BANK WRITTEN MATH COLLECTION
MOHAMMAD ZAHIRUL ISLAM Courtesy: Shaila Afrin Mousumi
1. In a two digit number, the digit in the units place is more than twice the digit in ten's place by 1. If the digits in the units place and the ten's place are interchanged, the difference between the newly formed number and the original number is less than the original number by 1, what is the original number? (JBL AEO 15) Solution: Let, ten’s digit = x, so unit’s digit = 2x + 1, the number is = 10x + 2x + 1 & the alternate number = 10(2x + 1) + x According to question, 10(2x + 1) + x – {10x + (2x + 1)} = 10x + (2x + 1) - 1 Or, 20x + 10 + x - 10x - 2x - 1 = 12x Or, 9x + 9 = 12x Or, x = 3 So, the number is = 10*3 + 2*3 + 1 = 37 Ans. 2. A, B & C, started a business investing tk 20,000 each. After five months A withdraw tk 5000, B withdraws tk 4000 and C invests tk 6000 more. At the end of the year a total profit of tk 69,900 was recorded. Find the share of each. (JBL AEO 15) Solution: Let us consider, k = 1000tk Then, ratio is, A : B : C = 20k*5 + 15k*7 : 20k*5 + 16k*7 : 20k*5 + 26k*7 = 205k : 212k : 282k = 205 : 212 : 282 Sum of the ratio = 205 + 212 + 282 = 699 Profit of A = (205*69900)/699 = 20500 Profit of B = (212*69900)/699 = 21200 Profit of C = (282*69900)/699 = 28200 Ans. 3. A machine P can print 1 lakh books in 8 hour, Q can same in 10 hour & R can print them in 12 hour. All the machine r started at 9 AM, while machine P in closed at 11 am and the remaining two machine complete the work. Approximately at what time the work will be finished? (JBL AEO 15) Solution: (P + Q + R)'s 1 hour's work = (1/8 + 1/10 + 1/12) = 37/120 Work done by P, Q and R in 2 hours = 2*37/120 = 37/60 Remaining work = (1 - 37/60) = 23/60 (Q + R)'s 1 hour's work = (1/10 + 1/12) = 11 /60. Now, 11/60 part work is done by Q and R in 1 hr. 1 part work is done by Q and R in = 1/(11/60) = 60/11 hrs So, 23/60 part work will be done by Q and R in = (60 /11)*( 23/60) = 23/11 hours = (23*60)/11 mins. = 125 mins (approx.) So, the work will be finished approximately 2 hours 5 mins. after 11 A.M., i.e., around 1.05 P.M. Ans. 4. a, b, c, d, e are 5 consecutive numbers in increasing order, deleting one of them from the set decreased the sum of the remaining numbers by 20% of the sum of 5. Which one of the number is deleted from the set? (BB AD 14) Solution: Let, the consecutive numbers are, a =1, b = 1 + 1 = 2, c = 1 + 2 = 3, d = 1 + 3 = 4 & e = 1 + 4 = 5 So, total = 1 + 2 + 3 + 4 + 5 = 15 Deleting 1 of the 5 numbers from the set then decreased 20% of the sum. 20% of the sum = (15 x 20)/100 = 3 So, the deleted number is the 3rd as c from the set Ans. 5. Rahim bought 2 varieties of rice costing tk 5 & 6 per kg each. Then he sold the mixture at tk7/kg, making profit
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Solution: Let, Rahim bought x kg rice at tk 5 & y kg rice at tk 6/kg So, cost price = 5x + 6y 20% profit = (5x + 6y)20/100 = (5x + 6y)/5 Again, selling price = 7(x + y) According to question, 7(x + y) - (5x + 6y) = (5x + 6y)/5
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of 20%. What was the ratio of the mixture? (BB AD 14, JBL EO 12)
Or, 7x + 7y – 5x - 6y = (5x + 6y)/5 Or, 2x + y = (5x + 6y)/5 Or, 10x + 5y = 5x + 6y Or, 10x + 5y – 5x – 6y = 0 Or, 5x – y = 0 Or, 5x = y Or, x/y = 1/5 Ans. 6. A team of 2 men and 5 women completed 1/4th of a job in 3ds. After that another man joined them and they all complete the next 1/4th of the job in 2ds. How many men can complete the whole job in 4 days? (Pubali SO 13) Solution: Here, in 3ds, 2 men & 5 women do 1/4 part So, in 1 d 2 men & 5 women do 1/12 part Again, in 2ds, 3 men & 5 women do 1/4 part Or, in 1d 3 men & 5 women do 1/8 part 1 mans’ 1 day work = 1/8 – 1/12 = 1/24 part So, in 24ds the whole job can be done by 1 man In 4ds the whole job can be done by = 24/4 = 6 men’ Ans. 7. A trader bought some mangoes for tk150/dozon and equal number of apples for tk 100/doz. If he sells all the fruits tk 140/dozon, what will be his profit/loss in percentage? (RKUB SO, 2014) Solution: Here, cost price of 1dozon mango = 150tk & cost price of 1dozon apple = 100tk So, cost price (1dozon mango + 1dozon apple) = 150 + 100 = 250tk Sale price (1dozon mango + 1dozon apple) = 140*2 = 280tk Profit = 280 – 250 = 30tk Percentage of profit = (30*100)/250 = 12% Ans. 8. Two partners A & B have 70% and 30% share in a business. After sometimes, a third partner C joined by investing tk 10 lakh and thus having 20% share in the business. What is the percentage of share of A now in the business? (RKUB Senior, 2014) Solution 20% share = 10lakh, 100% share = (10 x 100)/20 = 50 lakh Here, share of A&B = 50 – 10 = 40lakh A’s investment = (7*40)/10 = 28 lakh So, A’s percentage after joining C = (28*100)/50 = 56% Ans. 9. The average weight of three men A, B and C is 84kg. Another man D joins the group and the average becomes 80kg. Another man E, his wt is 3 kg more than that of D, replaces A. Now average wt of B, C, D & E becomes 79kg. What is the weight of A? (RKUB Senior, 2014)
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Solution A + B + C = 84 x 3 = 252kg A + B + C + D = 80 x 4 = 320kg So, D = 320 – 252 = 68kg, E = 68 + 3 = 71kg B + C + D + E = 4 x 79 = 316 Again, A + B + C + D + E = 320 + 71 = 391 So, A = 391 – 316 = 75 kg Ans.
10. A rectangle PQRS inscribed in a circle and PQ = 6. If the area of the rectangular region is 48, what is the area of the circular region? (RKUB Senior, 2014) Solution Here, area of PQRS = 48, PQ = 6, so, PQ x QR = 48 or, QR = 48/6 = 8 Now, if we connect P to R through a line, a right triangle PQR is formed. From Pythagoras theory we found, PR2 = PQ2 + QR2 Or, PR = (62 + 82) = 10 PR = Diameter of the circle = 10 So, area of the circle = π x (10/2)2 = 25π Ans. 11. If 5 is added to the sum of two digits of a number consisting of two digits, the sum will be three times the digits of the tens place. Moreover, if the place of the digits is interchanged, the number thus found will be 9 less than the original number. Find the number. (RB SO, 2013) Solution: Let, ten's digit = x & unit's digit = y, the number is = 10x + y. If the ten’s and unit’s digit interchanged the number will be = 10y + x According to the question, x + y + 5 = 3*x or, y = 2x – 5 ……… (1) again, according to the 2nd condition, 10y + x = 10x + y - 9 or, 9y = 9x - 9 or, y = x - 1 or, 2x - 5 = x – 1 [from (1)], so, x = 4, & y = 4 - 1 = 3. The number is = 10*4 + 3 = 43 Ans. 12. If a2 - 3a + 1 = 0, what is the value of a3 + 1/a3? (RB SO, 2013) Solution: a2 - 3a + 1 = 0 or, a(a + 1/a - 3) = 0 or, a + 1/a - 3 = 0 or, a + 1/a = 3..............(1) now, a3 + 1/ a3 = (a + 1/a)3 - 3a. 1/a (a + 1/a) = (3)3 - 3*3 [from (1)] = 33 - 33 = 0. Ans. 13. একটি আয়তকার ক্ষেত্রের পররসীমা আত্ররকটি বর্গ ত্রেত্রের পররসীমার সমান। আয়তেত্রের দৈর্ঘগ য তার প্রত্রের 3 গুন এবং ক্ষেেফল 1200sq.m। 50cm বর্গ াকার পাথর রৈত্রয় বর্গ ত্রেেটি পূরন করত্রত কতটি পাথর লার্ত্রব? (SBL SO 14) Solution: ধরি, আয়তকাি ক্ষেত্রেি প্রস্থ = x, দৈর্ঘ্য = 3x, তাহত্রে ক্ষেেফে, 3x2 = 1200, or, x2 = 4000 or, x = 20m & দৈর্ঘ্য = 60m আয়তকাি ক্ষেত্রেি পরিসীমা = 2(60+20) = 160m = বর্্ত্রেত্রেি পরিসীমা সুতিাাং, বর্্ত্রেত্রেি পরিসীমা, 4A = 160 or, A = 40m & ক্ষেেফে = 402 = 1600sq. m এখন পাথত্রিি দৈর্ঘ্য = 50cm = 0.5m, ক্ষেেফে = 0.5x0.5 = 0.25sq.m পাথি োর্ত্রব = 1600/0.25 = 6400 Ans. 14. এক বযরির ক্ষবতন প্ররত বছর রনরৈগ ষ্ট হাত্রর বাত্রে। ৪ বছর পত্রর তার ক্ষবতন ৩৫০০টাকা & ১০ বছর পত্রর ৪২৫০টাকা হল। ওই বযরি
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Solution: ৬ বছত্রি ক্ষবতন বাত্রে = ৪২৫০ – ৩৫০০ = ৭৫০টাকা ১ বছত্রি ক্ষবতন বাত্রে = ৭৫০/৬ = ১২৫টাকা ১০ বছত্রি ক্ষবতন বাত্রে = ১২৫*১০ = ১২৫০টাকা সুতিাাং, মূে ক্ষবতন = ৪২৫০ – ১২৫০ = ৩০০০টাকা
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কত টাকা ক্ষবতত্রন চাকরী শুরু কত্ররন এবং ক্ষবতন বাোর হার কত? (SBL SO 14)
তাহত্রে ৩০০০ টাকায় বাত্রে ১২৫ টাকা ১০০ টাকায় বাত্রে = (১২৫*১০০)/৩০০০ = ৪.১৭ টাকা সুতিাাং ক্ষবতন বাোি হাি মূে ক্ষবতত্রনি ৪.১৭% Ans. 15. যুরথ ও অরি একইরৈন একই বযাংক ক্ষথত্রক 20% সরল সুত্রৈ রিন্ন রিন্ন পররমান টাকা ক্ষলান কত্ররন। একই পররমান সুৈাসল যুরথ 3 বছত্রর এবং অরি 2 বছত্রর পররত্র াধ কত্ররন। তাত্রৈর ক্ষলাত্রনর অনুপাত কত? (SBL SO 14) Solution: মত্রন করি, যুরথি ক্ষোন = x এবাং অরিি ক্ষোন = y যুরথি ১ বচ্ছত্রিি সুৈ = x*20*1/100 = x/5 3 বছত্রি সুৈাসে = x + 3x/5 = 8x/5 আবাি অরিি ১ বছত্রিি সুৈ = y*20*1/100 = y/5 অরিি 2 বছত্রি সুৈাসে = y + 2*y*1/5 = 7y/5 প্রশ্নমত্রত, যুরথি সুৈাসে = অরিি সুৈাসে 8x/5 = 7y/5 বা, x/y = 7/8 Ans. 16. 13.5% হাত্রর কত টাকা 5 বছত্রর সুৈাসত্রল 8375 টাকা হত্রব? কত বছত্রর ওই টাকা মুনাফা আসত্রল 10400 টাকা হত্রব? (SBL JO 14) Solution: আমিা জারন, আসে P = 100*A/100+rt, A=সুৈাসে, r=সুত্রৈি হাি, t=সময় সুতিাাং, আসে = (100*8375)/(100+13.5*5) = 5000টাকা এখন, সুৈাসত্রে 10500 টাকা হত্রে সুৈ = 10400 – 5000 = 5400 5000 টাকাি 1 বছত্রিি সুৈ = 5000*13.5*1/100 = 675 টাকা 675 টাকা সুৈ হয় 1 বছত্রি 5400 টাকা সুৈ হয় = 5400/675 = 8বছত্রি Ans. 17. ২১ রমটার দৈর্ঘগ য এবং ১৫ রমটার প্রে রবর ষ্ট একটি পাত্রকগর চারপাত্র
২ রমটার চওো একটি রাস্তা আত্রছ, প্ররত বর্গ রমটার ২৫ টাকা হাত্রর
ওই রাস্তায় র্ঘাস লার্াত্রত কত খরচ পেত্রব? (SBL JO 14) Solution: িাস্তা ছাো পাত্রক্ি ক্ষেেফে = ২১*১৫ = ৩১৫ ব.রমটাি িাস্তাসহ বার্াত্রনি ক্ষেেফে = (২১+২*২)*(১৫+২*২)= ২৫*১৯ = ৪৭৫ ব.রমটাি িাস্তাি ক্ষেেফে = ৪৭৫-৩১৫ = ১৬০ ব.রমটাি র্ঘাস োর্াত্রত খিচ হত্রব = (১৬০*২৫)টাকা = ৪০০০ টাকা Ans. 18. Mr. Zaman defeated Mr. Younus in a vote where the ratio of their vote was 4:3, total number of voters was 581 of which 91 didn’t vote. Find out the margin of defeat. Solution: Casting of vote = 581-91=490, vote for Zaman = 490*4/7 = 280, Yonus = 490 – 280 = 210 So, margin of defeat = 280-210=70 Ans. 19. A mixture of 20kg spirit and water contains 10% water. How much water is to be added to increase the water up to 25%?
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Solution: Here, in 20kg solution amount of water = 20*10/100 = 2kg & amount of spirit = 20-2 = 18kg Now, in new solution, amount of water will be changed but amount of spirit will be the same. Percentage of water will be = 25% & percentage of spirit will be = 100-25 = 75% So, 75% spirit = 18kg, or, 100% spirit = 18*100/75 = 24kg. Water should be added = 24-20 = 4kg. Ans.
20. 15 years hence (after) A will be twice of his son, 5 years ago he was 4 th times of his son, what is their present age? Solution: Let us consider, present age of father = x & son = y 5 yrs ago fathers age, x – 5 = 4(y – 5) Or, x = 4y – 15 ……………….. (1) Again, after 15yrs fathers age, x + 15 = 2(y + 15) Or, 4y – 15 + 15 = 2y + 30 Or, 4y – 2y = 30 Or, y = 15, so father x = 4*15 – 15 = 45 Ans. 21. 3 partners A, B & C start a business. Twice the investment of A is equal to thrice the capital of B is 4 times the capital of C. Find the share of each out of a profit of tk. 297000. (Rupali SO 13) Solution let, 2A = 3B = 4C = x. A = x/2, B = x/3 & C = x/4. Ratio = A:B:C = x/2 : x/3 : x/4 = 1/2 : 1/3 : 1/4 = 6:4:3. A's share of profit = 297000*6/13 = tk. 137077. B's share of profit = 297000*4/13 = tk. 91385. C's share of profit = 29000*3/13 = tk. 68538. Ans. 22. in a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr & the time of flight increased by 30 minutes. Find out the duration of the flight. (Rupali SO 13) solution let, usual speed = x km/hr, so expected flight duration = 600/x hr Now, reduced speed = (x – 200) km/hr, so actual flight duration = 600/(x – 200) hr Time of flight increased = 30 mins = 30/60 hr = 1/2 hr According to question, 600/(x – 200) - 600/x = 1/2 Or, (600x – 600x + 120000)/x(x – 200) = 1/2 Or, 240000 = x2 – 200x Or, x2 – 200x – 240000 = 0 Or, x2 – 600x + 400x – 240000 = 0 Or, (x – 600)(x + 400) = 0 So, x = 600 Ans. 23. 2 pipes A & B can fill a tank in 36 minutes & 45 minutes respectively. Waste pipe C can empty the tank in 30 minutes. 1st A & B are opened. After 7 minutes, C is also opened. In how much time, the tank is full? (Rupali O 13)
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solution In 1 min A fills 1/36 part of tank In 1 min B fills 1/45 part of tank They together fill in 1 min = (1/36 + 1/45) = (5 + 4)/180 = 9/180 = 1/20 part In 7 min they together fill = 7/20 part of tank. Rest of the tank that is empty = 1 - 7/20 = 13/20 Again in 1min C empties = 1/30 part So, in 1 min A, B and C together fill = 1/20 – 1/30 = (30 - 20)/600 = 10/600 = 1/60 part Now, 1/60 part filled in 1 min 13/20 part filled in = (60*13)/20 = 39 min Ans.
24. ৬৪ক্ষকরি বালু ও পাথত্ররর রমশ্রত্রন বালুর পররমান ২৫%। কতটু কু বারল রম াত্রল পাথত্ররর পররমান ৪০% হত্রব? Solution ৬৪ ক্ষকরজ রমশ্রত্রন বােুি পরিমান = ৬৪*২৫/১০০ = ১৬ ক্ষকরজ, অতএব পাথত্রিি পরিমান = ৬৪ – ১৬ = ৪৮ ক্ষকরজ
এখন নতু ন রমশ্রত্রন শুধুমাে বােুি পরিমান বােত্রব এবাং পাথত্রিি পরিমান অপরিবরত্ ত থাকত্রব। নতু ন রমশ্রত্রন পাথত্রিি পরিমান = ৪০% ৪০% পাথি = ৪৮ ক্ষকরজ ১০০% পাথি = ৪৮*১০০/৪০ = ১২০ ক্ষকরজ সুতিাাং, বােু রমশাত্রত হত্রব = ১২০ – ৬৪ = ৫৬ ক্ষকরজ Ans. 25. A train traveling at 48 kmh completely crossed another train having half its length and traveling in opposite direction at 42 kmh in 12 seconds. It also passed a railway platform in 45 seconds. What is the length of the platform? Solution Let, length of the train is x Total length of two train = x + x/2 = 3x/2 Total speed of two train = (48+42) km/hr = 90 km/hr 1 km = 1000 meter 90 km = (100090) meter = 90,000 meter In 3600 seconds two train goes 90,000 meter “
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‘‘ 90000/3600
“ 12 “ “ “ ‘ (90000*12)/3600 = 300 meter So, 3x/2 = 300 or, 3x = 600 or, x = 200 Again, In 3600 seconds two train goes = 48,000 meter “
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“ = 48000/36
“ 45 “ “ “ “ = (48000*45)/36 = 600 meter Length of the train (600 200) meter = 400 meter Ans. 26. একটি আয়তাকার র্ঘর এর দৈর্ঘগ য প্রে অত্রপো ১৪রম. ক্ষব ী। র্ঘরটির পররসীমা ৭২ রম। হত্রল র্ঘরটির পররসীমার সমান পররসীমারবর ষ্ট বর্গ ত্রেত্রের ক্ষেেফল কত? (Sonali SO 13) Solution আয়তত্রেত্রেি পরিসীমা = বর্গ ত্রেত্রের পরিসীমা = ৭২ রম. সুতিাাং, বর্গ ত্রেত্রের এক বাহুর দৈর্ঘগ য = ৭২/৪ = ১৮ রম. বর্গ ত্রেত্রের ক্ষেেফল = ১৮২ = ৩২৪ রম২ Ans. 27. সুৈাসত্রল ৫ বছত্রর ৫২৫ টাকা এবং ৮ বছত্রর ৬৬০ টাকা হত্রল সুত্রৈর হার এবং আসল? (Sonali SO 13)
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Solution ক্ষৈওয়া আত্রছ, সুৈাসত্রে ৫ বছত্রি ৫২৫ টাকা এবাং ৮ বছত্রি ৬৬০ টাকা হয় সুতিাাং, ৩ বছত্রিি সুৈ = ৬৬০ – ৫২৫ = ১৩৫ টাকা ৫ বছত্রিি সুৈ = ১৩৫*৫/৩ = ২২৫ টাকা, আসে = ৫২৫ – ২২৫ = ৩০০ টাকা এখন, ৩০০ টাকাি ৫ বছত্রিি সুৈ ২২৫ টাকা ৩০০ টাকাি ১ বছত্রিি সুৈ ২২৫/৫ টাকা ১ টাকাি ১ বছত্রিি সুৈ ২২৫/(৫*৩০০) টাকা ১০০ টাকাি ১ বছত্রিি সুৈ (২২৫*১০০)/(৫*৩০০) = ১৫% Ans.
28. A simple interest rate of a bank was reduced to 5% from 7%. As a consequences Mr. B’s income was reduced by tk 2100 in 5 yrs. How much is Mr. B’s initial deposit in the bank? Solution: Here, reduced interest rate = 7% - 5% = 2% In 5 yrs interest reduced = 2100 tk In 1 yr interest reduced = 2100/5 = 420 tk So, 2% = 420 tk. or, 100% = (420*100)/2 = 2100 Tk. Ans. 29. A can do a piece of work in 10 days, while B alone can do it in 15 days. They work together for 5 days and the rest of the work is done by C in 2 days. If they get Tk.450 for the whole work, how should they divide the money? (BASIC-2014) Solution: A's 1 day work=1/10, his 5 days work = 5/10 = 1/2 of the work B's 1 day work =1/15, his 5 days work = 5/15 = 1/3 of the work 1 day work of (A+B) =1/10+1/15 = 1/6 so, 5 days work = 5/6 Remaining work=1-5/6=1/6 So, C has done 1/6 of the work So, A’s money = 450/2 = 225 tk, B = 450/3 = 150 tk, C = 450/6 = 75 tk Ans. 30. A sum of Tk. 1260 is borrowed from a money lender at 10% p.a compounded annually. If the amount is to be paid in two equal annual installments, find the annual installments. (BASIC-2014) Solution: We know, compound amount = P(1+r/100)t = 1260(1+10/100)2 = 1260(1.1)2 = 1260*1.21 = 1524.6 So, annual installment = 1524.6÷2 = 762.3tk. Ans. 31. ররহম, কররম এবং র্ািী ৩ িত্রন ১ টি কাি করত্রত পাত্রর যথাক্রত্রম ১৫, ১৬ এবং ১০ রৈত্রন। তারা একত্রে ঐ কািটি কত রৈত্রন ক্ষ ষ করত্রব? (SBL O 13) Solution: িরহম ১ রৈত্রন কত্রি কাজটিি ১/১৫ অাংশ অনুরুপভাত্রব করিম কত্রি ১/৬ এবাং র্াজী কত্রি ১/১০ অাংশ ৩ জত্রন একত্রে ১ রৈত্রন কত্রি = ১/১৫ + ১/৬ + ১/১০ = (২ + ৫ + ৩)/৩০ = ১/৩ অাংশ িরহম, করিম ও র্াজী ১ রৈত্রন কত্রি = ১/৩ অাংশ সুতিাাং, িরহম, করিম ও র্াজী একত্রে কাজটি ক্ষশষ কিত্রব = ৩ রৈত্রন Ans. 32. How much interest will tk 1000 earn in 1 year @ an annual interest rate of 20% if interest rate is compounded every 6 months? (Rupali SO 13)
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Solution: We know, compound amount = P(1+r/100)t = 1000 (1+10/100)2 [t=2 bcz interest rate is compounded in every 6 months] = 1000 * (1 + 1/10)2 = 1000*(1.1)2 = 1000*1.21 = 1210 tk. So, interest = 1210 – 1000 = 210 tk. Ans.
33. A train went 300 miles from X to city Y @ an average speed of 80mph. At what speed did it travel on the way back if it its average speed for the whole trip was 100mph? Solution: The train, started its journey from X to Y, than came back to X again So, total distance covered by it = 300*2 = 600 mile Here, average speed of total journey is 100 mph, so, total time spent = 600/100 = 6 hr. Time req. to go from X to Y at 80mph = 300/80 = 3.75 hrs. Time req. to comeback from Y to X = 6 – 3.75 = 2.25 hrs. So, speed of travelling on way back = 300/2.25 = 133.33 mph Ans. 34. If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be? Solution : Let 1 man's 1 day's work = x and 1 boy's 1 day's work = y. Then, 6x + 8y = 1/10 and 26x + 48y = 1/2. Solving these two equations, we get: x =1/ 100 and y = 1/200 (15 men + 20 boys) 1 day's work =15/100 + 20/200 = 1/ 4 15 men and 20 boys can do the work in 4 days. Ans 35. An article is sold at 20% profit. It its cost price is increased by Tk 50 and at the same time if its selling price is also increases by Tk 30. The percentage of profit decrease by10/3%. Find the cost price. (Pubali SO 13) Solution: Let cost price = x Tk. Then Selling price = x + x*20/100 = 6x/5 Profit = Selling price - Cost price = 6x/5 – x = (6x - 5x)/5 = x/5 --------->(1) Now, the increased cost price = x + 50 and selling price 6x/5 + 30 = (6x + 150)/5 So, profit is (6x + 150)/5 - (x + 50) = (6x + 150 – 5x – 250)/5 = (x - 100)/5 ------->(2) Now, the profit percentage decreases by 10/3% means {x/5 - (x - 100)/5}*100/(x/5) = 10/3 Or, {x/5 - (x - 100)/5} = 10/3*x/5*1/100 Or, (x – x + 100)/5 = x/150 Or, x = 100/5*150 = 3000 So, cost price is 3000 Tk. Ans. 36. A picnic was attended by 240 persons. There were 20 more men than women and 20 more adults than children. How many men were there in the picnic?
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Solution: let us consider, men = x, women = y & children = z According to the question, x + y + z = 240 …….. (1) x = y + 20 ……………… (2) x + y = z + 20 ………. (3) we can put value of x in equation (3) y + y + 20 = z + 20 or, z = 2y from equation (1) we can find y + 20 + y + 2y = 240 or, 4y = 240 – 20 or, y = 220/4 = 55 so, men = 55 + 20 = 75 Ans.
37. The ratio of the number of boys and girls in a school is 3:2. If 20% of the boys and 25% of girls are scholarship holders, what % of the students does not get scholarship? Solution: Let total Student 100 Boys = 100*3/5 = 60 Scholarship holder = 60*20% = 12 Girls = 100*2/5 = 40 Scholarship Holder = 40*25%= 10 Total Scholarship holder = 12 + 10 = 22 Does not get Scholarship: 100 – 22 =78% Ans. 38. A salesman's commission is 5% on all sales up to tk 10000 and 4% on all sales exceeding this. He remits tk 31100 to his parent company after deducting his commission. Find the total sales. Solution : Here, sales commission for 10000 = (10000*5)/100 = 500 tk. So, payable amount = 10000 – 500 = 9500 tk Rest amount = 31100 – 9500 = 21600 At 4% commission, If paid amount is 96 tk then sales is 100 tk If paid amount is 21600 tk then sales is (100*21600)/96 = 22500 tk So, total sales = 22500 + 10000 = 32500 tk Ans. 39. If sugar price reduced 25/4%, then one can buy 1kg more sugar at 120tk. Find the rate of original and reduced price. (JBL Cash 15) Solution: Let, the original rate = x tk/kg, so in 120 tk sugar can be found = 120/x kg Now, 25/4% discount in x tk = 25x/4*100 = x/16 tk So, reduced rate = x - x/16 = 15x/16 tk per kg and in 120 tk we can found = 120*16/15x = 128/x kg sugar According to question, 128/x - 120/x = 1 Or, 8/x = 1, or x = 8 tk/kg So, discount price is = 15*8/16 = 7.5 tk/kg Ans. 40. If 2 men and 3 boys can do a piece of work in 10 days; and if 3 men and 2 boys can do the same piece of work in 8 days, then 2 men and 1 boy can do that work in how many days? (JBL Cash 15) Solution: In 1 day 2 men and 3 boy do = 1/10 part ………………… (1) 3 men and 2 boy do = 1/8 part ………………….. (2) _____________________________________ 5 men and 5 boy do = 9/40 part So, 1 man and 1 boy do = 9/200 part in 1 day …………………. (3) We find from (2) – (3) 2 men and 1 boy do = 1/8 – 9/200 = (50 – 18)/400 = 32/400 = 2/25 part So, 2 men and 1 boy do 2/25 part work in 1 day 2 men and 1 boy do 1 part or whole work in 25/2 days = 12.5 days Ans.
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Solution: Let, 8% invest is = x 6% invest is = 1550 – x The interest on tk x = 8x/100 tk And interest on tk (1550 – x) = 6(1550 – x)/100 tk According to the question,
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41. A total amount of 1550 tk was invested in two parts. One part is 8% rate and the other part is 6% rate. If the annual income is tk 106, then how much money was invested in each part? (JBL Cash 15)
(1550 - x)*6/100 + 8x/100 = 106 => (1550 – x)*6 + 8x = 106*100 =>9300 - 6x + 8x = 10600 =>2x = 10600 - 9300 =>x = 650 8% invest is = 650 6% invest is = 1550 – 650 = 900 Ans. 42. A,B and C enter into partnership by investing in the ratio of 3:2:4.After one year, B invests another tk 270000.At the end of three years profits are shared in the ratio of 3:4:5. Find the initial investment of each. Solution: Say, A,B,C initial investment=3x,2x,4x So, ratio of their investment 3x*3:[2x*1 + (2x + 270000)2]:4x*3 = 9x:2x + 4x + 540000:12x = 9x:6x + 540000:12x According to the question, 9x:6x + 540000:12x = 3:4:5 Now, 9x/6x + 540000 =3/4 Or, 36x = 18x + 1620000 Or, 18x = 1620000 So, x= 90000 so A's initial investment = 3*90000 = 270000 Tk B = 2*90000 = 180000 Tk. C = 4*90000 = 360000 Tk Ans. 43. A & B can do a work in 12 days where B & C can do the job in 16 days. A worked for 5 days, B worked for 7 days and C did the remaining job in 13 days. How many days would require for C to do the job alone? (BCBL 2015) Solution: A & B complete the work in 12 days So, in 5 days they do 5/12 part Again B & C do the work in 16 days In 2 days they do = 2/16 = 1/8 part Work remaining =1 – (5/12 + 1/8) = 11/24 part 11/24 part work completed by C in 11 days Whole work or 1 part work completed by C in 11/(11/24) = 24 days Ans. 44. Mr. Nader drove for Mymensingh to Dhaka @ 60 miles/hr. Returning on the same route there was a lot of traffic and he was only able to drive @ 40 miles/hr. If the return trip took 1hr longer, what is the distance between Dhaka & Mymensingh? (BASIC AM 12) Solution: Let the distance between Dhaka & Mymensingh = x km We know, Speed = Distance/Time or, Time = Distance/Speed According to question, x/40 – x/60 = 1 or, (60x – 40x)/2400 = 1 or, 20x = 2400 or, x = 120 miles Ans.
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Solution: Here, 95% of stolen money = 1615 tk So, 100% of the stolen money = (1615*100)/95 = 1700tk According to question, 0.5x + 1x + 2x + 5x = 1700 Or, x(0.5 + 1 +2 + 5) = 1700 Or, x = 1700/8.5 = 200 Ans.
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45. A businessman before closing his shop, counted the money kept in the cash box and found there were X number 50 paisa coin, X number of 1 tk notes, X number of 2 tk notes and X number of 5 tk notes. Apart from this there is nothing in the box. The next day when he opened the shop he founds that the cash box had been stolen. As he was insured, he got tk 1615 which is 95% of the stolen money from the insurance company. How many 2 tk notes were in the box? (BASIC AM 12)
46. একটি করিউটার রবজ্ঞান পরীোয় ৩০% পরীোথী পা
কত্ররত্রছ। যারা পা
করত্রত পাত্রররন তাত্রৈর ১২ িন করিউটার রবজ্ঞান
ক্ষকাত্রসগ অং গ্রহন কত্ররত্রছ এবং ৩০ িন উি ক্ষকাত্রসগ অং গ্রহন কত্রররন। কয়িন পরীোথী পরীোয় অং গ্রহন কত্ররত্রছ? (BKB CO 12) Solution: এখাত্রন পিীোয় পাশ কত্রিত্রছ ৩০%, সুতিাাং ক্ষফে কত্রিত্রছ = ১০০ – ৩০ = ৭০% পাস না কিা পিীোথীি ক্ষমাট সাংখযা = ১২ + ৩০ = ৪২ সুতিাাং, ৭০% পিীোথী = ৪২ জন ১০০% পিীোথী = (৪২*১০০)/৭০ = ৬০ জন Ans. 47. ৩৬০০ টাকা কত্রর ৈুটি ক্ষচয়ার রবরক্র করা হত্রয়ত্রছ। একটি ২০% লাত্রি এবং অনযটি ২০% লত্রস রবক্রয় করা হল। সব রমরলত্রয় কত লাি বা ক্ষলাকসান হল? (BKB CO 12) Solution: এখাত্রন ৈুইটি ক্ষচয়াত্রিি রবক্রয় মূেয = ৩৬০০ + ৩৬০০ = ৭২০০ টাকা মত্রন করি, প্রত্রতযকটি ক্ষচয়াত্রিি রবক্রয়মূেয = ১০০ টাকা প্রথম ক্ষচয়ািটি ২০% োত্রভ রবরক্র হত্রে রবক্রয়মূেয ১২০ টাকা হত্রে ক্রয়মূেয = ১০০ টাকা রবক্রয়মূেয ৩৬০০ টাকা হত্রে ক্রয়মূেয = (১০০*৩৬০০)/১২০ = ৩০০০ টাকা তাহত্রে ২য় ক্ষচয়ািটি ২০% ক্ষোকসাত্রন রবরক্র হয় রবক্রয়মূেয ৮০ টাকা হত্রে ক্রয়মূেয = ১০০ টাকা রবক্রয়মূেয ৩৬০০ টাকা হত্রে ক্রয়মূেয = (১০০*৩৬০০)/৮০ = ৪৫০০ টাকা সুতিাাং, ৈুইটি ক্ষচয়াত্রিি ক্রয়মূেয = ৪৫০০ + ৩০০০ = ৭৫০০ টাকা অতএব েস = ৭৫০০ – ৭২০০ = ৩০০ টাকা Ans. 48. A person spends 1/3rd of the money with him on food, 1/5th of the remaining on education, 1/4th of the remaining on treatment. Now he is left with tk 200. How much did he have with him in the beginning? (One Bank Officer 12) Solution: Let us consider initial money = x tk 1/3rd of the money is spent for food, so remaining money = x – x/3 = 2x/3 1/5th of the remaining is spent for education = 1/5*2x/3 = 2x/15, Remaining money = 2x/3 – 2x/15 = 8x/15 1/4th of the remaining money spent on treatment = ¼ * 8x/15 = 2x/15 Remaining money = 8x/15 – 2x/15 = 2x/5 Acq. to ques. 2x/5 = 200 or, x = (200*5)/2 = 500 tk Ans. 49. Suppose 81p + 62q = 138 and 62p + 81q = 5, find out the value of p & q. (One Bank Officer 12)
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Solution: Here, 81p + 62q = 138…………… (1) 62p + 81q = 5………………. (2) We find from (1)*62 and 2*81 5022p + 3844q = 8556 …………. (4) 5022p + 6561q = 405 …………… (5) Now, we can find from (4) – (5) - 2717q = 8151 or, q = -3 So, p = 4 Ans.
50. in a business Piku invested tk 6500 for 6 months, Qazi invested 8400 for 5 months, Raj invested 10000 for 3 months. Piku wants to be working member for which he will receive 5% of profit. If the total profit earned is tk 7400, what is the share of Qazi in profit? (Jamuna PO 12) Solution: Ratio of Piku, Qazi & Raj’s investment = (6500*6):(8400*5):(10000*3) = 39000:42000:30000 = 13:14:10 As a working member Piku will receive = 5% of 7400 = (5*7400)/100 = 370 tk So, remaining = 7400 – 370 = 7030 tk So, Qazi’s share in profit = (7030*14)/37 = 2660 tk Ans. 51. A bag contains tomatoes that are green or red. The ratio of green tomatoes in the bag is 4:3. When 5 green and 5 red tomatoes are removed, ratio becomes 3:2. How many red tomatoes were originally in the bag? (Jamuna MTO 14) Solution: Let, Green tomatoes = 4x, Red tomatoes = 3x According to Q, 4x-5/3x-5=3/2 or, 8x – 10 = 9x – 15 or, x=5 Red tomatoes originally in the bag are 15 Ans. 52. Kalim is asked to write a study guide for a text book. For his work the publishing company is giving him a choice of a onetime payment of tk 13375 or tk 2000 plus 10% royalty per copy sold. If the proposed royalty rate of tk 3.25 per copy sold, how many study guide to be sold for the total income received by Kalim to be the same from either choice? (Jamuna MTO 14) Solution: Here, 10 % of royalties =3.25tk/copy Amount of royalties kalim has to receive = 13375-2000 =11375tk So, copy to be sold = 11375/3.25 = 3500 nos. Ans. 53. In a certain store, the profit was 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit now? (Jamuna O 12) Solution: Let, cost price is = 100tk, profit = 320 tk, so selling price = 100 + 320 = 420tk Then, new cost price = 125 tk, selling price = 420 tk, so profit = 420 – 125 = 295 tk Profit is the percentage of new selling price = (295*100)/420 = 1475/21 = 70% approximately Ans. 54. A can do a work in 10 days, B can do it in 15 days. They work together for 5 days, rest of the work is done by C in 2 days. If they get tk 4500 for whole work, how should they divide money? (Basic Officer 14)
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Solution: Here, A do in 1 day = 1/10, so in 5 days = 1/2 Similarly, B don in 5 days = 1/3 (A+B) 5 days work = 1/2 + 1/3 = 5/6 So, 2 days work of C = 1 – 5/6 = 1/6 So, share of A = 4500*1/2 = 2250 tk Share of B = 4500*1/3 = 1500 tk Share of C = 4500*1/6 = 750 tk Ans.
55. There are 2 examination room A and B. if 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent B to A, then the number of students in A is double of the number of the students in B. Number of the students in A? (NRB O 14) Solution: Let us consider, there are x students in the room A and y students in the room B. According to the question, x – 10 = y + 10 or, y = x – 20 …………………. (1) Again, x + 20 = 2(y – 20) …………………. (2) Solving for x on (2) X + 20 = 2(x – 40) Or, x + 20 = 2x – 80 Or, x = 100 So, there are 100 students in room A Ans. 56. Amin and Sajal are friends. Each has some money. If Amin gives Tk 30 to Sajal, then Sajal will have twice the money left with Amin. But, if Sajal gives Tk 10 to Amin, then Amin will have thrice as much as is left with Sajal. How much money does each have? (NRB SEO 14) Solution: Let, Amin has Tk x and Sajal has Tk y Sajal’s money will be twice of Amin, if Amin gives Tk 30 to Sajal. So according to the question, 2(x - 30) = y + 30 or, 2x – 60 = y + 30 or, y = 2x – 60 – 30 = 2x – 90 ……….. (i) If Sajal gives taka 10 to Amin, according to the question, 3(y - 10) = x + 10 or, 3y – 30 = x + 10 or, 3(2x – 90) – 30 = x + 10 or, 6x – 270 -30 = x + 10 Or, 6x – x = 10 + 300 or, 5x = 310 or, x = 62 So, y = 2*62 – 90 = 124 – 90 = 34 So, Amin has Taka 62 & Sajal has Taka 34. Ans. 57. A contract is to be completed in 46 days and 117 men were set to work, each working 8 hours a day. After 33 days, 4/7 of the work is completed. How many additional men may be employed so that the work may be completed in time, each man now working 9 hours a day? (NRB SEO 14) Solution: After 33 days time remains = (46 - 33) = 13 days 4/7th of the work is completed, so works remains=1 - 4/7 = (7 - 4)/7 = 3/7 In 33 days 8 hours 4/7th of work is done by 117 men ” 1 ” 1 ” 1 ” ” ” ” (117*7*33*8)/4 ” 13 ” 9 3/7 ” ” ” ” (117*7*33*8*3)/(4*19*9*7) = 198 Men Additional fund required = 198 - 117 = 81 men. Ans. 58. Mr. Jones gave 40% of the money he had, to his wife. He also gave 20% of the remaining amount to each of his three sons. Half of the amount now left was spent on miscellaneous items and the remaining amount of Tk. 12000 was deposited in the Bank. How much money did Mr. Jones have initially? (NRB SEO 14)
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Solution: Let, Mr. Jones has initially = x tk Amount given to his wife = 40x/100 = 2x/5 tk and balance = x - 2x/5 = 3x/5 tk Amount given to sons = 20/100 of 3x/5 = 3x/25 tk, balance = 3x/5 - 3x/25 = 12x/25 tk Amount spent on miscellaneous = 1/2 of 12x/25 = 12x/50 = 6x/25 According to the question, 6x/25 = 12000 or, 6x = 12000*25 = 300000 or, x = 300000/6 = 50000 Ans.
59. Robi was 4 times as old as his son 8 years ago. After 8 years, Robi will be twice as old as his son. What are their present ages? (NRB O 14) Solution: Let, Robi’s present age x Robi’s son’s present age is y 1st condition, (x – 8) = 4(y – 8) or, x = 4y – 24 ..……… (1) 2nd condition, (x + 8) = 2(y + 8) or, x = 2y + 8 …………… (2) From 1 and 2 , 4y – 24 = 2y + 8 or, 2y = 32 or, y = 16 so, x = 2*16 + 8 = 40 Ans 60. In an examination, 80% of the students passed in English, 85% in Mathematics, and 75% in both English and Mathematics. If 40 students failed in both the subjects, find the total number of students? (NRB O 14) Solution: Here, passed in both subject = 75% Passed only in English = 80 – 75 = 5% and passed only in Mathmatics = 85 – 75 = 10% So, students who have passed at least in one subject = 75 + 5 + 10 = 90% Failed in both subject = 100 – 90 = 10% According to question, 10% = 40 or, 100% = 100*40/10 = 400 Ans. 61. A, B and C enter into a partnership by investing in ratio 3:2:4. After one year, B invests another Tk 270000. At the end of three year profits are shared in the ratio of 3:4:5. Find the initial investment of each. (NRB O 14) Solution: Let us consider, invest of A, B & C for 1 year is 3x, 2x & 4x 3 year invest of A = 3x*3 = 9x, B= 3*2x + 270000*2 = 6x + 540000, C = 3x*4 = 12x Now, according to the question, (6x + 540000)/9x = 4/3 Or, 36x = 18x + 1620000 Or, 18x = 1620000 Or, x = 90000 So, investment of A = 3*90000 = 270000 tk Investment of B = 2*90000 = 180000 tk Investment of C = 4*90000 = 360000 tk Ans. 62. In a country, 60% of the male citizen and 70% of the female citizen are eligible to vote. 70% of the male & 60% of female citizen is eligible to cast their vote. What fraction of citizens voted during their election? (JBL EO 12). Solution: Let, total number of male = x Total number of female = y So, male voted = x*(60/100)*(70/100) And female voted = y*(70/100)*(60/100) So, fraction of the citizen voted ={x*(60/100)*(70/100) + y*(70/100)*(60/100)}/(x+y) = {(x+y)(60*70)/100*100}/(x+y) = 60*70/10000 = 21/50 Ans.
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Solution: Let, the price of one piece chocolate in 1st shop = x So, total chocolate = 120/x Price of one piece chocolate in 2nd shop = x – 3 Total chocolate 120/(x – 3) According to the question, 120/(x – 3) - 120/x = 2
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63. A boy purchased some chocolates from a shop for tk 120. In the next shop he found that the price of per piece chocolate is tk3 less than that charged @the previous shop, as such he could have purchased 2 more chocolates. How many chocolates did he buy from the first shop? (JBL EO 12, Premier MTO 12).
Or, (120x – 120x + 360)/x(x – 3) = 2 Or, 360/x2 – 3x = 2 Or, 2x2 – 6x – 360 = 0 Or, x2 – 3x – 180 = 0 Or, x2 – 15x + 12x – 180 = 0 Or, (x + 12)(x – 15) = 0 So, x = 15 Number of chocolates bought in the first shop = 120/15 = 8 nos. Ans. 64. Ripon, Liton and Pintu started a business jointly with a total amount of taka 280. Ripon paid taka 45 more than Liton & Liton paid tk 70 less than Pintu. If the company made a profit of tk 56, how much profit should Liton receive? (PJO Cash 12) Solution: Let investment of Liton = x tk So, investment of Ripon = (x + 45) tk & Pintu = (x + 70) tk According to the question, x + (x + 45) + (x + 70) = 280 or, 3x + 115 = 280 or, 3x = 165 or, x = 55 so, share of Liton = 45 tk, Ripon = 55 + 45 = 100 tk, Pintu = 55 + 70 = 125 tk L:R:P = 55:100:125 = 11:20:25 Profit will be received by Liton = (11/56)56 = 11 tk Ans. 65. A borrower pays 8% interest/yr on the first 600tk he borrows and 7%/yr on the part of the loan in excess of tk 600. How much interest will the borrower pay on a loan of tk 6,000 for one year? (PJO Cash 12) Solution: Out of tk 6000, the first 600 is charged with 8% interest & the rest amount is charged with 7% interest. So, 8% interest for first 600 tk = 600*8/100 = 48 tk And, 7% interest for next (6000 – 600) = 5400tk is = 5400*7/100 = 378tk So, total interest for 6000tk for 1 year = 48 + 378 = 426 tk Ans. 66. A worker is paid taka “x” per hour for the first 5 hrs he works each day. He is paid tk “y” per hour for each hr he works in excess of 5 hrs. During one week, he works 8 hrs on Saturday, 11 hrs on Sunday, 12 hrs on Monday, 10 hrs on Tuesday, 9 hrs on Wednesday. What is the average daily wage for five days of week? (PJO Cash 12) Solution: Day wise earning of the worker: Saturday = 5x + 3y Sunday = 5x + 6y Monday = 5x + 7y Tuesday = 5x + 5y Wednesday = 5x + 4y Sum of five days = 5x*5 + y(3 + 6 + 7 + 5 + 4) = 25x + 25y Average earning = 25(x + y)/5 = 5(x + y) Ans.
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Solution: We know in case of simple interest, I = rpt/100 Here, interest: I = 432tk, principal: P = 1200tk, t = time, r = rate of interest According to question, rate = time or, r = t So, I = rpt/100 Or, rt = 100*I/p Or, r2 = (100*432)/1200 = 36 Or, r = 6 Ans.
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67. Reena took a loan of tk 1200 at simple interest for as many years as the rate of interest. If she paid tk 432 as interest at the end of the loan period, what was the rate of interest? (IFIC O 12)
68. Two students Rahim & Karim, appeared at an examination. Rahim scored 9 marks more than Karim did and Rahim’s mark was 56% of the sum of their marks. What was Karim’s score? (IFIC O 12) Solution: Let, Karim’s marks = x So, Rahim’s mark = x + 9 Their total marks = x + x + 9 = 2x + 9 According to question (2x + 9)56/100 = x + 9 Or, 112x + 504 = 100x + 900 Or, 112x – 100x = 900 – 504 Or, 12x = 396 Or, x = 33 Ans. 69. Abu started a business investing tk 70,000. Robu joined him after 6 months with an amount of tk 1,05,000 and Sabu joined them with tk 1,40,000 after another 6 months. In what ratio the amount of profit earned should be distributed among them after 3 years? (DBBL PO 12) Solution: Here, ratio of investment of Abu, Robu & Sabu is (70,000*36):(1,05,000*30):(1,40,000*24) = 25,20,000:31,50,000:33,60,000 = 252:315:336 = 12:15:16 So, the profit earned by Abu, Robu & Sabu should be distributed to the following proportion 12:15:16 Ans. 70. A trader mixes 26kg of rice priced at tk20/kg with 30kg rice of another variety priced at tk36/kg & sells the mixture at tk30/kg. What is the percentage of profit or loss? (DBBL PO 12) Solution: Here, cost price = 26*20 + 30*36 = 520 + 1080 = 1600 Selling price = 30*(26 + 30) = 30*56 = 1680 Profit = 1680 – 1600 = 80 tk So, percentage of profit = (80*100)/1600 = 5% Ans. 71. If zakir loses 8 pounds, he will weight twice as much his sister. Together they are now weighs 278 pounds. What is zakir’s present weight? (SEBL PO 12) Solution: Let, sisters’ wt. = x lb So, zakirs’ wt. = (2x + 8) lb According to question, 2x + 8 + x = 278 or, 3x = 278 – 8 = 270 or, x 270/3 = 90 So, zakirs’ wt. = 2*90 + 8 = 188 lb Ans. 72. A seller incurs a loss of 15% when a table is sold @ tk10,200. At what price the table should be sold to make a profit of 35% (STBL AO 12)
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Now, at 35% profit, If tk 100 is the buying price then the sold price is tk 135 Tk 1 ,, ,, ,, ,, ,, ,, ,, ,, ,, 135/100 12,000 ,, ,, ,, ,, ,, ,, ,, ,, ,, 135*12,000/100 = 16,200 tk Ans.
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Solution: At 15% loss, If tk 85 is the sold price then the buying price is tk 100 Tk 1 ,, ,, ,, ,, ,, ,, ,, ,, ,, 100/85 10200 ,, ,, ,, ,, ,, ,, ,, ,, ,, 100*10200/85 = 12,000 tk
73. You want to make a garden in front of your house. The length of the rectangular region is greater than its breadth by 20 meters. If the perimeter of the land is 200m and gardening cost is tk20 for each square meter of land, how much will be the total gardening cost? (STBL AO 12) Solution: Let, the breadth = x meter & length = x + 20 According to question, the perimeter 2(x + 20 + x) = 200 Or, 4x = 200 – 40 Or, x = 160/40 = 40 So, breadth = 40m & length = 40 + 20 = 60m Area of the garden = 40*60 = 2400sqm So, money required to make the garden = 2400*20 = 48000 tk Ans. 74. A basketball team has won 15 games and lost 9. If these games represent 50/3% of the total games to be played, then how many more games must the team win to average 75% for the season? (Premier MTO 12) Solution: Here, games already played = 15 + 9 = 24 These are the 50/3% of the total games have to be played 50/3% games = 24 or, 100% games = (24*3*100)/50 = 144 And 75% of 144 = (144*75)/100 = 108 So, the team has to win more = 108 – 15 = 93 games Ans. 75. A person earns yearly interest of tk 920 by investing tk x at 4% & tk y at 5% simple interest. If he invest tk x at 5% & tk y at 4% his yearly interest earning will be reduced by tk 40. Find out his investment. (Premier MTO 12) Solution: Here, interest in first case = 920 Tk The reduced interest in second case = 920 – 40 = 880 Tk We know, interest, I = rpt/100 According to 1st condition, 4x/100 + 5y/100 = 920 Or, (4x + 5y)/100 = 920 Or, 4x + 5y = 92000 ………. (1) Again, 2nd condition, 5x/100 + 4y/100 = 880 Or, 5x + 4y = 88000 ……… (2) By solving this two equation with (1)*5 - (2)*4 we find 20x + 25y = 46,000 20x + 16y = 35,200 9y = 10,800 Or, y = 12,000 We can put this figure of “y” in equation (1) 4x + 60000 = 92000 Or, 4x = 92000 – 60000 = 32000 Or, x = 8000 So, his investment at 4% rate is 8000tk and 12000tk at 5% rate of interest Ans. 76. A red light flashes 3 times per minute and green light flashes 5 times in two minutes at regular intervals. If
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Solution: Red light flashes 3 times in 1 minute = 60/3 = every 20 seconds Green light flashes 5 times in 2 minutes = (60*2)/5 = every 24 seconds Therefore, they will flash together after LCM of 20 and 24 = every 120 seconds In every hour they will flash together (60*60)/120 = 3600/120 = 30 times Ans.
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both lights start flashing at the same time, how many times do they flash together in each hour? (NBL PO 14)
Alternate way: Red light 1minute flash = 3 times 60 minutes flash = 3*60 =180 Green light 2 minutes flash = 5 times 60 minutes flash = 5*60/2 = 150 180 and 150 HCF is = 30 So, in every hour they will flash = 30 times Ans. 77. A company paid tk5,00,000 in merit raises to employees whose performances were rated A, B or C. Each employee rated A received twice the amount of the raise that was paid to each employee rated C, each employee rated B received 150% of the amount of the raise that was paid to each employee rated C. If 50 workers were rated A, 100 were rated B and 150 were rated C, how much was the raise paid to each employee rated B? (BASIC AM 12) Solution: Let raise of each employee rated C = x So, raise of each employee rated A = 2x And, raise of each employee rated B = (150*x)/100 = 3x/2 According to question, 50*2x + 100* 3x/2 + 150x = 5,00,000 Or, 100x + 150x + 150x = 5,00,000 Or, 400x = 5,00,000 Or, x = 1250 So, raise of each employee rated B = (3*1250)/2 = 1875 tk Ans. 78. The ratio between the perimeter and the breadth of a rectangle is 5:1. If the area of the rectangle is 216 sqm, what is the length of the rectangle? (Exim MTO 12) Solution: Let the length is “a” and breadth is “b” So, area of the rectangle, ab = 216 …………. (1) Now, the perimeter = 2(a + b) According to question, 2(a + b)/b = 5/1 Or, 2a + 2b = 5b Or, 2a = 3b Or, a = 3b/2 From equation (1) we find that b*3b/2 = 216 or, 3b2 = 432 or, b2 = 144 or, b =12, so, a = 3*12/2 = 18 Ans. 79. Mr. Khan invested an amount of tk. 13,900 divided into two different schemes A and B at a simple interest rate of 14% per annum and 11% per annum respectively. If the total amount of simple interest earned in 2 years be tk
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Solution: Let, amount invested at 14% in scheme A = x So, interest in 2 years = (14*x*2)/100 = 28x/100 Again, amount invested at 11% in scheme B = 13900 – x So, interest in 2 years = {11*(13900 – x)*2}/100 According to the question, 28x/100 + {11*(13900 – x)*2}/100 = 3508 Or, 28x/100 + (305800 – 22x)/100 = 3508 Or, 28x + 305800 -22x = 3508*100 Or, 6x = 350800 – 305800 = 45000 Or, x = 45000/6 = 7500 So, amount invested in scheme B = 13900 – 7500 = 6400 Ans.
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3,508. What was the amount invested in scheme B? (Exim MTO 12)
80. Nine years ago the age of P & Q were in the ratio of 2:3. After 7 years the ratio of their age will be 3:4. What is the present age of P? (Prime MTO 11) Solution: Let nine years ago, Age of P = 2x Age of Q = 3x Current age of P = 2x + 9 & Q = 3x + 9 After seven years age of P = 2x + 9 + 7 = 2x + 16 Q = 3x + 9 + 7 = 3x + 16 According to the question, (2x + 16)/(3x + 16) = 3/4 Or, 8x + 64 = 9x + 48 Or, 9x – 8x = 64 – 48 Or, x = 16 So, present age of P = 2*16 + 9 = 41 yrs Ans. 81. Simon purchased brand X pens for tk 4.0 apiece and brand Y pens for tk 2.80 apiece. If Simon purchased a total of 12 of these pens for tk 42.00, how many brand X pens did she purchase? (EBL O 05) Solution: Let, number of X branded pen purchased = a So, number of Y branded pencil purchased = 12 – a According to question, 4a + 2.8(12 – a) = 42 Or, 4a + 33.6 – 2.8a = 42 Or, 1.2a = 42 – 33.6 = 8.4 0r, a = 8.4/1.2 = 7 Ans. 82. In a group the ration of male to female was 1 to 4. After 140 additional males were added, the ratio of male to female became 2 to 3. How large was the entire group after the additional males were added? (EBL O 05) Solution: Let, number of male = x, female = 4x According to question, (x + 140)/4x = 2/3 Or, 8x = 3x + 420 Or, 5x = 420 0r, x = 84, so number of male = 84 Now, female = 4*84 = 336 So, total group member after addition = 84 + 140 + 336 = 560 Ans. 82. Mr. Mamun deposited tk 1,000 in a bank for one year. At the end of the year the bank gave him back tk 1,300. What was the rate of interest thet Mr. Mamun received on his deposit? (Jamuna PO 09) Solution: Here, interest = 1300 – 1000 = 300 tk We know, I = rpt/100 or, r = I*100/pt = (300*100)/1000*1 = 30% Ans. 83. A rope 40 feet long is cut into two pieces. If one piece is 18 feet longer than the other, what is the length in
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Solution: Let, the shorter piece is = x feet So, longer piece is = (x + 18) feet According to question, x + x +18 = 40 or, 2x = 22 or, x = 11 so, shorter piece is 11 feet long Ans.
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feet of the shorter piece? (Jamuna PO 09)
84. At a certain company, the average salary of 10 of the employees is tk. 30,000, the average salary of the 30 other employee is tk 40,000 and the average salary of rest 20 employees is tk 60,000. What is the average salary of the 60 employee of the company? (Jamuna PO 09) Solution: Here, total salary of the first 10 = 30,000*10 = 3,00,000 tk Total salary of the next 30 = 40,000*30 = 1,20,000 tk Ans total salary of the rest 20 = 60,000*20 = 1,20,000 tk So, salary of (10 + 30 + 20) = 60 employees is = 3,00,000 + 1,20,000 + 1,20,000 = 5,40,000 tk So, average salary = 5,40,000/60 = 45,000 tk Ans. 85. এক ক্ষলািী বযাবসায়ী ক্ষকান ক্ষপ্রাডাক্ট ক্ষকনার সময় ১০% লাি কত্রর এবং রবক্রয় করার সময় আত্ররা ১০% লাি কত্রর। একটি ক্ষপ্রাডাক্ট রবরক্র কত্রর ক্ষমাট ১০৫০ টাকা লাি করত্রল ক্রয়মূলয কত? (SEBL 15) Solution by Arif Rahman: That shopkeeper will be termed "greedy" only when he falsified weight in both buying and selling. Here is the solution. Let, Initial Volume = 100 kg Total Price = 100 tk 1st step: In buying Goods, By false balace, he defrauds 10%, i.e. he bought 110 kg (instead of 100 kg) by paying 100 tk. 2nd step: In selling Goods, By false balance, he again defrauds 10%, i.e. He sold 90 kg (instead of 100kg) for 100 tk. Now, 90 kg's selling price 100 tk So, 110 kg's selling price = 100*110 /90 = 122.22 So, required profit % = (122.22-100) = 22.22% Therefore, CP = 100/22.22*1050 = 4725.47 tk. Ans 86. One person invests some amount in simple interest for 5 years at 13.5%. At the end he got total 33500 tk. Find his initial investment? How many year need to be the total amount 40600? (SEBL 15) Solution: We know, principle amount, P = 100*A/100 + rt Here, A = Total amount, r = Rate. T = Time So, initial investment = (100*33500)/(100 + 13.5*5) = 3350000/167.5 = 20000 Now, profit (40600 – 20000) = 20600 tk @13.5% rate, interest for 1yr for 20000 is = 13.5*20000/100 = 2700 Tk. So, 20600 Tk interest for = 20600/2700 = 7.63 yrs Ans
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Solution: Let, no. of tk 2 notes = x and tk 5 notes = 30 – x According to the question, 2x + 5(30 – x) = 120 Or, 2x + 150 – 5x = 120 Or, - 3x = 120 – 150 = - 30
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87. Mr. X lost a wallet containing tk 120. Incidentally, he had only notes of tk 2 and tk 5 denominations in the wallet. If the total number of notes was 30, how many tk 5 denominations notes he have? (Uttara AO 2011)
Or, x = 10 So, tk 5 notes = 30 – 10 = 20 Ans. 88. A person bought an article and sold it at a loss of 10%. If he had bought it for 20% less and sold it for Tk. 55 more, he would have made a profit of 40%. What was the cost of article in taka? Solution: Let, cost price is 100 Tk. So @10% loss sale price is 90 Tk. Now, if the cost price is 20% less = 100 – 20 = 80 Tk. then profit would be 40% So, at 40% profit if the cost price is 100 then sale price is 140 If the cost price is 80 then sale price is (140*80)/100 = 112 Tk. Difference of cost price and sale price = 112 – 90 = 22 Tk. If sale price is 22 Tk more than cost price is 100 Tk. If sale price is 55 Tk more than cost price is (100*55)/22 = 250 Tk. Ans 89. Mr. Shahin worked 3/4th of his job at his lawn in 1¼ hours. Mr. Ashik who works twice as fast, finished his job. How many minutes did Mr. Ashik work? (Premier MTO 12) Solution: Here, Shahin do 3/4th of the work in = 5/4 hrs = (5*60)/4 mins = 75 mins The rest of the work = 1 – 3/4 = 1/4 is done by Ashik Now, Shahin do 3/4th of work in = 75 minutes So, ,, ,, 1/4th of ,, ,, = (75*4)/(3*4) = 25 mins. As Ashik works twice as fast as Shahin, so to do 14th of the same job he will take = 25/2 = 12.5 minutes Ans. 90. Mr. Babar travelled from A to B at a speed of 3km/hr and returned from B to A at a speed of 5 km/hrs. if the round trip was 8 hrs, what was the distance between A and B in km? (Premier MTO 12) Solution: Let us consider, distance between A & B = D km So, time required for travelling A to B = D/3 hr Again, time required for travelling B to A = D/5 hr According to the question, D/3 + D/5 = 8 Or, 8D/15 = 8 Or, D = 15 km Ans. 91. Tk 1105 was divided among A, B & C. the amount of money received by A & B were the ratio of 1:2. Amount of money received by B & C were in the ratio of 3:4. How much did A receive? (Premier MTO 12) Solution: A:B = 1:2 ………. (1) & B:C = 3:4 …………. (2) We can find from (1)*3 & (2)*2 A:B:C = 3:6:8 So, A will receive = 1105*3/17 = 195 tk. Ans.
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Solution: Buying price of one pen = 360/12 = 30 tk At 20% profit, sell price of one pen = 30 + (30*20)/100 = 30 + 6 = 36 tk. At 10% discount, If price is 90tk then original price is 100 If price is 36tk then original price is (36*100)/90 = 40tk Initial selling price of each pen quoted by him is 40tk Ans.
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92. A shopkeeper buys pens which are tk.360 a dozen. He quotes a selling price the customer and offers a discount of 10% realizing that even with the discount he can earn a profit of 20%, what was the initial selling price of each pen quoted by him?
93. Three taps A, B and C can fill a tank in 12, 15, 20 hours respectively. If A is open all time and B and C are open for 1 hour each alternatively, the tank will be full in? Solution: Part filled in 1 hour by A & B = 1/12 + 1/15 = 3/20 Part filled in 1 hour by A & C = 1/12 + 1/20 = 2/15 Part filled in 2 hours = 3/20 + 2/15 = 17/60 Part filled in 6 hours = 3*17/60 = 17/20 Remaining part = 1- 17/20 = 3/20 Now in 7th hour it's (A+B)'s turn, 3/20 part = 1 hour we know Total time =6 + 1 = 7 hours Ans. 94. A shopkeeper has certain number of eggs of which 5% are found to be broken, he sells 93% of the reminder still has 266 eggs left. How many eggs did he originally have? Solution: let total no of eggs = x, so, broken egg = 5x/100 = x/20 Remainder = x - (x/20) = 19x/20 According to the question, (19x/20) - (19x/20)*(93/100) = 266 Or, (19x/20)(1 – 93/100) = 266 Or, x = 266*20/19*100/7 Or, x = 4000. Ans. 95. Sheela & meela r 4 miles apart. If shela starts walking towards meela at 3 miles per hr and at the same time meela starts walking toward sheela at 2 miles per hour, how much time will pass before they meet? (NBL MTO 13) Solution: Their relative speed = 3 + 2 = 5 m/hr They go together 5 mile in 60 min They go together 1 mile in 60/5 min They go together 4 mile in (4*60)/5 min = 48 mins. Ans. 96. Abul, Kalam and Bashar had lunch together. Bashar’s meal cost 50% more than Kalam’s meal and Abul’s meal cost 5/6 as much as Bashar’s meal. If Kalam paid tk. 1000 for his meal, what was the total that Abul and Kalam paid together for lunch? (SBAC MTO-2014) Solution: Here, Kalam paid = 1000 Tk. So, Bashar paid= 1000 + 50% of 1000 = 1500 Tk. Abul paid = 5/6 of 1500 = 1250 Tk. Abul and Kalam paid = 1000 + 1250 = 2250 Tk. Ans. 97. A man sells two commodities for tk. 4000 each, neither losing nor gaining in the deal. If he sold one commodity at a gain of 25%, the other commodity is sold at a loss of? (Megna Bank MTO- 2014)
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Solution: In the first case, at 25% profit, If sold price is 125 Tk then cost price is 100 Tk If sold price is 4000 Tk then cost price is (100*4000)/125 = 3200 Tk Profit = 4000 – 3200 = 800 tk so, in the second case, there was 800 Tk loss because there was no loss or gain by selling both cost price = 4000 + 800 = 4800 Tk. percentage of loss = (800*100)/4800 = 16.66% Ans.
98. When a producer allows 36% commission on the retail price of his product, he earns a profit of 8.8%. What would be his profit percent it the commission is reduced by 24%? (PBL SO/O-2014) Solution: Let the retail price is = 100 At 36% commission sales price = (100 – 36) = 64 At 8.8% profit, if sale price is 108.8 so cost price = 100 If sales price is 64 then cost price is = (64*100)/108.8 = 58.82 New commission = 36 – 24 = 12% New sale price at 12% commission = 100 – 12 = 88, and profit = 88 - 58.82 = 29.18 Profit percent= (29.18*100)/58.81 = 49.6% Ans. 99. Two tanks x and y are filled to capacity with jet fuel. Tank x holds more than 600 gallon than y. if 100 gallon of fuel were to be pumped from each tank, tank x would contain then 3 times as much fuel as tank y. what is the total number of gallons in the two full tanks? (NBL PO-2014) Solution: Let, capacity of gallon x = a gallon and gallon y = b gallon According to question, a = 600 + b ……………………………. (1) a – 100 = 3(b – 100) ……………… (2) If we put value of a in eq. (2) from eq. (1), then600 – b – 100 = 3b – 300 Or, 3b – b = 500 + 300 Or, 2b = 800 or, b = 400 gallon. So, tank y contains = 400 gallon, tank x contains = 400 + 600 = 1000 gallons Ans. 100. A female worker of a factory serves on the basis of monthly salary. At the end of every year she gets a fixed increment. Her monthly salary becomes tk.4500 after 4 years of service and tk. 5000 after 8 years of service. Find the salary at the beginning of her service and the amount of annual increment of salary. (Union Bank SO-2014) Solution: Here, the basic salary will be the same, only increment will be added in every year. So, increment for 4 years = 5000 – 4500 = 500 Increment for 1 years = 500/4 = 124 Tk. Ans.
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