Quantitative Aptitude fully covered topics

Time & Work Work is always taken as one unit. Construction of a building, filling water in the tank, painting a room etc. To complete a job, a man will do the same amount of work on each day of the total number of days he takes to complete that job. If a man can do a piece of work in 8 days, then his one day’s work is If a man’s one day’s work is

.

then he can complete the total work in 4 days.

If a man can complete a piece of work in A days and another man can complete the same work in B days, then they together can complete the work in

days.

Similarly, three persons A, B and C together can complete in

days.

Ex: A can complete a piece of work in 6 days and B in 8 days. In what time they complete if they work together?

MAN DAYS: If 6 men can complete a work in 4 days, then the number of man days required to complete that work is 6 × 4 = 24. Whatever may be the number of persons working on that, the total number of man days required for that work will be 24.

Ex:18 men together can complete a work in 14 days. In how many days 12 men finish that work? A) 5 dyas

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B) 6 days

C) 10 days

D) 8 days

E) None of these

Men and Time are inversely proportional i.e., when more men work, they take less time to complete the work. Similarly when less men work, they take more time to complete the work. Men and Work are directly proportional i.e., when more men are there, they do more work and less men are there, they do less work. Similarly, Time and Work are also directly proportional. If men work for more time then they do more work and less time then the work is also less. The relation among these variables can be shown in a formula If M is Men, D is Days (Time) and W is work then

12 men to complete a work in 15 days. But after 8days he notices that only 30% work has

Ex: A contractor employs

been completed. In order to finish the work in the given time, how many more men he has to recruit? A) 32

B) 24

C) 20

D) 9

E) None of these

To complete the remaining work 32 men are required Additional men required 32 - 12 = 20 Hence answer is (c)

PIPES AND CISTERNS The problems in pipes and cisterns are almost the same as those of Time and Work problems. Ex: If a pipe fills a cistern in 12 hours and a leak in the bottom empties it in15 hours. In what time can the tank be filled when both function simultaneously? Sol: Pipe’s one hour’s work = Leak’s one hour’s work =

(negative work)

Pipe and leak together can do Net work is Positive. Tank will be filled in 60 hours.

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work in one hour.

Some more models: 1. A can complete a piece of work in 240 days. A and B together can complete the same work in 144 days. In what time does B alone complete that work? A) 15 days

2.

B) 96 days

C) 148 days

D) 360 days

E) 240 days

A can do a piece of work in 12 days. B is 60% more efficient than A. In how many days B

alone finish the work?

3. 14 men can do a piece of work in 12 days. 5 days after they started the work, 6 more men joined them. How many days will they now take to complete the remaining work? A) 10 days

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B) 12 days

C) 14 days

D) 8 days

E) None of these

4. A can do a piece of work in 20 days. B can do the same work in12days. If B worked on the work for 9 days, how many days A worked on it?

5. Two pipes A and B can fill a tank in 3 and 4 hours respectively. A drain pipe C can empty a full tank in 6 hours. In what time can the tank be filled when all the three are open simultaneously? A) 7.5 hours

B) 2.4 hours

Sol: A = 3 (+) , B = 4 (+) , C = 6(-)

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C) 1 hours

D) 6 hours

E) None of these

AVERAGE Examples Example 1. A batsman has a certain average runs for 16 innings. In the 17th inning he made a score of 85 runs thereby his average is increased by 3. What is his average after 17th inning? Sol: The average for 17th inning has been increased by 3. The total increase in the runs for 17th inning = 17 × 3 = 51 But the batsman scores 85. Average runs in his 16th innings = 85 – 51 = 34. Hence the average of runs after 17th innings = 34 + 3 = 37 Example 2. A man has 7 children. When their average age was 12 years, the child who was 6 years of age, died. What was the average age of surviving children 5 years after the death of the above child? Sol: Average age of 7 children = 12 years Total age of 6 children = 12 × 7 = 84 years Total age of 6 children after the death of a child aged 6 years = 84 – 6 = 78 Hence the average age of the surviving children = After 5 yrs. = 13 + 5 = 18 yrs. Example 3. If the weights of 5 students of a class are 49.6 kg, 39.8 kg, 45.2 kg and 24.6 kg respectively then what is their average weight? Sol: Total weight of 5 students = 49.6 + 39.8 + 40.8 + 45.2 + 24.6 = 200 kg. Their average weight = = 40 kg. Example 4. The average temperature for Monday, Tuesday and Wednesday was 36ºC . The average temperature for Tuesday, Wednesday and Thursday was. If the temperature for Thursday was 37ºC, what was the temperature on Monday? Sol:

Average temperature for Monday, Tuesday and Wednesday = 36ºC Total temperature for Monday, Tuesday and Wednesday = 36 × 3 = 108ºC

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Average temperature for Tuesday, Wednesday and Thursday = 38ºC Total temperature for Tuesday, Wednesday and Thursday = 38 × 3 = 114ºC Total temperature for Tuesday and Wednesday only = 114 – 37 = 77ºC Temperature for Monday only = 108 – 77 = 31ºC Example 5. A train covers the first 16 km at a speed of 20 km per hour another 20 km at 40 km per hour and the last 10 km at 15 km per hour. Find the average speed for the entire journey.

Example 6. A vehicle travels from A to B at the speed of 40 km/hr, but from B to A at the speed of 60km/hr. what is its average speed during the whole journey? Sol: Let the distance from A to B be x km

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Example 7. The average age of a class of 40 boys is 16.95 years. A new boy joins the class and the average age now is 17 years. What is the age of the new boy? Sol: The average age of 40 boys = 16.95 years Total are of 40 boys = 16.95 × 40 = 678 years The average age of 41 boys = 17 years Total age of 41 boys = 17 × 41 = 697 years Age of the new boy = 697 – 678 = 19 years

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PROFIT & LOSS-I The money spent to purchase an article is cost price of the article, and the money received for selling an article is called selling price. Profit = selling price – cost price Loss = cost price – selling price

Profit or loss percent is always calculated on C.P unless it is required to calculate on S.P.

DISCOUNTS The price at which the article is marked is the marked price or listed price or labeled price or catalogue price. The discount is allowed on marked price for the cash payments.

When discount is deducted from the marked price, the remainder becomes the selling price. To find Marked price from Cost Price we can use the following formula

EXAMPLES Example 1. Ram sold a cow for Rs.136 at a loss of 15%. At what price should he have sold it to gain 15%? Sol: Let the cost price be Rs.100 then for 15% loss the S.P = 85 and for 15% profit it should be Rs. 115. Rs. 85 is the first S.P, then second S.P. = Rs. 115 Rs. 136 is the first S.P =

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Example 2. A sells a radio to B at a gain of 10% and B sells it to C at a gain of 5%.If C pays Rs. 462 for it, What did it cost to A? Sol: Let the cost price of A be Rs. 100 Then the cost price of B be Rs. 110

Example 3. A dealer allows 10% discount on the list price of a certain article and yet makes a profit of Rs. 25% on each article. Find the cost price of the article when list price is Rs. 50. Sol: Let the cost price of articles be Rs. 100 Then for 25% profit, S.P. = Rs. 125 If list price is Rs. 100, S.P. = Rs. 90 If S.P. is Rs. 90, list price = Rs. 100

Example 4. A person purchases 90 clocks and sells 40 clocks at a gain of 10% and 50 clocks at a gain of 20%. Had he sold all of them at a uniform profit of 15% he would have got Rs.40 less. Find the cost price of each clock. Sol: Let the C.P of clock is Rs. 100 each. By the profit of 10% S.P. of 40 clocks = = Rs. 4, 400 By the profit of 20% S.P of 50 clocks = = Rs. 6, 000 Total S.P. = Rs. 4400 + Rs. 6, 000 = Rs. 10, 400 C.P of 90 clocks = = Rs. 9000 By the profit of 15% S.P. of 90 clocks = = Rs. 10, 350

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Difference = Rs. 10, 400 – Rs. 10, 350 = Rs. 50 If the difference is Rs. 50 then C.P. = Rs. 100

If the difference is Rs. 40 then C.P. =

= Rs. 80

Example 5. A man buys 5 horses and 10 cows for Rs. 1,600. He sells horses at a profit of 15% and cows at a Loss of 10% if his over all profit was Rs. 90, what was the cost price of a horse and a cow? Sol: Let x be the cost price of a horse and y be the cost price of a cow C.P. of 5 horses = Rs. 5x and C.P of 10 cows = Rs. 10y Hence

5x + 10y = 1,600……..(i)

Since the profit is 15% on the horses

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PIPES AND CISTERNS The problems in pipes and cisterns are almost the same as those of Time and Work problems, but with a bit difference as negative work comes into play due to leakage or drain pipes in the questions. Let us understand this with examples. Example 1. If a pipe fills a tank in 10 hours and a leak in the bottom empties it in15 hours, when both function simultaneously then

Net work is Positive. Thus tank will be filled in 30 hours.

Example 2. If a pipe fills a tank in 5 hours and a leak empties in 4 hours, when both function simultaneously then

Net work is Negative. Thus tank will be emptied in 20 hours.

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TIME AND DISTANCE The relation among distance, time and speed is Distance = Speed x Time.

If a car covers 50 kilometres in each hour then the speed of the car is 50Kmph. If a man travels 10 metres in one second’s time the the speed of the man is 10 mps. Convertion of speeds: KMPH to MPS

MPS to KMPH

When a certain distance is traveled with a speed of x kmph and another equal distance is traveled at y kmph, then

the average speed =

TRAINS If a train has to cross a pole or a man or any particular point then it has to cover a distance equal to its own length. If a train has to cross a platform or bridge or tunnel etc. then it has to cover a length equal to its length + platform/bridge/tunnel length. Relative speed When two speeds are in opposite direction, Relative speed = sum of the speeds. When two speeds are in the same direction, Relative speed = difference of the speeds. BOATS AND STREAMS Still water: water without motion. A boat can travel or a man can swim in that water with their original speeds. Stream / current : water flow with some speed. Down stream: along with the flow. Upstream: against the water flow

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If the speed of the boat in still water is b kmph and speed of the current is c kmph respectively, then Downrate ( x ) = b + c Uprate ( y ) = b–c

Average speed

If a moving object travels from A to B at the speed of x km/hr, and from B to A at the speed of y km/hr, then

Example 1: A train 100 metre long is running at the speed of 21 km/hr and another train 150 meter long is running at the speed of 36 km/hr in the same direction. How long will the faster train take to pass the first train? Solution: Sum of the length of both the train = x1 + x2 = 100 + 150 = 250 m Difference of their speeds = y1 - y2 = 21 - 36 = 15 km/hr

Question with office concern Example 1: A person walking at x km/hr reaches his office t1 minutes late. If he walks at y km/hr, he reaches there t2 minutes earlier, then

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Example 5: A man covers a distance of 160 km at 64 km/hr and next 160 km at 80 km/hr. what is his average speed for his whole journey of 320 km?

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Example 6: What will be the length of the train P when it is running at 60 km/hr and crosses another train Q running in opposite direction, in 18 seconds? In order to answer this question which of the statements (a) and (b) is/are sufficient? (a) Length of the train Q is 80 meter (b) Speed of the train Q is 90 km/hr Solution: Both statements (a) and (b) together are necessary '.' The trains are running in opposite directions

Example 7: A boat takes 3 hours to go from P to Q downstream and from Q to P up stream. What is the speed of the boat in still water? to know the answer of this question, the knowledge of which of the statements (a) and (b) is/are sufficient? The distance between P and Q is 6 km.

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The speed of the river is 2 km/hr. Solution: Both statements (a) and (b) together are necessary to the question. Let the speed of the boat in still water be x km/hr. .'. Speed of the boat down stream = (x + 2) km/hr And Speed of the boat upstream = (x -2) km/hr

Now x can be calculated.

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PERCENTAGE A fraction with denominator 100 is called percentage. Numerator of that fraction is called rate percent. The term percent means per every 100. 50 percent means 50 out of 100. The symbol % is used to denote percent. Ex: 20%, 35%, 80% etc. Percent to fraction: To convert percent into fraction, divide the percent by 100.

Fraction to percent: To convert fraction into percent, multiply the fraction by 100.

Increment percent = Ex: increment % in 5 and 8 is 3/5 × 100 = 60%. Decrement percent =

Ex. Decrement % in 5 and 8 is 5/8 × 100 = 37.5%. Some important percent /fractions.

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Following rules should be kept in mind for solving the questions on percentage — 1. For converting a given percentage into a fraction, divided it by 100 or multiply by 2. For converting a given fraction into a percentage multiply it by 100. 3. For converting one given quantity, say, x as a percentage of another given quantity, say, y find Note - percentage is never expressed in any unit like rupees or kilos. 4. When there are two articles ‘a’ and ‘b’ in a group and in the ratio of a : b then to express any one article as a percentage of the group, divide the article by the total articles of the group and multiplies the whole by 100. Percentage of ‘a’ = And, Percentage of ‘b’ =

EXAMPLES

Example 4. How many percent is 15 cm of 1 metre?

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Note: Both quantities are converted into same unit. Example 5. A person spent 85% of his monthly income and thereafter saves Rs.360 per month. Find out his total income. Sol: Suppose his monthly income is Rs. 100 Expenditure = 85% = Rs. 85 Saving = Rs. 100 – 85 = Rs. 15 If saving is Rs. 15 income is Rs. 100 If saving is Rs. 360 income is = Rs. 2400 Example 6. If a student gets 38 marks out of 50, what is the percentage of his obtained marks? Sol: Percentage of obtained marks = 76% Example 7. If two numbers are respectively 30% and 40% more then a third number, what percentage is the first of the second? Sol: Let the third number be 100 First number = 100 + 30% of 100 = 130 and Second number = 100 + 40% of 100 = 140 Percentage of the first number to the second number to the second number

Example 8. A student scores 20% marks and fails by 30 marks while another student who scores 32% marks, gets 42 marks more than the minimum required pass marks. What are the maximum marks for the examination? Sol: 32% - 20% = 12%

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And 42–(-30) = 72 12% marks = 72 100% marks = Example 9. An engineering student has to secure 50% marks to pas. He gets 163 marks and fails by 37 marks. Find the maximum marks Sol: Minimum marks required = 163 + 37 = 200 If 50 are the minimum marks then maximum marks = 100 200 are the minimum marks then maximum marks =

PROFIT & LOSS-II The price at which an article is purchased is cost price of the article, and the price at which the article is sold is called selling price. Profit = Selling Price − Cost Price Loss = Cost Price − Selling Price To calculate profit/loss percent we have to use the following formulae

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Note: Profit or loss percent is always calculated on Cost Price unless it is required to calculate on Selling Price. e.g: A man buys an umbrella for Rs.120 and sells for Rs.150. Find his profit percent. Sol:

To calculate cost price or selling price, the following formulae are to be used.

e.g: Raghu sells an article for Rs.14250 and gains 14%. What is the cost price of that article? Sol: Selling Price = Rs.14250 Profit% = 14%

Short cut: Cost price is 100%, Profit =14% therefore S.P. = 114% If 114% =14250,

DISCOUNTS The price at which the article is marked is the marked price or displayed price or listed price or labelled price or catalogue price or printed price. The discount is allowed on marked price.

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Note: Discount percents are always calculated on Marked Price. When discount is deducted from the marked price, the remainder becomes the selling price. e.g: A dealer marked an article at Rs. 740 and sells for Rs.629. Find the discount percent he offered. Sol:

The relation between Marked Price and Cost Price can be expressed in the following formula.

e.g: A shopkeeper marks an article at Rs.990. He allows 12% discount on it and yet gains 21%. Find the rate at which he bought the article. Sol: M.P. = 990, Profit% = 21%, Discount% = 12%

1. By selling an article for Rs.450, a man loses 25%. At what price he will sell in order to gain 25%? Sol: S.P. = 450, Loss% = 25%,

2. The cost price of 18 articles is equal to the selling price of 15 22

articles. Find the gain percent. Sol: C.P. of 18 = S.P. of 15 Let the selling price of one article be Re.1 ... S.P. of 15 articles = Rs.15 ... C.P. of 18 articles = Rs.15 S.P. of 18 articles = Re.1 × 18 = Rs.18

(+is for Profit & − for Loss) 3. A reduction of 20% in the price of oranges enables a man to buy 5 oranges more for Rs.100. Find the price of an orange before reduction. Sol: With 20% money he buys 5 oranges, and with 100% money he buys 25 oranges. ... before reduction he gets 20 oranges only = Rs. 5

4. Two successive discounts 40% and 30% are equivalent to a single discount of Sol: We can use a formula

5. A man buys two articles for Rs.2100. He sells one at a gain of 16% and other at a loss of 12%. On the whole he neither gains nor loses. What does the first article cost?

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Sol: Let the cost price of first article be Rs.x . . . 16% of x = 12% of (2100 − x) 16x = 12 × 2100 −12x 28x = 12 × 2100

6. A dishonest dealer professes to sell his goods at 25% profit and uses only 800 grams in place of a kilogram weight. Find his net profit percent. Sol: 25% profit means 250 grams profit. By using 800 grams for a kilogram he gains 200 grams . . . His net profit is 250 + 200 = 450 grams

7. A sells a bicycle to B at a profit of 20% and B sells it to C at a gain of 25%. If C pays Rs.225 for it, what is the cost price of the bicycle for A? Sol: Let the cost price of A be Rs.100, then the selling price = Rs.120 Now, the cost price of B be Rs.120

.. C.P. of C = Rs.150 . But the cost price of C = Rs.225 If the cost price of C is Rs.150, then the C.P. of A = Rs.100 If the cost price of C is Rs.225, then the C.P for

Short cut: Let the C.P. of A be Rs.x then 120% of 125% of x = 225

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1. The cost of 10 pens and 15 books is Rs. 525. What is the cost of 8 pens and 12 books ? A: Rs.420 2. A man buys 120 kgs rice for Rs. 4560. He sells 80 kgs for Rs.42 per kg and the remaining at Rs.28 per kg. Find his overall profit/loss%? A: 1.75% profit 3. The cost price of 30 apples is same as selling price of 24 apples. Find the profit/ loss% A: 25% 4. An article is marked at Rs.625 and 16% discount is offered on it. How many such articles can be bought for Rs.9450? A: 18 5. A trader charges his customer 23% more than the cost price. If a customer paid Rs. 7,011 for an article, then what was the cost price of the article? A: Rs. 5700 6. A dealer allows 10% discount on the list price of a certain article and yet makes a profit of 25% on each article. Find the cost price of the article when list price is Rs.875. A: Rs.630

SIMPLE AND COMPOUND INTEREST ¤ Interest is the additional money paid for the usage of a certain amount. ¤ The amount borrowed is called the principal. ¤ The sum of interest and principal is called the amount. Simple Interest If the interest is calculated on same amount of money it is called the simple interest (S.I.). Simple Interest will be the same for all the years. If P is the principal, R is the rate of interest, T is time and S.I. the simple interest, then

Note: Simple interest is always calculated on principal. Therefore simple interest is equal for every period.

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e.g: 1. What would be the simple interest obtained on an a amount of Rs. 6850 at the rate of 6 p.c.p.a. after 3 years? (A) Rs. 2423 (B) Rs. 1233 (C) Rs. 1633 (D) Rs. 1525 (E) None of these Sol: Here P = Rs. 6850, T= 3 years and R = 6%

Hence answer is (B) 2. How long will it take for Rs. 1250 to become Rs. 1600 at 7% per annum simple interest? (A) 5 years (B) 3 years (C) 4 years (D) 6 years (E) None of these Sol:

Hence Answer is (C) 3. What would be the amount on Rs. 8250 for 4 years at 15% per annum simple interest? (A) Rs. 13200 (B) Rs. 12300 (C) Rs. 10450 (D) Rs. 11200 (E) None of these SOl:

Shortcut: for one year, rate of interest is 15% and for 4 years it is 15×4 = 60% The Amount will become 160% If 100% Money = 8250, Compound Interest: Interest which is calculated not only on the initial principal but also the accumulated interest of prior periods. If A is the amount, C.I. is the compound interest, P is the principal, R is the rate, and T is the time, then

e.g: What is the compound interest accrued on an amount of Rs. 8000, at the rate of 6% p.a. at the end of 2 years? (A) Rs.2545 (B) Rs.2,257.20 (C) Rs.2986 (D) Rs.2775.40 (E) None of these

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Sol:

Shortcut: Amount = 106% of 106% 8000 = 8988.80 C.I. = 8988.80 - 8000 = Rs. 988.80 Note: 1. If the interest is paid half yearly, time is doubled and the rate is halved. 2. If the interest is paid quarterly, time becomes 4 times and the rate becomes onefourth e.g: What is the interest accrued on Rs.12000 for one and half year at 4% p.a. compounded half yearly?

Difference between Simple and Compound Interest Difference between Simple Interest and Compound Interest can be calculated by using formulae. There is no difference between Simple and Compound Interest for one year. For 2 years

e.g: 1. What is the difference between Simple and Compound Interest for two years on Rs. 24000 at 7% rate? Sol:

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= Rs. 117.60 2. On what sum does the difference between Simple and Compound Interest for 3 years at 5% rate will be Rs. 244? Sol:

Some other Models of Questions: 1. A sum of money will become Rs. 8060 in 4 years at 6% per annum simple interest. Find the sum. Sol:

Shortcut: For one year, rate of interest is 6% and for 4 years it is 4 × 6% = 24% Then the amount will become 124% If 124% money is Rs. 8060, 100% money will be

× 8060 = Rs. 6500

2. Find the simple interest on Rs. 17500 at 7% per annum from August 5th to October 17th in the same year. Sol: Time from August 5th to October 17th = 26 days of August + 30 days of September and 17 days of October

3. A sum was put at simple interest at a certain rate for 2 years. Had it been put at 3% higher rate, it would have fetched Rs. 72 more. What is that sum? Sol: Interest for 2 years is Rs. 72. For one year it is Rs. 36 3% interest money is Rs. 72, then 100% money

× 36 = Rs. 1200

Sum is Rs. 1200 4. A certain sum of money invested at compound interest doubles in 3 years. In how many years will it become 6 times itself? Sol: Let the money be Rs. x It becomes Rs. 2x in 3 years As this is compound interest, 2x will be the principal for next period. Therefore, 2x will become 4x in next 3 years hence Rs. x will become 4x i.e. 4 times in 3 + 3 = 6 years

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5. A man deposits Rs. 12600 in a bank at 5% annual interest. After 8 months he withdraws Rs. 5400 together with interest and after 4 months the remaining money. How much does he get as interest at the end of the year? Sol: S.I. of Rs. 12600 for 8 months

He withdrew Rs. 5400 together with interest, the remaining amount = 12600 - 5400 = Rs. 7200 S.I. on Rs. 7200 at the rate of 5% for 4 months

Total interest = 420 + 120 = Rs. 540

NUMBER SYSTEM A golden opportunity for Bank exams aspirants. Thousands of posts in SBI and other Banks. Out of five papers in the SBI clerk exam, Quantitative Aptitude plays major role. Arithmetic and Data Interpretation questions will be there in this paper. Arithmetic is the maths, which we use in our day-to-day life. As arithmetic chapters are there upto 10th class only all the aspirants inclusive of non-maths might have studied this. By understanding all the chapters thoroughly and with a lot of practice it is not difficult to solve all the questions of the paper. Also important is simplifications. Out of 40 questions in Q.A., 20 are simplification questions. Students need to mug up tables upto 20, squares of numbers upto 25 and cubes of numbers upto 15 to make simplifications easy. They also learn Speed Maths for solving simplification questions in seconds without pen. All the chapters in arithmetic have to be practiced thoroughly. Let us see all the chapters of arithmetic in detail. Digit: 0 to 9 are the digits. Number: By using digits we write numbers. Face Value: The real value of the digit. Place Value: The value of the digit according to the place. e.g.: In 2794, the face value of 7 is '7'and the place value is '700'. Classification of Numbers: Odd Numbers Numbers which are not divisible by 2 e.g.: 1, 3, 5, 7, .......

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Even Numbers Numbers which are divisible by 2 e.g.: 2, 4, 6, 8, ....... Natural Numbers

All counting numbers

e.g.: N = {1, 2, 3, ........}

Sum of first 'n' natural numbers is Sum of squares of first 'n' natural numbers is Sum of cubes of first 'n' natural numbers is

Whole Numbers 0 and all natural numbers e.g.: W = { 0, 1, 2, 3, ...} Integers Natural numbers, zero and negative of natural numbers e.g.: I = { ........, -3, -2, -1, 0, +1, +2, +3, .......} Rational Numbers These are of the form of p/q, where q π 0 e.g.:

etc

Irrational Numbers Can't be written in p/q form, where q ≠ 0 π etc Composite Numbers Numbers having more than two factors e.g.:

Prime Numbers Numbers divisible by one and the number itself or numbers having only two factors. e.g.: 2, 3, 5, 7, 11, 13, ...... etc Some points about prime numbers ¤ All prime numbers end with 1, 3, 7 or 9 except 2 and 5. ¤ 2 is the only even prime number. ¤ '1' is neither a prime nor composite number. It is a unitary. ¤ There are 25 prime numbers upto 100. Twin prime numbers The difference between two prime numbers is '2'. e.g.: 3 - 5, 5 - 7, 11 - 13 etc Co prime numbers Numbers not having any common factors. e.g.: 3, 5 and 8 are co primes to each other.

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Some important Identities: 1. (a+b)2 = a2 + b2 + 2ab 2. (a-b)2 = a2 + b2 - 2ab 3. (a+b)(a-b) = a2 - b2 4. (a+b)3 = a3 + b3 + 3a2b + 3ab2 5. (a-b)3 = a3 - b3 - 3a2b + 3ab2 6. a3 + b3 = (a+b)(a2− ab + b2) 7. a3 - b3 = (a− b) (a2 + ab+ b2) 8. (a+b+c)2 = a2+b2+c2 + 2ab + 2bc+2ac 9. a3+b3+c3− 3abc = (a+b+c) (a2 +b2 + c2 - ab -bc-ac) e.g.:

1) 2.01

2) 2

3) 1

4) 1.93

Sol: It is based on the formula

a3

+

b3

5) None of these

= (a+b)(a2−ab+b2)

1.07 + 0.93 = 2, hence answer (2).

To solve the equations based on numbers, the number is written in algebraic form. For example let the number be x. The numerator of a fraction is written in the following form Three-fourth of a number = Two-third of a number =

Double of a number = 2x, Thrice of a number = 3x 20% of a number = 30% of 3/4 of one-third of a number =

e.g.: 1. 4/5 of 4/7 of 5/6 of 1218 = ? (SBI 2008) (A) 415 (B) 384 (C) 492 (D) 346 sol:

(E) None of these

hence answer (E)

2. One-seventh of a number is 51. What will be 64% of that number? (SBI 2008) (A) 248.12 (B) 228.48 (C) 238.24 (D) 198.36 (E) None of these Sol:

hence answer (E)

3. If (16)3 is subtracted from the square of a number, the answer so obtained is 3825.

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What is the number ? (SBI 2008) (A) 69 (B) 59 (C) 89 (D) 79 (E) None of these Sol: Let the number be x x2− 163= 3825 x = 89 hence answer (C). 4. If (92)2 is added to the square of a number, the answer so obtained is 10768. What is the number? (SBI 2008) (A) 46 (B) 2304 (C) 48 (D) 2116 (E) None of these Sol: Let the number be x x2+ 922= 3825 x = 48 hence answer (C). 5. If an amount of Rs. 97,836 is distributed equally amongst 31 children. How much amount would each child get? (SBI 2008) (A) Rs.3,756 (B) Rs.3,556 (C) Rs. 3,356 (D) Rs. 3,156 (E) None of these Sol:

hence answer (D).

Three consecutive odd numbers are x, (x+2) and (x+4), or (x-2), x and (x+2), where x is an odd number. Similarly, three consecutive even numbers are x, (x+2) and (x+4) or (x-2), x and (x+2) where x is an even number If a number 'a' is multiplied by itself for n times, the product is called n th power of 'a'. This is written as an. In an, a is called base and n is called index or exponent or power.

Laws of Indices

1. am × an = a m+n 4+3= 7 7

e.g.: 7 4× 7 3= 7

2.

3. (am)n= amn 4. (ab)m= am× bm

e.g.: (93)2= 9 3×2 e.g.: (3×5)2= 32× 52

6. 7. a0 = 1

e.g.: 258 ÷ 25 3 = (5) ? sol:

32

e.g.:

30 =

e.g.: 1

(SBI 2003)

(52) 5 = 5 ? ? = 10

5 10 = 5 ?

All simplification questions have to be solved based on BODMAS rule only. B - Bracket O - Of D - Division M - Multiplication A - Addition S - Subtraction e.g. 1: What is 394 times 113? (SBI 2008) (A) 44402 (B) 44632 (C) 44802 (D) 44522 (E) None of these Sol: 394 × 113 = 44522, hence answer (D) 2. What approximate value should come in place of the question mark (?) in the following question? 4780 ÷ 296 × (23)2 = ? (A) 9870 (B) 6760 (C)7590 (D) 3430 (E) 8540 Sol:

× 529 = 8542.63

Approximately 8540, hence answer(E) 3. When 40% of first number is added to the second number the second number becomes 1.2 times of itself. What is the ratio between the first and second numbers? (A) 2:3 (B) 4:3 (C) 1:2 (D) 5:7 (E) None of these Sol: Let the first and second number be x and y

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PARTNERSHIP When two or more persons start a business jointly and share the profit or loss thereof in an agreed proper portion, it is known as partnership business and the persons carrying on such business are called Partners. Generally partners share the profit or loss in the ratio of the capitals invested by them. Partnership may be (1) simple; or (2) compound. Simple Partnership - When the capitals of the partners are invested for the same time, then this type of partnership is called simple partnership. In such a case, the profit or loss is distributed in proportional to the capital invested. Compound Partnership - When the capital, which is equal or unequal, of the partners, is invested for different times, this type of partnership is called compound partnership. In such a caste the profit or loss is distributed in proportional to the products of the capital and the periods of their investment. An important formula for solving the problems of partnership is -

Working Rule -

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1. If the ratio of investment by three persons is a: b: c and ratio of time invested in their capital is x: y: z then the ratio of their profit will be ax: by: cz. 2. If the ratio of investment by three persons is a:b:c and ratio of their profit is p:q:r then, the ratio of time invested in their capital will be

EXAMPLES Example 1: A, B and C enter into partnership. A contributes one-third of the capital while B contributes as much as A and C together contribute. If the profit at the end of the year amounts to Rs. 840 what would each receive? Sol: As A contributes one-third of the capital A’s profit =

= Rs. 280

Now as B contributes as much as A and C So Profit of B = Profit of A + Profit of C = Rs. 280 + Profit of C Profit of B – Profit of C = Rs. 280 And Profit of B + Profit of C = Rs. 840 – Rs. 280 Adding 2 Profit of B = Rs. 840 Profit of B = Rs. 420 Hence Profit of C = 840 – 420 - 280 = Rs. 140 Example 2: A is working and B is a sleeping partner in a business. A puts Rs. 5, 000 and B puts in Rs. 6, 000. A receives 12 ½ % of the profit for Managing the business and the rest is divided in proportion of their capitals. What does each get out of a profit of Rs. 880? Sol: The amount, which A receives for managing = 12 ½% of Rs.880

The amount left = 880 – 110 = Rs. 770 The amount left is to be divided in the ratio = 5,000 : 6,000 = 5: 6

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Total share received by A = 110 + 350 = Rs. 460 Share received by B = Rs. 420 Example 3: A and B enter into a partnership. A contributes Rs. 5000 while B contributes Rs. 40000. After 1 month B withdraws 1/4 part of his contribution and after 3 months from the starting A puts Rs. 2000 more. When B withdraws his money at the same C also joins them with Rs. 7000. If at the end of 1 year there is a profit of Rs. 1218, what will be share of C in the profit? Sol: Since the contributions of three partners are different and their times also differ. Therefore, their contributions should be converted for equal durations. For this, contribution is multiplied by time. Contribution of A = Rs. 5000 for 12 months + Rs. 2000 for 9 months Contribution of A for 1 month = 5000 × 12 + 2000 × 9 = 60000 + 18000 = Rs. 78000 Contribution of B = Rs. 4000 for 1 month + of Rs. 4000 for 11 months Contribution of B for 1 month = 4000 × 1 + 3000 × 11 = 4000 + 33000 = Rs. 37000 Contribution of C = Rs. 7000 for 11 months Contribution of C for 1 month = 7000 × 11 = Rs. 77000 Ratio in their contributions = 78000:37000:77000 = 78:37:77 Sum of their ratios = 78 + 37 + 77 = 192 Share of C in the profit = = Rs. 488.47

Example 4: Alok started a business by investment of Rs.90000 after 3 months Pranav joined him with an investment of Rs. 120000. If they had a profit of Rs. 96000 after 2 years

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then what is the difference in the shares of two? Sol: Alok’s investment for 1 month = 9000 × 24 = 2160000 Pranav’s investment for 1 month = 120000 × 21= 25200 Ratio of their investment = 6:7

Example 5: A, B and C started a business in partnership. A invested Rs. 25 lacks and after 1 year he invested Rs. 10 lacks more. B invested Rs. 35 lacks in the beginning and withdrew Rs. 10 lacks after 2 years. C invested Rs. 30 lacks. What is the ratio of their profit after 3 years? Sol: A’s investment = 25 × 3 + 10 × 2 = Rs. 95 lacks B’s investment = 35 × 2 + 25 × 1 = Rs. 95 lacks C’s investment = 30 × 3 = Rs. 90 lacks Ratio of their investment = 19:19:18 Ratio of their profit = 19:19:18 (because time period is same, i.e., for 3 years)

Example 6: A, B and C investment in a partnership in the ratio of 5:6:8. Ratio of their profit is 5:3:12. Find the ratio of time for their investment.

Example 7: Three people A, B and C invested money in a partnership in the ratio of 4:2:8 ratio of their time of investment is 3:3:2. What is the ratio of their profit?

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Sol: Required ratio = 4 × 3 : 2 × 3 : 8 × 2 = 12 : 6 :16 = 6:3:8

RATIO AND PROPORTION RATIO: Ratio means Comparison. The number of times one quantity contains another quantity of the same kind. Thus the ratio between 5 litres of oil and 15 litres of oil can be possible, but not between 10 apples and 25 kg of rice.

*

The ratio between one quantity to another is measured by a : b or a/b

Ex: 8 : 9 or 5 : 7 etc.

*

The two quantities in the ratio are called its terms. The first is called the antecedent and the second term is called

consequent.

*

The terms of the ratio can be multiplied or divided by the same number.

Types of Ratios:

1. Duplicate ratio: The ratio of the squares of the two numbers. Ex: 9 : 16 is the duplicate ratio of 3 : 4. 2. Triplicate Ratio: The ratio of the cubes of the two numbers. Ex: 27 : 64 is the triplicate ratio of 3 : 4. 3. Sub-duplicate Ratio: The ratio between the square roots of the two numbers. Ex: 4 : 5 is the sub-duplicate ratio of 16 : 25. 4. Sub-triplicate Ratio: The ratio between the cube roots of the two numbers. Ex: 4 : 5 is the sub-triplicate ratio of 64 : 125. 5.Inverse ratio: If the two terms in the ratio interchange their places, then the new ratio is inverse ratio of the first.

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Ex: 9 :5 is the inverse ratio of 5 : 9. 6. Compound ratio: The ratio of the product of the first terms to that of the second terms of two or more ratios. Ex: The compound ratio of PROPORTION: If two ratios are equal, then they make a proportion. Thus Each term of the ratios

is called proportional.

The middle terms 5 and 8 are called means and the end terms 4 and 10 are called extrems.

Product of Means = Product of Extremes Continued Proportion: In the proportion

8, 12, 18 are in the continued

proportion. Fourth proportion: If a : b = c : x, then x is called fourth proportion of a,b and c. There fore fourth proportion of a, b, c = Third proportion: If a : b = b : x, then x is called third proportion of a and b. Therefore third proportion of a, b = Second or mean proportion: If a : x = x : b , then x is called second or mean proportion of a and b. Therefore mean proportion of a and b =

EXAMPLES 1. a : b = 3: 4; b : c = 6 : 7. Find a : b : c. Sol: a b c 3 4 6 7 a : b : c = 3 × 6 : 6 × 4 : 4 × 7 = 9 : 12 : 14 2. A sum of Rs.4960 has been divided among A, B and C in the ratio of 5:4:7. Find the share of B.

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Sol: B's share = = Rs.1240 3. 36% of first number is 28% of the second number. What is the respective ratio of the first number to the second number? Sol: Let the numbers be x and y. 36% of x = 28% of y x:y=7:9 4. Two numbers are in 4:7 ratio. The difference between them is 27. What is the bigger number? Sol: Let the numbers be 4x and 7x. 7x − 4x = 27 3x = 27 x=9 Bigger number is 7x = 7 × 9 = 63 Short cut: The difference of the terms of the ratio = 7 − 4 = 3. But the actual difference between the numbers is 27 3 parts is equal to 27 7 parts (Bigger number) = × 27 = 63 5. The ratio of the ages of a man and his son is 7: 3. The average of their ages is 30 years. What will be the ratio of their ages after 4 years? Sol: Average age = 30 years Total age = 2 × 30 = 60 years. Let their present ages be 7x and 3x years 7x + 3x = 60 x= =6 Their present ages are 7 × 6 and 3 × 6 = 42 and 18. Their ages after 4 years = 42 + 4 and 18 + 4 = 46 and 22 years ratio = 46 : 22 = 23 : 11 6. Two numbers are in the ratio of 3:4. If 4 is subtracted from each, the remainders are in the ratio of 5:7. What are the numbers? Sol: Let the numbers be 3x and 4x. If 4 is subtracted from each, the numbers will be (3x −4) and (4x −4). (3x −4) : (4x −4) = 5: 7 Product of means = Product of extremes (3x –4) 7 = (4x – 4) 5 21x – 28 = 20x – 20 x=8 The numbers are 3 × 8 and 4 × 8 = 24 and 32 7. In a bowl there is 30 litre mixture of milk and water. The ratio of milk and water is 7:3. How much water must be added to it so that the ratio of milk to the water be 3:7? Sol : Milk quantity in the mixture =

×30 = 21 litres Water = 30 - 21 = 9 litres

40

New ratio = 3 : 7 3 parts of milk is 21 litres (There is no difference in the milk quantity of new mixture) Water quantity in the mixture =

× 21 = 49 litres 49 - 9 = 40 litres water is to be added in the new mixture 8. A bag contains of one rupee, 50 paise and 25 paise coins. if these coins are in the ratio of 5 : 6 : 8, and the total amount of coins is Rs. 210, find the number of 50 paise coins in the bag. Sol : Let the number of one rupee, 50 paise, 25 paise coins be 5, 6 and 8 respectively The value of one rupee coins = Rs. 1 × 5 = Rs. 5

The value of fifty paise coins = Rs. 0.50 × 6 = Rs. 3 The value of twenty five paise coins = Rs. 0.25 × 8 = Rs. 2 Total value = 5 + 3 + 2 = Rs. 10 If the total value is Rs. 10 there are 6 coins of fifty paise if the total value is Rs. 210, then the number of 50 coins = × 6 = 126 9. If a sum of Rs.3150 were distributed among Ravi, Vijay and Suresh in the ratio of 12:9:14 respectively, then find the share of Vijay. Ans: Rs.810 Sol: Vijay's Share =

× 3150 = Rs.810

10. The total number of students in a school is 2850. If the number of boys in the school is 1650, then what is the respective ratio of the total number of boys to the total number of girls in the school? Ans: 11:8 Sol: Total number of students = 2850 Number of boys = 1650 Number of girls = 2850-1650 = 1200 Ratio between boys and girls =1650 : 1200 = 11 : 8 11. A sum of money is divided among A, B, C and D in the ratio of 5 : 6 : 12 : 15 respectively. If the share of C is Rs. 4092, then what is the total amount of money? Ans: Rs. 12958 Sol: Let the share of A, B, C and D be Rs. 5x, 6x, 12x and 15x respectively. C's share is Rs.4092 12x = 4092 x= = 341 Total money = 38x = 38 × 341= Rs.12958 12. Asum of Rs. 2820 has been distributed among A, B and C in the ratio respectively. What is the share of B?

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Ans: Rs. 900 Sol: LCM of 3, 4 and 5 is 60 ratio =

: 20 : 15 : 12

B's share = =

× 2820

×2820 = Rs. 900

13. A, B and C divide an amount of Rs. 6300 amongst themselves in the ratio of 7:6:8 respectively. If an amount of Rs.300 is added to each of their shares, what will be the new respective ratio of their shares of amount? Ans: 8 : 7 : 9 Sol: Total shares = 7 + 6 + 8 = 21 21 parts = 6300 each part =

= 300

Their shares are 7 × 300, 6 × 300 and 8 × 300 2100, 1800 and 2400 If 300 is added to each of them then their shares are 2400, 2100 and 2700 Their ratio = 2400 : 2100 : 2700 =8:7:9 14. Find out the two quantities whose difference is 30 and the ratio between them is 5/11. Sol: The difference of quantities, which are in the ratio 5:11, is 6. To make the difference 30, we should Multiply them by 5. Therefore 15. A factory employs skilled workers, unskilled workers and clerks in the ratio 8:5:1 and the wages of a skilled worker, an unskilled worker and a clerk are in the ratio 5:2:3 when 20 unskilled workers are employed the total daily wages fall amount to Rs. 318. Find out the daily wages paid to each category of employees. Sol: Number of skilled worker: unskilled worker: clerks = 8:5:1 and the ratio of their respective Wages = 5:2:3 Hence the amount will be paid in the ratio 8 × 5 : 5 × 2 : 3 × 1 = 40 : 10 : 3 Hence total amount distributed among unskilled workers

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But the number of unskilled workers is 20, so the daily wages of unskilled worker

The wages of a skilled worker, an unskilled worker and a clerk are in the ratio = 5:2:3 Multiplying the ratio by

we get = 7.50 : 3 : 4.50

So, if an unskilled worker gets Rs.3 a day then a skilled worker gets Rs. 7.50 per day a clerks Rs. 4.50 a day 16.Two numbers are in the ratio of 11:13. If 12 be subtracted from each, the remainders are in the ratio of 7:9 Find out the numbers.

Sol: Since the numbers are in the ratio of 11:13. Let the numbers be 11x and 13x. Now if 2 is subtracted from each, the numbers become (11x -12) and (13x-12). As they are in the ratio of 7:9

(11x-12): (13x-12):: 7: 9 (11x – 12) 9 = (13x – 12) 7 99x – 108 = 91x – 84 9x = 24 or x = 3

17. In what ratio the two kinds of tea must be mixed together one at Rs. 48 per kg. and another at Rs. 32 per kg. So that the mixture may cost Rs. 36 per kg. ? Sol: Mohan and Sohan = 5:6 or Sohan and Rakesh = 8:5 or Mohan and Rakesh = =4:3 18. What should be subtracted from 15, 28, 20 and 38 so that the remaining numbers may be proportional?

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19. If Rs. 279 were distributed among Ram, Mohan and Sohan in the ratio of 15:10:6 respectively, then how many rupees did Mohan obtain? Sol: Ratio in which Ram, Mohan and Sohan got = 15 :10 : 6 Sum of ratios = 15 + 10 + 6 = 31 Share of Mohan = Rs. 90 20. A bag contains of one rupee, 50 paise and 25 paise coins. If these coins are in the ratio of 2:3:10, and the total amount of coins is Rs288, find the number of 25 paise coins in the bag. Sol: Ratio of one rupee, 50 paise and 25 paise coins = 2:3:10 Ratio of their values = 8:6:10 = 4:3:5 And sum of the ratios of their values = 4 + 3 + 5 = 12 Value of 25 paise coins

= Rs. 120

No. of 25 paise coins = 120 × 4 = 480

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SIMPLIFICATIONS Directions (Q. 1 - 10): What will come in place of question mark (?) in the following questions? 1. 3463 × 295 − 18611 = ? + 5883 Ans: 997091 Exp: 3463 × 295 − 18611 = ? + 5883 ? = 1021585 − 18611 − 5883 = 997091 2. 2652 ÷ 39 ÷ 17 = ? Ans: 4 Exp:

Ans:

Exp:

4. (4444 ÷ 40) + (645 ÷ 25) + (3991 ÷ 26) = ? Ans: 290.4 Exp: (4444 ÷ 40) + (645 ÷ 25) + (3991 ÷ 26) 111.1 + 25.8 + 153.5 = 290.4 Ans:

382

45

Exp:

Ans: 3721 Exp:

Ans: 64 Exp:

8. (2746 × 49) + 150180 = ? × 118 Ans: 2413 Exp:

Ans: 3 Exp:

10. 362.3 × 61.4 ÷ 60.8 = 36? Ans: 2.6

46

Exp:

Directions (Q. 11 - 20): What approximate value should come in place of question mark (?) in the following questions? Note: You are not expected to calculate the exact value. 11. 39.897% of 4331 + 58.779% of 5003 = ? Ans: 4700 Exp: 39.897% of 4331 + 58.779% of 5003 40% of 4330 + 60% of 5000 1732 + 3000 = 4732 ≈ 4700

Ans: 10 Exp:

13. 43931.03 ÷ 2111.02 × 401.04 = ? Ans: 8300 Exp: 43931.03 ÷ 2111.02 × 401.04 43930 ÷ 2110 × 400 20.8 × 400 = 8320 ≈ 8300 Ans: 180 Exp:

47

Ans: 4890 Exp:

16. 1339.28% of 49.67 = ? Ans: 670 Exp: 1339.28% of 49.67 1340% of 50 = 50% of 1340 = 670 17. 335.01 × 244.99 ÷ 55 = ? Ans: 1490 Exp: 335.01 × 244.99 ÷ 55 335 × 245 ÷ 55 335 × 4.45 ≈ 1490

Ans: 51 Exp:

19. 124.945 + 7.894 × 1.493 = ? Ans: 137 Exp: 124.945 + 7.894 × 1.493 = ? 125 + 8 × 1.5 = 137 20. (352 % of 49.6) − 74.638 = ? Ans: 100 Exp: (352% of 49.6) − 74.638 350% of 50 − 75 = 100 Directions (Q. 21 - 25): What should come in place of question mark (?) in the following questions? 21. (56 × 280) ÷ (14 × 112) = ? Ans: 10 Exp: (56 × 280) ÷ (14 × 112)

48

22. 9207 ÷ 27 × (0.2)2 = ? 1) 136.4 2) 2492 3) 1.364 4) 126.7 5) None of these Ans: 5 Exp: 9207 ÷ 27 × (0.2) 2

Ans: 3 Exp:

Ans: 616 Exp:

Ans:

Exp:

49