Class 8 - Maths - Algebraic Expressions and Identities (NCERT Ex 9.2 Answers)

Algebraic Expressions and Identities 

(NCERT Ex 9.2 Answers)

Q1: Find the product of following pair of monomials.

(i) 4 , 7p

Answer:

4   ☓   7p = 4  ☓ 7   ☓ p = 28p

(ii) -4p , 7p

Answer:

-4p   ☓   7p = -4 ☓ p ☓ 7  ☓ p = (-4☓7)☓ (p☓ p) = -28p²

(iii) 4p , 7pq

Answer:

-4p  &#9747 7p = -4 ☓ p ☓7  ☓ p  ☓ q = (-4☓ 7)☓ (p☓ p☓ q) = -28p²q

(iv) 4p³ , -3p

Answer:

4p³  ☓  -3p = 4  ☓(-3)   ☓ p ☓ p☓ p☓ p = -12p⁴

(v) 4p , 0

Answer:

4p ☓ 0 = 4☓ p☓ 0 = 0


Q2: Find the areas of rectangles with the following pairs of monomials as their lengths
and breadths respectively.

(p , q);(10m , 5n);(20x² , 5y²);(4x , 3x²);(3mn ,4np)

Answer:

We know that,

Area of rectangle = Length ☓ Breadth

Area of 1^st rectangle = p ☓ q = pq

Area of 2^nd rectangle = 10m ☓ 5n = 10☓ 5 ☓ m ☓ n = 50mn

Area of 3^rd rectangle = 20x² ☓ 5y² = 20☓ 5 ☓ x² ☓ y² = 100x²y²

Area of 4^th rectangle = 4x ☓ 3x² = 4☓3 ☓ x ☓ x² = 12x³

Area of 5^th rectangle = 3mn ☓4np = 3☓ 4 ☓ m ☓ n ☓ n ☓ p = 12mn²p


Q3: Complete the table of products.

First Monomial (⇨) Second Monomial(⇩)

2x

-5y

3x2

-4xy

7x2y

-9x2y2

2x	4x2	...	...	...	...	...
-5y	...	...	-15x2y	...	...	...
3x2	...	...	...	...	...	...
-4xy	...	...	...	...	...	...
7x2y	...	...	...	...	...	...
-9x2y2	...	...	...	...	...	...

Answer:

CBSE Class 8 - Maths - CH 6 - Square and Square Roots (Ex 6.4)

Square and Square roots

Exercise 6.4

Q1: Find the square root of each of the following numbers by division method.

(i)2304  (ii) 4489  (iii)3481  (iv) 529   (v)3249 (vi) 1369

(vii)5776 (viii) 7921  (ix)576 (x) 1024 (xi)3136 (xii) 900

Answer:

(i)The square root of 2304 is calculated as follows.

	48
4	23 04 -16
88	704 704
	0

∴√ 2304  = 48

(ii)The square root of 4489 is calculated as follows.

	67
6	44 89 -36
127	889 889
	0

∴√ 4489  = 67

(iii)The square root of 3481 is calculated as follows.

	59
5	34 81 -25
109	981 981
	0

∴√ 3481  = 59

(iv)The square root of 529 is calculated as follows.

	23
5	5 29 -4
43	129 129
	0

∴√ 529  = 23

(v)The square root of 3249 is calculated as follows.

	57
5	32 49 -25
107	749 749
	0

∴√ 3249  = 57

(vi)The square root of 1369 is calculated as follows.

	37
3	13 69 -9
67	469 469
	0

∴√ 1369  = 37

(vii)The square root of 5776 is calculated as follows.

	76
7	57 76 -49
146	876 876
	0

∴√ 5776  = 76

CBSE Class 8 - Maths - Cube and Cube Roots (Ex 7.1)

Cube and Cube Roots 

(NCERT Ex 7.1)

Q1: Which of the following numbers are not perfect cubes?
(i)  216
(ii) 128
(iii) 1000
(iv) 100
(v) 46656

Answer:

(i) The prime factorisation of 216 is as follows:

2	216
2	108
2	54
3	27
3	9
3	3
	1

∴ 216 = 2 ☓ 2 ☓ 2 ☓ 3 ☓ 3 ☓ 3 = 2³ ☓ 3³

Here, each prime factor appears as many times as a perfect multiple of 3, therefore, 216 is a
perfect cube.

(ii) The prime factorisation of 128 is as follows:

2	128
2	64
2	32
2	16
2	8
2	4
2	2
	1

∴  128 = 2 ☓ 2 ☓ 2 ☓ 2 ☓ 2 ☓ 2 ☓ 2

Here, each prime factor does not appear as many times as a perfect multiple of 3. Therefore, 128 is not a perfect cube.

(iii)The prime factorisation of 1000 is as follows.

2	1000
2	500
2	250
5	125
5	25
5	5
	1

∴ 1000 = 2 ☓ 2 ☓ 2 ☓ 5 ☓ 5 ☓ 5 =  2³ ☓ 5³

Here, as each prime factor appears as many times as a perfect multiple of 3, therefore, 1000 is a perfect cube.

(iv)The prime factorisation of 100 is as follows.

2	100
2	50
5	25
5	5
	1

∴ 100 = 2 ☓ 2 ☓ 5 ☓ 5

Here, each prime factor does not form triplet group(s). Therefore, 100 is not a perfect cube.

(iv) The prime factorisation of 46656 is as follows.

2	46656
2	23328
2	11664
2	5832
2	2916
2	1458
3	729
3	243
3	81
3	27
3	9
3	3
	1

46656 = 2 ☓ 2 ☓ 2 ☓ 2 ☓ 2 ☓ 2 ☓ 3 ☓ 3 ☓ 3 ☓ 3 ☓ 3 ☓ 3
 = 2³ ☓ 2³ ☓ 3³ ☓ 3³
Here, each prime factor appears as many times as a perfect multiple of 3,
∴ 46656 is a perfect cube.

Q2: Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.

CBSE Class 8 - Maths - Comparing Quantities (Ex 8.2)

Comparing Quantities 

NCERT Chapter (Ex 8.2)

Q1:A man got a 10% increase in his salary.If his new salary is 1,54,000,find his original salary.
Answer:

Let original salary be x. Given that new salary is 1,54,000.

Original salary + increment= New salary

But it is given that increment is 10% of the original salary.

Therefore,







Thus,the original salary was Rs 1,40,000.

Q2: On Sunday 845 people went to the zoo.On Monday only 169 people went.What is the percent decrease in the people visiting zoo on Monday?

Answer:

Given that on Sunday,845 people went to the zoo and on Monday,169 people went.

Decrease in number of people = 845-169=676

Percentage Decrease

CBSE Class 8 - Maths - Comparing Quantities (Ex 8.1)

Comparing Quantities 

NCERT Chapter (Ex 8.1)

Q1: Find the ratio of the following:

(a)Speed of cycle 15 km per hour to speed of scooter 30 km per hour

Answer:
Ratio of speed of cycle to speed of scooter

(b) 5m to 10 km

Answer:

Since 1 Km = 1000 m,

Required Ratio

(c) 50 paise to Rs 5

Answer:

Since Re 1 = 100 paise ,

Required Ratio

Q2: Convert the following ratios to percentages.

(a)3:4

CBSE Class 8 - Maths - CH13 - Direct and Inverse Proportions (Ex 13.2)

Direct and Inverse Proportions

NCERT Exercise and Other Q & A

If there is an increase (↑) [decrease (↓)] in one quantity produces a proportionate decrease (↓) [increase (↑)] in another quantity, then we say that the two quantities are in inverse variation or inverse proportion.

If two quantities vary inversely, their product is a constant.
e.g. if  and y are two quantities which are in inverse proportion then,
       x × y = constant
or   x₁ × y₁ = x₂ × y₂

NCERT Exercise 13.2

Q1: Which of the following are in inverse proportion?
(i) The number of workers on a job and the time to complete the job.
(ii) The time taken for a journey and the distance travelled in a uniform speed.
(iii) Area of cultivated land and the crop harvested.
(iv) The time taken for a fixed journey and the speed of the vehicle.
(v) The population of a country and the area of land per person.

Answer:
(i) With an increase in number of workers, the job will complete in lesser time. ∴ The number of workers on a job and the time to complete the job are in inverse proportion.

(ii) At constant (uniform) speed, more time means more distance will be covered. ∴ The time taken for a journey and the distance travelled in a uniform speed are in direct proportion.

(iii) An increase (↑) in the cultivated area means an increase  (↑) in the crop harvested. ∴ Area of cultivated land and the crop harvested are in direct proportion.

(iv) An  increase (↑) in speed of the vehicle means the distance will be covered in lesser  (↓) time. ∴ The time taken for a fixed journey and the speed of the vehicle. are in inverse proportion.

CBSE Class 8 - Maths - CH13 - Direct and Inverse Proportions (Ex 13.1)

Direct and Inverse Proportions

An increase (↑) in one quantity brings about an increase (↑) in the other quantity and similarly a decrease (↓) in one quantity brings about a decrease (↓) in the other quantity, the these two quantities are in direct proportion or direct variation.

In direct proportion the ratio of two quantities (say x and y) is a constant called constant of proportionality.
⇒ y/x = k          or       y = k × x



NCERT Chapter Solutions and other Q & A

NCERT Ex. 13.1

Q1: Following are the car parking charges near a railway station upto
       4 hours Rs 60
       8 hours Rs 100
     12 hours Rs 140
     24 hours Rs 180

Check if the parking charges are in direct proportion to the parking time

Answer: Calculating the ratio of parking charging wrt parting time:

No. of Hours (a)	Parking Charges(B)	Ratio (B/A)
4	60	60/4 = 15:1
8	100	100/8 =25:2
12	140	140/12 =35:3
24	180	180/24=15:2

Since the ratio of various parking charging vs parking time is not the same, the parking charges are NOT in direct proportion to the parking time.


Q2: A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base.
In the following table, find the parts of base that need to be added.

Parts of red pigment	1	4	7	12	20
Parts of base	8	...	...	...	...

CBSE Class 8 - Maths - CH16 - Playing with Numbers

Playing with Numbers

Pascal Triangle
Can you guess the next layer values?

NCERT Chapter Answers and other Q & As

Q1: Write the following numbers in generalised form.
(i) 25 (ii) 73 (iii) 129 (iv) 302

Answer:
(i) 25 = 10 × 2 + 5         (ab = 10 ×a + b)
(ii) 73 = 10 × 7 + 3
(iii) 129 = 100 × 1 + 10 × 2  + 9             (abc = 100 × a + 10 × b + c)
(iv) 302 = 100 × 3 + 10 × 0  + 2


Q2: Write the following in the usual form.
(i) 10 × 5 + 6
(ii) 100 × 7 + 10 × 1 + 8
(iii) 100 × a + 10 × c + b

Answer:
(i) 10 × 5 + 6  = 50 + 6 = 56
(ii) 100 × 7 + 10 × 1 + 8
(iii) 100 × a + 10 × c + b


Q3: Find A and B in the addition.

    A
  + A
  + A
 ------
  B A
 ------

Numbers Quiz (CTET/Class6/Class7/Class 8)

Numbers Quiz

(CTET/Class6/Class7/Class 8)

Q1: 2, 4, 6, 8, 10 ... are

(a) prime numbers
(b) even numbers
(c) odd numbers
(d) none of these

Q2: The roman number CD in decimal form is

(a) 100
(b) 500
(c) 400
(d) 0

CBSE - Class 8 - Maths - Cubes and Cube Roots

Important Points 
& NCERT Solutions

1. If n is a perfect cube then n = m³ or m is the cube root of n. i.e. (n = m × m × m) 

2. A cube root is written as ∛n or n^1/3.

3. ∛2, ∛3, ∛4 etc. all are irrational numbers.

4. The cube root of negative perfect cube is negative. i.e.

    (-x)³=  -x³ 

NTSE SAT Quiz-13 (Maths) / Class 8 Maths

NTSE SAT Quiz-13 (Maths)

Q1: One third of a number is greater then one fourth of its successor by 1, find the number.

(a) 5
(b) 15
(c) 20
(d) 25

Class 8, NTSE - Maths - Square and Square Roots #class8Maths #eduvictors

Class 8, NTSE - Maths - Square and Square Roots 

Perfect Square

1. A number is called a perfect square if it is expressed as the square of a number.

2. E.g. 1, 4, 9, 16, 25, ... are called perfect squares (1x1 = 1, 2x2 = 4, 3 x 3 = 9...)

3. In square numbers, the digits at the unit’s place are always 0, 1, 4, 5, 6 or 9.

4. The numbers having 2, 3, 7 or 8 at their units' place are not perfect square numbers. 

Q1: Which one of the following numbers is a perfect square:

a) 622

b) 393

c) 5778

d) 625

Answer: d.

5. If a number ends with the odd number of zeros then it is not a perfect square.

Q2: Check which of the following is not a perfect square.

a) 81000

b) 8100

c) 900

d) 6250000

Answer: a) 81000 (= 9² x 10² x 10)

6. The square of an even number is an even number while the square of an odd number is an odd number.

7. If n is a positive whole number then (n+1)² - n² = 2n + 1 
or 2n numbers in between the squares of the numbers n and (n + 1)

Rules of Divisibility

Rules of Divisibility

 
2: A number is divisible by 2 if the last digit (unit's digit) is even. e.g. 32, 459992

3: A number is divisible by 3, if the sum of digits of a number is divisible by 3.
    e.g. 252 = 2 + 5 + 2 = 9 ÷ 3 = 3 ∴ 252 is divisible by 3

4: A number is divisible by 4, if the last two digits of the number is divisible by 4.
    e.g. 81924 = since last two digits 24 is divisible by 4, hence the number.

5: A number is divisible by 5, if the last digit of the number is 0 or 5.
    e.g. 35, 200, 1005

6: A number is divisible by 6, if the number is divisible by both 2 and 3.

7: A number is divisible by 7, to check for this follow these steps:

CBSE Class 8 - (Ch11) - Mensuration

Q1. The area of the floor of a rectangular hall of length 40m is 960 m². Carpets of size 6m x 4m are available. Find how many carpets are required to cover the hall. (Unsolved Exercise 16.1 from RD Sharma)
Answer:
     length = 6m    breadth = 4m  and Area = l x b
$\therefore$ Area of carpet is = 6 x 4 = 24 m^2.
Floor area = 960 m²
Number of carpets =

Q2: Find the area of a square the length of whose diagonal is 2.9 meters. (Unsolved Ex 16.1 from RDS)
Answer: Diagonal (d) = 2.9 m. Let x be the length of each side of the square.
Applying Pythagoras theorem,
$ x^ + x^ = 2x^ = (2.9)^2$
$\Rightarrow x^2 = \text {area of square} = \frac{(2.9)^2}{2} = \frac{8.41}{2} = 4.205 m^2$

Q3: The area of a square field is 0.5 hectares. Find the length of its diagonal in meters.
Answer: 1 hect = 10000 m2
$\therefore$ 0.5 hec = 5000 m2 $
Applying Pythagoras theorem,
$ \frac{diagonal^2}{2} = 5000$
$\Rightarrow diagonal = \sqrt{10000} = 100 m$

Q4: The diameter of a semi-circular field is 14 meters. What is the cost of fencing the plot at Rs. 10 per meter.

Answer: Diameter of field (d) = 14m
$\Rightarrow r = 14 \div 2 = 7m$
$ \text{Perimeter of semicircle} = (\pi + 2) \times r$
$ \Rightarrow = (\frac{22}{7} + 2) \times 7 = 36m $
Cost of fencing per meter = Rs 10
$\therefore \text{cost of fencing for 36m} = 36 \times 10 = \text{Rs }360$

NTSE MAT Quiz-1 (Series completion)

Complete the following series:

1.  6,12,21,___,48,66
 (a) 33      (b) 38       (c) 40       (d) 45

2.  125,80,45,20,___
 (a)  5        (b)    8     (c) 10        (d)12

3.  22,24,28,___,52,84

 (a)  30        (b)   36      (c)    42     (d) 46

4. 1,3,3,7,6,13,10,___,15,31


(a)  25        (b)   27      (c)    21     (d) 30

5. 3,1,7,3,13,7,21,15,31,31,___,63,57


(a)    39      (b) 41        (c)  38       (d) 43

NTSE SAT Quiz-7 (Maths)

Q1: The sides of a right triangle are a, a+d, a+2d with a and d both positive. What is the value of ratio a:d ?
(a) 1: 3
(b) 1:4
(c) 2:1
(d) 3:1

Q2: In a group of cows and hens, the number of legs was 14 more than twice the number of heads. the number of cows is:
(a) 5
(b) 7
(c) 12
(d) 14

Q3: Each edge of cube is increased by 50%. The percent increase in surface area of the cube will be:
(a) 50
(b) 125
(c) 150
(d) 300

NTSE Maths Challenges

Following are the typical problems asked in NTSE, Oympiad Maths,  and other Mathematical based aptitude tests
(best viewed in Chrome or Firefox).

Q1: Ramesh walks from his home to Railway station at the rate of 5 kmph, he misses the train by 7 minutes. However, if he walks at the rate of 6 kmph, he reaches the station 5 minutes before the arrival of the train. Find the distance of the station from his home.

Answer: Let 'd' be the distance between Ramesh's house to Station.
We know that distance = speed x time.

If he walks at speed 5 kmph, let time taken be t1
⇒ t1 = d/5

Let the time t2 is taken when he walks with a speed of 6 kmph.

⇒  t2 = d/6

Since the time gap between two cases is 12 mins = 1/5 hrs
⇒ t1 - t2 = d/5 - d/6 = 1/5 hrs
⇒  6d - 5d = 6
⇒ d = 6km

Q2: Roma travels from place A to B and then from B to C.
From A to B Distance is 55 km which she covers with an average speed of 22 km per hour
It takes her 1 hour 30 minutes moving from B to C with an average speed 30 km per hour
Calculate her average speed over the whole of her journey from A to C.

Answer: We know that $ Time = Distance ÷ Speed

∴ Time taken from A to B = 55/22 = 2.5 hours  = 2 Hrs and 30 mins

Distance from A to B = Speed x Time = 20 x 1.5  = 45 km.  (note time taken from B to C is given as 1 Hrs 30 minutes i.e. 1.5 Hrs)

Average Speed = Total Distance  ÷ Total Time
∴  = (55+45) ÷ (1.5+2.5) = 100  ÷ 4 = 25 kmph

Maths Challenges (Answers)

Maths Challenges (Answers)

See Maths Challenges (Questions)...

Answer1: When two distinct lines cut each other, they always meet at single point. These two lines can cut the circle maximum at four different points. Therefore, maximum number of intersection points is 5 i.e (d). Following figure explains this.


Answer 2:
Total possible outcomes are 16 i.e.
HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT
The catch here is at least as many heads as tails, such possibilities are 11.
$\Rightarrow$ the probability is $\frac{11}{16}$

Answer 3:
Here product is a prime number. Since we know prime number is divisible by self and number 1 only. So the only choice for m is 1.

Maths Challenges (Questions)

Maths Challenges (Questions)

Q1:A circle and two distinct lines are drawn on a sheet of paper. What is the largest possible number of points of intersection of these figures?
    (a) 2        (b) 3      (c)  4    (d) 5
 
Q2: Reena tosses a coin four times. The probability that she gets at least as many heads as tails is
    (a) 5/6      (b) 3/8     (c) 11/16      (d) 5/8

Q3: Given that the product of two whole numbers m × n is a prime number, and the value of m is
smaller than n, find the value of m.

Q4: Jane has 9 boxes with 9 accompanying keys. Each box can only be opened by its accompanying
key. If the 9 keys have been mixed up, find the maximum number of attempts Jane must make
before she can open all the boxes.                     (Singapore Asia-Pacific Mathematical Olympiad Competition)

Q5: If 32 people were to enter a statewide singles tennis tournament, how many matches would be played, including the championship?

Q6: I have 2 vats each containing 2 mats. 2 cats sat on each of the mats. Each cat wore 2 funny old hats. On each hat lay 2 thin rats. On each rat perched 2 black bats. How many things are in my vats? (NCERT class 7 Maths Brain Teasers)

Q7: How many positive integers less than and equal to 999 are divisible by 7 but not by 11?

Q8(SMOPS): How many digits are there before the hundredth 9 in the following number
        9797797779777797777797777779......?

Q9(Olympiad): Suppose today is Tuesday. What day of the week shall it be 100 days from now?

Follow the link to see the answers Maths Challenges (Answers)