Maths - Quardiatic Equations for Competition

Solving Quadratic Equations by Factorisation

Illustrative Examples

Example;Solve for x: (x +3)(x -3) = 40.

Solution:(x +3)(x -3) = 40 => x² -9 = 40

=> x² -49 = 0 => (x -7)(x +7) = 0,        => x -7 = 0 or x +7 = 0 => x = 7 or x = -7.
Hence the roots of the given equation are 7, -7.

Example:Solve(x -3)/(x +3) + (x +3)/(x -3) = 5/2

Solution:Given (x -3)/(x +3) + (x +3)/(x -3) = 5/2

To clear the fractions, multiply both sides by L.C.M. of fractions i.e. by 2(x +3)(x -3) to get
2(x -3)² +2(x +3)² = 5(x -3)(x +3),           => 2(x² -6x +9) +2(x² +6x +9) = 5(x² -9)
=> 2x² -12x +18 +2x² +12x +18 -5x² +45 = 0
=> -x² +81 = 0 => x² -81 = 0 => (x -9)(x +9) = 0
=> x -9 = 0 or x +9 = 0
Hence the values x = 9, x = -9 satisfy the given equation.

Example:Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60. Assume the middle number to be x and form a quadratic equation satisfying the above statement. Hence find the three numbers.

Solution:Since the middle number of the three consecutive numbers is x, the other two numbers are x -1 and x +1.

According to the given condition, we have
    x² = [(x +1)² -(x -1)²] +60
=> x² = (x² +2x +1) -(x² -2x +1) +60 = 4x +60
=> x² -4x -60 = 0 => x² - (10-6)x -60=0, x² - 10x +6x -60=0, 
x(x-10)+6(x-10)=0,  (x -10)(x +6) = 0, => x = 10 or -6
Since x is a natural number, we get x = 10
Hence the three numbers are 9, 10, 11
                                       Excercise:
 

Q1. x² - 5 x = 0 a. 5 b. 7 c. 9 d. 11

Q2. x (2x +1) = 6 a. -2, 3/2 b. ½, 3/2 c. 1, 2 d. 2, 2

Q3. x² - 3x -10 = 0 a. 5, 2 b. 2, 4 c. 5, 4 d. 5, -2

Q4. 3x² - 5x - 12 = 0 a. 3, -4/3 b. 3, 4 c. 2, 3 d. 3/5, 5/3

Q5. 3x² = x + 4 a. -1, -3 b. -1, 4/3 c. 4/3, 1 d. 1, 1/2

Q6. (x + 2)(x - 3) = 6 a. -3, 4 b. 3, -4 c. 4, 4 d. 3, 3

Q7. 3x – 8 / x = 2 a. 2, -4/3 b. 2, 4/3 c, 1, 2 d, 2, 3

Q8. (x + 2) / (x +3) = (2 x -3) / (3 x - 7) a. -1, 5 b. 1, -5 c. 1, 1 d. 5, 5

Q9. 8/(x +3) - 3/(2 -x) = 2 a. 5, -1/2 b. 5, ½ c. 5, 5 d, 1, 1

Q10. J and K has total 45 balls. If both lost 5-5 balls, than the multiplication of the balls remaining with them is 124. How much they have initially?

          a. 36, 9 b. 36, 36 c. 9, 9 c. Cant find e. None

Q11. The area of a ractangle fig is 506 sqmt. If length is one more than the width. Find the length.

          a. 22, 23 b. 23, 24 c. 24, 25 c. Cant Find e. None

Q12. Rohan's mother is 26 year older than him. The product of their ages after 3 years will be 360. Find Rohan's current age.

          a. 5 b. 7 c. 9 c. 11

Q13. The diagonal of a ractangle field is 60 m more than the smaller side. If longer side is 30 m more than the shorter side. Find the shorter side.

            a. 90 b. 150 c. 120 c. Cant Find e. None

Q14. The differance of the squares of two numbers is 180. The sq of smaller no is eight times the big no. Find the smaller one.

          a. 11 b. 12 c. 13 d. 14

Q15. The sum of the area of two squares is 468 sqm. If the diffrence of their perimeter is 24m. Find the side of smaller area.

         a. 10 b. 11 c. 12 d. 13

Q16. The perimeter of a ractangle field is 82 m and its area is 400 sq m. Find the breadth of the field is:

          a.14 m b. 16 m c. 18 m d. 20 m

Answers: 1a, 2a, 3d, 4a, 5b, 6a, 7a, 8a,9a, 10a, 11a, 12b, 13a, 14b, 15c, 16b

Maths - Volume and Surface area

Solid Cylinder: let r be the radius and h be height of a solid cylinder, then

(i) curved (lateral) surface area = 2 rh, (ii) total surface area = 2 r(h +r)
(iii) volume = r²h
Cone: Let r, h and l be the radius, height and slant height respectively of a cone, then, (i) slant height = r² +h², (ii) curved (lateral) surface area = rl
(iii) total surface area = r(l +r), (iv) volume = (1/3) r²h
Solid sphere: Let r be the radius of a solid sphere, then
(i) surface area = 4r², (ii) volume = (4/3) r³
Volume of cube: Side x side x side, Surface area= 6 x side x side
Volume of cuboid: L x B x H, Surface area= 2 (LB+LH+BH)
                                               Exercise

1. Find the volume and surface area of a cuboid 16 m long, 14 m broad and 7 m high.

        a. 866 b. 868 c. 686 d. 888 e. None

2. Find the length of the longest pole that can be placed in a room 12 mt long, 8 mt board and 9 m high.

        a. 17 b. 71 c. 77 d. 11 e. None

3. The surface area of a cube is 1734 sq cm. Find its volume.

        a. 9134 b. 1349 c. 4913 d. 9933 e. None

4. Three solid cubes of sides 1cm, 6cm & 8cm are melted and formed a new cube. Find the surface area of new cube.

         a. 486 b. 468 c. 684 d. 448 e. None

5. Find the volume, curved surface area and total surface area of a cyclinder with diameter of base 7cm and height 40cm.

         a. 1540, 880, 957 b. 1550, 880, 957 c. 1540, 880, 950 c. Cant find e. None

6. If the capacity of a cyclindrical tank is 1848 cube mt and the diameter of its base is 14 m, than find the debth of the tank is:

         a. 10 b. 11 c. 12 c. 13 e. None

7. How many iron rods, each of length 7 m and diameter 2cm can be made out of 0.88 cubic meter of iron?

        a . 400 b. 4 c. 40 c. 4000 e. None

8. The radii of two cylinders are in ratio 3:5 and their heights are in ratio of 2:3. Find the ratio of their curved surface areas.

         a. 2:5 b. 5:3 c. 3:3 c. Cant find e. None

9. Find the slant hight of a cone of radius 21 cm and hieght 28cm.

         a. 35 b. 53 c. 0.35 d. 0.53 e. None

10. The radii of the bases of a cylinder and a cone are in the ratio of 3 : 4 and their heights are in the ratio of 2 : 3. Find the ration of their volume.

        a. 9 : 8 b. 8 : 9 c. 8 : 8 d. Data inadequate e. None

11. How many spherical bullets can be made out of a lead cyclinder 28 cm high and with base radius 6 cm, each bullet being 1.5cm in diametre?

        a. 1792 b. 1729 c. 1722 d. 1799 e. None

12. A cone and a sphare have equal radii and equal volumes. Find the ratio of the diametre of the sphare to the height of the cone.

        a. 1 : 2 b. 1 : 1 c. 2 : 2 d. 2 : 1 e. None

13. A cone, a hemisphare and a cylinder stand on equal bases and have the same height. Find the ratio of their volumes.

       a. 1 : 2 : 3 b. 1 : 2: 2 c. 3 : 2 : 1 d. Cant find e. None

14. The capacity of a tank of dimentions (8 m x 6 m x 2.5 m) is:

        a. 120 lt b. 1200 lt c. 12000 lt d. 120000 lt
Answer: 1b, 2b, 3c, 4a, 5a, 6c, 7a, 8a, 9a, 10a, 11a, 12a, 13a, 14d

Maths - Permutaion and Combination

Combination:Each of the groups or selections which can be made by taking some or all of a number of things without reference to the order of things in each group is called a combination.

The number of combinations of n different things taken r at a time is
                   ⁿC_r = n!/[ r! (n -r)!]

For example, choosing two fruits out of a basket containing 4 oranges and 5 bananas - three combinations are possible - two oranges, two bananas or one orange and one banana.

Permutation:Each of the arrangements which can be made by taking some or all of a number of things is called permutation.

For example, the letters of word CAT can be arranged in six different ways
     CAT, CTA, ACT, ATC, TAC, TCA
Sometimes, students may be confused whether combination or permutation is being talked about. Certain key words/phrases can be helpful in deciding.
P(n, r) = n. (n -1). (n -2) ... (n -r +1) = n! / (n -r)

Combinations: Selection, choose, make group, distribute, committee, geometric problems.
Permutations: Arrangements, standing in a line, seated around a table, problems on digits or letters of a word.

                                                             Excersice

1. Evaluate: 30 ! / 28 !.                      a. 850 b. 860 c. 870 d. 880 e. None

2. Evaluate ⁶⁰P₃ :                                a. 205320 b. 202020 c. 535353 d. Cnat find e. None

3. Evaluate ¹⁰C₃ :                               a. 120 b. 12 c. 1.20 d. 12.2 e. None

4. How many words can be formed by using all the letters of the word “BIHAR”.

              a. 120 b. 12 c. 122 d. 112 e. None

 5. How many words can be made from the letters of the word “ EXTRA”, so that the wovels are  never together?

             a. 72 b. 27 c. 87 d. Cant find e. None

6. How many words can be made from the letters of the word “ DIRECTOR”, so that the wovels are always together?

               a. 2160 b. 1602 c. 2266 d. Cnat find e. None

7. In how many ways can a cricket team of 11 can be choosen out of a batch of 15 players?

              a. 1365 b. 1333 c. 1444 d. Cnat find e. None

8 In how many ways, a committee of 5 members can be selected from 6 men and 5 ladies of 3 men and 2 ladies.

              a. 100 b.200 c. 300 d. 400 e. None

9. How many words with or without meaning can be formed by using all the letters of the word “ APPLE”, using one letter exactly one time?

              a. 720 b. 120 c. 60 d. 180 e. None

10. In how many different ways can the letter of the word “ RUMOUR” can be arranged?

              a. 180 b. 90 c. 30 d. 720 e. None

11. How many words can be made from the letters of the word “ SIGNATURE”, so that the wovels are always together?

               a. 720 b. 1440 c. 2880 d. 3600 e. 17280

12. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?

                a. 32 b. 48 c. 64 d. Cant Find e. None

Answer:     1b, 2a, 3a, 4a, 5a, 6a, 7a, 8b, 9c, 10a, 11e, 12c

Maths - Percentage

• Percentages are special types of fractions, where the denominator is always hundred.
• To convert a ratio into a percent, write it as a fraction and multiply it by 100 and put a % sign.
• To convert a percent into fraction, remove the % sign, divide by 100 and reduce the fraction to lowest terms.
• To convert a decimal into a percent, we shift the decimal two places right and add a % sign.
• To convert a percent into a decimal, remove the % sign and shift the decimal two places to left.
• Percentage increase/decrease is given by
              (change / original value) × 100%

                                              Exercise

1. The ratio 5 : 4 expressed as a percent equals:           a. 12.5% b. 40% c. 80% d. 125%

2. 3.5 can be expressed in term of percentage is:          a. 0.35% b. 3.5% c. 35% d. 350%

3. 88% of 370 + 24% of 210 - ? = 118                           a. 256 b. 258 c. 268 d. 358

4. 60 % of 264 is the same as:                                       a. 10 % of 44 b. 15 % of 1056 c. 30 % of 132 d. none

5. What % of 7.2 kg is 18 gms?                                     a. 0.025% b. 0.25 % c. 2.5 % d. 25 %

6. It costs Rs. 1 to photocopy a sheet of paper. However 2% discount is allowed on photocopies done after first 1000 sheets. How much will it cost to copy 5000 sheets of paper?

          a. Rs 3920 b. Rs 3980 c. Rs 4900 d. Rs 4920

7. x % of 932 + 30 = 309.6

          a. 25 b. 30 c. 35 d. 40

8. 15 % of a % of 582 = 17.46

          a. 2 b. 10 c. 20 d. none

9. Two-fifth of one-third of three-seventh of a number is 15. What is the 40 % of that number?

          a. 72 b. 84 c. 136 d. 140 e. none

10. A number when 35 is subtracted from it, reduces to its 80 %. What is four-fifth of that number?

           a. 70 b. 90 c. 120 d. 140

11. The difference of two numbers is 20 % of the larger number. If the smaller number is 20, then the larger number is:

           a. 25 b. 45 c. 50 d. 80

12. A fruit seller had some apples. He sells 40 % of apples and still has 420 apples. Originally he had:

          a. 588 b. 600 c. 672 d. 700

13 A student has to obtain 33% of total topass. He got 125 and failed by 40 marks. The maximum markes are:

         a. 300 b. 500 c. 800 d. 1000

14. In an election A gets 84% vote and is elected by a majority of 476. What is the total number of votes?

          a. 672 b. 700 c. 749 d. 848

15. At an election in two candidates, 68 votes were invalid. The winning candidate secure 52% and wines by 98 votes. Total no of votes polled:

          a. 2382 b. 2450 c. 2518 d. none
Answer: 1d, 2d, 3b, 4b, 5d, 6d, 7b, 8c, 9e, 10d, 11a, 12d, 13b, 14b,15c

Maths - Problems on Ages

                                                     Exersice

1. Rajiv's age after 15 years will be 5 times his age 5 years back. What is the present age of Rajeev?

              a. 5 b. 10 c. 15 d. 20

2. The ages of two persons differ by 16 years. If 6 years ago the elder one be 3 times as ols as the younger one, find their persent ages?

                     a. 14, 30 b. 14, 14 c. 30, 30 d. 14, 15

3. The product of the ages of Ankit and Niketa is 240. if twice the age of Niketa is more than Ankit's age by 4 years, Niketa's age is:

                   a. 10 b. 12 c. 14 d. 16

4. Rohit was 4 times as old as his son 8 years ago. After 8 years, Rohit will be twice as old as his son. Rohit's present age is:

                     a. 20 b. 30 c. 40 d. 50

5. One year ago, the ratio of Gaurav's and Sachin's age was 6:7 respectivaly. Four year hence, the ratio will be 7:8. Sachin's age is:

                     a. 33 b. 34 c. 35 d. 36

6. Sachin is younger than Rahul by 7 years. If their ages are in the respective ratio of 7:9, how old is Sachin/

                      a. 16 b. 18 c. Cant find d. None

7. Hitesh is 40 years old and Ronnie is 60 years old. How many years ago was the ratio of their ages 3:5?

                       a. 5 b. 10 c. 20 d. 37

8. A is two year older than B who is twice as old as C. If the total of the ages of A, B & C is 27 year. Age of B is;

                        a. 7 b. 8 c. 9 d. 10

9. A person's present age is two-fifth of the age of his mother. Afetr 8 years, he will be one-half of the age of his mother. Mother's present age?

                       a. 32 b. 36 c. 40 d. 48

10. Q is as much younger than R as he is older than T. If the sum of the ages of R & T is 50. Than diffrence of ages of R and Q is:

                      a. 1 b. 2 c. Cant find d. None

11. The sum of the ages of 5 children born at the intervals of 3 years is 50 years. What is the age of the youngest one?

                         a. 4 b. 8 c. 10 d. None

Direction: Give ans (a) if data in I is sufficient to answer the Q, not II, give (b) if data in II is sufficient to answer the Q, not I, give (c) if data in either statement is sufficient to answer the Q, give (d) if data in both is not sufficient to answer the Q, give (e) if both needed to answer the Q.

12. The sum of the ages of P, Q & R is 96 years. The age of Q is:

               I) P is 6 years older than R. II) The total of the ages of Q & P is 56.

13. How old is C.

              I) 3 yrs ago average of A & B was 18 yrs. II) With joining C them now average becomes 22 yrs.

14. Divya is twice as old as Shruti. What is the diffrence of their ages?

            I) 5 yrs hence, the ratio of their ages will be 9:5. II) 10 yrs back, the ratio of their ages was 3:1.
Answers: 1b, 2a, 3b, 4c, 5d, 6d, 7d, 8d, 9c, 10c, 11a, 12e, 13e, 14c,

Maths - Average

Average: The average of any set of quantities can be found by adding all the numbers together then dividing by the number of quantities in the set.i.e. average = total of quantities / number of quantities e.g. A darts player throws the following scores during a match:  80, 100, 62, 180, 21, 55
Her average score per throw = (80 + 100 + 62 + 180 + 21 + 55) ÷ 6 = 83
Arithmetic Mean Definition:      Arithmetic mean is commonly called as average.Mean or Average is defined as the sum of all the given elements divided by the total number of elements.
  Formula: Mean = sum of elements / number of elements
              = a1+a2+a3+.....+an/n
  Example: To find the mean of 3,5,7.
      Step 1: Find the sum of the numbers.           3+5+7 = 15
      Step 2: Calculate the total number.   there are 3 numbers.
      Step 3: Finding mean.           15/3 = 5
                                                Excersie

1. David obtained 76, 65, 82, 67 and 85 marks out of 100 each in respective subjects. His average marks are:

             a. 65 b. 69 c. 72 d. 76 e. none

2. The average of first five multiples of 3 is:

                a. 3 b. 9 c. 12 d. 15

3. The average of 2, 7, 6 & x is 5, and the average of 18, 1, 6, x & y is 10. what is the value of y?

               a. 5 b. 10 c. 20 d. 30

4. The average of first 50 natural numbers is:

               a. 12.25 b. 21.25 c. 25 d. 25.50

5. The average age of the boys in the class is 16 years and that of the girls is 15 years. The average age for the whole class is:

                 a. 15 b. 15.5 c. 16 d. cannt find

6. The average of 5 numbers is 27. If one number is excluded the average becomes 25. The excluded number is:

                  a. 25 b. 27 c. 30 d. 35

7. The average age of 35 students in a class is 16 years. The average age of 21 students is 14. The average age of remaining 14 students is:

                  a. 15 b. 17 c. 18 d. 19

8. The average of six numbers is 3.95. The average of first two is 3.4 and last two is 3.85. What is the average of remaining two:

                  a. 4.5 b. 4.6 c. 4.7 d. 4.8

9. The average of a cricket player of 10 innings was 32. How many runs must he make in his next innings so as to increase his average by 4 runs?

                  a. 2 b. 4 c. 70 d. 76

10. Of the four numbers whose average is 60, the first is one-fourth of the sum of the last three. The first number is:

                  a. 15 b. 45 c. 48 d. 60.25
Answer: 1e, 2b, 3c, 4d, 5d, 6d, 7d, 8b, 9d, 10c

Maths - Simlification

BODMAS: use this formula to solve remove in the below sequence

B= Bracket,     O= Of,        D= Divide,        M= Multiplication,         A=Add,                  S=Subtract

1. 100 + 50 x 2                                                      (A) 75 (B) 150 (C) 200 (D) 300 (E) None
2. (4 86%of 6500) ÷ 36 =?                                    (A) 867.8 (B) 792.31 (C) 877.5 (D) 799.83 (E) None
3. (3080 + 6160) ÷ 28 = ?                                     (A)320 (B) 440 (C) 3320 (D) 3350 (E) None
4. 576÷ ? x 114=8208                                           (A)8 (B)7 (C)6 (D)9 (E) None
5. (1024—263—233)÷(986—764— 156) =?         (A)9 (B)6 (C)7 (D)8 (E) None
6. 125÷5 x ? = 625                                                  (A)5 (B) 25 (C)20 (D) 15 (E) None
7. 5004 ÷ 139 - 6 =?                                               (A) 10 (B) 20 (C) 30 (D) 40 (E) None
8. 384×12×2=?                                                       (A)9024 (B) 9216 (C)6676 (D) 6814 (E) None
9. 6534÷40÷33=?                                                   (A)3.06 (B) 5.25 (C)4.82 (D) 6.12 (E) None
10. 2704 x 209=?                                                  (A) 177996 (B)28844 (C)299452 (D)566181 (E)None
11.2536+4851—?=3450+313                              (A)3961 (B)4532 (C)3624 (D)4058 (E) None
12. (2560 x 1.4) + (7400 x 0.6) =?                        (A)7512 (B) 9746 (C)6523 (D) 8024 (E) None
13. 36%of 850+? %of 592 = 750                          (A)73 (B)89 (C)82 (D)75 (E) None
14.64%of 2650+40% 0f 320=?                              (A)1824 (B) 1902 (C)1829 (D) 1964 (E) None
15. 486+32×25—59=?                                           (A) 1514 (B) 1528 (C) 1227 (D) 507 (E) None
16. 1827 ÷ 36 x ? = 162.4                                       (A)4.4 (B)3.2 (C)2.1 (D) 3.7 (E) None
17. 1008÷36=?                                                        (A)28 (B) 32.5 (C)36 (D) 22.2 (E) None
18. 56.21 +2.36+5.41 —21.4+1.5=?                      (A)40.04 (B) 46.18 (C)44.08 (D) 43.12 (E) None
19. 65%of 320+?=686                                            (A) 480 (B) 452 (C)461 (D) 475 (E) None
20. 83250÷?=74×25                                               (A)50 (B) 45 (C)40 (D) 55 (E) None
Answer: 1c, 2c, 3e, 4a, 5d, 6b, 7c, 8e, 9e, 10d, 11c, 12d, 13d, 14a, 15c, 16b, 17a, 18c, 19e, 20b

Maths - Perimeter and Area

• Perimeter of a closed plane figure is the length of its boundary.
• Area of a closed plane figure is the measure of the region (surface) enclosed by its boundary.
• Area of a triangle: Area of a triangle = (1/2) × base × height
  Also, Area of a triangle = [s (s -a)(s -b)(s-c)] where a, b, c are the lengths of the sides and s = (a +b +c)/2
  Area of an equilateral triangle = 3a²/4, where a is the side.
• Rectangle: If l = length and b = breadth of a rectangle, then perimeter = 2 (l +b)
  length of diagonal = l²+b², area = l×b.
• Square: If a is the length of side of a square, then perimeter = 4a
  length of diagonal = 2 a, area = a²
• Circle: If r is the radius of a circle, then length of a diameter = 2r
  circumference = 2r, area = r²
  Take = 22/7 (unless given otherwise)
• Circular ring (track): If R and r are the radii of two concentric circles then
  area of the circular ring (track) = (R²-r²).

                                         Exercise

1. The length of a room is 5.5 m and width is 3.75 m . Find the cost of paving the floor by slabs at the rate of Rs 800 per sq m(in Rs).

          a. 15000 b. 15550 c. 15600 d. 16500 e. None

2. The length of a ractangle plot is 60% more than its breadth. If the diffrence between the length and breadth of that reactangle is 24 cm, what is the area of that ractangle?

          a. 2400 sq cm b. 2480 sq cm c. 2560 sqcm d. Data inadequate e. None

3. The length of a ractangle plot is 20 m more than its breadth. If the cost of fencing the plot @ Rs 26.50 per meter is Rs. 5300, what is the length of the plot in meters.

         a. 40 b. 50 c. 120 d. Data inadequate e. None

4. The ratio between the length and the perimeter of a ractangle is 1 : 3. What is ratio between the length & breadth of the plot?

          a. 1:2 b. 2:1 c. 3:2 d. Data inadequate e. None

5. A ractangular field is to be fenced on three sides leaving a side of 20 ft uncovered. If the area of the field is 680 sq feet, how many feet of fencing will be required?

          a. 34 b. 40 c. 68 d. 88 e. None

6. The ratio between the perimeter and breadth of a ractangle is 5:1. If the area of the ractangle is 216 sq cm. The length is:

          a. 16 cm b. 18 cm c. 24 cm d. Data inadequate e. None

7. A towel when bleached, was found to have lost 20% of its length and 10% of width. The % decrease in area is:

          a. 10 b.15 c. 20 d. 28 e. None

8. What will be the length of the diagonal of that square plot whose area is equal to the area of a ractangular plot of length 45 mt and breadth 40 mt?

          a. 42.5 b. 60 c. 75 d. Data inadequate e. None

9. If the side of a square is increased by 5 cm, the area increases by 165 sq cm. The side of the square is:

        a. 12 b. 13 c. 14 d. 15 e. None

10. The base of a tringle is 15 cm height is 12 cm. The height of the another tringle of the double the area having base 20cm is:

        a. 8 b. 9 c. 12.5 d. 18 e. None

11. The area of a right angle tringle is 40 times its base. What is its height?

        a. 45 b. 60 c. 80 d. Data inadequate e. None

12. The sides of a tringle are in ratio ½:1/3:1/4. if the perimeter is 52cm. Than the length of the smallest side is:

        a. 9 b. 10 c. 11 d. 12 e. None

13. The area of a circle of radius 5 is numerically what % of its circumference?

        a. 200 b. 225 c. 240 d. 250 e. None

14. A circle and a ractangle have the same perimeter. The sides of the ractangle are 18 & 26 cm. The area of circle is:

        a. 88 b. 154 c. 1250 d. Cant find e. None

15. A circular wire of radius of 42 cmis bent in the form of a ractangle whose sides are in ratio 6:5. Smaller side of ractangle is:

        a. 25 b. 30 c. 36 d. 60 e. None

  Ans: 1d, 2c, 3e, 4b, 5d, 6b, 7d, 8b, 9c, 10d, 11c, 12d, 13d, 14e, 15d

Maths - Compound Interest

• Amount = Principal + Interest
• Simple Interest = (Principal x Rate x Time)/100
• A = P Here A = amount, P = principal, r = rate percent yearly (or every fixed period) and n is the number of years (or terms of the fixed period).
• C.I. = P , where C.I. = compound interest
• S.I. (simple interest) and C.I. (compound interest) are equal for the first year (or the first term of the fixed period) on the same sum and at the same rate.
• C.I. of 2nd year (or the second term of the fixed period) is more than the C.I. of 1st year or the first term of the fixed period), and C.I. of 2nd Year -C.I. of 1st year = S.I. on the interest of the first year.

                                                    Exercise

1. Albert invested an amount of Rs. 8000 in afixed deposit scheme for 2 years at compound interest rate 5% pa. How much amount will Albert get on maturity of the fixed deposit?

      a. 8600 b. 8620 c. 8800 d. 8840 e. None

2. What will be the compound interest on a sum of Rs. 25000 after 3 years at the rate of 12 % pa?

     a. 9000.30 b. 9720 c. 10123.20 d. 10483.20 e. None

3. Sam invested Rs. 15000 at the rate of 10% pa for one year. If the interest is compouned half yearly, than the amonut received by Sam at the end of one year will be:

     a. 16500 b. 16525.50 c. 16537.50 d. 18150 e. None

4. Find the compoune interest on Rs. 15625 for 9 months at 16% per annum compounded quarterly.

     a. 1851 b. 1941 c. 1951 d. 1961

5. If the simple interest on asum of money for 2 years at 5% pa is Rs. 50, what is the Compound interest on the same sum at the same rate and for the same period?

     a. 51.25 b. 52 c. 54.25 d. 60

6. The difference between simple interest and compound interest on Rs 1200 for one year at the rate of 10% per annum reckoned half yearly is Rs. Is:

     a. 2.50 b. 3 c. 3.75 d. 4 e. None

7. The present worth of Rs. 169 due in 2 years at 4% pa componud interest is:

     a. 150.50 b. 154.75 c. 156.25 d. 158

8. The compound interest on a certain sum for 2 years at 10% pa is Rs. 525. The simple interest on the same sum for double the time at half the rate percent per annum is:

    a. 400 b. 500 c. 600 d.800

9. The difference between compound interest and simple interest on an amount of Rs. 15000 for 2 years is Rs. 96. What is the rate of interest per annum?

    a. 8 b. 10 c. 12 d. Can't find e. None

10. The effective annual rate of interest corrosponding to a nominal rate of 6% pa payable half yealy is;

    a. 6.06 % b. 6.07% c. 6.08% d. 6.09%
Ans: 1e, 2c, 3c, 4c, 5a, 6b, 7c, 8b, 9a, 10d, 11c