Author name: Bhagya

Haryana State Board HBSE 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.2 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Exercise 9.2

Question 1.
Find the product of the following pairs of monomials :
(i) 4, 7p
(ii) -4p, 7p
(iii) -4p, 7pq
(iv) 4p³, -3p
(v) 4p, 0.
Solution:
(i) 4 × 7p = (4 × 7) × p
= 28p.

(ii) -4p × 7p = (-4 × 7) × (p × p)
= -28p².

(iii) -4p × 7pq = (-4 × 7) × (p × pq).
= -28p²q.

(iv) 4p3, -3p = [(4) × (-3)] × (p3 × p)
= -12 p⁴.

(v) 4p × 0 = (4 × 0) × p = 0 × p = 0.

Question 2.
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)
Solution:
Area of rectangle = l × b sq. unit
(i) If l = p, b = q
Area of rectangle = p × q= pq sq.\unit.

(ii) If l = 10m, b = 5n
Area of rectangle = 10m × bn
= 50mn sq. unit.

(iii) If l = 20x², b =-5y²
Area of rectangle = 20x² × 5y²
= 100x²y² sq. unit

(iv) If l = 4x, b = 3x²
Area of rectangle = 4x × 3x² = 12x³sq. unit

(v) If l = 3mn, b = 4np
Area of rectangle = 3mn × 4np
= 12mn²p sq. unit.

Question 3.
Complete the table of products.

Solution:
Complete the table of products as shown in Table :

Question 4.
Obtain the volume of rectangular boxes with the following length, breadth and height respectively:
(i) 5a, 3a², 7a⁴,
(ii) 2p, 4p, 8r,
(iii) xy, 2x²y, 2xy²,
(iv) a, 2b, 3c.
Solution:
Volume of cuboid = l × b × h

(i) l = 5a, b = 3a², h = 7a⁴
Volume of rectangular box
= l × b × h cubic unit
= 5a × 3a² × 7a⁴
= (5 × 3 × 7) × (a × a2 × a4)
= 105a⁷ cubic unit.

(ii) l = 2p, b = 4q, h = 8r
v = l × b × h, = 2p × 4q × 8r
= (2 × 4 × 8) × (p × q × r)
= 64pqr.

(iii) l = xy, b = 2x²y, h = 2xy²
v = xy × 2x²y × 2xy²
= (1 × 2 × 2) × (xy × x²y × xy²)
= 4x⁴y⁴.

(iv) l = a, b = 2b, h = 3c
v = l × b × h = a × 2b × 3c
= (1 × 2 × 3) (a × b × c) = 6abc.

Question 5.
Obtain the product of
(i) xy, yz, zx
(ii) a, -a², a³
(iii) 2, 4y, 8y², 16y³
(iv) a, 2b, 3c, 6abc
(v) m, -mn, mnp.
Solution:
(i) xy × yz × zx = x²y²z².
(ii) a × (-a²) × (a³) = -a⁶.

(iii) 2 × 4y × 8y² × 16y³
= (2 × 4 × 8 × 16) × (y × y² × y³)
= 1024 y⁶.

(iv) a × 2b × 3c × 6abc
= (2 × 3 × 6) x (a x b × c × abc)
= 36a²b²c².

(v) mx (-mn) × mnp = -m³n²p.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.1 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Exercise 9.1

Question 1.
Identify the terms, their coefficients for each of the following expressions:
(i) 5xyz² – 3zy
(ii) 1 + x + x²
(iu) 4x²y² – 4x²y²z² + z²
(iv) 3 – pq + qr – rp
(v) \(\frac{x}{2}\) + \(\frac{y}{2}\) – xy
(v) 0.3a – 0.6ab + 0.5b.
Solution:
(i) There are two terms; 5xyz² and -3zy
Coefficient of 5xyz² = 5
Coefficient of-3zy = -3

(ii) There are three terms; 1, x and x²
Coefficient of x = 1
Coefficient of x² = 1

(iii) There are three terms; 4x²y², -4x²y²z², z².
Coefficient of 4x²y² = 4
Coefficient of -4x²y²z² = -4
Coefficient of z² = 1

(iv) There are four terms; 3, -pq, qr and – rp.
Coefficient of -pq = -1
Coefficient of qr = 1
Coefficient of -rp = -1.

(v) There are three terms; \(\frac{x}{2}\) , \(\frac{y}{2}\), -xy
Coefficient of \(\frac{x}{2}\) = \(\frac{1}{2}\)
Coefficient of \(\frac{y}{2}\) = \(\frac{1}{2}\)
Coefficient of-xy = -1

(vi) There are three terms; 0.3a, -0.6ab, 0.5b.
Coefficient of 0.3a = 0.3
Coefficient of-0.6a6 = -0.6 .
Coefficient of 0.56 = 0.5.

Question 2.
Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories ?.
x + y, 1000, x + x² + x³ + x⁴, 7 + y + 5x, 2y – 3y², 2y – 3y² + 4y³, 5x – 4y + 3xy, 4z – 15z², ab + bc + cd + da, pqr, p²q + pq², 2p + 2q.
Solution:
Monomial = 1000, par
Binomial = x + y, 2y – 3y², 4z – 15z², p²q + pq², 2p + 2q
Trinomial = 7 + y + 5x, 2y – 3y² + 4y³, 5x – 4y + 3xy
Polynomial = x + x² + x³ + x⁴, ab + bc + cd + da,

Question 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 – 3p²q²
(iv) l² + m², m² + n², n² + l², 2lm + 2mn + 2nl.
Solution:
(i) Add :

Question 4.
(a) Subtract 4a – 7ab + 36 + 12 from 12a – 9ab + 5b – 3
(6) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
(c) Subtract 4p²q – 3pq + 5pq² – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq² + bp²q.
Solution:
(a)

Haryana State Board HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities InText Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 8 Comparing Quantities InText Questions

Try These (Page 119)

Question 1.
In a primary school, the parents were asked about the number of hours they spend per day in helping their children to do homework. There were 90 parent who helped for \(\frac{1}{2}\) hour to 1\(\frac{1}{2}\) hours. The distribution of parents according to time for which, they said they helped is given in the adjoining figure; 20% helped for more than 1\(\frac{1}{2}\) hours per day; 30% helped for \(\frac{1}{2}\) hour to 1\(\frac{1}{2}\) hours; 50% did not help at all. It

Using this answer the following :
(i) How many parents were surveyed ?
(ii) How many said that they did not help ?
(iii) How many said that they helped for more than 1\(\frac{1}{2}\) hours ?
Solution:
(i) Let total no. of parents be x 30% of x helped for \(\frac{1}{2}\) hour to 1\(\frac{1}{2}\) hour
So 30% of x = 90
\(\frac{30}{100}\) × x = 90 or x = \(\frac{90 \times 100}{30}\) = 300
Number of parents = 300.

(ii) 50% parents did not help,
So number of parents did not help
= 50% of 300 = \(\frac{50}{100}\) × 300 = 150

(iii) 20% parents helped for more than 1\(\frac{1}{2}\) hours
So, number of such parents
= 20% of 300 = \(\frac{20}{100}\) × 300 = 60

Try These (Page 121)

Question 1.
A shop gives 20% discount. What would the sale price of each of these be ?
(a) A dress marked at Rs. 120.
(b) A pair of shoes marked at Rs. 750.
(c) A bag marked at Rs. 250.
Solution:
(a) 20% discount means On Rs. 100, discount is Rs. 20
On Re. 1 = Rs. \(\frac{20}{100}\)
∴ On 120 = \(\frac{20}{100}\) × 120 = 24
Sale price = M.P. – Discount
= 120 – 24 = 96
∴ Sale price = Rs. 96

(b) 20% discount means
20% discount on M.P.
⇒ Selling price = 100 – 20
= 80% on M.P.
∴ Sale price of shoe of MP Rs. 750
= 80% of 750
= \(\frac{80}{100}\) × 750 = 600
∴ Sale price of shoe = Rs. 600.

(c) 20% discount means
Selling price = 80% of M.P.
∴ Selling price of bag of M.P. Rs. 250
= 80% of 250
= \(\frac{80}{100}\) × 250 = 200
∴ Sale price of bag = Rs. 200.

Question 2.
A table marked at Rs. 15,000 is available for Rs. 14,400. Find the discount given and the discount per cent.
Solution:
M.P. = Rs. 15000
S.P. = 1440
Discount = ?
Discount % = ?
Discount = M.P. – S.P.
= 15000 – 14400 = 600
Discount % = \(\frac{Discount}{M.P.}\) × 100
= \(\frac{600}{15000}\) × 100 = 4%
Discount – Rs. 600, discount % = 4.

Question 3.
An almirah is sold at Rs. 5,225 after allowing a discount of 5%. Find its marked price.
Solution:
M.P. = ?
S.P. = Rs. 5,225
Discount % = 5 %

Try These (Page 123)

Question 1.
Find selling price (S.P.) if a profit of 5% is made on
(a) a cycle of Rs. 700 with Rs. 50 as overhead charges.
(b) a lawn mower bought at Rs. 1150 with Rs. 50 as transportation charges.
(c) a fan bought for Rs. 560 and expenses of Rs. 40 made on its repairs.
Solution:
(a) Cycle of worth Rs. 700
Overhead charges on cycle = 50
∴ C.P. of cycle = 700 + 50 = 750
Profit % = 5
∴ Profit of Rs. 5 on C.P. of Rs. 100
⇒ S.P. = (100 + 5)% of C.P.
∴ S.P. of cycle of C.P. Rs. 750 = 105% of 750
= \(\frac{105}{100}\) × 750 = 787.50
∴ S.P. of cycle = Rs. 787.50.

(b) Lawn mower bought in Rs. 1150
Transportation charge = Rs. 50
∴ C.P. of lawn mower = 1150 + 50 = 1200
5% profit on lawn mower is made
∴ S.P. = 105% of C.P.
= \(\frac{105}{100}\) × 1200 = 1260
∴ Sale price of lawn mower = Rs. 1260.

(c) Fan bought in Rs. 560
Expenses on its repairs = Rs. 40
∴ C.P. of fan = 560 + 40 = 600
5% profit on fan is made
∴ S.P. of fan = 105% of C.P. of fan
= \(\frac{105}{100}\) × 600 = 630
S.P. of fan = Rs. 630.

Question 2.
A shopkeeper bought two T.V. sets at Rs. 10,000 each. He sold one at a profit of 10% and the other at a loss of 10%. Find whether he made an overall profit or loss.
Solution:
Overall C.P. of each T.V. set = Rs. 10,000
One is sold at a profit of 10%
⇒ If C.P. is Rs. 100, S.P. is Rs. 110
Therefore, when C.P. is Rs. 10,000,
Then, S.P. = Rs. \(\frac{110}{100}\) × 10000 = 11000
In second case, T.V. at a loss of 10%
⇒ If C.P. is Rs. 100, S.P. is Rs. 90
Therefore, when C.P. is Rs. 10,000
Then, S.P. = Rs. \(\frac{90}{100}\) × 10000 = 9000
Total C.P. of two sets of T.V.
= Rs. 10000 + Rs. 10000
= Rs. 20,000
Total S.P. of two sets of T.V.
= Rs. 11000 + Rs. 9000
= Rs. 20,000
Since Total S.P. = Total C,P.
⇒ Neither profit nor loss.

Try These (Page 124-125)

Question 1.
Find the buying price of each of the following when 5% S.T. is added on the purchase of
(a) A towel at Rs. 50.
(b) Two bars of soap at Rs. 35 each.
(c) 5 kg of flour at Rs. 15 per kg.
Solution:
(a) S.T. of Rs 50 = 50 × \(\frac{10}{100}\) = Rs. 5
Bill amount = Cost of towel + sales tax
= 50 + 5 = Rs. 55.

(b) S.T. of Rs. 35 = 35 × \(\frac{10}{100}\) = Rs. 3.50
∴ Bill amount = cost of two bars of soap + S.T.
= 70 + 7 = Rs. 77.

(c) Cost of 5 kg flour = 5 × 15 = Rs. 75
Sales tax = Rs. 75 × \(\frac{10}{100}\)
= Rs. 7.5
Bill amount = Cost of flour + S.T.
= 75 + 7.5
= Rs. 82.50.

Question 2.
If 8% VAT is included in the prices, find the original price of
(a) A TV bought for Rs. 13,500.
(b) A shampoo bottle bought for Rs. 180.
Solution:
(a) 8% VAT mean an article of Rs. 100
will costs Rs. 100 + 8 = Rs. 108
Therefore, when price including VAT is Rs. 108, original price = Rs. 100
∴ When price including VAT is Rs. 13500
Original price = \(\frac{100}{108}\) × 13500 = 12500
∴ Original price of T.V. = Rs. 12500.

(b) 8% VAT included in the price
⇒ When price including VAT is Rs. 108, original price = Rs. 100
∴ When price including VAT is Rs. 180, original price
= \(\frac{100}{108}\) × 180 = 166.67
∴ Original price of shampoo = Rs. 166.67.

Try These (Page 126)

Question 1.
Find interest and amount to be paid on Rs. 15000 at 5% per annum after 2 years.
Solution:
P = Rs. 15000, R = 5% per annum, T = 2 year, I = ?, A = ?
I = \(\frac{P \times R \times T}{100}\) × \(\frac{5}{100}\) × 2 = Rs. 1500
A = P + I
= Rs. 15000 + Rs. 1500
= Rs. 16500.

Try These (Page 129)

Question 1.
Find C.I. on a sum of Rs. 8000 for 2 years at 5% per annum compounded annually.
Solution:
P = Rs. 8000, R = 5% per annum, T = 2 year, C.I. = ?

= 20 × 21 × 21 = Rs. 8820
C.I. = A – P
= Rs. 8820 – Rs. 8000
= Rs. 820.

Try These (Page 130)

Question 1.
Find the time period and rate for each.
1. A sum taken for 1\(\frac{1}{2}\) years at 8% per annum is compounded half yearly.
2. A sum taken for 2 years at 4% per annum compounded half yearly.
Solution:
1. If the time period yearly and the interest is compounded half yearly, then the time, T will be T × 2
∴ T = 1\(\frac{1}{2}\) year
= \(\frac{3}{2}\) × 2 = 3 years.
and, if the rate of interest is R% per annum and the interest is compounded half yearly, then the rate of interest will be \(\frac{R}{2}\) % per half hear.
R = 8% = \(\frac{R}{2}\)
= 4% per half year.

2. If the time period yearly and the interest is compounded half yearly then the time, T will be T × 2
∴ T = 2 years = 2 × 2 = 4 years
and, if the rate of interest is R% per annum and the interest is compounded half yearly, then the rate of interest will be \(\frac{R}{2}\) % per half year.
∴ R = 4% = \(\frac{4}{2}\)%
= 2% per half year.

Try These (Page 131)

Question 1.
Find the amount to be paid
(i) At the end of 2 years on Rs. 2,400 at ty5% per annum compounded annually.
(ii) At the end of 1 year on Rs. 1,800 at 8% per annum compounded quarterly.
Solution:
(i) A = ?, T = 2 years, P = Rs. 2400, R = 5% per annum compounded annually.

= 6 × 441 = Rs. 2646
∴ Amount = Rs. 2646.

(ii) R = 8% per annum compounded quarterly = 2% per quarterly
T = 1 year = 4 quarter
P = Rs. 1800, A = ?

= 1800 × \(\frac{51}{50}\) × \(\frac{51}{50}\)
= Rs. 1872.7
Amount = Rs. 1872.72.

Try These (Page 133)

Question 1.
A machinery worth Rs. 10,500 depreciated by 5%. Find its value after one year.
Solution:
P = Rs. 10500, R = 5%, T = 1 year
Cost of machinery after one year
= P(1 – \(\frac{R}{100}\))¹ = 10500(1 – \(\frac{5}{100}\))¹
= 10500(1 – \(\frac{1}{20}\))¹
= 10500 \(\frac{19}{20}\)
= 525 × 19 = Rs. 9975.

Question 2.
Find the population of a city after 2 years, which is at present 12 lakh, if the rate of increase is 4%.
Solution:
T = 2 year, present population, P = 12 lakh
R = 4%, A = ?
= 1200000 (1 + \(\frac{4}{100}\))²
= 1200000(1 + \(\frac{1}{25}\))²
= 1200000 × \(\frac{26}{25}\) × \(\frac{26}{25}\)
= 1297920
∴ Population after 2 years
= 1297920.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 8 Comparing Quantities Exercise 8.3

Question 1.
Calculate the amount and compound interest on :
(a) Rs. 10,800 for 3 years at 12\(\frac{1}{2}\)% per annum compounded annually.
(b) Rs. 18,000 for 2\(\frac{1}{2}\) years at 10% per annum compounded annually.
(c) Rs. 62,500 for 1\(\frac{1}{2}\) years at 8% per annum compounded half yearly.
(d) Rs. 8,000 for 1 year at 9% per annum compounded half yearly. (You could use the year by year calculation using S.I. formula to verify).
(e) Rs. 10,000 for 1 year at 8% per annum compounded half yearly.
Solution:
(a) A = ?, C.I. = ?, P = Rs. 10,800,
T = 3 years, R = 12 \(\frac{1}{2}\)% per annum
A = P(1 + \(\frac{R}{100}\))^T
= 1080o(1 + \(\frac{25}{2}\) × \(\frac{1}{100}\))³
= 10800(\(\frac{17}{16}\))³
= 10800 × \(\frac{17}{16}\) × \(\frac{17}{16}\) × \(\frac{17}{16}\)
= Rs. 12954.20
C.I. = A – P
= Rs. 1295.4.20 – Rs. 10800
= Rs. 2154.20.

(b) P = Rs. 18000, T = 2\(\frac{1}{2}\) years, R = 10% per annum compounded annually, A = ?, C.I. = ?
A = P(1 + \(\frac{R}{100}\))^T
= 18000(1 + \(\frac{10}{100}\))^{2\(\frac{1}{2}\)}
So amount for 2 years is given by
A = Rs. 18000(1 + \(\frac{10}{100}\))²
= 18000(1 + \(\frac{1}{10}\))²
= 18000 × \(\frac{11}{10}\) × \(\frac{11}{10}\)
= 180 × 121 = Rs. 21780
Rs. 21780 would act as principal for next \(\frac{1}{2}\) year. We find S.I. on Rs. 21780 for \(\frac{1}{2}\) year
S.I = Rs. \(\frac{21780 \times \frac{1}{2} \times 10}{100}\)
= Rs. 1089
Interest for two years
= Rs. 21780 – Rs. 18000
= Rs. 3780
Interest for next \(\frac{1}{2}\) year = Rs. 1089
Therefore, total compound interest
= Rs. 3780 + Rs. 1089
= Rs. 4869
A = P + C.I.
= Rs. 18000 + Rs. 4869
= Rs. 22869.

(c) P = Rs. 62500, T = 1\(\frac{1}{2}\) years = 3 half years, R = 8% per annum = 4% per half yearly, A = ?, C.I. = ?

= Rs 70304
C.I. = A – P
= Rs. 70304 – Rs. 62500
= Rs. 7804.

(d) P = Rs. 8000, T = 1 year = 2 half years, R = 9% per annum = \(\frac{9}{2}\) % per half year, A = ?, C.I. = ?

= Rs. 8736.20
C.I. = A – P
= Rs. 8736.20 – Rs. 8000
= Rs. 736.20.

(e) P = Rs. 10000, T = 1 year = 2 half years, R = 8% per annum = 4% per half year, A = ?, C.I. = ?

= Rs. 10816
C.I. = A – P
= Rs. 10816 – Rs. 10000
= Rs. 816.

Question 2.
Kamla borrowed Rs. 26,400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan ?
[Hint : Find A for 2 years with interest is compounded yearly and then find S.I. on the 2nd year amount for 4/12 years]
Solution:
P = Rs. 26400, R = 15% per annum, T = 2 years and 4 months, A = ?


= 66 × 23 × 23
= Rs. 34914
Rs. 34914 would act as principal for next 4 months i.e. \(\frac{1}{3}\) year
So S.I. = P × R × T
= 34914 × \(\frac{15}{100}\) × \(\frac{1}{3}\)
= Rs. 1745.70
Amount paid by Kamla after 2 years
= Rs. 34914 + Rs. 1745.70
= Rs. 36659.70.

Question 3.
Fabina borrows Rs. 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much ?
Solution:
In case of Fabina :
P = Rs. 12500,
R = 12% per annum, T = 3 years
S.I = P × R × T
= 12500 × \(\frac{12}{100}\) × 3 = RS. 4500.
In case of Radha :
P = Rs. 12500,
R = 10% per annum,
T = 3 years, C.I. = ?

= Rs. 16637.50
C.I. = A – P
= Rs. 16637.50 – Rs. 12500
= Rs. 4137.50
It is obvious that Fabina would pay more interest by Rs. 4500 – Rs. 4137.50 = Rs. 362.50 So,
Fabina paid Rs. 362.50 more than Radha.

Question 4.
I borrowed Rs. 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay ?
Solution:
P = Rs. 12000, R = 6% per annum, T = 2 years, S.I. = ?, C.I. = ?
In case of simple interest:
S.I = P × R × T
= 12000 × \(\frac{6}{100}\) × 2
= Rs. 1440
In case of compound interest:

= Rs. 13483.20
So, Amount = Rs. 13483.20
∴ C.I. = A – P
= Rs. 13483.20 – Rs. 12000
= Rs. 1483.20
It is obvious that C.I. > S.I
So extra amount paid in case of C.I. is
= Rs. 1483.20 – Rs. 1440
= Rs. 43.20.

Question 5.
Vasudevan invested Rs. 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get
(i) after 6 months ?
(ii) after 1 year ?
Solution:
P = Rs. 60000,
R = 12% per annum compounded half yearly
= 6% half yearly
A₆ = ?
A₁₂ = ?
T₆ = 6 months = 1 half year
T₁₂ = 1 year = 2 half years

(i) Amount after 6 months

= Rs. 67416.
∴ A₁₂ = Rs. 67416.

Question 6.
Arif took a loan of Rs. 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he
would be paying after 1\(\frac{1}{2}\) years if the interest is
(i) Compounded annually.
(ii) Compounded half yearly.
Solution:
P = Rs. 80000,
R = 10% per annum,
T = 1\(\frac{1}{2}\) years
= 5% half yearly = 3 half years
In 1st Case : When amount calculated annual compounding

Rs. 88000 would be principal for half year to calculate S.I. for \(\frac{1}{2}\) year
S.I. = P × R × T
= 88000 × \(\frac{10}{100}\) × \(\frac{1}{2}\)
= Rs. 4400
∴ C.I. for 1\(\frac{1}{2}\) on given principal
= (Rs. 88000 – Rs. 80000) + Rs. 4400
= Rs. 12400
So, A = P + C.I.
= Rs. 80000 + Rs. 12400
= Rs. 92400
In 2nd Case : When amount calculated half yearly compounding

= Rs. 92610
Difference in amount in both the cases
= Rs. 92610 – Rs. 92400
= Rs. 210.

Question 7.
Maria invested Rs. 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find
(i) The amount credited against her name at the end of the second year.
(ii) The interest for the 3rd year.
Solution:
(i) P = Rs. 8000, R = 5% per annum, T = 2 year, T = 3 year, A₂ = ?, A₃ = ?
Amount after 2 years,

Amount credited after 2 years = Rs. 8820.
Amount after 3 years,

= Rs. 9261
∴ Interest for 3rd year = A₃ – A₂
= Rs. 9261 – Rs. 8820
= Rs. 441.

Question 8.
Find the amount and the compound interest on Rs. 10,000 for 1\(\frac{1}{2}\) years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually ?
Solution:
P = Rs. 10000
T = 1\(\frac{1}{2}\) years = 3 half years
R = 10% per annum = 5% half yearly
A_h =?, A_a = ?

= Rs. 11576.25
So, Amount = Rs. 11576.25
∴ C.I. = A – P
= Rs. 11576.25 – Rs. 10000
= Rs. 1576.25
If n = 1\(\frac{1}{2}\) = \(\frac{3}{2}\) years
A_a = P(1 + \(\frac{R}{100}\))^T

= 1000 × 1.05 = Rs. 11550
∴ C.I. = A_a – P = 11550 – 10000
= Rs. 1550.
or, Rs. 11000 would be principal for \(\frac{1}{2}\) year to calculate S.I. for \(\frac{1}{2}\) year.
S.I. = P × R × T
= 11000 × \(\frac{10}{100}\) × \(\frac{1}{2}\) = Rs. 550
∴ C.I. = (Rs. 11000 – Rs. 10000) + Rs. 550 = Rs. 1550
It is obvious that C.I. calculated half yearly is greater than that when calculated yearly.
Extra amount of interest = Rs. 1576.25 – Rs. 1500 = Rs. 76.25.

Question 9.
Find the amount which Ram will get on Rs. 4096, if have gave it for 18 months at 12\(\frac{1}{2}\) % per annum, interest being compounded half yearly.
Solution:
P = Rs. 4096,
T = 18 months = 3 half years,
R = 12\(\frac{1}{2}\) % per annum
= \(\frac{25}{4}\) % per half year
A = ?

= Rs. 4913

Question 10.
The population of a place increased to 54,000 in 2003 at a rate of 5% per annum
(i) find the population in 2001.
(ii) what would be its population in 2005 ?
Solution:
(i) P = 54000, R = 5% per annum, A = ?, T = 2 years
Population before 2 years

(ii) For population in 2005 i.e. 2 years after 2003
P = 54000, T = 2 years, R = 5% per annum, A = ?

∴ Population in 2005 be 59535.

Question 11.
In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.
Solution:
Initial count of bacteria P = 506000, T = 2 hour
Rate of increasing bacteria R = 2.5% per hour, A = ?

∴ Count of bacteria after 2 hour = 531616.

Question 12.
A scooter was bought at Rs. 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
Solution:
P = Rs. 42000, R = 8% per annum, T = 1 year, A = ?

Haryana State Board HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.2 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 8 Comparing Quantities Exercise 8.2

Question 1.
A man got a 10% increase in his salary. If his new salary is Rs. 1,54,000, find his original salary.
Solution:
10% increase in salary means
If previous salary is Rs. 100, increased salary = Rs. 110
If new salary is Rs. 110, original salary is Rs. 100
New salary is Rs. 1,54,000, original salary is = \(\frac{100}{110}\) × 154000 = 1,40,000
Original salary = Rs. 1,40,000.

Question 2.
On Sunday 845 people went to the Zoo. On Monday only 169 people went. What is the percent decrease in the people visiting the Zoo on Monday.
Solution:
On Sunday Zoo visitor = 845 persons
On Monday Zoo visitor = 169 persons
Decrease in people visiting Zoo on Monday
= 845 – 169 = 676
Percent decrease on Monday
= \(\frac{Decrease}{Visitor on Sunday}\) × 100
= \(\frac{676}{845}\) × 100 = 80%

Question 3.
A shopkeeper buys 80 articles for Rs. 2,400 and sells them for a profit of 16%. Find the selling price of one article.
Solution:
C.P. of all 80 articles = Rs. 2400
16% profit means if C.P. is Rs. 100, S.P. is Rs. 116
Therefore, if C.P. is Rs. 2400,
S P = \(\frac{116}{100}\) × 2400 = 2784
S.P. of 80 articles = Rs. 2784
∴ S.P. of one article = Rs. 34.80.

Question 4.
The cost of an article was Rs. 15,500. Rs. 450 were spent on its repairs. If it is sold for a profit of 15%, find the selling price of the article.
Solution:
C.P. of article = Rs. 15500
Charge on its repairing = Rs. 450
So, new C.P. = Rs. 15500 + Rs. 450
= 15950
At profit of 15% is sold
So, S.P. = 115% of C.P,
= \(\frac{115}{100}\) × 15950
= 18342.50
∴ S.P. = Rs. 18342.50.

Question 5.
A VCR and TV were bought for Rs. 8,000 each. The shopkeeper made a loss of 4% on the VCR and a profit of 8% on the TV. Find the gain or loss percent on the whole transaction.
Solution:
C.P. ofT.V. = Rs. 8000
C.P. of VCR = Rs. 8000
4% loss on VCR means
If C.P. is Rs. 100, S.P. is Rs. 96
So, if C.P. is 8000, S.P. will be
= \(\frac{96}{100}\) × 8000 = 7680
∴ S.P. of VCR = Rs. 7680
8% profit on TV means
If C.P. is Rs. 100, S.P. is Rs. 108
So if C.P. is Rs. 8000, S.P. will be
\(\frac{108}{100}\) × 8000 = 8640
∴ S.P. of T.V. = Rs. 8640
Overall C.P. of TV and VCR
= 8000 + 8000 = 16,000
Overall S.P. of TV and VCR
= 7680 + 8640 = 16320
∵ Overall S.P. > Overall C.P.
⇒ Profit of Rs. (16320 – 16000)
= Rs. 320.

Question 6.
During a sale, a shop offered a discount of 10% on the marked prices of all the items. What would a customer have to pay for a pair of jeans marked at Rs. 1450 and two shirts marked at Rs. 850 each ?
Solution:
10% discount on marked price (M.P.)
⇒ Article of M.P. Rs. 100, has S.P. = Rs. 90
So, if article of M.P. Rs. 1450 has S.P.
= Rs.\(\frac{90}{100}\) × 1450
= Rs. 1305
and if the article of M.P. 1700 has S.P.
= Rs. \(\frac{90}{100}\) × 1700 = Rs. 1530
Two shirts M.P. = 850 × 2 = Rs. 1700
So customer have to pay for Jeans and two shirts
= Rs. 1305 + Rs. 1530 = Rs. 2835.

Question 7.
A milkman sold two of his buffaloes for Rs. 20,000 each. On one he made a gain of 5% and on the other a loss of 10%. Find his overall gain or loss.
[Hint: Find C.P. of each]
Solution:
S.P. of each buffalo = Rs. 20000
5% gain means
⇒ a profit of Rs. 5 on C.P. of Rs. 100
⇒ If S.P. is of Rs. 105, C.P. is Rs. 100
∴ If S.P. is Rs. 20,000
C.P. = Rs. \(\frac{100}{105}\) × 20,000 = 19048
∴ C.P. of one of buffalo = Rs. 19048
10% loss on other buffalo means
If S.P. is 90, C.P. is Rs.. 100
So, if S.P. is Rs. 20,000,
C.P. = Rs. \(\frac{100}{90}\) × 20,000
= Rs. 22222
∴ C.P. of other buffalo = Rs. 22222
∴ Overall S.P. = Rs. 20,000 + Rs. 20000
= Rs. 40,000
Overall C.P. = Rs. 19048 + Rs. 22222
= Rs. 41270
∵ Overall S.P. > Overall C.P.
⇒ Profit of Rs. (41270 – 40000) = Rs. 1270
∴ Net profit = Rs. 1270.

Question 8.
The price of a TV is Rs. 13,000. The sales tax charged on it is at the rate of 12%. Find the amount that Vinod will have to pay if he buys it.
Solution:
Price of TV = Rs. 13,000
Sale tax charge = 12%
∴ Sale tax on T.V.= 12% of Rs. 13,000
= Rs. \(\frac{12}{100}\) × 13000
= Rs. 1560
∴ Amound paid by Vinod for TV
= Rs. 13000 + Rs. 1560
= Rs. 14560.

Question 9.
Arun bought a pair of skates of a sale where the discount given was 20%. If the amount he pays is Rs. 1,600, find the marked price.
Solution:
20% discount means
For M.P. of Rs. 100, S.P.
= Rs. (100 – 20) = Rs. 80
So, if S.P. = Rs. 80, M.P. is Rs. 100
If S.P. = Rs. 1600, M.P. will be Rs. \(\frac{100}{80}\) × 1600
= 2000.
M.P. of skates Rs. 2000.

Question 10.
I purchased a hair-dryer for Rs. 5,400 including 8% VAT. Find the price before VAT was added.
Solution:
8% VAT included means.
Rs. 8 is added to original price of Rs. 100
⇒ If including VAT price is Rs. 108, original price = Rs. 100
So including VAT price is Rs. 5400, original price
= RS \(\frac{100}{108}\) × 5400
= Rs. 5000
∴ Price before VAT = Rs. 5000.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.1 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 8 Comparing Quantities Exercise 8.1

Question 1.
Find the ratio of the following :
(a) Speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.
(b) 5 m to 10 km.
(c) 50 paise to Rs. 5.
Solution:
(a) Ratio of speed of cycle to speed of scooter = 15 : 30 = 1 : 2

(b) Ratio of 5 m to 10 km
⇒ Ratio of 5 m to 100000 m = 5 : 100000 = 1 : 20000

(c) Ratio of 50 paise to Rs. 5
⇒ Ratio of 50 paise to 500 paise
= 50 : 500 = 1 : 10.

Question 2.
Convert the following ratios to percentages.
(а) 3 : 4
(b) 2 : 3
Solution:
(a) Ratio = 3 : 4
Fraction = \(\frac{3}{4}\)
Percentage = \(\frac{3 \times 25}{4 \times 25}\) = \(\frac{75}{100}\) = 75%

(b) Ratio = 2 : 3
Fraction = \(\frac{2}{3}\)
Percentage = \(\frac{2}{3} \times \frac{100}{100}\) = \(\frac{200}{3} \times \frac{1}{100}\)
= 66\(\frac{2}{3}\)%.

Question 3.
72% of 25 students are good in mathematics. How many are not good in mathematics ?
Solution:
Total percent good in mathematics and not good in mathematics = 100
72 + percentage of student not good in mathematics = 100
∴ Percentage of students not good in mathematics = 100 – 72 = 28
So number of students hot good in mathematics = 28% of 25
= \(\frac{28}{100}\) × 25 = 7 students.

Question 4.
A football team won 10 matches out of the total number of matches they played. If their win percentage was 40, then how many matches did they play in all ?
Solution:
Let the total number of matches played by football team be x their win was 40%.
So, 40% of x = 10
\(\frac{40}{100}\) × x = 10
x = \(\frac{10 \times 100}{40}\) = 25
Number of played matches = 25.

Question 5.
If Chameli had Rs. 600 left after spending 75% of her money, how much did she have in the beginning ?
Solution:
Let she had Rs. x in the beginning
Chameli spent 75% of her money.
So she left 100 – 75 = 25% of her moriey.
So, 25% of x = Rs. 600
\(\frac{25}{100}\) × x = 600
x = \(\frac{600 \times 100}{25}\) = 2400
Chameli had Rs. 2400 in the beginning.

Question 6.
If 60% people in a city like cricket, 30% like football and the remaining like other games, then what percent of the people like other games ? If the total number of people are .50 lakh, find the exact number who like each type of game.
Solution:
Total percent = 100
60% people like cricket
30% people like football
remaining like other game
So 100 – (60 + 30) = 10% like other game.
Total no. of people = 50 lakh
No. of people like cricket
= 60% of 50 lakh
= \(\frac{60}{100}\) × 50 lakh = 30 lakh
No. of people like football
= 30% of 50 lakh on
= \(\frac{30}{100}\) × 50 lakh = 15 lakh
No. of people like other game
= 10% of 50 .lakh
= \(\frac{10}{100}\) × 50 lakh = 5 lakh.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots Ex 7.2 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 7 Cube and Cube Roots Exercise 7.2

Question 1.
Find the cube root of each of the following numbers by prime factorisation method s
(i) 64
(ii) 512
(iii) 10648
(iv) 27000
(v) 15625
(vi) 13824
(vii) 110592
(viii) 46656
(ix) 175616
(x) 91125.
Solution:
(i) 64 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\)

= 2³ × 2³
= 2 × 2
∴ \(\sqrt[3]{64}\) = 4

(ii) 512 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\)

= 2³ × 2³ × 2³
= 2 × 2 × 2
∴ \(\sqrt[3]{512}\) = 8

(iii) 10648 = \(\underline{2 \times 2 \times 2}\) × \(\underline{11 \times 11 \times 11}\)

= 2³ × 11³
= 2 × 11
∴ \(\sqrt[3]{10648}\) = 22

(iv) 27000 = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5

= 2³ × 3³ × 5³
∴ \(\sqrt[3]{27000}\) = 2 × 3 × 5
= 30

(v) 15625 = 5 × 5 × 5 × 5 × 5 × 5

= 5³ × 5³
= 5 × 5
∴ \(\sqrt[3]{15625}\) = 25

(vi) 13824 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) ×  \(\underline{2 \times 2 \times 2}\) × \(\underline{3 \times 3 \times 3}\)

= 2³ × 2³ × 2³ × 3³
= 2 × 2 × 2 × 3
∴ \(\sqrt[3]{13824}\) = 24

(vii) 110592 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) ×  \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{3 \times 3 \times 3}\)

= 2³ × 2³ × 2³ × 2³ × 3³
= 2 × 2 × 2 × 2 × 3
∴ \(\sqrt[3]{110592}\) = 48

(viii) 46656 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{3 \times 3 \times 3}\) × \(\underline{3 \times 3 \times 3}\)

= 2³ × 2³ × 3³ × 3³
= 2 × 2 × 3 × 3
∴ \(\sqrt[3]{46656}\) = 36

(ix) 175616 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{7 \times 7 \times 7}\)

= 2³ × 2³ × 2³ × 7³
= 2 × 2 × 2 × 7
∴ \(\sqrt[3]{175616}\) = 56

(x) 91125 = \(\underline{3 \times 3 \times 3}\) × \(\underline{3 \times 3 \times 3}\) × \(\underline{5 \times 5 \times 5}\)

= 3³ × 3³ × 5³
= 3 × 3 × 5
∴ \(\sqrt[3]{91125}\) = 45

Question 2.
State true or false.
(i) Cube of any odd number is even
(ii) A perfect cube does not end with two zeros
(iii) If square of a number ends with 5, then its cube ends with 25.
(iv) There is no perfect cube which ends with 8.
(v) The cube of a two digit number may be a three digit number.
(vi) The cube of a two digit number may have seven or more digits.
(vii) The cube of a single digit number may be a single digit number.
Solution:
(i) False,
(ii) True,
(iii) True,
(iv) False,
(v) False,
(vi) False,
(vii) True

Question 3.
You ar told that 1,331 is a perfect cube. Can you guess without factorisation what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.
Solution:
The given number is 1331
Step 1. From groups of three starting from the rightmost digit of 1331 is 1331.In this case one group 331 has three digits whereas 1 has only two digits.

Step 2. Take 331.
The digit 1 is at its one’s place.
We take the one’s place of the required cube root as 1.

Step 3. Take the other group is 1.
Cube of 1 is 1 and cube of 2 is 8.1 lies between 0 and 8.
The smaller number among 1 and 2 is 1. The one’s place of 1 is 1 itself. Take 1 as ten’s place of the cube root of 1331.
Thus, \(\sqrt[3]{1331}\) = 11

Similarly, 4913
Step 1. 4913
Step 2. Take 913
The digit 3 is at its one’s place we take the one’s place of the required cube root as (3 × 3 × 3 = 27)7.

Step 3. Take the other group is 4. Cube of 1 is 1 and cube of 2 is 8. 4 lies between 1 and 8.
The smaller number among 1 and 2 is 1. The one’s place of 1 is 1 itself. Take 1 as ten’s place of the cube root of 4913.
Thus, \(\sqrt[3]{4913}\) = 17

Similarly, 12167
Step 1. 12167
Step 2. 7 × 7 × 7 = 343 i.e. one’s place is 3
Step 3. 12, 2 × 2 × 2 = 8 and 3 × 3 × 3 = 27 8 < 12 < 27
The smaller number among 2 and 3 is 2. The one’s place of 2 is 2 itself. Take 2 as tens place of the cube root of 12167.
Thus, \(\sqrt[3]{12167}\) = 23

Similarly, 32768
Step 1. 32768
Step 2. 8 × 8 × 8 = 512
i.e. one’s place is 2
Step 3. 32, 3 × 3 × 3 = 27
4 × 4 × 4 = 64
The smaller number along 3 and 4 is 3. The one’s place of 3 is 3 itself. Take 3 as ten’s place of the cube root 32768.
Thus \(\sqrt[3]{32768}\) = 32

Haryana State Board HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots Ex 7.1 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 7 Cube and Cube Roots Exercise 7.1

Question 1.
Which of the following numbers are not perfect cubes:
(i) 216
(ii) 128
(iii) 1000
(iv) 100
(v) 46656.
Solution:
(i) 216 = \(\underline{2 \times 2 \times 2}\) × \(\underline{3 \times 3 \times 3}\)

= 2³ × 3²
= (2 × 3)³
= (6)³
Which is a perfect cube.

(ii) 128 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × 2

= 2³ × 2³ × 2
∴ 2 does not appear in a group of three.
Hence, 128 is not a percent cube.

(iii) 1000 = 2 × 2 × 2 × 5 × 5 × 5

= 2 × 5
= 10
Which is a perfect cube.

(iv) 100 = 2 × 2 × 5 × 5

Prime factor of 100 is
2 × 2 × 5 × 5
So, 100 is not a perfect cube.

(v) 46656 = 2 × 2 × 2 × 2 × 2 × 2 × 9 × 9 × 9

= 2 × 2 × 9
= 36
Which is a perfect cube.

Question 2.
Find the smallest number by which each of the following number must be multified to obtain a perfect cube.
(i) 243
(ii) 256
(iii) 72
(iv) 675
(v) 100.
Solution:
(i) 243 = 3 × 3 × 3 × 3 × 3
The prime factor 3 × 3 = 9 does not appear in a group of three. Therefore 243 is not a perfect cube. To make it a cube, we need one more 3.
In that case,
243 × 3 = \(\underline{3 \times 3 \times 3}\) × \(\underline{3 \times 3 \times 3}\)
= 729,
which is a perfect cube.
Hence the required smallest number by which 243 should be multiplied to make a perfect is 3.

(ii) 256 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × 2 × 2
The prime factor 2 × 2 = 4 does not appear in a group of three. Therefore 256 is not a perfect cube. To make it cube, we need one more 2., In that case,
256 × 2 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\)
= 512
Hence the required smallest number by which 256 should be multiplied to make a perfect cube is 2.

(iii) 72 = \(\underline{2 \times 2 \times 2}\) × \(\underline{3 \times 3}\)
The prime factor 3 does not appear in a group of three. Therefore 72 is not a perfect cube. To make it a cube, we need one more 3. In that case
72 = \(\underline{2 \times 2 \times 2}\) × \(\underline{3 \times 3 \times 3}\)
or, 72 × 3
= 216,
which is a perfect cube.
Hence, the required smallest number by which 72 should be multiplied to make a perfect cube is 3.

(iv) 675 = \(\underline{3 \times 3 \times 3}\) × 5 × 5
The prime factor 5 does not appear in a group of three. Therefore 675 is not a perfect cube. To make it a cube, we need one more 5. In that case
675 × 5 = \(\underline{3 \times 3 \times 3}\) × \(\underline{5 \times 5 \times 5}\)
= 3375.
which is a perfect cube.
Hence, the required smallest number by which 675 should be multiplied to make a perfect cube is 5.

(v) 100 = \(\underline{2 \times 2}\) × \(\underline{5 \times 5}\)
The prime factor 2 and 5 do not appear in a group of three. Therefore 100 is not a perfect cube. To make it a cube, we need one more 2 and 5 respectively. In that case
100 × 10 = \(\underline{2 \times 2 \times 2}\) × \(\underline{5 \times 5 \times 5}\)
= 1000,
which is a perfect cube. Hence the required smallest number by which 100 should be multiplied to make a perfect cube is (2 × 5) = 10.

Question 3.
Find the smallest whole number by which each of the following numbers must be divided to obtain a perfect cube :
(i) 81
(ii) 128
(iii) 135
(iv) 192
(v) 704
Solution:
(i) 81 = \(\underline{3 \times 3 \times 3}\) × 3
The prime factor 3 does not appear in a group of three. So 81 is not a perfect cube. In the factorisation 3 appears only one time. So if we divide 81 by 3, then the prime factorisation of the quotient will not contain 3.
81 ÷ 3 = 3 × 3 × 3
Further the perfect cube in that case is 81 , 3 = 27. Hence the smallest whole number by which 81 should be divided to make it perfect cube is 3.

(ii) 128 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × 2
The prime factor 2 does not appear in a group of three. So 128 is not a perfect cube. In the factorisation 2 appear only one time. So if we divide 128 by 2, then the prime factorisation of the quotient will not contain 2
128 ÷ 2 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\)
Further the perfect cube in that case is
128 ÷ 2 = 64.
Hence the smallest whole number by which 128 should be divided to make it perfect cube is 2.

(iii) 135 = \(\underline{3 \times 3 \times 3}\) × 5
The prime Factor 5 × 5 = 25 does not appear in a group of three. So, 135 is not a Perfect Cube. In the factorisation 5 appears only one times. So if we divide 135 by 5, then the prime factorisation of the quotient will not contain 5.
135 ÷ 5 = 3 × 3 × 3
Further then perfect cube in that case is
135 ÷ 5 = 27
Hence the smallest whole number by which 135 should be divided to make it perfect cube is 5.

(iv) 192 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × 3
The prime factor 3 does not appear in a group of three, so, 192 is not a perfect cube. In the factorisation 3 appears only one times. So, if we divide 192 by 3, then the prime factorisation of the quotient will not contain 3.
192 ÷ 3 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\)
Further the perfect cube in that case is
(192 ÷ 3 = 64)
Hence, the smallest whole numbers by which 192 should be divided to make it perfect cube is 3.

(v) 704 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\) × 11
The prime factor 11 does not appear in a group of three. So 704 is not a perfect cube. In the factorisation 11 appears only one time. So if we divide 704 by 11, then the prime factorisation of the quotient will not contain 11.
704 ÷ 11 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\)
Further the perfect cube in that case is
704 ÷ 11 = 64.
Hence the smallest whole number by which 704 should be divided to make it perfect cube is 11.

Question 4.
Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboides will he need to form a cube?
Solution:
Parikshit’s cuboid = l × b × h
= 5 cm × 2 cm × 5cm
Now Parikshit’s volume of cube
= 5 cm × 5 cm × 25 cm
= 5 cm × 5 cm × 5 cm
The prime of factor of cuboid, 5 does not appear in a group of three. 50 cm3 is not a perfect cube. To make it cube, we need one more 5 cm. In that case,
Volume of cube = 5 cm × 5 cm × 5 cm
= 125 cm²
Hence the required smallest number by which 25 cm² should be multiplied to make a perfect cube is 5 cm.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 7 Cube and Cube Roots InText Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 7 Cube and Cube Roots InText Questions

Try These (Page 111)

Question 1.
Find the one’s digit of the cube of each of the following numbers :
(i) 3331
(ii) 8888
(iii) 149
(iv) 1005
(v) 1024
(vi) 77
(vii) 5022
(viii) 53
Solution:
We know, that the cube of a number ending in,
(a) 0, 1, 4, 5, 6 and 9 ends in 0, 1, 4, 5, 6 and 9 respectively.
(b) 2 ends in 8 and vice-versa.
(c) 3 or 7 ends in 7 or 3 respectively.
(i) ∵ 3331 ends in 1, therefore, its cube ends in 1.
Hence, the required unit’s digit of the cube is 1.

(ii) ∵ 8888 ends in 8, therefore, its cube ends in 2.
Hence, the required unit’s digit of the cube is 2.

(iii) ∵ 149 ends in 9, therefore, its cube ends is 9.
Hence, the required unit’s digit of the cube is 9.

(iv) ∵ 1005 ends in 5, therefore, its cube ends is 5.
Hence, the required unit’s digits of the cube is 5.

(v) ∵ 1024 ends is 4, therefore, its cube ends is 4.
Hence, the required unit’s digit of the cube is 4

(vi) ∵ 77 ends is 7, therefore, its cube ends is 3.
Hence, the required unit’s digits of the cube is 3.

(vii) ∵ 5022 ends is 2, therefore, its cube . ends is 8.
Hence, the required unit’s digits of the cube is 8.

(viii) ∵ 53 ends is 3, therefore, its cube ends is 7.
Hence, the required unit’s digits of the cube is 7.

Question 2.
Express the following numbers as the sum of odd numbers using the above pattern ?
(a) 6³
(b) 8³
(c) 7³
Solution:
Observe the following pattern of sums of odd numbers :
1 = 1 = 1³
3 + 5 = 8 = 2³
7 + 9 + 11 = 27 = 3³
13 + 15 + 17 + 19 = 64 = 4³
21 + 23 + 25 + 27 + 29 = 125 = 5³
31 + 33 + 35 + 37 + 39 + 41 = 216 = 6³
— — — — — — = 343 = 7³
— — — — — — = 512 = 8³
— — — — — — = 729 = 9³
— — — — — — = 1000 = 10³
According to above pattern, we have
(a) 6³ = 31 + 33 +35 + 37 + 39 + 41
= 216
(b) 8³ = 57 + 59 + 61 + 63 + 65 + 67 + 69+71
= 512
(c) 7³ = 43 + 45 + 47 + 49 + 51 + 53 + 55
= 343

Question 3.
Consider the following pattern.
2³ – 1³ = 1 + 2 × 1 × 3
3³ – 2³ = 1 + 3 × 2 × 3
4³ – 3³ = 1 + 4 × 3 × 3
Using the above pattern, find the value of the following:
(i) 7³ – 6²
(ii) 12³ – 11³
(iii) 20³ – 19³
(iv) 51³ – 50³
Solution:
(i) 7³ – 6² = 1 + 7 × 6 × 3
= 1 + 126 = 127

(ii) 12³ – 11³ = 1 + 12 × 11 × 3
= 1 + 396 = 397

(iii) 20³ – 19³ = 1 + 20 × 19 × 3
= 1 + 1140= 1141

(iv) 51³ – 50³ = 1 + 51 × 50 × 3
= 1 + 7650 = 7651

Try These (Page 112)

Question 1.
Which of the following are perfect cubes ?
1. 400
2. 3375
3. 8000
4. 15625
5. 9000
6. 6859
7. 2025
8. 10648
Solution:
1. 400 = 2 × 2 × 2 × 2 × 5 × 5
Prime factorisation of 400 is

2 × 2 × 2 × 2 × 5 × 5
So, 400 is not a perfect cube.

2. 3375 = \(\underline{3 \times 3 \times 3}\) × \(\underline{5 \times 5 \times 5}\)

= 3³ × 5³
= (3 × 5)³
= (15)³
which is a perfect cube.

3. 8000 = \(\underline{2 \times 2 \times 2}\) × \(\underline{2 \times 2 \times 2}\)

= 2³ × 2³ × 5³
= (2 × 2 × 5)³
= (20)³
which is a perfect cube.

4. 15625 = \(\underline{5 \times 5 \times 5}\) × \(\underline{5 \times 5 \times 5}\)

= 5³ × 3³
= (5 × 5)³
= (25)³
which is a perfect cube.

5. 9000 = 2 × 2 × 2 × 3 × 3 × 5 × 5 × 5

Prime factorisation of 9000 is
2 × 2 × 2 × 3 × 3 × 5 × 5 × 5
So, 9000 is not a perfect cube.

6. 6859 = 19 × 19 × 19

Prime factorisation of 6859
19 × 19 × 19
So, 6859 is a perfect cube.

7. 2025 = 3 × 3 × 3 × 3 × 5 × 5

Prime factorisation of 2025 is
3 × 3 × 3 × 3 × 5 × 5
So, 2025 is not a perfect cube.

8. 10648 = \(\underline{2 \times 2 \times 2}\) × \(\underline{11 \times 11 \times 11}\)

= 2³ × 11³
= (2 × 11)³
= (22)³
which is a perfect Cube.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 6 Square and Square Roots Ex 6.4 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 6 Square and Square Roots Exercise 6.4

Question 1.
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
Solution:

Question 2.
Find the number of digits in the square root of each of the following numbers (without any calculation).
(i) 64
(ii) 144
(iii) 4489
(iv) 27225
(v) 390625.
Solution:
(i) 64, Here n = 2 (even)
Number of digits in the square root of 64
= \(\frac{n}{2}\) = \(\frac{2}{2}\) = 1

(ii) Here, n = 3 (odd)
No. of digits in the square root of 144
= \(\frac{n+1}{2}\) = \(\frac{3+1}{2}\) = 2

(iii) 4489, Here n = 4 (even)
∴ No. of digits in the square root of 4489
= \(\frac{n}{2}\) = \(\frac{4}{2}\) = 2

(iv) 27225, Here n = 5 (odd)
No. of digits in the square root of 27225 = n+1 5+1
= \(\frac{n+1}{2}\) = \(\frac{5+1}{2}\) = 3

(v) 390625, Here n = 6 (even)
∴ The number of digits in the square root of
390625 = \(\frac{n}{2}\) = \(\frac{6}{2}\) = 3

Question 3.
Find the square root of the following decimal numbers :
(i) 2.56
(ii) 7.29
(iii) 51.84
(iv) 42.25
(v) 31.36
Solution:

Question 4.
Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 402
(ii) 1989
(iii) 3250
(iv) 825
(v) 4000.
Solution:
(i)
We get remainder 2. It shows that 20² is less than 402 by 2.
Thus, the number to be subtracted so as to make it a perfect a square is 2.
∴ Required perfect square
= 402 – 2 = 400
and \(\sqrt {400}\) = 20

(ii)
We get remainder 53. It shows that 44² is less than 1989 by 53.
Thus, the number to be subtracted so as to make it a perfect square is 3.
∴ Required perfect square
= 1989 – 53 = 1936
and \(\sqrt {1936}\) = 44.

(iii) We get the remainder 1.
It shows that 57² is less than 3250 by 1.
Thus, the number to be subtracted so as to make 3250 a perfect square is 1.

∴ Required number
= 3250 – 1 = 3249
and \(\sqrt {3249}\) = 57

(iv) We get the remainder 41.
It shows that 28² is less than 825 by 41.
Thus, the smallest number that should be subtracted to make it a perfect square is 41.

∴ Required number = 825 – 41 = 784
and \(\sqrt {784}\) = 28

(v) Since remainder 31.
∴ The smallest number that should be subtracted to make it a perfect square is 31.

and Required number
= 4000 – 31
= 3969
and \(\sqrt {3669}\) = 63.

Question 5.
Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 525
(ii) 1750
(iii) 252
(iv) 1825
(v) 6412.
Solution:
(i) Since remainderis 41.
∴ 22² < 525

Next perfect square number is 23² = 529
Hence, the number to be added is
23² – 525 = 529 – 525 = 4
∴ The required no. = 525 + 4 = 529
and \(\sqrt {529}\) = 23

(ii) It is clear by long division method of \(\sqrt {1750}\) that remainder is 69.

It shows that 41² < 1750
Next perfect square is 42²
= 1764
Hence, the number to be added is 42² – 1750
= 1764 – 1750 = 14.
∴ Required number
= 1750 + 14 = 1764
and \(\sqrt {1764}\) = 42

(iii) ∵ Remainder = 27
∴ 15² < 252 < 16².

Hence, the required number to be added to make it a perfect square
= 16² – 252
= 256 – 252 = 4
and \(\sqrt {256}\) = 16.

(iv) ∵ Remainder = 61

∴ 42² < 1825 < 43²
43² = 1849
Hence, the required number to be added to make it a perfect square
= 43² – 1825
= 1849 – 1825 = 24
∴ \(\sqrt {1849}\) = 43.

(v) Since remainder = 12.

∴ 80² < 6412 < 81²
∴ Required no. to be added to make it a perfect square is
81² – 6412 = 6561 – 6412
= 149
∴ \(\sqrt {6561}\) = 81.

Question 6.
Find the length of the side of a square whose area is 441 m².
Solution:

Let the side of square be x.
then, area of a square = x²
∴ x² = 441
⇒ x = \(\sqrt {441}\)
= 21 m.

Question 7.
In a right triangle ABC, ∠B = 90°.
(а) If AB = 6cm, BC = 8cm, findAC.
(б) If AC = 13 cm, BC = 5 cm, find AB.
Solution:
(i) Using Pythagoras theorem

AC² = AB² + BC²
= 6² + 8²
= 36 + 64
= 100
⇒ AC = \(\sqrt {100}\)
= 10 cm

(ii) AC = 13 cm
BC = 5 cm
AC² = AB² + BC²
(Using Pythagoras theorem)

(13)² = x² + (5)²
⇒ 169 = x² + 25
⇒ 169 – 25 = x²
⇒ 144 = x²
⇒ x = \(\sqrt {144}\) = 12 cm

Question 8.
A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this.
Solution:
By prime factorisation, we get
1000 = 2 × \(\underline{2 \times 2}\) × 5 × \(\underline{5 \times 5}\)
We see that 2 and 5 are still unpaired.

∴ We have to multiply 1000 by 2 × 5 i.e., 10 to make it a perfect square.
Required number of plants
∴ 1000 × 10 = 10000
Thus, he needed 10 more plants.

Question 9.
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 ?
Solution:
By prime factorisation, we get

500 = \(\underline{2 \times 2}\) × \(\underline{5 \times 5}\) × 5
Since 5 is not in pair.
∴ 5 children would be left out in this arrangement.