Distance is directly proportional to Velocity when time is constantTwo cars leave simultaneously from points A & B (100km apart) & they meet at a point 40 km from A. What is Va/Vb?Time is constant so V1/V2=S1/S2=40/60=4/6A train meets with an accident and moves at (3/4)th its original speed. Due to this, it is 20 min late. Find the original time for the journey beyond the point of accident?Method1:Think about 2 diff. situations, 1st with accident and another w/o accident. As distance in both the cases is constantSo V1/V2=T2/T1=>V1/[(3/4)V1]=T1+20/T1=>43=T1+20/T1=>T1=60Method 2:Velocity decreases by 25% (3/4 of original speed => decrement by 1/4) so time will increase by 33.3% (4/3 of original time => increment by 1/3)now, 33.3%=20 min=>100%=60 minCase 1:Two bodies are moving in opposite directions at speed V1&V2 respectively. The relative speed is defined as Vr=V1+V2Case 2:Two bodies are moving in same directions at speed V1&V2 respectively. The relative speed is defined asVr=|V1V2|When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object.The distance to be covered when crossing an object, whenever trains crosses an object will be equal to: Length of the train + Length of the objectLet,U= Velocity of the boat in still waterV= Velocity of the stream.Upstream:While moving in upstream, distance covered,S=(UV)TDownstream:In case of downstream, distance covered ,S=(U+V)TFor clock problems consider the clock as a circular track of 60km.Min. hand moves at the speed of 60km/hr (think min. hand as a point on the track) and hour hand moves at 5km/hr and second hand at the speed of 3600 km/hr.Relative speed between HOUR hand and MINUTE hand = 55 Time taken by a train of length lmetres to pass a pole or standing man or a signal post is equal to the time taken by the train to cover lmetres. Time taken by a train of length lmetres to pass a stationery object of length bmetres is the time taken by the train to cover (l + b) metres. Suppose two trains or two objects bodies are moving in the same direction at u m/s and v m/s, where u>v, then their relative speed is = (u - v) m/s. Suppose two trains or two objects bodies are moving in opposite directions at u m/s and v m/s, then their relative speed is = (u + v) m/s. If two trains of length ametres and bmetres are moving in opposite directions at u m/s and v m/s, then:The time taken by the trains to cross each other =(a + b)/(u + v)sec.

If two trains of length ametres and bmetres are moving in the same direction at u m/s and v m/s, then:The time taken by the faster train to cross the slower train =(a + b)/(u - v)sec.

If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then:(A's speed) : (B's speed) = (b : a)

Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?

A.9B.10

C.12D.20

Explanation:Due to stoppages, it covers 9 km less.Time taken to cover 9 km =9x 60min= 10 min.

9.A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. The average speed for the first 320 km of the tour is:

A.35.55 km/hrB.36 km/hr

C.71.11 km/hrD.71 km/hr

Explanation:Total time taken =160+160hrs.=9hrs.

64802

Average speed =320 x2km/hr= 71.11 km/hr.

9

It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the cars is:

A.2 : 3B.3 : 2

C.3 : 4D.4 : 3

Explanation:Let the speed of the train be x km/hr and that of the car be y km/hr.Then,120+480= 8 1+4=1....(i)

xyxy15

And,200+400=25 1+2=1....(ii)

xy3xy24

Solving (i) and (ii), we get: x = 60 and y = 80.Ratio of speeds = 60 : 80 = 3 : 4.A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is:

A.35B.362

3

C.371

2

D.40

Explanation:Let distance = x km and usual rate = ykmph.Then,x-x=40 2y(y + 3) = 9x ....(i)

yy + 360

And,x-x=40 y(y - 2) = 3x ....(ii)

y -2y60

On dividing (i) by (ii), we get: x = 40.Two cars pass each other in opposite direction. How long would they take to be 500 km apart?

I.The sum of their speeds is 135 km/hr.

II.The difference of their speed is 25 km/hr.

A.I alone sufficient while II alone not sufficient to answer

B.II alone sufficient while I alone not sufficient to answer

C.Either I or II alone sufficient to answer

D.Both I and II are not sufficient to answer

E.Both I and II are necessary to answer

Explanation:I gives, relative speed = 135 km/hr.Time taken =500hrs.

135

II does not give the relative speed.I alone gives the answer and II is irrelevant.Correct answer is (A).Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:

A.1 : 3B.3 : 2

C.3 : 4D.None of these

Explanation:Let the speeds of the two trains be x m/sec and y m/sec respectively.Then, length of the first train = 27xmetres,and length of the second train = 17ymetres.27x + 17y/x+ y = 23

27x + 17y = 23x + 23y4x = 6yx=3.

A jogger running at 9 kmph alongside a railway track in 240 metres ahead of the engine of a 120 metres long train running at 45 kmph in the same direction. In how much time will the train pass the jogger?

A.3.6 secB.18 sec

C.36 secD.72 sec

Explanation:Speed of train relative to jogger = (45 - 9) km/hr = 36 km/hr. =36 x5m/sec

18

= 10 m/sec.Distance to be covered = (240 + 120) m = 360 m.Time taken =360sec= 36 sec.

10

Two goods train each 500 m long, are running in opposite directions on parallel tracks. Their speeds are 45 km/hr and 30 km/hr respectively. Find the time taken by the slower train to pass the driver of the faster one.

A.12 secB.24 sec

C.48 secD.60 sec

Explanation:Relative speed == (45 + 30) km/hr

=75 x5m/sec

18

=125m/sec.

6

We have to find the time taken by the slower train to pass the DRIVER of the faster train and not the complete train.So, distance covered = Length of the slower train.Therefore, Distance covered = 500 m.Required time =500 x6= 24 sec.

125

Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 a.m. and travels towards B at 20 kmph. Another train starts from B at 8 a.m. and travels towards A at a speed of 25 kmph. At what time will they meet?

A.9 a.m.B.10 a.m.

C.10.30 a.m.D.11 a.m.

Suppose they meet x hours after 7 a.m.Distance covered by A in x hours = 20x km.Distance covered by B in (x - 1) hours = 25(x - 1) km.20x + 25(x - 1) = 11045x = 135x = 3.So, they meet at 10 a.m.1. Trigonometry:In a right angled OAB, where BOA = ,

i. sin =Perpendicular=AB;

HypotenuseOB

ii. cos =Base=OA;

HypotenuseOB

iii. tan =Perpendicular=AB;

BaseOA

iv. cosec =1=OB;

sin AB

v. sec =1=OB;

cos OA

vi. cot =1=OA;

tan AB

2. Trigonometrical Identities:i. sin2+ cos2= 1.ii. 1 + tan2= sec2.iii. 1 + cot2= cosec2.3. Values of T-ratios:0(/6)

30(/4)

45(/3)

60(/2)

90

sin 01

root2

root3

2

1

cos 1root3

2

1

root2

0

tan 01

root3

1root3not defined

4. Root 2 = 1.4145. Root 3 = 1.7326. 7. Angle of Elevation:

Suppose a man from a point O looks up at an object P, placed above the level of his eye. Then, the angle which the line of sight makes with the horizontal through O, is called the angle of elevation of P as seen from O.Angle of elevation of P from O = AOP.8. Angle of Depression:

Suppose a man from a point O looks down at an object P, placed below the level of his eye, then the angle which the line of sight makes with the horizontal through O, is called the angle of depression of P as seen from O.

In what ratio must a person mix three kind of tea each of which has a price of 70, 80 and 120 rupees per kg, in such a way that the mixture costs him 100 rupees per kg?Solution:Here the prices of tea are 70, 80 and 120 And mean price is 100.Here, prices LOWER than the mean are 70 and 80. And prices HIGHER than the mean is 120.Thus possible pairs which can give mean value of 100 is: {70, 120} and {80, 120}Let us denote tea of Rs. 70 with t70, tea of Rs. 80 with t80 and tea of Rs. 120 with t120We apply the old alligation rule to ALL (two in this case) the pairsFor the 1st pair (t70, t120)

t70:t120=20:30Similarly, for the 2st pair (t80, t120)

t80:t120=20:20t80:t120=20:20Thus, Final ratio:t70:t80:t120=20:20:(30+20)t70:t80:t120=2:2:5Note:It's best to simplify the intermediate ratios at the end, else you may get wrong answer.Here, if we had simplified the ratio to be, t70:t120=2:3 after first pair and t80:t120=1:1 after second pair.We could have easily gone to calculate ratio to be, t70:t80:t120=2:1:(3+1) or t70:t80:t120=2:1:4 which is different from the final ratio (t70:t80:t120=2:2:5)

How must a shop owner mix 4 types of rice worth Rs 95, Rs 60, Rs 90 and Rs 50 per kg so that he can make the mixture of these rice worth Rs 80 per kg?Solution:Here the prices of sugars are 50, 60, 90 and 95.And the mean price is 80.Here, prices LOWER than the mean are 50 and 60. And prices HIGHER than the mean are 90 and 95.Thus possible pairs which can give mean value of 80 are: {50, 95} and {60, 90}Let us denote rice of Rs. 50 with r50, rice of Rs. 60 with r60, rice of Rs. 90 with r90 and rice of Rs. 90 with r95We apply the old alligation rule to ALL (two in this case) the pairsFor the 1st pair (r50, r95)

r50:r95=15:30r50:r95=15:30Similarly, for the 2st pair (r60, r90)

r60:r90=10:20r60:r90=10:20Thus, Final ratio:r50:r60:r90:r95=15:10:20:30r50:r60:r90:r95=3:2:4:6

Let the quantity of ingredient to be added = Q liters Quantity of ingredient in the given mixture = x% of P = x/100 * P Percentage of ingredient in the final mixture = Quantity of ingredient in final mixture / Total quantity of final mixture. Quantity of ingredient in final mixture = [x/100 * P] + Q = [ P*x + 100 * Q] / 100Total quantity of final mixture = P + Q y/100 = [[ P*x + 100 * Q] / 100]/[P + Q] y[P + Q] = [P*x + 100 * Q]The quantity of ingredient to be added5) Quantity of ingredient to be added to change the ratio of ingredients in a mixture

In a mixture of x liters, the ratio of milk and water is a : b. If the this ratio is to be c : d, then the quantity of water to be further added is:In original mixture Quantity of milk = x * a/(a + b) litersQuantity of water = x * b/(a + b) litersLet quantity of water to be added further be w litres.Therefor in new mixture:Quantity of milk = x * a/(a + b) liters Equation(1)Quantity of water = [x * b/(a + b) ] + w liters Equation (2) c / d = Equation (1) / Equation (2)Quantity of water to be added further, 1. Work from Days:If A can do a piece of work in n days, then A's 1 day's work =1.

n

2. Days from Work:If A's 1 day's work =1,then A can finish the work in n days.

n

3. Ratio:If A is thrice as good a workman as B, then:Ratio of work done by A and B = 3 : 1.Ratio of times taken by A and B to finish a work = 1 : 3. If M1 men can do W1 work in D1 days working H1 hours per day and M2 men can do W2 work in D2 days working H2 hours per day (where all men work at the same rate), thenM1 D1 H1 / W1 = M2 D2 H2 / W2

If A can do a piece of work in p days and B can do the same in q days, A and B together can finish it in pq / (p+q) daysP,Q and R can completely solve a problem together in 4 hrs. P and R together take 15 hrs less than Q working alone. Q works on the problem for first 2 hrs and then P and R joins him. After another two hours, Q quits. In how many hours is the problem actually solved ?Let Q alone can answer the problem in x hoursThen P&R together needs (x-15) hours to solve the problem

In 1 hr, Q can complete 1/x of the question, In 1 hr, P&R can complete 1/(x-15) of the question In 1 hr,P,Q and R can complete 1/4th of the question

1/x + 1/(x-15) = 1/44(x-15)+4x =x(x-15)4x - 60 + 4x = x^2 - 15xx^2 - 23x + 60 = 0(x-20)(x-3)=0x = 20 or 3since (x-15) is positive, x = 20

In 1 hr, Q can complete 1/20 of the question.In 1 hr, P and R can complete 1/5 of the questionIn 1 hr, P, Q and R can complete 1/4 of the question

In the first two hours, 2* 1/20 = 1/10 of the question is completed by QP,Q and R work in next 2 hrs and completes 2 * 1/4 = 1/2 of the question

Remaining part = 1 - 1/10 - 1/2 = 2/5Let P&R work for n hours and completes this partn = (2/5)/(1/5) = 2

Total time taken to solve the question = 2+2+2 = 6 hoursP, Q and R can do a work in 20, 30 and 60 days respectively. How many days does it need to complete the work if P does the work and he is assisted by Q and R on every third day?

A. 10 daysB. 14 days

C. 15 daysD. 9 days

Explanation :Amount of work P can do in 1 day = 1/20

Amount of work Q can do in 1 day = 1/30

Amount of work R can do in 1 day = 1/60

P is working alone and every third day Q and R is helping him

Work completed in every three days = 2 (1/20) + (1/20 + 1/30 + 1/60) = 1/5

So work completed in 15 days = 5 1/5 = 1

Ie, the work will be done in 15 days6 men and 8 women can complete a work in 10 days. 26 men and 48 women can finish the same work in 2 days. 15 men and 20 women can do the same work in - days.

A. 4 daysB. 6 days

C. 2 daysD. 8 days

Explanation :Let work done by 1 man in 1 day = m and work done by 1 woman in 1 day = b

Work done by 6 men and 8 women in 1 day = 1/10

=> 6m + 8b = 1/10

=> 60m + 80b = 1 --- (1)

Work done by 26 men and 48 women in 1 day = 1/2

=> 26m + 48b =

=> 52m + 96b = 1--- (2)

Solving equation 1 and equation 2. We get m = 1/100 and b = 1/200

Work done by 15 men and 20 women in 1 day

= 15/100 + 20/200 =1/4

=> Time taken by 15 men and 20 women in doing the work = 4 daysA completes 80% of a work in 20 days. Then B also joins and A and B together finish the remaining work in 3 days. How long does it need for B if he alone completes the work?

A. 37 daysB. 22 days

C. 31 daysD. 22 days

Explanation :Work done by A in 20 days = 80/100 = 8/10 = 4/5

Work done by A in 1 day = (4/5) / 20 = 4/100 = 1/25 --- (1)

Work done by A and B in 3 days = 20/100 = 1/5 (Because remaining 20% is done in 3 days by A and B)

Work done by A and B in 1 day = 1/15 ---(2)

Work done by B in 1 day = 1/15 1/25 = 2/75

=> B can complete the work in 75/2 days = 37 daysFather is aged three times more than his son Sunil. After 8 years, he would be two and a half times of Sunil's age. After further 8 years, how many times would he be of Sunil's age?

A. 4 timesB. 4 times

C. 2 timesD. 3 times

Explanation: Assume that Sunil's present age = x. Then father's present age = 3x + x = 4x

After 8 years, father's age = 212 times of Sunils' age=> (4x+8) = 212(x+8)=> 4x + 8 = 52(x + 8)=> 8x + 16 = 5x + 40=> 3x = 40 - 16 = 24=> x = 243 = 8

After further 8 years, Sunil's age = x + 8 + 8 = 8 + 8 + 8 = 24Father's age = 4x + 8 + 8 = 4 8 + 8 + 8 = 48

Father's age/Sunil's age = 4824 = 2The present ages of A,B and C are in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 56. What are their present ages (in years)?

A. Insufficient dataB. 16, 30, 40

C. 16, 28 40D. 16, 28, 36

Explanation: Let's take the present age of A,B and C as 4x, 7x and 9x respectively

(4x - 8) + (7x 8) + (9x 8) = 56=> 20x = 80=> x = 4

Hence the present age of A, B and C are 44, 74 and 94 respectively ie, 16,28 and 36 respectively.Pi is a mathematical constant which is the ratio of a circle's circumference to its diameter. It is denoted by 3.142274.1. Properties of Triangle Sum of the angles of a triangle = 180 Sum of any two sides of a triangle is greater than the third side. The line joining the midpoint of a side of a triangle to the positive vertex is called the median The median of a triangle divides the triangle into two triangles with equal areas Centroid is the point where the three medians of a triangle meet. Centroid divides each median into segments with a 2:1 ratio Area of a triangle formed by joining the midpoints of the sides of a given triangle is one-fourth of the area of the given triangle. An equilateral triangle is a triangle in which all three sides are equal In an equilateral triangle, all three internal angles are congruent to each other In an equilateral triangle, all three internal angles are each 60 An isosceles triangle is a triangle with (at least) two equal sides In isosceles triangle, altitude from vertex bisects the base.4.2. Properties of Quadrilaterals

4.2.1. Rectangle The diagonals of a rectangle are equal and bisect each other opposite sides of a rectangle are parallel opposite sides of a rectangle are congruent opposite angles of a rectangle are congruent All four angles of a rectangle are right angles The diagonals of a rectangle are congruent4.2.2. Square All four sides of a square are congruent Opposite sides of a square are parallel The diagonals of a square are equal The diagonals of a square bisect each other at right angles All angles of a square are 90 degrees. A square is a special kind of rectangle where all the sides have equal length4.2.3 Parallelogram The opposite sides of a parallelogram are equal in length. The opposite angles of a parallelogram are congruent (equal measure). The diagonals of a parallelogram bisect each other. Each diagonal of a parallelogram divides it into two triangles of the same area4.2.4. Rhombus All the sides of a rhombus are congruent Opposite sides of a rhombus are parallel. The diagonals of a rhombus bisect each other at right angles Opposite internal angles of a rhombus are congruent (equal in size) Any two consecutive internal angles of a rhombus are supplementary; i.e. the sum of their angles = 180 (equal in size) If each angle of a rhombus is 90, it is a square4.2.5 Other properties of quadrilaterals Sum of the interior angles of a quadrilateral is 360 degrees If a square and a rhombus lie on the same base, area of the square will be greater than area of the rhombus (In the special case when each angle of the rhombus is 90, rhombus is also a square and therefore areas will be equal) A parallelogram and a rectangle on the same base and between the same parallels are equal in area. Of all the parallelogram of given sides, the parallelogram which is a rectangle has the greatest area. Each diagonal of a parallelogram divides it into two triangles of the same area A square is a rhombus and a rectangle.4.3 Sum of Interior Angles of a polygon Sum of the interior angles of a polygon = 180(n - 2) degrees where n = number of sides

Example 1 : Number of sides of a triangle = 3. Hence, sum of the interior angles of a triangle = 180(3 - 2) = 180 1 = 180

Example 2 : Number of sides of a quadrilateral = 4. Hence, sum of the interior angles of any quadrilateral = 180(4 - 2) = 180 2 = 360

The square root of a number is just the number which when multiplied by itself gives the first number. So 2 is the square root of 4 because 2 * 2 = 4.Start with the number you want to find the square root of. Let's use 12. There are three steps:1. Guess 2. Divide 3. Average.... and then just keep repeating steps 2 and 3. First, start by guessing a square root value. It helps if your guess is a good one but it will work even if it is a terrible guess. We will guess that 2 is the square root of 12.In step two, we divide 12 by our guess of 2 and we get 6.In step three, we average 6 and 2: (6+2)/2 = 4Now we repeat step two with the new guess of 4. So 12/4 = 3Now average 4 and 3: (4+3)/2 = 3.5 Repeat step two: 12/3.5 = 3.43Average: (3.5 + 3.43)/2 = 3.465We could keep going forever, getting a better and better approximation but let's stop here to see how we are doing. 3.465 * 3.465 = 12.006225That is quite close to 12, so we are doing pretty well. Cost Price:The price, at which an article is purchased, is called its cost price, abbreviated as C.P.Selling Price:The price, at which an article is sold, is called its selling prices, abbreviated as S.P.Profit or Gain:If S.P. is greater than C.P., the seller is said to have a profit or gain.Loss:If S.P. is less than C.P., the seller is said to have incurred a loss.IMPORTANT FORMULAE1. Gain = (S.P.) - (C.P.)2. Loss = (C.P.) - (S.P.)3. Loss or gain is always reckoned on C.P.4. Gain Percentage: (Gain %) Gain % =Gain x 100

C.P.

5. Loss Percentage: (Loss %) Loss % =Loss x 100

C.P.

6. Selling Price: (S.P.) SP =(100 + Gain %)x C.P

100

7. Selling Price: (S.P.) SP =(100 - Loss %)x C.P.

100

8. Cost Price: (C.P.) C.P. =100x S.P.

(100 + Gain %)

9. Cost Price: (C.P.) C.P. =100x S.P.

(100 - Loss %)

10. If an article is sold at a gain of say 35%, then S.P. = 135% of C.P.11. If an article is sold at a loss of say, 35% then S.P. = 65% of C.P.12. When a person sells two similar items, one at a gain of say x%, and the other at a loss of x%, then the seller always incurs a loss given by: Loss % =Common Loss and Gain %2=x2.

1010

13. If a trader professes to sell his goods at cost price, but uses false weights, then Gain % =Errorx 100%.

(True Value) - (Error)

The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then the value of x is:

A.15B.16

C.18D.25

Explanation:Let C.P. of each article be Re. 1 C.P. of x articles = Rs. x.S.P. of x articles = Rs. 20.Profit = Rs. (20 - x).20 - xx 100 = 25

x

2000 - 100x = 25x125x = 2000x = 16.If selling price is doubled, the profit triples. Find the profit percent.

A.662

3

B.100

C.1051

3

D.120

Explanation:Let C.P. be Rs. x and S.P. be Rs. y. Then, 3(y - x) = (2y - x) y = 2x.Profit = Rs. (y - x) = Rs. (2x - x) = Rs. x.Profit % =xx 100% = 100%

x

In a certain store, the profit is 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?

A.30%B.70%

C.100%D.250%

Explanation:Let C.P.=Rs. 100. Then, Profit = Rs. 320, S.P. = Rs. 420.New C.P. = 125% of Rs. 100 = Rs. 125New S.P. = Rs. 420.Profit = Rs. (420 - 125) = Rs. 295.Required percentage =295x 100%=1475% = 70% (approximately).

42021

Suresh started a business, investing Rs.18000. After 3 months and 4 months respectively, Rohan and Sudhir joined him with

capitals of 12000 and 9000. At the end of the year the total profit was Rs.3982. What is the difference between Rohans and

Sudhirs share in the profit? A.) 354 B.) 370

C.) 362 D.) 121

Suresh : Rohan : SudhirRatio of their investments = 18000 12 : 12000 9 : 9000 8 = 6 : 3: 2The difference between Rohans and Sudhirs share = 1 share:.i.e. = Rs. 3982 1/11 = Rs.362.

The word percent can be understood as follows:Per cent for every 100.If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is:=[(R/(100+R))100]%If the price of the commodity decreases by R%, then to maintain the same expenditure by increasing the consumption is:=[(R/(100R))100]%Let the population of the town be P now and suppose it increases at the rate of R% per annum, then:1. Population after n years =P[1+(R/100)]n2. Population n years ago =P/[1+(R/100)]nLet the present value of a machine be P. Suppose it depreciates at the rate R% per annum. Then:1. Value of the machine after n years =P[1(R/100)]n2. Value of the machine n years ago =P/[1(R/100)]n5. If A is R% more than B, then B is less than A by=[(R/(100+R))100]%If A is R% less than B , then B is more than A by=[(R/(100R))100]%When a value is increased by 20%, by what percent should it be reduced to get the actual value?Solution:(It is equivalent to 1.2 reduced to 1 and we can use % decrease formula)%decrease=((1.21)/1.2)100=16.66%When a value is subjected multiple changes, the overall effect of all the changes can be obtained by multiplying all the individual factors of the changes.The population of a town increased by 10%, 20% and then decreased by 30%. Thenew population is what % of the original?Solution:The overall effect =1.11.20.7 (Since 10%, 20% increase and 30% decrease)=0.924=92.4%.Two successive discounts of 10% and 20% are equal to a single discount of ___Solution:Discount is same as decrease of price.So, decrease =0.90.8=0.7228% decrease (Since only 72% is remaining)In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid.If the total number of votes was 7500, the number of valid votes that the other candidate got, was:

A.2700

B.2900

C.3000

D.3100

Solution:Option(A) is correctNumber of valid votes = 80% of 7500 = 6000.Valid votes polled by other candidate = 45% of 6000=(45/100)6000 = 2700.If a quantity is increased by a% and then further by b% these percent change given are equivalent to a single percent change given by%

Similarly if there is a successive decrease of a% followed by b% then effective percentage decrease is

Marked price.The price which is displayed on the tag of the article is known as marked price.Generally the SP is less then the marked price (MP) the difference MP SP is known as discount DDiscount % , D% = Ratio is a pure number it does not have a unit. The 2 numbers used in ratio are known as the 'terms' of the ratio. The first term is known as 'antecedent' and the second term is known as 'consequent'.Fresh grapes contain 90% water by weight while dried grapes contain 20% water by weight . What is the weight of dry grapes available from 20 kg of fresh grapes ?1. 2 kg2. 2.4 kg3. 2.5 kg4. none of theseSolutionLet x kg be weight of dry grapes obtained from 20 kg of fresh grapes then weight of solid substance in fresh grapes will be same as in dried grapes hence we get

x = 2.5 kgWhen processing flower-nectar into honeybees' extract, a considerable amount of water gets reduced. How much flower-nectar must be processed to yield 1kg of honey, if nectar contains 50 water, and the honey obtained from this nectar contains 15 water?

A.1.5 kgs

B.1.7 kgs

C.3.33 kgs

D.None of these

Solution:Option(B) is correctFlower-nectar contains 50% of non-water part.In honey this non-water part constitutes 85%(10015).Therefore 0.5X Amount of flower-nectar = 0.85X Amount of honey = 0.851 kgTherefore amount of flower-nectar needed=(0.85/0.5)1=1.7kg30% of the men are more than 25 years old and 80% of the men are less than or equal to 50 years old. 20% of all men play football. If 20% of the men above the age of 50 play football, what percentage of the football players are less than or equal to 50 years?

A.15%

B.20%

C.80%

D.70%

Solution:Option(C) is correct20% of the men are above the age of 50 years. 20% of these men play football. Therefore, 20% of 20% or 4% of the total men are football players above the age of 50 years.

20\% of the men are football players. Therefore, 16% of the men are football players below the age of 50 years.

Therefore, the % of men who are football players and below the age of50=16/20100

A student multiplied a number by 3/5 instead of 5/3. What is the percentage error in the calculation?

A.34%

B.44%

C.54%

D.64%

It is not a bad idea to assume some number and then proceed with the given instructions. Let's assume the number to be 15.Now, if the student had multiplied it correctly by (53), he/she would have gotthe result as =(53)15=25. But instead the student ended up multiplying by =(35) and got=(35)15=9 as a result.Here, we can see that error percentage,% error=(25925)=0.64 or 64%if(x+yxy=43)and x0, then what percentage (to the nearest integer) of x+3y is x3y ?

A.20%

B.25%

C.30%

D.40%

Solution:Option(D) is correctDividing both the numerator and the denominator of the given equation :x+y/xy=43x/y+1/x/y1=43Cross-multiplying this equation yields3x/y+3=4x/y4Solving for x/y yieldsx/y=7Now, the percentage of x+3y the expression x3y makes isx3y/x+3y100Dividing both the numerator and the denominator of the expression by y yields=x/y3/x/y+3100=73/7+3100=40%A man purchased a bag for Rs.360 and sold it the same day for Rs. 360, allowing the buyer a credit of 9 years.If the interest be 712% then the man has gain %?

A.159

B.173

C.243

D.287

Solution:Option(C) is correctInterest rate =152=7.5%.CP=Rs.360.SP=360+intrest on 360 for 9 yearsInterest = 3601592100=Rs. 243.Gain = Interest =Rs. 243.A shepherd has 1 million sheep at the beginning of Year 2000. The numbers grow by x during the year. A famine hits his village in the next year and many of his sheep die. The sheep population decreases by y during 2001 and at the beginning of 2002 the shepherd finds that he is left with 1 million sheep. Which of the following is correct?

A.x>y

B.y>x

C.x=y

D.Cannot be determined

Solution:Option(A) is correctLet us assume the value of x to be 10%.Therefore, the number of sheep in the herd at the beginning of year 2001 (end of 2000) will be 1 million + 10% of 1 million = 1.1 million

In 2001, the numbers decrease by y% and at the end of the year the number sheep in the herd = 1 million.i.e., 0.1 million sheep have died in 2001.

In terms of the percentage of the number of sheep alive at the beginning of 2001,it will be (0.1/1.1)100%=9.09%.

From the above illustration it is clear thatx>yA candidate who gets 20% marks fails by 10 marks but another candidate who gets 42% marks gets 12% more than the passing marks.Find the maximum marks.

A.50

B.100

C.150

D.200

Solution:Option(B) is correctFrom the given statement pass percentage is 42%12%=30%By hypothesis, 30% of x20% of x=10 (marks)i.e., 10% of x=10Therefore, x =100 marks.Vikash bought a suitcase with 15% discount on the labelled price. He said the suitcase for Rs.2880 with 20% profit on the labelled price. At what price did he buy the suitcase? Rs.2040 Rs.2400 Rs.2604 Rs.2640

Answer: A

Solution : Let the labelled price be Rs.x. Then, 120% of x = 2880 Therefore x=(2880100/120) = 2400. C.P = 85% of Rs.2400 = Rs(85/100x2400) =Rs.2040.

A shopkeeper gives a discount of 12, whereas a customer makes cash payment. Let 'p' denotes the percentage, above the cost price, that the shopkeeper must mark up the price of the articles ['p' is an integer] in order to make a profit of x. Which of the following is the possible value(s) of x?

A.54

B.76

C.96

D.32

Solution:Option(C) is correctLet the cost price of one article is Rs r then:Marked price of the article=r(1+p/100)Selling price=r(1+p/100)(112/100)Profit percentage =x%=r(1+p/100)(112/100)=r(1+x/100)x=25(12+x)/22As, p is an integer, x must be multiple of 22.All the possible values x, less than 100 are k=10,32,54,76,98Exceptoption 'C'all the options are possible.The volume of the sphere Q is(dfrac{37}{64}%)less than the volume of sphere P and the volume of sphere R is(dfrac{19}{27}%)less than that of sphere Q. By what is the surface area of sphere R less than the surface area of sphere P?

A.77.77%

B.87.5%

C.75%

D.67.5%

Solution:Option(C) is correctLet the volume of sphere P be 64 parts.Therefore volume of sphere Q=6437/64%of 64=6437=27 parts.The volume of R=2719/2727=2719=8 parts.Volume ratio:=P:Q:R=64:27:8Radius ratio:=P:Q:R=4:3:2The surface area will be 16:9:5Surface area of R is less than the surface area of sphere P16k4k=12kNow,=12k/16k100=75%Thus surface area of sphere R is less than the surface area of sphere P by 75%23. Prasanth bought a car and paid 10 % less than the original price. He sold it with 30% profit on the price he had paid. What percentage of profit did he earn on the original price?

A. 17%B. 16%

C. 18%D. 14%

Explanation : Let the original price=100Then the price at which he purchased (CP)=90% of 100=90Profit=30%SP=((100+Profit%)/100)CP=((100+30)/100)90=(130/100)90=139=117Required%=((117100)/100)100=17%If a seller reduces the selling price of an item from Rs.400 to Rs.380, his loss increases by 2%. What is the cost price of the item?

A. 1000B. 800

C. 1200D. 1100

Sunil purchases two books at Rs.300 each. He sold one book 10% gain and other at 10% loss. What is the total loss or gain in percentage?

A. 10% gainB. 1% loss

C. No loss or no gainD. 1% gain

A reduction of 10% in the price of a pen enabled a trader to purchase 9 more for Rs.540. What is the reduced price of the pen?

A. 8B. 6

C. 5D. 4

What is the present worth of Rs. 132 due in 2 years at 5% simple interest per annum1. 1102. 1203. 1304. 140Answer And ExplanationAnswer: Option BExplanation:Let the present worth be Rs.xThen,S.I.= Rs.(132 - x)

= (x*5*2/100) = 132 - x

= 10x = 13200 - 100x

= 110x = 13200

x= 120 A financier claims to be lending money at simple interest, But he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes.1. 10.25%2. 10%3. 9.25%4. 9%Answer And ExplanationAnswer: Option AExplanation:Let the sum is 100.

As financier includes interest every six months., then we will calculate SI for 6 months, then again for six months as below:

SI for first Six Months = (100*10*1)/(100*2) = Rs. 5

Important: now sum will become 100+5 = 105

SI for last Six Months = (105*10*1)/(100*2) = Rs. 5.25

So amount at the end of year will be (100+5+5.25)= 110.25

Effective rate = 110.25 - 100 = 10.25

If 8men 8hrs per day works for 8days get 45/- then how many men required if the work is 5hrs per day for10days they get 60/-?