How to find the number of factors of a Number

The method to find the Numbers of factors of a number can be illustrated with an example:

Q 1. Find the number of different divisors of the number 1150?

Solution:
1150 can be written as a multiplication of prime numbers as

1150 = 2 x 5 x 5 x 23
or
1150 =   2 × 5² x 23
we have to add 1 to the power of each of prime factors and then multiply them to get the required answer.
Power of 2 is 1. Add 1 to 1 to get (1+1) =2
Power of 5 is 2. Add 1 to 2 to get (2+1) = 3
Power of 23 is 1. Add 1 to 1 to get (1+1)  = 2

Now, the number of different divisors of 1150 are  2 x 3 x 2=12

Q2. Find the number of divisors of the number 144500 excluding itself and unity.

Solution:
144500 = 2²×5³×17²
Now,
add 1 to the power of 2 to get (2+1)  = 3
add 1 to the power of 3 to get (3+1)  = 4
add 1 to the power of 2 to get (2+1)  = 3

Therefore number of divisors of 144500 are (3)(4)(3) = 36
And excluding unity and itself the number of divisors comes out to be 36-2 = 34

Solved problems on Number System- Part2

1. what could be the maximum value of q in the following equation ?
            5 p 9 + 3 r 7+ 2 q 8 = 1114
(a) 81                                  (b) 5
(c) 9                                    (d) 18

Answer:  (c) 9     
Solution: first add the digits at the unit's place - 9+7+8 = 24. Carry the 2.
Adding digits at the ten's place plus carry 2 from the addition of units place -
we get
         p + r+ q+ 2 
 it is evident from the above equation that the sum should be 11. Now for getting the maximum value of q, r and p should be zero.
        => q+2 = 11
       => q =9

2. Find the value of 1+ 2 + 3 + ........+ 105
(a) 5565                              (b) 2782
(c) 4268                              (d) 3254
 
   Answer: (a) 5565    
  Solution: Applying the formula for arithmetic progression, we get
     
    1+ 2 + 3 + ........+ 105 = (105)(1+105)/2=5565

3. Find the value of 1+3+5......+ 49
(a) 298                                (b) 625
 (c) 525                               (d) 400

Answer: (b) 625
Solution:
It form an AP with 1 as the first term and 49 as the last and 2 as the common difference.
Therefore, 1+3+5... ...+ 49= (25)(1+49)/2=(25)(25) = 625

4. Find the value of 2+4+6+8+.......+100
(a) 1125                             (b) 625
(d) 2190                             (d) 2550

Answer: (d) 2550
Solution: It is an Arithmetic progression with first AP as 2 and the last AP =100
Applying the formula for the sum of the AP to get
2+4+6+8+.......+100= (50)(2+100)/2 = (50)(102)/2 = 25(102) = 2550


5. Find the sum of all 2 digit number divisible by 3
(a) 835                                (b) 1665
(c) 1533                              (d) 1683

Answer:
2 digits numbers that are divisible by 3 are 3,6,9,12,...99 or 3x1,3x2,3x3,....3x33.
Therefore there are 33 such numbers beginning with 3 till 99.
Sum of 2 digit numbers divisible by 3 = 3+6+9+...99 = (33)(3+99)/2 = 33 x 102/2 = 33x51 = 1683

6. How many number between 11 and 90 are divisible by 7 ?
(a)11                                   (b) 55
(c) 27                                  (d) 39

Answer:(a)11      
Solution:
After 11, the first number divisible by 7 is 14. And the last number divisible by 7 before the number 90 is 84.
Therefore the numbers between 7 and 90 that are divisible by 7 are
          14,21,28,,,,84 or 7x2,7x3,....7x12
Therefore there are 12-2 +1=11 such numbers.


7. Find the sum of all odd number up to 100
(a) 1175                               (b) 1250
(c) 2500                               (d) 2650

Answer: (c) 2500    
Solution: Odd number between 1 and 100 are 50.
First term = 1 and last term =99
therefore, sum = (50)(1+99)/2 = (50)(100)/2 = 50 x 50 = 2500

8. Find the sum of all even number up to 100
(a) 1365                               (b) 2550
(c) 1790                               (d) 4820

Answer: (b) 2550 
Solution: Even numbers between 1 and 100 are 50.
First term = 2 and last term =100
therefore, sum = (50)(2+100)/2 = (50)(102)/2 = 50 x 51 = 2550

9. Find the value 12+ 22+32+....... 102
(a) 570                                  (b) 450
(c) 132                                  (d) 582

Answer: (a) 570  
Solution
This is an AP with the first term as 12 and the last term as 102 and common difference as 10.
Number of terms is         102 = 12 + (n-1)10
                                  => 90 = 10n - 10
                                  =>100 = 10n
                                  => n = 10   
Therefore the sum of 12+ 22+32+....... 102 = (10)(12+102)/2 = (10)(114)/2=5x114=570

10. How many terms are there in 2,4,8,16 ........1024?
(a) 8                                      (b) 10
(c) 12                                    (d) 6

Answer: (b) 10
Solution:
This is a GP with 2 as the common ratio.
therefore, 2,4,8,16 ........102 is the same as 2^1, 2^2, 2^3, 2^4,.....2^10
1024 = 2^10
Therefore there are 10 terms.

11. How many number up to three digits are divisible by 17 ?
(a) 58                                   (b) 47
(c) 52                                   (d) 31

Answer:(a) 58         
Solution:

The first number divisible by 17 is 17 itself.
To get the last number before 999 that is divisible by 17 we divide 999 by 17.
we get ,
                   999/17 = 58x17 + 13
Subtract 13 from 999 to get  986 which is the last three  number divisible by 17
Therefore there are 58 numbers that are divisible by 17.


13. Find the no of divisors of 8064
(a) 44                                  (b) 48
(c) 84                                  (d)74

Answer: (b) 48
Solution:
8064 = 2^7 x 3^2 x 7^1
The number of divisors of 8064 = (7+1)(2+1)(1+1) = (8)(3)(2)=48

14. Find the sum of the divisors of a number 8064
(a) 26520                         (b) 58140
(c) 62195                          (d) 29870

Answer: 26520
Solution:
8064 = 2^7 x 3^2 x 7^1
Sum of divisors = (2^0 +2^1 +2^2 +2^3 +2^4 +2^5 +2^6 +2^7 )(3^0 + 3^1+3^2)(7^0+7^1)
                          = (1+2+4+8+16+32+64+128)(1+3+9)(1+7)
                          = (26520)

15. The sum of the digits of a two digit number is 8 .If the digits are reversed the number is decreased by 54 find the number .
(a) 97                               (b) 71
(c) 33                               (d) 49

Answer: (b) 71
Solution: Let the number be 10x +y
then given that x+y =8  ----------------------i
Again if number is reveresed then it is decreases by 54.
=>  (10x+y ) - (10y +x ) = 54
=> 10x + y - 10y -x =54
=> 9x - 9y = 54
=>  x-y = 6 ----------------------------- ii
Adding i and ii , we have
           2x = 14
=> x= 7
And from  i the value of y is 8-7  =1
Therefore the number is 71.

16. The ratio of the sum and the difference of two number are 7: 1 find the ratio of those two number
(a) 4:3                              (b) 2:3
(c) 3:2                              (d) 3:4

Answer:(a) 4:3             
Solution:
option (a)  4:3 
 Sum is 4+3 = 7
Difference is 4-3 =1
And ratio of sum and difference is 7:1.

17. The ratio between a two digit number and the sum of the digits of that number is 4:1 .If the digit in the  units place is three more than digit in the ten place. what is the number ?
(a) 44                               (b) 36
(c) 24                               (d) 18

Answer: (b) 36
Solution:
Let the number be 10x+y with x at the tens place and y at the ones place.
Then according to the question
(10x+y) /(x+y) = 4/1
=> (10x+y) = 4(x+y)
=> 10x + y = 4x +4y
=>6x=3y
=> 2x=y -----------------------------(i)
again given that
y= x+3 --------------------- (ii)
Solving (i) and (ii)
2x = x+3
=> x=3
and y=2x = 2.3 =6
Therefore the number is 36

Shortcut: Lets check the options
Given that the digit in the units place is 3 more than the digit at the tens place. Only option (b) fits the bill and is the answer.

18. A number consists of two digit the sum of digits is 9. If 27 is subtracted from the number ,its digits are interchanged. find the number .
(a) 81                               (b) 63
(c) 51                               (d) 91

Answer:
Solution:
We will not solve it by forming equations and then solving them.
We will go through the options to zero in on the right answer.
According to questions the sum of the digits is 9. Only option (a)and option (b) has the sum of digits as 9.
Now , if we subtract 27 from 81 we get 54. The digits are not interchanged.
That leaves us the second option. From 63 if we subtract 27 we get 36. Digits are interchanged. Hence option (b) is the correct answer.

19. If three numbers are added in pairs the sum equal 10, 19 and 21 find the number
(a) 2,10,21                       (b) 3,9,27
(c) 6,8,3                           (d) 6,4,15

Answer:(d) 6,4,15
Solution:
Try each option. only option (d) satisfies the condition.

20 Find the greatest number of 3 digits which is exactly divisible by 35 ?
(a) 980                             (b) 220
(c) 880                             (d) 1015

Answer: (a) 980        
Solution;
Starting by taking the greatest three digit number 999.
Divide this number by 35. we have,
999= 35x28 + 19
Now subtract 19 from the 999 to get 980 as the answer.

21. What least number must be added to 49123 to get a number exactly divisible by 263 ?
(a)10                                 (b)58
(c)9                                   (d)18

Answer: (b)58
Solution:
Divide 49123 by 263 to get
49123 = 186.263 + 205
Now we have to add 58  (263-205)a number to 49123 completely divisible by 263.

23. A number when divided by 899 given a remainder of 63. what remainder will obtained by dividing the same number by 29 ?
(a) 13                               (b) 3
(c) 5                                 (d) 21

Answer: (c) 5     
Solution:The number could be written as 899xQ + 63 where Q is the quotient of dividing the number by 899. Since 899 is completely divided by 29then the number 899xQ is also divided by 29. what remains is 63. Now if we divide 63 by 29 we get 5 as the remainder,

24. The sum of squares of three consecutive odd number is 2531 find the number
(a) 32,24 12                     (b) 27, 29, 31
(c) 8,13,19                       (d) 9,15, 25

Answer: (b) 27, 29, 31
Solution:
option(a) : The unit digit in the squares of number 32,24,12 are 4,6and 4 and their sum would have a unit digit of 4. but our sum has 1 in its unit digit. option rejected.
option(b) : The unit digit in the squares of number 27,29,31 are 9,1and 1 and their sum would have a unit digit of 1. and so is our answer. But we would check more options.
option(c) : The unit digit in the squares of number 8,13,19 are 4,9and 1 and their sum would have a unit digit of 4. but our sum has 1 in its unit digit. option rejected.
option(d) :Number are not consecutive.


25. If a number is divided by 84 the remainder is 37. what will be the remainder if it is divided by 21 ?
(a) 16                                (b) 18
(c) 12                                (d) 9

Answer:(a) 16 
Solution:
just divide the remainder by 21 to get the required answer. Here dived 37 by 21 to 16 as the remainder.

26. What is the sum of all odd number between 1 to 50 ?
(a) 180                              (b)625
(c) 840                              (d) 820

Answer:(b)625
Solution:
There are 25 odd number from 1 to 50 with a common difference as 2.
therefore the sum = (25)(1+49) /2 = 25x25 = 625

27. On converting the binary number 1101101 in to decimal form what will we get ?
(a) 109                             (b) 208
(c) 207                             (d) -111

Answer: (a) 109     
Solution:
1.2^6 +1. 2^5 +0. 2^4+1.2^3+1.2^2+0.2^1+2^0 = 64+32+0+8+4+0+1 = 96+13 = 109

28. What is the unit of digit is the product of 207 X 39 X 94 ?
(a) 9                                 (b) 1
(c) 7                                 (d) 2

Answer: (d) 2
Solution:
The unit digit of  207 X 39 is 3 and the unit digit in the 207 X 39 X 94 is 2.

29. How many digits are required to write the number from 1 to 100 ?
(a) 100                             (b) 192
(c) 99                               (d) 198    

Answer: (b) 192
Solution:
From 1 to 9, there are 9 digits.
From 10 to 99 there are 10x2x9 =180 digits.
and to write 100 we need 3 digits.
Therefore, the total numbers of digits are 9+180+3=183+9 = 192

30. Find the largest number of five digits which is divisible by 17 ?
(a) 99999                         (b) 99960
(c) 99994                         (d) 10013

Answer: (c) 99994 
Solution:
First the largest 5 digit number is 99999. Now divide this number by 17 to get the quotient and the remainder.
We get,      99999 = 5882x17 + 5
Therefore the largest number completely divisible by 17 is 99994. 

Problems on Ages

Most of the problems on ages could be solved by 

• forming linear equations from the relationships or conditions given in the questions
• solving those linear equations to get the required answer.

Here are some solved problems on the ages:

Q1. A got married 8 years ago. A’s 1 present age is 1(1/4) times his age at the time of marriage.A’s son’s age is 1/10 times his present age. His son’s age in years, is
(a) 2        (b) 3
(c) 4       (d) 5

Answer:(c) 4   
Solution:
Let A's present age be x age, then according to the conditions given in the questions, we have
            x = 1(1/4) (x-8)
     =>  x = 1.25(x-8)
     =>  x = 1.25x - 10 
     => 1.25x -x  = 10
            0.25 x   =10
and x = 40 years
Thus, A's son age = (1/10)40 = 4 years

Q2.Three years ago the average age of a family of 5 members was 17 years . A baby having been born , the average age of the family remains the same today. The age of the baby is
(a) 3 years           (b) 2 years
(c) 1 year            (d) 1.5 years

Answer: (b) 2 years
Solution:
Let the age of the baby today is y years.
Then average of the family member today = (20x5+y) /6
But this average is equal to 17.
Thus,    (100 + y) = 17x6
      =>  100 + y = 102
      => y =2 years

Q3. The sum of ages of two brothers , having a difference of 8 years between them, will double after 10 years. What is the ratio of the younger brother to that of the elder brother?
(a) 3:7             (b) 8:9
(c) 10:11         (d) 7:11

Answer: (a) 3:7 
Solution:
Let the age of elder brother be x years and smaller brother be y years. Then given that
                                x - y = 8 ----- (i)
                         and sum is x+y
Again, it is given that the sum of their ages will double after 10 years. The equations thus formed is
        2(x+y) = (x+10) + (y+10)
Thus, 2x + 2y = x+y+20
        => x+y=20 ------ (ii)
adding (i) and (ii), we get
                    2x = 28
                => x = 14 years
and y is 14- 8 = 6 years
Therefore, ratio of younger to elder brother is 6/14 = 3/7          

Trigonometry

• Pythagoras theorem

• Definition of Trigonometric Ratio

• Trigonometric ratios of commonly used angles

• Angle of elevation

• Angle of depression

 Pythagoras Theorem

The relationship between the hypotenuse, base and perpendicular in a right angled triangle is given by
 
h² = p² + b²






Definition of Trigonometric angles

1. Sine - Sine of an angle is the ratio of the height of a right angled triangle to the hypotenuse of a right
               angled triangle.
               or mathematically,   Sin θ = p/h

2.Cosine - Cosine of an angle is the ratio of the base of a right angled triangle to the hypotenuse.
               or mathematically,   Cosθ = b/h

3.Tangent - Tangent of an angle is the ratio of the height of a right angled triangle to hypotenuse of a right
                angled triangle ..
               or mathematically,   Tan θ = p/b



Trigonometric ratios of commonly used angles

Angle (θ in degrees)	Sinθ	Cosθ	Tanθ
0	0	1	0
30	1/2	√3/2	1/√3
45	1/√2	1/√2	1
60	√3/2	1/2	√3
90	1	0	Not Defined

Angle of elevation

Angle of elevation as appeared in the figure, on the right, is the angle formed when looking from a horizontal platform to a point that is above the horizontal platform.

Angle of elevation is used to measure height of poles, mountains and measure distance between two objects if height of one is known.   

 
Angle of depression

Angle of depression is the angle formed when looking down from a higher point to a point on a ground or on a horizontal surface.

 

Angle of elevation is used to measure distances or heights between two objects if one is known. 

Solved Questions on Trigonometry

1. Two poles of equal heights are standing opposite to each other on either side of a road which is 100m wide. From a point between them on road, angles of elevation of their tops are 30° and 60°. The height of
each pole in metre, is
(a) 25 √3          (b) 20 √3
(c) 28 √3          (d) 30 √3

Answer:(a) 25 √3
Solution:
Assume height of pole = p meter
Let the distance of  the point from the first pole be (X) meter with angle of elevation 30° and then the distance from the second becomes (100-X) with angle of elevation 60° . 
Then,    tan30° = p/x 
      =>  1/√3 = p/x
      => x = p√3 ------ (i)
Again, for the second pole, we have 
             tan60°= p/(100-x)
       => √3  = p/(100-x)
       =>  p= √3 (100-x) 
  Using (i), we have
             p = √3(100 -p√3)
             p = 100√3 - 3p
           4p = 100√3
             p = 25√3  

Short Cut:

Since both the poles are of equal length, then the distance between the two poles and the point is inversely proportional to the respective angles of elevation.    
Therefore,      (100-x)/x = tan30° /tan60° = (1/√3/√3/1)
                => (100-x)/x = 1/3
                => 300 - 3x = x          = >  4x = 300    => x= 75 m
Let the height of pole be p meter. Then, tan60°= p/75
                                                       => 75 √3 = p
                                                      => p = 25√3

2.The shadow of a tower standing on a level ground is found to be 40 m longer when the sun's altitude is 30° than when it is 60°. Find the length of the tower.
(a) 10√3        (b) 20 m
(c)  20√3       (d) 10 m  

Answer:20√3  = p
Solution: The distance of a point from the tower is inversely proportional to the tangent of the angle.
Let the height of the pole be p meter and the distance from the tower is x meter at 60° .  
Thus,      tan60°/tan30° = (x+40)/x
       =>   √3/1/√3 = (x+40)/x            => (√3)(√3) = (x+40)/x
       =>           3x  =  x+40                => 2x = 40
       =>            x = 20 meter
Using this to get
                         tan60° = p/20
                  =>   20√3  = p

3. The length of the shadow of the tower is 9 meters when the sun's altitude is 30°. What is the height of the tower?
(a)  9√3  m                  (b)  9√3/2  m
(c)  3√3  m                  (d)  4(1/2) m

Answer:(a)  9√3  m   
Solution:
"when the sun's altitude is 30°" means the angle of the depression is 30°.
Therefore, the angle of elevation is 60°. Thus,
                                                                       Tan 60° = p/9
                                                              =>         9√3  = p
                                         
4. The value of (secθ + cosecθ) when θ = 45°, is
(a) 4√2           (b)  2√2
(c) 5√2           (d)   3√2

Answer: (b)  2√2
Solution:
We know, sec 45° = √2 and cosec 45° = √2
Thus, (sec 45° + cosec 45°) = (√2 +√2) = 2√2      

Profit and Loss

Import Concepts

Cost price: Cost price is the price at which the seller has acquired the article in question.

Selling Price: Selling price is the price at which the seller is willing to sell the product.

Profit: Profit is the amount the seller earns by selling an article at a price greater than the cost price.

                     
                Profit = Selling Price - Cost price

Loss: Loss is the amount the seller loses by selling an article at a price less than the cost price.

                Loss =  Cost price - Selling Price 

Profit Percentage: Profit percentage is the profit percent that the seller earns on the sell the article. It is calculated on the cost Price.

Overheads: overheads is the expenses that the seller has to incur to sell the products/articles. Overhead includes the rent of the shop, salaries of the employees, utilities bills, advertisement bill or any kind of expenses the seller has to pay to keep his business running.

Fixed Costs: Fixed cost is the cost that the seller has to bear in spite of quantity he is selling. Fixed cost include the rent , utility bills, employee salary and so

Variable Costs: Variable cost or direct cost is the cost the seller has to bear to sell the article. The cost price of the articles, cost price of the raw material and others.

 

Formula for profit and loss:

1.Profit Percentage =  {(S.P. - C.P)/C.P.}x100

2.Loss Percentage  =  {(C.P. - S.P.)/C.P.}x100

3.To find the Cost Price when Selling Price (S.P.) and Percentage profit (x) is given

                  Cost Price =   {S.P/(100 + x)} x 100

4.To find the Selling Price when the cost price (C.P.) and percentage profit (x) is   

   given by

                  Selling Price = {(100 + x)(C.P.)/100}                                                                    

5.To find the Cost Price when Selling Price (S.P.) and Percentage loss (x) is given

                  Cost Price =   {S.P/(100 - x)} x 100

6.To find the Selling Price when the cost price (C.P.) and percentage loss (x) is   

   given by

                  Selling Price = {(100 - x)(C.P.)/100}

7. If an article is sold at a profit of x% which is further sold at a profit of y% then 
    the final Selling price is 

                  Final Selling Price = Initial Cost Price{(100+x)/100}{(100+y)/100)  

8. If two successive profits x and y are made on the same article then the net 
    profit is 
                  (x + y + xy/100)   
    Note: above formula could also be used for loss. Just make sure to use +ve 
    sign for profit and -ve sign for the loss.  

9. If a false weight is used instead of the true weight then, the profit 
    percentage  is 
                         {(True Weight - false weight)/False Weight} x 100



                      

Solved Examples:

1. An article is sold for 75 at a loss of 25%. Then find the cost price of the article.

Solution:

 Using the formula 5 mentioned above we have ,

              Cost Price =   {S.P/(100 - x)} x 100

                             =   {75/(100 - 25)} x 100

                             =   {75/75}x 100

                             =   (1/1)x100

                             =    100

2. An article with a cost price of Rs. 840 is sold for a profit of 10% and then 

   again sold for a profit of 20%. Find the final selling price.

Solution:

Using the formula 7 mentioned above, we have 

           Final Selling price = 840{(100+10)/100}{(100+20)/100}

                                    = 840(110/100)(120/100)         

                                    = 840(11/10)(12/10)

                                    = 840(1.1)(1.2)

                                    = 1108.80

3. A sells a mobile to B worth Rs. 10,000. He makes a profit of 15%. Again B 
   sold it back to A at a loss of 15%. Find the A's profit or loss.

Solution:
Using the formula 8, taking Y as negative, We have
Net profit/loss  = ( 15 - 15 -225/100)
                      = (-2.25)
 First, A made a profit of 15% and again total loss of  (-2.25)% in the entire deal was profit to A.
Therefore, total Profit A made = 15+2.25 = 17.25 %

4. I gain 0.70 paisa on Rs. 70. My gain percent is : 

(a)0.1%                         (b)1%

(c)7%                            (d) 10% 

Solution:

Total profit = 0.70 paisa
And C.P. = Rs 70
Therefore, Gain percent = (0.70/70)x100 = 1%

 5. In terms of percentage profit , which is the best transaction ? 

C.P (in Rs.)                                  profit (in Rs.)

(a)36                                              17

(b)50                                              24

(c)40                                             19 

(d) 60                                             29 

Solution:
We have to calculate the percentage profit in each case, the table becomes

C.P (in Rs.)                                  profit (in Rs.)   profit %

(a)36                                              17                 47.22

(b)50                                              24                 48

(c)40                                              19                 47.5

(d) 60                                             29                 48.33

Therefore, the best transaction is (d)

6. A shopkeeper purchased 70 kg of potatoes for Rs.420 and sold the whole lot at the rate of Rs. 6.50 per kg. What will be his gain percent ?
(a) 4(1/6) %                                (b)6(1/4)
(c)8(1/3)%                                   (d)20%

Solution:
Rate per kg at which the potatoes are bought = 420/70 = Rs 6 per kg
And they are sold at 6.5 kg per kg.
Therefore total profit = (0.5/6)x100 = 8(1/3) %

7. The price of the a radio is raised by 40% above the cost price and sold at a discount of 10%. What will be percentage of profit?
 (a) 14%     (b)  30%
 (c) 25%     (d)  26%

Answer:  (d)  26%
Solution:
Let the Cost price was 100 then the selling price at 40% profit will be 140. 
And at this price a discount of 10% is offered then the final selling price will be  
         140 - (0.10)(140) = 140 - 14 = 126
And net profit is   126 - 100 = 26 and in percentage it will be 26%

8. Profit after selling a commodity for 425 is the same as the loss after selling it for 355.The cost of the commodity is 
(a) 390      (b) 395
(c) 400      (d)  385

Answer:(a) 390
Solution:
Since the profit after selling the commodity at 425 and loss after selling it at 355 is the same.Then the Cost Price for the commodity lies exactly in the middle of the two numbers. It will be found by finding the averages of the these two numbers.
    There the Cost Price   = (425 + 355)/2 = 390

9. A watch is purchased for 400 and sold for 460.The profit percentage is 
   (a) 15.5%        (b) 12%
   (c) 13%           (d)  15%

Answer:
Solution:
Total profit = 60 and cost price is 400. Therefore the percentage profit is 
(60/400)x100 = 15 %

10. By selling an article for 450 , a man loses 10%. The gain or loss percent if he sells it for 540 is
(a) gain 9%   (b)  loss 9%
(c) gain 8%   (d)  loss 8%

Answer: (a) gain 9%
Solution:
Here, Cost  Price = 450(110/100) = 495
         If he sold it for 540, the profit would have been (540 -495) =     45
         And therefore the profit = {(45)/(495)} x100 =  9%      

11. The cost price : selling price of an article is a : b. If b is 200% of a then the percentage of profit on  cost price is
(a) 75%    (b) 125%
(c) 100%   (d) 200%

Answer:(c) 100%  
Solution:
Given that b is 200% of a. This implies that if a = 1 then b = 2. 
This Implies cost price : Selling price = 1:2
And profit percentage = {(2-1)/1}x 100 = 100%  

12. The successive discounts of 10% and 20% are equivalent to a single discount of 
(a) 30%           (b) 28%
(c) 25%          (d) 27%

Answer:   (b) 28%
Solution:
Let the Selling price be 100. Then the discount of 10% is 10. The selling price
will become 90. Again a 20% discount is offered then the selling price becomes 90 - (0.2)(90) =  72.
Hence the net discount becomes 100 - 72 = 28% 

13. A dealer marks his goods at 40% above the cost price and allows a discount of 20% on the marked price. The dealer has a
(a) loss of 20% (b) gain of 25%
(c) loss of 12% (d) gain of 12% 

Answer:(d) gain of 12%
Solution:
Let the Cost price be 100. Then after it was marked up 40% , the selling price would be 100 + (0.40)(100) = 140.
Again a discount of 20% was offered on the marked price then the final selling price become
                          140 - (0.20)(140) = 140 - 28 = 112
And the profit is 112 - 100 = 12% or gain of 12%

14. A person sells 400 mangoes at the cost price of 320 mangoes. His percentage of loss is 
 (a) 10    (b) 15
 (c) 20    (d) 25

Answer: (c) 20 % 
Solution:
Out of 400 mangoes he sold,he was able to recover the cost price of 320 mangoes. And his lost is the cost price of the 80 mangoes.
Thus, total loss = {(80)/400} x 100 = 20 %

15. A shoe company sold 50 pairs of shoes on a day costing Rs. 189.50 each 
for Rs. 10,000. Then the profit obtained in rupees is
(a) 522        (b) 525
(c) 573        (d) 612

Answer:
Solution:
Cost of 50 pairs of shoes = (189.5)(50)  = 9475
And he sold them for     10000
Therefore , total profit =  10000 - 9475
                                 =  525

List of Important Formulas for Competitive Exams

Shortcut Formulas/Formulas for Competitive exams like CTET, SSC and others.

1.  (a+b)² = a² + b² + 2ab

2.  (a-b)²=a² + b² - 2ab 

3.   a² - b²= (a - b)(a + b)

4.  a³ - b³= (a-b)(a² + ab + b²)

5.  a³ + b³= (a+b)(a² - ab + b²)

6.  (a + b)² = (a - b)² +4ab

7.  (a - b)² = (a + b)² -4ab

8.  a³ + b³ + c³ - 3abc =(a + b + c)(a² + b² + c² - ab - bc - ca)

9.   If (a + b + c) = 0, then a³+b³+c³=3abc
  
10. (a+b+c)³ = (a + b + c)³= a³ + b³ + c³ + 3(a + b)(b + c)(c + a)

11. (a + b + c )²  = a² + b² + c² + 2ab + 2bc + 2ca

12. x² + x(a+b) + ab = (x + a)(x + b)


13. a²(b+c) + b²(c+a)+c²(a+b) + 3abc = (a+b+c)(ab + bc + ca)

14. a²(b-c) + b²(c-a)+c²(a-b) + 3abc = -(a - b)(b - c)(c - a)
                      
                                (Number System)

15. For list of shortcut formulas for time, speed and distance click here.

16. Sum of first n natural numbers is {(n)(n+1)/2 }

17. Sum of first n even natural numbers is {(n)(n+1)}

18. Sum of first n odd natural numbers is n²

19. Sum of first even number upto n is {(n/2)(n/2+1)}

20. Sum of cubes of first natural numbers is {(n)(n+1)/2 }² 

21. Sum of squares of first n natural numbers is {(n)(n + 1)(2n +1)/6}

22. a³-b³ = (a-b)³ + 3ab(a-b)

23.  a³+b³ = (a+b)³ - 3ab(a+b)

24. Average of n multiples of x is x(1+n)/2

25. Average of first n natural numbers even numbers is (n+1)

26. Average of first n natural odd number is n.
                       
                                  (Profit and Loss formulas)

27. Profit Percentage =  {(S.P. - C.P)/C.P.}x100

28. Loss Percentage  =  {(C.P. - S.P.)/C.P.}x100

29.To find the Cost Price when Selling Price (S.P.) and Percentage profit (x) is 
     given
              Cost Price =   {S.P/(100 + x)} x 100

30.To find the Selling Price when the cost price (C.P.) and percentage profit (x) is 

     given by
                   Selling Price = {(100 + x)(C.P.)/100}                                                               

31.To find the Cost Price when Selling Price (S.P.) and Percentage loss (x) is given

                  Cost Price =   {S.P/(100 - x)} x 100

32.To find the Selling Price when the cost price (C.P.) and percentage loss (x) is   

     given by

                  Selling Price = {(100 - x)(C.P.)/100}

33. If an article is sold at a profit of x% which is further sold at a profit of y% then the final Selling price is 

                  Final Selling Price = Initial Cost Price{(100+x)/100}{(100+y)/100)  

34. If two successive profits x and y are made on the same article then the net 
    profit is 
                  (x + y + xy/100)  

.
35. If a false weight is used instead of the true weight then, the profit 
      percentage  is 
                         {(True Weight - false weight)/False Weight} x 100
  
                              (Time and Work formulas)

36.  If A can do a piece of work in x days and B can do the work in y days, then working together they can complete the work in  (xy)/(x+y) days.

37.  If A can do a piece of work in x days and (A and B) can do the same work in y days, then B can complete the work in  (xy)/(x-y) days.