SELINA Solution Class 9 Compound interest (without using formula) Chapter 2 Exercise 2C

Question 1

A sum is invested at compound interest, compounded yearly. If the interest for two successive years is Rs. 5,700 and Rs. 7,410. calculate the rate of interest.

Sol:

Rate of interest = ${\displaystyle \frac{\text{Difference in the interest of the two consecutive periods}\times 100}{\text{C.I. of preceeding year}\times \text{Time}}\%}$

= ${\displaystyle \frac{(7410-5700)\times 100}{5700\times 1}\%}$

= 30 %

Question 2

A certain sum of money is put at compound interest, compounded half-yearly. If the interest for two successive half-years are Rs. 650 and Rs. 760.50; find the rate of interest.

Solution 1:

∵ Difference between the C.I. of two successive half-years
= Rs. 760.50 - Rs. 650= Rs. 110.50
⇒ Rs. 110.50 is the interest of one half-year on Rs. 650
∴ Rate of interest = Rs. ${\displaystyle \frac{100\times \text{I}}{\text{P}\times \text{T}}\%}$
= ${\displaystyle 650\times \frac{1}{2}}$ % 
= 34 %

Solution 2:

Let sum of money = P, Rate = r % half-yearly.

Simple interest for I half-year on sum = Rs. 650.

Simple interest for II half-year on sum = P + 650 = Rs. 760.50

∵ Difference between the C.I. of two successive half-years
= Rs. 760.50 - Rs. 650= Rs. 110.50

⇒ Rs. 110.50 is the interest of one half-year on Rs. 650.

Simple interest (S.I.) = ${\displaystyle \frac{P\times R\times T}{100}}$

∴ ${\displaystyle 110.50=\frac{650\times R\times 1}{100}}$

∴ 110.50 = 6.5 × R

∴ ${\displaystyle \frac{110.50}{6.5}}$ = R

∴ R = 17% Per half-yearly

∴ Rate of interest = 17 × 2 = 34% p.a.

Question 3.1

A certain sum amounts to Rs. 5,292 in two years and Rs. 5,556.60 in three years, interest being compounded annually. Find : the rate of interest.

Sol:

Amount in two years= Rs. 5,292
Amount in three years= Rs. 5,556.60
Difference between the amounts of two successive years
= Rs. 5,556.60 - Rs. 5,292 = Rs. 264.60

⇒ Rs. 264.60 is the interest of one year on Rs. 5,292

∴ Rate of interest = Rs. ${\displaystyle \frac{100\times \text{I}}{\text{P}\times \text{T}}\%}$
= ${\displaystyle \frac{100\times 264.60}{5,292\times 1}\%}$ 
= 5%

Question 3.2

A certain sum amounts to Rs. 5,292 in two years and Rs. 5,556.60 in three years, interest being compounded annually. Find: the original sum.

Sol:

Let the sum of money = Rs. 100
Interest on it for 1^st year= 5% of Rs. 100 = Rs. 5
⇒ Amount in one year= Rs. 100 + Rs. 5 = Rs. 105
Similarly, amount in two years = Rs. 105 + 5% of Rs. 105
= Rs. 105+ Rs. 5.25
= Rs. 110.25
When amount in two years is Rs. 110.25, sum = Rs. 100
⇒ When amount in two years is Rs. 5,292,
sum = Rs. ${\displaystyle \frac{100\times 5,292}{110.25}}$ = Rs. 4,800.

Question 4

The compound interest, calculated yearly, on a certain sum of money for the second year is Rs. 1,089 and for the third year it is Rs. 1,197.90. Calculate the rate of interest and the sum of money.

Sol:

(i) C.I. for second year = Rs. 1,089
C.I. for third year = Rs. 1,197.90
∵ Difference between the C.I. of two successive years
= Rs. 1,197.90 - Rs. 1089 = Rs. 108.90
⇒ Rs. 108.90 is the interest of one year on Rs.1089.

∴ Rate of interest = Rs. ${\displaystyle \frac{100\times \text{I}}{\text{P}\times \text{T}}}$ %

                             = ${\displaystyle \frac{100\times 108.90}{1089\times 1}}$ % = 10%

(ii) Let the sum of money = Rs.100
∴ Interest on it for 1^st year = 10% of Rs.100= Rs.10

⇒ Amount in one year = Rs. 100 + Rs. 10 = Rs. 110
Similarly, C.I. for 2^nd year = 10% of Rs. 110 = Rs. 11
When C.I. for 2^nd year is Rs. 11, sum = Rs. 100
When C.I. for 2^nd year is Rs. 1089, sum = Rs. ${\displaystyle \frac{100\times 1089}{11}}$
= Rs. 9,900.

Question 5

Mohit invests Rs. 8,000 for 3 years at a certain rate of interest, compounded annually. At the end of one year it amounts to Rs. 9,440. Calculate : 
(i) the rate of interest per annum.
(ii) the amount at the end of the second year.
(iii) the interest accrued in the third year.

Sol:

For 1^st year
P = Rs. 8,000; A = 9,440 and T= 1 year
Interest = Rs. 9,440 - Rs. 8,000 = Rs. 1,440
Rate = ${\displaystyle \frac{\text{I}\times 100}{\text{P}\times \text{T}}\%}$

= ${\displaystyle \frac{1,440\times 100}{8,000\times 1}}$% = 18%

For 2^nd year
P= Rs. 9,440; R = 18% and T= 1 year

Interest = Rs ${\displaystyle \frac{9,440\times 18\times 1}{100}}$= Rs. 1,699.20

Amount = Rs. 9,440 + Rs. 1,699.20 = Rs. 11,139.20

For 3^rd year

P = Rs. 11,139.20; R = 18 % and T= 1year

Interest = Rs. ${\displaystyle \frac{11,139.20\times 18\times 1}{100}}$= Rs. 2,005.06

Question 6

Geeta borrowed Rs. 15,000 for 18 months at a certain rate of interest compounded semi-annually. If at the end of six months it amounted to Rs. 15,600; calculate :
(i) the rate of interest per annum.
(ii) the total amount of money that Geeta must pay at the end of 18 months in order to clear the account.

Sol:

For 1^st half - year :
P= Rs. 15,000; A= Rs. 15,600 and T = ½ year
Interest = Rs. 15,600 - Rs. 15,000= Rs. 600

Rate= `["I" xx 100 ]/["P" xx "T"] %

= [600 xx 100]/[15,000 xx 1/2]` % = 8% .

For 2^nd half - year : 
P = Rs. 15,600; R = 8% and T = ${\displaystyle \frac{1}{2}}$ year

Interest = Rs. ${\displaystyle \frac{15,600\times 8\times \frac{1}{2}}{100}}$ = Rs. 624

Amount = Rs. 15,600 + Rs. 624 = Rs. 16,224

For 3^rd half - year :

P = Rs. 16,224; R = 8 % and T = ${\displaystyle \frac{1}{2}}$ year

Interest = Rs. ${\displaystyle \frac{16,224\times 8\times \frac{1}{2}}{100}}$ = Rs. 648.96

Amount = Rs. 16,224 + Rs. 648.96 = Rs. 16,872.96.

Question 7

For 1^st half - year :
P= Rs. 15,000; A= Rs. 15,600 and T = ½ year
Interest = Rs. 15,600 - Rs. 15,000= Rs. 600

Rate= `["I" xx 100 ]/["P" xx "T"] %

= [600 xx 100]/[15,000 xx 1/2]` % = 8% .

For 2^nd half - year : 
P = Rs. 15,600; R = 8% and T = ${\displaystyle \frac{1}{2}}$ year

Interest = Rs. ${\displaystyle \frac{15,600\times 8\times \frac{1}{2}}{100}}$ = Rs. 624

Amount = Rs. 15,600 + Rs. 624 = Rs. 16,224

For 3^rd half - year :

P = Rs. 16,224; R = 8 % and T = ${\displaystyle \frac{1}{2}}$ year

Interest = Rs. ${\displaystyle \frac{16,224\times 8\times \frac{1}{2}}{100}}$ = Rs. 648.96

Amount = Rs. 16,224 + Rs. 648.96 = Rs. 16,872.96.

Sol:

For 1^st year :
P = Rs. 12,800; R = 10 % and T = 1 year

Interest = Rs. ${\displaystyle \frac{12,800\times 10\times 1}{100}}$ = Rs. 1,280.

Amount = Rs. 12,800 + Rs. 1,280 = Rs. 14,080.

For 2^nd year :
P = Rs. 14,080; R = 10 % and T = 1 year

Interest = Rs. ${\displaystyle \frac{14,080\times 10\times 1}{100}}$ = Rs. 1,408.

Amount = Rs. 14,080 + Rs. 1,408 = Rs. 15,488

For 3^rd year :
P = Rs. 15,488; R = 10 % and T = 1 year

Interest = Rs. ${\displaystyle \frac{15,488\times 10\times 1}{100}}$ = Rs. 1,548.80

Amount = Rs. 15,488 + Rs. 1,548.80 = Rs. 17,036.80

Question 8

Rs. 8,000 is lent out at 7% compound interest for 2 years. At the end of the first year Rs. 3,560 are returned. Calculate :
(i) the interest paid for the second year.
(ii) the total interest paid in two years.
(iii) the total amount of money paid in two years to clear the debt.

Sol:

(i) For 1^st year : 
P = Rs. 8,000; R = 7 % and T = 1 year

Interest = Rs. ${\displaystyle \frac{8,000\times 7\times 1}{100}}$ = Rs. 560.

Amount = Rs. 8,000 + Rs. 560 = Rs. 8,560
Money returned = Rs. 3,560
Balance money for 2^nd year= Rs. 8,560 - Rs. 3,560 = Rs. 5,000

For 2^nd year :
P = Rs. 5,000; R = 7 % and T = 1 year.

Interest paid for the second year = Rs. ${\displaystyle \frac{5,000\times 7\times 1}{100}}$ 
= Rs. 350

(ii) The total interest paid in two years= Rs. 350 + Rs. 560 = Rs. 910

(iii) The total amount of money paid in two years to clear the debt

= Rs. 8,000+ Rs. 910 = Rs. 8,910

Question 9

The cost of a machine depreciated by Rs. 4,000 during the first year and by Rs. 3,600 during the second year. Calculate :

1. The rate of depreciation.
2. The original cost of the machine.
3. Its cost at the end of the third year.

Sol:

(i) Difference between depreciation in value between the first and second years Rs. 4,000 - Rs. 3,600 = Rs. 400.
⇒ Depreciation of one year on Rs. 4,000 = Rs. 400

⇒ Rate of depreciation = ${\displaystyle \frac{400}{4000} \times 100\%}$ = 10%

(ii) Let Rs.100 be the original cost of the machine.
Depreciation during the 1^st year = 10% of Rs.100 = Rs.10
When the values depreciates by Rs.10 during the 1^st year, Original cost = Rs.100
⇒ When the depreciation during 1^st year = Rs. 4,000

Original Cost = ${\displaystyle \frac{100}{10}\times 4000}$ = Rs. 40,000

The original cost of the machine is Rs. 40,000.

(iii) Total depreciation during all the three years
= Depreciation  in value during(1^st year + 2^nd year + 3^rd year)
= Rs. 4,000 + Rs. 3,600 + 10% of (Rs. 40,000 - Rs. 7,600)
= Rs. 4,000 + Rs. 3,600 + Rs. 3,240
= Rs.10,840

The cost of the machine at the end of the third year
= Rs. 40,000 - Rs.10,840 = Rs. 29,160

Question 10

Find the sum, invested at 10% compounded annually, on which the interest for the third year exceeds the interest of the first year by Rs. 252.

Sol:

Let the sum of money be Rs.100.
Rate of interest = 10% p.a.
Interest at the end of 1^st year = 10% of Rs. 100 = Rs. 10
Amount at the end of 1^st year = Rs. 100 + Rs. 10 = Rs. 110
Interest at the end of 2^nd year = 10% of Rs. 110 = Rs. 11
Amount at the end of 2^nd year = Rs. 110 + Rs. 11 = Rs.121
Interest at the end of 3^rd year =10% of Rs. 121= Rs. 12.10
Difference between interest of 3^rd year and 1^st year
= Rs. 12.10 - Rs. 10 = Rs. 2.10
When difference is Rs. 2.10, principal is Rs. 100
When difference is Rs. 252, principal = ${\displaystyle \frac{100\times 252}{2.10}}$ = Rs.12,000.

Question 11

A man borrows Rs.10,000 at 10% compound interest compounded yearly. At the end of each year, he pays back 30% of the sum borrowed. How much money is left unpaid just after the second year ?

Sol:

For 1^st year :
P = Rs. 10,000; R = 10% and T = 1 year

Interest = Rs. ${\displaystyle \frac{10,000\times 10\times 1}{100}}$= Rs.1,000

Amount at the end of 1^st year = Rs. 10,000 + Rs. 1,000 = Rs. 11,000

Money paid at the end of 1^st year = 30% of Rs. 10,000 = Rs. 3,000

∴ Principal for 2^nd year = Rs. 11,000 - Rs. 3,000 = Rs. 8,000

For 2^nd year :

P = Rs. 8,000; R = 10% and T = 1 year

Interest = Rs. ${\displaystyle \frac{8,000\times 10\times 1}{100}}$ = Rs. 800

Amount at the end of 2^nd year = Rs. 8,000 + Rs. 800 = Rs. 8,800

Money paid at the end of 2^nd year = 30% of Rs. 10,000 = Rs. 3,000

∴ Principal for 3^rd year = Rs. 8,800 - Rs. 3,000 =Rs. 5,800.

Question 12

A man borrows Rs.10,000 at 10% compound interest compounded yearly. At the end of each year, he pays back 20% of the amount for that year. How much money is left unpaid just after the second year ?

Sol:

For 1^st year :

P = Rs. 10,000; R = 10% and T = 1 year

Interest = Rs. ${\displaystyle \frac{10,000\times 10\times 1}{100}}$ = Rs. 1,000

Amount at the end of 1^st year = Rs. 10,000 + Rs. 1,000 = Rs. 11,000

Money paid at the end of 1^st year = 20% of Rs. 11,000 = Rs. 2,200

∴ Principal for 2^nd year = Rs. 11,000 - Rs. 2,200 = Rs. 8,800

For 2^nd year :

P = Rs. 8,800; R = 10% and T= 1 year

Interest = Rs. ${\displaystyle \frac{8,800\times 10\times 1}{100}}$= Rs. 880

Amount at the end of 2^nd year = Rs. 8,800 + Rs. 880 = Rs. 9,680

Money paid at the end of 2^nd year = 20% of Rs. 9,680 = Rs.1,936

∴ Principal for 3^rd year =Rs. 9,680 - Rs. 1,936 = Rs. 7,744.

SChand New Learning Composite Mathematics Class 6 Chapter 2 Whole Numbers Exercise 2C

 Exercise 2C

Question 1

SChand Composite Mathematics Class 7 Chapter 2 Fractions Exercise 2C

 Exercise 2C

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Question 1

$\frac{3}{5}$ of a road is to be resurfaced. If the road is $2 \frac{3}{5} \mathrm{~km}$ long, what length in kilometres is to be resurfaced?

Sol :

$=\frac{3}{5} \times 2 \frac{3}{5}$

$=\frac{3}{5} \times \frac{2 \times 5+3}{5}=\frac{3}{5} \times \frac{13}{5}$

$=\frac{39}{25}=1\frac{14}{25}$

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Question 2

You spent $1 \frac{1}{2}$ hours on homework last night and $\frac{1}{4}$ of that time was spent on mathematics. What fraction of hour did you spent on math's?

Sol :

$=\frac{1}{4} \text{ of } \frac{3}{2}$

$=\frac{1}{4} \times \frac{3}{2}=\frac{3}{8}$ hour

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Question 3

Cost of 1 litre Petrol is $₹ 42 \frac{4}{7}$. What is the cast of $10 \frac{1}{2}$ litres petrol?

Sol :

$=₹ 42 \frac{4}{7}=\frac{42 \times 7+4}{7}$

$=\frac{298}{7} \times \frac{21}{2}$

=149×3=447

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Question 4

After covering $\frac{3}{4}$ of my Journey, I find that 15 km is still left. How much distance in the total journey?

Sol :

Let the total journey be x.

Journey covered$=\frac{3}{4}$ of total journey is $\frac{3x}{4}$

A.T.Q

Total journey-Journey covered=15 km

$x-\frac{3x}{4}=15$

$\frac{4x-3x}{4}=15$

$\frac{x}{4}=15$

x=15×4=60 km

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Question 5

In a school, $\frac{4}{9}$ of the students are boys and the number of girls is 775 . Find the number of boys in the school.

Sol :

Let the total students be x.

Number of boys$=\frac{4}{9} \text{ of } x=\frac{4x}{9}$

Number of girls = 775

Total students=Number of boys + Number of girls

$x=\frac{4x}{9}+775$

$x-\frac{4x}{9}=775$

$frac{9x-4x}{9}=775$

5x=775×9

$x=\frac{6975}{5}$

Total students=x=1395

Number of boys$=\frac{4}{9} \text{ of } 1395=\frac{4}{9} \times 1395$

=620

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Question 6

A tank is $\frac{3}{5}$ full of water. 50 litres more are required to fill it up. How many litres can the tank hold?

Sol :

Let the tank hold x litres of water.

Tank is filled= $\frac{3}{5}$ of x $=\frac{3x}{5}$

A.T.Q

Tank filled+50 litres=Total 

$\frac{3x}{5}+50=x$

$x-\frac{3x}{5}=50$

$\frac{5x-3x}{5}=50$

2x=50×5

$x=\frac{250}{2}=125$ litres

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Question 7

A Sum of money was divided between Rajeev and Ashish on so that Rajeev gets $\frac{7}{19}$ of the whole amount. If he gets 5600 , how much was the total sum.

Sol :

Let the total money be x

Rajeev receives$=\frac{7}{19}$ of total$=\frac{7x}{19}$

Also,$\frac{7x}{19}=5600$

7x=5600×19

$x=\frac{5600 \times 19}{7}$

Total money=x=15200

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Question 8

Mr Joshi bequeath $\frac{1}{3}$ of his money to his son, $\frac{1}{5}$ to his daughter and the remaining to his wife who got 42000. What was the total amount?

Sol :

Total money be x

Also, Total money=Money received by son + Money received by daughter + Money received by wife

x=$\frac{1}{3}$ of x + $\frac{1}{5}$ of x + 42000

$x=\frac{1x}{3}+\frac{1x}{5}+42000$

$x-\frac{1x}{3}-\frac{1x}{5}=42000$

$\frac{15x-5x-3x}{15}=42000$

7x=42000×15

$x=\frac{42000 \times 15}{7}$

Total money=x=6000×15=90000

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Question 9

A boy spends $\frac{3}{4}$ of his pocket money and then $\frac{4}{5}$ of the remainder is given to his sister. If he has 40 left , what did he have at first?

Sol :

Total pocket money be x.

Boy spends$=\frac{3}{4}$ of x$=\frac{3x}{4}$

Remaining=$x-\frac{3x}{4}=\frac{4x-3x}{4}=\frac{x}{4}$

Sister receives$\frac{4}{5}$ of remaining $=\frac{4}{5} \times \frac{x}{4}=\frac{x}{5}$

Remaining - Sister received amount=40

$\frac{x}{4}-\frac{x}{5}=40$

$\frac{5x-4x}{20}=40$

$\frac{x}{20}=40$

Total pocket money be x=40×20=800

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Question 10

A magazine has 60 Pages of which $\frac{7}{12}$ are for ad. , $\frac{4}{15}$ have only photographs on them and the rest are for articles and stories. How many pages have articles and stories on them?

Sol :

Total pages=Pages for ads+ Photographs + Articles and stories

60=$\frac{7}{12}$ of 60+$\frac{4}{15}$ of 60 + x

$60=\frac{7}{12} \times 60+ \frac{4}{15} \times 60$ +x

60=7×5+4×4+x

60=35+16+x

60=51+x

x=60-51=9 pages for Articles and stories

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Question 11

Steve is planning to take 3 loaves of bread. Each loaf cost for $5\frac{1}{4}$ cups of flour. He knows he has 20 cups on hand . Will he have enough flour left for a cake recipe that requires $3 \frac{5}{4}$ cups?

Sol :

Loaf cost $5\frac{1}{4}=\frac{5 \times 4+1}{4}=\frac{21}{4}$ cups of flour

To make 3 loaves of bread it requires $=3 \times \frac{21}{4}=\frac{63}{4}$

Steve has 20 cups, so what is left is:

$20-\frac{63}{4}=\frac{20 \times 4-63}{4}=\frac{80-63}{4}=\frac{17}{4}$

Cake recipe that requires $\frac{3 \times 4+5}{4}=\frac{17}{4}$

Answer is yes.

S.chand publication New Learning Composite mathematics solution of class 7 Chapter 2 Fractions and Decimals Exercise 2C

Exercise 2C

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Question 1

Simplify:

(a) $\frac{3}{4}\times \frac{12}{17}$

Sol :

$=\frac{9}{17}$

(b) $\frac{7}{9}\times \frac{18}{19}$

Sol :

$=\frac{14}{19}$

(c) $\frac{8}{15}\times \frac{5}{16}$

Sol :

$=\frac{1}{6}$

(d) $\frac{5}{26}\times \frac{39}{40}$

Sol :

$=\frac{3}{16}$

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Question 2

Multiply and write each answer in simplest form.

(a) $\frac{9}{16}$ by $\frac{8}{15}, \frac{10}{27}, 1 \frac{7}{9}, 3 \frac{5}{9}$

(i)

$\frac{9}{16}\times \frac{8}{15}$

$=\frac{3}{10}$

(ii)

$\frac{9}{16}\times \frac{10}{27}$

$=\frac{5}{24}$

(iii)

$\frac{9}{16}\times 1\frac{7}{9}$

$=\frac{9}{16}\times \frac{16}{9}$

=1

(iv)

$\frac{9}{16}\times 3\frac{5}{9}$

$=\frac{9}{16}\times \frac{32}{9}$

=2

(b) $1 \frac{3}{5}$ by $\frac{5}{8}, 1 \frac{1}{4}, \frac{15}{16}, 3 \frac{3}{4}$

(i)

$1\frac{3}{5}\times \frac{5}{8}$

$=\frac{8}{5}\times \frac{5}{8}$

=1

(ii)

$1\frac{3}{5}\times 1 \frac{1}{4}$

$=\frac{8}{5}\times \frac{5}{4}$

=2

(iii)

$1\frac{3}{5}\times \frac{15}{16}$

$=\frac{8}{5}\times \frac{15}{16}$

$=\frac{3}{2}=1\frac{1}{2}$

(iv)

$1\frac{3}{5}\times 3\frac{3}{4}$

$=\frac{8}{5}\times \frac{15}{4}$

=6

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Question 3

Simplify the following.

(a) $2\frac{3}{8}\times 5\frac{1}{5}$

Sol :

$=\frac{19}{8}\times \frac{26}{5}$

$=\frac{247}{20}=12\frac{7}{20}$

(b) $2\frac{2}{9}\times \frac{3}{5}$

Sol :

$=\frac{20}{9}\times \frac{3}{5}$

$=\frac{4}{3}=1\frac{1}{3}$

(c) $1\frac{1}{2}\times 2\frac{2}{5}$

Sol :

$=\frac{3}{2}\times \frac{12}{5}$

$=\frac{18}{5}=3\frac{3}{5}$

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Question 4

(a) $2 \times \frac{4}{5} \times 1 \frac{2}{3}$

$2\frac{4}{5}\times \frac{5}{3}$

$=\frac{8}{3}$

(b) $2\frac{5}{6}\times \frac{18}{85}\times 3\frac{8}{9}$

$=\frac{17}{6}\times \frac{18}{85}\times \frac{35}{9}$

$=\frac{7}{3}=2\frac{1}{3}$

(c) $1\frac{2}{7}\times 3\times 2\frac{5}{8}$

$=\frac{9}{7}\times 3\times \frac{21}{8}$

$=\frac{81}{8}=10\frac{1}{8}$

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Question 5

Ravi used $1 \frac{2}{5}$ bags of soil for his garden. He is digging another garden that will need $\frac{1}{5}$ as much soil a original. How much soil will he use in total?

Sol :

Ravi used soil in total$=\left(1\frac{2}{5}\times \frac{1}{5}\right)+1\frac{2}{5}$

$=\left(\frac{7}{5}\times \frac{1}{5}\right)+\frac{7}{5}$

$=\frac{7}{25}+\frac{7}{5}$

$=\frac{7+35}{25}=\frac{42}{25}$

$=1\frac{17}{25}$

RS Aggarwal solution class 8 chapter 2 Exponents Exercise 2C

Exercise 2C

Page-37

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Question 1:

Tick (✓) the correct answer
The value of ${\left(\frac{2}{5}\right)}^{-3}$ is
(a) $-\frac{8}{125}$
(b) $\frac{25}{4}$
(c) $\frac{125}{8}$
(d) $-\frac{2}{5}$

Answer 1:

(c) $\frac{125}{8}$

${\left(\frac{2}{5}\right)}^{-3}={\left(\frac{5}{2}\right)}^3=\frac{5^3}{2^3}=\frac{125}{8}$

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Q2 | Ex-2C | Exponents | Class 8 | RS AGGARWAL | Chapter 2 | myhelper

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Question 2:

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The value of (−3)⁻⁴ is
(a) 12
(b) 81
(c) $-\frac{1}{12}$
(d) $\frac{1}{81}$

Answer 2:

(d) $\frac{1}{81}$

${\left(-3\right)}^{-4}=\frac{1}{{\left(-3\right)}^4}=\frac{1}{{\left(-1\right)}^4\times {\left(3\right)}^4}=\frac{1}{{\left(3\right)}^4}=\frac{1}{81}$

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Q3 | Ex-2C | Exponents | Class 8 | RS AGGARWAL | Chapter 2 | myhelper

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Question 3:

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The value of (−2)⁻⁵ is
(a) −32
(b) $\frac{-1}{32}$
(c) 32
(d) $\frac{1}{32}$

Answer 3:

(b) $\frac{-1}{32}$

${\left(-2\right)}^{-5}=\frac{1}{{\left(-2\right)}^5}=\frac{1}{-32}=\frac{1\times \left(-1\right)}{-32\times \left(-1\right)}=\frac{-1}{32}$

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Q4 | Ex-2C | Exponents | Class 8 | RS AGGARWAL | Chapter 2 | myhelper

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Question 4:

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(2⁻⁵ ÷ 2⁻²) = ?
(a) $\frac{1}{128}$
(b) $\frac{-1}{128}$
(c) $-\frac{1}{8}$
(d) $\frac{1}{8}$

Answer 4:

(d) $\frac{1}{8}$

$\left(2^{-5} \div 2^{-2}\right)=\left(\frac{1}{2^{5}} \div \frac{1}{2^{2}}\right)$

$=\left(\frac{1}{32} \div \frac{1}{4}\right)=\left(\frac{1}{32} \times 4\right)$

$=\frac{4}{32}=\frac{1}{8}$

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Q5 | Ex-2C | Exponents | Class 8 | RS AGGARWAL | Chapter 2 | myhelper

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Question 5:

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The value of (3⁻¹ + 4⁻¹)⁻¹ ÷ 5⁻¹ is
(a) $\frac{7}{10}$
(b) $\frac{60}{7}$
(c) $\frac{7}{5}$
(d) $\frac{7}{15}$

Answer 5:

(b) $\frac{60}{7}$

$\left(3^{-1}+4^{-1}\right)^{-1} \div 5^{-1}$

$=\left(\frac{1}{3}+\frac{1}{4}\right)^{-1} \div \frac{1}{5}$

$=\left(\frac{4+3}{12}\right)^{-1} \div \frac{1}{5}=\left(\frac{7}{12}\right)^{-1} \div \frac{1}{5}$

$=\left(\frac{12}{7}\right) \div \frac{1}{5}=\frac{12}{7} \times 5=\frac{60}{7}$

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Q6 | Ex-2C | Exponents | Class 8 | RS AGGARWAL | Chapter 2 | myhelper

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Question 6:

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${\left(\frac{1}{2}\right)}^{-2}+{\left(\frac{1}{3}\right)}^{-2}+{\left(\frac{1}{4}\right)}^{-2}=?$
(a) $\frac{61}{144}$
(b) $\frac{144}{61}$
(c) 29
(d) $\frac{1}{29}$

Answer 6:

(c) 29

$$\begin{gathered} {\left(\frac{1}{2}\right)}^{-2}+{\left(\frac{1}{3}\right)}^{-2}+{\left(\frac{1}{4}\right)}^{-2}= {\left(\frac{2}{1}\right)}^2+{\left(\frac{3}{1}\right)}^2+{\left(\frac{4}{1}\right)}^2 \\ =2^2+3^2+4^2 \\ =4+9+16 \\ =29 \end{gathered}$$

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Q7 | Ex-2C | Exponents | Class 8 | RS AGGARWAL | Chapter 2 | myhelper

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Question 7:

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$\left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3}=?$
(a) $\frac{19}{64}$
(b) $\frac{27}{16}$
(c) $\frac{64}{19}$
(d) $\frac{16}{25}$

Answer 7:

(a) $\frac{19}{64}$

$$\begin{gathered} \left\{{\left(\frac{1}{3}\right)}^{-3}-{\left(\frac{1}{2}\right)}^{-3}\right\}÷{\left(\frac{1}{4}\right)}^{-3} \\ =\left\{3^3-2^3\right\}÷4^3 \\ =\left\{27-8\right\}÷64 \\ =19÷64 \\ =\frac{19}{64} \end{gathered}$$

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Q8 | Ex-2C | Exponents | Class 8 | RS AGGARWAL | Chapter 2 | myhelper

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Question 8:

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${\left[{\left\{{\left(-\frac{1}{2}\right)}^2\right\}}^{-2}\right]}^{-1}=?$
(a) $\frac{1}{16}$
(b) 16
(c) $-\frac{1}{16}$
(d) −16

Answer 8:

(a) $\frac{1}{16}$

$$\begin{gathered} {\left[{\left\{{\left(-\frac{1}{2}\right)}^2\right\}}^{-2}\right]}^{-1} \\ ={\left[{\left\{-\frac{1}{2}\right\}}^{-4}\right]}^{-1} \\ ={\left(-\frac{1}{2}\right)}^{(-4\times -1)} \\ ={\left(-\frac{1}{2}\right)}^4 \\ =\frac{1}{16} \end{gathered}$$

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Q9 | Ex-2C | Exponents | Class 8 | RS AGGARWAL | Chapter 2 | myhelper

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Question 9:

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The value of x for which ${\left(\frac{7}{12}\right)}^{-4}\times {\left(\frac{7}{12}\right)}^{3x}={\left(\frac{7}{12}\right)}^5,$ is
(a) −1
(b) 1
(c) 2
(d) 3

Answer 9:

(d) 3

$$\begin{gathered} {\left(\frac{7}{12}\right)}^{-4}\times {\left(\frac{7}{12}\right)}^{3x}={\left(\frac{7}{12}\right)}^5 \\ \Rightarrow {\left(\frac{7}{12}\right)}^{-4+3x}={\left(\frac{7}{12}\right)}^5 \\ \Rightarrow 3x-4=5 \\ 3x=9 \\ \mathrm{or} x=\frac{9}{3}=3 \end{gathered}$$

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Q10 | Ex-2C | Exponents | Class 8 | RS AGGARWAL | Chapter 2 | myhelper

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Question 10:

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If (2^{3x − 1} + 10) ÷ 7 = 6, then x is equal to
(a) −2
(b) 0
(c) 1
(d) 2

Answer 10:

(d) 2

$$\begin{gathered} \left(2^{3x-1}+10\right)÷7=6 \\ \Rightarrow \frac{\left(2^{3x-1}+10\right)}{7}=\frac{6}{1} \\ \mathrm{On} \mathrm{cross} \mathrm{multiplying}: \\ \left(2^{3x-1}+10\right)\times 1=6\times 7=42 \end{gathered}$$

⇒ 2^3x-1  = 42 -10
⇒2^3x-1 = 32
⇒ 2^3x-1  = 2⁵
⇒ 3x-1 = 5
⇒ 3x = 6
Therefore, x =  2

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Q11 | Ex-2C | Exponents | Class 8 | RS AGGARWAL | Chapter 2 | myhelper

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Question 11:

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${\left(\frac{2}{3}\right)}^0=?$
(a) $\frac{3}{2}$
(b) $\frac{2}{3}$
(c) 1
(d) 0

Answer 11:

(c) 1

Using the law of exponents ${\left(\frac{a}{b}\right)}^0=1$:
$\therefore$ ${\left(\frac{2}{3}\right)}^0=1$

Page-38

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Q12 | Ex-2C | Exponents | Class 8 | RS AGGARWAL | Chapter 2 | myhelper

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Question 12:

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${\left(\frac{-5}{3}\right)}^{-1}=?$
(a) $\frac{5}{3}$
(b) $\frac{3}{5}$
(c) $\frac{-3}{5}$
(d) none of these

Answer 12:

(c) $\frac{-3}{5}$

${\left(\frac{-5}{3}\right)}^{-1}={\left(\frac{3}{-5}\right)}^1=\frac{3}{-5}=\frac{3\times \left(-1\right)}{-5\times \left(-1\right)}=\frac{-3}{5}$

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Q13 | Ex-2C | Exponents | Class 8 | RS AGGARWAL | Chapter 2 | myhelper

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Question 13:

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${\left(\frac{-1}{2}\right)}^3=?$
(a) $\frac{-1}{6}$
(b) $\frac{1}{6}$
(c) $\frac{1}{8}$
(d) $\frac{-1}{8}$

Answer 13:

(d) $\frac{-1}{8}$

${\left(\frac{-1}{2}\right)}^3=\frac{-1^3}{2^3}=\frac{-1}{8}$

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Q14 | Ex-2C | Exponents | Class 8 | RS AGGARWAL | Chapter 2 | myhelper

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Question 14:

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${\left(\frac{-3}{4}\right)}^2=?$
(a) $\frac{-9}{16}$
(b) $\frac{9}{16}$
(c) $\frac{16}{9}$
(d) $\frac{-16}{9}$

Answer 14:

(b) $\frac{9}{16}$

${\left(\frac{-3}{4}\right)}^2=\frac{{\left(-3\right)}^2}{{\left(4\right)}^2}=\frac{9}{16}$

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Q15 | Ex-2C | Exponents | Class 8 | RS AGGARWAL | Chapter 2 | myhelper

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Question 15:

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3670000 in standard form is
(a) 367 × 10⁴
(b) 36.7 × 10⁵
(c) 3.67 × 10⁶
(d) none of these

Answer 15:

(c) $3.67 \times 10^{6}$

$3670000=367 \times 10^{4}$

$=3.67 \times 100 \times 10^{4}$

$=3.67 \times 10^{2} \times 10^{4}$

$=3.67 \times 10^{(2+4)}$

$=3.67 \times 10^{6}$

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Q16 | Ex-2C | Exponents | Class 8 | RS AGGARWAL | Chapter 2 | myhelper

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Question 16:

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0.0000463 in standard form is
(a) 463 × 10⁻⁷
(b) 4.63 × 10⁻⁵
(c) 4.63 × 10⁻⁹
(d) 46.3 × 10⁻⁶

Answer 16:

(b) $4.63 \times 10^{-5}$

$0.0000463=\frac{463}{10^{7}}=\frac{4.63 \times 10^{2}}{10^{7}}$ 

$=4.63 \times 10^{(2-7)}=4.63 \times 10^{-5}$

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Q17 | Ex-2C | Exponents | Class 8 | RS AGGARWAL | Chapter 2 | myhelper

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Question 17:

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0.000367 × 10⁴ in usual form is
(a) 3.67
(b) 36.7
(c) 0.367
(d) 0.0367

Answer 17:

(a) 3.67

$0.000367 \times 10^{4}=\frac{367}{10^{6}} \times 10^{4}$

$=367 \times 10^{(4-6)}=367 \times 10^{-2}$

$=\frac{367}{10^{2}}=\frac{367}{100}=3.67$