Find the required value of time.In the question, it is given that the rate of interest is 712% per annum and the principal amount is Rs.7500 and the final amount is Rs.8625.Using the Simple Interest Formula SimpleInterest=PrincipleAmount×Rate×Time100. Rate=7×2+12%⇒Rate=152%So, the rate of interest is 152% per annum. And, SimpleInterest=FinalAmount-PrincipalAmount SimpleInterest=8625-7500⇒SimpleInterest=1125Let, time be t years and then apply the Simple Interest Formula.1125=7500×15× t100×2⇒t=22501125⇒t =2yearsHence in 2Years Rs.7500 amount to Rs.8625 at rate 7.5% per annum. Show
(ii) 16 2/3% of 600 gm- 33 1/3% of 180 gm Sol. (i) 28% of 450 + 45% of 280 =[(28/100)*450 + (45/100)*280] = (126+126) =252. (iii) 16 2/3% of 600 gm –33 1/3% of 180 gm = [ ((50/3)*(1/100)*600) – Ex. 5. Sol. Required percentage = [ (40/(2 * 10)) * 100]% = 200%. Ex. 6. (i) ?% of 25 = 20125 (ii) 9% of ? = 63 (iii) 0.25% of ? = 0.04 Sol. (i) Let x% of 25 = 2.125. Then , (x/100)*25 = 2.125 (ii) Let 9% of x =6.3. Then , 9*x/100 = 6.3 (iii) Let 0.25% of x = 0.04. Then , 0.25*x/100 = 0.04 Ex. 7. Sol. 16 (2/3)% =[ (50/3)* )1/100)] = 1/6 = 0.166, 2/15 = 0.133. Clearly, 0.17 is the Ex. 8. = [(7/20-(10/3)]% of Rs.8400 =1/6 % of Rs.8400 Ex. 9. An inspector rejects 0.08% of the meters as defective. How many will be Sol. Let the number of meters to be examined be x. Ex. 10. Sixty five percent of a number is 21
less than four fifth of that number. What Sol. Let the number be x. Ex.11. Difference of two numbers is 1660. If 7.5% of the number is 12.5% of the X = 125*y/75 = 5*y/3. Ex. 12. Required percentage = [(0.028/81.472)*100]% = 0.034%. Ex. 13. Ex.14. Shobha’s mathematics test had 75 problems i.e.10 arithmetic, 30 algebra and Sol. Number of questions attempted correctly=(70% of 10 + 40% of 30 + 60% 0f 35) questions to be answered correctly for 60% grade=60% of 75 = 45 therefore required number of questions= (45-40) = 5. Ex.15. if 50% of (x-y) = 30% of (x+y) then what percent of x is y? Sol.50% of (x-y)=30% of(x+y) (50/100)(x-y)=(30/100)(x+y) Ex.16. Mr.Jones gave 40% of the money he had to his wife. he also gave 20% of the Sol. Let the
initial amount with Mr.jones be Rs.x then, Therefore 3x/25=12000 x= ((12000 x 35)/3)=100000 So Mr.Jones initially had Rs.1,00,000 with him. Short-cut Method : Let the initial amount with Mr.Jones be Rs.x Then,(1/2)[100-(3*20)]% of x=12000 (1/2)*(40/100)*(60/100)*x=12000 Ex 17 10% of the inhabitants of village having died of cholera.,a panic set in , during x=((4050*40)/27)=6000. Ex.18 A salesman`s commission is 5% on
all sales upto Rs.10,000 and 4% on all x-500-((x-10000)/25)=31,100 Ex .19 Raman`s salary was decreased
by 50% and subsequently increased by Ex.20 Paulson spends 75% of his income. His income is increased by 20% and he Ex21. The salary of a person was reduced by 10% .By what percent should his Ex.22 When the
price fo a product was decreased by 10% , the number sold Ex 23 . If the numerator of a fraction be increased by 15% and its denominator be Ex.24 In the new budget , the price of kerosene oil rose by 25%. By how much Ex.25 The population of a town is 1,76,400 . If it increases at the
rate of 5% per Ex.26 The value of a machine depreiates at the rate of 10% per annum. If its present Ex27. During one year, the population of town increased by 5% . If the total Ex.28 If A earns 99/3% more than B,how much percent does B earn less then A ? Ex. 29 If A`s salary is 20% less then B`s salary , by how much percent is B`s salary Ex30 .How many kg of pure salt must be added to 30kg of 2% solution of salt and Ex 31. Due to reduction of 25/4% in the price of sugar , a man is able to buy 1kg Ex.32 In an examination , 35% of total students failed in Hindi , 45% failed in Ex33. In an examination , 80% of the students passed in English , 85% in 11. PROFIT AND LOSS IMPORTANT FACTS COST PRICE: THE PRICE AT WHICH ARTICLE IS SELLING PRICE: THE PRICE AT WHICH ARTICLE IS SOLD. PROFIT OR GAIN:IF SP IS GREATER THAN
CP,THE SELLING PRICE LOSS: IF SPIS LESS THAN CP,THE SELLER IS SAID TO INCURED A FORMULA SOLVED PROBLEMS ex.1 A man buys an article for rs.27.50 and sells it for rs.28.50. find his gain %. Ex.2. If the a radio is sold for rs 490 and sold for rs 465.50.find loss%. Ex.3.find S.P when (ii)CP=rs 80.40,loss=5% ex.4 find cp
when: (i) cp=rs{(100/116)*40.60}=rs 35. ex.5 A person incures loss for by selling a watch for rs1140.at what price should the sol. let the new sp be rsx.then ex.6 A book was sold for rs 27.50 with a profit of 10%. if
it were sold for rs25.75, sol. SP=rs 27.50: profit =10%. Profit%={(0.75/25)*100}%=25/6%=3% Ex7 .If the cost price is 96% of sp then whqt is the profit % Ex.8. The cp of 21 articles is equal to sp of 18 articles.find gain or loss % Ex.9 By selling 33 metres of cloth , one gains the selling price of 11 metres . Find the Ex10 A vendor bought bananas at 6 for Rs.10 and sold them at Rs.4 for Rs.6 .Find Ex.11. A man brought toffees at for a rupee. How many for a rupee must he sell to For Rs.3/2, toffees sold =3, for Re.1, toffees sold = [3*(2/3)] = 2. Ex. 12.A grocer purchased 80 kg of sugar at Rs.13.50 per kg and mixed it with S.P =116% 0f Rs.3000 =Rs.[(116/200) *3000]=Rs.3480. Ex.13. Pure ghee cost Rs.100 per kg. After adulterating it with vegetable oil costing By the rate of allegation : C.P of 1kg oil 100 Mean price 30 20 Required ratio =30:20 =3:2. Ex. 14. A dishonest dealer professes to sell his goods at cost price but uses a weight of 960 gms for a kg weight . Find his gain percent. Sol .Gain% =[ Error *100 ]% = [(40/960)*100] % = 4 1 % (error value)-(error) 6 Ex 15. If the manufacturer
gains 10%,the wholesale dealer 15% and the retailer Sol: Ex16 . Monika purchesed a pressure cooker at 9/10th of its selling price and sold it at Sol: Gain%=(18x/100*10/9x*100)%=20% Ex .17 An article is sold at certain price. By selling it at 2/3 of its price one losses sol: now C.P=Rs20x/27*27/20x*100)%=35% Ex .18. A tradesman sold an article at a loss of 20%.if the selling price has
been Sol: x/4=100 or x=400 Sol: =>5x/4-26x/25=10.50x=(10.50*100)/21=50 Sol: Let the orginal price of the jewel be Rs p and let the profit earned by the thrid seller be x% ((100+X)/100*125/100*120/100*P)=(165/100*P) (100+X)=(165*100*100)/(125*120)=110=>X=10% Ex21 . A man 2 flats for Rs 675958 each.on one he gains 16% while on the other he Sol: In this case there will be alwalys loss. The selling price is immaterial Hence, loss % = (common loss and gain%)2 /10=(16/10)%=(64/25)%=2.56% Ex.22. A dealer sold three-fourth of his article at a gain
of 20% and remaining at a Let the C.P of the whole be Rs x Let the C.p of the horse be Rs.x, then C.P of the carriage =Rs(3000-x) let the marked price be Rs 100 =Rs(95/100*90/100*80/100*100)=Rs68.40 Ex .25 After getting 2 successive discounts,
a shirt with a list price of Rs 150 is Let the first discount be x% Let C.P =Rs 100.then ,marked price =Rs100 let the market price of each pen be Rs 1 Ex 28 . At
what % above C.P must an article be marked so as to gain 33% after Let C.P be Rs 100.then S.P be Rs 133 Let the retail price =Rs
100.then, commission=Rs 36 12. RATIO AND PROPORTION IMPORTANT FACTS AND FORMULAE I. RATIO: The ratio of two quantities a and b in the same units, is the fraction a/b and we write it as a:b. Ex. The ratio 5: 9 represents 5/9 with antecedent = 5, consequent = 9. 2. PROPORTION: The equality of two ratios is called proportion. 3. (i) Fourth Proportional: If a : b = c: d, then d is called the fourth proportional (ii) Third Proportional: If a: b = b: c, then c is called the third proportional to (iii) Mean Proportional: Mean proportional between a and b is square root of ab 4. (i) COMPARISON OF RATIOS: 5. (i) Duplicate ratio of (a : b) is (a2 : b2). (Componendo and dividendo) (ii) Sub-duplicate ratio of (a : b) is (√a : √b). (iv) Sub-triplicate ratio of (a : b) is (a ⅓ : b ⅓ ). 6. VARIATION: we write, x y. we write, x∞(1/y) SOLVED PROBLEMS Ex. 1. If a : b = 5 : 9 and b : c = 4: 7, find a : b : c. a:b:c = 5:9:63/4 =20:36:63. Ex. 2. Find: Sol. Then, 4 : 9 : : 12 : x 4 x x=9x12 X=(9 x 12)/14=27; (iii) Mean proportional between 0.08 and 0.18 Ex. 3. If x : y = 3 : 4, find (4x + 5y) : (5x - 2y). =(3+5)/(7/4)=32/7 Ex. 4. Divide Rs. 672 in the ratio 5 : 3. First part = Rs. (672 x (5/8)) = Rs. 420; Second part = Rs. (672 x (3/8)) = Rs. 252. Ex. 5. Divide Rs. 1162 among A, B, C in the ratio 35 : 28 : 20. A's share = Rs. (1162 x (35/83))= Rs. 490; B's share = Rs. (1162 x(28/83))= Rs. 392; C's share = Rs. (1162 x (20/83))= Rs. 280. Ex. 6. A bag contains 50 p, 25 P and 10 p coins in the ratio 5: 9: 4, amounting to Sol. Let the number of 50 p, 25 P and 10 p coins be 5x, 9x and 4x respectively. Ex. 7. A mixture contains
alcohol and water in the ratio 4 : 3. If 5 litres of water is added to the Sol. Let the quantity of alcohol and water be 4x litres and 3x litres respectively 13. PARTNERSHIP !IMPORTANT FACTS AND FORMULAE I called partners and the deal is known as partnership. 2. Ratio of Division of Gains: distributed a among the partners in the ratio of their investments. ii) When investments are for different time periods, then
equivalent capitals are Suppose A invests Rs. x for p months and B invests Rs. y for q months, then 3. Working and Sleeping Partners: A partner who manages the business is known . as a SOLVED EXAMPLES Ex. 1. A, B and C started a business by investing Rs. 1,20,000, Rs. 1,35,000 and Sol. Ratio of shares of A, Band C = Ratio of their investments A’s share = Rs. (56700 x (8/27))= Rs. 16800. B's share = Rs. ( 56700 x (9/27)) = Rs. 18900. C's share = Rs. ( 56700 x (10/27))=Rs. 21000. Ex. 2. Alfred started a
business investing Rs. 45,000. After 3 months, Peter joined Sol. Clearly, Alfred invested his capital for 12 months, Peter for 9 months and Ronald So, ratio of their capitals = (45000 x 12) : (60000 x 9) : (90000 x 3) Alfred's share = Rs. (16500 x (2/5)) = Rs. 6600 Peter's share = Rs. (16500 x (2/5)) = Rs. 6600 Ronald's share = Rs. (16500 x (1/5)) = Rs. 3300. Sol. Ratio of the capitals of A, Band C A’s share = Rs. 69900 x (205/699) = Rs. 20500 I B's share = Rs. 69900 x (212/699) = Rs. 21200; C's share = Rs. 69900 x (282/699) = Rs. 28200. Ex. 4. A, Band C enter into partnership. A invests 3 times as much as B Sol. Let C's capital = Rs. x. Then, B's capital = Rs.
(2/3)x Ratio of their capitals = 2x : (2/3)x :x = 6 : 2 : 3. Hence, B's share = Rs. ( 6600 x (2/11))= Rs. 1200. Ex. 5. Four milkmen rented a pasture. A grazed 24 cows for 3 months; B 10 for 5 Sol. Ratio of shares of A, B, C, D = (24 x 3) : (10 x 5) : (35 x 4) : (21 x 3) = 72 : 50 : 140 Let total rent be Rs. x. Then, A’s share = Rs. (72x)/325 Hence, total rent of the field is Rs. 3250. Ex.6. A invested Rs. 76,000 in a business. After few months, B joined him Rs. Sol. Suppose B joined after x months. Then, B's money was invested for (12 - x) 114 (12 - x) = 91212-x=8x=4 Hence, B joined after 4 months. Ex.7. A, Band C enter into a partnership by investing in the ratio of 3 : 2: 4. After 1 Sol. Let the initial investments of A, Band C be Rs. 3x, Rs. 2x and Rs. 4x respectively. 1O8x : (72x + 6480000) : (144x + 3240000) = 3 : 4 : 5 Hence, A’s initial investment = 3x = Rs. 2,70,000; 14. CHAIN RULE _IMPORTANT FACTS AND FORMULAE 1. Direct Proportion: Two quantities
are said to be directly proportional, if on the increase (or decrease) (More Articles, More Cost) Ex. 2. Work done is directly proportional to the number of men working on it 2. Indirect Proportion: Two quantities are said to be indirectly proportional,if on the increase of the one, (More speed, Less is the time taken to cover a distance) Ex. 2. Time taken to finish a work is inversely proportional to the num of persons working at it. Remark: In solving questions by chain rule, we compare every item with the term to be found out. SOL VED EXAMPLES Ex. 1. If 15 toys cost Rs, 234, what do 35 toys cost? Sol. Let the required cost be Rs. x. Then, More toys, More cost (Direct Proportion) . 15 : 35 : : 234 : x (15 x x) = (35 x 234) x=(35 X 234)/15 =546 Hence, the cost of 35 toys is Rs. 546. Ex. 2. If 36 men can do a piece of work in 25 hours, in how many hours will 15 men Less men, More hours (Indirect Proportion) 15 : 36 : : 25 : x (15 x x) = (36 x 25) (36 x 25)/15 = 60 Hence, 15 men can do it in 60 hours. Ex. 3. If the wages of 6 men for 15 days be Rs.2100, then find the wages of Sol. Let the required wages be Rs. x. More men, More wages (Direct Proportion) Men 6: 9 : :2100:x Days 15:12 Hence the required wages are Rs. 2520. Ex. 4. If
20 men can build a wall 66 metres long in 6 days, what length of a similar can be built by 86 Sol. Let the required length be x metres More men, More length built (Direct Proportion) Less days, Less length built (Direct Proportion) Men 20: 35 Therefore (20 x 6 x x)=(35 x 3 x 56)x=(35 x 3 x 56)/120=49 Hence, the required length is 49 m. Ex. 5. If 15 men, working 9 hours a day, can reap a field in 16 days, in
how many Sol. Let the required number of days be x. Men 18 : 15 Hence, required number of days = 15. Ex. 6. If 9 engines consume 24 metric tonnes of coal, when each is working 8 hours Sol. Let 3 engines of former type consume 1 unit in 1 hour. Therefore 1 engine of former type consumes(1/3) unit in 1 hour. 1 engine of latter type consumes(1/4) unit in 1 hour. Let the required consumption of coal be x units. Number of engines 9: 8 Hence, the required consumption of coal = 26 metric tonnes. Ex. 7. A contract is to be completsd in 46 days sad 117 men were said to work 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? Sol. Remaining work = (1-(4/7) =(3/7) Remaining period = (46 - 33) days = 13days Let the total men working at it be x. (Direct Proportion) Work (4/7): (3/7) Additional men to be employed = (198 - 117) = 81. Ex. 8. A garrison of 3300 men had provisions for 32 days, when given at the rate of 860 gns per head. Sol. The problem becomes: 3300 men taking 850 gms per head have provisions for (32 - 7) or 25 days, How many men taking 825 gms each have provisions for 17 days? Less ration per head, more men (Indirect Proportion) Less days, More men (Indirect Proportion) Ration 825 : 850 Days 17: 25 } : : 3300 : x Strength of reinforcement = (5500 - 3300) = 1700. 15. TIME AND WORK 1. If A can do a piece of work in n days, then A's 1 day's work = (1/n). 2. If A’s 1 day's work = (1/n),then A can finish the work in n days. 3. A is thrice as good a workman as B, then: SOLVED EXAMPLES Ex. 1. Worker A takes 8 hours to do a job. Worker B takes 10 hours to do the same Job.How long should it take both A and B, working together but independently, to do the same job? (IGNOU, 2003) Sol. A’s 1 hour's work = 1/8 B's 1 hour's work = 1/10 (A + B)'s 1 hour's work = (1/8) +(1/10)=9/40 Both A and B will finish the work in 40/9 days. Ex. 2. A and B together can complete a piece of work in 4 days. If A alone can complete the Sol. (A + B)'s 1 day's work = (1/4). A's 1 day's work = (1/12). B's 1 day's work =((1/4)-(1/12))=(1/6) Hence, B alone can complete the work in 6 days. Ex. 3. A can do a piece of work in 7 days of 9 hours each and B can do it in 6 days of 7 bours each. How long will they take to do it, working together 8 hours a day? Sol. A can complete the work in (7 x 9) = 63 hours. Both
will finish the work in (126/5) hrs. Ex. 4. A and B can do a piece of work in 18 days; Band C can do it in 24 days A and C can do Sol. (A + B)'s 1 day's work = (1/18) (B + C)'s 1 day's work = (1/24) Adding, we get: 2 (A + B + C)'s 1 day's work =(1/18 + 1/24 + 1/36) (A +B + C)'s 1 day's work =1/16 Thus, A, Band C together can finish the work in 16 days. =(1/16 – 1/24)= 1/48 A alone can finish the work in 48 days. Ex. 6. A is twice as good a workman as B and
together they finish a piece Sol. (A’s 1 day’s work):)(B’s 1 days work) = 2 : 1. (A + B)'s 1 day's work = 1/18 Divide 1/18 in the ratio 2 : 1. :. A’s 1 day's work =(1/18*2/3)=1/27 Hence, A alone can finish the work in 27 days. Ex. 6. A can do a certain job in 12 days. B is 60% more efficient than A. How many Sol. Ratio of times taken by A and B
= 160 : 100 = 8 : 5. Then, 8 : 5 :: 12 : x = 8x = 5 x 12 =x = 7 1/2 days. Ex. 7. A can do a piece of work in 80 days. He works at it for 10 days B alone finishes the Now,7/ 8 work is done by B in 42 days. (A+B)'s 1 day's work = (1/80+1/48)=8/240=1/30 Ex. 8. A and B undertake to do a piece of work for Rs. 600. A alone can do it in 6 days while B Sol :C's 1 day's work = 1/3-(1/6+1/8)=24 A : B : C = Ratio of their 1 day's work = 1/6:1/8:1/24= 4 : 3 : 1. Ex. 9. A and B working separately can do a piece of work in 9 and 12 days respectively, If Remaining work =(1-35/36)=1/36 On 11th day, it is A’s turn. 1/9 work is done by him in 1 day. 1/36 work is done by him in(9*1/36)=1/4 day Total time taken = (10 + 1/4) days = 10 1/4days. Ex 10 .45 men can complete a work in 16 days. Six days after they started working, 30 more (45 x 16) men can complete the work in 1 day. 1 man's 1 day's work = 1/720 45 men's 6 days' work =(1/16*6)=3/8 Remaining work =(1-3/8)=5/8 75 men's 1 day's work = 75/720=5/48 Now,5 work is done by them in 1 day. 5work is done by them in (48 x 5)=6 days. Ex:11. 2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the 10 8 200 100 (2 men + 1 boy)’s 1 day’s work = (2 x 7 +
1 x 1 ) = 16 = 2 So, 2 men and 1 boy together can finish the work in 25 =12 1 days 16. PIPES AND CISTERNS IMPORTANT FACTS AND FORMULAE 1. Inlet: A pipe connected with a tank or a cistern or a reservoir, that fills it, is known as an (iii) If a pipe can .fill a tank in x hours and another pipe can empty the full tank in y hours SOLVED EXAMPLES Ex.
1:Two pipes A and B can fill a tank in 36 bours and 46 bours respectively. If both Part filled by B in 1 hour = (1/45); Ex. 3: If two pipes function simultaneously, tbe reservoir will be filled in 12 hours. One Sol:let the reservoir be filled by first pipe in x hours. Then ,second pipe fill it in (x+10)hrs. Therefore (1/x)+(1/x+10)=(1/12) (x+10+x)/(x(x+10))=(1/12). x^2 –14x-120=0 (x-20)(x+6)=0 x=20 [neglecting the negative value of x] so, the second pipe will take (20+10)hrs. (i.e) 30 hours to fill the reservoir Ex. 4: A cistern has two taps which fill it in 12 minutes and 15minutes Sol: Workdone by the waste pipe in 1min =(1/20)-(1/12)+(1/15) = -1/10 [negative sign means emptying] therefore the waste pipe will empty the full cistern in 10min Ex. 5: An electric pump can fill a tank in 3 hours. Because of a leak in ,the tank it took Sol: work done by the leak in 1 hour=(1/3)-(1/(7/2))=(1/3)-(2/7)=(1/21). The leak will empty .the tank in 21 hours. Ex. 6. Two pipes can fill a cistern in 14 hours and 16 hours respectively. The pipes Sol: Work done by the two pipes in 1 hour =(1/14)+(1/16)=(15/112). Time taken by these pipes
to fill the tank = (112/15) hrs = 7 hrs 28 min. Work done by (two pipes + leak) in 1 hour = (1/8). Ex. 7: Two pipes A and B can fill a tank in 36 min. and 45 min. respectively. A water Sol:Part filled in 7 min. = 7*((1/36)+(1/45))=(7/20). Ex.8: Two pipes A,B can fill a tank in 24 min. and 32 min. respectively. If both the pipes Sol: let B be closed after x min. then , 17. TIME AND DISTANCE IMPORTANT FACTS AND FORMULAE Distance Distance 1. Speed = Time , Time= Speed , Distance = (Speed * Time) 2. x km / hr = x * 5 3. x m/sec = (x * 18/5) km /hr 4. If the ratio of
the speeds of A and B is a:b , then the ratio of the times taken by them to ab x+y SOLVED EXAMPLES Ex. 1. How many minutes does Aditya take to cover a distance of 400 m, if he runs 18 9 50 60 5 Ex. 2. A cyclist covers a distnce of 750 m in 2 min 30 sec. What is the speed in km/hr 150 5 Ex. 3. A dog takes 4 leaps for every 5 leaps of a hare but 3 leaps of a dog are equal to Then , 3x = 4y => x = 4 y => 4x = 16 y. Ratio of speeds of dog and hare = Ratio of distances covered by them in the = 4x : 5y = 16 y : 5y =16 : 5 = 16:15 Ex. 4.While covering a distance of 24 km, a man noticed that after walking for 1 hour and 40 minutes, the distance covered by him was 5 of the remaining distance. What was his speed in metres per second? 7 Sol. Let the speed be x km/hr. Then, distance covered in 1 hr. 40 min. i.e., 1 2 hrs = 5x km 33 Remaining distance = { 24 – 5x } km. 3 5x = 5 { 24 - 5x } 5x = 5 { 72-5x } 7x =72 –5x 37 3 37 3 Hence speed = 6 km/hr ={ 6 * 5 } m/sec = 5 m/sec = 1 2 18 3 3 Ex. 5.Peter can cover a certain distance in 1 hr. 24 min. by covering two-third of the distance at 4 kmph and the rest at 5 kmph. Find the total
distance. 2x 1x 4 5 5 6 15 5 Ex. 6.A man traveled from the village to the post-office at the rate of 25 kmph and walked back at the rate of 4 kmph. If the whole journey took 5 hours 48 minutes, find the distance of the post-office from the village. Sol. Average speed = { 2xy } km/hr ={ 2*25*4 } km/hr = 200 km/hr x+y 25+4 29 Distance traveled in 5 hours 48 minutes i.e., 5 4 hrs. = { 200 * 29 } km = 40 km 5 29 5 Distance of the post-office from the village ={ 40 } = 20 km 2 Ex. 7.An aeroplane files along the four sides of a square at the speeds of 200,400,600 and 800km/hr.Find the average speed of the plane around the field. Sol. : Let each side of the square be x km and let the average speed of the plane around the field by y km per hour then
, hence average speed =384 km/hr Ex. 8.Walking at 5 of its usual speed, a train is 10 minutes too late. Find its usual 7 Sol. :New speed =5/6 of the usual speed So,( 6/5 of the usual time )-( usual time)=10 minutes. usual time=10 minutes Ex. 9.If a man walks at the rate of 5
kmph, he misses a train by 7 minutes. However, Ex. 10. A and B are two stations 390 km apart. A train starts from A at 10 a.m. and Sol. Suppose they meet x hours after 10 a.m. Then, hrs]=390. 65x + 35(x-1) = 390 => 100x = 425 => x = 17/4 So, they meet 4 hrs.15 min. after 10 a.m i.e., at 2.15 p.m. Ex. 11. A goods train leaves a station at a certain time and at a fixed speed.
After Sol. Let the speed of the goods train be x kmph. train in 4 hours Ex. 12. A thief is spotted by a policeman from a distance of
100 metres. When the Sol. Relative speed of the policeman = (10-8) km/hr =2 km/hr. 1000 2 20 20 20 5 Ex.13. I walk a certain distance and ride back
taking a total time of 37 minutes. I Sol. Let the distance be x km. Then, But, the time taken to walk 2x km = 55 min. 18. PROBLEMS ON TRAINS IMPORTANT FACTS AND FORMULAE 1. a km/hr= (a* 5/18) m/s. 2. a m / s = (a*18/5) km/hr. 3 Time taken by a train of length 1 metres to pass a pole or a standing man or a 4. Time taken by a train of length 1 metres to pass a stationary object of length b 5. Suppose two trains or two bodies are moving in the same direction at u m / s and 6. Suppose two trains or two bodies are moving in opposite directions at u m / s 7. If two trains of length a metres and b metres are moving in opposite directions at 8.If two trains of length a metres and b metres are moving in the same direction 9. If two trains (or bodies) start at the same time from points A and B towards each SOLVED EXAMPLES Ex.I. A train 100 m long is running at the speed of 30 km / hr. Find the time taken by it to pass a man standing near the railway line. (S.S.C. 2001) Sol. Speed of the train = (30 x 5/18_) m / sec = (25/3) m/ sec. Required time taken = 100/(25/3) = (100 *(3/25)) sec = 12 sec Ex. 2. A train is moving at a speed of 132 km/br. If the length of the train is (Section Officers', 2003) Sol. Speed of train = 132 *(5/18) m/sec = 110/3 m/sec. Time taken =275 *(3/110) sec =15/2 sec = 7 ½ sec Ex. 3. A man is standing on a railway bridge which is 180 m long. He finds Sol. Let the length of the train be x metres, Then, the train covers x metres in 8 seconds and (x + 180) metres in 20 sec Length of the train = 120 m. Speed of the train = (120/8) m / sec = m / sec = (15 *18/5) kmph = 54 km Ex. 4. A train 150 m long is running with a speed of 68 kmph. In what time will Sol: Speed of the train relative to man = (68 - 8) kmph = (60* 5/18) m/sec = (50/3)m/sec Time taken by the train to cross the man I = Time taken by It to cover 150 m at 50/3 m / sec = 150 *3/ 50 sec = 9sec Ex. 5. A train 220 m long is running with a speed of 59 kmph.. In what will it pass a man who is running at 7 kmph in the direction opposite to that in sol. Speed of the train relative to man = (59 + 7) kmph at (55/3) m / sec = (220 *3/55) sec = 12 sec Ex. 6. Two trains 137 metres and 163 metres in length are running
towards Sol. Relative speed of the trains = (42 + 48) kmph = 90 kmph Time taken by the trains to'pass each other Ex. 7. Two trains 100 metres and 120 metres long are running in the same Sol: Relative speed of the trains = (72 - 54) km/hr = 18 km/hr Time taken by the trains to cross each other Ex. 8. A train 100 metres long takes 6 seconds to cross a man walking at 5 train.? Speed of the train relative to man = (x + 5) kmph = (x + 5) *5/18 m/sec. Therefore 100/((x+5)*5/18)=6 <=> 30 (x + 5) = 1800 <=> x = 55 Length of train = (Relative speed * Time) = ( 40/3)*12 m = 160 m. Also, speed of the train = 54 *(5/18)m / sec = 15 m / sec. (x+y)/15 = 20 <=> x + y = 300 <=> Y = (300 - 160) m = 140 m. Ex10. A man sitting in a train which is traveling
at 50 kmph observes that a Sol: Relative speed = 280/9 m / sec = ((280/9)*(18/5)) kmph = 112 kmph. Speed of goods train = (112 - 50) kmph = 62 kmph. 19.BOATS AND STREAMS IMPORTANT FACTS AND FORMULAE 1.In water ,the direction along the stream is called downstream and ,the direction against 2.If
the speed of a boat in still water is u km/hr and the speed of the stream is v speed downstream=(u+v)km/hr. 3.If the speed downstream is a km/hr and the speed upstream is b km/hr,then : SOLVED EXAMPLES EX.1.A man can row upstream at 7 kmph and downstream at 10kmph.find man’s Sol. Rate in still
water=1/2(10+7)km/hr=8.5 km/hr. EX.2. A man takes 3 hours 45 minutes to row a boat 15 km downstream of a river Sol. rate downstream=(15/3 ¾)km/hr=(15*4/15)km/hr=4km/hr. EX.3. a man can row 18 kmph in still water.it takes him thrice as
long to row up as Sol. Let man’s rate upstream be x kmph.then ,his rate downstream=3xkmph. EX.4. there is a road beside a river.two friends started from a place A,moved to a on a cycle at a speed of 12 km/hr,while the other
sails on a boat at a speed of 10 =42/5 km/hr=8.4 km/hr. EX.5. A man can row
7 ½ kmph in still water.if in a river running at 1.5 km/hr an Sol. Speed downstream =(7.5+1.5)km/hr=9 km/hr; EX.6. In a stream running at 2kmph,a motar boat goes 6km upstream and back Sol.let the speed of the motarboat in still water be x kmph.then, EX.7.A man can row 40km upstream and 55km downstream in 13 hours also, he Sol.let rate upstream=x km/hr and rate downstream=y
km/hr. 20. ALLIGATION OR MIXTURE IMPORTANT FACTS AND FORMULAE 1. Alligation: It is the rule that enables us to find the ratio in which two or more ingredients at the (Quantity of cheaper) = (C.P. of dearer) - (Mean price) We present as under: C.P. of a unit quantity of dearer (m) (d-m) (m-c) (Cheaper quantity) : (Dearer quantity) = (d - m) : (m - c). 4. Suppose a container contains x units of liquid from which y units are taken out and replaced by [ ]water. After n operations the quantity of pure liquid= x(1-y/x)^n units. . SOLVED EXAMPLES Ex. 1. In what ratio must rice at Rs. 9.30 per kg be mixed with rice at Rs. 10.80 per kg so that the Sol. By the rule of alligation, we have: C.P. of 1 kg rice of 2nd kind (in paise) 930 1080
. 80 Ex. 2. How much water must be added to 60 litres of milk at 1 ½ litres for Rs. 2 Sol. C.P. of 1 litre of milk = Rs. (20 x 2/3) = Rs. 40/3 c.p of 1 litre of milk c.p of 1 litre of milk 0 Rs.40 3 Mean price 3 (40/3-32/3)=8/3 (32/3-0)=32/3 Ratio of water and milk =8 : 32 = 8 : 32 = 1 : 4 33 [ ]Quantity of water to be added to 60 litres of milk = 1/4 X 60 litres =15 litre Ex. 3. In what ratio must water be mixed with milk to gain 20 % by selling the mixture at cost Sol. Let C.P. of milk be Re. 1 per litre. Then, S.P. of 1 litre of mixture = Re. 1. [ ] C.P. of 1 litre of mixture = Rs. (100/120)* 1 =Rs.5/6 By the rule of alligation, we have: C.P. of 1 litre of water C.P. of1litre ofmilk (Re. 5/6) (1- (5/6))= 1/6 ((5/6)-0)=5/6 Ex. 4. .How many kgs. of wheat costing Rs. 8 per kg must be mixed with 86 kg of rice costing Rs. Sol. S.P. of 1 kg mixture = Rs. 7.20,Gain = 20%. [ ] C.P. of 1 kg mixture = Rs. (100/120)*7.20 =Rs. 6. By the rule of alligation, we have: C.P. of 1 kg wheat of 2nd kind (800p) . (540 p) Mean price 60 200 Ex. 5. The milk and water in two vessels A and B are in the ratio 4 : 3 and 2: 3 respectively. In what ratio, the liquids in both the vessels be mixed to obtain a new mixture in vessel C Sol. Let the C.P. of milk be Re. 1 per litre Milk in 1 litre mixture of A = 4/7litre; Milk in 1 litre mixture of B = 2/5 litre; Milk in 1 litre mixture of C = ½ litre C.P. of 1 litre mixture in A = Re .4/7; C.P. of 1 litre mixture in B = Re.2/5 Mean price = Re.1/2 By the rule of alligation, we have: C.P. of 1 litre mix. in A C.P. of 1 litre mix. in B (4/7) (2/5) (1/2) (1/14) (1/10) 21. SIMPLE INTEREST IMPORTANT FACTS AND FORMULAE 1.. Principal: The money borrowed or lent out for a certain period is called the 2. Interest: Extra money paid for using other's money is called interest. (i) S.I.
= (P*R*T )/100 SOLVED EXAMPLES Ex. 1. Find the simple interest on Rs. 68,000 at 16 2/3% per annum for 9 months. Sol. P = Rs.68000,R = 50/3% p.a and T = 9/12 years = 3/4years. ( ) S.I. = (P*R*T)/100 = Rs. 68,000*(50/3)*(3/4)*(1/100) = Rs.8500 Ex. 2. Find the simple interest on Rs. 3000 at 6 1/4% per annum for the period from P = Rs.3000 and R = 6 ¼ %p.a = 25/4%p.a ( )S.I. = Rs. 3,000*(25/4)*(1/5)*(1/100) = Rs.37.50. Remark : The day on which money is deposited is not counted while the day on which ( )Sol. Let sum be Rs. x then , S.I.=Rs. x*(27/2) *4*(1/100) = Rs.27x/50 ( )amount = Rs. x+(27x/50) = Rs.77x/50 77x/50 = 2502.50 x = 2502.50 * 50 = 1625 Hence , sum = Rs.1625. Ex. 4. A sum of Rs. 800 amounts to Rs. 920 in 8 years at simple intere Sol. S.l. = Rs. (920 - 800) = Rs. 120; p = Rs. 800, T = 3 yrs. _ ( ). R = (100 x 120)/(800*3) % = 5%. New rate = (5 + 3)% = 8%. Ex. 5. Adam borrowed some
money at the rate of 6% p.a. for the first two years , at the Sol. Let the sum borrowed be x. Then, (x*2*6)/100 + (x*9*3)/100 + (x*14*4)/100 = 11400 ( ) 3x/25 + 27x/100 + 14x / 25 = 11400 95x/100 = 11400 x = (11400*100)/95 = 12000. Hence , sum borrowed = Rs.12,000. Ex. 6. A certain sum of money amounts to Rs. 1008 in 2 years and to Rs.1164 in 3 ½ Sol.. S.I. for 1 ½ years = Rs.(1164-1008) = Rs.156. S.l. for 2 years = Rs.(156*(2/3)*2)=Rs.208 Principal = Rs. (1008 - 208) = Rs. 800. Now, P = 800, T = 2 and S.l. = 208. Rate =(100* 208)/(800*2)% = 13% Ex. 7. At what rate percent per annum will a sum of money double in 16 years. Sol.. Let principal = P.
Then, S.l. = P and T = 16 yrs. Ex. 8. The simple interest on a sum of money is 4/9 of the principal .Find the rate Sol. Let sum = Rs. x. Then, S.l. = Rs. 4x/9 Rate = 6 2/3 % and Time = 6 2/3 years = 6 years 8 months. Ex. 9. The simple interest on a certain sum of money for 2 l/2
years at 12% per ( ) – ( )Sol. Let the sum be Rs. x Then, (x*10*7)/(100*2) (7x/20)-(3x/10)=40 Hence, the sum is Rs. 800. Ex. 10. A sum was put at simple interest at a certain rate for 3 years. Had it been Sol. Let sum = P and original rate = R. [ ] – [ ]( )/Then, P*(R+2)*3 100 3PR + 6P - 3PR = 36000 6P=36000 P=6000 Hence, sum = Rs. 6000. Ex. 11. What annual instalment will discharge a debt of Rs. 1092 due in 3 years . Sol .Let each Instalment be Rs. x ( ) + ( )Then, x+ ((x*12*1)/100) In what time will 7500 amount to ₹ 8625 if simple interest is reckoned at 7So, in 2 years Rs. 7500 becomes Rs. 8625. So, this is the required answer.
In what time will Rs 6000 amount to Rs 7500 if the simple interest is calculated at 10% per annum?Time = 2.5 years
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At what rate per cent per annum simple interest will 66000 amount to 72720 in 2 years?66000 amount to Rs. 72720 in 2 years? UPLOAD PHOTO AND GET THE ANSWER NOW! Answer : `5""1/3% ` p.a.
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