Monday 13 June 2011

class vii test paper

1. Answer these questions (any five) (2x5=10)

(i) (-6) x (---) = 6

(ii) (---) by 36 = -2

(iii) By how much does -3 exceeds -5

(iv) How much less than -8 is -3 ?

(v) Define: Fraction , Vulgar fraction

(vi) Of 2/3 and 5/9, which is greater and by how much?

2. Solve any five questions (4x5=20)

1. By what number should 3/2be divided to get 2/3

2. If the cost of a note book is 7 . How many notebook can be purchased for Rs. 69

3. Express 45mm into cm , km

4. Nine boards are stacked on the top of each other. The thickness of each board is 3 cm. How is the stack?

5. Amit’s weight 35 kg. His sister kavita’s weight is of Amit’s weight. How much does Kavita’s weight?

6. By What number should 4/5  be divided to get 1

7. How many pieces of ply wood each 0.35 cm thick are required to make a pile 1.89 m high?

8. Write five rational number between -1 and -2


CBSE CLASS VII MATHS  and SCIENCE PAGE LINKS
CBSE SCIENCE TEST PAPERS
CBSE SCIENCE SOLVED PAPERS
CBSE MATHS TEST PAPERS
CBSE MATHS SOLVED PAPERS

Practice CBSE Math for class 7

CBSE CLASS VII MATHS  and SCIENCE PAGE LINKS
CBSE SCIENCE TEST PAPERS
CBSE SCIENCE SOLVED PAPERS
CBSE MATHS TEST PAPERS
CBSE MATHS SOLVED PAPERS
1. Simplify { - 13 - ( - 27 ) } + { - 25 – ( - 40 ) }
2. Subtract - 1 3 4 from the sum of 3 8 and – 8 7
3. Write pairs of integers whose sum gives an integer smaller than both the integer?
4. Sum of two integers is 14. If one of them is – 8 find other?
5. Of 2/3 and 5/9 which is greater and buy how much ?
6. Represent on number line (a ) -1/4 (b) 35/6
7. There are 42 students in class and 55/7 of the students are boys Find numbers of girls?
8. By what number should 6 be divided to get 2
9. Lalit read a book for 1 hrs everyday and read entire book in 6 days. How many hours does he take to read entire book?
10.By what number should 3/2 be divided to get 2/3

7th cbse maths FRACTIONS & DECIMALS


FRACTIONS & DECIMALS
1 The fraction ( p+q)/q equals 

a) p

b) p/q +q

c) p/q +1

d) p/q +p
2 One side of a square 2/a ,its perimeter is 

a) 4/a

b) 4/a2

c)8/a

d)8/a2

3 Product of 5/7× 3/5× 2/5 ×5 is equal to a 

a) 3

b)11/7

c)36/7

d)6/7

Write in descending order 4/5, 2/3, 1/2, ¾ 

a)4/5>2/3> 1/2 3/4

b)1/2>2/3>4/5>3/4

c)3/4>1/2>4/5>2/3

d)4/5>3/4>2/3>1/2 

5 Reciprocal of 1 3/11 

a)11/13

b)11/10

c)11/9

d)11/14

6 Which of the fraction is not equal to the other three
a)2/5

b)26/65 

c)14/35

d)34/68

7 Simplify 3 ¼ + 1/2÷ 3/4 - 1/2× 3 ½ 

a)2 1/6

b) 3 1/6

c) 2 1/5

d)3 1/5

(2/5 +3/5)÷ (7/5 -4/10) 

a) 1

b) 0

c)3/5

d)5/7
9  When simplified the product ( 2- 1/3) (2- 3/5 )(2-5/7) (2- 17/19) 

a)21/9 

b)23/3 

c)19/17

d)none

10)The value of ½ of (3/4÷ 2/3 )is 
a) 3/16

b)9/16

c)3/6

d) 9/4

11 If 25 ×32 =800 then 2.5 × .32 = 

a)0.800

b)8.00

c)80.0

d).0800

12 0.0302×0.52 = 

a)0.051704

b)0.015407 

c)0.15704 

d)none
13 0.04+ 404/1000= 

a)0.0404

b)0.0444

c)444

d)0.444
14  16.96÷ 400 = 

a)420

b)0.0424

c)420.04

d) 0.420

15 The value of 1.8/(0.4×0.3) is 

a)0.3/0.2

b)0.4/0.2

c)6/0.3

d)none
16 The value of (0.5)2×(0.1)3 is 

a)25

b)0.00025 

c)0.25

d)0.0025

17) If 43m =0.086 then m has the value
a)0.002

b)0.02

c)0.2

d)2

18) The result of adding the difference of 3.003 and 2.05 to their sum 

a)6.006

b)60.06

c)600.6

d)0.6060
19) Find the average of 0.3, 3, 0.03 , and 0.002 is 

a)0.833

b)0.803

c)83.3

d)833

20) 0.089m in cm is 

a)89

b)8.9

c)890

d)0.89

Answer Key 

Monday 6 June 2011

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? .

Algebraic Expressions and Identities soved


1. Identify the terms, their coefficients for each of the following expressions.
(i) 5xyz2 – 3zy
Answer: Term: xyz2 -- Coefficient = 5
Term: zy –- Coefficient = 3
(ii) 1 + x + x2
Answer: Term: x – Coefficient = 1 & Term: x2 – Coefficient = 1
(iii) 4x2y2 – 4x2y2z2 + z2
Answer: 4 is the coefficient for x2y2z2
1 is the term for z2
(iv) 3 – pq + qr – rp
Answer: For each term the coefficient is 1
(v) 0.3a – 0.6ab + 0.5b
Answer: 0.3 is the coefficient for a, for b there are two coefficients 0.6 and 0.5
2. Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?
Answer:
x + y: Binomial
1000: Mononomial
x + x2 + x3 + x4: Polynomial
7 + y + 5x: Binomial
2y – 3y2: Binomial
2y – 3y2 + 4y3: Trinomial
5x – 4y + 3xy: Trinomial
4z – 15z2: Binomial
ab + bc + cd + da: Polynomial
pqr: Mononomial
p2q + pq2: Binomial
2p + 2q: Binomial
3. Add the following.
(i) ab – bc, bc – ca, ca – ab (ii) a – b + ab, b – c + bc, c – a + ac
(iii) 2p2q2 – 3pq + 4, 5 + 7pq – 3p2q(iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl
Answer: (ab - bc) + (bc - ca) + (ca-ab)
ab + bc + ca - bc - ca - ab =
= 0
(ii) a – b + ab, b – c + bc, c – a + ac
Answer: (a - b + ab) + (b - c + bc) + (c - a + ac)
= a + b + c + ab + bc + ca - b - c - a
= ab + bc + ca
(iii) 2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2
= (2p2q2 - 3pq + 4) + (5 + 7pq - 3p2q2)
= 2p2q2 - 3p2q2 - 3pq + 7pq + 4 + 5
= - p2q2 + 4pq + 9
(iv) l2 + m2, m2 + n2, n2 + l2, 2lm + 2mn + 2nl
Answer: (l2 + m2) + (m2 + n2) + (n2 + l2) + (2lm + 2mn + 2nl)
= l2 + l2 + m2 + m2 + n2 + n2 + 2lm + 2mn + 2nl
= 2l2 + 2m2 + 2n2 + 2lm + 2mn + 2nl
4. (a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3
Answer: (12a - 9ab + 5b - 3) - (4a - 7ab + 3b + 12)
= 12a - 9ab + 5b - 3 - 4a + 7ab - 3b - 12
(signs are reversed after –sign once bracket is opened)
= 8a - 2ab + 2b - 15
(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
Answer: (5xy - 2yz - 2zx + 10xyz) - (3xy + 5yz - 7zx)
= 5xy - 2yz - 2zx + 10xyz - 3xy - 5yz + 7zx
= 2xy - 7yz + 5zx + 10xyz
(c) Subtract 4p2q – 3pq + 5pq2 – 8p + 7q – 10
from 18 – 3p – 11q + 5pq – 2pq2 + 5p2q
Answer: (18 - 3p - 11q + 5pq - 2pq2 + 5p2q) - (4p2q - 3pq + 5pq2 - 8p + 7q - 10)
= 18 - 3p - 11q + 5pq - 2pq2 + 5p2q - 4p2q + 3pq - 5pq2 + 8p - 7q + 10
= 28 + 5p - 18q + 8pq - 7pq2 - p2

5. (a) Add: p ( p – q), q ( q – r) and r ( r – p)
Answer: (p2 - pq) + (q2 - qr) + (r2 - pr)
= p2 + q2 + r2 - pq - qr - pr
(b) Add: 2x (z – x – y) and 2y (z – y – x)
Answer: (2xz - 2x2 - 2xy) + (2yz - 2y2 - 2xy)
= 2xz - 4xy + 2yz - 2x2 - 2y2
(c) Subtract: 3l (l – 4 m + 5 n) from 4l ( 10 n – 3 m + 2 l )
Answer: (40ln - 12lm + 8l2) - (3l2 - 12lm + 15ln)
= 40ln - 12lm + 8l2 - 3l2 - 12lm + 15ln
= 55ln - 24lm + 5l2
(d) Subtract: 3a (a + b + c ) – 2 b (a – b + c) from 4c ( – a + b + c )
= (-4ac + 4bc + 4c2) - (3a2 + 3ab + 3ac)
= -4ac + 4bc + 4c2 - 3a2 - 3ab - 3ac
= -7ac + 4bc + 4c2 - 3a2 - 3ab
6. Multiply the binomials.
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) (2.5l – 0.5m) and (2.5l + 0.5m)
Answer: (2.5l - 0.5m)(2.5l + 0.5)
Using (a+b)(a-b) = a2 - b2
We get = 6.25l2 - 0.25m2
(iv) (a + 3b) and (x + 5)
Answer: = ax + 5a + 3bx + 15b
(v) (2pq + 3q2) and (3pq – 2q2)
Answer: = 2pq x 3pq - 2pq x 2q2 + 3q2 x 3pq - 3q2 x 2q2
= 6p2q2 - 4pq3 + 9pq3 - 6q4 
= 6p2q2 - 5pq3 - 6p4

7. Find the product.
(i) (5 – 2x) (3 + x) 
(ii) (x + 7y) (7x – y) 

iii) (a2+ b) (a + b2
(iv) (p2– q2) (2p + q)




Answer: = 15 + 5x - 6x - 2x2
= 15 - x - x 2
(ii) (x + 7y) (7x – y)
Answer: = 7x2 - xy + 49xy - 7y2
= 7x2 - 7y2 + 48xy
iii) (a2+ b) (a + b2)
Answer: a2 x a + a2 x b + a x b + b3
= a3 + a2b + ab + b3
= a3 + b3 + a2b + ab
(iv) (p2– q2) (2p + q)
Answer: = 2p3 + p2q - 2pq2 - q3
= 2p3 - q3 + p2q - 2pq2
8. Simplify.
(i) (x2– 5) (x + 5) + 25
Answer: = x3 + 5x2 - 5x - 25 + 25
= x3 + 5x2 -5x

(ii) (a2+ 5) (b3+ 3) + 5
Answer: a2b3 + 3a2 + 5b3 + 15 + 5
= a2b3 + 5b3 + 3a2 + 20

(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)
Answer: = (ac - ad + bc - bd) + (ac + ad - bc - bd) + (2ac + 2bd)
= ac - ad + bc - bd + ac + ad - bc - bd + 2ac + 2bd
= 4ac

(v) (x + y)(2x + y) + (x + 2y)(x – y)
Answer: 2x2 + xy + 2xy + y2 + x2 - xy + 2xy - 2y2
= 3x2 + 4xy - y2

(vi) (x + y)(x2– xy + y2)
Answer: = x3 - x2y + xy2 + x2y - xy2 + y3
= x3 + y3

(vii) (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

(viii) (a + b + c)(a + b – c)
Answer: = a2 + ab - ac + ab + b2 - bc + ac + bc - c2
= a2 + b2 - c2 + 2ab

9 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 = a2 - 2ab + b2 we get the following equation:
= 4a2 - 28a + 49

10. Use the identity (x + a) (x + b) = x2 + (a + b) x + ab to find the following products.
(i) (x + 3) (x + 7)
Answer: x2 + (3+7)x + 21
= x2 + 10x + 21

(ii) (4x + 5) (4x + 1)
= 16x2 + (5 + 1)4x + 5
= 16x2 + 24x + 5

(iii) (4x – 5) (4x – 1)
= 16x2 + (-5-1)4x + 5
= 16x2 - 20x + 5

(iv) (4x + 5) (4x – 1)
= 16x2 + (5-1)4x - 5
= 16x2 +16x - 5

(v) (2x + 5y) (2x + 3y)
= 4x2 + (5y + 3y)4x + 15y2
= 4x2 + 32xy + 15y2

(vi) (2a2+ 9) (2a2+ 5)
= 4a4 + (9+5)2a2 + 45
= 4a4 + 28a2 + 45

(vii) (xyz – 4) (xyz – 2)
= x2y2z2 + (-4 -2)xyz - 8
= x2y2z2 - 6xyz - 8
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8th Cube and Cube root Test paper

CBSE TEST PAPER
8th Cube and cube root
1 Mark Each
1. Find the one.s digit of the cube of 3031
2. Without actually finding the cubes find the value of 21- 203
3. Is 32 a perfect cube?
4. Express 132 as the sum of two consecutive integers.
5. What will be the unit digit of the square of 3873?
6. Why 2332 is not perfect square ?
7. Cube of any odd number is even. Yes/No Why? 
2 Mark Each
8. Find the cube root of 27000.
9. Find the cube root of 17576 through estimation. 
3 Mark Each
8. Anubhav makes a cuboid of plasticine of sides 5 cm, 3 cm, 5 cm. How many such cuboids will he need to form a cube?
9. Is 5488 a perfect cube? If not, find the smallest natural number by which 5488 must be multiplied so that the product is a perfect cube
10. Is 5324 a perfect cube? If not, then by which smallest natural number should 5324 be divided so that the quotient is a perfect cube?

Cube and cube root Solved for class 8

1. Find the cube root of each of the following numbers by prime factorisation method.


(i) 64  
(ii) 512  
(iii) 10648
(iv) 27000
(v) 15625
(vi) 13824
(vii) 110592
(viii) 46656
(ix) 175616
(x) 91125

2. State true or false.

(i) Cube of any odd number is even.

FALSE: Odd multiplied by odd is always odd

(ii) A perfect cube does not end with two zeros.

TRUE: A perfect cube will end with odd number of zeroes


(iii) If square of a number ends with 5, then its cube ends with 25.


TRUE: 5 multiplied by 5 any number of times always gives 5 at unit’s place


(iv) There is no perfect cube which ends with 8.


False: 23 = 8


(v) The cube of a two digit number may be a three digit number.


FALSE: The smallest two digit number is 10 and 103 = 1000 is a three digit number


(vi) The cube of a two digit number may have seven or more digits.


FALSE: 99 is the largest 2 digit number; 993 = 989901 is a 6 digit number


(vii) The cube of a single digit number may be a single digit number.


TRUE: 23 = 8 is a single digit number


3. You are told that 1,331 is a perfect cube. Can you guess without factorisation what is its cube root? 
Similarly, guess the cube root of 4913.


Answer: Let us divide 1331 in two groups of 31 and 13 for extreme right half and extreme left half of the 
number.


As you know 13 = 1 so there would be 1 at unit’s place in cube root of 1331.


Now 23 = 8 and 33 = 27


It is clear that 8<13<27, so the 10s digit of cube root of 1331 may be 2


So, cube root of 1331 may be 21 but 213 = 9261 is not equal to 1331


So, let us test the 10s digit as 1


113 = 1331 satisfies the condition


4913:


Right group = 13


Left group = 49


73 gives 3 at unit’s place so unit digit number in cube root of 4913 should be 7


33 = 27 and 43 = 64


27<49<64


So, 10s digit in cube root of 4913 should be 3


Test: 373 = 50653 is not equal to 4913


Let Us test 273 = 19683 ¹ 4913


Let us test 173 = 4913 gives the answer