Four numbers are in a proportion if the product of extremes = product of means. E. g. if a : b :: c : d, then ad = bcIf the ratio of first and second equals to the ratio of second and third (with three or more quantities), they are said to be in continued proportion. E. g. if a/b = b/c, then a, b and c are in a continued proportion. If a number, say Z, is divided into three parts, whose ratio is a: b: c, then: First part = Second part = Third part = Illustration 1: The ratio of number of ladies to gents in a party was 1: 2, but when 2 ladies and 2 gents left, the ratio became 1: 3. How many people were originally present at the party? 48
12
16Solution: Let the number of ladies be x& the number of gents be 2x3x-6 = 2x -2Total number of persons at the party = 3x = 3 × 4 = 12Illustration 2: The sum of the present ages of A, B and C is 30 years. Six year ago, their ages were in the ratio 1: 2: 3. What is the present age of C? 36
42
5024Solution: Sum of the ages of A, B, C six years ago = 90 – 3= 72 yearsNow, Six years ago, the ages of A, B & C were in ratio 1: 2: 3Age of C six years ago == 36 yearsPresent age of C = 36 +6= 42 yearsIllustration 3: If bc : ac: ab = 1: 2: 3, find : 1 : 39 : 14 : 9
4 : 1
Solution: : / =
Also bc : ac = 1: 2
=
: = = 2 = Illustration 4: The sum of the squares of three numbers is 532 and the ratio of the first to the second as also of the second to the third is 3: 2. What is the second number?
12
181521
Solution: = , =
A = B & C = BNow, A2 + B2 + C2 = 532B2 + B2 + = 532= 532133B2 = 532 × 36B = 12Illustration 5: A person distributes his pens among four friends A, B, C and D in the ratio 1/3: 1/4: 1/5: 1/6. What is the minimum number of pens that the person should have to perform this exercise? 60
57
4852Solution: LCM of 3, 4, 5, 6 = 60So, the pens are distributed among A, B, C & D in the ratio15 : 12 : 10Total number of pens = 20x + 15x + 12x + 10xTo gauge the minimum number of pens, the value of x should be taken as 1= 20 + 15 + 12 + 10 = 57
Practice Exercise
If half of one number is equal to 0. 07 of another, the ratio of the numbers is: 50: 75: 7
7: 50
1: 14Solution: x = 0. 07 y= 0. 07 × 2 = × 2 = 7/50If = 2: 3 then =? 4/94/316/9
4/27
Solution: = {given}= 2 × = A man divides his property so that ratio of his son’s share to his wife’s share and the ratio of the wife’s share to his daughter’s share are both 3: 1. If the daughter gets Rs. 10, 000 less than the son, then the total worth of his property is: 25, 000
16, 250
65, 00027, 350Solution: Son : Wife = 3 : 1 & Wife : Daughter = 3 : 1Son : Wife : Daughter = 9 : 3 : 1Now, Son’s share – Daughter’s share = 10, 0009x – x = 10, 0008x = 10, 000x = 1250Total property = 13x= 13 × 1250= 16, 250The incomes of A & B is are in the ratio 3: 2 and their expenditures are in the ratio 5: 3. If each one of them saves Rs. 1000, then, A’s income can be: Rs. 3000Rs. 4000
Rs. 6000
Rs. 9000Solution: Let the income of A & B be 3x and 2x& the total expenditures of A & B be 5y & 3y3x – 5y = 1000…… (I)2x – 3y = 1000………….. (II)Solving (I) & (II)x = 2000y = 1000A’s income = 3x = 3 × 2000 = 6000If Rs. 58 is divided among 150 children such that each girl and each boy gets 25 p and 50 p respectively, then how many girls are there in total? 72
68
50110Solution: Let the number of boys & girls be x & y respectivelyx + y = 150 …….. (I)x + y = 582x + y = 232 ……. (II)Solving (I) & (II)y = 6860 kg of an alloy A is mixed with 100 kg of alloy B. If alloy A has lead and tin in the ratio 3: 2 and alloy B has tin and copper in the ratio 1: 4, then the amount of tin in the new alloy is: 36 kg
44 kg
53 kg80 kgSolution: Tin in Alloy ‘ A’ = = 24 kgTin in Alloy ‘ B’ = = 20 kgTin in new Alloy C = 24 + 20= 44 kg15 liters of mixture contains 20% alcohol and the rest water. If 3 liters of water is mixed with it, the percentage of alcohol in the new mixture would be: 15%17 %18Solution: Amount of alcohol in mixture = 20 % × 15= = 3 litersPercentage of alcohol after adding 3 liters water
=
= %
The speeds of the three cars are in the ratio 5: 4: 6. The ratio between the time taken by them to travel the same distance is: 5: 4: 66: 4: 510: 12: 15
12: 15: 10
Solution: Speeds are in ratio of 5: 4: 6For same distance time will be in ratio: : LCM (5, 4, 6) = 60Required ratio = : : : 60= 12 : 15 : 10Seats for Mechanical, Civil and Electrical branch in an engineering college are in the ratio of 5: 7: 8. There is a proposal to increase these seats by 40%, 50% & 75% respectively. What will be the ratio of increased seats?
2: 3: 4
6: 7: 86: 8: 9None of theseGiven ratio 5: 7: 8After increment 5×140: 7×150: 8×175700: 1050: 1400= 2 : 3 : 4Hence, the answer is (a)The sides of the triangles are in the ratio of : : and its perimeter is 104 cm. The length of the longest side is: 52 cm
48 cm
32 cm26 cmSolution: Let the sides be , and+ + = 1046 x + 4x + 3x = 104 × 1213x = 104 × 12x = x = 96Longest side = ½ x= x × 96 = 48
Simple Interest/Compound Interest
Important Formulas/Concepts
Simple Interest = From the above-mentioned formula, the following formulas can be derived: P = R = T = Let P = Principal, R = R% per annum, Time = T years and number of times the interest is compounded in a fixed interval = N, then (Compound Interest): Amount = P×TTo find number of years it will take for money to double (in case of compound interest): Number of years = E. g. If rate = 18%, then number of years for money to double = = 4 yearsIllustration 1: A man investsRs. 3000 at the rate of 5% per annum. How much more should he invest at the rate of 8%, so that he can earn a total of 6% per annum? Rs. 1200Rs. 1300
Rs. 1500
Rs. 2000Solution: 30, 000×+ (3000+ x) ×15000+8x = 18000+6xThus, x= 1500Illustration 2: A sum of money lent out at simple interest amounts to Rs. 720 after 2years and to Rs. 900 after a further period of 5 years. The sum and the rate % are: 500, 10%
600, 10%
500, 15%600, 15%Solution: A-P = 720-P = …..(i)Similarly, 900 –p =………. (ii)Dividing (i) by (ii)
=
5(720-P) = 2(900-P)3600 -5P = 1800 -2PP = 600Now, Rate% 720 -600 == R10% = RIllustration 3: Out of a certain sum, 1/3rd is invested at 3%, 1/6th at 6% and the rest at 8%. If the simple interest for 2 years from all these investments amount to Rs. 600, then the original sum is: Rs. 3000Rs. 4500
Rs. 5000
Rs. 6000Solution: Let the initial amount be xNow,
+
12x = 60, 000Thus, x = 5000Illustration 4: If Rs. 85 amounts to Rs. 95 in 3years, what Rs. 102 will amount to in 5 years at the same rate %? 117
122
132142Solution: A – P = 95-85 =…… (I)Similarly, A -102 =…… (II)Dividing equation (I) by (II)= 85/102 3/5Thus, A = 122Illustration 5: If simple interest on a certain sum is 16 over 25 of the sum, then the rate percent and time, if both are equal, are: 567
8
Solution: Simple Interest = Given, Simple Interest = 16/25P& R = T16/25P = 64 = R2R = T = 8R = 8
Practice Exercise
If the difference between the Compound Interest and Simple Interest on a certain sum of money is Rs. 72 at 12 percent per annum for 2 years, then the amount is: Rs. 6000
Rs. 5000
Rs. 6500Rs. 5500Solution: Compound Interest – Simple Interest = 72tP – P – = 722P= 72P = 72P = 72Thus, P = 5000A sum of money placed at a compound interest doubles itself in 3 years. In how many years will it amount to 8 times of itself?
9 years
8 years27 years7 yearsSolution: Given A = 2PA = PnOr 2 = 3……………..(i)Similarly 8P= P n8 = n23 = n …………(ii)From (i) & (ii)[()3]3= nThus, n = 9If x is the simple interest on y and y is the simple interest on z, what is the relation between x, y and z, the rate % and the time being the same in both cases? x2 = y z
y2 = x z
x y z = 1x = 2y+zSolution: x ==…..(I)y…..(II)Dividing (I) & (II)
=
xz= y2A sum of money invested triples itself in 8 years at simple interest. Find in how many years will at become 8 times itself at the same rate? 24 years
28 years
30 years21 yearsSolution: 2P =…….(i)7P = ()………..(ii)Dividing (I) by (II)Thus, T = 28 yearsFind the compound interest at the rate of 10% for 3 years on the principle which in 3 years at the rate of 10% per annum gives Rs. 300 as simple interest.
331
310330333Solution: Simple Interest = 300= Thus, P = 1000Now, Compound Interest = P [()n- 1]= 1000[()3- 1]= 1000Thus, Compound Interest = 331A man wants to invest Rs. 1, 50, 000among two schemes at the rate of 5% and 10% to earn interest at Rs. 10, 000 in 1 year. Thus, the amount invested in 5% and 10% schemes respectively is:
1, 00, 000 & 50, 000
50, 000 & 1, 00, 00075, 000 & 75, 0001, 25, 000 &25, 000Solution: Let amount invested in 5% scheme be xThus, += 100005x +1500000 -10x = 1000000500000= 5×10000 = xIf Rs. 1100 is obtained after lending out same amount at 5% per annum for 2 years and Rs. 1800 is obtained after lending the remaining amount at 10% per annum for 2 years, then the total amount lent out is:
2500
300020002200Solution: A1 = P1 + I1I1 = P1 ×1100 = P1 = 1000Similarly, 1800-P2P2 = 1500Total principle P = P1+P2= 1000+1500= 2500Shamit invested Rs 6000 in a company at a compound interest compounded semi-annually. He receives Rs 7986 after 18 months from the company. Find the rate of interest per annum. 10%
20%
5%12. 5%Solution: 7986 = 6000 (1 + )3/2×2R = 10%Yearly rate = 2 × 10% = 20%A sum was invested at a simple interest at a certain interest for 2 years. It would have fetched Rs 60 more had it been invested at 2% higher rate. What was the sum?
1500
130025001000Solution: Condition…… (I)I = P×R×2/100Condition …… (II)I+60 = 1500 = PA sum of Rs. 3, 800 is lent out in two parts in such a way that the interest on one part at 8% for 5 years is equal to that on the another part at ½% for 15 years. Find the sum lent out at 8%. 500400
600
700Solution: Let the amount invested at 8% be x
= ×
16x = 11400-3x19x = 11400x = 600
Time and Work / Pipes and Cisterns
Important Formulas/Concepts
Work = Working units TimeIf A can do a piece of work in b days, then his one day’s work = Illustration 1: Varun can do a piece of work in 20 days. Tarun is 25% more efficient than Varun. The number of days taken by Tarun to do the same piece of work is: 15
16
1825Solution: Amount of work done by Varun in one day= Amount of work done by Tarun in one day = 125% × = = Tarun will complete the same piece of work in 16 daysIllustration 2: Sam can do a piece of work in 40 days. He works on it for 8 days and then John finished it in 16 days. How long will they take to complete the work together?
days
15 days20 days56 daysSolution: Amount of work completed by Sam in 8 days 8 × = Work left to be completed by John = 1 – = Let the amount of work completed by John in one day = Using work equationAmount of work they can complete in one day = = Required time = daysIllustration 3: Mukesh, Anil and Sumit are three civil engineers. Mukesh can design a plan alone of a multistoried apartment in 5 hours and Anil in 4 hours. They together can do it in 2 hours. In what time can Sumit do it alone? 7 hours15 hours
20 hours
18 hoursSolution: Using work equation,
= –
= =
Sumit can design the plan alone in 20 hours. Illustration 4: If 4/7th of a piece of work is completed in 7/4th days, in how many days can rest of the work be completed? 7/33/7
21/16
16/21Solution: of a work is completed in daysTotal work is completed in = daysWork remaining to be completed = 1- = Time taken to complete work = = daysIllustration 5: One man can paint a house in ‘ r’ days and another man can do it in‘ t’ hours. If they can together do it in ‘ d’ days, then ‘ d’ is given by: Solution: Work completed by a man in one day= 1/rWork completed by another man in one day = 24/tUsing work equation,
=
d =
Practice Exercise
Mary has ‘ m’ minutes of home work in each of her ‘ s’ subjects. In one hour, she completes what part of her homework? 1/ms
60/ms
m/sm/60sSolution: Mary completes home work in each of the subjects in ‘ m’ minutesAnd there are in total‘ s’ subjectsTotal homework of Mary is completed in ms minutesi. e. ms/60 hoursHence, in one hour, Mary completed 60/ms workFederer can do a job in 40 days. He worked on it for 5 days and then Paes finished it in 21 days. In how many days Federer and Paes can finish the work together? 10
15
2012Solution: Work completed by Federer in 5 days = = Work remaining to be completed = 1- = Paes completed the remaining work in 21 days. Let the work completed by Paes in one day be xUsing work equation, Amount of work completed by both of them together in one day = Thus, together they will take 15 days. Rohit can copy 80 pages in 20 hours. Rohit and Rahul can copy 135 pages in 27 hours. How many pages Rahul can copy in 20 hours?
20
275535Solution: Pages copied by Rohit in 1 hour = = 4 pagesPages copied by Rohit and Rahul in 1 hour = = 5 pagesNumber of pages Rahul can copy in an hour = 5 – 4 = 1 pageHence, pages completed by Rahul in 20 hours = 20 × 1 = 20 pagesA and B can do a piece of work in12 days, B and C in 15 days, C and A in 20 days. How long would each take separately to do the same amount of work? 30, 60, 20
30, 20, 60
20, 15, 3015, 20, 35Solution: ………………………… (I)…………………………….. (II)……………………………..(IIIAdding (I), (II) & (III)2== ……….(IV)Solving (I), (II), (III) & (IV)Time taken by A = 30 daysTime taken by B = 20 daysTime taken by C = 60 daysPipes P & Q can fill up a tank in 4 hours & 8 hours respectively and pipe R can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in: 10 hours7 hours16/7 hours
24/7 hours
Solution: Using time and work equation: Thus, T = 24/7 hoursThree taps A, B & C can fill a tank in 12, 15& 20 hours respectively. If A is open all the time, B and C are open for one hour each alternately, the tank will be full in: 6 hours6hours. 5 hours
7 hours
Solution: Work completed by A & B in 1 hour = Work completed by A & C in 1 hour = Work completed in 2 consecutive hours = Work completed in 6 hours = Remaining work = 1-Now A & B can complete the remaining work in 1 hourTotal work is completed in 6 + 1 = 7 hours12 buckets of water fill a tank when the capacity of each bucket is 13. 5 litres. How many buckets will be needed to fill the same tank, if the capacity of each tank is 9 litres?
18
15106Solution: Using concept of variation: 12 × 13. 5 = x × 9x = 18A cistern is filled in 9 hours but it takes 10 hours when there is a leak in its bottom. If the cistern is full, in what time shall the leak empty it?
90 hours
94 hours92 hours91 hoursSolution: Let the time take by leak to empty the tank be x hours. Using work equationx = 90 hoursFrom a leaking tap ‘ a’ drop comes out in ‘ b’ minutes. If there are ‘ c’ drops in a litre, then in how many hours one litre of water will be wasted? Solution: Given ‘ a’ drop falls in b minutes or hours1 drop falls in hoursHence, C drops will fall in hoursIf a tanker is normally filled by a tap in 8 hours, but suddenly a leak develops and it empties the full tanker in 24 hours, the cistern will be filled (with a leak) in: 10 hours
12 hours
15 hours18 hoursSolution: Using time equation,
=
t = 12 hours
Speed and Distance
Important Formulas/Concepts
Speed = Time = Distance = Speed × TimeIf ratio of speeds of A and B is x : y, then ratio of time taken by them to cover the same distance = : Illustration 1: Walking at 5/7th of his usual rate, a boy reaches his school 6 minutes late. Find his usual time to reach the school. 6 minutes8 minutes12 minutes
15 minutes
Solution: Let the usual speed of the boy be x, then his slower speed= x, Also let the school be at a distance of DNow using SDT equation,= 6= 6D = 15xUsual time to reach the school = 15 minutesIllustration 2: A train completes a journey without stopping in 8 hours. If it had travelled 5 km an hour faster, it would have completed the journey in 6 hours 40 minutes. What is its slower speed? 23 km/hr30 km/hr
25 km/hr
42 km/hrSolution: Using distance relation for equal travelSlow speed × More time= Fast speed × Less timex × 8= (x × 5)Thus x= 25 km/hr. Illustration 3: Without stoppages, a train travels certain distance with an average speed of 80 km/hr, and with stoppages; it covers the same distance with an average speed of 60 km/hr. How many minutes per hour does the train stop? 20 minutes/hours
15 minutes/hours
12 minutes/hours8 minutes/hoursSolution: Using the formula, Stoppage time/hours = hours= hours= 15 minutesIllustration 4: If a car moves from A to B at a speed of 60 km/hr, and comes back from B to A at a speed of 40 km/hr, then its average speed is: 46 km/hr50 km/hr56 km/hr
48 km/hr
Solution: Average speed == 48 km/hrIllustration 5: A car during its journey travels 30 minutes at a speed of 40 km/hr, and another 45 minutes at a speed of 60 km/hr, and another 2 hours at a speed of 70 km/hr. Find its average speed. 65 km/hr56 km/hr48 km/hr
63 km/hr
Solution: Average speed =
=
= 63 km/hr
Practice Exercise
Two friends X and Y walk from A to B, a distance of 39 km, at 3 km/hour and 3 ½ km/hour respectively. Y reaches B, returns immediately and meets X at C. Find the distance from A to C. 35 km
36 km
28 km32 kmSolution: Let the distance of C from B be xFor same time, equation will be: 2………………..(i)On solving equation (i)x = 3Thus, distance A to C = 39-x = 39 – 3 = 36kmA long distance runner runs 9 laps of a 400 meters track every day. His timings for four consecutive days are 88, 96, 89 and 87 respectively. On an average how many meters/minutes does the runner cover? 36 meters/minute12 meters/ minute
40 meters/minute
48 meters/minuteSolution: = 40 meters/minuteAverage speed = = 40 m/minA thief runs away with a car at a speed of 40 km/hr. The theft was discovered after half an hour and the owner sets off in another car at 50 km/hr. When will the owner overtake the thief from the time of theft? 2 hours
hours
3 hours1 ½ hoursSolution: Let the time taken by owner be tBoth thief and owner cover equal distances50 t = (t + ) 40Thus, T = 2 hoursHe will be a caught after 2 ½ hoursThe speed of the boat in still water is 12 km/hr and the speed of the stream is 3 km/hr. A distance of 27 km, going upstream, will be covered in:
3 hours
5 hours4/3 hours3/2 hoursSolution: Time taken to row upstream == = 3 hoursA boat rows 20 km upstream and 30 km downstream in 5 hours each. The velocity of the stream is: 3km/hr2 km/hr
1 km/hr
4 km/hrSolution: Let the velocity of the stream be y & velocity of the boat be xFor upstream = 5, = 5x – y = 4 ….(I)x + y = 6…..(II)Solving (I) & (II)y = 1 km/hr. A 270 meter long train running at the speed of 120 km/hr crosses another train running in opposite direction at the speed of 80 km/hr in 9 seconds. What is the length of the other train? 230 meters240 meters260 meters320 metersSolution: Relative speed of train = (120 + 80)km/hr200km/hr = 200 ×500/9 m/sLength of the trainThus, x = 230A train of length 150 meters takes 40. 5 seconds to cross a tunnel of length 300 meters. What is the speed of the train in km/hr.? 13. 3326. 67
40
66. 67Solution: Speed== 40 km/hoursA farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot at 4 km/hr and partly on bicycle at the rate of 9 km/hr. The distance travelled on foot is: 14 km15 km
16 km
17 kmSolution: Let the distance travelled on foot be ‘ x’ kmUsing time equation,+ = 99x + 4(61 – x) = 9 × 369x + 244 – 4x = 3245x = 80x = 16 kmThus, distance travelled by farmer on foot is 16kmTwo cars P & Q start at the same time from A & B which are 120 km apart. If the two cars travel in opposite directions, they meet after 1 hour and if they travel in the same direction, then P meets Q after 6 hours. What is the speed of the car P? 60 km/hr
70 km/hr
100 km/hr120 km/hrSolution: Let the speed of car P be ‘ x’ & speed of car Q be ‘ y’Then, the two equations will be: x + y = 120….(I)6x – 6y = 120… (II)Solving (I) & (II)x = 70 km/hr. A bus while moving at an average speed of 40 km/hr reaches its destination on time. When its average speed becomes 35 km/hr, then it reaches 15 minutes late. Find the length of Journey. 30 km40 km
70 km
80 kmSolution: Using time equation, x = x = 70 km
Number System
Important Formulas/Concepts
(a + b)2 = a2 + 2ab + b2a2 + b2 = (a + b)2 – 2ab(a – b)2 = a2 – 2ab + b2a2 + b2 = (a – b)2 + 2aba2 – b2 = (a + b)(a – b)(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)(a + b)3 = a3 + b3 + 3ab(a + b)(a – b)3 = a3 – b3 -3ab (a-b)a3+ b3 = (a + b)3 – 3ab(a + b)a3 + b3 = (a + b)(a2 – ab + b2)a3 – b3 = (a – b)3 + 3ab (a – b)a3 – b3 = (a – b)(a2 + ab + b2)A number is divisible by
2
ifIts unit digit is any of 0, 2, 4, 6, 8
3
The sum of its digits is divisible by 3
4
The sum of its last two digits is divisible by 4
5
Its unit digit is either 5 or 0
6
It is divisible by both 2 and 3
8
The number formed by its last 3 digits is divisible by 8
9
The sum of its digits is divisible by 9
10
Its unit digit is 0
11
The difference of the sum of its digits at odd places and that at even places is either 0 or a number divisible by 11. Illustration 1: The sum of two numbers is 25 and their product is 114. What is the sum of the reciprocals of these numbers? 5/1919/5114/25
25/114
Solution: Given- x + y = 25&xy = 114Required answer = = Illustration 2: Which of the following is not the perfect square? 1, 99, 8091, 41, 3766, 59, 344
17, 82, 362
Solution: Since the square of natural number never ends with 2, 3, 7, 81782362 is not a perfect square, as it ends with 2. Illustration 3: If the number 812×74 is completely divisible by 11, then smallest whole number in place of x will be:
1
357Solution: Divisibility of 11 = (Sum of number at odd place) – (Sum of number at even place)Thus, it should be either O or a number divisible by 11= (8+2+7) – (1+x+4)= (12-x)Hence, minimum value of x should be 1 so as to make the number divisible by 11. Illustration 4: The difference between place values of both 6s in the numerical 376982604 is: 44404
59, 99, 400
5, 99, 400Solution: Place values of two 6s are 6000000 & 600Difference = 6, 00, 0000 – 600 = 5999400. Illustration 5: If the number78xis divisible by 6, then the value of x is: 45
6
7Solution: For any number to be divisible by 6, it must be divisible by both 2 and 36 is the correct answer. Since, 786 is divisible by: 2(the last digit is even)3(7+8+6 = 21 which is divisible by 3)
Practice Exercise
How many prime numbers are less than 30? 8
10
1214Solution: Prime numbers below 30 are ‘ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29’. Thus, 10 is the correct answer.
=
1
105
10 -5510 -12
Solution:
106+5-7= 104 = 105The value of 102×98 is: 996986
9996
9986Solution: (100+2) (100-2) = 1002 -22= 10000 – 4= 9996
=
Solution: = =
If f(x) = x2-6x – 5, then f(2) –f(-2)= 2448
-24
-48Solution: f (x) = x2-6x – 5f (2) = 22 -6(2) -5 = 4 -12 -5 =-13f (-2) = (-2)2 -6(-2) -5 = 4 +12 -5 = 11f(2) – f (-2) = -13 – 11 =-24Which of the following fractions has a non-terminating repeating decimal? 33/5017/625171/800
17/90
Solution: To get a non-terminating repeating decimal expansion: Denominator 2m 5n50 = 21×52, 625 = 20× 54, 800 = 25 ×52, 90 = 21×32×5Therefore, the answer isRational form of 0. is:
2/3
3/533/50333/500Solution: Let x = 0.……..(i)x = 0. 6…….(ii)Multiplying equation (ii) by 1010x = 6.…..(iii)Subtracting (iii) from (i)9x = 6x = 6/9 = 2/3is: An integerA rational number
An irrational number
None of these
Solution: = =
Here, 2 is a rational number& 3 isirrational. Now, we know that Rational ×Irrational = IrrationalHence, is an irrational number. Rational form of 2. 4 is: 12/5
22/9
11/76/5Solution: Let, x = 2.….. (i)x = 2. 4…… (ii)On multiplying equation (ii) by 1010x = 24.…… (iii)Now, subtract (i) from (iii)9x = 22x = The decimal expansion of the number 27683/625 will terminate after: One decimal placesTwo decimal placesThree decimal places
Four decimal places
Solution: =
=
=
The correct answer is (d) as the number ends after four decimal expressions.
Average
Important Formulas/Concepts
Average = Let average age of existing A people be x, and the number of new people be n, and increase in average age be y years. Then,
If n new people join,
If n people join and the average age increases, the average age of n people = x +If n people join and the average age decreases, the average age of n people = x –
If n people leave,
If n people leave and average age increases, then average age of n people = x +If n people leave and average age decreases, then average age of n people = x -Illustration 1: If the mean of x, x + 3, x + 5, x + 7 & x + 10 is 9, then the mean of last three observations is: 10
11
11Solution: Average of 5 numbers = 9= 9Thus, x = 4Hence, the last 3 numbers are (4+5), (4+7) and (4+10)Average (9, 11, 14) =
=
= 11 . Illustration 2: If the mean of five observations is 4, then the mean of same observations after increasing 5 is: 68
9
10Solution: Original sum of five observations = (x1+x2+x3+x4+x5) = 5×4 = 20Increase in sum after adding 5 to each observation= (x1+5) + (x2+5)+(x3+5)+(x4+5)+(x5+5)= (x1+x2+x3+x4+x5) + (5×5)= 20 + 25= 45New mean = = 9Illustration 3: The mean of the marks scored by 50 students was found to be 39. Later on it was discovered that a score of 43 was misread as 23. The corrected mean is: 38. 6
39. 4
39. 839. 2Solution: Wrong sum of 50 observations = 50× wrong mean= 50×39= 1950Corrected sum = wrong sum –wrong reading +correct reading= 1950 – 23+ 43 = 1970Corrected mean = = 39. 4Illustration 4: The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is: 2634
38
46Solution: Sum of five numbers = 5×mean= 5×30= 150Sum of remaining four numbers = 4× New mean= 4× 28= 112The excluded number = sum of five observations – sum of four observations= 150-112= 38Illustration 5: The mean of 31 results is 60. If the mean of first 16 results is 58 and that of the last 16 results is 62, the 16th result is:
60
403530Solution: Sum of first sixteen results = 16 × mean= 16 × 58= 928Sum of last sixteen results = 16 × mean= 16 × 62 = 992Sum of all thirty one results = 31 × mean= 31 × 60= 1860Sixteenth observation = (928 + 992 – 1860)= 60
Practice Exercise
The average cost of 5 jeans & 7 shirts is Rs. 600 and the average cost of 5 jeans is Rs. 740 , then the average cost of 7 shirts is:
500
560450480Solution: Total cost of 5 jeans & 7 shirts = Average (5J, 7S) ×12= 600×12= 7200Total cost of 5 jeans = Average (5J) ×5= 740×5= 3700Total cost of 7 shirts = 7200-3700= 3500Thus, average cost of 7 shirts = = 500At a shop ‘ book A’ was priced thrice as ‘ book B’ and ‘ book B’ was priced 9 more than twice the price of ‘ book C’. If average price of the three books is Rs. 360, the price of ‘ book C’ is: 310312256
116
Solution: Let the prices of books A, B and C be x, y and z respectivelyx = 3yy = 9 + 2zAccording to given condition= 360y = 241z = Z = (y-9)/2 = (241-9) /2 = 116Thus, the cost of book C = 116Three years ago, the average age of Marsh & Mathew was 16. With Steve joining them now, the average age becomes 20. The present age of Steve is: 16
22
1820Solution: Sum of ages of Marsh and Mathew presently = 16×2+3×2 = 38Sum of ages of Marsh, Mathew and Steve = 20×3 = 60Therefore, Steve’s age = 60-38 = 22 yearsA cricketer whose bowling average is 12. 4 runs per wicket takes 5 wickets for 26 runs and thereby decreases his average by 0. 4. The number of wickets taken by him till the last match was: 6575
85
95Solution: Let the number of wickets till the last match be ‘ x’New average after 5 more wickets for 26 runs is decreases by 0. 4= 12The average of 5 subjects increased by 5, when marks in English were increased in rechecking. If the marks in English before rechecking were 45, then the marks after rechecking in English are: 60
70
7585Solution: Total increase in marks after rechecking = 5×5 = 25Therefore, corrected marks in English = 45+25= 70If the average of three numbers a, b and c is A, then the average of a, b, c and A is: 2AA/24A
A
Solution: Given, Thus, a+b+c = 3AAverage of a, b, c & A =
=
= = AThe total production of 10 tea estates is 550 tonnes. By opening two new tea estates the average increases by 3 tonnes. The average cash production of these two new tea estates (in tonnes)is: 573570
568
564Solution: Total production of 10 tea estates = 550×10= 5500 tonnesTotal production of 12 tea estates = 553×12= 6636 tonnesProduction by 2 new estates = 6636 -5500= 1136Average production by each new estate == 568Find the average increase rate if increase in the population in the first year is 30% and that in the second year is 40%. 35%38%40%
41%
Solution: Population initially = 100%Population after 1 year = 130%Population after 2 years = 130%+40% (of 130%)= 182%Thus, increase in population in 2 years = 82%Average increase = 82%/2 = 41%A person travels three equal distances at a speed of x km/hr, y km/hr and z km/hr respectively. What was his average speed during the whole journey? xyz / (xy+yz+zx)(xy+yz+zx) / xyz
3xyz / (xy+yz+zx)
None of theseSolution: Let, the equal distance be ‘ D’ kmsNow, Average speed =
=
=
=
=
The average salary of the entire staff in an office is Rs. 3200 per month. The average salary of officers is Rs. 6800 and that of non-officers is Rs. 2000. If the number of officers is 5, then find the number of non-officers in the office. 812
15
5Solution: Let the number of non-officers in the office be xNow, average salary, 3200= 3200(x+5) = 3400+2000xThus, x= 15
HCF/LCM
Important Formulas/Concepts
Product of 2 numbers = HCF of 2 numbers × LCM of 2 (same) numbersHCF of fractions = LCM of fractions = Illustration 1: If the HCF of 27 and 63 is of the form (63 – 27m), then value of m is: 1
2
34Solution: HCF [27, 63] = 9Also, HCF [27, 63] = 63-27m (given)63-27 m = 9Thus, m = 2Illustration 2: What is the largest number that divides 70, 97, 125, leaving remainder 5, 6, & 8 respectively? 369
13
Solution: H C F [70-5, 97-6, 125-8]H C F [65, 91, 117] = 13Illustration 3: If r & s are positive integer such that r = a3 b2 & s = a2 b3 , then their LCM is: a2 b2
a3 b3
a6 b6a bSolution: L C M [r, s] = L C M [a3b2, a2b3]= a3 b3Illustration 4: Which among the following is a pair of co-primes?(14, 21)
(18, 25)
(31, 62)(32, 62)Solution: We know that, co-primes have their H C F= 1(18, 25) is pair of co-prime becauseits H C F= 1Illustration 5: What is the least number that is divisible by all prime numbers from 1 to 6? 30
60
90120Solution: LCM [1, 2, 3, 4, 5, 6]LCM [1, 2, 3, 2 = 1 = 60
Practice Exercise
Three measuring rods are 30, 45 & 60 cm in length. Find the least length of cloth that can be measured an exact number of times, using any one of the rods.
180
240360540Solution: L C M [30, 45, 60] is the least length of cloth that can be measured exact number of timesL C M [30, 45, 60] =[2×3×5, 3×3×5, 2×2×3×5]= 2×3×5×3×2= 180cmThe HCF & LCM of two numbers is 16 & 2304. If one number is 256 then other number is: 169
144
12596Solution: We know that: Product of numbers = H C F ×L C MThus, 256× x = 16×2304Thus, x= 144By what number should 162 be divided to get 10 as a quotient and 12 as a remainder? 6912
15
Solution: Let the divisor be ‘ x’It is known that: Dividend = Divisor × Quotient + Remainder162 = x×10+ 12Thus, x = 15The least number of four digits which is exactly divisible by 12, 15 & 18 is: 11201060
1080
1180Solution: L C M [12, 15, 18] = [2×2×3, 3×5, 2×3×3]2×2×3×3×5= 180Now, least four digits number is 1000, when divided by 180 gives remainder as 100Least four digit number divisible by12, 15, 18 is (1000 +180-100)= 1080What is the least number which when doubled will be exactly divisible by 6, 9, 12 and 15? 304560
90
Solution: Let the number be xL C M [6, 9, 12, 15] = [2×3, 3×3, 2×2×3, 3×5]= 2×2×3×3×5= 180Now 2x = 180x = 90The HCF of (23×33×52),(22×33×52) & (24×3×53×7) is: 304860
300
Solution: HCF of (23×33×52), (22×33×52) & (24×3×53×7)= 22×3×52= 4×3×25= 12×25= 300Richard takes 18 minutes to complete a circular track while John takes 12 minutes to complete the same track. If both start together in same direction, they both will meet again after: 12 minutes18 minutes24 minutes
36 minutes
Solution: Richard and John will meet at the L C M (18, 12) minutesL C M (18, 12) = 2×2×3×3= 36minutesThe numbers nearest to 10, 000 but greater than 10, 000 which is exactly divisible by 5, 6, & 8 is: 100201040
10080
1025Solution: We know that: L C M (5, 6, 8) = 2×2×2×3×5= 120Now, to obtain the number nearest but greater than 10, 000 divisible exactly by 5, 6, & 8, We will divide 10, 000 by L C M (5, 6, 8)= 10000/120 = Required number = 120 × 84= 10, 080The smallest fraction, which each of 2/7, 3/5, 4/21 will exactly divide is:
12/7
6/1436/724/14Solution: To obtain a number that is divisible by 2/7, 3/14 &4/21:= 12/7The maximum number of students among whom 960 books and 720 copies can be distributed under ” Sarv Shiksha Abhiyan” government scheme in such a way that each student gets the same number of books and copies is: 120
240
360420Solution: For distribution of equal number of books and copies, We will have to take out the H C F of 960 and 720.= 240
Percentage
Important Formulas/Concepts
Percentage equivalents of oft-appearing fractions: 1/1010%1/911. 11%1/812. 5%1/714. 28%1/616. 66%1/520%1/425%1/333. 33%1/250%2/366. 66%
¾
75%If P is the current population of a town, and it increases by R% per annum, then population after n years: P nIf P is the current population of a town, and it increases by R% per annum, then population n years ago: nIf A is R% more than B, then B is less than A by: If A is R% less than B, then B is more than A by: Illustration 1: 30% of a number subtracted from 91, gives the number itself. Find the number. 60 kg65 kg
70 kg
75 kgSolution: Let x be the numberAccording to given condition, 91 – 30% x = x91 – x = x91 = x = 70Illustration 2: Stuti spends 20% of her monthly income on her household expenditures, 15% of the rest on books, 30% of the rest on clothes and saves the rest. On counting, she came to know that she finally saved Rs. 9520. Find her monthly income. 10, 00025, 000
20, 000
12, 000Solution: Let her total income be xx -20% x – = 9520- -9520= 9520Illustration 3: 30% of x% of y is 150% of y% of z. Which of the following is z? 0. 25 x1. 2 x5/6 y
0. 2 x
Solution: 30% × x% × y = 150% × y% × z= z0. 20x = zIllustration 4: Two years ago, Ravi used to purchase 2 mangoes more than today which he can afford at Rs. 40. If the price is raised by 20%, then what is the cost of a dozen mangoes today? 5462
32
45Solution: Let the cost of a mango 2 years ago be ‘ x’Then, the cost of a mango presently after 20%: Price rice will beAccording to given condition,- = 3= 3x = Rs. Cost of a mango today will beCost of a dozen mangos = 12 × = 32Illustration 5: Two numbers are less than the third number by 25% and 40% respectively. How much percent is the second number less than the first? 15%
20%
25%40%Solution: Let the third number be ‘ x’, then the first number = 75% xSecond number = 60% xDifference in the first & second number = 15% xRequired percentage = ×100 = 20%
Practice Exercise
5% of income of A is equal to 15% income of B and 10% of income of B is equal to 20% of income of C. if C’s income is Rs. 2000, then the total income of A, B &C is:
18, 000
16, 00010, 0008000Solution: 5 % A = 15% B….. (I)10% B = 20% C….. (II)From (II)B = × 20000B = 4000From (I)A = × BA = 3 × 4000 = 12, 000Therefore total income of A, B, C = 12000 + 4000 + 2000 = 18, 000A building worth Rs. 12, 16, 700 is constructed on a land worth Rs. 4, 91, 300. After how many years will the values of both will be the same if the land appreciates at 15% per annum and the building depreciates at 15% per annum? 1 year1 ½ years2 ½ years
3 years
Solution: According to the given condition, 4, 91, 300 × n = 12, 16, 700 × nn ×3n = 3If the price of petrol is increased by 20%, by how much percent a car owner must reduce his consumption in order to maintain the same budget?
16 %
33 %40%23%Solution: We know that: Decrease in consumption = 1 ×100
=
37 ½% of the candidates in examination were girls, 75% of the boys and 62 ½% of the girls passed and 342 girls failed. The total number of boys failed were: 350360370
380
Solution: Let the total population of the class be x, Number of girls failed: 37. 5% × 37. 5% × x = 342(375/100)2 x = 342x = 2432Number of boys who failed:= 380If the price of a cricket bat is first decreased by 15% and then increased by 25%, then the net change in price of bat will be: 5%
6. 25%
7. 5%10%Solution: Let the original price of the bat be Rs 1000Price after decrement of 15% = 85% × 1000 = Rs 850Further, price after increment of 25%:= 125% × 850 = 1062. 5Net change in price = The difference between one fifth of 1000 and one fifth percent of 1000 is: 0998
198
800= 198If price of sugar is increased by 7%, then by how much percent should a housewife reduce her consumption of sugar so as to incur no extra expenditure? 7%93%
6
12%Solution: Decrease in consumption:
%=
The daily wage is increased by 15% and a person now gets Rs. 23 per day. What was his daily wage before the increase? 1518
20
Solution: Original daily wage= ×100= Rs. 20The ratio of salary of a worker in July to that in June was. By what % was the salary of July more than salary of June? 10%76 %Solution: ×100× 100= 11The population of a village is 1, 00, 000. The rate of increase is 10% per annum. Find the population increase in the third year?
12, 100
10, 00033, 10021, 000Solution: Population after 2nd year = 100000 × 2= 100000 ×= 121000Population after 3rd year = 100000 × 3100000 × = 133100Increase in population:= 133100 – 121000= 12, 100
Profit, Loss and Discount
Important Formulas/Concepts
Profit = Selling Price – Cost PriceLoss = Cost Price – Selling PriceProfit % = Loss % = Selling Price = Cost PriceSelling Price = Cost PriceCost Price = Selling PriceCost Price = Selling PriceDiscount = Marked Price – Selling PriceRate of discount = Illustration 1: A man purchased two watches for Rs. 560. He sold one at a 15% profit and the other at a 10% loss, and thus he neither gains nor loses. Find the cost price of each watch. 320, 240240, 320
224, 336
336, 224Solution: CP1 ×0. 15 = CP2 × 0. 10CP1/CP2 = 2/3…….(I)CP1 + CP2 = 560………..(II)On comparing (I) & (II)CP1= 224CP2= 336Illustration 2: A man buys 10 pencils for Rs. 3 and sold 8 of them for Rs 3. His gain percent is: 20%
25%
30%27%Solution: CP of 1 pencil = 3/10SP of 1 pencil = 3/8Gain% = × 100= × 100= 25%Illustration 3: A radio costing Rs. 500 is available on 10% discount on cash purchase. The shopkeeper gives sequential discounts. Asha paid Rs 427. 50 for the radio. Find the rate of the discount on cash price. 20%15%10%
5%
Solution: Price after 1st discount of 10% = 90%= 450If final cash price is Rs 427. 50Discount % == 5%Illustration 4: A person marks his goods 20% higher than cost price and allows a discount of 5%. His percentage profit is: 15%20%5%
14%
Solution: Let the cost price be 100, Marked Price = 120% Cost Price= × 100= 120Selling Price = 95% of Marked Price
=
= 19= 114
=
= 14%Illustration 5: A fruit seller has 24 kg of apples. He sells a part of these at a gain of 20% and the remaining at a loss of 5%. If on the whole he earns a profit of 10%, the amount of apples sold at a loss is: 4. 6 kgs6 kgs
9. 6 kgs
11. 4 kgsSolution: Let the number of apples sold at a loss be x and the cost price of an apple because be Rs 1. 1× (24-x) ×120% – 1×x×95% = 1×24×110%(24-x)×6/5 – x×19/20 = 24× 11/10Thus, x = 9. 6 kgs
Practice Exercise
A retailer buys 30 articles from a wholesaler at the price of 27. If he sells them at their marked price the gain percent in the transactions is: 90%
11
16Solution: Let the marked price of one article be Rs. 1C P of 30 articles = 27& S P/ M P of 30 articles = 30Profit %= 3/27×100= 100/9
=%
A dishonest shopkeeper professes to sell potatoes at the cost price, but he weighs 875 grams instead of one kg. What is his percentage of profit? 6. 5%
12. 5%
18. 75%15. 25%Solution: Profit% == 12. 5%A man sold his book for Rs. 891, thereby gaining 1/10th of its costprice. The cost price is: 850
810
851840Solution: Selling Price = Cost Price + 1/10th Cost Price891 = Cost PriceCost Price == 810A shopkeeper increased the price of a product by 50% and later on reduced the price by 50%. The shopkeeper’s loss was: 0%2. 5%
25%
10%Solution: Let the actual price of the article be Rs. 100Price after increasing 50% = 150% × 100= 150Price after reducing 50% = 50% × 150= 75Loss % = × 100 = 25%A CD music system when sold at a certain price gives a gain of 20%. If sold for thrice that price, the gain percent will be: 260%
200%
360%300%Solution: Let the selling price of CD music system be xCost Price = = If the new Selling Price is thrice that of Cost Price, then = 3
=
Profit % =× 100× × 100= 200%A shopkeeper earns a profit of 15% on selling a book at 10% discount on the printed price. The ratio of the cost price and the printed price of the book is. 23: 18
18: 23
3: 22: 3Solution: Let the Selling Price be xThen CP =%& Marked Price = %
= =
The cost price of a shirt and a pair of trousers is Rs. 371. If the shirt costs 12% more than the trousers, than the cost price of the trousers is: Rs. 125Rs. 150
Rs. 175
Rs. 200Solution: Let the Cost Price of the trousers be xThen, x + 112% x = 371212 x = 37, 100x = 175The cost of manufacturing an article is made up of material, labour and overheads in the ratio 4: 3: 2. If the cost of laborers is Rs. 45, find the profit percent if the article is sold for Rs. 180. 50%33. 33%
25%
20%Solution: Let the total cost of manufacturing be xThen labour cost, x = 45x = 135Selling Price = 180 (given)Gain% =× 100= 25%A trader purchases apples at Rs. 60 per hundred. He spends 15% on the transportation. What should be the selling price per dozen to earn a profit of 20%? Rs. 8. 21
Rs. 9. 936
Rs. 10. 2Rs. 3. 362Solution: Total Cost Price of 100 Apples = 60 = 69of 100 apples earning profit 20% = Selling Price of 1 dozen apples = = 9. 936A shopkeeper purchases 10 kg of rice at Rs. 600 and sells at a loss as much as the selling price of 2 kg of rice. Find the sale of rice purchases/kg.
Rs. 5
Rs. 10Rs. 12. 5Rs. 15Solution: Let the Selling Price be Rs. x / kgLoss = Cost Price –Selling Price2x = 600 – 10 xx = Rs. 5/ kg
Square Roots and Cube Roots
Illustration 1: If = than x= 1516
17
18
Solution: =
On squaring both sides – 1+-1Thus, x = 17Illustration 2: If x+ 1/x = 5, then the value of x2 + 1/x2 = 21
23
2527Solution: x += 5On squaring both sides, we get:(x+) 2 = 52×2+ + 2 = 25×2 + = 23Illustration 3: 1+1. 7321. 414
2
None of these(Take = 1. 414, = 1. 732, = 2)Solution: 1 + + +1 – (1 –2) – (2 – 3) – ( 3 – 4)1 – 1 + 2 – 2+ 3 – 3 + 4= 4= 2Illustration 4: If (x-7)2 + ( y–5 )2 +( z – 3)2 = 0, then value of x, y & z are …………… respectively.√7,√ 5,√349, 25, 9
7, 5, 3
Can’t be determinedSolution: Since, x – 7 = 0 = x = 7Also, y – 5 = 0 = y = 5z – 3 = 0 = z = 3
Illustration 5:
– 0. 4472
+ 0. 4472– 0. 5773+0. 5773(Take = 1. 732, = 2. 236)
Solution: +
=
= = -2. 236 / 5 = – 0. 4472
Practice Exercise
The greatest four digit perfect square is: 99629921
9801
9726Solution: 999 99999 81181 189917011989999 – 198 = 9801The smallest three digit perfect square is:
100
120125144Solution: 100 = 102100 is the smallest perfect squareIf x then values of x are: 2, 3
-2, 3
-5, 15, 1Solution: x = x = x2= 6 + xx2– x- 6 = 0x2– x – 6×2– 3x + 2x -6x (x – 3) +2 (x – 3)(x+2) (x-3)x = -2, 3The cube root of 0. 000000729 is: 0. 00270. 0270. 0009
0. 009
Solution:
3
=
= 0. 009If x = 5 – 2√6 , then x2 – 1/x2 = 36 √6– 27 √6
– 40 √6
72 √6Solution: x= 5 – 2 √6×2 = 25 + 24 – 20 √6 = 49 – 20 √6== 5 + 2 √6= 25 + 24 + 20 √6 = 49 + 20√6×2 – = 49 – 20 √6 -49-20= – 40√6The value of is not equal to: 366√363√144
18 √12
Solution: = = 36= 6= 318What is the least number which must be multiplied in 4851 to form a perfect square? 7
11
1213Solution: Factor of 4851 = 3×3×7×7×11So, 11 must be multiplied in 4851 to get a perfect square. What is the least number that must be divided in864 to form a perfect square? 23
6
9Solution: Factor of 864 = 2×2×2×2×2×3×3×3So, 6 must be divided in 864 to get a perfect square.
:
Solution:
If a = 13 , b = 5 , find value of 31896
1728
343216Solution: 3= 3= 3= 3= 1728
Partnership
Illustration 1: A, B and C enter into partnership with a total of Rs. 8200. A’s capital is Rs. 1000 more man B’s capital and Rs. 2000 less man C’s capital. What is B’s share of the year’s profit of Rs. 2460? 560920820
420
Solution: A + B + C = 8200…..(i)A = 1000+B => B = A – 1000A = C – 2000 => C = A + 2000Substituting in equation (i)A + A – 1000 + A + 2000 = 82003 A = 7200A = 2400B = 1400B’s share == 420Illustration 2: What amount of money is divided between Atul, Mradul, &Ishaan, if Mradul&Ishaan together get Rs. 3000 and Atul gets five times as much as Mradul while Ishaan and Atul together get Rs 6000? 875080007500
6750
Solution: M + I = 3000 ………(i)A = 5 M …………….(ii)I + A = 6000………..(iii)Substitute equation (ii) in (iii)5M + I = 6000 ……….(iv)Solving (i) & (iv)M = 750I = 2250A = 3750Hence, total money shared between the three is (750 + 2250 + 3750) = 6750
Practice Exercise
A, B & C enter into a partnership. A contributes Rs. 320 for 4 months, B contributes Rs. 510 for 3 months, and C contributes Rs. 270 for 5 months. If the total profit is Rs. 208, find the profit share of the partners. 78, 98, 32
64, 76. 50, 67. 50
90, 48, 7084, 45, 79Solution: 320 × 4 : 510 × 3 : 270 × 5128 : 153 : 135A’s share × 208 = 64B’s share × 208 = 76. 50C’s share × 208 = 67. 50A began a business with Rs. 1, 250 and was joined afterwards by B with Rs. 3, 750. When did B join if the profits at the end of the year are divided equally? After 6 months
After 8 months
After 4 monthsAfter 7 monthsSolution: Say B joined the business after ‘ x’ monthsIf the profits are shared equally, 1250 × 12 = 3750 ×(12 – x)= 3(12 – x) x = 8The cost of a music system and an LCD TV are in the ratio of 3: 8 and total price of both is Rs. 20, 900. The difference in their price is: 7, 6006, 050
9, 500
8, 100Solution: The ratio of the price is 3: 8Let the cost of the music system be 3x and the cost of LCD TV be 8xDifference in their prices = × 20, 900× 20, 900= 9500Four milkmen rented a pasture. ‘ A’ grazed 24 cows for 3 months; ‘ B’ grazed 10 cows for 5 months; C grazed 35 cows for 4 months and D grazed 21 cows for 3 months. If A’s share of rent is Rs. 720, find the total rent of the field.
3250
267045002750Solution: Let the total rent be ‘ x’Ratio of rent shared by four milkmen isA B C D24×3 : 10×5 : 35×4 : 21×372 : 50 : 140 : 63A’ share of rent = Rs. 720 (given)× x = 720Thus, x = 3250Hrithik and Ravish entered into partnership with capitals in the ratio 4: 5. After 3 months, Hrithik withdrew ¼ of his capital and Ravish withdrew 1/5th of his capital. If the gain at the end of 10 months was Rs. 760, Ravish’s share in this profit is: Rs. 330Rs. 360Rs. 380
Rs. 430
Solution: Let initially the sum invested by Hrithik and Ravish be 4x & 5xRatio of their profit share is4x × 3 + 4x ×7 : 5x × 3+ × 5x × 712x +21x: 15x +28x33x : 43xRavish share in profit× 760 = 430A workman earned Rs. 180 in a certain number of days. If his daily wages had been Rs. 2. he would take one more day to earn the same amount, find how many days he worked at the higher rate. 18 days
9 days
6 days12 daysLet the no of days initially be xUsing many equation,= 2= 2×2+ x – 90 = 0x2+ 10x – 9x – 90 = 0x (x + 10) – 9 (x + 10) = 0x = 9 daysAyush, Sujoy&Kartik hired a taxi for Rs. 780 and used it for 17, 8 & 14 hours respectively. The charges paid by Karthik are: 320
280
460300Solution: Kartik’s share = = Rs. 280In a partnership A invested 1/6th of the capital for 1/6th of the time, B invest 1/3rd of the capital for 1/3rd of the time and C, the rest of the capital for the whole time. Out of a profit of Rs. 4600, B’s share is: 650
800
9601000Solution: Let x be the total amount & y be the total time period investmentRatio of the profit for A, B, & C is× x × 1 y1 : 4 : 18Profit share of B = 4600 ×= 800Adil & Akram invest in a grocery business in the ratio 4: 3. If 9% of the total profit goes to certain taxes and Akram’s share is Rs. 35, 100, the total profit is: 1, 15, 0001, 09, 000
90, 000
68, 000Solution: Net profit after paying all taxes = 91% xGiven ratio 4: 3× 91% x = 35, 100x = x = 90, 000A and B are partners in a business. A contributes ¼ of the capital for 15 months and B received 2/3rd of the profit. For how long B’ money was used? 6 months9 months
10 months
1 yearSolution: The invested ratio is 1: 3& the profit ratio is 1: 2Equally born,
=
y = 10 months
Probability
Important Formulas/Concepts
Probability = Illustration 1: If two dices are thrown at random, probability of getting a sum more than 9 is:
1/6
5/62/3
½
Solution: Favorable outcomes – (5, 5) (4, 6) (6, 4) (5, 6) (6, 5) (6, 6)Also, total outcomes = 36Probability = 6/ 36 = 1/6Illustration 2: Probability of getting a black face card, when a card is drawn randomly from a 52 card deck is: 3/52
3/26
1/132/13Solution: Favorable outcomes = 6Total number of outcomes = 52Probability = Illustration 3: Find the probability of getting a prime number on throwing a dice.
¼
1/3
½
1Solution: Favorable outcomes = 3; {2, 3, 5}Total number of outcomes = 6Probability = 3/6 = 1/2Illustration 4: Find the probability of getting a composite number on throwing a dice.
1/3
½
2/31Solution: Favorable outcomes = 2 ;{ 4, 6}Total number of outcomes = 6Probability = 2/6 = 1/3Illustration 5: The probability of getting atleast 2 heads, on tossing a coin three times is:
½
3/8
¾
¼
Solution: The various possibilities (of getting at least 2 heads) are: H HHH H TH T HH T TT H HT H TT T HT TTThus, total number of favorable outcomes = 4Total number of outcomes = 8Thus, probability = 4/8 = ½
Practice Exercise
The probability of getting a doublet on throwing a pair of dice is: 1/21/3
1/6
5/6Solution: Favorable outcomes = 6; {(1, 1),( 2, 2),(3, 3),(4, 4),(5, 5),(6, 6)}Total number of outcomes = 36Probability = = The probability of getting 53 Sundays in a leap year is: 1/7
2/7
3/74/7Solution: Total complete weeks in a leap year = 52Also, remaining days in a leap year = 2 daysNow, favorable outcomes = {(sat, sun), (sun, mon)}Thus, 2/7 is the correct answer. The probability of getting 53 Mondays in a non-leap year is:
1/7
2/73/74/7Solution: Total number of complete weeks in a non-leap year = 52Also, remaining days in a leap year = 1Number of favorable outcomes = 1Total number of outcomes = 7Hence, probability = Among a group of 9 males & 6 females, a President & a Vice-President is to be appointed. Find the probability of occupying either both places by a female or one each by a male and a female. 12/10525/10537/105
54/105
Solution:
=
=
A class consists of 12 boys and 8 girls. If a panel of 3 students is to be made, then the probability of choosing two girls and one boy is:
28/95
7/2011/7512/19Solution: Number of ways of choosing 2 girls and 1 boys = 12c1×8c2Total number of ways to choose = 20c3To select 3 such students: Required probability =
=
=
A lady wardrobe consists of five black pants, three brown pants and six white tops and four red tops. The probability of choosing a black pant with a red top for a party is: 4/51/5
1/4
2/5Solution: Number of ways of choosing a black pant = 5c1Number of ways of choosing a red top = 4c1Total number of ways of choosing a pant and a top = 8c1×10c1Required probability =
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If two dices are thrown at random, find the probability of getting an odd number on both faces. 1/21/3
1/4
1/6Solution: Number of favorable cases = {(1, 1) (3, 3) (5, 5), (1, 3) (3, 1)(5, 3), (1, 5)(3, 5)(5, 1)}= 9Total number of cases = 36Required probability = = If three coins are tossed simultaneously, find the probability of getting at most one tail.
½
3/8
¼
5/8Solution: Number of favorable cases = 4 ;{(H, H, H,)(H, T, H)(H, H, T)(T, H, H)}Total number of outcomes = 8Required probability = 4/8 = 1/2From the deck of 52 cards, if a card is drawn at random, find the probability of getting either a Black jack or a Red king. 1/263/26
1/13
2/13Solution: Number of ways of getting a black jack= 2c1Number of ways of getting a red king = 2c1Total number of ways of getting a card = 52c1Required probability = = If out of a sample of 40 pens, 12 are defective, find the probability of choosing a defective and a perfect pen. 12/30
84/195
2/197/40Solution: Number of ways to select a defective pen= 12c1Number of ways to select a correct/perfect pen = 28c1Total number of ways of selecting two pens = 40c2Required probability == 84/195
