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A seller bought two items P and Q for Rs 1200 and sold it such that selling price of item P is _____ than cost price of item Q. He marked item P 60% above its cost price and sold it after giving a discount of Rs _____ . If average of marked price of item P and Q together is Rs 1500 and item Q is sold at a discount of 20%, then profit/loss earned on selling item Q is Rs ______ . The values given in which of the following options will fill the blanks in the same order in which is it given to make the statement true: I.50% more, 60, 1032 II. 60% less, 120, 1071 III.30% more, 615, 990 Only I Only II Only I and II Only I and III All I, II and III

The average marks obtained by all the students of a class in Science is 150. If the marks of 2 students had been less by 36 and 52 respectively, then the average marks had been 4 less. If the marks of all the students are arranged in A.P., then the difference between the consecutive terms came out to be same. If the highest marks obtained is 50% more than the lowest marks obtained, then which of the following can be determined using the given data? I.Number of students in the class II.The highest marks obtained by a student III. The marks obtained by the 8th student Only I Only III Only I and II Only II and III All I, II and III

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10. Answer: e) x = y or no relationship can be established Solution: From (I) x² - 37x - 4x + 148 = 0 (x - 37) (x - 4) = 0 x = 37, 4 From (II) y² + 29y - 9y - 261 = 0 (y + 29) (y - 9) = 0 y = -29, 9 9. Answer: b) x < y Solution: From (I) x² + 48x – 153 = 0 x² + 51x – 3x - 153 = 0 (x + 51) (x - 3) = 0 x = -51, 3 From (II) y² - 23y +132 = 0 y² - 11y – 12y + 28 = 0 (y - 11) (y - 12) = 0 y = 11, 12 8. Answer: a) x > y Solution: From (I) x² - 19x - 17x + 323 = 0 (x - 19) (x - 17) = 0 x = 19, 17 From (II) y² - 13y +7y - 91 = 0 (y - 13) (y + 7) = 0 y = 13, -7 7. Answer: e) x = y or no relationship can be established Solution: From (I) x² + 57x - 2x - 114 = 0 (x + 57) (x - 2) = 0 x = -57, 2 From (II) y² - 15y + 11y - 165 = 0 (y - 15) ( y + 11) = 0 y = 15, -11 6. Answer: a) x > y Solution: From (I) x² - 31x +198 =0 x² - 22x - 9x + 198 = 0 (x - 22) (x - 9) = 0 x = 22, 9 From (II) y² + 35y + 124 = 0 y² + 31y + 4y +124 = 0 (y + 31) (y + 4) = 0 y = -31, -4 5. Answer: c) x ≤ y Solution From (I) 2x² - 6x – 80 = 0 2x² - 16x + 10x - 80 = 0 2x(x - 8) + 10(x - 8) = 0 x = 8, - 5 From (II) 3y² - 54y + 240 = 0 3y² - 24y – 30y + 240 = 0 3y(y - 8) - 30(y - 8) = 0 (3y – 30) (y - 8) = 0 y = 8, 10 4. Answer: b) x < y Solution: From (I) 2x² + 31x - 51 = 0 2x² + 34x – 3x – 51 = 0 2x(x + 17) – 3(x + 17) = 0 x = - 17, 3/2 From (II) y² - 40y + 144 = 0 y² - 36y – 4y + 144 = 0 (y – 36) (y – 4) = 0 y = 36, 4 3. Answer: d) x ≥ y Solution From (I) 5x² + 18x = 35 5x² - 7x + 25x - 35 = 0 x (5x – 7) + 5(5x – 7) = 0 (5x – 7) (x + 5) = 0 x = + 7/5, - 5 From (II) 2y² + 21y + 55 = 0 2y² + 10y + 11y + 55 = 0 2y (y + 5) +11(y + 5) = 0 y = – 5, – 11/2 2. Answer: b) x < y Solution: From (I) x² + 29x + 2x + 58 = 0 (x + 29) (x + 2) = 0 x = -29, -2 From (II) y² - 9y – 4y + 36 = 0 (y - 9) (y - 4) = 0 y = 9, 4 1. Answer: e) x = y or no relationship can be established Solution: From (I) x² + 11x = - 28 x² + 7x + 4x + 28 = 0 (x + 7) (x + 4) = 0 x = -7, -4 From (II) y² + 19y – 92 = 0 y² + 23y – 4y - 92 = 0 (y + 23) (y - 4) = 0 y = -23, 4

Quadratic Eqn ( SBI | IBPS ) a) x > y b) x < y c) x ≤ y d) x ≥ y e) x = y or no relationship can be established 1. I. x² + 11x = - 28 II. y² + 19y – 92 = 0 2. I. x² + 31x + 58 = 0 II. y² - 13y +36 = 0 3. I. 5x² + 18x = 35 II. 2y² + 21y + 55 = 0 4. I. 2x² + 31x - 51 = 0 II. y² - 40y + 144 = 0 5. I. 2x² - 6x – 80 = 0 II. 3y² - 54y + 240 = 0

From the given we can find speeds of the car as 40 km/hr and 20 km/hr, and we cannot find which car is faster. So the given data is insufficient

In the first statement we can find the speed of car A 60=3 part 2 parts= 40 In the second statement the ratio is given for speed of car B and the sum of the speed of car A and B By subtracting the ratio we get the speed of car A 1x=40 2x=80 Speed of car B Required difference = 80-40=40Km/hr Therefore, both statements required to answer the questions

From i) For 450 rupees we can buy 4.5 liter of petrol Required mileage = 900/4.5 =200 km/lit From ii) we don’t know about honda bike So, statement I alone sufficient

From the statement i) we get the sum of the speed =60 km/hr From the statement ii) we get the Aspeed -BSpeed = 20 km/hr From these two-equation, speed of A = 40 km/hr So, both the statements needed to answer the question

From statement I we got only speed ratio and time ratio, from that we cant able to find time From both statement I and II we got speeds of A and B Speed of A is 60 km/hr Speed of B = (60/4)*3=45 km/hr Required time = (900/105) hr Both statements required to answer the question.

📚 Sbi Clerk 2022 | Ibps Po 2022 1) Find the time taken to cover a 300 km distance by car B. i) Sum of the speeds of the car A and car B is 60 km/hr ii) Difference between the speed of the car A and B is 20 km/hr. a) Statement I alone is sufficient. b) Statement II alone is sufficient. c) Both the Statement I and Statement II are required to answer the question. d) Both Statement I and Statement II together are not enough to answer the question. e) Either statement I or statement II is enough to answer the question. 2) Find the difference between the speed of car A and B and also given that speed of car C is 60km/hr? i) The ratio between the speed of car A and C is 2:3 ii) Ratio between the speed of car B and sum of the speeds of car A and B is 2:3. a) Statement I alone is sufficient. b) Statement II alone is sufficient. c) Both the Statement I and Statement II are required to answer the question. d) Both Statement I and Statement II together are not enough to answer the question. e) Either statement I or statement II is enough to answer the question 3) Find the mileage of Yamaha bike in km/litre, Bajaj bike gives 60 km/litre mileage and the petrol price per litre is 100/litre. i) Yamaha bike spends Rs. 450 to cover 900 km . ii) Mileage of Yamaha bike is 30% more than the mileage of Honda bike. a) Statement I alone is sufficient. b) Statement II alone is sufficient. c) Both the Statement I and Statement II are required to answer the question. d) Both Statement I and Statement II together are not enough to answer the question. e) Either statement I or statement II is enough to answer the question 4) The distance between A and B is 600 km. If the speed of A is greater than B then find the speed of A. i) If they started at same time travelled in opposite direction A and B meets in 10 hours ii) If they started at same time travelled in same direction A meets B in 30 hours a) Statement I alone is sufficient. b) Statement II alone is sufficient. c) Both the Statement I and Statement II are required to answer the question. d) Both Statement I and Statement II together are not enough to answer the question. e) Either statement I or statement II is enough to answer the question. 5) The distance between A and B is 900 km. Find the time taken to meet each other if they start at the same time. i) Speed of A is 33.33% more than the speed of B. ii) Speed of A is 60km/hr a) Statement I alone is sufficient. b) Statement II alone is sufficient. c) Both the Statement I and Statement II are required to answer the question. d) Both Statement I and Statement II together are not enough to answer the question. e) Either statement I or statement II is enough to answer the question. Join: @Quant_Genius

? × 350.01 ÷ 3.992 = 983 ÷ 14.001 After approximation, we get ? × 350 ÷ 16 = 980 ÷ 14 ⇒ ? × 350 ÷ 16 = 70 ⇒ ? = (70 × 16) ÷ 350 = 3.2

(1110.02 + 89.81) ÷ ? – √15 = 15.98 After approximation, we get (1110 + 90) ÷ ? – 4 = 16 ⇒ 1200 ÷ ? = 16 + 4 = 20 ⇒ ? = 1200 ÷ 20 = 60.

? ÷ 2.997 + 799.98 × 8.9 = 9199.978 After approximation, we get ? ÷ 3 + 800 × 9 = 9200 ⇒ ? = (9200 – 7200) × 3 = 2000 × 3 = 6000.

?= 49.99% of 6400.002 ÷ 999.99 ≈ 50% of 6400 ÷ 1000 = 3200 ÷ 1000 = 3.2

?^2 = 468.02 + 79.91 ÷ 5.01 ≈ 468 + 80 ÷ 5 = 468 + 16 = 484 = 22^2 ⇒ ? = 22