BASE INSTITUTE - NAMAKKAL | www.ibpsguide.com | PH: 900 37 111 66 SBI PO Prelims 2017 – Practice Mock Test-19 (Based on Previous Year Pattern) Solution REASONING ABILITY (Direction 1-5):.

No box is a pencil (E) + Some pencils are buses (I) = E + I = O* = Some buses are not boxes. So there exists a possibility that all boxes are buses. Hence I follows but II does not follow. 20) Answer: A) There is no negative statement. Hence I follows and II does not follow. 21) Answer: D)

1) A 2) E 3) B 4) A 5) C 6) A 7) A 8) E 9) C 10) D Direction(11-15): SHOP NO PERSON SHOP

1 2 3 4 5 6 7

Sweets Watches Electronics Cloths Mobiles Crockery Shoes

Mohan Roshant Rakesh Santosh Rupesh Rajesh Kamlesh

11) C 12) B 13) A 14) D 15) E 16) Answer: B) All bottles are cups (A) + No cup is a desk (E) = A + E = E = No bottle is a desk. Now, No bottle is a desk (E) + All desks are stickers (A) = E + A = O* = Some stickers are not bottles. Hence, I does not follow. Again, No cup is a desk (E) + All desks are stickers (A) = E +A= O* = Some stickers are not cups, But, there exists a possibility in conclusion II. Hence conclusion II follows.

Vertical distance = HG – [AB – (FE – CD)] = 17 – [7 – (7-5)] = 12 m Horizontal distance = BC + [DE – (IH – FG)] = 8 + [3 – (8 – 6)] = 9 So IA = √(122 + 92) = 15 m 22) Answer: C) He took 2 turns to reach to point G, so after 8 m south he took a left then, walked (3+6) = 9 m and then turned left again to reach point G (here he covered (3+7) = 10 m) So total distance = 8 + 9 + 10 = 27 m 23) Answer: D)

17) Answer: E) All bottles are cups (A) + No cup is a desk (E) = A+ E = E = No bottle is a desk. Hence conclusion I follows. No bottle is a desk + All desks are stickers = E +A = O* = Some stickers are not bottles. But this does leave the possibility in II. 18) Answer: B) Some pencils are buses (I) +All buses are toys (A) = I + A= I = Some pencils are toys. Hence, conclusion I does not follow. And there exists a possibility of all toys being pencils.

24) Answer: C) P is brother of D. So D is female (since M has 1 daughter and one son). So if E is married to D, he is brother-in-law of P.

19) Answer: A)

25) Answer: D)

BASE INSTITUTE – NAMAKKAL | www.ibpsguide.com | PH: 900 37 111 66 The order is – F, D/A, A/D, C, B, E (in descending order of height) There are 2 people shorter than C. B is shorter than A and E is shorter than C, So B and E both are shorter than C. Since B is not the shortest, so E is the shortest. E = 145 cm, A = 157 cm 26) C

53) E 54) D 55) B 56) D 57) B 58) A 59) D 60) C 61) A 62) D 63) B 64) A 65) C 66). Answer: e) 8 2/3 + 6 4/5 = (8 + 6) + (10/15 + 12/15) = 14 + 1 7/15 = 15 7/15

27) Answer: C) Forward – RT, IL

67). Answer: b) 2 + [2 + {2 + (2 + (2 / 23)}] = 2 + 6 + 1/4 = 33/4

28) Answer: B) MAHARASHTRA changes to LBGBQBRISSZ There are 2 S’s so makes a pair

68). Answer: a) ? = 118500 - 33/4 * 4500 = 118500 - 37125 = 8.1375 * 104

29) C 30) A

69). Answer: c) (492 * 324) / 324 = 492

31) Answer: A) Q ≥ S ≥ D ≥ I. So Q ≥ I 32) Answer: E) K > A ≥ D , so K > D C > A ≥ D ≥ S so C > S 33) Answer: E) E > S > N = P, so E > P F ≤ N < S < E, so F < E 34) Answer: B) P > E ≥ L > F < D, so relation cant be determined between P and D E ≥ L > F, so E > F 35) Answer: C) E < A > S ≤ X, so relation cant be determined between E and X, but is for sure that either E is greater than X or is less than equal to X, so either or between the two ENGLISH LANGUAGE 36). D) Replace 'performance' with `performer' 37). E) 38). D) Replace 'over' with 'on' 39). C) Replace 'in' with 'among' 40). C) Replace 'him' with 'them' Directions (41-45): CGFAEBD 41) C 42) B 43) E 44) E 45) D 46) C 47) A 48) B 49) D 50) A 51) A 52) C

70). Answer: a) 20% of 40 + ? % of 80 = 120  800/100 + ?/100 * 80 = 120  ? = 140 71). Answer: d) Required ratio = (75 + 65) / (85 + 95) = 140 / 180 = 7 / 9 . 72). Answer: c) Required percentage = (70 + 80) / (95 + 110) 100% = 150 / 205 x 100% = 73.17%

x

73). Answer: e) Average sales (in thousand number) of branches B1, B3 and B6 in 2000 = (80 + 95 + 70) / 3 = 245 / 3 Average sales (in thousand number) of branches B1, B2 and B3 in 2001 = (105 + 65 + 110) / 3 = 280 / 3 Required percentage = (245/3) / (280/3) x 100 % = 245 / 280 x 100% = 87.5% 74). Answer: b) Average sales of all the six branches (in thousand numbers) for the year 2000 = 1 / 6 x [80 + 75 + 95 + 85 + 75 + 70] = 80. 75). Answer: d) Total sales of branches B1, B3 and B5 for both the years (in thousand numbers) = (80 + 105) + (95 + 110) + (75 + 95)

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BASE INSTITUTE – NAMAKKAL | www.ibpsguide.com | PH: 900 37 111 66 = 560. 76). Answer: c) Formula: If an article is sold at a price of S1, a person loses a certain amount. If the article is sold at a price of S2, he gains the same amount. Then, cost price = (S1+S2) / 2 According to that, Cost price = (1920+1280) / 2 = 1600 Required selling price = 1600 + 1600×1/4 = 2000 77). Answer: d) Speed of boat in still water = 25 km/hr Speed upstream =10 / 1 = 10 km/hr Speed of the stream = (25-10) = 15 km/hr Speed downstream = (25+15) = 40 km/hr Time taken to travel 10 km downstream = 10 / 40 hours = 10 × 60 / 40 = 15 minutes 78). Answer: a) A : B : C = 20000 × 24: 15000 × 24: 20000 × 18 = 20 × 4 : 15 × 4 : 20 × 3 = 4 × 4 : 3 × 4 : 4 × 3 =4:3:3 B's share = 25000 × 3 / 10 = 7500 79). Answer: c) The denominator be P, the numerator will be (P - 8). The fraction will be (P - 8)/P. Adding 3 to the numerator and subtracting 3 from the denominator, (P - 8 + 3)/(P - 3) = 3/4. (P - 5)/(P - 3) = 3/4 P = 20 - 9  P = 11. The fraction is: 3/11. 80). Answer: b) Let the age of father and son 10 years ago be 3x and x years respectively. Then, (3x + 10) + 10 = 2[(x + 10) + 10] 3x + 20 = 2x + 40 => x = 20. Required ratio = (3x + 10): (x + 10) = 70: 30 = 7: 3

83). Answer: a) I. 9a2 + 3a + 15a + 5 = 0  (3a + 5)(3a + 1) = 0  a = -5/3, -1/3 II. 2b2 + 8b + 5b + 20 = 0  (2b + 5)(b + 4) = 0  b = -5/2, -4 a > b. 84). Answer: e) I. (a - 5)(a - 4) = 0  a = 5, 4 II. (2b + 3)(b - 4) = 0  b = 4, -3/2 a ≥ b 85). Answer: b) I. (a + 6)(a + 5) = 0  a = - 6, -5 II. (b + 5)(b + 1) = 0  b = -5, -1 a ≤ b 86). Answer: e) 8×2 – 2 = 14 14×4 – (2+6) = 48 48×6 – (8+10) = 270 270×8 – (18+14) = 2128 2128×10 – (32+18) = 21230 87). Answer: b) 2×6 + 62 = 48 48×5 + 52 = 265 265×4 + 42 = 1076 1076×3 + 32 = 3237 3237×2 + 22 = 6478 88). Answer: c) 1010 – 152 = 785 785 – 132 = 616 616 – 112 = 495 495 – 92 = 414 414 – 72 = 365

81). Answer: b) a3 = 1331 => a = 11 b2 = 121 => b = ± 11 a≥b

89). Answer: d) (970-30) ÷ 2 = 470 (470-26) ÷ 2 = 222 (222-22) ÷ 2 = 100 (100-18) ÷ 2 = 41 (41-14) ÷ 2 = 13.5

82). Answer: d) I. a2 - 13a + 42 = 0  (a - 6)(a - 7) = 0  a = 6, 7 II. b2 - 15b + 56 = 0  (b - 7)(b - 8) = 0  b = 7, 8 a ≤ b

90) Answer: b) 8×1 + 1×8 = 16 16×2 + 2×8 = 48 48×3 + 3×8 = 168 168×4 + 4× 8 = 704 704 ×5 + 5×8 = 3560 91). Answer: a)

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BASE INSTITUTE – NAMAKKAL | www.ibpsguide.com | PH: 900 37 111 66 Percentage of money spend on Tennis = (45 / 360 x 100) % = 12 ½ %

4500(R1 - R2) = 1350 R1 - R2 = 0.3%

92). Answer: d) Let the total spending on sports be Rs. x. Then, Amount spent on Golf = Rs. (36/360 * x) = Rs. x / 10 Amount spent on Hockey = Rs (63/360 * x) = Rs. 7x / 40 Difference = Rs. (7x/40 – x / 10) = Rs 3x / 40 Therefore, Required percentage = Rs. ((3x/40 ) / (x/10) x 100) % = 75%.

97). Answer: c) Let the speed of the train be x km/hr and that of the car be y km/hr. Then, 120/x + 480/y = 8 or 1/x + 4/y = 1/15 --- (i) And, 200/x + 400/y = 25/3 or 1/x + 2/y = 1/24 --- (ii) Solving (i) and (ii), we get x = 60 and y = 80 Ratio of speeds = 60: 80 = 3: 4

93). Answer: a) Amount spent on Basketball exceeds that on Tennis by: = Rs. ((50 - 45) / 360 x 1,80,00,000) = Rs. 2,50,000. 94). Answer: c) Let the total spending on sports be Rs. x. Then, Amount spent on Cricket = Rs. (81/360 * x) = Rs. 9x/40 . Amount spent on Football = Rs. (54/360 * x) = Rs. 3x/20 . Difference = Rs. (9x/40 – 3x/20) = Rs. 3x /40 . Therefore, Required percentage = Rs. ((3x/40) / (9x/40) x 100)% = 33 1/3 %. 95). Answer: b) Amount spent on Cricket and Hockey together = Rs. ((81 + 63) / 360 x 2) crores = Rs. 0.8 crores = Rs. 80,00,000. 96). Answer: e) (1500 * R1 * 3)/100 - (1500 * R2 * 3)/100 = 13.50

98). Answer: a) Perimeter of the sector = length of the arc + 2(radius) = (135/360 * 2 * 22/7 * 21) + 2(21) = 49.5 + 42 = 91.5 cm 99). Answer: c) (A + B + C)'s 1 day work = 1/6; (A + B)'s 1 day work = 1/8 (B + C)'s 1 day work = 1/12 (A + C)'s 1 day work = (2 * 1/6) - (1/8 + 1/12) = (1/3 - 5/24) = 1/8 So, A and C together will do the work in 8 days. 100). Answer: b) 4/12 + x/15 = 1 x = 10

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