IBPS PO Previous Year Question Paper 2018

Quantitative Aptitude (Questions & Solutions)

1. Ratio of present ages of A and B is 16:7. After 12 years, A’s age is twice of B’s age then find present ages of A and B?

A. 64 years; 28 years B. 80 years; 35 years C. None of these D. 96 years; 42 years E. 102 years; 49 years

Answer D

Solution 1: Present ages of A and B are in the ratio 16:7, hence the ages can be represented as 16x and 7x respectively. After 12 years, A’s age is twice of B’s age, hence it can be put in equation as (16x + 12) / (7x + 12) = 2/1 16x + 12 = 14x + 24 2x = 12 x=6

Hence A’s present age = 16 ✕ 6 = 96 years

B’s present age = 7 ✕ 6 = 42 years

2. A man invested a certain sum in scheme A at 15% p.a. for 2 years and earned Rs 1950 as simple interest. He increased his sum by Rs. ‘x’ and invested in another scheme B at 10% p.a. C.I. for 2 years and received Rs. 1680 as compound interest. Find the value of ‘x’ ?

A. Rs. 1750 B. Rs. 1500 C. Rs. 1250 D. None of these E. Rs. 1850

Answer B

Solution 2:

SI = PTR / 100

1950 = (P ✕ 2 ✕ 15 ) / 100

P = (1950 ✕ 100 ) / (2 ✕ 15) P = Rs 6500 Compound Interest in 2 years at 10% per annum = 10+ 10+ [(10*10)/100] = 21% (6500+X) * (21/100) = 1680 X = 1500

3. In a class there are 30 girls and 15 boys and total average weight of class is 47 kg. Total average weight of boys is 58 kg. Find the average approximate weight of girls?

A. 32 Kg B. 42 Kg C. 52 Kg D. 35 Kg E. 50 Kg

Answer B

Solution 3:

Total weight of students in class = 47 ✕ (30+15) = 47 ✕ 45 = 2115 Kg

Total weight of Boys = 58 ✕ 15 = 870 Kg Total weight of girls = 2115 - 870 = 1245 Kg Average weight of Girls = 1245 / 30 = 41.5 Kg Hence approx. weight would be 42 Kg.

4. Ram bought a bike at 20% discount on MRP. After 1 year Ram sell the bike to Ramesh at 10% loss. After 1 year more Ramesh sell the bike at 20% profit to Ranjan. If Ranjan paid Rs. 1,29,600, then find the M.R.P. of the bike ?

A. 1,50,000 B. 2,25,000 C. 1,40,000 D. 2,00,000 E. 1,80,000

Answer A.

Solution 4: As the bike is bought at a discount of 20%

Rams Cost Price = Market Price ✕ (80/100) As the bike is sold at 10% discount

Ramesh’s Cost Price = Market Price ✕ (80/100) ✕ (90/100)

Ranjan’s Cost price = Market Price ✕ (80/100) ✕ (90/100) ✕ (120/100)

129600 = Market Price ✕ (80/100) ✕ (90/100) ✕ (120/100) Market Price = Rs 1,50,000/- 5.Directions: What value will come in place of question mark (?) in the number series given below?

14, 28, 84, 336, (?)

A. 1800 B. 1780 C. 1728 D. 1680 E. None of these

Answer. D.

Solution 5:

The pattern of series is every number is multiplied by 2. 14 × 2 = 28 28 × 3 = 84 84 × 4 = 336 336 × 5 = 1680 6. Directions: What should come in place of question mark (?) in the following number series? 3, 7, 19, 39, (?), 103 A. 49 B. 51 C. 53 D. 67 E. 63 Answer. D. Solution 6:

7 = 4 ✕ 1 + 3

19 = 4 ✕ 3 + 7

39 = 4 ✕ 5 + 19

X= 4✕7 + 39 = 28 + 39 = 67

7. Direction: What should come in place of question mark (?) in the following question? (You do not

have to calculate the exact value.)

5030.05 ÷42.93 + 24.49 % of 5049.93 100 =? A. 150 B. 170 C. 130 D. 90 E. 110 Answer. C. Solution 7: 5030.05 ÷ 42.93 + 24.49 % of 5049.93 ÷ 100 = ? = 5030 ÷ 43 + 24.5 % of 5050 ÷ 100 = 116.9764 + 1237.25 ÷100 = 130

8.Direction: What approximate value should come in place of question mark (?) in the following questions? (39.99)2 - (9.9)2 - (15.1)2 = ? A. 1675 B. 1500 C. 1725 D. 1375 E. 1275 Answer E Solution 8: (39.99) 2 – (9.9)2 - (14.9)2 The question says find approx value so we can round off the numbers to next whole number = (40)2 – (10)2 - (15)2 = 1275 9. Direction: What should come in place of question mark (?) in the following question? (You do not have to calculate the exact value.) 1325×√17 + 20% of 508.24 – ¾ of 85.39=? A. 5500 B. 5338 C. 5860 D. 4900 E. 5950 Answer. B. Solution 9: 1325√17 + 20% of 508.24 – ¾ of 85.39=? Once again we just have to find an approximate value so we can find square root of 16 for easier calculation. And we can round off 85.39 to 84. Hence, =1325 * 4 + 101 – 63 = 5300 +101-63 = 5338 10. Direction: You are required to calculate approximate value of the sums given below (7.02)2 × (360.85)1÷2 × (31.98)2 ÷ [(7.99)3 - (16.02)2] = ? A. 3152 B. 3955 C. 4251 D. 4584 E. 3724 Answer. E.

Solution 10: Once again the job is to find approximate value, Hence round off 7.02 to 7 Round of 360.85 to 361 because square of 19 is 361. Round of 31.98 to 32 Round off 7.99 to 8 Round off 16.02 to 16 = 72 × √361 × 322 ÷ (83-162) = 49×19×32×32÷(512-256) = 49×19×32×32÷(256) = 3724 11. A) x2 – 10x + 25 = 0 B) y2 = 25 A. x > y B. x < y C. x ≤ y D. x ≥ y E. x = y or no relation can be established. Answer. D. Solution 11. A. x2 – 10x + 25 = 0 x2 – 5x - 5x + 25 = 0 x = +5 B. y2 = 25 Y = +5,-5 So x ≥ y 12. A) x2 – 36x + 324 = 0 B) y2 – 42y + 441 = 0 A. x > y B. x < y C. x ≤ y D. x ≥ y E. x = y or no relation can be established. Answer B Solution 12: x2 – 36x + 324 = 0 x2 – 18x – 18x + 324 = 0 x=18 B. y2 – 42y + 441 = 0 Y2 – 21y – 21y + 441 = 0 y=21 x<y 13.Two trains leave Delhi for Kolkata at 4:00 am and 4:30 am and travel at a speed of 50 Kmph and 75 Kmph respectively, after how many kilometres from Delhi will both trains be together?

A. 85 Km B. 75 Km C. 45 Km D. 55 Km E. None of these Answer. B. Solution 13. In 30 minutes duration the train which departs at 4:00 AM from Delhi, with 50 Km speed will reach a distance of 25 Km And their relative speed is 25 Km/h Hence time taken 25/25 =1Hr Distance from Delhi when the two trains will be together 75*1 =75 KM 14.John bought a machine for Rs. 50,000 and spent Rs. 2000 on repairs and Rs. 500 on transport and sold it with 20% profit. What price (in Rs.) did he sell the machine? A. 62000 B. 60000 C. 61000 D.63000 E. None of these Answer. D Solution 14. Cost Price (50000 + 2000 + 500) = Rs. 52,500 Profit = 20% Hence, Selling price = (120/100) 52500 = Rs. 63,000 15. The average age of a group of some persons is 16.75 years. By joining 20 new persons with an average age of 13.25 years, the average age of the group becomes 15 years. Find out the number of people in the group initially. A. 20 B. 21 C. 23 D. 24 E. 26 Answer. A Solution 15: Let the number of persons in the group Initially be x, then x × 16.75 + 20 × 13.25 = (x + 20) × 15 Solve the above equation to find the value of X. x = 20

16. If the ratio of incomes of A and B in 2001 is 2 : 3 and the ratio of incomes of A in 2001 and 2002 is 4 : 5. Find the expenditure of A in 2002, if saving in the same year is Rs. 4000. It is given that in 2001 the sum of income of A and B is Rs. 25000. A. Rs. 5000 B. Rs. 10500 C. Rs. 9500 D. Rs. 7500 E. Rs. 8500 Answer. E Solution 16: Ratio of incomes of A and B in 2001 = 2:3 Ratio of incomes of A in 2001 and B in 2002 = 4:5 Multiply the first ratio by 2. Then A’s income in 2001 = 4x B’s income in 2001 = 6x A and B income in 2001 = 25000 4x + 6x =10x But 10x = 25000 Hence x = 2500 A’s income in 2001 = 4x = 4*2500 = Rs 10000 B’s income in 2001 = 6x = 6*2500 = Rs 15000 A’s income in 2002 = 5x = 5*2500 = Rs 12500 Savings of A in 2002 = Rs 4000 Expenditure = Income – Savings 12500 – 4000 = Rs 8500 17.The ratio of ages of Ram and Shyam is 2:6 and after 5 years the ratio of their age becomes 6 : 8. What will be their average age (in years) after 10 years? A. 12 B. 13 C. 14 D. 15 E. 16 Answer. A. Solution 17:

Let the current ages be y and 3y Their ages after 5 years will be y+5 and 3y+5 respectively. (y+5)/(3y+5) = ¾ Y= 1 So, their current ages are 1 & 3 years and after 10 years the average age be 12 years

18.One container contains a mixture of spirit and water in the ratio 2: 3 and another contains the mixture of spirit and water in the ratio 3: 2. How much quantity from the second should be mixed with 10 litres of the first so that the resultant mixture has ratio of 4: 5? A.2.86 litres B. 3.45 litres C. 4.31 litres D. 5.67 litres E. 8.94 litres Answer. A Solution 18: Ratio of mixture of spirit and water in Container 1 = 2: 3 Amount of mixture taken = 10 litres Amount of spirit =( ⅖ )×10 = 4 litres Amount of water = (⅗ )×10 = 6 litres Ratio of mixture of spirit and water in Container 2 = 3: 2 Amount of mixture taken = x litres Amount of spirit = (⅗)× (x) = 3x/5 litres Amount of water = (⅖)× (x) = 2x/5 litres Ratio of mixture of spirit and water in resultant mixture = 4: 5 Hence, (4+3x/5) / (6+2x/5) = 4/5 (20/5+3x/5) / (30/5+2x/5)= 4/5 (20+3x) / (30+2x) = 4/5 100+15x =120+8x X = 2.86 litres 19.Direction: Find the wrong term in the following number series? 0.5 , 2, 1, 4, 32, 512 A. 0.5 B. 2 C. 4 D. 32 E. 512 Answer B Solution 19: 0.5, 2, 1, 4, 32, 512 512÷ 24 = 32 32 ÷ 23 = 4 4 ÷ 22 = 1 1 ÷ 21= 0.5 ≠ 2 0.5 ÷ 20 = 0.5 Hence 2 is wrong term, because in place of 2 the right answer for the series is 0.5

Directions (20-24): Line chart given below shows number of labors (men and women) working in 6 different years. Study the data carefully and answer the following questions.

20. Total number of Men working in 2012 and 2013 together is what percent of the total number of labourers (Men + Women) working in 2014?

A. 60% B. 70% C. 80% D. 90% E. 40%

Answer. D Solution 20: Required percentage = [ (120+240) / (160+240) ] × 100 = (360 / 400) × 100 = 90% 21. Average number of Women working in 2014, 2015 and 2016 together is how much more/less than the average number of Men working in 2011, 2014 and 2016 together?

A. 100 B. 80 C. 90 D. 70 E. None of the given options

Answer A Solution 21:

Average number of women working in 2014, 2015, 2016 together = (240+360+300) / 3 = 900/3 = 300 Average number of men working in 2011, 2014, 2016 together = (80+160+360) / 3 = 600/3 = 200 Therefore 300 - 200 = 100. Hence the average number of women working in 2014, 2015, 2016 together is more than the average number of men working in 2011, 2014, 2016 together by 100. 22. Number of men working in 2017 is 15% more than that of 2015 while number of women working in 2017 is 40% less than that of 2014. Find the total number of labours (Men + Women) working in 2017? A. 561 B. 456 C. 489 D. 594 E. 630 Answer C Solution 22: Number of men working in 2017 is 15% more than 2015. Number of men in 2015 is 300. = (115/100) × 300 = 345 Number of women working in 2017 is 40% less than in 2014. Number of women in 2014 is 240. = (60/100) × 240 = 144 Total number of labourers working in 2017 = 345+144 = 489 23. Find the ratio between total number of Labourers working in 2012 and 2013 together to total number of labourers working in 2015 and 2016 together? A. 2:1 B. 1:2 C. 35:66 D. 11:10 E. None of the above

Answer B Solution 23: Required ratio = (120+180+240+120) / (300+360+360+300) = 660/1320 = 1:2 24. Total number of Men working in all six years is how much more/less than total number of Women working in all six years together?

A. 140 B. 160 C. 180 D. 200

Answer D Solution 24:

Total number of men working in all 6 years = 80+120+240+160+300+360 =1260 Total number of women working in all 6 years = 260+180+120+240+360+300 = 1460 Required difference = 1460 - 1260 = 200

Direction (25-29): There are 450 coupons which can be used in Pedicure and Hair cutting. Ratio between Males to Females who use their coupons in Hair cutting is 13 : 7 Number of males who use their coupons in Pedicure is 72 more than number of females who use their coupon in Hair cutting. Total number of males who use their coupon in Pedicure and Haircutting together is 174 more than total number of females who use their coupon in Pedicure and Haircutting together. 25. Males who use their coupon in pedicure is what percent of the Males who use their coupons in haircutting? A. 200% B. 100% C. None of the given options D. 0% E. 150%

Answer B Solution 25: Number of males who use their coupon in haircutting = 13x Number of females who use their coupon in haircutting = 7x Number of females who use their coupons in pedicure = 450 - 13x - 7x - 7x - 72 = 378- 27x Total number of males who use their coupon on pedicure and haircutting 7x + 72 + 13x = 174 + 378 - 27x + 7x x = 12 For Pedicure Number of males = 156 Number of females = 54 For Haircutting Number of males = 156 Number of females = 84 Required percentage = (156/156) × 100 = 100%

26. Find the ratio between Total number persons who use their coupons in Pedicure to the total number of persons who use their coupons in Haircutting? (a) 52 : 23 (b) None of the given options (c) 8 : 9 (d) 8 : 7 (e) 7 : 8 Answer E Solution 26:

Required Ratio = (156+64) / (156+84) = 7:8 27. Females who use their coupon in Haircutting is how much more than Females who use their coupon in Pedicure?

A. 15 B. 45 C. 30 D. None of the given options E. 60

Answer C Solution 27: Required difference = 84-54 = 30 28. Out of males who use their coupons in Haircutting, 25% belongs to city A, then find number of males who use their coupons in Haircutting which doesn’t belongs to city A?

A. None of the give options B. 108 C. 126 D. 117 E. 135

Answer D Solution 28: Number of males who use their coupons in Haircutting which doesn’t belong to City A = 156 × (75/100) = 117 29. Ratio between Males who use their coupon in Pedicure to that of in Spa is 4 : 5, while ratio between Females who use their coupon in Haircutting to that of in Spa is 6 : 11. Find total number of people who use their coupons in Spa? (a) 349 (b) 481 (c) 300 (d) 440 (e) None of the given options Answer A Solution 29: Number of Males who use their coupons in spa = 156 × (5/4) = 195 Number of Females who use their coupon in spa = 84 × (11/6) = 154 Total number of people who use their coupon in spa = 195 + 154 = 349

Directions (30-35): In each of these questions, two equations (i) and (ii) are given, you have to solve both the equations and give answer accordingly

30. (I) 2x2 + 9x + 9 = 0 (II) 15 y2 + 16 y + 4 = 0 A. X>Y B. X<Y C. X≥Y D. X≤Y E. X=Y or no relation can be established between X & Y.

Answer B Solution 30: (i) 2x2 + 9x + 9 = 0 2x2 + (6+3)x + 9 = 0 2x (x+3) +3( x+3) = 0 X = -3/2, -3 (ii) 15 y2 + 16 y + 4 = 0 15 y2 + 10 y + 6y + 4 = 0 5y (3y+2) + 2(3y+2) = 0 Y = - ⅖, - ⅔ X<Y 31. (I) 2x3 = √256 (II) 2Y 2 - 9Y + 10 = 0 A. X=Y or no relation can be established between X & Y B. X<Y C. X≤Y D. X≥Y E. X>Y

Answer C Solution 31: (i) 2x3 = 16 X3 = 8 x= 2 (ii) 2y2 - 9y +10 = 0 2y2 - (5+4)y + 10 = 0 2y2 - 5y - 4y + 10 = 0 y(2y-5) - 2(2y-5) = 0 y= 2, 5/2 X ≤ Y 32. (I) 6x2 - 11x + 4 = 0 (II) 3 y 2 - 5 y + 2 = 0 A. X≤Y B. X<Y

C. X≥Y D. X>Y E. X=Y or no relation can be established between x & y.

Answer E Solution 32: (i) 6x2 - 11x + 4 = 0 6x2 - (8+3) x + 4 = 0 6x2 - 8x - 3x + 4 = 0 2x(3x-4) - 1(3x-4) = 0 X= ½, 4/3 (ii) 3y2 - 5y + 2 = 0 3y2 - (3+2) y + 2 = 0 3y2 - 3y - 2y + 2 = 0 3y(y-1) -2( y-1) = 0 Y = ⅔, 1 No relation between x and y 33. (I) 3x2 + 11x + 10 = 0 (II) 2y2 + 11y + 14 =0 A. X≥Y B. X≤Y C. X>Y D. X<Y E. x = y or no relation can be established between x & y

Answer A

Solution 33: 3x2 + 11x + 10 = 0 3x 2 + 6x + 5x + 10 = 0 3x (x+2) + 5 (x+2) = 0 X = -2 , -5/3 (ii) 2y2 + 11y + 14 = 0 2y2 + 7y + 4y + 14 = 0 y(2y+7) + 2(2y+7) = 0 Y = -2, -7/2 X ≥ Y 34. (i) 12x² + 11x + 2 = 0 (ii) 12y²+ 7y + 1 = 0

A. x ≥ y B. x = y or no relation can be established between x & y. C. x < y D. x ≤ y E. x > y

Answer B

Solution 34: (i) 12x2 + 8x + 3x + 2 = 0 4x(3x+2) +1(3x+2) = 0 X = -⅔, -¼ (ii) 12y2 + 7y+ 1 = 0 12y2 + 4y + 3y + 1 = 0 4y(3y+1) +1( 3y+1) = 0 Y = -⅓, -¼ No relation between x and y 35. (I) 21x² + 10x + 1 = 0 (II) 24y²+ 26y + 5 = 0

A. x ≤ y B. x = y or no relation can be established between x & y. C. x ≥ y D. x > y E. x < y

Answer B Solution 35: (i) 21x2 + 10x + 1 = 0 21x2 + 7x + 3x+ 1= 0 7x(3x+1) +1( 3x+1) = 0 X = -⅓, -1/7 (ii) 24y2 + 26y + 5 = 0 24y2 + (20+6)y + 5 = 0 24y2 + 20y + 6y + 5 = 0 4y (6y+5) +1(6y+5) = 0 Y = -⅚, -¼ No relation between X and Y.