Class 8 Maths Compound interest Assignemnt

Compound Interest by Using Formula, when it is calculated annually

Case I:

When the interest is compounded annually

Let principal = $ P, rate = R % per annum and time = n years.

Then, the amount A is given by the formula

A = P (1 + R/100)ⁿ

Therefore, compound interest = (amount) - (principal).

1. Find the amount of $ 8000 for 3 years, compounded annually at 5% per annum. Also, find the compound interest.

Solution:

Here, P = $ 8000, R = 5 % per annum and n = 3 years.

Using the formula A = $ P(1 + R/ 100)ⁿ

amount after 3 years = $ {8000 × (1 + 5/100)³}

= $ (8000 × 21/20 × 21/20 × 21/20)

= $ 9261.

Thus, amount after 3 years = $ 9261.

And, compound interest = $ (9261 - 8000)

Therefore, compound interest = $ 1261.

2. Find the compound interest on $ 6400 for 2 years, compounded annually at 7¹/₂ % per annum.

Solution:

Here, P = $ 6400, R % p. a. and n = 2 years.

Using the formula A = P (1 + R/100)ⁿ

Amount after 2 years = [6400 × {1 + 15/(2 × 100)}²]

= $ (6400 × 43/40 × 43/40)

=$ 7396.

Thus, amount = $ 7396

and compound interest = $ (7396 - 6400)

Therefore, compound interest = $ 996.



Case 2:

When the interest is compounded annually but rates are different for different years

Let principal = $ P, time = 2 years, and let the rates of interest be p % p.a. during the first year and q % p.a. during the second year.

Then, amount after 2 years = $ {P × (1 + P/100) × (1 + q/100)}.

This formula may similarly be extended for any number of years.

1. Find the amount of $ 12000 after 2 years, compounded annually; the rate of interest being 5 % p.a. during the first year and 6 % p.a. during the second year. Also, find the compound interest.

Solution:

Here, P = $12000, p = 5 % p.a. and q = 6 % p.a.

Using the formula A = {P × (1 + P/100) × (1 + q/100)}

amount after 2 years = $ {12000 × (1 + 5/100) × (1 + 6/100)}

= $ (12000 × 21/20 × 53/50)

=$ 13356

Thus, amount after 2 years = $ 13356

And, compound interest = $ (13356 – 12000)

Therefore, compound interest = $ 1356.



Case 3:

When interest is compounded annually but time is a fraction

For example suppose time is 2³/₅ years then,

Amount = P × (1 + R/100)² × [1 + (3/5 × R)/100]

1. Find the compound interest on $ 31250 at 8 % per annum for 2 years. Solution Amount after 2³/₄ years

Solution:

Amount after 2³/₄ years

= $ [31250 × (1 + 8/100)² × (1 + (3/4 × 8)/100)]

= ${31250 × (27/25)² × (53/50)}

= $ (31250 × 27/25 × 27/25 × 53/50)

= $ 38637.

Therefore, Amount = $ 38637,

Hence, compound interest = $ (38637 - 31250) = $ 7387.

Compound Interest by Using Formula, when it is calculated half-yearly

Interest Compounded Half-Yearly

Let principal = $ P, rate = R% per annum, time = a years.

Suppose that the interest is compounded half- yearly.

Then, rate = (R/2) % per half-year, time = (2n) half-years, and

amount = P × (1 + R/(2 × 100))²ⁿ

Compound interest = (amount) - (principal).

1. Find the compound interest on $ 15625 for 1¹/₂ years at 8 % per annum when compounded half-yearly.

Solution:

Here, principal = $ 15625, rate = 8 % per annum = 4% per half-year,

time = 1¹/₂ years = 3 half-years.

Amount = $ [15625 × (1 + 4/100)³]

=$ (15625 × 26/25 × 26/25 × 26/25)= $ 17576.

Compound interest = $ (17576 - 15625) = $ 1951.

2. Find the compound interest on $ 160000 for 2 years at 10% per annum when compounded semi-annually.

Solution:

Here, principal = $ 160000, rate = 10 % per annum = 5% per half-year, time = 2 years = 4 half-years.

Amount = $ {160000 × (1 + 5/100)⁴}

=$ (160000 × 21/20 × 21/20 × 21/20 × 21/20)

compound interest = $ (194481- 160000) = $ 34481.

Compound Interest by Using Formula, when it is calculated Quarterly

Interest Compounded Quarterly

Let principal = $ P. rate = R % per annum, time = n years.

Suppose that the interest is compounded quarterly.

Then, rate = (R/4) % Per quarter, time = (4n) quarters, and

amount = P × (1 + R/(4 × 100))⁴ⁿ

Compound interest = (amount) - (principal).

1. Find the compound interest on $ 125000, if Mike took loan from a bank for 9 months at 8 % per annum, compounded quarterly.

Solution:

Here, principal = $ 125000,

rate = 8 % per annum = (8/4) % per quarter = 2 % per quarter,

time = 9 months = 3 quarters.

Therefore, amount = $ {125000 × ( 1 + 2/100)³}

=$ (125000 × 51/50 × 51/50 × 51/50)= $ 132651

Therefore, compound interest $ (132651 - 125000) = $ 7651.

Compound Interest Test sample paper for class 8

1. You invest Rs 5000 at 12% interest compounded annually. How much is in the account after 2 years, assuming that you make no subsequent withdrawal or deposit?

2. Find the amount and the compound interest on Rs 4000 at 10% p.a. for 2½ years.

3. A man invests Rs 5000 for three years at a certain rate of interest, compounded annually. At the end of one year it amounts to Rs 5600. Calculate(i) the rate of interest per annum,(ii) the interest accrued in the second year,(iii) the amount at the end of the third year.

4. A sum of Rs 9600 is invested for 3 years at 10% per annum at compound interest.(i) What is the sum due at the end of the first year?(ii) What is the sum due at the end of the second year?(iii) Find the compound interest earned in two years.(iv) Find the difference between the answers (ii) and (i) and find the interest on this sum for one year.(v) Hence write down the compound interest for the third year.

5. Find the difference between the S.I. and C.I. on Rs 2500 for 2 years at 4% p.a., compound interest reckoned semi-annually.

6. Find the amount and the compound interest on Rs 8000 in 2 years if the rate is 10% for the first year and 12% for the second year.

7. A man invests Rs 6500 for 3 years at 4·5% p.a. compound interest reckoned yearly. Income tax at the rate of 20% is deducted at the end of each year. Find the amount at the end of the third year.

8. Calculate the compound interest for the second year on Rs 8000 invested for 3 years at 10% p.a.

9. Find the sum which amounts to Rs 9261 at 10% p.a. compound interest for 18 months, interest payable half-yearly.

10. On what sum will the compound interest for 2 years at 5% p.a. be Rs 246?

11. On what sum will the compound interest (reckoned yearly) for 3 years at 6¼% per annum be Rs 408·50?

12. A man invests Rs 1200 for two years at compound interest. After one year his money amounts to Rs 1275. Find the rate of compound interest. Also find the amount which the man will get after 2 years correct to the nearest paise.

13. At what rate percent per annum compound interest will Rs 2000 amount to Rs 2315·25 in 3 years?

14. If Rs 50000 amounts to Rs 73205 in 4 years, find the rate of compound interest payable yearly. In what time will Rs 15625 amount to Rs 17576 at 4% per annum compound interest? .

Answers

1. Rs 6272 2. Rs 5082; Rs 1082


3. (i) 12% (ii) Rs 672 (iii) Rs 6952·64


4. (i) Rs 10560 (ii) Rs 11616 (iii) Rs 2016


(iv) Rs 1056, Rs 105·60 (v) Rs 1161·60


5. Rs 6·08 6. Rs 9856; Rs 1856 7. Rs 7227·56


8. Rs 880 9. Rs 8000 10. Rs 2400 11. Rs 2048


12. 6¼%; Rs 1354·69 13. 5% 14. 10%


15. 3 years

Linear Equation in One Variable Solved questions Amend Education Academy Poonam Dua

AMEND EDUCATION ACADEMY

Worked-out problems on linear equations in one variable:

1. The sum of three consecutive multiples of 4 is 444. Find these multiples.

Solution:

If x is a multiple of 4, the next multiple is x + 4, next to this is x + 8.

Their sum = 444

According to the question,

x + (x + 4) + (x + 8) = 444

⇒ x + x + 4 + x + 8 = 444

⇒ x + x + x + 4 + 8 = 444

⇒ 3x + 12 = 444

⇒ 3x = 444 - 12

⇒ x = 432/3

⇒ x = 144

Therefore, x + 4 = 144 + 4 = 148

Therefore, x + 8 - 144 + 8 – 152

Therefore, the three consecutive multiples of 4 are 144, 148, 152.
2. The denominator of a rational number is greater than its numerator by 3. If the numerator is increased by 7 and the denominator is decreased by 1, the new number becomes 3/2. Find the original number.

Solution:

Let the numerator of a rational number = x

Then the denominator of a rational number = x + 3

When numerator is increased by 7, then new numerator = x + 7

When denominator is decreased by 1, then new denominator = x + 3 - 1

The new number formed = 3/2

According to the question,

(x + 7)/(x + 3 - 1) = 3/2

⇒ (x + 7)/(x + 2) = 3/2

⇒ 2(x + 7) = 3(x + 2)

⇒ 2x + 14 = 3x + 6

⇒ 3x - 2x = 14 - 6

⇒ x = 8

The original number i.e., x/(x + 3) = 8/(8 + 3) = 8/11


3. The sum of the digits of a two digit number is 7. If the number formed by reversing the digits is less than the original number by 27, find the original number.

Solution:

Let the units digit of the original number be x.

Then the tens digit of the original number be 7 - x

Then the number formed = 10(7 - x) + x × 1

                  = 70 - 10x + x = 70 - 9x

On reversing the digits, the number formed

                  = 10 × x + (7 - x) × 1

                  = 10x + 7 - x = 9x + 7

According to the question,

New number = original number - 27

9x + 7 = 70 - 9x - 27

9x + 7 = 43 - 9x

9x + 9x = 43 – 7

18x = 36

x = 36/18

x = 2

Therefore, 7 - x

        = 7 - 2

        = 5

The original number is 52



4. A motorboat goes downstream in river and covers a distance between two coastal towns in 5 hours. It covers this distance upstream in 6 hours. If the speed of the stream is 3 km/hr, find the speed of the boat in still water.

Solution:

Let the speed of the boat in still water = x km/hr.

Speed of the boat downstream = (x + 3) km/hr.

Time taken to cover the distance = 5 hrs

Therefore, distance covered in 5 hrs = (x + 3) × 5   (D = Speed × Time)

Speed of the boat upstream = (x - 3) km/hr

Time taken to cover the distance = 6 hrs.

Therefore, distance covered in 6 hrs = 6(x - 3)

Therefore, the distance between two coastal towns is fixed, i.e., same.

According to the question,

5(x + 3) = 6(x - 3)

⇒ 5x + 15 = 6x - 18

⇒ 5x - 6x = -18 – 15

⇒ -x = -33

⇒ x = 33

Required speed of the boat is 33 km/hr.


5. Divide 28 into two parts in such a way that 6/5 of one part is equal to 2/3 of the other.

Solution:

Let one part be x.

Then other part = 28 - x

It is given 6/5 of one part = 2/3 of the other.

⇒ 6/5x = 2/3(28 - x)

⇒ 3x/5 = 1/3(28 - x)

⇒ 9x = 5(28 - x)

⇒ 9x = 140 - 5x

⇒ 9x + 5x = 140

⇒ 14x = 140

⇒ x = 140/14

⇒ x = 10

Then the two parts are 10 and 28 - 10 = 18.


6. A total of $10000 is distributed among 150 persons as gift. A gift is either of $50 or $100. Find the number of gifts of each type.

Solution:

Total number of gifts = 150

Let the number of $50 is x

Then the number of gifts of $100 is (150 - x)

Amount spent on x gifts of $50 = $ 50x

Amount spent on (150 - x) gifts of $100 = $100(150 - x)

Total amount spent for prizes = $10000

According to the question,

50x + 100 (150 - x) = 10000

⇒ 50x + 15000 - 100x = 10000

⇒ -50x = 10000 - 15000

⇒ -50x = -5000

⇒ x = 5000/50

⇒ x = 100

⇒ 150 - x = 150 - 100 = 50

Therefore, gifts of $50 are 100 and gifts of $100 are 50.


The above step-by-step examples demonstrate the solved problems on linear equations in one variable.

Class 8 Maths Square Square Root Examples and assignment

Amend Education Academy 9999908238

Square /Square  Root Examples and assignment

1 - What will be the unit digit of the squares of the following numbers?

(i) 81

Answer: 1 Since, 1² ends up having 1 as the digit at unit’s place so 81² will have 1 at unit’s place.

(ii) 272

Asnwer: 4 Since, 2² = 4, therefore, square of 272 will have 2 at it's unit place.

 2. The following numbers are obviously not perfect squares. Give reason.

(i) 1057 (ii) 23453 (iii) 7928 (iv) 222222 (v) 64000 (vi) 89722 (vii) 222000 (viii) 505050

Answer: (i), (ii), (iii), (iv), (vi) don’t have any of the 0, 1, 4, 5, 6, or 9 at unit’s place, so they are not be perfect squares.

(v), (vii) and (viii) don’t have even number of zeroes at the end so they are not perfect squares.

3. Write a Pythagorean triplet whose one member is:

(i) 6

Solution : As we know 2m, m ² + 1 and m² - 1 form a Pythagorean triplet for any number, m > 1.

Let us assume 2m = 6

Therefore, m = 3

And, m² + 1 = 3 ² + 1= 9 + 1 = 10

And, m ² - 1 = 3 ² - 1 = 9 - 1 = 8

Test: 6 ² + 8 ² = 36 + 64 = 100 = 10²

Hence, the triplet is 6, 8, and 10 Answer

There are 16 primitive Pythagorean triples with c ≤ 100:

(3, 4, 5 )	(5, 12, 13)	(8, 15, 17)	(7, 24, 25)
(20, 21, 29)	(12, 35, 37)	( 9, 40, 41)	(28, 45, 53)
(11, 60, 61)	(16, 63, 65)	(33, 56, 65)	(48, 55, 73)
(13, 84, 85)	(36, 77, 85)	(39, 80, 89)	(65, 72, 97)

4. What could be the possible ‘one’s’ digits of the square root of each of the following numbers?

(i) 9801

Answer: 1 and 9.

Explanation: Since 1² and 9² give 1 at unit’s place, so these are the possible values of unit digit of the square root.

5. For the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained.

(i) 252

Solution:

By prime factorisation we get,

252 = 2 x 2 x 3 x 3 x 7

Here, 2 and 3 are in pairs but 7 needs a pair. Thus, 7 can become pair after multiplying 252 with 7.

So, 252 will become a perfect square when multiplied by 7.

Thus, Answer = 7

6. For the following number, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained.

(i) 252

Solution:

By prime factorisation of 252, we get

252 = 2 x 2 x 3 x 3 x 7

Here, 2 and 3 are in pair, but 7 has no pair, which can be eliminated after dividing 768 by 7.

Hence, 252 needs to be divided by 7 to become a perfect square

Thus, Answer = 7

7. The students of Class VIII of a school donated Rs 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.

Solution:

We need to calculate the square root of 2401 to get the solution.

By prime factorisation of 2401, we get

2401 = 7 x 7 x 7 x 7

There are 49 students, each contributing 49 rupees

Thus, Answer = 49

8. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.

Solution: Let us find LCM of 4, 9 and 10

4 = 2 x 2

9 = 3 x 3

10 = 5 x 2

So, LCM = 2 ² x 3 ² x 5 = 180

Now the LCM gives us a clue that if 180 is multiplied by 5 then it will become a perfect square.

The Required number = 180 x 5 = 900

Examples on square root of a perfect square by using the long division method

1. Find the square root of 784 by the long-division method.
Solution:
Marking periods and using the long-division method,

Therefore, √784 = 28

2. Evaluate: √10609.

Solution:

Marking periods and using the long-division method,

Therefore, √10609 = 103

3. What least number must be subtracted from 7250 to get a perfect square? Also, find the square root of this perfect square.

Solution:

Let us try to find the square root of 7250.

This shows that (85)² is less than 7250 by 25.

So, the least number to be subtracted from 7250 is 25.

Required perfect square number = (7250 - 25) = 7225

And, √7225 = 85.

4. Find the greatest number of four digits which is a perfect square.
Solution
Greatest number of four digits = 9999.
Let us try to find the square root of 9999.

This shows that (99)² is less than 9999 by 198.

So, the least number to be subtracted is 198.

Hence, the required number is (9999 - 198) = 9801.

Unsloved Questions

1.     What least number must be added to 5607 to make the sum a perfect square? Find this perfect square and its square root

2.     Find the least number of six digits which is a perfect square. Find the square root of this number.

3.     Find the smallest number by which 1100 must be divided so that the quotient is a perfect square.

4.     Write a Pythagorean triplet whose one member is:18

5.     There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement.

Ans

1.     18

2.     489, 317

3.     11

4.     18, 82, 80

5.     16 children will be left out in the arrangement.

Class 8 Maths Algebraic Expressions assignment

Amend Education Academy 9999908238

Solved Assignment Algebric expressions

Basic Algebra Formulas:

Algebra is the part of mathematics which involves in manipulating equations or algebraic expressionsA list of given  basic algebra formulas can be used for the  application process, helps in understanding of the concept.

List of Algebraic Formulas :
The following are some of the important algebraic identities or expression used in class 8^th and 9^th maths

      1.  (a + b)² = a² + 2ab + b²

      2.  (a - b)²  = a² - 2ab + b²

      3.  (a + b) (a - b)  = a2 - b2

      4.  (x + a)(x + b)  = x2 + (a + b)x + ab

      5.  (x + a)(x - b)  = x² + (a - b)x - ab

      6.  (x - a)(x + b)  = x² + (b - a)x - ab

      7.  (x - a)(x - b)  = x²  - (a + b)x + ab

      8.  (a + b)³  = a³ + b³ + 3ab(a + b)

      9.  (a - b)³  =  a³ - b³ - 3ab(a - b)

     10.  (x + y + z)²  = x² + y² + z² + 2xy + 2yz + 2xz

     11.  (x + y - z)²  =  x² + y² + z² + 2xy - 2yz - 2xz

     12.  (x - y + z)²  = x² + y² + z² - 2xy - 2yz + 2xz

     13.  (x - y - z)²  = x² + y² + z² - 2xy +  2yz - 2xz

     14.  x³ + y³ + z³ - 3xyz  = (x + y + z)(x² + y² + z² - xy - yz -xz)

     15.  x² + y²  = 12 [(x + y)² + (x - y)²]

     16.  (x + a) (x + b) (x + c) = x³ + (a + b +c)x² + (ab + bc + ca)x + abc

     17.  x³ + y³ = (x + y) (x² - xy + y²)

     18.  x³ - y³  = (x - y) (x² + xy + y²)

     19.  x² + y² + z² -xy - yz - zx = 12 [(x-y)² + (y-z)² + (z-x)²]

Questions

1. Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

(i) 5a, 3a², 7a⁸ (ii) 2p, 4q, 8r (iii) xy, 2x²y, 2xy²(iv) a, 2b, 3c

2. Multiply following

i) (2x + 5) and (4x – 3) (ii) (y – 8) and (3y – 4) (iii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y

3. Use a suitable identity to get each of the following products.

(i) (x + 3) (x + 3) (ii) (2y + 5) (2y + 5) (iii) (2a – 7) (2a – 7)

4. Using identities, evaluate.

(i) 71² (ii) 99² (iii) 102²

5. Using (x + a) (x + b) = x²+ (a + b) x + ab, find

(i) 103 x 104

Factorization

1. Factorise the following expressions.

(i) a² + 8a + 16  (ii) p² – 10 p + 25 (iii) 25m² + 30m + 9

2. Factorise.

(i) 4p² – 9q²  (ii) 63a² – 112b²  (iii) 49x² – 36 (iv) 16x⁵ – 144x³ (v) 25a² – 4b² + 28bc – 49c²

3. Factorise the following expressions.

(i) p² + 6p + 8   (ii) q² – 10q + 21  (iii) p² + 6p – 16

Answers

1: Volume = length x  breadth x  height

(i) 5a x 3a² x 7a⁸ = 105a¹¹

(ii) 2p x 4q x 8r = 64pqr

(iii) xy x 2x²y x 2xy² = 4x⁴y⁴

(iv) a x  2b x 3c = 6abc

2. 1. Multiply following

i) (2x + 5) and (4x – 3)

Answer: (2x + 5)(4x - 3)

= 2x x 4x - 2x x 3 + 5 x 4x - 5 x 3

= 8x² - 6x + 20x -15

= 8x² + 14x -15

(ii) (y – 8) and (3y – 4)

Answer: ( y - 8)(3y - 4)

= y x 3y - 4y - 8 x 3y + 32

= 3y² - 4y - 24y + 32

= 3y² - 28y + 32

(iii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y

Answer: = 2.25x² + 6xy + 4.5x - 6xy - 16y² - 12y - 4.5x + 12y

= 2.25x² - 16y²

3. Use a suitable identity to get each of the following products.

(i) (x + 3) (x + 3)

Answer: Using (a + b)² = a² + 2ab + b² we get the following equation:

= x² + 6x + 9

(ii) (2y + 5) (2y + 5)

Answer: 4y² + 20y + 25

(iii) (2a – 7) (2a – 7)

Answer: Using (a - b)² = a^{2 -} 2ab + b² we get the following equation:

= 4a² - 28a + 49

4(i)  71² = (70+1)²

Using (a + b)² = a² + 2ab + b²

= 70² + 140 + 1²

= 4900 + 140 +1= 5041

(ii) 99²

= (100 -1)²

= 100² - 200 + 1²

= 10000 - 200 + 1

= 9801

(iii) 102²

= (100 + 2)²

= 100² + 400 + 2²

= 10000 + 400 + 4

= 10404

5. (i) 103 x 104

= (100 + 3)(100 + 4)

= 100² + (3 + 4)100 + 12

= 10000 + 1200 + 12

= 11212

Factorization

1. Factorise the following expressions.

(i) a² + 8a + 16

(ii) p² – 10 p + 25

(iii) 25m² + 30m + 9

2. Factorise.

(i) 4p² – 9q²

(ii) 63a² – 112b²

(iii) 49x² – 36

(iv) 16x⁵ – 144x³

Answer:16x⁵-144x³

= x³(16x²-144)

= x³(4x+12)(4x-12)

(v) 25a² – 4b² + 28bc – 49c²

3. Factorise the following expressions.

(i) p² + 6p + 8

Asnwer: p²+6p+8

=p(p+6)+8

(ii) q² – 10q + 21

Answer: q²-10q+21

=q(q-10)+21

(iii) p² + 6p – 16

Answer: p²+6p-16

=p(p+6)-1