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Multiplying 5 times any number


Base Multiplication or Close-together Method


Multiplying Numbers Close to the base 10, 100, 1000 and so on

 

    A)   When both the numbers are below the base

 

To multiply numbers that are close to the bases of powers of 10 i.e. 10, 100, 1000 and so on easily at extremely fast speed, just follow the steps as below:

 

Step-1:  Find deficits of both the numbers from base (10, 100 or 1000).

 

Step-2:  Subtract the difference between one multiplicand with the deviation of the other (from base). This is the first part of the answer. It can be arrived at in any one of the four ways.

i) Cross-subtract deviation on the second row from the first multiplicand.

ii) Cross-subtract deviation on the first row from the second multiplicand.

iii) Subtract the base from the sum of the given numbers.

iv) Subtract the sum of the two deviations from the base.

 

Step-3:  The last part of the answer is the product of the deviations of the numbers from base. It contains the number of digits equal to number of zeroes in the base i.e. for numbers near base 10; 1 place to go, since 10 has 1 zero, for numbers near base 100; 2 places to go since 100 has 2 zeros, accordingly since 1000 has 3 zeros, 3 places to go for numbers near base 1000 and so on. So, carry forward or put extra zero(s) if necessary to place the digits in accurate number.

 

The general form of the multiplication:

 

Let N1 and N2 be two numbers near to a given base in powers of 10, and D1 and D2 are their respective deviations from the base. Then N1 X N2 can be represented as

 

N1        D1

N2        D2

------------------------------------

(N1+D2) OR (N2+D1) / (D1xD2)

 

Suppose you need 8 x 7.

 

Here base is 10. See how far the numbers are below 10, subtract one number's deficiency from the other number, and multiply the deficiencies together.

8 is 2 below 10 and 7 is 3 below 10.

The diagram below shows how you get it.

You subtract crosswise 8-3 or 7-2 to get 5, the first figure of the answer.

And you multiply vertically: 2 x 3 to get 6, the last figure of the answer.

 

So, 7 x 6 =?

 

Here there is a carry: the 1 in the 12 goes over to make 3 into 4.

 

Suppose you want to multiply 96 by 92

 

Both the numbers are near 100, so the base here is 100.

 

96 is 4 below the base and 92 is 8 below.

 

We can cross-subtract either way: 96-8=88 or 92-4=88. This is the first part of the answer and multiplying the "differences" vertically 4x8=32 gives the second part of the answer.                   

So, 88 × 98 =?

 

Here base is 100. 88 is 12 below 100 and 98 is 2 below 100.

 

You can imagine the sum set out like this:

          

86 comes from subtracting crosswise: 88 - 2 = 86 (or 98 - 12 = 86: you can subtract either way, you will always get the same answer).

 

And, multiplying vertically both the differences (12 and 2) from 100 results in 24.

Find 75 × 95.

 

Deviation of 75 from 100 is -25

Deviation of 95 from 100 is -05

 

75           -25       [BASE 100]

 

95           -05

-------------------

(75-05) or (95-25) / (25X5)

--------------------

70 / 125

 

Since the base is 100, we write down 25 and carry 1 over to the left giving us 70 / 125 = (70+1) / 25

 

So, the final answer is 7,125.

 

786 × 998 =?

 

Here base is 1000

 

Complement of 786 is 214.

 

Complement of 998 is 002 (7 from 9 is 2 and 8 from 9 is 1 and 6 from 10 is 4).

 

786         -214     [BASE 1000]

 

998         -002

-------------------

(786-002) or (998-214) / (214X2)

-------------------

784 / 428

 

The answer is 784,428.

 

Find 994 × 988.

 

994         -006     [BASE 1000]

 

988         -012

-------------------

(786-002) or (998-214) / (214X2)

-------------------

982 / 072

 

Answer is 982,072.

Find 750 × 995.

 

750        -250                 [BASE 1000]

 

995        -005

-------------------

(750-005) or (995-250) / (250X005)

-------------------

745 / 1250

 

Since the base is 1000, we write down 250 and carry 1 over to the left giving us 745 / 1250 = (745+1) / 250

 

So, the final answer is 746,250.

 

568 × 998 =?

 

Complement of 568 is 432

Complement of 998 is 002.

 

568        -432     [BASE 1000]

 

998        -002

______________

568 - 2 / 864

 

=> 566 / 864

 

Answer is 566864

 

    B)   When both the numbers are above the base

 

The technique works equally well for numbers above the base. Here we add the differences in step-2.

 

Find 13 × 12.

 

Here, base is 10 and both the numbers are above 10. So instead of subtracting we will add the difference.

 

13        +3        [BASE 10]

 

12        +2

-------------------

(13 + 2) or (12 + 3) / (3 X 2)

-------------------

15 / 6

 

The answer is 156.

18 × 14 =?

 

18        +8        [BASE 10]

14        +4

-------------------

(18 + 4) or (14 + 8) / (8 X 4)

-------------------

22 / 32

 

Since the base is 10, we write down 2 and carry 3 over to the left giving us 22 / 32 = (22+3) / 2

Answer is 252.

 

104 × 102 =?

 

104      +04      [BASE 100]

102      +02

-------------------

(104 + 02) or (102 + 04) / (04 X 02)

-------------------

106 / 08

 

Answer is 10,608.

 

103 x 104 =?

 

The answer is 107 and 12,


107 is just 103 + 4 (or 104 + 3), and 12 is just 3 × 4.

So, 103 x 104 = 10,712

 

Similarly 107 × 106 = 11,342

107 + 6 = 113 and 7 × 6 = 42

 

And, 105 x 111 = 11,655

 

Find 1275 × 1004.

 

1275    +275                [BASE 1000]

1004    +004

-------------------

(1275 + 004) or (1004 + 275) / (275 X 004)

-------------------

1279 / 1100

 

Since the base is 1000, we write down 100 and carry 1 over to the left giving us 1279 / 1100 = (1279+1) / 100

 

So, the answer is 1,280,100.

 

 

    C)   When one number is above and the other is below  the base

 

For multiplying such numbers:

Ø  Subtract (from the number which is above the base) or add (with the number which is below the base) crosswise. Then append as many ‘0’s to the right as there are ‘0’ in base and subtract the product of both the deviation from this result.

 

You can also do this calculation in a very simple way: calculate ‘Vertically and Cross-wise’ as before. Subtract 1 from the first part to get the first part of the answer and find the complement of the product (last part) with base to get the last part of the answer.

 

Example: Find 13 × 7.

 

13        +3        [BASE 10]

7           -3

----------------

10    /    -9

 

One deviation is positive and the other is negative. So the product of deviations becomes negative. So the right hand side of the answer obtained will therefore have to be subtracted.

 

Since the base 10 has one ‘0’ so put one ‘0’ after the first part of the answer 10 to get 100. Now subtract 9 from 100 to get the final answer (100-9=) 91.

 

Suppose you want to multiply 102 by 97

 

102 is 2 more than 100 and 97 is 3 less than 100.

So, subtract (102-3) or add (97+2) crosswise to get 99.

Append 2 zeros to 99 to get 9900.

 

Now, subtract the product of both the differences that is 6 (2x3=6) from 9900. You get 9900-6 = 9,894

 

So, 102 x 97 = 9894.

Another example, 107 x 98 =?

 

107 is 7 more than 100 and 98 is 2 below 100.

So, subtract (107-2) or add (98+7) crosswise to get 105.

Append 2 zeros to 105 to get 10500.

Now, subtract the product of both the differences (7x2=) 14 from 10500.

You get 10500-14 = 10,486

 

Find 108 × 94.

 

108         +08      [BASE 100]

 

  94         -06

----------------

102   /   -48

 

Complement of 48 is 52 and 102 is decreased by 1

(102-1) / Complement of 48 = 10152

Answer is 10,152.

 

So, 998 × 1025 =?

 

  998      -002    [BASE 1000]

 

1025      +025

-------------------

1023 / -050

 

Complement of 50 is 950 and 1023 is decreased by 1

(1023-1) / Complement of 50 = 1022950

 

So, the final answer is 1,022,950.

 

A Lesson in Probability


A Lesson in Probability - Solve Accurately

Probability comes for like 1 mark or 2 … the good is it is easy … the bad is it comes only for a mark or two!

On requests, I have decided to take the plunge and do a piece on probability – all dice, coins and cards come out and play!

 

So,

1. What is probability?

Probability is the chance of the happening or non-happening of event; denoted by ‘P’.


 

2. What are events and sample space?

Sample space is the total number of occurrences that can happen.

Event is the occurrence of ‘something’ which we are concerned with.

For example: Jai and Veeru did that coin toss to gamble away their lives – awesome – but it has got an important lesson of probability too.

 

One coin – what are the possible out comes? – Two, as there can be a Head or a Tail.

Therefore, our sample space (S) = 2 = total number of possible outcomes (either head or tail).

Say, Veeru wanted Heads – how many heads is possible in one coin? – One. Thus 1 is our Event (E)!

 

Get Quantitative Aptitude Shortcuts

3. How to find Probability

Probability is the chance of the occurrence of an event; [P = E/S]

Thus, with E = 1, and S = 2,

Probability of Veeru going to die = E/S = ½ = 0.5!    [But it was different in the film - I know, I know!]

4. Hold on now! If the chances of Veeru’s death is 0.5; what could be the chance of Jai being the one dying?

Again, ½ =0.5!                            [Think the film makers calculated only this probability!]


 

5.  The ‘Non – event’.

Every event has its corresponding ‘non-event’; which can be denoted as E'. If the ‘event’ is happening, then non-event will not happen and vice versa.

If Veeru is going to die (E), then Jai won’t (E'); if Jai’s (E) going to die then Veeru (E') won’t!

Thus, P(E) + P(E') = 1

In words, probability of event and probability of a non-event add up to 1.

Therefore, 1- P(E) = P(E'),
                  1- P(E') = P(E).


 

6. AND ‘n’ OR:

First off – AND is multiplying.
OR is for addition.

If a question is worded like this – ‘if the probability of A hitting the target is 1/3 and B hitting the target is ½, what is the probability of A and Bboth, hitting the target if a shot is taken by both.’

which means, P(A) AND P(B) = P(A and B hitting the target); P(A) x P(B)

P(A) x P(B) = 1/6.

Now, if the question was worded - ‘if the probability of A hitting the target is 1/3 and B hitting the target is ½, what is the probability of A or B hitting the target?’

which means, P(A) OR P(B) = P(A or B hitting the target); P(A) + P(B)

P(A) + P(B) = 5/6.


 

7. Some common sample space(s)!


For Coins


One Coin


Two Coins


Three Coins

Sample Space (S) =

2

2 x 2 = 4

2 x 2 x 2 = 8

and so on…

       

For Dice

One Dice

Two Die

Three Die

(S) =

6

6 x 6 = 36

6 x 6 x 6 = 216
and so on…

       

For Cards

Cards in one suit (Either Spade, Clubs, Hearts or Diamonds)

One Pack of Cards/ Deck = Total number of cards

Face Cards (King, Queen, Jack and Ace) of all the fours suits together

(S)

13

13 x 4 = 52

4 x 4 = 16

       


 

8. Concept of Odds:

Sometimes probability is viewed in terms of ‘odds for’ or ‘odds against’ an event.

Odds in favour of an event = P(E)/P(E')

Odds against an event,
 or,
Odds in favour of the non-event
 = P(E')/P(E)               

 

… fairly simple, right? All you got to do is calculate the P(E) and the P(E'); then use the above formulae, if and only if the word ‘odds’ is in the question! Otherwise we calculate the normal probabilities as asked in the question.

 

A simple trick for writing multiplication table of numbers ending in 9


Base Multiplication or Close-together Method


Multiplying Numbers Close to the base 10, 100, 1000 and so on

 

    A)   When both the numbers are below the base

 

To multiply numbers that are close to the bases of powers of 10 i.e. 10, 100, 1000 and so on easily at extremely fast speed, just follow the steps as below:

 

Step-1:  Find deficits of both the numbers from base (10, 100 or 1000).

 

Step-2:  Subtract the difference between one multiplicand with the deviation of the other (from base). This is the first part of the answer. It can be arrived at in any one of the four ways.

i) Cross-subtract deviation on the second row from the first multiplicand.

ii) Cross-subtract deviation on the first row from the second multiplicand.

iii) Subtract the base from the sum of the given numbers.

iv) Subtract the sum of the two deviations from the base.

 

Step-3:  The last part of the answer is the product of the deviations of the numbers from base. It contains the number of digits equal to number of zeroes in the base i.e. for numbers near base 10; 1 place to go, since 10 has 1 zero, for numbers near base 100; 2 places to go since 100 has 2 zeros, accordingly since 1000 has 3 zeros, 3 places to go for numbers near base 1000 and so on. So, carry forward or put extra zero(s) if necessary to place the digits in accurate number.

 

The general form of the multiplication:

 

Let N1 and N2 be two numbers near to a given base in powers of 10, and D1 and D2 are their respective deviations from the base. Then N1 X N2 can be represented as

 

N1        D1

N2        D2

------------------------------------

(N1+D2) OR (N2+D1) / (D1xD2)

 

Suppose you need 8 x 7.

 

Here base is 10. See how far the numbers are below 10, subtract one number's deficiency from the other number, and multiply the deficiencies together.

 

8 is 2 below 10 and 7 is 3 below 10.

 

The diagram below shows how you get it.

 

 

You subtract crosswise 8-3 or 7-2 to get 5, the first figure of the answer.

And you multiply vertically: 2 x 3 to get 6, the last figure of the answer.

 

 

So, 7 x 6 =?

 

Here there is a carry: the 1 in the 12 goes over to make 3 into 4.

 

Suppose you want to multiply 96 by 92

 

Both the numbers are near 100, so the base here is 100.

 

96 is 4 below the base and 92 is 8 below.

 

We can cross-subtract either way: 96-8=88 or 92-4=88. This is the first part of the answer and multiplying the "differences" vertically 4x8=32 gives the second part of the answer.

 

                        

So, 88 × 98 =?

 

Here base is 100. 88 is 12 below 100 and 98 is 2 below 100.

 

You can imagine the sum set out like this:

          

86 comes from subtracting crosswise: 88 - 2 = 86 (or 98 - 12 = 86: you can subtract either way, you will always get the same answer).

 

And, multiplying vertically both the differences (12 and 2) from 100 results in 24.

Find 75 × 95.

 

Deviation of 75 from 100 is -25

Deviation of 95 from 100 is -05

 

75           -25       [BASE 100]

 

95           -05

-------------------

(75-05) or (95-25) / (25X5)

--------------------

70 / 125

 

Since the base is 100, we write down 25 and carry 1 over to the left giving us 70 / 125 = (70+1) / 25

 

So, the final answer is 7,125.

 

786 × 998 =?

 

Here base is 1000

 

Complement of 786 is 214.

 

Complement of 998 is 002 (7 from 9 is 2 and 8 from 9 is 1 and 6 from 10 is 4).

 

786         -214     [BASE 1000]

 

998         -002

-------------------

(786-002) or (998-214) / (214X2)

-------------------

784 / 428

 

The answer is 784,428.

 

Find 994 × 988.

 

994         -006     [BASE 1000]

 

988         -012

-------------------

(786-002) or (998-214) / (214X2)

-------------------

982 / 072

 

Answer is 982,072.

 

 

Find 750 × 995.

 

750        -250                 [BASE 1000]

 

995        -005

-------------------

(750-005) or (995-250) / (250X005)

-------------------

745 / 1250

 

Since the base is 1000, we write down 250 and carry 1 over to the left giving us 745 / 1250 = (745+1) / 250

 

So, the final answer is 746,250.

 

568 × 998 =?

 

Complement of 568 is 432

Complement of 998 is 002.

 

568        -432     [BASE 1000]

 

998        -002

______________

568 - 2 / 864

 

=> 566 / 864

 

Answer is 566864

 

    B)   When both the numbers are above the base

 

The technique works equally well for numbers above the base. Here we add the differences in step-2.

 

Find 13 × 12.

 

Here, base is 10 and both the numbers are above 10. So instead of subtracting we will add the difference.

 

13        +3        [BASE 10]

 

12        +2

-------------------

(13 + 2) or (12 + 3) / (3 X 2)

-------------------

15 / 6

 

The answer is 156.

18 × 14 =?

 

18        +8        [BASE 10]

14        +4

-------------------

(18 + 4) or (14 + 8) / (8 X 4)

-------------------

22 / 32

 

Since the base is 10, we write down 2 and carry 3 over to the left giving us 22 / 32 = (22+3) / 2

Answer is 252.

 

104 × 102 =?

 

104      +04      [BASE 100]

102      +02

-------------------

(104 + 02) or (102 + 04) / (04 X 02)

-------------------

106 / 08

 

Answer is 10,608.

 

103 x 104 =?

 

The answer is 107 and 12,


107 is just 103 + 4 (or 104 + 3), and 12 is just 3 × 4.

So, 103 x 104 = 10,712

 

Similarly 107 × 106 = 11,342

107 + 6 = 113 and 7 × 6 = 42

 

And, 105 x 111 = 11,655

 

Find 1275 × 1004.

 

1275    +275                [BASE 1000]

1004    +004

-------------------

(1275 + 004) or (1004 + 275) / (275 X 004)

-------------------

1279 / 1100

 

Since the base is 1000, we write down 100 and carry 1 over to the left giving us 1279 / 1100 = (1279+1) / 100

 

So, the answer is 1,280,100.

 

 

    C)   When one number is above and the other is below  the base

 

For multiplying such numbers:

Ø  Subtract (from the number which is above the base) or add (with the number which is below the base) crosswise. Then append as many ‘0’s to the right as there are ‘0’ in base and subtract the product of both the deviation from this result.

 

You can also do this calculation in a very simple way: calculate ‘Vertically and Cross-wise’ as before. Subtract 1 from the first part to get the first part of the answer and find the complement of the product (last part) with base to get the last part of the answer.

 

Example: Find 13 × 7.

 

13        +3        [BASE 10]

7           -3

----------------

10    /    -9

 

One deviation is positive and the other is negative. So the product of deviations becomes negative. So the right hand side of the answer obtained will therefore have to be subtracted.

 

Since the base 10 has one ‘0’ so put one ‘0’ after the first part of the answer 10 to get 100. Now subtract 9 from 100 to get the final answer (100-9=) 91.

 

Suppose you want to multiply 102 by 97

 

102 is 2 more than 100 and 97 is 3 less than 100.

So, subtract (102-3) or add (97+2) crosswise to get 99.

Append 2 zeros to 99 to get 9900.

 

Now, subtract the product of both the differences that is 6 (2x3=6) from 9900. You get 9900-6 = 9,894

 

So, 102 x 97 = 9894.

Another example, 107 x 98 =?

 

107 is 7 more than 100 and 98 is 2 below 100.

So, subtract (107-2) or add (98+7) crosswise to get 105.

Append 2 zeros to 105 to get 10500.

Now, subtract the product of both the differences (7x2=) 14 from 10500.

You get 10500-14 = 10,486

 

Find 108 × 94.

 

108         +08      [BASE 100]

 

  94         -06

----------------

102   /   -48

 

Complement of 48 is 52 and 102 is decreased by 1

(102-1) / Complement of 48 = 10152

Answer is 10,152.

 

So, 998 × 1025 =?

 

  998      -002    [BASE 1000]

 

1025      +025

-------------------

1023 / -050

 

Complement of 50 is 950 and 1023 is decreased by 1

(1023-1) / Complement of 50 = 1022950

 

So, the final answer is 1,022,950.

Percentages


Percentages - Aptitude Test Tricks & Shortcuts & Formulas

Basic Definition:
Percent implies “for every hundred” and the sign % is read as percentage and x % is read as x per cent. In other words, a fraction with denominator 100 is called a per cent. For example, 20 % means 20/100 (i.e. 20 parts from 100). This can also be written as 0.2.

Basic Formula:
In order to calculate p % of q, use the formula:
(p/100) x q = (pxq)/100

2. Percentage – Ratio Equivalence:
formulas-and-tricks-11
formulas-and-tricks-21
N is Numerator
D is the Denominator

13. Product Stability Ratio:
A × B = P
If A is increased by a certain percentage, then B is required to be decreased by a certain percentage to keep the product P stable.

Expressing the percentage figures in ratios:

formulas-and-tricks-3

Calculation of Percentage

If we have to find y% of x, then
 

Calculation of Percentage

1. To express x% as a fraction : 

We know

x% = x/100

Thus 10% = 10/100 (means 10 parts out of 100 parts)

1/10 (means 1 part out of 10 parts)

 

2. To express x/y as a percentage :

We know that x/y  = (x/y× 100 )

Thus 1/4 = ( 1/4 ×100 )% = 25%

and 0.8 = ( 8/10 ×100 )% = 80%

 

3. To increase a number by a given percentage(x%): 

Multiply the number by the following factor

 increase a number by a given percentage

4. To decrease a number by a given percentage(x%): 

Multiply the number by the following factor

To decrease a number by a given percentage

5. To find the % increase of a number:

To find the % increase of a number

6. To find the % decrease of a number:

To find the % decrease of a number:

 

Some Observation

#1 

If 20% candidate failed in an exam then observations are

  • 80% represent passed in exam
  • 100% represent total appeared in exam
  • (80%-20%) = 60% represent difference between passed and failed candidate in exam

percentage

#2

 If a number is increased by 25% then observations are 

  • 100% represent the old number
  • 125% represent the new number.

 

percentage concept

 

#3

 Remember that Base in the given sentence (Question) is always 100%

Eg. Income of Ram is increased by 20% 

In this sentence 

100% - represent the income of Ram

20% - represent increment

120% - represent new income of Ram.

Remember it :

1 = 100%

1/2 = 50%

1/3 = 33 1/3%

1/4 = 25%

1/5 = 20%

1/6 = 162/3%

1/7 = 142/7%

1/8 = 121/2%

1/9 = 111/3%

1/10 = 10%

1/11 = 91/11%

1/10 = 81/3%

1/13% = 79/13%

percentage problems

25% = 1/4

6.25% = 1/16

125% = 5/4

150% = 3/2

200% = 2

350% = 7/2

 

#4

If of A is equal to y% of B then -

If of A is equal to y% of B then -

Example: - If 10% of A is equal to 12% of B, then 15% of A is equal to what percent of B?

percentages

#5

If A is more than B,

percentage tricks

Example: - If income of Ravi is 20% more than that of Ram, then the income of Ram is how much percent less than that of Ravi?

percentage tricks

#6

If the passing marks in an examination is P%. If a candidate scores S marks and fails by F marks then–

percentage tricks problems

Example: - Pankaj Sharma has to score 40% marks to get through. If he gets 40 marks and fails by 40, then find the total marks set for the examination?

percentage tricks problems examples

#7

If a candidate scores marks and fails by a marks while an another candidate scores y% marks and gets b marks more than minimum passing marks, then –

percentage tricks problems examples



Example: - Raj scores 30% and fails by 60 marks, while Rohan who scores 55% marks, gets 40 marks more than the minimum required marks to pass the examination. Find the maximum marks for the examination?

percentage problems examples


 

#8

If due to decrement in the price of an item, a person can buy Kg more in y rupees, then actual price of that item -

percentage problems examples

Example: - Ram can buy 5 Kg more sugar in rupees 100 as the price of sugar has decreased by 10%. Find the actual price of the sugar?

percentage problems examples


 

#9

If in an election, a candidate got of total votes cast and still lose by y votes, the total number of votes cast –

percentage problems examples

Example: - In an election contested by two candidates, one candidate got 40% of total votes and still lost by 500 votes, find the total number of votes casted?

percentage problems examples

#10

If the population of a town is P and it increases or decreases at the rate of R% per annum then –
I. Population after ‘n’ years :
percentage population

II. Population ‘n’ years ago :
percentage population
Example: - The population of a town is 352800. If it increases at the rate of 5% per annum, then what will be its population after 2 years and 2 years ago?

percentages


 

#11

If the value of a number is first increased by and again decreased by the net effect is always decreased by x2/100%
Example: -The salary of a worker is first increased by 5% and then it is decreased by 5%. What is the change in his salary?

percentages

 

Examples

#1 

Q. If the difference between 62% of a number and 3/5th of that number is 36. what is the number ?

Sol:

Let the number be x.

Then x × 62% - x × 3/5 = 36

x ×62% -x V 60% = 36 (60% = 3/5)

x ×2% = 36

x ×2/100 =36

x = 36 ×100/2 = 1800

#2

Q. 40% of Ram's income Rs. 1200 Then Find 

  1. 75% of Ram's income ?
  2. 1/4 part of Ram's income ?
  3. 1/3 part of Ram's income ?

Sol :

(1)

40% = 1200 Rs.

75% = 1200/40 ×75 = 2250 Rs.

Trick : 1200 / 40  × 75 = Rs. 2250/-

 

(2)

40% of income = Rs. 1200

Then 1/4 part (i.e. 25% ) of Ram's

income = 1200/40 ×25 

= Rs. 750/- Ans

 

(3)

40% of Ram's income

= Rs. 1200

i.e. 2/5 part of Ram's income

= Rs. 1200

Then total income of Ram 

= Rs. 1200 ×5/2

1/3 part of Ram's income

= Rs. 1200  × 5/2  × 1/3 

= Rs. 1000 Ans.

Trick :

1200/2/5  × 1/3

1200/2 × 5/3 = 100

Volume Surface Area and Perimeter Shortcuts


Volume Surface Area and Perimeter Shortcuts and Formulas

This Module describes various standard geometrical structures that exist and methods/formulas or calculating its characteristic values that can be used for various applications/calculations. The structures include circle, triangle, cylinder etc., that exists within or as a shape, in all the practical entities we see in our daily life. So, calculating its area, Volume and perimeter gives the mathematical insight about the structure.

Prerequisites (Related Formulas) Area Calculations

Area is a quantity that expresses the extent of a two – dimensional surface or shape in the plane. Area can be understood as the amount of material with a given thickness that would be necessary to fashion a model of the shape, or the amount of paint necessary to cover the surface with a single coat.
1 square kilometer = 1,000,000
square meters, 1 square meter = 10,000 square centimetres = 1,000,000 square millimetres
1 square centimetre = 100 square millimeters
1 square yard = 9 square feet
1 square mile = 3,097,600 square yards = 27,878,400 square feet

Volume is the quantity of three – dimensional space enclosed by some closed boundary, for example, the space that a substance (solid, liquid, gas, or plasma) or shape occupies or contains. Volume is often quantified numerically using the SI derived unit , the cubic metre.

1 litre = (10 cm)3 = 1000 cubic centimeters = 0.001 cubic metres,
1 cubic metre = 1000 liters.
Small amounts of liquid are often measured in millilitres,
Where 1millilitre = 0.001 litres = 1 cubic centimetre

 

Surface Area Calculations

Surface area is the measure of how much exposed area a solid object has, expressed in square units. Mathematical description of the surface area is considerably more involved than the definition of arc length of a curve.

a. Surface Area of a Cube = 6a(a is the length of the side of each edge of the cube).

In words, the surface area of a cube is the area of the six squares that cover it. The area of one of them is a*a, or a2 . Since these are all the same, you can multiply one of them by six, so the surface area of a cube is 6 times one of the sides squared.

b. Surface Area of a Rectangular Prism = 2ab + 2bc + 2ac (a, b, and c are the lengths of the 3 sides).

In words, the surface area of a rectangular prism is the area of the six rectangles that cover it. But we don't have to figure out all six because we know that the top and bottom are the same, the front and back are the same, and the left and right sides are the same.
The area of the top and bottom (side lengths a and c) = a*c. Since there are two of them, you get 2ac. The front and back have side lengths of b and c. The area of one of them is b*c, and there are two of them, so the surface area of those two is 2bc. The left and right side have side lengths of a and b, so the surface area of one of them is a*b. Again, there are two of them, so their combined surface area is 2ab.

c. Surface Area of Any Prism (b is the shape of the ends)

Surface Area = Lateral area + Area of two ends
(Lateral area) = (perimeter of shape b) * L
Surface Area = (perimeter of shape b) * L+ 2*(Area of shape b)

d. Surface Area of a Sphere = 4 pi r2(r is radius of circle)

e. Surface Area of a Cylinder = 2 pi r2+ 2 pi r h (h is the height of the cylinder, r is the radius of the top)

Surface Area = Areas of top and bottom +Area of the side
Surface Area = 2(Area of top) + (perimeter of top)* height
Surface Area = 2(pir2) + (2pir)* h

In words, the easiest way is to think of a can. The surface area is the areas of all the parts needed to cover the can. That's the top, the bottom, and the paper label that wraps around the middle. You can find the area of the top (or the bottom). That's the formula for area of a circle (pir2).

Since there is both a top and a bottom, that gets multiplied by two. The side is like the label of the can. If you peel it off and lay it flat it will be a rectangle. The area of a rectangle is the product of the two sides. One side is the height of the can, the other side is the perimeter of the circle, since the label wraps once around the can.
So the area of the rectangle is (2pir)* h. Add those two parts together and you have the formula for the surface area of a cylinder

Perimeter Calculations

A perimeter is a path that surrounds an area. The word comes from the Greek peri (around) and meter (measure). The term may be used either for the path or its length - it can be thought of as the length of the outline of a shape.

circle = pi d (where d is the diameter). The perimeter of a circle is more commonly known as the circumference.

Properties of various Geometrical structures

Triangle

In an equilateral triangle all sides have the same length. An equilateral triangle is also a regular polygon with all angles measuring 60°.
In an isosceles triangle , two sides are equal in length. An isosceles triangle also has two angles of the same measure; namely, the angles opposite to the two sides of the same length; this fact is the content of the Isosceles triangle theorem.

In a scalene triangle , all sides are unequal. The three angles are also all different in measure. Some (but not all) scalene triangles are also right triangles.

A right triangle (or right - angled triangle , formerly called a rectangled triangle) has one of its interior angles measuring 90° (a right angle). The side opposite to the right angle is the hypotenuse ; it is the longest side of the right triangle. The other two sides are called the legs of the triangle.

Right triangles obey the Pythagorean theorem : the sum of the squares of the lengths of the two legs is equal to the square of the length of the hypotenuse: a2 +b2=c2, where a and b are the lengths of the legs and c is the length of the hypotenuse.

Triangles that do not have an angle that measures 90° are called oblique triangles.

A triangle that has all interior angles measuring less than 90° is an acute triangle or acute - angled triangle.

A triangle that has one angle that measures more than 90° is an obtuse triangle or obtuse - angled triangle.

A "triangle" with an interior angle of 180° (and collinear vertices) is degenerate.

Quadrilateral

In Euclidean plane geometry, a quadrilateral is a polygon with four sides (or 'edges') and four vertices or corners . Sometimes, the term quadrangle is used, by analogy with triangle, and sometimes tetragon for consistency with pentagon (5- sided), hexagon (6- sided) and so on. The Word quadrilateral is made of the words quad (meaning "four") and lateral (meaning "of sides").

The diagonals of parallelogram bisect each other
Each diagonal of a parallelogram divides it into two triangles of the same area
The diagonals of rectangle are equal and bisect each other
The diagonals of square are equal and bisect each other at right angles
The diagonals of rhombus are unequal and bisect each other at right angles

Some Important Metrics
1. 10,000 sq meters = 1 hectare
2. 100 hectares = 1 sq kilo meter
3. 1000 millimeters = 1 meter
4. 100 centimeters = 1 meter
5. 1000 metres = 1 kilometer
6. 1000 kilograms = 1 mega gram or 1 tonne
7. 3.6 kilometers per hour = 1 meter per second
8. 3600 kilometers per hour = 1 kilometer per second

 

 

Profit and loss shortcuts tricks


PROFIT AND LOSS SHORTCUTS TRICKS

  • Cost Price (CP)
  • Selling price (SP)
  • Profit (P),  Loss(L), Profit percent, Loss percent, Gain , Gain percent
  • Variable cost, Fixed cost,
  • Mark Up, Marked price,
  • Discounts, Successive discounts,
  • Break even point,
  • Faulty balances,  faulty measure , Chain rules … etc

Now we see one by one definitions, its related formulas with example off profit and loss sums.

Cost Price ( CP ) : The price at which a person buys a product is the cost price of the product for that person. The cost price can be also defined as the amount paid or expended in either purchasing or producing an article.

Selling price (SP) : The Price at which a person sells a product is the sales price of the product for that person. It can be also defined as the amount got when an product is sold is called as the selling price (SP).

Profit or gain : When a person able to sell a article at a price higher then its cost price, then we say that he got profit or gain.

i.e The selling price is more than the cost price then called profit or gain. (CP < SP )

Loss : When a person able to sell a article at a price lower then its cost price, then we say that he incurred loss.

i.e The selling price is lower than the cost price then called loss. (CP > SP

Formulas on Profit and Loss :

  • Profit (Gain ) =  SP – CP  ( i.e SP = Profit + CP )
  • Loss =  CP – SP ( i.e CP = Loss + SP )
  • Profit Percentage = [ Profit x  100] / CP
  • Profit = [ profit% x CP ] / 100
  • Loss Percentage =  [ Loss x  100] / CP
  • Loss = [ Loss % x CP ] / 100
  • Sell Price = CP + Gain or Profit
     

Profit and loss formulas for quantitative aptitude | profit and loss shortcut tricks for bank exams, ssc cgl | profit and loss problems with solutions for all types of competitive exams

  • Sell Price = CP – Loss

Profit and loss formulas for quantitative aptitude | profit and loss shortcut tricks for bank exams, ssc cgl | profit and loss problems with solutions for all types of competitive exams

 

. One can generate a profit only if Selling Price> Cost Price

2. One generates a loss when Selling Price < Cost Price.

3. Profit = Selling Price – Cost Price
%profit = {(Selling Price – Cost Price)/Cost Price} x 100

4. Loss = Cost Price – Selling Price
%Loss = {(Cost Price – Selling Price)/Cost Price} x 100

5. Sale price :- If there is a profit of P %,
Cost Price = C
Then SP = {(100+P)/100}xC

6. If there is a loss of L %,
Cost Price = C
Then
SP = {(100-L)/100}xC

7. Cost price :-
If there is a profit of P %,
Cost Price = C
Sale price= SP
Then C = {100/(100+p)} x SP
If there is a loss of L %,
Then
C = {100/(100-L)}xSP

8. A dishonest dealer claims to sell his goods at cost price ,but he uses a weight of lesser weight .Find his gain%.

Profit-and-Loss-Formulas-and-Tricks-2

9. A shopkeeper sells an item at a profit of x % and uses a weight which is y % less .find his total profit

Profit-and-Loss-Formulas-and-Tricks-3

10. When dealer sells goods at loss on cost price but uses less weight .

11. A dishonest dealer sells goods at x % loss on cost price but uses a gm instead of b gm . his profit or loss percent :-

Profit-and-Loss-Formulas-and-Tricks-5

Note :- profit or loss will be decided according to sign .if +ive it is profit ,if –ve it is loss .

12.  If the price of an item increases by r%  , then the reduction in consumption so that  expenditure remains the same  ,is

Profit-and-Loss-Formulas-and-Tricks-6

13. If the price of a commodity decreases by r% then increase in consumption , so as not to decrease expenditure on this item is

Profit-and-Loss-Formulas-and-Tricks-8

14. A reduction of x% in price enables a person to buy y kg more for Rs. A. Then the

Profit-and-Loss-Formulas-and-Tricks-7

15. When there are two successive profits of x% and y% then the net percentage profit =[x+y+xy/100]

When there is a profit of x% and loss of y% then net percentage profit or loss = [x – y – xy/100]

Note: If the final sign in the above expression is positive then there is net profit but if it is negative then there is net loss.

16. A sells goods to B at a profit of x% andB sells it to C at a profit of y%. If C pays RsP for it,then the cost price for A is

Note:- for loss replace plus sign with minus .

17. When each of the two things is sold at the same price,and a profit of p% is made on the first and a loss of L% is made on the second,then the percentage gain or loss is .

18. If profit percentage and loss percentage are equal, put P=L

=>   %loss = p2 /100

Dishonest Dealer: A dishonest dealer is one who claims to sell his product in welfare of customer but either he alters weight or he marks up price too high and then gives discount to attract customers.
Let us assume that the shopkeeper makes a gain of G% in selling his product.
Therefore,
(100+G)/(100+x) = True weight/ False weight
Here,
G = Overall gain percentage
X = %loss or % gain
Also, another formula that you can remember, in case the shopkeeper sells at the cost price, is:
Gain Percentage

profit-and-loss-Discounts-and-Marked-Price-2
Let us check this formula by taking an example.
Let us say that I went to a shop to by 1kg almonds. The shopkeeper told me that he is selling the almonds at the cost price but he actually used a faulty weight and gave me only 800gm. Let the cost which the shopkeeper paid for 1 kg was Rs 1000 and as he told me that he is selling at the cost price so I will pay him Rs 1000 for 1kg. But what I am getting instead 1kg, only 800gm. Now try to understand the logic. The shopkeeper sold me 800 gm almonds for Rs 1000 for which he actually paid Rs 800 (cost price of 800gm). So he gained 1000 – 800 = Rs 200 and his percentage profit = (200/800)x100
which is same as given above in the formula. Here 200 is the error and 800 = 1000 – 200 = True weight – Error.

So how to proceed with the problems when the shopkeeper is selling the articles at a price different from that of cost price while using faulty weight? In this case the shopkeeper will make two profits. First profit is because of the wrong weight and the second is because of the actual difference between the cost price and the selling price. The final profit percentage is arrived by using the result {P + Q + (PQ/100)} where P and Q are the two profits.
Let us understand it by a example.

Example 1: Let a dishonest shopkeeper sells sugar at Rs 18/kg which he has bought at Rs 15/kg and he is giving 800gm instead of 1000gm. Find his actual profit percentage.
Solution: Here the cost price of the sugar = Rs 15/kg
The selling price = Rs 18/kg.
The profit made by the shopkeeper is of Rs 3 and the profit percentage = 3/15 x 100 = 20%
This will be his total profit if he has actually sold 1 kg sugar. But that is not the case here as he is using the false weight.
Now the profit due to wrong weight =
profit-and-loss-Discounts-and-Marked-Price-2
= (200/800)x100 = 25%
The overall profit percentage = {P + Q + (PQ/100)} = [20+25+{(20 x 25)/100}] = 50%
You can also attempt this problem by another method.
The cost price of 1kg sugar is Rs 15. As the shopkeeper is giving only 800gm so the cost price of the 800 gm sugar is Rs 12. Here we have calculated the cost price of 800gm sugar because the shopkeeper is actually selling only 800gm. He is selling 800 gm sugar for Rs 18 for which he had paid Rs 12. So he gained Rs 6 and the profit percentage = 6/12 x 100 = 50%
The answer in this case is same as calculated above but the first method is more easy than the second one. Just find the individual profits and put the values in {P + Q + (PQ/100)}
The above formula is also valid if the shopkeeper is making losses due to some reason. In that case you will put the negative value for the loss. Let us take an example.

Example 2: A shopkeeper claims that he is selling sugar at Rs 23/kg which cost him Rs 25/kg but he is giving 800gm instead of 1000gm. What is his percentage profit or loss?

Solution: Here the cost price of the sugar is Rs 25/kg and the selling price is Rs 23/kg. So the loss is Rs 2/kg and the loss percentage = 2/25 x 100 = 8%
The profit due to wrong weight = 200/800 x 100 = 25%
Hence the overall profit and loss is given by  {P + Q + (PQ/100)}
But as he is making loss of 8% in the first case so we put -8 in the above expression. If the final value is positive then he is making profit otherwise loss. So the net profit and loss =
{25 – 8 + {25 x (-8)}/100} = 25 – 8 – 2 = 15%
As the final value is positive so he is making a profit of 15%.

Example-3: Lalit marks up his goods by 40% and gives a discount of 10%. Apart from this, he uses a faulty balance also, which reads 1000 gm for 800 gm.  What is his net profit percentage?
Solution:
Let us assume his CP/1000 gm = Rs 100
So, his SP/kg (800 gm) = Rs 126
So, his CP/800 gm = Rs 80
So, profit = Rs 46
So, profit percentage = 46/80 x 100 = 57.5%

Shortcuts for Faulty Weights Questions:

Case-1: When dealer reduces weight in terms of percentage and earns profit 

Example: A shopkeeper sells an item at a profit of 20 % and uses a weight which is 20% less. Find his total profit.

Solution:

profit-and-loss-Discounts-and-Marked-Price-4

Case-2: When dealer sells goods at loss on cost price but uses less weight.

Note :- profit or loss will be decided according to sign . If +ive it is profit ,if –ve it is loss.
Example: A dishonest dealer sells goods at 10% loss on cost price but uses 20%  less weight. Calculate profit or loss percent.

Solution:
Apply formula:

{(20-10)/(100-20)} x 100 = 25/2%
Here sign is positive so there is a profit of 12.5%

Case-3: When dealer sells product at loss but alters weight :-
profit-and-loss-Discounts-and-Marked-Price-7

Note :- Profit or loss will be decided according to sign .if +ive it is profit ,if –ve it is loss .

Example: A dishonest dealer sells products at 10% loss on cost price but uses 2 gm instead of 4 gm . what is his profit or loss percent?

Solution:

Apply formula :

profit-and-loss-Discounts-and-Marked-Price-7

[100-10] 4/2-100=80%

Case-4: When dealer sells product at profit but alters weight

Profit% or loss% =  [100+gain%][1000/altered weight ] – 100

Note :- Profit or loss will be decided according to sign. If +ive it is profit ,if –ve it is loss .

Example: A shopkeeper uses 940 gm in place of one kg. He sells it at 4% profit. What will be the overall profit or loss?

Solution:

profit-and-loss-Discounts-and-Marked-Price-8

Profit and Loss: Dishonest Dealers and Faulty Weights Exercise:

Question 1:
A dishonest dealer professes to sell his good at cost price but uses a false weight and thus gains 20%. For a kilogram he uses a weight of
(a) 700g
(b) 750g
(c) 800g
(d) None of these

Answer and Explanation

Question 2:
A grocery dealer cheats to the ex¬tent of 10% while buying as well as selling by using false weight. What is his increase in the profit % ?
(a) 20%
(b) 21%
(c) 22%
(d) None of these

Answer and Explanation

Question 3:
A shopkeeper wants to make some profit by selling rice. He contemplates about various methods. Which of the following would maximise his profit ?
I. Sell rice at 15% profit.
II. Use 850 g of weight instead of 1 kg.
III. Mix 15% impurities in rice and selling rice at cost price.
IV. Increase the price by 7.5% and reduce weights by 7.5%.
(a) I or III
(b) II
(c) II, III and IV
(d) Profits are same

Answer and Explanation

Question 4:
A dealer buys apples at Rs.50, Rs. 40 and Rs. 30 per kilogram. He mixes them in the ratio 2 :4 : 9 by weight, and sells at a profit of 30%. At what approximate price per kilogram does he sell the dry fruit ?
(a) Rs 54
(b) Rs 77
(c) Rs 67
(d) Rs 46

Answer and Explanation

Question 5:
A dishonest shopkeeper, using a faulty balance makes a profit of 5% while buying as well as while selling his goods. His ac¬tual gain percent in the whole process amounts to
(a) 11
(b) 10
(c) 10.25
(d) 10.5

 

 

Progressions Formulas


Progressions Formulas

Arithmetic Progressions 
Geometric Progressions 
Progressions Important Questions 

Sequences, following specific patterns are called progressions. Arithmetic progression (A.P). In this Chapter, besides discussing more about A.P.; arithmetic mean, geometric mean, relationship between A.M. and G.M., special series in forms of sum to n terms of consecutive natural numbers, sum to n terms of squares of natural numbers and sum to n terms of cubes of natural numbers will also be studied.

Sequences

Let the number of person’s ancestors for the first, second, third, ..., tenth generations are 2, 4, 8, 16, 32, ..., 1024. These numbers form what we call a sequence. Consider the successive quotients that we obtain in the division of 10 by 3 at different steps of division. In this process we get 3,3.3,3.33,3.333, ... and so on. These quotients also form a sequence. The various numbers occurring in a sequence are called its term . We denote the terms of a sequence by a1,a2,a3,.....,ana1,a2,a3,.....,an etc, the subscripts denote the position of the term. The nthnth term is the number at the position of the sequence and is denoted by anan The nthnth term is also called the general term of the sequence.

A sequence is called infinite, if it is not a finite sequence. For example, the sequence of successive quotients mentioned above is an infinite sequence, infinite in the sense that it never ends .Often, it is possible to express the rule, which yields the various terms of a sequencein terms of algebraic formula. Consider for instance, the sequence of even natural number 2, 4, 6,...
Here, 
a1=2=2×1,a2=4=2×2,a3=6=2×3,........a23=46=2×23,a24=48=2×24a1=2=2×1,a2=4=2×2,a3=6=2×3,........a23=46=2×23,a24=48=2×24 , and so on. In some cases, an arrangement of numbers such as 1, 1, 2, 3, 5, 8,.. has no visible pattern, but the sequence is generated by the recurrence relation given by.
a1=a2=1,a3=a1+a2,an=an+2+an1,n>2a1=a2=1,a3=a1+a2,an=an+2+an−1,n>2
Above sequence is called Fibonacci sequence

Series

a1,a2,a3,.....,ana1,a2,a3,.....,an be a given sequence. Then, the expression a1+a2+a3+.....+ana1+a2+a3+.....+an s called the series associated with the given sequence. The series is finite or infinite according as the given sequence is finite or infinite. Series are often represented in compact form called sigma notation as means indicating the summation involved .Thus , the series a1+a2+a3+.....+ana1+a2+a3+.....+an is abbreviated . as nk1∑k−1n

Note: When the series is used, it refers to the indicated sum not to the sum itself. For example, 1 + 3 + 5 + 7 is a finite series with four terms. When we use the phrase “ sum of a series,” we will mean the number that results from adding the terms, the
sum of the series is 16.

Arithmetic Progression (A.P)

a1,a2,a3,.....,ana1,a2,a3,.....,an called arithmetic sequence or arithmetic progression if an+1=an+d,nNan+1=an+d,nN , where a1a1 is called the first term and the constant term d is called the common difference of the A.

Let us consider an A.P. (in its standard form) with first term a and common difference d , i.e., a , a + d, a+ 2d...,
Then the 
nthnth term (general term) of the A.P. is an=a+(n1)dan=a+(n−1)d

We can verify the following simple properties of an AP:
(i) If a constant is added to each term of an A.P., the resulting sequence is also an A.P.

(ii) If a constant is subtracted from each term of an A.P., the resulting sequence is also an A.P.

(iii) If each term of an A.P. is multiplied by a constant, then the resulting sequence is also an A.P.

(iv) If each term of an A.P. is divided by a non-zero constant then the resulting sequence is also an A.P.
Here, we shall use the following notations for an arithmetic progression :
a = the first term , l = the last term , d = common difference, n = the some of the series s _{n} = the sum to n terms of AP
Let a, a + d, a + 2d, ..., a + (n –1)dbe an A.P.
Then , l = a + (n– 1)d and 
sn=n2[2a+(n1)d]sn=n2[2a+(n−1)d]

Arithmetic mean Given two numbers a and b. We can insert a number A between them so that a, A, bis an A.P. Such a number A is called the arithmetic mean (A.M.) of the numbers a and b. Note that, in this case, we have.

A – a = b – A , i.e A=a+b2A=a+b2 We may also interpret the A.M. between two numbers a and b as their average generally, given any two numbers a and b, we can insert as many numbers as we like between them such that the resulting sequence is an A.P. Let A1,A2,A3,......AnA1,A2,A3,......An b is an A.P
here, b is the 
(n+2)th(n+2)th term , i.e, b = a+ [(n + 2) – 1] d = a+ (n + 1)d

This gives d=ban+1d=b−an+1
Thus, n numbers between a and b are as follows:
A1=a+d=a+ban+1,A2=a+2d=a+2(ba)n+1,A3=a+3d=a+3(ba)n+1,......,An=a+nd=a+n(ba)n+1A1=a+d=a+b−an+1,A2=a+2d=a+2(b−a)n+1,A3=a+3d=a+3(b−a)n+1,......,An=a+nd=a+n(b−a)n+1

 

Geometric Progression (G . P.)

a1,a2,a3,.....,ana1,a2,a3,.....,an is called geometric progression , non zero and ak+1ak=r,k1ak+1ak=r,k≥1
By letting a1=aa1=a we obtain a geometric progression, a , ar , ar2,ar3ar2,ar3 ........where a is called the first term and r is called the common ratio of the G.P.

As in case of arithmetic progression, the problem of finding the nth term or sum of n terms of a geometric progression containing a large number of terms would be difficult without the use of the formula which we shall develop in the next Section. We shall use the following notations with these formula: a = the first term, r = the common ratio, l = the last term
n = the numbers of terms, 
snsn = the sum of last term

General term of a G .P. Let us consider a G.P. with first non-zero term ‘ a ’ and common ratio ‘ r.Write a few terms of it. The second term is obtained by multiplying a by r thus a2a2 = ar similarly, third term is obtained by multiplying a2a2 by r thus a3=a2ra3=a2r , and so on

Thus, a , G.P. can be written as a,ar,ar2,ar3,....an1a,ar,ar2,ar3,....an−1 , according as G.P is finite or infinite , respectively . Thus the series a+ar+,ar2+ar3+....arn1a+ar+,ar2+ar3+....arn−1 , are called finite or infinite geometric series , respectively.
Sum to n terms of a G .P. Let the first term of a G.P. be a and the common ratio be r .
Let us denote by S the sum to first n terms of G.P.

Then
sn=a+ar+,ar2+ar3+....arn1sn=a+ar+,ar2+ar3+....arn−1 , .............(1)
Case 1: If r = 1, we have S n = a + a + a + ... + a (n terms) = na

case 2: If r ≠ 1, multiplying by r, we have
rsn=a+ar+,ar2+ar3+....arnrsn=a+ar+,ar2+ar3+....arn ,..............(2)
Subtracting (1) and (2) , we get (1 - r) s_{n}=a-ar^{n}=a(1-r^{n}
This gives , 
sn=a(1rn)1rsn=a(1−rn)1−r

 

Geometric Mean (G .M)

The geometric mean of two positive numbers a and b is the number ab . Therefore, the geometric mean of 2 and 8 is 4. We observe that the three numbers 2,4,8 are consecutive terms of a G.P. This leads to a generalisation of the concept of geometric means of two numbers. Given any two positive numbers a and b , we can insert as many numbers as we like between them to make the resulting sequence in a G.P.

Let G1,G2,G3,.....GnG1,G2,G3,.....Gn be n numbers between positive numbers a and b such that aG1,G2,G3,.....GnbaG1,G2,G3,.....Gnb is a G.P . Thus b being (n+2)th(n+2)th term , we have
b=arn+1,orr=(ba)2n+1b=arn+1,orr=(ba)2n+1
G1=ar=a(ba)1n+1G1=ar=a(ba)1n+1 , G21=ar2=a(ba)2n+1G21=ar2=a(ba)2n+1 , G3=ar3=a(ba)3n+1G3=ar3=a(ba)3n+1 , Gn=arn=a(ba)nn+1Gn=arn=a(ba)nn+1

Relationship Between A.M. and G.M.: Let A and G be A.M. and G.M. of two given positive real numbers a
and b , respectively Thus we have A-G= 
a+b2ab−−√a+b2−ab
Thus we have
AG=a+b2ab−−√=a+b2ab2=(ab)220A−G=a+b2−ab=a+b−2ab2=(a−b)22≥0
We obtain the relationship AGA≥G

 

Sum to n Terms of Special Series

We shall now find the sum of first n terms of some special series, namely ;

1. 1 + 2 + 3 + ............+n (sum of first n natural numbers)
2. 
12+22+32+........+n212+22+32+........+n2 (sum of squares of the first n natural numbers)
3. 
13+23+33+........+n313+23+33+........+n3 (sum of cubes of the first n natural numbers).

Let us take them one by one:
(a). 
sn=1+2+3+......n,thensn=n(n+1)2sn=1+2+3+......n,thensn=n(n+1)2
(b). sn=12+22+32+........+n2sn=12+22+32+........+n2

Point to be Remember
1. By a sequence , we mean an arrangement of number in definite order according to some rule. Also, we define a sequence as a function whose domain is the set of natural numbers or some subsets of the type {1, 2, 3, .... k}. A sequence containing a finite number of terms is called a finite sequence. A sequence is called infinite if it is not a finite sequence.

2. a1,a2,a3,......a1,a2,a3,...... be the sequence, then the sum expressed as a1+a2+a3+......a1+a2+a3+...... is called series. A series is called finite series if it has got finite number of terms.

3. n arithmetic progression (A.P.) is a sequence in which terms increase or decrease regularly by the same constant. This constant is called common difference of the A.P. Usually, we denote the first term of A.P. by a , the common difference by d and the last term by l . The general term or the nthnth term of the A.P. is given by an=a+(n1dan=a+(n−1d
The sum Sn of the first n terms of an A.P. is given by
sn=n2[2a+(n1)d]=n2(a+l)sn=n2[2a+(n−1)d]=n2(a+l)

4. The arithmetic mean A of any two numbers a and b is given by a+b2a+b2 i.e the sequence a, A ,b is in A.P

5. A sequence is said to be a geometric progression or G. P., if the ratio of any term to its preceding term is same throughout. This constant factor is called the common ratio. Usually, we denote the first term of a G.P. by a and its common ratio by r. The general or the nthnth term of G.P. is given by an=arn1an=arn−1 .The sum SnSn of the first n terms of G.P. is given by sn=a(rn1)r1ifr0sn=a(rn−1)r−1ifr≠0

6. The geometric mean (G.M.) of any two positive numbers a and bis given by ab−−√ab i.e the sequence a, G, b is GP .

Series Some Important Power
1. 1+2+3+......+n=n=n(n+1)21+2+3+......+n=∑n=n(n+1)2
2. 12+22+32+......+n2=n2=n(n+1)(2n+1)612+22+32+......+n2=∑n2=n(n+1)(2n+1)6
3. 13+23+33+......+n3=n3=n2(n+1)2413+23+33+......+n3=∑n3=n2(n+1)24
4. 1+1+1+1+....+1nterms=1=n1+1+1+1+....+1nterms=∑1=n

 

 

Quick Divisibility tricks


Quick Divisibility tricks

Posted by Pooja Gupta on Friday, April 28, 2017

Divisibility by 2: If last digit of the number is even i.e. 0,2,4,6,8
Ex : 876999578 is divisible by 2.


Divisibility by 3: If the sum of digits is divisible by 3.
Ex: 327 is divisible by 3, since sum of its digits = (3+2+7) = 12 , which is divisible by 3.


Divisibility by 4: If the last two digits of the number is divisible by 4
Ex: 2648 is divisible by 4, since the number formed by the last two digits is 48 which is divisible by 4.


Divisibility by 5: If the last digit of the number ends with 0 or 5.
Ex: 20870 ends in a 0, so it is divisible by 5.


Divisibility by 6: If the number is divisible by both 2 & 3.
Ex: 558 is divisible by 6, because it is divisible by 2(number is even) as well as 3 (5+5+8=18,which is divisible by 3).


Divisibility by 7: Multiply the last digit by 5 and add the product to the remaining truncated number. Continue doing these steps until you reach a 2 digit number.
If the result is divisible by 7, we say that the original dividend is divisible by 7.
Example: 33803--> 3380+(3*5)=3380+15=3395 -->339+(5*5)=339+25 = 364 --> 36+(4*5)=36+20=56 (since this number is divisible by 7, you can say 3185 is also divisible by 7)
* There are many more approaches to check the divisibility by 7.


Divisibility by 8: If the last three digits of the number are divisible by 8.
Ex: 3652736 is divisible by 8 because last three digits (736) is divisible by 8.
Note: Rule of divisibility by 2 & 4 on the last three digit number will not be applicable here. You have to check the divisibility manually.
Ex: 516 is divisible by 2 & 4 but not by 8.


Divisibility by 9: If the sum of the digits is divisible by 9.
Ex: 672381 is divisible by 9, since sum of digits = (6+7+2+3+8+1) = 27 is divisible by 9.


Divisibility by 10: If the digit at units place is 0 it is divisible by 10.
Ex: 697420, 243540 is divisible by 10.


Divisibility by 11: If the difference of 'sum of its digits at odd places' and 'sum of its digits at even places' is either 0 or a number divisible by 11.
Ex: 4832718 is divisible by 11, since:
(Sum of digits at odd places) and (sum of digits at even places)
= (8+7+3+4)-(1+2+8) = 11


Divisibility by 12: A number is divisible by 12 if it is divisible by both 4 and 3.
Ex: 34632
(i) The number formed by last two digits is 32, which is divisible by 4
(ii) Sum of digits = (3+4+6+2) = 18, which is divisible by 3.


Divisibility by 13: Multiply the last digit by 4 and add the product to the remaining truncated number. Continue doing these steps until you reach a 2 digit number.
If the result is divisible by 13, we say that the original dividend is divisible by 13.
Example: 3185--> 318+(5*4)=318+20=338 -->33+(8*4)=33+32 = 65 (since this number is divisible by 13, you can say 3185 is also divisible by 13)


Divisibility by 14: If a number is divisible by both 2 & 7.

Divisibility by 15: If a number is divisible by both 3 & 5.

TIP: If a number is divisible by two different prime numbers, then it is divisible by the products of those two numbers.
Ex: 30 is divisible by both 3 and 5, it is also divisible by 15.

 

Races and Games


Races and Games : Explanations

 

 

IMPORTANT TERMS

(i) Race: A competition in running, riding, swimming, cycling etc.

(ii) Race Course: The plane on which competitions of races are contested.

(iii) Starting Point: The point from which the race starts .

(iv) Winning Point (Post) : The last point at which race is completed.

(v) Dead Heat: A contest of race in which all the competitors set the same time and no one is winner.

(vi) A Game of Hundred: A game in which the players agree that whoever first score hundred points is the winner.

(vii) while A scores 100 points, B scores only 75 points, then we say that 'In a game of 100 A can give B 25 points'.

(viii) In a race of 100 metres A beats B by 20 metres, means while A runs 100 metres B runs 80 metres.

Question 1.

A is times quicker than B. If A gives B a Start of 75 meters, Find the length of racecourse so that both of them reach the winning post at the same time.

(a) 100 m

(b) 150 m

(c) 200 m
(d) 180 m


 

Question 2.

In a game of Snooker, A can Give B 16 Points in 80 and A can Give C 15 point in 90. How many points can C give B in a game of 50? 

(a) 10 Points

(b) 12 Points

(c) 5 Points

(d) 2 Points

 

Question 3.

P can run 100m in 20 second and Q in 25 seconds. P beats Q by
(a) 10 m
(b) 20 m
(c) 25 m
(d) 12 m


 

Question 4.

In a 2 Km race, P can give Q 200 m and R 560 m. In the same race, Q can give R
(a) 400 m
(b) 300 m
(c) 350 m
(d) 500 M


 

Question 5.

A runs 1.5 Times as fast as B can. If A gives B a start of 50 m, how far must the winning post be in order that A and B reach at the same time
(a) 150 m
(b) 120 m
(c) 125 m
(d) 180 m 


 

Question 6.

P and Q run a Kilometers and P win by 1 minutes. P and R run a kilometre and P wins by 375 m. Q and R run a kilometre and Q wins By 30 Seconds. Find the time taken by each runner to run a kilometre. (a) 150 sec, 210 sec, 240sec
(b) 120 sec, 160 sec, 200sec
(c) 100 sec, 160 sec, 90sec
(d) 140 sec, 200 sec, 230sec 


 

Question 7.

A can give B 25 point, A can give C 40 points and B can give C 20 Points. How many points make the game?
(a) 120
(b) 100
(c) 150
(d) 80


 

Question 8.

At a game of billiard, A can give B 12 points in a game of 40 and A can give C 10 Points in game of 50. How many points can C and B in a game of 80?
(a) 10
(b) 20
(c) 12
(d) 18


 

Question 9.

A can run 1 km in 5 minutes 48 seconds and B in 6 minutes. How many metres start can A give B in 1 km race so that the race may end in a dead heat?

 

Question 10.

A and B run a km race A wins by 1 minute. A and C run a km race and A wins by 375 metres. B and C run a km race and B win By 30 seconds.Find the time that C takes to run a km race.

 

Question 11.

In a race of 800 m, A can beat B by 74 m and in a race of 600 m, B beat C by 50 m. By how many metres will A beat C in a race of 500 m?

 

Question 12.

In a game of billiards, A can give B 10 points in 60 and he can give C 15 in 60.How many points can B give C in a game of 90?
 

Question 13.

A can gives B 40 points, A can give C 64 points, and B can give C 30 points.How many points to make the game?
 

Question 14.

Two men A and B walk around a circle 1500 metres in circumference. A walks at the rate of 140 metres, and B at the rate 80 metres per minute .if they start from the same point, and walk in the same direction, when will they be together again?
 

Question 15.

Two men, A and B run a 10 km race on a course 400 m round.If their rates are 5: 4, how often does the winner pass the other?
 

Question 16.

Three girls Seema, Kiran and Nitu walk around a circle 3 km in circumference at the rates of 200, 150 and 125 metres per minute respectively. if they all start together and walk in the same direction, when will they first be together again?
 

Question 17.

A and B start from the same point and travel in the same direction around a circular track 3 km in circumference, If their speed is 200 and 150 metres a minute, when will they first together again at the starting point?
(a) 40 min
(b) 55 min
(c) 1 hour
(d) 45 min


 

Question 18.

Two men, A and B run a 500 m race, A having 140 m start their speeds are 3 :4. Then, A wins By:
(a) 10 m
(b) 20 m
(c) 40 m
(d) 60 m


 

Question 19.

in a race of 200 metres, A can give a start of 10 metres to A C give a start of 20 metres to B. The start that C can give to A, in the same race is
(a) 30m
(b) 29 m
(c) 27 m
(d) 25 m


 

Question 20.

In a race of 300 metres A rates B by 15 metres or 5 seconds, Hoe much time A take to complete the race?
(a) 105 sec.
(b) 100 sec.
(c) 95 sec.
(d) 90 sec.

 

Solutions:

Ans 1: 200 m

 

Ans 2:  2 Points

 

Ans 3: 20 m

 

Ans 4: 400 m

When Q run (2000 – 200)m, R runs(2000 – 560)m. 

 

Ans 5: 150 m

 

Ans 6: 150 sec, 210 sec, 240sec 

 

Ans 7: 100

Let the game be of x points. 

When A scores x points, B scores X – 25 points, and C scores X – 40 Ponts

 

Ans 8:  10

When A scores 40 points, B scores (40 – 12) points.
When A scores 50 points, C scores (50 – 10) points.

 

Ans: 9

A can give B (6 minutes - 5 minutes 48 seconds) or 12 seconds start. Now we must find the distance B can run in 12 seconds.
The distance run by B in 6 minutes = 1000 m

The distance run by B in 12 seconds = 1000 x 12 / 6 x 60 = 100/3 = 33 1/3 m
A can give B 33 1/3 metres' Start.

Ans: 10

A beats B by 1 minute (60 seconds) and B beats C by 30 Seconds
A beats C by ( 60 + 30) or 90 seconds
But A beats C by 375 m

But A runs C by 375 m.
C runs 1000 m in 90/375 x 1000 = 240 seconds = 4 minutes.

Ans: 11

While B runs 726 m, A runs 800 m. While C runs 550 m, B runs 600 m.
. . While B runs 726m, C runs 550/600 x 726 = 665.5 m.
In a race of 800 m, A gives ( 800- 665.5) i.e. 134.5 m start to C, so in a race of 500 m A
gives 134.5/800x 500 m start to C i.e. 84.06 m start to C.

Ans: 12

A scores 60 while B scores 50, and C scores 45.
. . B scores 90 while C scores 45/50 x 90 = 81
Hence, in a game of 90, B can give C ( 90-81) or 9 points .Let x points make the game, so that, according to the question,
if A has x points, then B has ( x-40) points.
B has x points, then C has ( x- 30) Points.
C has (x -64) points , then A has x points
. . By chain rule
x x ( x-64) = ( x- 40) ( x- 30)x
=> x2 - 64x = x2 - 70x + 1200
=> x = 2000
Hence, 200 points make the game

 

Ans: 13

Let x points make the game, so that , according to the question,
if A has x points, then B has ( x-40) points.
B has x points, then C has ( x- 30) Points.
C has (x -64) points , then A has x points
. . By chain rule
x x ( x-64) = ( x- 40) ( x- 30)x
=> x2 - 64x = x2 - 70x + 1200
=> x = 2000
Hence, 200 points make the game

Ans: 14

A and B will be together again for the first time when A has gained one complete round on B.
Now A gains ( 140- 80) or 60 metres on B in 1 minute
. . A gains 1500 metres in 1500/60 or 25 minutes.

Ans 15

Since the rates are 5 :4 , a runs 5 rounds while B runs 4 rounds
. . A passes B each time that a has runs 5 rounds i.6. 5 x 2/6 or 3 kms
. . ( 400m = 2/5 km)
Now 2 kms is contained 5 times in 10 kms
. . A passes B 5 times.

 

Ans 16 

Seema is quickest girl one complete round on Neetu, the slowest girl, in 3000/200-125 or 40 minutes.
Seema gains one complete round on Kiran, the next slowest girl in 3000/200-150 or 40 minutes.
Thus, Seema and Nitu are together after every 40 minutes, and Seema and Kiran are together after 60 minutes.
'Hence, Seema, Kiran and Nitu will be together in the time which is LCM of 40 and 60. The LCM of 40 and 60 is 120 minutes or 2 hours.

Ans 17. 1 hour


L.C.m of 15 min and 20 min = 60 min = 1 hr.
Hence, they will meet first together at the starting point after 1 hour.

 

Ans 18: 20 m

Here, greatest have to cover a distance of ( 500-140) = 360 m to reach the winning point while B 500 m Now, A covers 3m while B covers 4 m
. . A covers 360 m while B covers 4/3 x 360 = 480 m
Hence, B will be ( 500- 480) = 20 m behind the winning post.
. . A wins by 20 min.

Ans 19: 29 m

B covers 200 m while A 190 m
B covers 180 m while A 190/200x180 = 171 m
Hence, C covers 200 m while A = 171 m
So, C can give to A ( 200 - 171) = 29 m Start.

Ans 20: 95 sec.

From the question it is clear that speed of B = 15/5 = 3 m/s
the time taken by B to cover 300 m = 300 / 3 = 100 sec.
Hence, Time taken by A to complete the race = 100-5 = 95 sec

 

Ratio and Proportion


Ratio and Proportion

Concept of ratio.

'Ratio is the relation between two or more same kind of quantities”. More clearly ratio is the magnitude comparison of quantities in same nature.

For example; if the salaries of Ajay and Bhavan are Rs.5000 and Rs. 6000 respectively. Then we can say that their respective salaries are in the ratio of 5:6. Here 5 and 6 are most relatively simplified representation of the original values 5000 and 6000, or 5 and 6 are the relatively prime numbers (doesn't have any common factors other than 1) which are the factors of the given values. Means, we can't simplify these values further as integers.

In a ratio expression, the order of values is very important. In the above example the required ratio is 5:6 instead 6:5 is wrong.

The ratio of any two same kinds quantities x and y can be expressed as either x/y or x : y. Here x is called 'antecedent' and y is called 'consequent'.

In a reverse approach, if it is given that the ratio of the salaries of Ajay and Bhavan is 5:6, then it doesn't mean that the salaries of Ajay and Bhavan are 5 rupee and 6 rupee respectively. Instead their respective salaries are a certain multiple of 5 and 6. So, from the given data, we can express the salaries of Ajay and Bhavan are in the following way;

Salary of Ajay = 5k

Salary of Bhavan = 6k, where k is a positive real number.

And 'k' is called the 'Multiplicative Constant'.

Calculation as per ratio expression:

For finding the original quantities form a given ratio expression required at least one constant value related to the given data, which may be in any of the following manner;

  • Sum of their individual salaries is Rs.11,000.
  • Bhavan's salary is Rs.1,000 more than that of Ajay.

How to utilize these data for finding the individual salaries as per the pre mentioned example?

As per given ratio expression, Ajay's and Bhavan's salaries are in the ratio of 5:6.

Let Ajay's salary = 5k
And Bhavan's salary = 6k

From data (i);

Sum of salaries = 5k + 6k = 11k
11k = 11,000
K = 1,000
From the assumption,
Ajay's salary = 5k
= 5 * 1000
= Rs.5,000.
Bhavan's salary = 6k
= 6*1000
= Rs.6000.
From data (ii);
Bhavan's salary - Ajay's salary = 6k - 5k
= k =1,000
Therefore, Ajay's salary = 5k
= 5 *1000
= Rs.5,000.
Bhavan's salary = 6k
= 6*1000
= Rs.6000.

Example: (practical understanding of Ratio)

In Cahndu's birthday celebration, he cut a beautiful and delicious round chocolate cake and distributed it to his family members, consisting his father, mother and sister. The ratio of the quantity of cake received by Chandu, father, mother and sister is 3:1:2:4.

Means; if he divided the cake into ten equal pieces, Chandu got 3 out of the ten equal pieces.

His father got 1 out of the ten equal pieces.

His mother got 2 out of the ten equal pieces, and his sister got 3 out of the ten equal pieces.

We can express the same concept in another way.

Sum of the ratio values = 3 + 1 + 2 + 4 = 10

Consider the given full cake into ten equal pieces.

Then the share for Chandu = 3/10 of the cake.

Share for Chandu 's father = 1/10 of the cake.

Share for Chandu 's father = 2/10 of the cake.

Share for Chandu 's sister = 4/10 of the cake.

Example:

A total of 120 toffees are distributed among three friends A, B and C in a respective ratio of 3:4:5. Find the share of each member.

Solution:

Sum of ratio values = 3 + 4 + 5 = 12

Total 120 is divided in to 12 equal parts. Each part consists 10 toffees.

Share for A = 3/12 of 120 = 30 toffees.

Share for B = 4/12 of 120 = 40 toffees.

Share for C = 5/12 of 120 = 50 toffees.

 

Ratio:

- The ratio of two quantities a and b of same units is the fraction x/y, where b ≠ 0

- The fraction x/y can be represented as x:y

Different types of ratios are:

1) Duplicate ratio: It is the ratio of squares of two numbers.

Duplicate ratio of the fraction

x

is given as:

x

=

x2

or x : y = x2 : y2

y

y

y2


2) Sub-duplicate ratio: It is the ratio between square roots of two numbers.

Duplicate ratio of the fraction

x

is given as:

x

=

x

or x : y = x : y

y

y

y


3) Triplicate ratio: It is the ratio of cubes of two numbers.

Triplicate ratio of the fraction

x

is given as

x

=

x3

y

y

y3


4) Sub- Triplicate ratio: It is the ratio between cube roots of two numbers

Sub-Triplicate ratio of the fraction

x

is given as

x

=

x(1/3)

y

y

y(1/3)


5) Compound ratio: It is the ratio of product of first terms in every ratio to that of product of second term in every ratio.

For example:
Compound ratio of (a : x), (b : y), (c : z) is (abc : xyz)

6) Inverse ratio: The ratio formed by interchanging their old places in the ratio to new

The inverse ratio of 5 : 8 is 8 : 5.

Proportion:

1) Proportion is the equality of two ratios.

When (a : b = x : y) is represented as (a : b :: x : y), then a, b, x, y are said to be in proportion.

In 
(a : b :: x : y), a and y are called as extremes and b and x are called as mean terms.

Product of means = Product of extremes

2) Mean proportion: Mean proportion between x and y is given as xy

3) Third proportion: If p : q = q : s, then s is called as third proportional to p and q.

4) Fourth proportion: If u : v = x : y, then y is the fourth proportional of u, v and x.

Quick Tips and Tricks

1) Comparison of ratios:

If (x : y) > ( a : b) →

x

>

a

y

b


2) Proportion


4) Variation:
- If a = kb for some constant k, then we can say that a is directly proportional to b.
- If ba =k for some constant k, then we can say that a is inversely proportional to b.

5) If ratio between first and second quantity 
m : n = a : x, second and third quantity n : p = b : y, fourth and fifth quantity p : q = c : z, then m : n : p : q can be easily solved by using the trick shown below:


m : n : p : q = abc : xbc: xyc : xyz

6) If a number a is divided in the ratio x : y,

1) First part:

ax

(x + y)

 

2) Second part:

ay

(x + y)



Question Variety

Type 1: Proportion


Examples:

Q 1. If a : b = 2 : 3 and b : c = 5 : 7, then find a : b : c.

a. 12 : 15 : 9
b. 10 : 15 : 21
c. 14 : 12 : 21
d. 2 : 15 : 7

View solution
Q 2. If A : B : C = 3 : 4 : 7, then what is the ratio of (A / B) : (B / C) : (C / A)?

a. 63 : 48 : 196
b. 66 : 49 : 190
c. 56 : 40 : 186
d. 46 : 38 : 160

View solution

Type 2: Division and Distribution of objects into ratios


In this type of numericals, given amount or quantity is divided among two or more persons and amount or quantity retained by each person is calculated.

Examples:

Q 3. If Suresh distributes his pens in the ratio of 1/2 : 1/4 : 1/5 : 1/7 between his four friends A, B, C and D, then find the total number of pens Suresh should have?

a. 153
b. 150
c. 100
d. 125

View solution
Q 4. If Rs 1050 is divided into three parts, proportional to (1 / 3) : (3 / 4) : ( 4 / 6), then what is the first part?

a. 500
b. 300
c. 200
d. 100

View solution

Type 3: Mixture of different contents


Examples:

Q 5. In a mixture of 13 litres, the ratio of milk and water is 3 : 2. If 3 liters of this mixture is replaced by 3 liters of milk, then what will be the ratio of milk and water in the newly formed mixture?

a. 10 : 3
b. 8 : 5
c. 9 : 4
d. 1 : 1

View solution
Q 6. A mixture contains alcohol and water in the ratio of 7 : 5. If 8 liters of water is added to the mixture, then the ratio becomes 7 : 9. Find the quantity of alcohol in the given mixture?

a. 15 liters
b. 14 liters
c. 19 liters
d. 21 liters

View solution

Type 4: Income/ Expenditure and Salary


These questions are related to annual savings or expenditure. We are asked to find the salaries in ratio form or vice-versa.

Examples:

Q 7. The annual income of Puja, Hema and Jaya taken together is Rs. 46,000. Puja spends 70 % of income, Hema spends 80 % of her income and Jaya spends 92 % of her income. If their annual savings are 15 : 11 : 10, find the annual saving of Puja?

a. 10, 000 /-
b. 12, 000 /-
c. 17, 500 / -
d. 25, 000 /-

View solution
Q 8. A man, his wife and daughter worked in a graden. The man worked for 3 days, his wife for 2 days and daughter for 4 days. The ratio of daily wages for man to women is 5 : 4 and the ratio for man to daughter is 5 : 3. If their total earnings is mounted to Rs. 105, then find the daily wage of the daughter.

a. Rs. 15
b. Rs. 12
c. Rs. 10
d. Rs. 9

View solution
Q 9. Amit, Raju and Ram agree to pay their total electricity bill in the proportion 3 : 4 : 5. Amit pays first day's bill of Rs. 50, Raju pays second day's bill of Rs. 55 and Ram pays third day's bill of Rs. 75. How much amount should Amit pay to settle the accounts?

a. Rs. 15.25
b. Rs. 17
c. Rs. 12
d. Rs. 5

View solution
Q 10. Salaries of Ram and Sham are in the ratio of 4 : 5. If the salary of each is increased by Rs. 5000, then the new ratio becomes 50 : 60. What is Sham's present salary?

a. Rs. 20,000
b. Rs. 25,000
c. Rs. 30,000
d. Rs. 35,000

View solution

Type 5: Coins and Values


These are ratio based numericals, in which a bag containing different number of coins like 10 p, 50, 25 p are given along with the total amount and we are asked to find each type of coin in the bag or the value of coins.

Examples:

Q 11. A bag contains equal number of 25 paise, 50 paise and one rupee coins respectively. If the total value is Rs 105, how many types of each type are present?

a. 75 coins
b. 60 coins
c. 30 coins
d. 25 coins

View solution
Q 12. A purse contains 342 coins consisting of one rupees, 50 paise and 25 paise coins. If their values are in the ratio of 11 : 9 : 5 then find the number of 50 paise coins?

a. 180
b. 150
c. 162
d. 99

Relations and Functions Solutions and Shortcuts 1


Relations and Functions Solutions and Shortcuts

Relations and Functions Examples -
Relations and Functions Questions - 
Relations and Functions Video Lecture - 

Consider the two sets P = {a,b,c} and Q = {Ali, Bhanu, Binoy, Chandra, Divya}.