Hello Aspirants. Welcome to Online Quant Section in AffairsCloud.com. We are starting **IBPS Clerk course 2015 **and we are creating sample questions in **Quantitative Aptitude** section, type of which will be asked in IBPS Clerk Prelims Exam.

**Stratus – IBPS Clerk Course 2015 **

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**x**A) x < y B) x ≥ y C) x = y or relationship cannot be determined D) y ≥ x E) x > y^{2}+ 6x + 8 = 0, y^{2}– 2y – 8 = 0**D) y ≥ x****Explanation:**x^{2}+ 6x + 8 = 0 (x+4) (x+2) = 0 x = -4, -2 y^{2}– 2y – 8 = 0 (y-4)(y+2) = 0 Y = 4, -2 Put on number line as -4 -2 4 We see that when y = 4, then y is greater than both values of x And when y = -2, then y is greater than one value of x (-4) and equal to other value of x (-2) So overall we get**y ≥ x****2x**A) y ≥ x B) x = y or relationship cannot be determined C) y > x D) x > y E) x ≥ y^{2}– 10x = -12, 2y^{2}+ y – 15 = 0**B) x = y or relationship cannot be determined****Explanation:**2x^{2}– 10x = -12 2x^{2}– 10x + 12 = 0 2x^{2}-6x – 4x + 12 = 0 2x(x-3) + (-4)(x-3) = 0 (2x-4) (x-3) = 0 x = 2, 3 2y^{2}+ y – 15 = 0 2y^{2}+ 6y – 5y – 15 = 0 2y(y+3)+ (-5)(y+3) = 0 (2y-5)(y+3) = 0 Y = 2.5, -3 Put on number line as -3 2 2.5 3 When we see that on the number line, x and y have alternate values, then**relationship cannot be determined**between them.**3x**A) x = y or relationship cannot be determined B) y ≤ x C) x < y D) y < x E) x ≤ y^{2}+ 21x + 36 = 0, y^{2}+ 13y + 42 = 0**D) y < x****Explanation:**3x^{2}+ 21x + 36 = 0 3x^{2}+ 12x + 9x + 36 = 0 3x(x+4) + 9(x+4) = 0 (3x+9)(x+4) = 0 x = -4, -3 y^{2}+ 13y + 42 = 0 (y+7)(y+6) = 0 Y = -6, -7 Put on number line as -7 -6 -4 -3 We see both values of x as greater than both values of y, so**x > y****x**A) x ≤ y B) y < x C) y ≤ x D) x = y or relationship cannot be determined E) y > x^{2}+ 10x + 24 = 0, 3y^{2}– 6y – 72 = 0**A) x ≤ y****Explanation:**x^{2}+ 10x + 24 = 0 (x+6) (x+4) = 0 x = -6, -4 3y^{2}– 6y – 72 = 0 3y^{2}– 18y + 12y -72 = 0 3y(y-6)+ 12(y-6) = 0 (3y+12) (y-6) = 0 y = -4, 6 Put on number line as -6 -4 6 We see that when y = 4, then y is greater than both values of x And when y = -4, then y is greater than one value of x (-6) and equal to other value of x (-4) So overall we get**y ≥ x****x**A) x > y B) x ≤ y C) y > x D) y ≤ x E) x = y or relationship cannot be determined^{2}– x – 12 = 0, y^{2}+ 11y + 28 = 0A) x > y**Explanation:**x^{2}– x – 12 = 0 (x-4) (x+3) = 0 x = 4, -3 y^{2}+ 11y + 28 = 0 (y+7)(y+4) = 0 Y = -4, -7 Put on number line as -7 -4 -3 4 We see both values of x as greater than both values of y, so**x > y****Rs 1200 is divided among 2 men, 3 women and 4 children such that the share of a man, a woman and a child is in the ratio of 5 : 6 : 8. What is the share of a child?**A) Rs 40 B) Rs 60 C) Rs 50 D) Rs 80 E) None of these**A) Rs 40****Explanation:**The ratio of share are 2*5 : 3*6 : 4*8 5 : 9 : 16 So share of 16 children = 16/(5+9+16) * 1200 = 640 So 1 child receives 640/16 = Rs 40**2 men or 3 women or 5 children can do a work in 62 days. Then in how many days will 1 man, 1 woman and 1 child do the same work?**A) 35 days B) 50 days C) 40 days D) 60 days E) None of these**D) 60 days****Explanation:**For these type of questions we have a direct formula as 62*2*3*5/ (2*3 + 3*5 + 2*5) = 60 days Remember the formula can be used only when to find days for 1 man, 1 woman and 1 child**An article is sold at 30% profit. It its CP is increased by Rs 10 and SP decreased by Rs 11, the percentage of profit deceases by 15%. Find the cost price of article.**A) Rs 160 B) Rs 140 C) Rs 150 D) Rs 135 E) Rs 175**C) Rs 150****Explanation:**Let CP = x Then SP = 130/100 * x When CP = x+10, SP = 130/100 * x – 11, profit% = 30-15 = 15% So 115/100 * (x+10) = 130/100 * x – 11 Solve, x = 150**From 6 males and 5 females, a committee is to formed consisting of 5 members, Find the number of ways in which this can be done such that one male is always to be included.**A) 226 B) 210 C) 250 D) 462 E) 380**B) 210****Explanation:**Since 1 male is included always, we remain with 5 males and 5 females (total 10) from which a committee of 4 is to be formed now. So ways are^{10}C_{4}= 10*9*8*7/4*3*2*1 = 210**A mixture of certain quantity of milk with 15 litres of water is for Rs 50 per litre. The pure milk is worth Rs 70. What is the amount of milk present in the mixture?**A) 30 litres B) 33.5 litres C) 32 litres D) 37.5 litres E) None of these**D) 37.5 litres****Explanation:**CP of water is Rs 0 By allegation formula: Milk water 70 0 . 50 50-0=50 70-50=20 Ratio = 50 : 20 = 5 : 2 Let there is x litres of milk with 15 litres of water So 5/2 = x/15 x = 75/2 = 37.5 l

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