D km.

Speed upstream = Speed in still water + Speed of current = s + c.

Speed downstream = Speed in still water – Speed of current = s – c.

Time upstream = Distance/Speed upstream = D/(s –c) …(1) and

Time downstream = Distance/Speed downstream = D/(s+c) …(2)

By the condition given in question Time upstream/2 = Time downstream

Shortcut method: Try working backwards. For option (d), s = 3x, c = 1x. So downstream speed = 4x and upstream speed = 2x. Clearly for the same distance, ratio of times taken will be ratio of these speeds, and that comes out to 4x : 2x = 2:1. (upstream time : downstream time). Hence option (d) is correct.

[A simple question from Boats and Streams. Must be attempted.]

Sol. Ans.(a).

Natural numbers are integers starting with 1.

Let the two numbers be a and b then

a² – b² = 63 =>(a + b) (a – b) = 63. Now 63 = 9 × 7 = 21 × 3 = 63 × 1 So,

Case I: a + b = 9; a – b = 7. Here we will get a = 8 and b = 1

Case II: a + b = 21; a – b = 3. Here we will get a = 12 and b = 9.

Case III: a + b = 63; a – b = 1. Here we will get a = 32 and b = 31

Since 63 has only these 3 pairs of factors, so the correct option is (a).

[A good question based on basic number theory; may be attempted.]

Sol. Ans.(d).

Value of this percent will depend on ‘a’. As numerator is a constant value and ‘a’ is in denominator, the higher the value will be the shorter the percentage will be. Hence correct option is (d).

Otherwise you can try it by direct method. . How? Of all given fractions, the largest is 5/6 = 83.3%. Rest all are smaller. So the largest will show the minimum impact of this operation.

[A good but not so easy question on basic numbers.]

Sol. Ans.(d).

A digit n > 3 is divisible by 3 but not divisible by 6 is 9.

So n = 9. Of all given options 2n, 3n, 2n+4, and 3n + 1, only the last one ( = 3×9 + 1 = 28 ) will be divisible by 4.

[A direct question that must be solved.]

Sol. Ans.(c).

Given 1 litre = 1 kg.

We know that 1 m³ (1 cubic metre) is exactly 1000 Litres.

=>1000 litre = 109 cubic millimetres.

(How? 1 m x 1 m x 1 m = 100 cm x 100 cm x 100 cm = 1000 mm x 1000 mm x 1000 mm = 10⁹ mm³)

=>1 litre = 10⁶ cubic millimetres = 1 kg = 1000 gm

So 0.1 gram = 10² mm³ = 100 cubic millimetres.

[Slightly involved calculations, though from the first principles only.]

Sol. Ans.(a).

Regular method: Let the weight of empty vessel = v and weight of water = w, then by the conditions given we have

v + w = 40 …. (1) and

v + w/3 = 20 …. (2)

By solving equations (1) and (2) simultaneously we have w = 30 and v = 10.

Short-cut: Taking option (a), weight of empty vessel = 10, so weight of water when vessel is completely filled = 40 – 10 = 30. When water is one-third fill its weight = 30/3 = 10. So weight of vessel when one-third filled with water = 10 + 10 = 20. Which matches with the condition given in question. So answer is option (a).

[Short-cut method is much better here]

Sol. Ans.(b).

Total distance to be jumped by frog to come out = 4.5 m = 450 cm.

Frog jumps 30 cm and slips 15 cm, which means that, in a jump he ascends 30 – 15 = 15 cm.

In 28 such jumps he will ascend to 28 × 15 = 420 cm.

Now only 30 cm remain to come out, which the frog will jump in the next (29^th) jump.

Note – In the last jump frog will not slip because he will be outside the well.

[A nice question using logical reasoning.]

Sol. Ans.(d). Theme is “Crop dependence of nations”

Soil and climate conditions were not the reasons for destruction of the homogenous potato crop throughout Ireland. Soil and climate conditions were not the reasons for destruction of the homogenous potato crop throughout Ireland.

The passage does not talk about prevention of fungal infections, so (c) is not correct. Passage does not even talk about whether the tuber crops can replace the cereal crops or not, so (b) is not correct. Passage gives the fundamental, critically important message that “do not rely on a homogenous food source else a single disease can wipe out the entire food economy”.

[An easy question that must be attempted.]

Sol. Ans.(c). Theme is “Malnutrition in India”.

A very tricky question. Note that the passage is focused solely on “malnutrition” and no other topic at all – neither vaccination, nor water/sanitation nor breastfeeding, etc.

Hence, of all the statements given, if we have to focus on malnutrition alone, then it means food supply for all citizens, especially the weaker sections. And that means a better PDS system. Only statement 1 fits. If the Union govt. can monitor the PDS from the perspective of eliminating malnutrition, then the problem can be mitigated.

Statement 5 is not mentioned, so options (b) and (d) are eliminated.

And option (a) is eliminated due to 2 not being relevant instantly.

[A tricky question that needs some deep thinking.]

Sol. Ans.(c). Theme is “Paddy versus Pulses in North Indian farms”

Statement 5 is not correct as it is mentioned that paddy stalks are tough to collect or plough back into the ground, hence farmers choose to burn them (so its not mechanized agriculture that’s the culprit).

So options (a) and (b) are ruled out.

Now, statement 1 is not correct as we do not know really whether a direct comparison of pulse cultivation and paddy cultivation is even possible. It’s not explicitly mentioned.

So options (b) and (d) are ruled out.

So finally we are left with option (c), which is 2 and 3 only.

[A not so easy question, handle with care!]

Sol. Ans.(c). Theme is “Water reservoirs and monsoon”

Three options have statement 2 in them. So focus on 2 first.

Statement 2 says – “Storage capacity of dams should not be fully used before or during monsoon season”. But the passage says that reservoirs must be used during monsoon rains to store excess rainwater. So statement 2 does not seem right.

Hence, only option (c) is correct.

[A very easy question, is solved using options]

Sol. Ans.(a). Theme is “Economic liberalization in India”. There are two questions, so must be attempted.

We are asked to find which on is a rational and logical inference.

Statement 1 speaks about rethinking the economic role of State, in the quest for development. From the passage, it is clear that the Indian liberalization process was triggered by government’s economic problems and not people’s problems, and that led to many design flaws, which are now visible in multiple shortcomings across sectors.

So, 1 does seem a rational inference. The State’s economic role must be rethought and redefined if we want proper development. So answer can be (a) or (c).

Statement 2 talks about “implementation of policies” and “insufficient investments”. It is difficult to infer this from the para that speaks mostly of “conception and design” and not “implementation”. Also the term ‘social sectors’ is not defined. Hence final answer is (a).

[A difficult one, to be handled with care.]

Sol. Ans.(d). Second question on same set. Theme is “Economic liberalization in India”.

A mere glance at both the statements 1 and 2 shows that both are not valid at all.

The passage hasn’t talked about integration of Indian economy with the global economy. It also hasn’t talked about potentially high growth being the driver of bringing economic liberalization in India (which the second statement is hinting at). Hence 2 is also wrong.

Hence answer is (d).

[An easy question, must be attempted.]

Sol. Ans.(d). All the options are equally good for the customer.

Case 1 –Let the price of the article = 100, after first discount of 10% it will be of 90. After second discount of 20% it will be 90 – 18 (20% of 90) = 72. And after services tax, it will become 72 – 7.2 (10% of 72) = 64.8.

Case 2 Let the price of the article = 100, after first discount of 20% it will be of 80. After second discount of 10% it will be 80 – 8 (10% of 80) = 72. And after services tax, it will become 72 – 7.2 (10% of 72) = = 64.8.

Case 3 Let the price of the article = 100, after 10% service it will be of 90. After second discount of 20% it will be 90 – 18 (20% of 90) = 72. And after the last discount of 10% it will become 72 – 7.2 (10% of 72) = 64.8.

[A good question, based on simple profit and loss topic. Must be attempted.]

Sol. Ans.(d).

It is given that A to Z are numbered in same order as 1 to 26 (the word “respectively”).

Now, GHI must logically become 789. But we are given GHI = 1578.

Note that 789 x 2 = 1578.

Also given that DEF = 912. But 912 = 456 x 2 and 456 = DEF in the given coding scheme.

So logic is “doubling of give number code”.

So correct answer would be ABC = 123 = 246.

[A slightly tricky question on coding, but can be done.]

Sol. Ans.(d).

ACPQ : BESU :: MNGI : @

Here A + 1 = B; C + 2 = E; P + 3 = S; Q + 4 = U (So ACPQ becomes BESU)

Similarly, M + 1 = N; N + 2 = P; G + 3 = J; I + 4 = M.

So the correct option is (d) as MNGI will become NPJM.

[A direct question on coding, and must be attempted.]

Sol. Ans.(c).

(1/2)^{– 6} = 64;

(1/4)^{– 3} = 64;

(1/3)^{– 4} = 81;

(1/6)–^{– 2} = 36;

So the correct option is (c).

[All you have to do is take the denominator to the numerator.]

Sol. Ans.(c). Theme is “Failing banks and resolution process”.

The passage is focused on the limitations of the way the Indian banking system is structured, as banks are not allowed to fail, and there is no formal resolution process for that.

Option (a) is too broad.

Option (b) is too generalized.

Option (c) is precise and the actual inference we can draw.

Option (d) is irrelevant.

[An easy question, must do.]

Sol. Ans.(c). Theme is “India’s emissions and renewables push”.

The passage says that ‘India has potential for solar energy. We have decided to shift to renewable sources of energy but they are costly and technologically not perfect. Indian government is committed to reduce emission by 33% till 2030 and for that it has made efforts in gas-based economy and renewable energy fields. Business-houses are reluctant to invest in renewable energy sources because of lack of technology.’

The passage does not provide comparison of gas over renewable sources, and definitely does not say that gas must be preferred over renewables. So (b) is clearly ruled out.

Similarly, option (d) sounds like a policy prescription that’s not intended, in the passage.

So between (a) and (c), it is a tough choice. We will go with (c).

[A tough one.]

Sol. Ans.(d). Second question on the same passage. Theme is “India’s emissions and renewables push”.

Passage does not talk about subsidies. In fact it is talking of investment from corporates. So statement 1 can be ruled out. So options (a) and (c) are ruled out.

Statement 2 is only slightly correct, because it sounds almost like a foregone conclusion. This is 2020, and we are talking about 2030. So things can change in the meantime. So, 2 may not be entirely right.

So we reject 2 also. Hence answer is (d).

[A tough one.]

Sol. Ans.(c). Theme is “Genome editing”.

Both the assumptions are valid.

Statement 1 is evident from ‘Researchers use ‘molecular scissors’ to dissect the genome and repair it..’.

Statement 2 looks like the summary of the passage. It is correct.

Hence answer is (c).

[A very easy one.]

Sol. Ans.(c). Theme is “Solid waste management”.

The passage does not talk about ‘resources of urban local bodies’ and ‘drinking water shortage’, so (b) and (d) are ruled out.

Between (a) and (c), option (c) is evident from ‘… when unscientific solid waste management combines with poor drainage..’. So correct answer is option (c).

[An easy one and must be attempted.]

Sol. Ans.(b). Theme is “Freedom of Religion u/Art 25”.

The passage wants to say that our constitution provides the fundamental right of religious freedom with certain conditions. If it is harmful for the society in any way, the state can intervene.

Options (c) and (d) are both extreme and can be immediately rejected.

Option (a) is not correct at all.

[A relatively easy question.]

Sol. Ans.(d).

D _ _ _ I

We have three places and three letters E, H, L which can be arranged in 3! = 6 ways.

So the correct option is (d). This is a question from permutations and combinations. If you don’t know the topic, you can try doing it manually as well which may take some time.

[An easy question if you knew the topic.]

Sol. Ans.(c). Successive replacement case – If y units of some other liquid is replaced in x units of original liquid n times then the ratio of the quantity of original liquid to the total quantity of liquid after n operation = ((x-y)/x)^{^n}.

Applying to our case x = 20, y = 4, n = 2.

So at the end of the 2nd operation the ratio of the quantity of liquid A to the total quantity of liquid =

It means out of the total quantity of 25 units liquid A is 16 units and liquid B is 26 – 16 = 9 units.

Hence required ratio is 16 : 9.

[Knowing the formula helps a lot in this.]

Sol. Ans.(c).

Regular method:Average of 50 innings = 46.4.

Sum of 50 innings = 50 × 46.4 = 2320 …(i)

Average of 60 innings = 46.4 + 2.6 = 49

Sum of 60 innings = 60 × 49 = 2940 …(ii)

So runs scored in last ten innings = 2940 – 2320 = 620.

Average = 620/10 = 62. So the correct option is (c).

Short-cut: The runs scored in last ten innings increase the average of all 60 innings. It means that in last ten innings 2.6×60 more runs were scored.

So, in last ten innings total runs scored = 10 × 46.4 (previous average) + extra runs which caused the increase in the average of all 60 innings

= (10 × 46.4) + (2.6 × 60) = 464 + 156 = 620.

So required average 620/10 = 62.

[A tough question on averages, which can be avoided in the first round of solving.]

Sol. Ans.(b).

Start directly from the options.

Option (a) – Rs.30. Hike it by 25% and it becomes Rs.37.5 per kg. Originally one could buy (1200 / 30) = 40 kg rice and now one can buy (1200 / 37.5) = 32 kg. Difference is 8 kg and not 6 kgs.

Option (b) – Rs.40. Hike it by 25% and it becomes Rs.50 per kg. Originally one could buy (1200 / 40) = 30 kg rice and now one can buy (1200 / 50) = 24 kg. . Difference is 6 kg and (b) is our answer.

Other method: Let the original price be P. So, 1.25 P is the new price.

Earlier quantity purchased = 1200 / P kg.

New quantity purchased = 1200 / 1.25P kg

Difference = 6 kg = 1200/P – 1200/1.25P.

Solve for P.

[A tricky question in CP/SP/Percentages but can be cracked from options easily.]