IBPS PO - MAINS - SPEED TEST - 3 -SOLUTIONS PO MAINS SPEED... · IBPS PO (MAINS) MAINS TEST -...

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IBPS PO (MAINS) MAINS TEST - 3(SOLUTIONS) CALL 8427029060

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IBPS PO - (MAINS) SPEED TEST - 3 DIRECTION (1-5):-

Above shown diagram will be final arrangement. 1. (A) Teacher is P and 3rd to right of S is sitting. As we can see in above diagram S like Egg Biryani Hence, 3rd to right of Teacher likes Egg Biryani 2. (D) AS we cans see Y is commissioner and 2nd to left of Y is W and Q is an IPS Officer. As per family tree, W is Uncle of Q. 3. (D) R, P and Z are female but S is male. Hence, S does not belong to group which all other belong. 4. (A) 1) Q – IPS - Kufta ⇒ False (as Q – IPS – Mattar - Paneer can also possible) 2) P – Teacher – Mutton - Kurma ⇒ True 3) W – Major – Dal - Makhni ⇒ True 4) S – Engineer – Egg - Biryani ⇒ True Hence, Q – IPS – Kufta is possibly false combination. 5. (C) Hence, third to left of IPS, Manager is sitting. 6. (D) Given Statements: W © F; F @ D; D * K; K $ J

© @ $ * %

≤ = ≥ < >

On decoding: W ≤ F; F = D; D < K; K ≥ J On combining: W ≤ F = D < K ≥ J Conclusion: I. K % W → It means K > W, hence true as it is given that W ≤ F = D < K. II. D $ W → It means D ≥ W, hence true as it is given that W ≤ F = D. III. F * K → It means F < K, hence true as it is given that F = D < K.

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Hence, all the conclusion I, II and III are true. 7. (E) Given Statements: D @ M, M $ B; B * R; R % T

© @ $ * %

≤ = ≥ < >

On decoding: D = M, M ≥ B; B < R; R > T On combining: D = M ≥ B < R > T Conclusion: I. B * D → It means B < D, hence not true as it is given that D = M ≥ B, so they can be equal as well. II. B @ D → It means B = D, hence not true as it is given that D = M ≥ B, so D can be greater than B as well. III. T * M → It means T < M, hence not true as it is given that M ≥ B < R > T. it show that there is not a definite relationship between them. Conclusion I and II are complementary to each other. Hence, either conclusion I or II is true. 8. (C) Given Statements: R * K; K © M; M % T; T $ J

© @ $ * %

≤ = ≥ < >

On decoding: R < K; K ≤ M; M > T; T ≥ J On combining: R < K ≤ M > T ≥ J Conclusion: I. J * M → It means J < M, hence true as M > T ≥ J. II. R * M → It means R < M, hence true as R < K ≤ M. III. K © J → it means K ≤ J, hence not true as no definite relationship between them. Hence, only Conclusion I and II are true. 9. (B) The standard ASCII character set users only 7 bits for each alphanumeric character. 10. (C) If sum of the weights are equal to 9, then the code is said to be self - complemented. ⇒ 6 + 6 + ( - 1) + ( - 2) = 9 (Yes) ⇒ 5 + 1 + 2 + 1 = 9 (Yes) ⇒ 6 + 4 + ( - 3) + ( - 1) = 6 (No) ⇒ 4 + 2 + 1 + 2 = 9 (Yes) DIRECTION (11-15):-

11. (E) Since, Bhusan likes Magenta colour and bhusan is not one of the given option. Hence, none of these is the correct option. 12. (A) Since, Alok sits second to the right of Rekha and Alok faces Sunil. Hence, sunil is the correct option.

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13. (A) Hence, Mahesh likes Arrow brand. 14. (A) Hence, Rekha likes red colour. 15. (D) Hence, there are four people sit between Priya and Bhusan. 16. (D) As we can see in above flow diagram person no. 1 is Unmarried with at least one child. (Married: no; children: yes; male child: yes) 17. (B) As we can see in above flow diagram person no. 1 is Unmarried with at least one child. (Married: no; children: yes; male child: yes) 18. (D) From Statement I: Some Tea are Cold. Some Cold are hot. Some hot are Coffee.

i) Some hot are Tea is a possibility → True (It is an possibility) ii) Some Coffee are Cold → False (Possible but not definitely true). iii) Some Coffee are hot → True iv) All Tea is hot is a possibility → True (It is an possibility) Therefore, Statement I will not result in the given conclusions. From Statement II: All Tea are Cold. Some Cold are not hot. Some hot are Coffee

or i) Some hot are Tea is a possibility → True ii) Some Coffee are Cold → False (Possible but not definitely true). iii) Some Coffee are hot → True iv) All Tea are hot is a possibility → False (as some cold are not Hot and all tea and cold). Therefore, Statement II will not result in the given conclusions. From Statement III: No Tea are Cold. All hot are Cold. All cold are Coffee

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i) Some hot are Tea is a possibility → False ii) Some Coffee are Cold → True iii) Some Coffee are hot → True iv) All Tea are hot is a possibility → False Therefore, Statement III will not result in the given conclusions. From Statement IV: All Tea are Cold. Some Cold are hot. All hot are Coffee.

i) Some hot are Tea is a possibility → True ii) Some Coffee are Cold → True iii) Some Coffee are hot → True iv) All Tea are hot is a possibility → True Therefore, Statement IV will result in the given conclusions. From Statement V: All Tea are Cold. No Cold are hot. All hot are Coffee.

i) Some hot are Tea is a possibility → False ii) Some Coffee are Cold → False (Possible but not definitely true). iii) Some Coffee are hot → True iv) All Tea are hot is a possibility → False Therefore, Statement V will not result in the given conclusions. 19. (A) From I: In that code language rem tez kullu pullu tullu means ‘Sher Singh is my son’ and ‘gullu sullu rullu pullu’ means ‘is he at home’.

Therefore, we can’t determine meaning of “kullu” is given coded language from statement I alone. From II: In that code language nel pullu kullu dela means ‘my daughter is Nirmala’ and setha gama lala means ‘sit with me’. There are no common word in both statements therefore we cannot determine meaning of “kullu” is given coded language from statement II alone. From III: In that code language nel dela pullu nillu means ‘her daughter is Nirmala’ and ‘kettu bala’ means ‘go home’

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There are no common word in both statements therefore we cannot determine meaning of “kullu” is given coded language from statement III alone. From I and II:

Therefore, meaning of code for “kullu” will be “My” Hence, data in statement I and II together is sufficient to answer the question. From II and III:

Therefore, meaning of code for “kullu” will be “My" only if we consider the conclusions from statements I and II. Hence, data in statement II and III together is sufficient to answer the question. From I and III:

Here we cannot determine code for “kullu”. Hence, data in statement I and III together is insufficient to answer the question. Therefore, option(1) is correct option as it implies both I and II together or both II and III together. 20. (E) From I: X and Y are brothers. B is the brother of A and A is the mother of X.

As per family tree shown above B is maternal uncle of Y. Therefore, statement I alone is sufficient. From II: A is the brother of B and father of X, who is the brother of Y.

As family tree shown above gender of B is not definite therefore we can’t determine how B related to Y. Therefore, statement II alone is insufficient.

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From III: B is the brother of A and father of X. X is the brother of Y.

As per family tree shown above B is Father of Y. Therefore, statement III alone is sufficient. Hence, Data in either statement I or Statement III alone is sufficient. As there are no such option as Either I or III therefore none of these is correct choice. Note- In data sufficiency question whenever there are two different answer from different statement then we choose either – or are correct choice. As in this question from statement I we get B is maternal uncle of Y and statement III we get B is father of Y. Therefore either – or will be correct combination. DIRECTION (21-25):-

Players Playing role Country IPL team Salary package

P S - fi Hong Kong Crack 25 cr

Q Action India Kick 10 cr

R Comedy Netherlands Golmal 20 cr

S Comedy UAE Dhamal 40 cr

T Si - fi Ireland Dangal 5 cr

U Horror Australia Robot 50 cr

V Action Nepal Housefull 30 cr

21. (C) Above combination will be final combination. Hence, P act in crack. 22. (A) Hence, V’s salary package is 30 cr. 23. (C) U is to shoot his movie in Australia and he is working in Horror movie. Hence, Horror movie being shoot in Australia. 24. (D) U is to shoot his movie in Australia and he is working in Horror movie. Hence, U is acting in a horror movie. 25. (A) Hence, T act in for Dangal. DIRECTION (26-30):- Given table:

Numbers/ symbols

Q E F T H X R I L A G C U N D Y S P M V O

Letter code 8 5 2 7 1 # ^ * ! ) @ 9 4 6 3 % + ~ ` $ /

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According to the given conditions, Rule No: Condition Result

1 Second letter and fourth letter are vowels Replace the second letter by the code number ‘8’ and fourth letter by the code symbol ‘$’

2 They are no vowels in the code Write the code in reverse order

3 Letters like ‘A’, ’W’, ‘V’, ‘S’ in third or fifth position

Replace the code by ‘5’.

4 Vowel is immediately followed by a consonant

Interchange the codes of the consonant and vowels

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Fourth element is a consonant and is immediately preceded by another consonant or immediately followed by vowel but not both

Code of the fourth element consonant is coded as ‘@’.

26. (B) Here 2nd rule is applied i.e. there are no vowels. And even 5th rule is also applied i.e. fourth element is a consonant and is immediately followed by another consonant. Therefore QXTYMN is coded as 6`@7#8. 27. (A) Here 4th rule is applied because E is a vowel and T is a consonant and U is a vowel and N is consonant. There we will Interchange the codes of the consonant and vowels. Therefore AETNUY will be coded as )756%4. 28. (E) Here 3 rules are applied 1st, 3rd and 4th rules. By 1st rule i.e. we are having vowels in second and fourth places. By 3rd rule i.e. 5th element is ‘S’ so need to replace by ‘5’. And finally 4th rule is vowel is immediately followed by a consonant. Therefore HILUSO is coded as 1!85$/ 29. (B) Given table: Here 1st rule is applied i.e. second and fourth elements are vowels and even 4th rule is also applied i.e. vowel is immediately followed by a consonant. Therefore XEIUNR is coded as #8*6$^. 30. (A) Here 3rd rule is applied i.e. the ‘A’ is the fifth element in the code. Therefore PFQIAO is coded as ~28*5/. DIRECTION (31-35):- In the given illustration: 1) Numbers are arranged in descending order. 2) In case of words first vowels are arranged in ascending order followed by consonants in ascending order as per their position in dictionary. 3) A word and number is arranged simultaneously in each step. In first step, the lowest number is arranged at leftmost position i.e. at first position and lowest vowel as per dictionary is arranged at rightmost position i.e. at last position of the sentence. Both word and number are arranged simultaneously and else all are shifted accordingly. In step 2 next lowest number is placed again at first position i.e. immediate left to the previously arranged number and next lowest vowel/consonant is placed at last position i.e. immediate right to previously arranged word and else all are shifted accordingly. The same process is followed till all the words and numbers are rearranged as all numbers at left side in decreasing order and all words first according to vowel than according to consonant at right side of the lowest number in increasing order as per dictionary. Now given input is: 86 youth hammer 18 34 user enter ocean ink 44 25 mock 63 98 Step I: 18 86 youth hammer 34 user ocean ink 44 25 mock 63 98 enter Step II: 25 18 86 youth hammer 34 user ocean 44 mock 63 98 enter ink Step III: 34 25 18 86 youth hammer user 44 mock 63 98 enter ink ocean Step IV: 44 34 25 18 86 youth hammer mock 63 98 enter ink ocean user Step V: 63 44 34 25 18 86 youth mock 98 enter ink ocean user hammer Step VI: 86 63 44 34 25 18 youth 98 enter ink ocean user hammer mock Step VII: 98 86 63 44 34 25 18 enter ink ocean user hammer mock youth Step VII is the last step of the given input.

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31. (D) Clearly, ‘youth’ is 7th from left and 8th from right in step VI. 32. (D) Clearly, four words/numbers are there in between ‘ink’ and ‘44’ in last step of the input. 33. (E) Clearly, there is no such step which matches the given step. 34. (B) Clearly, number ‘98’ is at 6th position from right in step V. 35. (E) Thus step VI is the penultimate step. Here 3rd number from left end is 44 and 6th number from left end is 18. Difference between 44 and 18 is 26. Thus difference between 3rd number from left end and 6th number from left end in penultimate step is 63-25= 38. 36. (C) From statement I: 1) Point D is 5m to the south of Point P. Point M is 8m to the west of point D. 2) Point S is 2.5m to the north of point M. Point O is 10m to the east of Points S. 3) Point K is 2.5m to the south of points O.

As we can see in above diagram point M is 10m to the west of K. From statement II: 1) Point K is 10m to the east of point M. Point U is 8m to the west of Point M. 2) Point D is to the east of M. Point M is the midpoint of the lines formed by joining points U and D.

Above diagram we can conclude that point M is 10m to the west of K. Hence, the data either in statement I alone or in statement II alone are sufficient to answer the question. 37. (B) From statement I: People: P, Q, R, S, T and U. 1) R is heavier than only two people. __>__> R > __> __ 2) P is heavier than U but lighter than Q. 3) S is not the heaviest. Case -1 → Q > S > R > P > U Case -2 → Q > P > R > U > S As we can see there are two cases hence either S or P is having 2nd heaviest among them. Therefore we can’t determine who is heaviest. From statement II: People: P, Q, R, S, T and U. 1) U is lighter than only two people. 2) R is lighter than U but heavier than S. 3) P is lighter than only Q. Q > P > U > R >S Hence, as we can see above P is 2nd heaviest among them. Hence, data in statement II alone are sufficient to answer the question, while the data in statement I alone are not sufficient to answer the question.

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38. (A) I. There are 4 Sundays, 4 Fridays and 4 Saturdays in a month. II. There are so many holidays III. There are 5 Mondays. Number of Holidays in a month: 4 Sunday + 5 Mondays + 2 Saturday = 11 days. There are 31 days in the month of March. So working days = 31 – 11 = 20 days. Therefore, only statement I and III are sufficient to answer the question. 39. (A) The question asks for an inference. The author makes it a point to mention that small plots of land sell for Rs. 50 lakhs. From this we can infer that small plots on the outskirts of a city don’t usually sell for Rs. 50 lakhs. 40. (C) The question asks which statement shows us that these regions gaining popularity in the Real Estate markets could be just a bubble. “Bubble” refers to temporary increased popularity that will “burst” or lessen considerably in some time. This is best reflected in statement (D). If a similar case happened in Kolkata, it could point to something like this happening in Mumbai as well. 41. (D) The question asks for an assumption. Of the given options, only statement (H) is an assumption that can be made on the basis of the facts/information given. We are told that these outskirts are hot properties and we are told that even small plots on the outskirts are sold for Rs. 50 lakhs. Thus we can make the assumption that rates such as these are sufficient for the area to become labelled hot property. 42. (E) One of the reasons people might prefer to buy properties on the outskirts of the city is because of the amenities provided at a short distance. Thus the best option is statement (F). 43. (B) The question asks for a probable cause for increased popularity of the outskirt regions of Mumbai. Among the options, only statement (G) could be a probable cause. When a popular and populated city like Mumbai starts to have unaffordable property rates, it could cause people to find areas close by with more affordable rates. 44. (B) Conclusion I is not mentioned anywhere in the statement, so we cannot infer it. We need to stick to the scope of the information mentioned and we should not assume anything on our own. Any conclusion which does not directly follow but may be indirectly implied or is suggested can be followed. Note that conclusion II follows from the given statement because the second sentence concerns itself about saving money which is possible through saving fuel. Thus option 2 is the correct answer. 45. (B) The fact in the first conclusion is quite contrary to the given statement. So, I does not follow. II mentions the direct implications of the state discussed in the statement. Thus, II follows. It is quite obvious that with the increase in population and the depletion of resources. The Government will have a tough time meeting the requirements of the people of the country. Hence, 2 is the correct answer. 121. (C) Interest = PRT/100 Where P = Principal, R = Rate of interest, T = Time For the first two years in Scheme II of company A Let the principal be P Rate of interest = 9% ⇒ S.I. = (P × 9 × 2)/100 = 18P/100 For the next three years in Scheme IV of company C Let the principal be P Rate of interest = 10% ⇒ S.I. = (P × 10 × 3)/100 = 3P/10 ⇒ Total interest = 18P/100 + 3P/10 = 48P/100 Total interest = Rs. 1200 48P/100 = 1200 P = Rs. 2500

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122. (C) The formula for annual compound interest, including principal sum, is:

⇒ 1

Where: C.I. = Compound Interest P = the principal investment amount (the initial deposit or loan amount) r = the annual interest rate n = the number of times that interest is compounded per year t = the number of years the money is invested or borrowed for For company D under Scheme II Principal = Rs. 1,40,00 Rate of interest = 10% Interest is calculated annually, n = 1 Time (t) = 2 years

⇒ 14000 1 14000 = Rs. 29400

For company B under Scheme II Interest = PRT/100 Where P = Principal, R = Rate of interest, T is the time for which money is borrowed/lent Principal = Rs. 1,40,00 Rate of interest = 8% Time (T) = 2 years ⇒ Interest = (140000 × 8 × 2)/100 = Rs. 22400 ⇒ Difference between interests earned = 29400 – 22400 = Rs. 700 123. (C) Under scheme IV for 1 year Rate of interest = 10% Principal = Rs. 20000 ⇒ Interest = (20000 × 10 × 1)/100 = Rs. 2000 ⇒ Total amount = 20000 + 2000 = Rs. 22000 Under scheme II for 1 year

⇒ 1

Where: A = Amount after the end of investment P = the principal investment amount (the initial deposit or loan amount) r = the annual interest rate n = the number of times that interest is compounded per year t = the number of years the money is invested or borrowed for Rate of interest = 8.5% n = 1, t = 1 year, P = Rs. 22000

⇒A = 22000 1.

= Rs. 25,899

⇒ A = 22000(1 + 8.5/100)2 = Rs. 25898.95 = Rs. 25,899 124. (A) For Harsh Rate of interest = 9.5% Interest = (40000 × 9.5 × 2)/100 = Rs. 7600 For Sagar Rate of interest = 8%

⇒ 1

P = Rs. 40000 t = 2 years, n = 1 ⇒ C.I. = 40000(1 + 8/100)2 – 40000 = Rs. 6656 ⇒ Harsh receives more interest i.e. = 7600 - 6656 = Rs. 944

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125. (B) The formula for annual compound interest, including principal sum, is:

⇒ 1

Where: C.I. = Compound Interest P = the principal investment amount (the initial deposit or loan amount) r = the annual interest rate n = the number of times that interest is compounded per year t = the number of years the money is invested or borrowed for For Company B under Scheme III, P = Rs. 25000 t = 2 years r = 9% n = 2 ⇒ C.I. = 25000(1 + 9/200)4 – 25000 = Rs. 4813 Company D under Scheme II, P = Rs. 30000 t = 2 years; r = 10%; n = 1 ⇒ C.I. = 30000(1 + 10/100)2 – 30000 = Rs. 6300 ⇒ Total interest earned = 4813 + 6300 = Rs. 11113 126. (C) For rural area ⇒ Candidates having experience in Public Sector Banks only = 500 ⇒ Candidates having experience in Private Sector Banks only = 400 ⇒ Candidates having experience in Public and private Sector Banks only = 450 ⇒ Total candidates = 500 + 400 + 450 = 1350 127. (B) ⇒ Candidates having experience in Public Sector Banks in rural area only = 500 ⇒ Candidates having experience in Public Sector Banks in urban area only = 2000 ⇒ Candidates having experience in Public and private Sector Banks in urban area only = 2100 ⇒ Candidates having experience in Public and private Sector Banks rural area only = 450 ⇒ Total candidates having experience in public sector banks = 500 + 2000 + 2100 + 450 = 5050 128. (C) ⇒ Candidates having experience in Public Sector Banks in urban area only = 2000 ⇒ Candidates having experience in Private Sector Banks in urban area only = 1000 ⇒ Candidates having experience in Public and private Sector Banks in urban area only = 2100 ⇒ Total number of candidates who have worked in urban areas = 2000 + 1000 + 2100 = 5100 ⇒ required ratio = 2000/5100 = 20/51 129. (C) ⇒ Candidates having experience in Private Sector Banks in urban area only = 1000 ⇒ Candidates having experience in Public and private Sector Banks in urban area only = 2100 ⇒ Total candidates having experience in private sector in urban area = 1000 + 2100 = 3100 130. (D) ⇒ Candidates having no experience = 3550 ⇒ Candidates having experience in Public and private Sector Banks in urban area only = 2100 ⇒ Candidates having experience in Public and private Sector Banks rural area only = 450 ⇒ Total candidates having experience working in both banks = 2100 + 450 = 2550 ⇒ Percentage = (2550/3550) × 100 = 71.83% = 72% 131. (E)

The avg of profits =

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Total profits in 6 years = 30 + 15 + 45 + 50 + 60 + 55 = 255 So, avg = 255/6 = 42.5 crores = Rs 425000000 132. (A)

Profit % =

100

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= This Year Profit−Previous Year ProfitPrevious Year Profit×100This Year Profit−Previous Year ProfitPrevious Year Profit×100 Profit in year Profit in crores Profit % Profit in 2000 30 - Profit in 2001 15 -50% Profit in 2002 45 200% Profit in 2003 50 11% Profit in 2004 60 20% Profit in 2005 55 -8% Comparing the magnitudes, we can say that the year of highest per cent increase / decrease in profit is 2002. 133. (B) Income of company = expenditure + Profit Expenditure = 45 crore Profit in 2001 = 15 crores (from line graph) Thus, income = 45 + 15 = 60 crores = Rs 60,00,00,000 134. (C) Income of company = expenditure + profit of a company Therefore, expenditure in 2004 = 136 – 60 = 76 .….. (1) {From line graph profit in 2004 = 60 }

Therefore profit percentage = 100 100 79%

135. (D) Expenditure of a company = Income of a company + Profit by a company Thus from the above formula, expenditure of a company in 2000 = 50 crores – 30 crores {∵ from line graph profit of company is 30 crores in 2000} Expenditure of company in 2000 = 20 crores Similarily, expenditure of company in 2003 = 95crores – 50 crores = 45 crores. So, ratio of expenditure of company in 2000 to 2003 = 20/45=4/9 DIRECTION (136-140):-

Universities Architecture Journalism Web Designing

Interior Designing

Nursing Total

A 140 150 125 130 135 680 B 95 95 150 90 70 500 C 90 130 140 140 85 585 D 85 115 145 160 75 580 E 80 120 105 130 140 575 F 110 110 135 120 125 600 Total 600 720 800 770 630 136. (A) ⇒ Number of boys in Web designing and Interior designing from University B = 240 ⇒ Number of boys in Architecture and Nursing from University B = 165 ⇒ Number of girls in Journalism from University B = 95 ⇒ Total number of boys learning different courses from University B = 500 ⇒ So, 77.5% of 240 + 80% of 165 + 60% of 95 got certificate from University B = 186 + 132 + 57 = ∴ required percentage = 375/500 × 100 = 75% 137. (E) ⇒ Total number of boys learning from University A = 680 ⇒ Average numbers of girls in each course of University A = 680/5 = 136 ⇒ Total number of boys learning from University F = 600 ⇒ Average numbers of boys in each course of University F = 600/5 = 120 ∴ Difference = 136 – 120 = 16 138. (C) ⇒ Total number of boys learning Architecture from all Universities = 600 ⇒ Number of boys learning Architecture in University C = 90 ∴ required percentage = 90/600 × 100 = 15% 139. (B) ⇒ Total number of boys of all Universities learning Nursing = 630 ⇒ Total number of boys of all Universities learning Interior designing = 770

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∴ respected ratio = 9 : 11 140. (C) ⇒ Total number of boys learning Journalism in University B, C and D = 95 + 130 + 115 = 340 ⇒ Total number of boys learning Nursing in University A, E and F = 135 + 140 +125 = 400 ∴ required percentage = 340/400 × 100 = 85% 141. (A) First we will find Quantity A, Quantity A: 10 × nC2 = 3 × n+1C3

∵ nCr = !

! !

⇒10 = 3

⇒ 10 = n + 1 ⇒ n = 9 ∴ n = 9 Now, Quantity B: nC4 = nC6 (∵ nCr = nCs then r = s or r+s = n) ⇒ n = 4 + 6 = 10 ∴ n = 10 ∴ Quantity B > Quantity A 142. (D) M and N are relatively prime. So, HCF of M and N will be 1. ⇒ HCF of 2M and 2N will be 2. Since, M > N, the least possible values they can have are 2 and 1. Then LCM will be 2. For any other values, M > 2 and hence LCM > 2. ∴ B ≥ A 143. (D) From statement I: The cost price is Rs. 20 less than selling price. With this, no relation with marked price can be established and hence its value cannot be found. ∴ Statement I alone is not sufficient to answer the question. From statement II: The discount offered is 15%. We know, Selling price = Marked price × (1 – discount percentage /100) We have two variables and one equation. Marked price cannot be found ∴ Statement II alone is not sufficient to answer the question. From statements I and II together: The cost price is Rs. 20 less than selling price. Selling price – 20 = Cost price The discount offered is 15%. We know, Selling price = Marked price × (1 – discount percentage /100) We have three variables and two equations. Marked price cannot be found. ∴ Using even both the statements together, we cannot answer the given question. 144. (C) From statement I: The ratio of length and breadth of rectangle is 12 : 7. Let the length be 12T and breadth be 7T. Area = 12T × 7T, which cannot be found if T is not known. ∴ Statement I alone is not sufficient to answer the question. From statement II: Length is 5 cm lesser than the diagonal. Let diagonal length be N and length be (N - 5). We know, diagonal length = √ (square of length + square of breadth) But, we have two variables, breadth and N, hence breadth and length cannot be found, and hence the area. ∴ Statement II alone is not sufficient to answer the question. From statements I and II together: The ratio of length and breadth of rectangle is 12 : 7. Let the length be 12T and breadth be 7T. Length is 5 cm lesser than the diagonal. Let diagonal length be (12T - 5).

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We know, diagonal length = √ (square of length + square of breadth) Using this equation, we can solve for variable T, and find area. Area = 12T × 7T ∴ Using both the statements together, we can answer the given question. 145. (D) From statement I: The average marks of all boys in class is 74. Only with this information, number of students cannot be found. ∴ Statement I alone is not sufficient to answer the question. From statement II: The average marks of all girls in class is 3 less than average marks of the class. If there are G girls and S students, then total marks of girls/G = (total marks of all students/S) – 3 Clearly, there are too many variables and S cannot be found. ∴ Statement II alone is not sufficient to answer the question. From statements I and II together: The average marks of all boys in class is 74. The average marks of all girls in class is 3 less than average marks of the class. If there are G girls and S students, then total marks of girls/G = (total marks of all students/S) – 3 Number of boys will be S - G. So, total marks of boys = 74(S - G) Still, there are too many variables and two equations and hence S cannot be found. ∴ Using even both the statements together, we cannot answer the given question. 146. (B) Total number of students who qualified the entrance exam from both the states together over all the years = 480 + 360 + 530 + 390 + 560 + 450 + 610 + 550 + 670 + 490 + 710 + 600 = 6400 Total number of girls who qualified the entrance exam from both the states together over all the years = 46% of 6400 = 0.46 × 6400 = 2944 ∴ Total number of boys who qualified the entrance exam from both the states together over all the years = 6400 – 2944 = 3456 147. (C) Total number of students who qualified in the entrance exam from state A over all the years together = 480 + 530 + 560 + 610 + 670 + 710 = 3560 Total number of students who qualified in the entrance exam from both the states together in the year 2013 and 2015 together = 610 + 550 + 710 + 600 = 2470 ∴ Required percentage = (3560/2470) × 100 = 144% 148. (B) The total number of students who qualified in the entrance exam in the year 2014 from both the states together = 670 + 490 = 1160 The total number of students from state B who qualified in the entrance exam over all the years together = 360 + 390 + 450 + 550 + 490 + 600 = 2840 ∴ Required % = (1160/2840) × 100 = 40.845% ≈ 41% 149. (E) From bar graph, The number of students who qualified in the entrance exam from state A in the year 2010 = 480 The number of students who qualified in the entrance exam from state B in the year 2012 = 450 ∴ Required ratio = 480 : 450 = 16 : 15 150. (C) % increase in the number of students who qualified in the exam from state B in the year 2013 as compared to the previous year = [(550 – 450)/450] × 100 = 22.22 ≈ 22% DIRECTION (151-153):- From the given data, The total number of students in the college = 19000 Given that the 25% of the total number of students belong to the discipline of Arts. ∴The total no. of students belong to the discipline of Arts = 19000 × (25/100) =4750 From the given data, it is clear that 6% of the students from Arts discipline have taken all the three subjects.

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∴The total no. of students from Arts discipline who took all the three subjects = 4750 × (6/100) =285 ……. (1) Given that the 35% of the total number of students belong to the discipline of Commerce. ∴The total no. of students belong to the discipline of Commerce = 19000 × (35/100) =6650 From the given data, it is clear that 14% of the students from Commerce discipline have taken all the three subjects. ∴The total no. of students from Commerce discipline who took all the three subjects = 6650 × (14/100) =931 ……. (2) Given that the 40% of the total number of students belong to the discipline of Science. ∴The total no. of students belong to the discipline of Science = 19000 × (40/100) =7600 151. (C) From the given data, it is clear that 18% of the students from Science discipline have taken all the three subjects. ∴The total no. of students from Science discipline who took all the three subjects = 7600 × (18/100) =1368 ……. (3) ∴The total no. of students from all disciplines who took all the three subjects = 285 + 931 + 1368 = 2584 [∵From (1), (2) and (3)]

We know that, to find `a’ is what percent of `b’ we have 100

∴The percentage of total number of students who have taken all the three subjects to the total number of students in the college= (2584/19000) × 100 =13.6 152. (D) Given that 4% of the students from Science discipline have taken only Cookery as a subject. ∴The total no. of students from Science discipline who took only Cookery as their subject = 7600 × (4/100) =304 ……. (3) ∴The total no. of students in the college who have taken only Cookery as their subject = 1900 + 931 + 304 = 3135 153. (A) Given that 35% of the students from the discipline of Science have taken Social Work ∴The total no. of students from the discipline of Science who took Social Work = 7600 × (35/100) = 2660 ………. (2) We know that, to find `a’ is what percent of `b’ we have 100

∴The percentage of the total no. of students from the discipline of Arts who took Social Work to the total no. of students taking the same subject from the discipline of Science = (1140/2660) ×100 = 42.857 ≈ 43 [∵From (1) and (2)] 154. (E) From the given data, The total number of students in the college having three disciplines = 19000 Given that the 25% of the total number of students belong to the discipline of Arts. ∴The total no. of students belong to the discipline of Arts = 19000 × (25/100) =4750 Given that 24% of the students from the discipline of Arts have taken Social Work ∴The total no. of students from the discipline of Arts who took Social Work = 4750 × (24/100) =1140 ………. (1) Given that 30% of the students from the discipline of Arts have taken Physical Training ∴The total no. of students from the discipline of Arts who took Physical Training = 4750 × (30/100)

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=1425 ………. (2) Given that the 35% of the total number of students belong to the discipline of Commerce. ∴The total no. of students belong to the discipline of Commerce = 19000 × (35/100) =6650 Given that 44% of the students from the discipline of Commerce have taken Social Work ∴The total no. of students from the discipline of Commerce who took Social Work = 6650 × (44/100) =2926 ………. (3) Given that 28% of the students from the discipline of Commerce have taken Physical Training ∴The total no. of students from the discipline of Commerce who took Physical Training = 6650 × (28/100) =1862 ………. (4) Given that the 40% of the total number of students belong to the discipline of Science. ∴The total no. of students belong to the discipline of Science = 19000 × (40/100) =7600 Given that 35% of the students from the discipline of Science have taken Social Work ∴The total no. of students from the discipline of Science who took Social Work = 7600 × (35/100) = 2660 ………. (5) Given that 43% of the students from the discipline of Science have taken Physical Training ∴The total no. of students from the discipline of Science who took Physical Training = 7600 × (43/100) =3268 ………. (6) ∴The total number of students who have taken Social Work and Physical Training from all the three disciplines = 1140 + 1425 + 2926 + 1862 + 2660 + 3268 = 13281 [∵From (1), (2), (3), (4), (5) and (6)] 155. (B) From the given data, The total number of students in the college having three disciplines = 19000 Given that the 35% of the total number of students belong to the discipline of Commerce. ∴The total no. of students belong to the discipline of Commerce = 19000 × (35/100) =6650 Given that the 40% of the total number of students belong to the discipline of Science. ∴The total no. of students belong to the discipline of Science = 19000 × (40/100) =7600 ∴The ratio of the total number of students from the discipline of Commerce to those from of Science = 6650:7600 = 7:8 Easy method: - The ratio of the total number of students from the discipline of Commerce to those from of Science= [19000 × (35/100)] : [19000 × (40/100)] = 35 : 40 = 7 : 8

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IBPS PO (MAINS) SPEED TEST - 3

ANSWER KEY

1(A) 2(D) 3(D) 4(A) 5(C) 6(D) 7(E) 8(C) 9(B) 10(C)

11(E) 12(A) 13(A) 14(A) 15(D) 16(D) 17(B) 18(D) 19(A) 20(E)

21(C) 22(A) 23(C) 24(D) 25(A) 26(B) 27(A) 28(E) 29(B) 30(A)

31(D) 32(D) 33(E) 34(B) 35(E) 36(C) 37(B) 38(A) 39(A) 40(C)

41(D) 42(E) 43(B) 44(B) 45(B) 46(A) 47(B) 48(B) 49(A) 50(A)

51(C) 52(D) 53(B) 54(C) 55(A) 56(B) 57(C) 58(B) 59(C) 60(B)

61(B) 62(A) 63(B) 64(C) 65(D) 66(B) 67(B) 68(B) 69(B) 70(C)

71(C) 72(A) 73(C) 74(E) 75(C) 76(D) 77(D) 78(B) 79(D) 80(C)

81(B) 82(A) 83(A) 84(D) 85(B) 86(B) 87(C) 88(C) 89(D) 90(E)

91(C) 92(E) 93(C) 94(C) 95(C) 96(A) 97(C) 98(B) 99(A) 100(E)

101(D) 102(C) 103(D) 104(E) 105(B) 106(B) 107(D) 108(E) 109(A) 110(B)

111(C) 112(D) 113(E) 114(A) 115(A) 116(D) 117(B) 118(C) 119(B) 120(C)

121(C) 122(C) 123(C) 124(A) 125(B) 126(C) 127(B) 128(C) 129(C) 130(D)

131(E) 132(A) 133(B) 134(C) 135(D) 136(A) 137(E) 138(C) 139(B) 140(C)

141(A) 142(D) 143(D) 144(C) 145(D) 146(B) 147(C) 148(B) 149(E) 150(C)

151(C) 152(D) 153(A) 154(E) 155(B)

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IBPS PO - MAINS - SPEED TEST - 3 -SOLUTIONS PO MAINS SPEED... · IBPS PO (MAINS) MAINS TEST -...

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