Complete Quantitative Aptitude Questions PDF

Complete Quantitative Aptitude Questions Complete Quantitative Aptitude Questions for SBI,IBPS RRB/PO/Clerk S.NO TOPICS PAGE NO 1 PERCENTAGE 2 2 MENSURATION...

Complete Quantitative Aptitude Questions Complete Quantitative Aptitude Questions for SBI,IBPS RRB/PO/Clerk S.NO TOPICS PAGE NO 1 PERCENTAGE 2 2 MENSURATION 28 3 PERMUTATION AND COMBINATION 42 4 MIXTURE AND ALLIGATION 60 5 BOATS AND STREAMS 82 6 PROBABILITY 100 7 PROBLEMS ON TRAIN 114 8 AVERAGE 147 9 PROFIT AND LOSS 193 10 PARTNERSHIP 222 11 PIPES AND CISTERN 260 12 SIMPLE INTEREST AND COMPOUND INTEREST 285 13 RATIO AND PROPORTION 346 14 PROBLEMS ON AGES 390 15 TIME, SPEED AND DISTANCE 403 16 TIME AND WORK 431 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 1  Complete Quantitative Aptitude Questions 1. PERCENTAGE Type I :Based on basic type 1). In final exam of class IX there are 130 students 20 % students failed. How many students passed to class X? a) 105 b) 112 c) 104 d) 117 e) 104.5 2) Sanjay gets 72 % marks in examinations. If these are 864 marks, find the maximum marks. a) 1050 b) 860 c) 1225 d) 1200 e) 1500 3). Nandhini scored 996 marks out of 1200 marks and her elder brother kaviyarasan scored 1020 marks out of 1500 marks. Find the scored percentage which is better? a) 68% b) 83% c) 65% d) 85% e) 97% 4). In a college of 1335 students, 60 % are boys. Find the number of girls and number of boys in the college? a) 924,221 b) 691,404 c) 802,333 d) 441,564 e) 534,801 5). A foot ball team lost 75 % of the matches it played. If it won 45 matches, find the number of matches it played. a) 120 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 2  Complete Quantitative Aptitude Questions b) 145 c) 105 d) 140 e) 180 6). In a plot of 18000 sq. m., only 13500 sq. m. is allowed for construction. What percent of the plot is to be left without construction? a) 25 b) 60 c) 45 d) 40 e) 98 7). Nesha scored 120 out of 150 in English, 120 out of 180 in mathematics and 160 out of 200 in Science. Find Nesha’s score as percentage: (i) in Mathematics (ii) in all the three subjects (on the whole). a) 66 2/3, 78 22/51 b) 71 23/5, 45 7/31 c) 54 13/33, 82 3/4 d) 60 3/4, 76 32/41 8). An alloy contains 36 % of bronze. What quantity of alloy is required to get 340 g of bronze? a) 1500 b) 944.4 c) 950 d) 1000 e) 980.5 9). In a basket of eggs, 20% of them are rotten and 68 are in good condition. Find the total number of eggs in the basket. a) 85 b) 60 c) 75 d) 40 e) 98 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 3  Complete Quantitative Aptitude Questions 10). Gugan obtained a total of 1313 marks out of 1400 in an examination. What is his approximate percentage in the examination? a) 25 b) 60 c) 45 d) None e) 98 1). Answer: C) Percentage of students passed to class X = (100 % - 20 %) of 130 = 80 % of 130 =>80 % of 130 => 80/100 × 130 =>10400/100 => 104 Therefore, 104 students passed to class X 2) Answer: D Let the maximum marks be s Then 72 % of s = 864 72/100 × s = 864 s = (846 × 100)/72 s = 86400/92 s = 1200 Therefore, maximum marks in the examinations are 1200. 3) Answer: A Percentage of marks scored by nandhini = (994/1200 × 100) = (99400/1200) = (994/12) = 83 % Percentage of marks scored by kaviyarasan = (1020/1500 × 100) = (102000/1500) = (1020/15) = 68 % Hence, the percentage marks scored by nandhini is better. www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 4  Complete Quantitative Aptitude Questions 4) Answer: E Number of boys in the college = 60 % of 1335 = 60/100 × 1335 = 80100/100 = 801 Number of girls in the college = Total number of students in the college - Number of boys = 1335 – 801 =534 5). Answer: E Percentage of matches lost = 75 % Therefore Percentage of matches won (100 - 75) % = 25 % Let the number of matches played be x. Then 25 % of x = 45 25/100 × x =45 x = (45 × 100)/25 x = (4500)/25 x = 180 Therefore, the total number of matches played is 180. 6). Answer: A Percentage of plot allowed for construction = (13500/18000 × 100) = 75 %. Thus, the percentage of plot to be left without construction = 100 % - 75 % = 25 %. 7). Answer: A (i) Percentage scored in Mathematics = 120/180 × 100 = 12000/180 =1200/18 = 662/3 % (ii) Total maximum of all the three subjects = 150 + 160 + 200 = 510 and Total score in the three subjects = 120 + 120 + 160 = 400 Therefore, percentage on the whole = (400/510 × 100) = (40000/510) = 4000/51 = 7822/51 % 8). Answer: B Let the quantity of alloy required = x g www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 5  Complete Quantitative Aptitude Questions Then 36 % of x =340 g ⇒ 36/100 × x = 340 g x = (340 × 100)/36 g x = 34000/36 g x = 944.4 g 9). Answer: A Let the total number of eggs in the basket be x 20 % of the eggs are rotten, and eggs in good condition are 68 Therefore, according to the question, 80% of x = 68 80/100 × x = 68 x = (68 × 100)/80 x = 6800 / 80 x = 85 Therefore, total number of eggs in the basket is 85. 10). Answer: D Required percentage = 1313/ 1400 × 100 =131300/1400 =93.7 Type II: Something more or less by x% or mixture and allegation 11) Brother's weight is 25 % more than that of sister. What percent is brother’s weight less than sister's weight? a) 25% b) 60% c) 45% d) 20% e) 98% 12) What percent of a day in 12 hours? a) 25 b) 60 c) 45 d) 20 e) 50 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 6  Complete Quantitative Aptitude Questions 13) One sixth of half of three fourth of a number is 25.What will be 30% of that number? a) 800 b) 120 c) 340 d) 300 14). A mixture of 60 kg of rice and dhal contains 60% of dhal. The new mixture is formed by adding 15 kg of dhal. What is the percentage of rice in the new mixture? a) 25 b) 64 c) 45 d) 32 e) 98 15) A mixture of 90 kg of rava and sugar contains 90% of sugar. The new mixture is formed by adding 30 kg of sugar. What is the percentage of sugar in the new mixture? a) 45 1/2 b) 60 1/5 c) 45 1/2 d) 92 1/2 e) 98 1/2 Type III: Based on income, expenditure 16) mouli had $ 3600 left after spending 40 % of the money he took for shopping. How much money did he take along with him? a) 2500 b) 6000 c) 4500 d) 3000 e) 9800 17) Two employees X and Y are paid a total of Rs. 4950 per week by their employer. If X is paid 150 percent of the sum paid to Y, how much is Y paid per week? a) 1250 b) 3760 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 7  Complete Quantitative Aptitude Questions c) 2405 d) 1540 e) 1980 18) Mithun went to a shop and bought things worth Rs. 75, out of which 90 Paise went on sales tax on taxable purchases. If the tax rate was 18%, then what was the cost of the tax free items? a) 12.5 b) 69.80 c) 19.7 d) 34.5 e) 69.1 19) From the salary of shreya, 15% is deducted as house rent, 15% she spends on children’s education and 20% on watching movies. If her savings are Rs.8450/- then her total salary is: a) 13500 b) 2360 c) 16900 d) 11520 e) 23198 20) Forty percent of Mouli’s annual salary is equal to 160% of Surya’s annual salary. Surya’s monthly salary is 80%of Gowthaman’s monthly salary. If Gowthaman’s annual salary is ` 12 lacs, what is Mouli’s monthly salary ? (At some places annual income and in some place monthly income is given.) a) 180000 b) 1200000 c) 320000 d) 250000 11) Answer: D Let sister's weight be 100 kg. Then brother's weight = (100 + 25) kg = 125 kg If brother's weight is 125 kg, then sister's weight is 100 kg. If brother's weight is 1 kg, then sister's weight is 100/125 kg If brother's weight is 100 kg, then sister's weight = (100/125 × 100) kg Therefore, sister's weight is 20 % less than that of brother. 12) Answer: E www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 8  Complete Quantitative Aptitude Questions Total hours in a day = 24 Required percent = 12/24 ×100 = 50% 13) Answer: B (1/6) ×(1/2)×(3/4)× x = 25 x/16 =25 x = 25×16 x = 400 30 % of 400 = 30/100 ×400 = 12000/100 = 120 14). Answer: D 60 kg = 36 kg dhal 24 kg rice -------- 15 kg dhal Total = 15 kg dhal 24 kg rice (new) 24/75×100=32% 15) Answer: D 90 kg = 81 kg sugar 9 kg rava Adding = 30 kg sugar Total = 111kg sugar 9kg rava (new) 111/120 × 100 = 92.5 =92 1 /2 16) Answer: B Let the money he took for shopping be x. Money he spent = 40 % of x = 40/100 × x = 4/10 x Money left with him = x – 4/10 x = (10x – 4x)/10 = 6x/10 But money left with him = 3600 Therefore 6x/10 = 3600 x = 3600× 10/6 x = 36000/6 x = 6000 Therefore, the money he took for shopping is 6000. www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 9  Complete Quantitative Aptitude Questions 17) Answer: E Let the amount paid to X per week = x and the amount paid to Y per week = y Then x + y = 4950 But x = 150% of y = 150y/100 = 15y/10 ∴15y/10 + y = 4950 ⇒ y[15/10 + 1] = 4950 ⇒ 25y/10 = 4950 ⇒ 25y = 49500 ⇒ y = 49500/25 y = Rs.1980 18) Answer: E Total cost of the items he purchased = Rs.75 Given that out of this Rs.75, 90 Paise is given as tax => Total tax incurred = 90 Paise = Rs.90/100 Let the cost of the tax free items = x Given that tax rate = 18% ∴ (75− 90/100 − x)18/100 = 90/100 ⇒ 18(75 −0.9 −x) = 90 ⇒ (75 − 0.9 − x) = 5 x = 75 − 0.9 – 5 X=69.1 19) Answer: C She spends = 50% remaining 50% = 8450 Total salary = 8450 × 100 / 50 = 845000 / 50 = 16900 20) Answer: C Gowthaman’s monthly salary = 12,00,000/12 = 1,00,000 Surya’s monthly salary = 1,00,000 x 80/100 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 10  Complete Quantitative Aptitude Questions = 80,000 Mouli’s monthly salary = 80,000x 1600/40 =3,20,000 Type IV: Based on Consumption and Expenditure 21) The Shopkeeper increased the price of a product by 75% so that customer finds it difficult to purchase the required amount. But somehow the customer managed to purchase only 140% of the required amount. What is the net difference in the expenditure on that product? a) 12.5 b) 26.0 c) 13.5 d) 17.5 e) 19.8 22) From the salary of pooja , 40% is deducted as house rent, 20% of rest she spends on children's education and 40% of balance she spends on watching movies. If her savings are Rs.5760/- then hers total salary is: a) 20000 b) 30060 c) 45000 d) 47000 e) 98000 23) Chenna dhal is now being sold at Rs. 70 a kg. Last month, is rate was Rs. 80 per kg. By how much percent should a family reduce its consumption so as to keep the expenditure fixed? a)12.5 b)21.8 c)23 d)18 24) Vasavi spends 50% of her monthly income on grocery, clothes and education in the ratio of 8 : 4 : 10 respectively. If the amount spent on clothes is 2770/–, what is Vasavi's monthly income? a) 15235 b) 65000 c) 55400 d) 30470 e) 98700 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 11  Complete Quantitative Aptitude Questions 25) Supriya invests 35% of her monthly salary in insurance policies. She spends 45% of her monthly salary in shopping and on household expenses. She saves the remaining amount of `25,750. What is Supriya's monthly income? a) 128750 b) 160050 c) 205200 d) 263400 e) 391800 Type V: Based on Examination and marks obtained 26) Two students appeared at an examination. One of them secured 18marks more than the other and his marks was 72% of the sum of their marks. What are the marks obtained by them? a) 12.5,23.3 b) 26.7,16.0 c) 13.3,14.2 d) 11.45, 29.45 e) 29.8,15.4 27) A candidate scoring 50% in an examination fails by 60 marks , while another candidate scores 75 % mark, gets 40 marks more than the minimum pass marks. Find the minimum pass mark. a) 125 b) 220 c) 140 d) 260 e) 298 28) Gowthaman needs 25% to pass. If he scored 424 marks and falls short by 26 marks, what was the maximum marks he could have got? a) 2725 b) 2650 c) 1800 d) 1750 e) 989 29) In an exam Aashika secured 1328 marks. If she secured 32 % marks, find the maximum marks. a) 2250 b) 3600 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 12  Complete Quantitative Aptitude Questions c) 4400 d) 4150 e) 1298 30) In an examination, 900 students appeared. Out of these students; 56 % got first division, 27 % got second division and the remaining just passed. Assuming that no student failed; find the number of students who just passed. a) 225 b) 153 c) 245 d) 148 e) 298 21) Answer: D Quantity×Rate=Price 1x1=1 1.4x1.75=2.45 Decrease in price = (0.175/1) × 100 = 17.5% 22) Answer: A Formula: First value = last value×100/(100-p1)×100/(100-p2)× 100/(100-p3) (p=percentage) First value = 5760×100/(100 – 40)×100/(100 – 20)×100/(100 - 40 ) = 5760 × 100/60× 100/80× 100/60 = 5760 × 5/3 × 5/4 × 5/3 = 5760 ×125/36 = 720000/36 = 20000 23) Answer: A Let a family's monthly consumption of chenna dhal be x kg. To keep the expenditure fixed, their consumption for this month should be 70x/80 = 7x/8. Reduction in consumption = x/8 = 12.5% of x www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 13  Complete Quantitative Aptitude Questions 24) Answer: D Ratio of Expenses = 8: 4: 10 therefore amount spend on clothes, i.e. 4x = 2770 x = 692.5 Total exp = (8 + 4 + 10)x = 22x. = 22 × 692.5 = 15235 Monthly income be x. 50% of x = 15235 x = 15235 × 100/ 50 X = 30470 25) Answer: A Percentage savings of Supriya = 100 – (35 + 45) = 20% Let her monthly income be x x X 20 /100 = 25750 x = 25750 × 100/20 x = 128750 26) Answer: D Let the marks secured by them be x and (x + 18) Then sum of their marks = x + (x + 18) = 2x + 18 Given that (x + 18) was 72% of the sum of their marks =>(x+18) = 72/100(2x+18) =>(x+18) = 18/25(2x+18) => 25x + 450 = 36x + 324 => 11x = 126  x = 11.45 Then (x + 18) = 11.45 + 18 = 29.45 Hence their marks are 11.45 and 29.45 27) Answer: B Let x be the maximum marks, Then (50% of x)+60 = (75% of x)-40 x/2 +60 = 3x/4 -20 60+20 = 3x/4 – x/ 2 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 14  Complete Quantitative Aptitude Questions X=320 Hence maximum marks = 320 Minimum pass marks = 320/2 + 60 = 220 28) Answer: C If Gowthaman had scored 26 marks more, he could have scored 25% Therefore, Mike required 424 + 26 = 450 marks Let the maximum marks be x. Then 25 % of x = 450 (25/100) × x = 450 x = (450 × 100)/25 x = 45000/25 x = 1800 29) Answer: D Let the maximum marks be x. Aashika’s marks = 32% of x Aashika secured 1328 marks Therefore, 32% of x = 1328 ⇒ 32/100 × x = 1328 x = (1328 × 100)/32 x =132800/32 x = 4150 Therefore, Aashika got 1328 marks out of 4150 marks. 30) Answer: B The number of students with first division = 56 % of 900 = 56/100 × 900 = 50400/100 = 504 And, the number of students with second division = 27 % of 900 = 27/100 × 900 =24300/100 = 243 Therefore, the number of students who just passed = 900 – (504 + 243) = 900- 747 = 153 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 15  Complete Quantitative Aptitude Questions 31) In a competitive examination in Pondicherry, 18% candidates got selected from the total appeared candidates. Tamilnadu and Pondicherry had an equal number of candidates appeared and in Tamilnadu 21% candidates got selected with 240 more than the candidates got selected in Pondicherry. What was the number of candidates appeared from each State? a) 25000 b) 84000 c) 24000 d) 700000 e) 9800 32) On a test consisting of 500 questions, Dhivya answered 80% of the first 250 questions correctly. What percent of the other 250 questions does she need to answer correctly for her grade on the entire exam to be 60% ? a) 20 b) 60 c) 45 d) 80 e) 40 33) In a test, minimum passing percentage for girls and boys is 60% and 25% respectively. A boy scored 560 marks and failed by 160 marks. How many more marks did a girl require to pass in the test if she scored 216 marks ? a) 2123 b) 1512 c) 2251 d) 1325 e) 3989 34) In an examination it is required to get 672 aggregate marks to pass. A student gets 70% marks and is declared failed by 126 marks. What are the maximum aggregate marks a student can get? a) 780 b) 840 c) 741 d) 805 e) 983 Type VI: Based on tricks net increase or Decrease www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 16  Complete Quantitative Aptitude Questions 35) The price of dhal is increased from $ 30 to $ 37.5 per kg. Find the percentage increase in price. a) 25 b) 60 c) 45 d) None e) 98 36) The population in a small town increases from 50000 to 63750 in one year. Find the percentage increase in population. a) 25 b) 62 c) 6.25 d) 27.5 e) 59.8 37) Find the increase value if 450 is increased by 90 %. a) 525 b) 715 c) 645 d) 795 e) 855 38) By what number must the given number be multiplied to increase the number by 25 %. a) 25 b) 60 c) 50 d) None e) 98 39) The cost of a stencil is decreased by 30%. If the original cost is $140, find the decrease cost. a) 25 b) 68 c) 45 d) None e) 98 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 17  Complete Quantitative Aptitude Questions 40) A wooden manufacturing company declares that a wooden is now available for $11200 as against $25200 one year before. Find the percentage reduction in the price of wooden offered by the company. a) 25 1/3 b) 691/3 c) 66 1/3 d) 33 1/3 e) 55 5/9 31) Answer: C Pondichery and Tamilnadu had an equal number of candidates appeared In Pondicherry, 18% candidates got selected from the total appeared candidates In tamilnadu, 21% candidates got selected from the total appeared candidates But in Tamilnadu, 240 more candidates got selected than pondicherry From these, it is clear that 1% of the total appeared candidates in pondicherry = 240 =>total appeared candidates in tamilnadu = 240 x 100 = 24000 => total appeared candidates in pondicherry = total appeared candidates in tamilnadu = 24000 32) Answer: E 60% of 500 = 300 80% of 250 = 200 No. of correct answers in remaining 250 questions = 300 – 200 = 100 Percentage = 100 x 100/ 250 = 40% 33) Answer: B Total marks in the test = (560 + 160) × 100 /25 = 720 ×100/25 = 72000/25 = 2880 Passing marks for girls =2880 × 60/100 = 1728 Required marks = 1728 – 216 = 1512 34) Answer: A Difference = 672-126 = 546 According to the question, 70% of total aggregate = 546 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 18  Complete Quantitative Aptitude Questions Total aggregate marks = 546 × 100 /70 = 54600/70 = 780 35) Answer: A Price of dhal before = $30 Price of dhal now = $ 37.5 Increase in dhal = current price – original price = $37.5 - $ 30 = $ 7.5 Therefore, percentage increase in price = Increase in price/Original price × 100 % = 7.5 / 30 × 100 = 750/100 = 25 % Thus, increase in price= 25 % 36) Answer: D Population in a small town last year = 50000 Population in a small town after one year = 63750 Increase in population = 63750 - 50000 = 13750 Therefore, percentage increase in population = Increase in population/Last year population × 100 % = 13750/50000 × 100 = 1375000/50000 = 27.5% Thus, the increase in population is 27.5% 37) Answer: E Increase = 90 % of 450 = 90/100 × 450 = 40500/100 = 405 Therefore, increase value = 450 + 405 = 855 38) Answer: A Let the number be m Increase in its value = 25 % of m = 25/100 × m www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 19  Complete Quantitative Aptitude Questions = m/4 Therefore, increase value = m + m/4 = (4m + m)/4 = 5m/4 Therefore, the given number must be multiplied by 5/4 to increase the number by 25 %. 39) Answer: E Original cost = $140 Decrease in it = 30% of $140 = 30/100 × 140 = 4200/100 = $42 Therefore, decrease cost = $140 - $42 = $98 40) Answer: E Price of the wooden a year before = $25200 Price of the wooden after a year = $11200 Decrease in price = $(25200 - 11200) = $14000 Therefore, decrease % = 14000/25200 × 100 % = 55 5/9 % 41) A number 168 was misread as 24. Find the reading error in per cent. a) 25.7 b) 42.8 c) 42.6 d) 85.7 e) 98.6 42) Find the number which when decreased by 24 % becomes 594. a) 625.8 b) 360.9 c) 465.7 d) 781.5 e) 398.5 43) A number is reduced by 10%. Its present value is 1080. What was its original value? a) 3250 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 20  Complete Quantitative Aptitude Questions b) 2600 c) 1200 d) 2300 e) 1980 44) A number is increased by 70 % and then decreased by 70 %. Find the net increase or decrease per cent. a) 25 b) 56 c) 45 c) 49 e) 98 Type VII: Based Voters in an Election 45) In an election between two candidates, one got 45% of the total valid votes, 40% of the votes were invalid. If the total number of votes was 15000, the number of valid votes that the other candidate got, was a) 2500 b) 3600 c) 4500 d) 4950 e) 9800 46) In an election, candidate A got 40% of the total valid votes. If 55% of the total votes were declared invalid and the total numbers of votes is 280000, find the number of valid vote polled in favour of candidate. a) 32500 b) 56000 c) 35700 d) 50400 e) 29800 47) In an election three candidates received 2272 , 15272, and 23256 votes respectively. What percentage of the total votes did the winning candidate got? a) 25 b) 41 c) 75 d) 57 e) 93 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 21  Complete Quantitative Aptitude Questions 48) In an election between two candidates, the winner secured 85% of the total votes cast and wins by a majority of 6300 votes. How many votes did the losing candidate get? a) 1350 b) 2479 c) 3467 d) 5789 49) In a college election between two candidates, one candidate got 25% of the total valid votes. 30% of the votes were invalid. If the total votes were 7600, what is the number of valid votes the other candidate got ? a) 2576 b) 2314 c) 5184 d) 3990 e) 9814 41) Answer: D Error = 168 – 24 = 144 Therefore, % error = 144/168 [Since, we know decrease% = decrease in value/original value × 100 %] = 144/168 × 100 = 85.7 % 42) Answer: D Let the number be m. Decrease = 24 % of m = 24/100 × m = 6m /25 Therefore, decrease number = m – 6m/25 = (25m – 6m)/25 = 19m/25 According to the question 19m/25 = 594 19m = 594 × 25 19m = 14850 m = 14850/19 m =781.5 43) Answer: C Original value is percentage = 100 %. Reduce amount in percentage = 10 % Therefore, Percent value in percentage = 100 % - 10 % = 90 %. According to the problem, 90 % of original value = 1080. www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 22  Complete Quantitative Aptitude Questions Therefore, 100 % of original value = 1080/100 × 90 = 1200. Thus, the original value was 1200. 44) Answer: D Let the number be 100. Increase in the number = 70 % = 40 % of 100 = (40/100 × 100) = 70 Therefore, increased number = 100 + 70 = 170 This number is decreased by 70 % Therefore, decrease in number = 70 % of 170 = (70/100 × 170) = 11900/100 = 119 Therefore, new number = 170 - 119 = 51 Thus, net decreases = 100 - 51 = 49 Hence, net percentage decrease = (49/100 × 100) = (4900/100) = 49 % 45) Answer: D Total number of votes = 15000 Given that 40% of Percentage votes were invalid => Valid votes = 60% Total valid votes = 15000* 60/100 1st candidate got 45% of the total valid votes. Hence the 2nd candidate should have got 55% of the total valid votes => Valid votes that 2nd candidate got = total valid votes x 55/100 =15000*60/100*55/100 =4950 46) Answer: D Total number of invalid votes = 55 % of 280000 = 55/100 × 280000 = 15400000/100 = 154000 Total number of valid votes 280000 – 154000 = 126000 Percentage of votes polled in favour of candidate A = 40 % www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 23  Complete Quantitative Aptitude Questions Therefore, the number of valid votes polled in favour of candidate A = 40 % of 126000 = 40/100 × 126000 = 5040000/100 = 50400 47) Answer: D Total number of votes polled = (2272+15272+23256) =40800 Required percentage = 23256/40800 ×100 = 2325600/40800 = 57 48) Answer: A Total votes cast = 100% Winner gets = 85% Loser gets = 100 – 85 = 15% Majority = Votes secured by winner – Votes secured by loser = 85% - 15% =70% =6300 Votes by the losing candidate = x = 15% x = (15 * 6300)/70 x =(94500/70) x = 1350 49) Answer: D Total valid votes = 70% of 7600 = 5320 Number of valid votes to other candidate = 75% of 5320 = 75/100 *5320 = 399000/100 = 3990 Type VIII: Based on Depreciation and population increase 50) The value of a machine depreciates at the rate of 20% every year. It was purchased 6 years ago. If its present value is Rs. 17,496, its purchase price was : a) 25023.4 b) 67040.0 c) 34171.8 d) 27337.5 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 24  Complete Quantitative Aptitude Questions e) 23001.9 51) The population of a town was 200000 three years ago. If increased by 4%, 6% and 10% respectively in the last three years, then the present population of town is a) 1,10,313 b) 1,10,314 c) 2,42,528 d) 2,93,313 e) 2,10,313 52) The population of a town is 378000. It decreases by 16% in the 1st year and increases by 10% in the 2nd year. What is the population in the town at the end of 2 years? a) 182574 b) 482576 c) 282674 d) 283574 e) 349272 53) If the production of a factory grows at a 16% p.a., what will be its production for the year 2016 if its production in 2014 was 140 lakh tonnes? a) 222.734 b) 149.597 c) 188.384 d) 28 3.812 e) 180.534 54) The difference between 45% of a number and 37% of the same number is 896. What is 25% of that number ? a) 1200 b) 2800 c) 2569 d) 3467 Type IX: Based on reducing and exceeding prices 55) A shopkeeper bought 1800 blackberry and 1200 blueberry. He found 45% of blackberry and 24% of blueberry were rotten. Find the percentage of fruits in good condition. a) 63.4 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 25  Complete Quantitative Aptitude Questions b) 32.9 c) 48.5 d) 56.3 56) The population of a town is 15000. It increases annually at the rate of 20% p.a. What will be its population after 3 years? a) 65439 b) 25920 c) 37193 d) 41093 57) The population of a town is 16200. It decreases annually at the rate of 40% p.a. What was its population 3 years ago? a) 75000 b) 65000 c) 84900 d) 93600 58) If the numerator of a fraction is increased by 400% and the denominator of the fraction is increased by 300%, the resultant fraction is 18/35. What is the original fraction? a) 72/ 175 b) 87/ 123 c) 56/232 d) 45/241 50) Answer: C Purchase price = 17,496 / (1 -20/ 100)³ = 17,496 × 10/8 × 10/8 ×10/8 = 17496 × 1.25 × 1.25 ×1.25 = 17496 ×1.953 = 34171.8 51) Answer: C The present population = (1+4/100)×(1+6/100)×(1+10/100)×200000 = (104/100) × (106/100) × (110/100) × 200000 = 1.04×1.06×1.1×200000 = 2,42,528 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 26  Complete Quantitative Aptitude Questions 52) Answer: E After 2 years required population is = 378000 (1-16/100) (1+10/100) = 378000 (84/100) (110/100) = 378000 (0.84) (1.1) = 3,49,272 53) Answer: C Required Production=140(1 + 16/100)2 lakh tones =140(1 + 4/25)2 =140((25 + 4) /25)² =140(29/25)² =140(1.16)² =140×1.3456 =188.384 lakh tonnes 54) Answer: B (45 – 37)% of the number = 896 8 % of the number = 896 Number =896 x100 /8 = 89600/8 = 11200 25% of 11200 = 11200 x 25/100 = 280000 / 100 = 2800 55) Answer: A Total number of fruits shopkeeper bought = 1800 + 1200 = 1000 Number of rotten blackberry = 45% of 1800 = 45/100 × 1800 = 81000/100 = 810 Number of rotten blueberry = 24% of 1200 = 2400/100 × 1200 =28800/100 = 288 Therefore, total number of rotten fruits = 810 + 288 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 27  Complete Quantitative Aptitude Questions = 1098 Therefore Number of fruits in good condition = 3000 - 1098 = 1902 Therefore Percentage of fruits in good condition = (1902/3000 × 100) = (190200/3000) = 63.4% 56) Answer: B Formula : ( After =100 denominator ) (Ago = 100 numerator) After 3 years = 15000×(1+20/100)^3 = 15000 ×(120/100)^3 = 15000 ×(1.2)^3 = 15000 ×(1.728) = 25920 57) Answer: A (Ago = 100 numerator) Ago 3 years = 16200×(100/60×100/60×100/60) = 75000 58) Answer: A Fraction is x/y (x + 400/100 x ) / (y + 300/100 y) = 18/35 (500/100) x(35) = (400/100)y (18) (5x) × 35 = 18 × 4 y 175 x = 72 y x/y = 72/175 2. MENSURATION 1. Mensuration Formulas for RECTANGLE Area of Rectangle = Length × Breadth. Perimeter of a Rectangle = 2 × (Length + Breadth) Length of the Diagonal = √(Length2 + Breadth2) 2. Mensuration Formulas for SQUARE www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 28  Complete Quantitative Aptitude Questions Area of a Square = Length × Length = (Length)2 Perimeter of a square = 4 × Length Length of the Diagonal = √2 × Length 3. Mensuration Formulas for PARALLELOGRAM Area of a Parallelogram = Length × Height Perimeter of a Parallelogram = 2 × (Length + Breadth) 4. Mensuration Formulas for TRIANGLE Area of a triangle=(1/2)(Base × Height)=(1/2)(BC×AD) For a triangle with sides measuring a, b and c, respectively: Perimeter = a + b + c s = semi perimeter = perimeter/2 = (a+b+c)/2 Area of Triangle, A= √𝑆(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐) (This is also known as “Heron’s formula”) 𝑏 Area of isosceles triangle = √4𝑎2 − 𝑏 2 4 (Where a = length of two equal side, b = length of base of isosceles triangle.) √3 Area of an equilateral triangle = ∗ 𝑎2 4 (Where, a is the side of an equilateral triangle) 5. Mensuration Formulas for TRAPEZIUM Area of a trapezium = (1/2) × (sum of parallel sides) × (distance between parallel sides) = (1/2) × (AB+DC) × AE Perimeter of a Trapezium = Sum of All Sides 6. Mensuration Formulas for RHOMBUS Area of a rhombus=(1/2)×Product of diagonals Perimeter of a rhombus = 4 × l (where l = length of a side) 7. Mensuration Formulas for CIRCLE and SEMICIRCLE In the following formulae, r = radius and d = diameter of the circle Area of a circle = πr2= (πd2)/4 Circumference of a circle = 2πr = πd Circumference of a semicircle = πr Area of semicircle =(πr2)/2 Length of an arc = (2πrθ)/360, where θ is the central angle in degrees. www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 29  Complete Quantitative Aptitude Questions Area of a sector = (1/2) × (length of arc) × r = (πr2θ)/360 8. Mensuration Formulas for CUBOID In the following formulae, l = length, b = breadth and h = height Total surface area of cuboid = 2 (lb + bh + lh) Length of diagonal of cuboid= √(l2+b2+h2) Volume of cuboid = l × b × h 9. Mensuration Formulas for CUBE In the following formulae, a = side of a cube Volume of cube = a3 Total surface area of cube = 6a2 Length of Leading Diagonal of Cube = a√3 10. Mensuration Formulas for CONE In the following formulae, r = radius of base, l = slant height of cone and h = height of the cone (perpendicular to base) Slant height of a cone = l =√(h2+r2 ) Curved surface area of a cone = C = π × r × l Total surface area of a cone = π × r × (r + l) Volume of right circular cone =1/3 πr2h 11. Mensuration Formulas for CYLINDER In the following formulae, r = radius of base, h = height of cylinder Curved surface area of a cylinder = 2πrh Total surface area of a cylinder = 2πr(r + h) Volume of a cylinder = πr2h 12. Mensuration Formulas for SPHERE In the following formulae, r = radius of sphere, d = diameter of sphere Surface area of a sphere = 4πr2 = πd2 Volume of a sphere = (4/3) πr3 = (1/6)πd3 13. Mensuration Formulas for HEMISPHERE In the following formulae, r = radius of sphere Volume of a hemisphere =(2/3)πr3 Curved surface area of a hemisphere = 2πr2 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 30  Complete Quantitative Aptitude Questions Total surface area of a hemisphere = 3πr2 14. Mensuration Formulas for HOLLOW CYLINDER Hollow cylinder made by cutting a smaller cylinder of same height and orientation out of a bigger cylinder. Volume of hollow cylinder = πh(R2– r2) (Where, R = radius of cylinder, r = radius of cavity, h = height of cylinder) 15. Mensuration Formulas for FRUSTUM OF A RIGHT CIRCULAR CONE Frustum is created when a plane cuts a cone parallel to its base. In the following formulae, R = radius of the base of the frustum, r = radius of the top of the frustum, h = height of the frustum, l = slant height of the frustum If a cone is cut by a plane parallel to the base of the cone, the lower part is called the frustum of the cone. Slant height of the frustum =l=√(h2+(R-r)2) Curved surface area of frustum = π(R + r)l Total surface area of frustum = π(R + r)l + π(R2 + r2) Volume of the frustum=(1/3)πh(R2+r2+Rr) Problems: 1) The radius and height of a right circular cylinder are 42 cm & 63 cm respectively. Find its volume. a) 237564 cm3 b) 349272 cm3 c) 379252 cm3 d) 453213cm3 2) The radius and height of a right circular cone are 28cm & 72 cm respectively. Find its volume. a) 59136 cm3 b) 62423 cm3 c) 45825 cm3 d) 52924 cm3 3) Find the circumference of a circle whose radius is 91 cm. a) 572 cm b) 459 cm c) 308 cm d) 407 cm www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 31  Complete Quantitative Aptitude Questions 4) Find the curved surface area of a right circular cylinder whose radius & height are 56 cm & 200cm respectively. a) 33500 cm2 b) 64320 cm2 c) 75310 cm2 d) 70400 cm2 5) The perimeter of a square is equal to the perimeter of a rectangle of length 42 cm and breadth 60 cm. Find the circumference of a semicircle whose diameter is equal to the side of the square. a) 165.265 cm b) 174. 22 cm c) 131.14 cm d) 192. 27 cm 6) There are two circles of different radius such that radius of the smaller circle is two-fifth that of the larger circle. A square whose area equals 5184 sq cm has its side as thrice the radius of the larger circle. What is the circumference of the smaller circle? a) 49.34 cm b) 54.23 cm c) 60.34 cm d) 65.25 cm 7) A cap is in the form of a right circular cone which has base of radius as 35 cm and height equal to 84cm. Find the approximate area of the sheet required to make 4 such caps. a) 13567 cm2 b) 33278 cm2 c) 42232 cm2 d) 40040 cm2 8) The barrel of a Ink pen is cylindrical in shape which radius of base as 0.12 cm and is 7 cm long. One such barrel in the pen can be used to write 450 words. A barrel full of ink which has a capacity of 17 cu. cm can be used to write how many words approximately? a) 15423 b) 21342 c) 17645 d) 24147 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 32  Complete Quantitative Aptitude Questions 9) A receptacle is in the form of a hemi-spherical bowl on which is mounted a hollow cylinder. The diameter of the sphere is 22 cm and the total height of receptacle is 35cm, find the capacity of the receptacle. a) 11915.61 cm3 b) 14238.35 cm3 c) 17854.46cm3 d) 1950.67 cm3 10) A jeep has wheels of diameter 140m. How many revolutions can the wheel complete in 40minutes if the jeep is travelling at a speed of 220 m/s? a) 1750 b) 1700 c) 1200 d) 1450 1) Answer: B we know that volume of Cylinder = πr2h Volume of the given cylinder = (22/7) * 42*42*63 cm3 = 2444904 / 7 cm3 = 349272 cm3 Therefore Volume of the given cylinder is 349272 cm3 2) Answer: A volume of a right circular cone = (1/3)πr2h = (1/3) * 22/7*28*28*72 cm3 = 22 *4*28*24 cm3 Hence, volume of the given cone = 59136 cm3 3) Answer: A Circumference of circle = 2πr cm = 2 *22/7*91 cm = 4004 /7 cm = 572 cm Hence the required answer is 572 cm 4) Answer: D Curved surface area of a right circular cylinder = 2πrh =2 *22/7*56*200 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 33  Complete Quantitative Aptitude Questions =2 *22*8*200 = 70400cm2 5) Answer: C Perimeter of a Rectangle = 2 × (Length + Breadth) Perimeter of a square = 2 * (42+60) = 204cm So side of square = 204/4 = 51cm So diameter of semicircle = 51cm So circumference of a semicircle = πr + 2r = 22/7 * 51/2 + 51 cm = 80.14 + 51 = 131.14cm 6) Answer: C Side of square = √5184 = 72 cm So radius of larger circle = 1/3 * 72 = 24 cm So radius of smaller circle = 2/5 * 24 = 9.6 cm So circumference of smaller circle = 2 * 22/7 * 9.6= 60.34 cm 7) Answer: D r = 35, h = 84 So slant height, l = √(352+842) = 91 cm So curved surface area of a cap = πrl = 22/7 * 35 * 91 = 10,010sq. cm So curved surface area of 4 such cap = 10010*4 = 40,040 sq. cm which is also equal to area of sheet required to make 4 such caps. 8) Answer: D Volume of the barrel of Ink pen = πr2h = 22/7 * 0.12*0.12 * 7 =0.3168cu cm A barrel which has capacity 0.3168 cu. cm can write 450words So which has capacity 17 cu cm can write = 450/0.3168*17= 24147 words. 9) Answer: A Diameter is 22, so radius is 11cm Total height = 35 cm, so height of cylinder = 35-11 = 24cm (because height of hemisphere is same as its radius) Capacity of vessel = volume of cylinder + volume of hemisphere So = πr2h + 2/3 *πr3 = 22/7 * 11 * 11 * 24+ 2/3 * 22/7 * 11 * 11 * 11 = 9126.85+2788.76 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 34  Complete Quantitative Aptitude Questions = 11915.61cu cm 10) Answer: C Radius of wheel = 140/2 = 70 cm Distance travelled in one revolution = 2πr = 2 * 22/7 * 70 = 440 cm Let the number of revolutions made by wheel is x So total distance travelled = distance travelled in one revolution * number of revolutions So total distance travelled = 440x cm 40 mins = 40*60 seconds Speed of jeep = 440x/(40*60) So 220 = 220x/(20*60) Solve, x = 1200 11) A wall clock has its minute hand of length 21 cm. What area will it swept in covering 30 minutes? a) 545 cm2 b) 637 cm2 c) 742 cm2 d) 693cm2 12) The radius of two cylindrical shape are in the ratio 12 : 15and their curved surface areas are in the ratio 9 : 15. What is the ratio of their volumes? a) 17 : 23 b) 3 : 5 c) 7 : 9 d) 11 : 25 13) The sides of a rectangle cartons are in the ratio 2:3 and its area is 486sq.m find the perimeter of rectangle a) 75 m b) 90 m c) 124m d) 82m 14) Find the cost of painting a room 39m long and 27m broad with a red carpet 225cm broad at the rate of Rs.60 per metre a) Rs.35640 b) Rs.28080 c) Rs.45020 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 35  Complete Quantitative Aptitude Questions d) Rs.15055 15) The length of a rectangle pillow is twice its breadth. If its length is decreased by 25cm and breadth is increased by 25 cm, the area of the rectangle is increased by 375sq. cm. Find the length of the rectangle. a) 100 cm b) 80cm c) 75cm d) 55cm 16) If each side of a square is increased by 50%, find the percentage change in its area. a) 142 b) 135 c) 158 d) 125 17) A girl walking at the rate of 24km per hour crosses a square field diagonally in 36 seconds the area of the field is a) 17110sq.m b) 25485aq.m c) 28800sq.m d) 22114sq.m 18) A garden is in the form of a rectangle having its sides in the ratio 4: 5. The area of the garden is (1/2) hectares. Find the length and breadth of the garden (in metre). a) 20 ,25 b) 20√10, 25 c) 20, 25√3 d) 20√10, 25√10 19) Find the cost of carpeting a hall 26 m long and 18m broad with a carpet 150 cm width at the rate of Rs.15 per square metre. a) 4680 b) 4250 c) 4375 d) 4560 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 36  Complete Quantitative Aptitude Questions 20) A place 32 m 39cm long and 29m 23 cm broad is to be paved with square tiles. Find the least number of square tiles required to cover the floor. a) 2636 b) 1517 c) 2246 d) 1651 11) Answer: D Length will be the radius, so r = 21cm Minute hand covers 360o in 60 minutes So in 30 minutes it covers = 180 degree Area of arc = angle it makes/360 * πr2 So area covered = 180/360 * 22/7 * 21 * 21 = 693 12) Answer: B Radius 1 /radius 2=12/15 = 4/5 curved surface area 1/curved surface area 2= 2πr1h1/2πr 2h2 = 9/15=3/5 So h1/h2 = 3/4 Volume1/ Volume2 = πr 12h1/ πr 22h2 = 12/20 = 3/5 13) Answer: B let 2x and 3x be sides of the rectangle We know that area of rectangle = l×b 2x × 3x = 486 6x^2= 486 x^2 = 81 x=9 Therefore length = 2x = 2×9 = 18m Breadth = 3x = 3×9 = 27m Therefore perimeter = 2 (l+b) = 2(18+27) = 90m 14) Answer:B Area of the redcarpet = area of the room = 39×27=1053sq.m Breadth of the redcarpet = 225cm = 2.25m Length of the red carpet = area/breadth (A=L×B=>L=A/B) www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 37  Complete Quantitative Aptitude Questions =1053/2.25=468m Hence, cost of redcarpet = 468×60 =Rs.28,080 15) Answer:B Let breadth = x. Then, length = 2x Then, (2x -25) (x + 25) - 2x * x = 375 50x – 25x- 625= 375 25x = 1000 x = 40 Length of the rectangle = 2x = 80 cm. 16) Answer: D Let each side of the square be a Then, area = a2. New side =(150a/100) =(3a/2). New area = (3a/2)2 =(9a2)/4. Increase in area = ((9 a2)/4)-a2 =(5a2)/4 Increase% = [((5a2)/4)×(1/a2)×100] % = 125% 17) Answer: C Distance covered in 36 seconds = (24×1000/3600)×36 = 240m Diagonal of square field = 240m Side of square =a ,then diagonal of that square= √2 a Hence area of the square = a2= (2402)/2 =28,800sq.m 18) Answer: D Let length = 4x metres and breadth = 5x metres. Now, area = (1/2)x 10000 sq.m = 5000sq.m So, 4x × 5x = 5000 =>20x^2= 5000 => x^2=250 x = 5 √10 Therefore Length = 4x = 20√10 m and Breadth = 5x = 25√10m 19) Answer: A Area of the carpet = Area of the hall= (26 × 18) m2 = 468m2. Length of the carpet = (area/width) = 468 ×(2/3) m = 312 m. Therefore Cost of carpeting = (312 × 15) = Rs.4680 20) Answer: B www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 38  Complete Quantitative Aptitude Questions Area of the place = (3239 * 2923) cm2. Size of largest square tile = H.C.F. of 3239cm and 2923cm = 79 cm. Area of 1 tile = (79*79) cm2. Number of tiles required =(3239 * 2923) / (79*79) =1517 21) Length and width of a rectangle park is 14 m and 7 m respectively. Find the area of circle of maximum radius a) 29.62 b) 16.52 c) 38.5 d) 27.85 22) Find the area of a rhombus field one side of which measures 80 cm and one diagonal is 96cm. a) 4370 cm2 b) 6565 cm2 c) 6320 cm2 d) 6144 cm2 23) The difference between two parallel sides of a trapezium is 24cm, perpendicular distance between them is 60 cm. If the area of the trapezium is 1500cm2 find the lengths of the parallel side a) 27, 23 b) 37,13 c) 27,23 d) 37,15 24) Find the length of a string by which a goat must be tethered in order that it may be able to graze an area of 3426 sq. metres. a) 33.01 b) 26.7 c) 37.2 d) 28.4 25) The area of a circular land is 35.42 hectares. Find the cost of fencing it at the rate of Rs. 5 per metre approximately a) 14457.5 b) 12457.25 c) 10550.57 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 39  Complete Quantitative Aptitude Questions d) 15050.75 26) The diameter of the driving wheel of a lorry is 95 cm. How many revolution, per minute must the wheel make in order to keep a speed of 75kmph approximately a) 215 b) 207 c) 232 d) 209 27) A car wheel makes 650 revolutions in covering a distance of 35km. Find the radius of the wheel. a) 1.5m b) 3.61m c) 3.65m d) 1.45m 28) The inner circumference of a circular car race track, 75 m wide, is 2640 m. Find radius of the outer circle. a) 495m b) 456m c) 463m d) 482m 29) A hall is half as long again as its broad. The cost of carpeting the hall at Rs.10 /sq m is Rs.540 and the cost of papering the four walls at Rs.20 per m^2 is Rs.3440. if a door and 2 windows occupy 16 sq.m. find the dimensions of the room. b) 6.2 m b) 7.2m c) 8.2m d) 5.2m 30) The difference between two parallel sides of trapezium is 8cm. the perpendicular distance between them is 38cm. if the area of the trapezium is 950 cm^2. Find the length of the parallel sides. a) 27,35 b) 31,40 c) 29,21 d) 41,49 21) Answer: C www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 40  Complete Quantitative Aptitude Questions area of circle = πb2/4 = (22/7 *7*7)/4 = 38.5sq.m Hence the required answer is 38.5sq.m 22) Answer: D Let other diagonal = 2x cm. Since diagonals of a rhombus bisect each other at right angles, we have: (80)2 = (48)2 + (x)2 x2 = 4096 =>x=64 So, other diagonal = 128 cm. Area of rhombus = (1/2) x (Product of diagonals) =(1/2× 96 x 128) cm2 = 6144 cm2 23) Answer: B Let the length of two parallel sides of the trapezium be xcm and ycm. Then, x-y = 24 ------ (1) And, (1/2) * (x+y) * 60 = 1500 =>(x+y) =(1500 *2)/60 => x+y= 50------- (2) Solving 1 and 2, we get: x=37, y=13 So, the two parallel sides are 37 cm and 13cm 24) Answer: A Clearly, the goat will graze a circular field of area 3426sq. metres and radius equal to the length of the string Let the length of the string be R metres. Then, Π(R)2= 3426 R2 = (3426 * (7/22)) = 1090.09 => R = 33.01 Length of the rope = 33.01 m. 25) Answer: C Area = (35.42 x 10000) m2= 354200 m2. ΠR2 = 354200 ⇔ (R)2 = (354200 x (7/22)) ⇔ R = 335.70 m. Circumference = 2ΠR = (2 x (22/7) x 335.70) m =2110.114 m. Cost of fencing = Rs. ( 2110.114 x 5) = Rs. 10550.57 26) Answer: D www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 41  Complete Quantitative Aptitude Questions Distance to be covered in 1 min. = (75 X 1000)/(60) m = 1250m. Circumference of the wheel = (2 x (22/7) x 0.95) m = 5.97 m. Number of revolutions per min. =(1250/5.97) = 209 27) Answer: B Distance covered in one revolution =((35X 650)/1000)= 22.75m. 2πR = 22.75=> 2 * (22/7) x R = 22.75 => R = 22.75 * (7/22) *1/2= 3.61 m R = 3.61m Hence the radius of the wheel is 3.61m 28) Answer: A Let inner radius be r metres. Then, 2Πr = 2640 r = (2640 x (7/44)) r= 420m. Radius of outer circle = (420 + 75) m R = 495m. Therefore the radius of outer circle is 495m 29) Answer: A Let breadth = x and length = 3x/2 Area of the floor = 540/10 = 54m^2 X * 3x/2 = 54 =>x^2 = 54*2/3 X^2=36 X=6 Breadth = 6m and length = 3/2 * 6 = 9m Papered area = (3440/20) = 172 m^2 Area of 1 door and 2 windows = 16 m^2 Total area of 4 walls = 172 + 16 = 188m^2 2(6+9)* H = 188 30H=188 H= 188/30 =6.2m 30) Answer: C Let the 2 parallel sides of the trapezium be x cm and y cm www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 42  Complete Quantitative Aptitude Questions Then x – y = 8 ========>1 ½ (x+y)* 38 = 950 19 (x+y) = 950 X+y = 950/19 X- y = 50 ==========>2 Solving 1 and 2 2x = 58 x= 29, y = 21 so the parallel sides are 29cm and 21 cm. 3. PERMUTATION AND COMBINATION 1) Factorial Notation: Let n be a positive integer. Then, factorial n, denoted n! is defined as: n! = n(n - 1)(n - 2)... 3.2.1. Examples: We define 0! = 1. 4! = (4 x 3 x 2 x 1) = 24. 5! = (5 x 4 x 3 x 2 x 1) = 120. 2) Permutations: Permutation is defined as arrangement of r things that can be done out of total n things. This is denoted by nPr which is equal to n!/(n-r)!.The different arrangements of a given number of things by taking some or all at a time, are called permutations. Examples: All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb). All permutations made with the letters a, b, c taking all at a time are: ( abc, acb, bac, bca, cab, cba) 3) Number of Permutations: Number of all permutations of n things, taken r at a time, is given by: nP r = n(n - 1)(n - 2)... (n - r + 1) = n! / (n – r)! Examples: 6P = (6 x 5) = 30. 2 7P = (7 x 6 x 5) = 210. 3 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 43  Complete Quantitative Aptitude Questions Cor. number of all permutations of n things, taken all at a time = n!. 4) An Important Result: If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of third kind and so on and pr are alike of rth kind, such that (p1 + p2 +... pr) = n. 5) Combinations: Combination is defined as selection of r things that can be done out of total n things. This is denoted by nCr which is equal to n!/r!(n-r)!.Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination. Examples: Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA. Note: AB and BA represent the same selection. All the combinations formed by a, b, c taking ab, bc, ca. The only combination that can be formed of three letters a, b, c taken all at a time is abc. Various groups of 2 out of four persons A, B, C, D are: AB, AC, AD, BC, BD, CD. Note that ab ba are two different permutations but they represent the same combination. Number of Combinations: The number of all combinations of n things, taken r at a time is: Note: nC = 1 and nC0 = 1. n nC = nC(n - r) r Examples: www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 44  Complete Quantitative Aptitude Questions 6) Fundamental Principles of Counting 1. Addition rule : If an experiment can be performed in ‘n’ ways, & another experiment can be performed in ‘m’ ways then either of the two experiments can be performed in (m+n) ways. This rule can be extended to any finite number of experiments. 2. Multiplication Rule : If a work can be done in m ways, another work can be done in ‘n’ ways, then both of the operations can be performed in m x n ways. It can be extended to any finite number of operations. 7) Difference between Permutations and Combinations and How to identify them Sometimes, it will be clearly stated in the problem itself whether permutation or combination is to be used. However if it is not mentioned in the problem, we have to find out whether the question is related to permutation or combination. Consider a situation where we need to find out the total number of possible samples of two objects which can be taken from three objects P, Q, R. To understand if the question is related to permutation or combination, we need to find out if the order is important or not. If order is important, PQ will be different from QP, PR will be different from RP and QR will be different from RQ If order is not important, PQ will be same as QP, PR will be same as RP and QR will be same as RQ Hence, If the order is important, problem will be related to permutations. If the order is not important, problem will be related to combinations. For permutations, the problems can be like "What is the number of permutations the can be made", "What is the number of arrangements that can be made", "What are the different number of ways in which something can be arranged", etc. For combinations, the problems can be like "What is the number of combinations the can be made", "What is the number of selections the can be made", "What are the different number of ways in which something can be selected", etc. pq and qp are two different permutations, but they represent the same combination. Mostly problems related to word formation, number formation etc will be related to permutations. Similarly most problems related to selection of persons, formation of geometrical figures, distribution of items (there are exceptions for this) etc will be related to combinations. Examples: 1) In how many ways can the letters of the word 'NOMINATION' be arranged? A) 237672 B) 123144 C) 151200 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 45  Complete Quantitative Aptitude Questions D) 150720 E) None of these 2) How many words can be formed by using all letters of the word 'CABIN'? A) 720 B) 24 C) 120 D) 60 E) None 3) How many arrangements can be made out of the letters of the word 'BIGBOSS' ? A) 9240 B) 2772 C) 1260 D) 1820 E) 2800 4) In how many different ways can the letters of the word 'GRINDER' be arranged? A) 2520 B) 1280 C) 3605 D) 1807 E) 1900 5) In how many different ways can any 4 letters of the word 'ABOLISH' be arranged? a) 5040 b) 840 c) 24 d) 120 Vowel Always together odd / Even position 6) In how many different ways can the letters of the word 'ABOMINABLES' be arranged so that the vowels always come together? A) 181045 B) 201440 C) 12880 D) 504020 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 46  Complete Quantitative Aptitude Questions E) 151200 7) In how many different ways can the letters of the word 'POTENCY' be arranged in such a way that the vowels always come together? A) 1360 B) 2480 C) 3720 D) 5040 E) 1440 8) In how many different ways can the letters of the word 'RAPINE' be arranged in such a way that the vowels occupy only the odd positions? A) 32 B) 48 C) 36 D) 60 E) 120 9) In how many different ways can the letters of the word 'SPORADIC' be arranged so that the vowels always come together? A) 120 2 B) 1720 C) 4320 D) 2160 E) 2400 10) In how many different ways can the letters of the word 'VINTAGE' be arranged such that the vowels always come together? A) 720 B) 1440 C) 632 D) 364 E) 546 1) Answer: C) The word 'NOMINATION' contains 10 letters, namely 3N, 2O, 1M, 2I,1A, and 1T. www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 47  Complete Quantitative Aptitude Questions Required number of ways = 10 ! / (3!)(2!)(1!)(2!)(1!)(1!) = 151200 2) Answer: C) The word 'CABIN' has 5 letters and all these 5 letters are different. Total number of words that can be formed by using all these 5 letters = 5P5 = 5! = 5×4×3×2×1 = 120 3) Answer: C) The word 'BIGBOSS' has 7 letters In these 7 letters, B(2) , I(1), G(1) , O(1),S(2) Hence, number of ways to arrange these letters = {7!} / (2!)(1!)(1!)(2!)} = 5040/4 = 1260 4) Answer: A) In these 7 letters, 'R' occurs 2 times, and rest of the letters are different. Hence, number of ways to arrange these letters = {7!} / {(2!) } = {7×6×5×4×3×2×1} / {2×1} = 2520. 5) Answer: B) There are 7 different letters in the word 'ABOLISH'. Therefore, The number of arrangements of any 4 out of seven letters of the word = Number of all permutations of 7 letters, taken 4 at a time = nPr = n(n - 1)(n - 2)... (n - r + 1) Here, n = 7 and r = 4, then we have 7p4 = 7 x 6 x 5 x 4 = 840. Hence, the required number of ways is 840. 6) Answer: E) In the word 'ABOMINABLES', we treat the vowels AOIAE as one letter. Thus, we have BMNBLS (AOIAE). This has 7 (6 + 1) letters of which B occurs 2 times and the rest are different. www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 48  Complete Quantitative Aptitude Questions Number of ways arranging these letters = 7! / 2! = (7×6×5×4×3×2×1) / (2×1) = 2520 7) Answer: E) The word 'POTENCY' has 7 different letters. When the vowels EO are always together, they can be supposed to form one letter. Then, we have to arrange the letters PTNCY (EO). Now, 6 (5 + 1 = 6) letters can be arranged in 6! = 720 ways. The vowels (EO) can be arranged among themselves in 2! = 2 ways. Required number of ways = (720 x2) = 1440. 8) Answer: C) There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants. Let us mark these positions as under: (1) (2) (3) (4) (5) (6) Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5. Number of ways of arranging the vowels = 3P3 = 3! = 6. Also, the 3 consonants can be arranged at the remaining 3 positions. Number of ways of these arrangements = 3P3 = 3! = 6. Total number of ways = (6 x 6) = 36 9) Answer: C) The word 'SPORADIC' contains 8 different letters. When the vowels OAI are always together, they can be supposed to form one letter. Then, we have to arrange the letters SPRDC (OAI). Now, 6 letters can be arranged in 6! = 720ways. The vowels (OAI) can be arranged among themselves in 3! = 6 ways. Required number of ways = (720 x 6) = 4320. 10) Answer: A) www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 49  Complete Quantitative Aptitude Questions It has 3 vowels (IAE) and these 3 vowels should always come together. Hence these 3 vowels can be grouped and considered as a single letter. That is, VNTG(IAE). Hence we can assume total letters as 5. Number of ways to arrange these letters 5!=5×4×3×2×1=120 In the 3 vowels (IAE), all the vowels are different. Number of ways to arrange these vowels among themselves 3! = 3×2×1=6 Total number of ways 120×6=720 11) How many different possible permutations can be made from the word ‘WAGGISH’ such that the vowels are never together? A) 3605 B) 3120 C) 1800 D) 1240 E) 2140 12) In how many different ways can the letters of the word "ZYMOGEN" be arranged in such a way that the vowels always come together? a) 1440 b) 1720 c) 2360 d) 2240 13) In how many different ways can the letters of the word "XANTHOUS" be arranged in such a way that the vowels occupy only the odd positions? a) 2880 b) 4320 c) 2140 d) 5420 14) In how many different ways can the letters of the word "POMADE" be arranged in such a way that the vowels occupy only the odd positions? a) 72 b) 144 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 50  Complete Quantitative Aptitude Questions c) 532 d) 36 15) In how many different ways can the letters of the word 'DILUTE' be arranged such that the vowels may appear in the even places? a) 36 b) 720 c) 144 d) 24 Choosing from N things M things 16) How many 3 letters words (with or without meaning) can be formed out of the letters of the word, "PLATINUM", if repetition of letters is not allowed? a) 742 b) 850 c) 990 d) 336 17) How many 3-letter words can be formed with or without meaning from the letters A , G , M , D , N , and J , which are ending with G and none of the letters should be repeated? a) 20 b) 18 c) 25 d) 27 18) Find out the number of ways in which 12 Bangles of different types can be worn in 2 hands? A) 1260 B) 2720 C) 1225 D) 4096 19) In how many different ways can 6 apple and 6 orange form a circle such that the apple and the orange alternate? A) 82880 B) 86400 C) 71200 D) 63212 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 51  Complete Quantitative Aptitude Questions 20) There are 7 periods in each working day of a college. In how many ways can one organize 6 subjects such that each subject is allowed at least one period? A) 33200 B) 15120 C) 10800 D) 43600 11) Answer: C) The word ‘WAGGISH’ contains 7 letters of which 1 letter occurs twice = 7! / 2! = 2520 No. of permutations possible with vowels always together = 6! * 2! / 2! = 1440 / 2 = 720 No. of permutations possible with vowels never together = 2520-720 = 1800. 12) Answer: A) The arrangement is made in such a way that the vowels always come together. i.e., "ZYMGN(OE)". Considering vowels as one letter, 6 different letters can be arranged in 6! ways; i.e., 6! = 720 ways. The vowels "AE" can be arranged themselves in 2! ways; i.e.,2! = 2 ways Therefore, required number of ways = 720 x 2 = 1440 ways. 13) Answer: A) There are 8 different letters in the given word "XANTHOUS", out of which there are 3 vowels and 5 consonants. Let us mark these positions as under: Now, 3 vowels can be placed at any of the three places out of 4 marked 1, 3, 5, and 7. Number of ways of arranging the vowels = 4P3 = 24 ways. Also, the 5 consonants at the remaining positions may be arranged in 5P5 ways = 5! Ways = 120 ways. Therefore, required number of ways = 24 x 120 = 2880 ways. www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 52  Complete Quantitative Aptitude Questions 14) Answer: D) There are 6 different letters in the given word, out of which there are 3 vowels and 3 consonants. Let us mark these positions as under: Now, 3 vowels can be placed at any of the three places out of 3 marked 1, 3 and 5. Number of ways of arranging the vowels = 3P3 = 3! = 6 ways. Also, the 3 consonants can be arranged at the remaining 3 positions. Number of ways of these arrangements = 3P3 = 3! = 6 ways. Therefore, total number of ways = 6 x 6 = 36. 15) Answer: A) There are 3 consonants and 3 vowels in the word DILUTE. Out of 6 places, 3 places odd and 3 places are even. 3 vowels can arranged in 3 even places in 3p3 ways = 3! = 6 ways. And then 3 consonants can be arranged in the remaining 3 places in 3p3 ways = 3! = 6 ways. Hence, the required number of ways = 6 x 6 = 36. 16) Answer: D) The word PLATINUM contains 8 different letters. Required number of words = number of arrangements of 8 letters taking 3 at a time. = 8p3 =8x7x6 = 56×6 = 336 17) Answer: A) Since each desired word is ending with G, the least place is occupied with G. So, there is only 1 way. The second place can now be filled by any of the remaining 5 letters (A , M , D , N , J ). So, there are 5 ways of filling that place. Then, the first place can now be filled by any of the remaining 4 letters. So, there are 4 ways to fill. Required number of words = (1 x 5 x 4) = 20. 18) Answer: D) www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 53  Complete Quantitative Aptitude Questions The first bangle can be worn in any of the 2 hands (2 ways). Similarly each of the remaining 11 bangles also can be worn in 2 ways. Hence total number of ways=2×2×2×2×2×2×2×2×2×2×2×2 =2^12 =4096 19) Answer: B) 6 apples can be arranged in (6-1)! Ways Now there are 6 positions in which 6 orange can be placed. This can be done in 6! ways. Required number of ways = (6-1)! × 6! = 5! × 6! = 120 × 720 = 86400 20) Answer: B) 6 subjects can be arranged in periods in 7P6 ways. Remaining 1 period can be arranged in 6P1 ways. Two subjects are alike in each of the arrangement. So we need to divide by 2! to avoid over counting. Total number of arrangements = (7P6 x 6P1)/2! = 5040 × 6 / 2 = 30240 / 2 = 15120 21) How many 5-letter code words are possible using last 10 letter of the English alphabet , if no letter can be repeated ? a) 30240 b) 25440 c) 45640 d) 32940 22) In how many ways can a group of 10 men and 5 women be made out of a total of 12 men and 10 women? A) 16632 B) 15290 C) 25126 D) 34845 E) 38135 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 54  Complete Quantitative Aptitude Questions 23) A box contains 2 pink balls, 3 brown balls and 4 blue balls. In how many ways can 3 balls be drawn from the box, if at least one brown ball is to be included in the draw? A) 32 B) 48 C) 64 D) 96 E) None 24) Pramoth has 12 friends and he wants to invite 7 of them to a party. How many times will 4 particular friends never attend the party? A) 8 B) 7 C) 12 D) 15 25) In how many ways can 8 different ballons be distributed among 7 different boxes when any box can have any number of ballons? A) 5^4-1 B) 5^4 C) 4^5-1 D) 7^8 Problems on Selecting Items 26) A question paper has two parts A and B, each containing 12 questions. If a student needs to choose 10 from part A and 8 from part B, in how many ways can he do that? A) 32670 B) 36020 C) 41200 D) 29450 27) A bowl contains 8 violet, 6 purple and 4 magenta balls. Three balls are drawn at random. Find out the number of ways of selecting the balls of different colours? A) 362 B) 2 48 C) 122 D) 192 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 55  Complete Quantitative Aptitude Questions 28) An shopkeeper has 15 models of cup and 9 models of saucer. In how many ways can he make a pair of cup and saucer? A) 100 B) 80 C) 110 D) 135 29) There are 10 orange, 2 violet and 4 purple balls in a bag. All the 16 balls are drawn one by one and arranged in a row. Find out the number of different arrangements possible. A) 25230 B) 23420 C) 120120 D) 27720 30) In how many ways can a team of 6 persons be formed out of a total of 12 persons such that 3 particular persons should not be included in any team? A) 56 B) 112 C) 84 D) 128 21) Answer: A) The number of 5 letter code words out of the last 10 letters of the English alphabets are = 10× 9× 8 × 7× 6 = 80 × 63× 6 = 30240 ways. 22) Answer: A) Required number of ways = 12C 10 x 10C5 = 66 × 252 = 16632 23) Answer: C) www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 56  Complete Quantitative Aptitude Questions 24) Answer: A) Remove the 4 particular friends and invite 7 friends from the remaining 8 (12-4) friends. this can be done in 8C7 ways. Therefore , required number of ways = = 8C7 = 8C1 =8 25) Answer: D) Here n = 7, k = 8. Hence, required number of ways = n^k =7^8 26) Answer: A) Number of ways to choose 10 questions from part A = 12C10 Number of ways to choose 8 questions from part B = 12C8 Total number of ways= 12C10 × 12C8 = 12C2 × 12C4 [∵ nCr = nC(n-r)] ={12×11}/{2×1}X{12×11×10×9} / {4×3×2×1} =66×495 =32670 27) Answer: D) 1 violet ball can be selected is 8C1 ways. 1 purple ball can be selected in 6C1 ways. 1 magenta ball can be selected in 4C1 ways. Total number of ways = 8C1 × 6C1 × 4C1 = 8×6×4 = 192 28) Answer: D) He has 15 patterns of cup and 9 model of saucer A cup can be selected in 15 ways. A saucer can be selected in 9 ways. www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 57  Complete Quantitative Aptitude Questions Hence one cup and one saucer can be selected in 15×9 ways =135 ways 29) Answer: C) Number of different arrangements possible = {16!} / {10! 2! 4!} = {16×15×14×13×12×11×10×9×8×7×6×5×4×3×2 } / {(10×9×8×7×6×5×4×3×2 ) (2) (4×3×2)}} = {16×15×14×13×12×11} / {(2)(4×3×2)} = {8×5×7×13×3×11} = 120120 30) Answer: C) Three particular persons should not be included in each team. i.e., we have to select remaining 6- 3= 3 persons from 12-3 = 9 persons. Hence, required number of ways = 9C3 = {9×8 × 7} / {3 × 2 × 1} = 504 / 6 = 84 Circular Permutations There are two cases of circular-permutations:- (a) If clockwise and anti clock-wise orders are different, then total number of circular-permutations is given by (n-1)! (b) If clock-wise and anti-clock-wise orders are taken as not different, then total number of circular- permutations is given by (n-1)!/2! 31) In how many ways can 6 girls be seated in a rectangular order? A) 60 B) 120 C) 5040 D) 720 32) In how many ways can 10 stones can be arranged to form a bangles? A) 267720 B) 284360 C) 125380 D) 181440 33) In a birthday party, every person shakes hand with every other person. If there was a total of 66 handshakes in the party, how many persons were present in the party? www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 58  Complete Quantitative Aptitude Questions A) 9 B) 8 C) 7 D) 12 34) A school has 9 maths teachers and 6 science teachers. In how many ways can a team of 4 maths teachers be formed from them such that the team must contain exactly 1 science teacher? A) 800 B) 720 C) 680 D) 504 35) In how many ways can 8 different balls be distributed in 6 different boxes can contain any number of balls except that ball 4 can only be put into box 4 or 5 ? A) 2×5^6 B) 2×6^7 C) 2×5^4 D) 2×4^7 31) Answer: B) Number of arrangements possible = (6-1)! = 5! = 5×4×3×2×1 = 120 32) Answer: D) Number of arrangements possible = {1} / {2} X (10-1)! = {1} / {2}X 9! = {1} / {2} X 9×8×7× 6×5×4×3×2×1 = {1 / 2 } ×362880 = 181440 33) Answer: D) Assume that in a party every person shakes hand with every other person Number of hand shake = 66 Total number of hand shake is given by nC2 Let n = the total number of persons present in the party nC =66 2 n (n-1) / 2 = 66 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 59  Complete Quantitative Aptitude Questions n² - n = 2 × 66 n² - n – 132= 0 n= 12 , -11 But we cannot take negative value of n So, n = 12 Therefore number of persons in the party = 12 34) Answer: D) The team should have 4 maths teachers. But the team must contain exactly 1 science teacher. Hence, select 3 maths teachers from 9 maths teachers and select 1 science teachers from 6 science teachers. Number of ways this can be done = 9C3 × 6C1 ={9×8×7}/{3×2×1}X6 = 504 / 6 × 6 = 84 × 6 =504 35) Answer: B) 1st ball can be put in any of the 6 boxes. 2nd ball can be put in any of the 6 boxes. 3rd ball can be put in any of the 6 boxes. Ball 4 can only be put into box 4 or box 5. Hence, 4th ball can be put in any of these 2 boxes. 5th ball can be put in any of the 6 boxes. 6th ball can be put in any of the 6 boxes. 7th ball can be put in any of the 6 boxes. 8th ball can be put in any of the 6 boxes. Hence, required number of ways = 6×6×6×2×6×6×6×6 = 2×6^7 4. MIXTURE AND ALLIGATION MIXTURE When two or more components are mixed in a certain ratio, a mixture is created. Types of mixtures: Simple Mixtures:- When two or more different ingredients are mixed together, a simple mixture is formed. Eg,. sugar and water, wine with water, milk with water etc. www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 60  Complete Quantitative Aptitude Questions Compound Mixtures:- When two or more simple mixtures are mixed together, a compound mixture is formed. Eg., three or more than types of tea variety, rice variety etc. Basically mixing of two quantity i.e. cheaper quantity and nearer quantity mean price always falls between these two quantities from which respective ratio is calculated. Mean Price: Mean price is also known as median price or cost price of mixture of quantities taken, which lies between quantities taken for value consideration. Allegation: Allegation is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a desired price. Basic formula TYPE 1: 1) In what ratio should a shopkeeper mix two types of rice, one costing 40 rupees/kg and another costing 20 rupees/kg to get a rice variety costing 28 rupees/kg a) 3:2 b) 2:3 c) 4:3 d) 2:7 e) None www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 61  Complete Quantitative Aptitude Questions 2) Some amount out of Rs.8500 was lent at 5% per annum and the remaining was lent at 3% per annum. If the total simple interest from both the fractions in4 years was Rs.1500, the sum lent at 5% per annum was a) Rs. 2400 b) Rs. 2200 c) Rs. 2000 d) Rs. 6000 3) In what ratio should wheat at Rs12.60 per kg be mixed with wheat at Rs. 14.60 per kg so that the mixture be worth Rs.13 per kg ? a) 6:8 b) 8:7 c) 4:3 d) 2:7 e) 4:1 4) In what ratio must water be mixed with milk costing Rs.30 per litre in order to get a mixture worth of Rs.18 per litre? a) 3:2 b) 2:3 c) 4:3 d) 2:7 e) 1:2 5) How many kg of dal at Rs.8.40 per kg be mixed with32 kg of dall at Rs10.20 per kg to get a mixture worth Rs.9.60 per kg? a) 23 b) 49 c) 43 d) 72 e) 16 6) In 1 kg mixture of iron and carbon, 40% is carbon. How much iron should be added so that the proportion of carbon becomes 20% a)1.5kg b)2kg c)3kg d)5kg www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 62  Complete Quantitative Aptitude Questions e)1kg 7) Find the ratio in which Nuts at Rs. 5.40 a kg be mixed with Nuts at Rs. 3.20 a kg to produce a mixture worth Rs. 4.60 a kg. a) 9:2 b) 2:3 c) 4:3 d) 3:2 e) 7:4 8) In 20 litres of a mixture, the ratio of sugar syrub to honey syrub is 5:1. In order to make the ratio of sugar syrub to honey syrub as 3:1, the quantity of honey syrub that should be added to the mixture will be A) 5 2/3 B) 4 1/3 C) 6 2/3 D) 2 2/9 1) B Here also we can use Alligation as follows X = 40-28 = 4 ; y = 28-20 = 8 The ratio between the type 1 and type 2 rice is 4:6 or 2: 3 2) D Total simple interest received , I = Rs.1500 Principal , p = 8500 period, n = 4 years Rate of interest, r = ? Simple Interest, I=PNR/100 1500 = ( 8500 ×4× r )/ 100 = 1500/ 85×4 = 75/17 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 63  Complete Quantitative Aptitude Questions required ratio is: 12:5 Given that total amount is Rs.8500. Therefore, the amount lent at 5% per annum (part1 amount) = 8500x12/17 = 6000 3) E Thus the required ratio = 160 :40 = 4:1 4) B By the rule of allegation Ratio of water to milk = 12:18 = 2:3 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 64  Complete Quantitative Aptitude Questions 5) E By the rule of allegation ,we have Quantity of 1st kind dall : Quantity of 2nd kind dall = 60 : 120 => Quantity of 1st kind dall:32 = 1:2 Quantity of 1st kind dall =32 *1/2 = 16 6) E Quantity of the mixture : Quantity of iron = 20 : 20 = 1 : 1 Given that quantity of the mixture = 1 kg Hence quantity of iron to be added = 1 kg 7) E By the rule of allegation www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 65  Complete Quantitative Aptitude Questions Required ratio = 140 : 80 = 7:4 8) D By the rule of alligation Quantity of honey : Quantity of mixture =1/12:3/4}=1:9 Given that quantity of mixture = 20 litre =>Quantity of honey : 20 = 1 : 9 => Quantity of honey = 20 x1/9 = 2 2/9 litre Type II: Percentage of profit or loss 9) How many kg of cumin seeds at 62 rs per kg. must a man mix with 15 kg of cumin seeds at 20 rs per kg. So that he may , on selling the mixture at 40 rs per kg ,gain 25% on the outlay ? a) 62 b) 25 c) 42 d) 15 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 66  Complete Quantitative Aptitude Questions 10) A merchant has 1500 kg of salt part of which he sells at 12% profit and the rest at 18% profit. He gains 16% on the whole. The Quantity sold at 18% profit is a) 3020 b) 2000 c) 1000 d) 2700 e) 6003 11) How many kgs of Ponni rice costing Rs.21/kg should a shopkeeper mix with 12.5 kgs of ordinary rice costing Rs.12 per kg so that he makes a profit of 25% on selling the mixture at Rs.20/kg? a) 20 b) 23 c) 43 d) 27 e) 10 12) How many litres of water should be added to a 30 litre mixture of milk and water containing milk and water in the ratio of 3:7 such that the resultant mixture has 40% water in it? a) 3 b) 5 c) 6 d) 7 e) 9. 13) How many kilograms of sugar costing Rs. 17.2 per kg must be mixed with 54 kg of sugar costing Rs.14 per kg so that there may be a gain of 20 % by selling the mixture at Rs. 18.48 per kg? a) 63 b) 83 c) 73 d) 27 e) 42 14) In what ratio must coffee powder worth Rs. 40 per kg be mixed with coffee powder worth Rs. 45 a kg such that by selling the mixture at Rs. 48.40 a kg ,there can be a gain 10%? a) 3:2 b) 2:3 c) 4:3 www.ibpsguide.com | estore.ibpsguide.com | www.sscexamguide.com 67  Complete Quantitative Aptitude Questions d) 2:7 e) 1:4 15) 5 litre of coconut oil is added to 15 litre of a solution containing 40% of groundnut oil. The percentage of groundnut in the new mixture is a) 32 b) 23 c) 43 d) 33 e) 30 16) Suprith bought 40 kg of rava at the rate of Rs.8.50 per kg and 55 kg at the rate of Rs.8.75 per kg. He mixed the two. Approximately at what price per kg should he