Showing posts with label of class 7. Show all posts
Showing posts with label of class 7. Show all posts

S.chand publication New Learning Composite mathematics solution of class 7 Chapter 8 Percentage and Its Applications Exercise 8A

 Exercise 8A


Q1 | Ex-8A |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper


Question 1

Express each of the following as a percentage.

(a) 

(i) $\frac{1}{5}$

Sol : $\frac{1}{5} \times 100$=20%


(ii) $\frac{3}{4}$

Sol : $\frac{3}{4} \times 100$=75%


(iii) $\frac{17}{50}$ 

Sol : $\frac{17}{50} \times 100$=34%


(iv) $\frac{11}{20}$ 

Sol : $\frac{11}{20} \times 100$=55%


(v) $2\frac{9}{10}$

Sol : $\frac{29}{10}\times 100$=290%


(vi) $2\frac{8}{25}$

Sol : $\frac{58}{25}\times 100$=232%


(b) 

(i) 0.3 

Sol : 0.3×100=30%

(ii) 0.47 

Sol : 0.89×100=89%

(iii) 0.89 

Sol : 0.89×100=89%

(iv) 1.05 

Sol : 1.05×100=105%

(v) 0.09 

Sol : 0.09×100=9%

(vi) 0.769 

Sol : 0.769×100=76.9%

(vii) 3 

Sol : 3×100=300%

(viii) 9.002 

Sol : 9.002×100=900.2%

(ix) 2.25 

Sol : 2.25×100%

(x) 0.75

Sol : 0.75×100=75%



Q2 | Ex-8A |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

Question 2

Express as a fraction in its simplest form.

(a) 80% 

Sol : $=\frac{80}{100}=\frac{4}{5}$

(b) 36% 

Sol : $\frac{36}{100}=\frac{9}{25}$

(c) 60% 

Sol : $\frac{60}{100}=\frac{3}{5}$

(d) 19% 

Sol : $\frac{19}{100}$

(e) $62\frac{1}{2}$% 

Sol : $\dfrac{\frac{125}{2}}{100}=\frac{125}{100}=\frac{5}{8}$

(f) $133\frac{1}{3}$% 

Sol : $\dfrac{\frac{400}{3}}{100}=\frac{400}{300}=1\frac{1}{3}$

(g) 120% 

Sol : $\frac{120}{100}=1\frac{1}{5}$

(h) 65% 

Sol : $\frac{65}{100}=\frac{13}{20}$

(i) 270% 

Sol : $\frac{270}{100}=2\frac{7}{10}$

(j) 350%

Sol : $\frac{350}{100}=\frac{7}{2}=3\frac{1}{2}$



Q3 | Ex-8A |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

Question 3

Express as a decimal.

(a) 7% 

Sol : $\frac{7}{100}$=0.07

(b) 80% 

Sol : $\frac{80}{100}$=0.8

(c) 72% 

Sol : $\frac{72}{100}$ =0.72

(d) 39% 

Sol : $\frac{39}{100}$=0.39

(e) $1\frac{1}{2}$% 

Sol : $\frac{3}{2}=\frac{15}{100}$

=0.01

(f) $6\frac{1}{4}$% 

Sol : $\frac{24}{4}=\frac{6.25}{100}$

=0.0625

(g) $27\frac{1}{2}$% 

Sol : $\frac{55}{2}=\frac{27.5}{100}$

=0.275

(h) $72\frac{3}{5}$% 

Sol : $\frac{363}{5}=\frac{72.6}{100}$

=0.726

(i) 189% 

Sol : $\frac{189}{100}$

=1.89

(j) $205\frac{2}{5}$%

Sol : $\frac{1027}{5}=\frac{205.4}{100}$

=2.054



Q4 | Ex-8A |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

Question 4

Compare and Write <,> or = in the box.

(a) $\frac{5}{8}$ ___ 70%

Sol :

$\frac{5}{8}\times 100$=62.5%

$\frac{5}{8} < 70\%$


(b) 64% ___ $\frac{3}{5}$

Sol :

$\frac{3}{5}\times 100$=60%

64% >$\frac{3}{5}$


(c) 92.7% ___ 0.927

Sol :

0.927=0.927×100%=92.7%

∴92.7% = 0.927


(d) 0.99 ____ 100%

Sol :

0.99=0.99×100%=99%

∴0.99<100%



Q5 | Ex-8A |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

Question 5

Arrange in ascending order.

(a) 38%, 0.4, $35\frac{1}{3}$%, $\frac{5}{8}$

Sol :

0.4=0.4×100%=40%

$35\frac{1}{3}=\frac{106}{3}$=35.33%

$\frac{5}{8}=\frac{5}{8}\times 100$%=62.5%


Ans : $35\frac{1}{3}$% , 38%, 0.4, $\frac{5}{8}$


(b) 0.93, 19.20, $1\frac{2}{3}$, 67%

Sol :



Q6 | Ex-8A |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

Question 6

Arrange in descending order.

(a) $\frac{4}{5}$, 0.5, 500%,$66\frac{2}{3}$%


Sol : 


(b) $\frac{7}{20}$, 0.038, 39%, 0.59

Sol :



Q7 | Ex-8A |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

Question 7

Use the following diagrams to complete the table

(a)
    
 
 
 

(b)
               
  
     


(c)










(d) 







Fraction shaded Percentage shaded Percentage unshaded
(a) $\frac{3}{4}$ 75% 25%
(b) $\frac{9}{15}=\frac{3}{5}$ 60% 40%
(c) $\frac{2}{5}$ 40% 60%
(d) $\frac{5}{8}$ $62\frac{1}{2}$% $37\frac{1}{2}$%



Q8 | Ex-8A |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

Question 8

A molecular of water is made up of 2 atoms of hydrogen and 1 atom of oxygen. What percent of the atoms a water molecular is oxygen?

Sol :

∴Percentage of oxygen in water $=\frac{1}{3}\times 100$%$=33\frac{1}{3}$%



Q9 | Ex-8A |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

Question 9

Sterling silver is an alloy combining 925 parts of pure silver and 75 parts of another metal, such as copper. What percent of sterling silver is not pure silver?

Sol :

∴Percentage of sterling silver$=\frac{75}{(925+75)}\times 100$%

$=\frac{75}{1000}\times 100$

=7.5%



Q10 | Ex-8A |Class 7 |S.Chand | New Learning Composite maths |Percentage and Its Applications |Ch-7 |myhelper

Question 10

Prashant spends 4 hours a day on his self study, 1 hour playing video games and 1 hour in doing workouts at fitness centre. What per cent of the day does Prahant spend on these activities?

Sol :

∴Prashant spend time on these activities 

$=\frac{(4+1+1)}{24}\times 100$%

$=\frac{6}{24}\times 100$%

=25%

S.chand publication New Learning Composite mathematics solution of class 7 Chapter 7 Ratio and Proportion and Unitary Method Exercise 7B

 Exercise 7B


Q1 |  Ex-7B | SChand | New Learning Composite Maths | Class 7 |  Ratio and Proportion and Unitary Method | myhelper

Question 1

Find out whether the following groups of numbers are in proportion or not.

(a) 18, 20, 9, 10

Sol :

Product of extremes=18×10=180

Product of means=20×9=180

So, product of extremes=product of means

∴They are proportional


(b) 21, 6, 35, 10

Sol :

Product of extremes=21×10=210

Product of means=6×35=210

So, product of extremes=product of means

∴They are proportional


(c) 12, 18, 28, 12

Sol :

Product of extremes=12×12=144

Product of means=18×28=504

So, product of extremes≠product of means

∴They are not proportional


(d) 5.2, 3.9, 3, 4

Sol :

Product of extremes=5.2×4=20.8

Product of means=3.9×3=11.7

So, product of extremes≠product of means

∴They are not proportional


(e) 16, $3 \frac{1}{3}$, 18, $3\frac{3}{4}$

Sol :

Product of extremes$=16\times 3\frac{3}{4}=16\times \frac{15}{4}$

=60

Product of means$=3\frac{1}{3} \times 18=\frac{10}{3} \times 18$

=60

So, product of extremes=product of means

∴They are proportional



Q2 |  Ex-7B | SChand | New Learning Composite Maths | Class 7 |  Ratio and Proportion and Unitary Method | myhelper

Question 2

Find x in the following proportions.

(a) x, 6, 55, 11

Sol :

$\frac{x}{6}=\frac{55}{11}$

11x=6×55

$x=\frac{6\times 55}{11}=30$


(b) 16, 18, x, 108

Sol :

$\frac{16}{18}=\frac{x}{108}$

$x=\frac{16 \times 108}{18}=96$


(c) 4, x, x, 16

Sol :

$x=\sqrt{4\times 16}$

x=8


(d) 7.5, 15, 5, x

Sol :

$\frac{7.5}{15} =\frac{5}{x}$

$x=\frac{15\times 5 \times 10}{75}$

∴x=10



Q3 |  Ex-7B | SChand | New Learning Composite Maths | Class 7 |  Ratio and Proportion and Unitary Method | myhelper

Question 3

Find the fourth proportional to

(a) 32, 64, 6

Sol :

$\frac{32}{64}=\frac{6}{x}$

$x=\frac{6\times 64}{32}=12$

∴Fourth proportional is 12


(b) 8, 9, 24

Sol :

$\frac{8}{9}=\frac{24}{x}$

$x=\frac{9\times 24}{8}$

∴x=27

∴Fourth proportional is 27


(c) $\frac{1}{5}, \frac{1}{9}, \frac{1}{10}$

Sol :

$\frac{9}{5}=\frac{x}{10}$

$x=\frac{9\times 10}{5}=18$

∴Fourth proportional is $\frac{1}{18}$



Q4 |  Ex-7B | SChand | New Learning Composite Maths | Class 7 |  Ratio and Proportion and Unitary Method | myhelper

Question 4

Find the mean proportional between

(a) 16 and 25

Sol :

$\frac{16}{x}=\frac{x}{25}$

$x^2=16\times 25$

$x=\sqrt{16 \times 25}$

∴x=4×5=20

∴Mean proportional is 20


(b) 9 and $\frac{1}{36}$

Sol :

$\frac{9}{x}=x\times \frac{36}{1}$

$x^2=\frac{9}{36}$

$x=\sqrt{\frac{1}{4}}$

∴$x=\frac{1}{2}$

∴Mean proportional is $\frac{1}{2}$


(c) 0.01 and 0.81

Sol :

$\frac{0.01}{x}=\frac{x}{0.81}$

$x^2=0.01\times 0.81$

$x=\sqrt{0.01 \times 0.81}$

∴x=0.09

∴Mean proportional is 0.09



Q5 |  Ex-7B | SChand | New Learning Composite Maths | Class 7 |  Ratio and Proportion and Unitary Method | myhelper

Question 5

Find the third proportional to

(a) 25, 40

Sol :

25×x=40×40

$x=\frac{40\times 40}{25}=64$

∴x=64

∴Third proportional is 64


(b) 18, 30

Sol :

18×x=30×30

$x=\frac{30\times 30}{18}$

∴x=50

∴Third proportional is 50


(c) 3.9, 11.7

Sol :

3.9×x=11.7×11.7

$x=\frac{11.7\times 11.7}{3.9}$

∴x=35.1

∴Third proportional is 35.1



Q6 |  Ex-7B | SChand | New Learning Composite Maths | Class 7 |  Ratio and Proportion and Unitary Method | myhelper

Question 6

Show that 5, 25, 125 are in continued proportions.

Sol :

Since 5×5=25

and 25×5=125

∴It can be said that they are in proportion



Q7 |  Ex-7B | SChand | New Learning Composite Maths | Class 7 |  Ratio and Proportion and Unitary Method | myhelper

Question 7

If the cost of 8 m of cloth is Rs. 600, find the cost of 15 m of cloth.

Sol :

The cost of 8m of cloth is 600

The cost of 15m of cloth is $=\frac{600\times 15}{8}$

=1125



Q8 |  Ex-7B | SChand | New Learning Composite Maths | Class 7 |  Ratio and Proportion and Unitary Method | myhelper

Question 8

When a man spends $\frac{5}{9}$ of his monthly income, he saves Rs. 3600 a month, find his income.

Sol :

Let , his monthly income be x

ATQ,

$x-\frac{5x}{9}=3600$

4x=3600×9

$x=\frac{3600 \times 9}{4}$

∴x=8100

His monthly income is 8100



Q9 |  Ex-7B | SChand | New Learning Composite Maths | Class 7 |  Ratio and Proportion and Unitary Method | myhelper

Question 9

After selling 85 copies of a book for every 100 copies printed, a publisher finds that 450 copies are left unsold. How many copies were printed?

Sol :

(100-85)=15 copies left

15 copies left for every 100 copies printed

450 copies left for every $\frac{100 \times 450}{15}= 3000$ or copies printed


Ans 3000 copies were printed



Q10 |  Ex-7B | SChand | New Learning Composite Maths | Class 7 |  Ratio and Proportion and Unitary Method | myhelper

Question 10

A truck requires 108 L of diesel for covering a distance of 594 km. How much diesel will be required by the truck to cover a distance of 1650 km?

Sol :

For covering 594 km distance diesel required 108L

For covering 1650 km distance diesel required $\frac{108\times 165}{594}$

=300 L


Ans 300L diesel will be required

S.chand publication New Learning Composite mathematics solution of class 7 Chapter 5 Algebraic Expression Exercise 5C

 Exercise 5C


Q1 | Ex-5C|Class 8 |S.Chand |Composite maths | Algebraic Expression |Ch-6 |myhelper

Question 1

Subtract

(a) 2x from 7x

Sol : 5x


(b) -5a from -2a

Sol : -2a-(-5a)

=3a


(c) 3b2 from -5b2

Sol :

=-8b2


(d) -2xy from 0

Sol :

=2xy


(e) -8cd from 8cd

Sol :

=8cd+8cd

=16cd


(f) 3(x + y) from 4(x + y)

Sol :

=4x+4y-3x-3y

=x+y



Q2 | Ex-5C|Class 8 |S.Chand |Composite maths | Algebraic Expression |Ch-6 |myhelper

Question 2

Subtract the lower expression from the upper expression.

(a) (7a + 5) – (2a + 1)

Sol :

$\begin{array}{r}7a+5 \\ 2a+1 \\ -\phantom{2a}-\phantom{1} \\ \hline 5a+4 \end{array}$


(b) (9x2 + 3y2) – (5x2 + y2)

Sol :

$\begin{array}{r}9x^2+3y^2\\ 5x^2+y^2 \\ -\phantom{5x^2}-\phantom{y^2} \\ \hline 4x^2+2y^2 \end{array}$

(c) (5ab – 7bc) – (2ab + bc)

Sol :

$\begin{array}{r}5ab-7bc\\2ab+bc\\ -\phantom{2ab}-\phantom{bc} \\ \hline 3ab-8bc\end{array}$

(d) (9x2y – 11) – (-2x2y – 5)

Sol :

$\begin{array}{r}9x^2y-11\\+2x^2y+5\\ \hline 11x^2y-6\end{array}$

(e) (2m3 – 9n2) – (4m3 – 11n2)

Sol :

$\begin{array}{r}2m^3-9n^2\\ 4m^3+11n^2\\ \hline -2m^3+2n^2\end{array}$

(f) (4x2y – 5xy2) – (4x2y – 5xy2)

Sol :

$\begin{array}{r}4x^2y-5xy^2\\ 4x^2y+5xy^2\\ \hline 0\end{array}$


Q3 | Ex-5C|Class 8 |S.Chand |Composite maths | Algebraic Expression |Ch-6 |myhelper

Question 3

Arrange the following in columns and subtract.

(a) 7p + 2q + 5c from 9p + 4q + 7c

Sol :

$\begin{array}{r}9p + 4q + 7c \\ 7p+2q+5c\\ -\phantom{7p}-\phantom{2q}-\phantom{5c} \\ \hline 2p+2q+2c\end{array}$

(b) 3x2 – 8xy – 9y2 from 3x2 – 8xy – 9y2

Sol :

$\begin{array}{r}3x^2-8xy-9y^2 \\ 3x^2+8xy+9y^2\\ -\phantom{3x^2}-\phantom{8xy}-\phantom{9y^2} \\\hline 0\end{array}$

(c) 8x2 – 9x + 5 from 4x3 – 3x2 + 7x + 11

Sol :

$\begin{array}{r}4x^3-3x^2+7x+11\\ 8x^2+9x+5\\ -\phantom{8x^2}-\phantom{9x}-\phantom{5} \\ \hline 4x^3-11x^2+16x+6\end{array}$

(d) -2p3 + 7p2 – 20 + 3p from 0

Sol :

=2p3-7p2+20-3p


(e) 4ax2 + 7bx – 9t + 6 from 8ax2 – 2bx + 10t -4

Sol :

$\begin{array}{r} 8ax^2-2bx+10x-4 \\ 4ax^2+7bx+9x+6\\ -\phantom{4ax^2}-\phantom{7bx}-\phantom{9x}-\phantom{6} \\ \hline 4ax^2-9bx+19x-10\end{array}$



Q4 | Ex-5C|Class 8 |S.Chand |Composite maths | Algebraic Expression |Ch-6 |myhelper

Question 4

Subtract the following without writing in vertical form.

(a) 2x + 3y from 7x + 9y

Sol : 5x+6y

(b) -2x – y from 2x + y

Sol : 4x+2y

(c) – 5m2 + 4mn – n2 from 2m2 + 8mn + 3n2

Sol : 2m2 + 8mn + 3n2-(– 5m2 + 4mn – n2)

(d) – 2p3 + 4p2 – 7 – p from 16 – 8p

Sol : 16-8p+2p3–4p2+7+p

(e) 4xy – 3x2y + 7xy2 from 27 – 7xy2 + 11xy

Sol : 27 – 7xy2 + 11xy-(4xy – 3x2y + 7xy2)

=23-7p+2p3-4p2


(f) The sum of x2 – 2xy + y2 and x2 + 3xy + y2 from x2 – 6xy + 4y2

Sol :

=2x2+xy+2y2


∴x2-6xy+4y2-2x2-xy-2y2

=-x2-7xy+2y2


(g) The sum of 9b2 – 3c2 and 2b2 + bc – 2c2 from the sum of 2b2 – 2bc – c2 and c2 + 2bc – b2

Sol :

2b2 – 2bc – c2 and c2 + 2bc – b2-(9b2 – 3c2 and 2b2 + bc – 2c2)

=b2-(11b2-5c2+bc)

=b2-11b2+5c2-bc

=-10b2+5c2-bc



Q5 | Ex-5C|Class 8 |S.Chand |Composite maths | Algebraic Expression |Ch-6 |myhelper

Question 5

Simplify:

(a) 3x2 – 5x + 7 – 5x2 – 11 + 8x + 20 + 9x2 – 6x

Sol :

=7x2-3x+16


(b) 5x3 – 8 + 7x2y – 7xy + 10xy2 + 11 – 2x3 – 5x2y + 9xy – 6xy2

Sol :

=3x3+3+2x2y+2xy+4xy2



Q6 | Ex-5C|Class 8 |S.Chand |Composite maths | Algebraic Expression |Ch-6 |myhelper

Question 6

The area of a square 4x2 – 2xy – 6y2 square units. A triangle inside has an area of -5xy + 4y2 – 3x2 square units. Find the area of the shaded region.






Sol :

Area of shaded region=(4x2 – 2xy – 6y2+5xy-4y2+3x2) sq unit

=(7x2+3xy-10y2)

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

 Exercise 2L

Convert:

Question 1

(a) 3 km = How many meter

Sol : = (3×1000)m=3000m

(b) 2 m = How many centimeter

Sol : = (2×100)cm=200cm

(c) 5 cm = How many milimeter

Sol : (5×10)mm=50mm

(d) 2.9 m = How many centimeter

Sol : (2.9×100)cm=290cm

(e) 0.89 km = How many meter

Sol : (0.89×1000)m=890m

(f) 0.3 m = How many milimeter

Sol : (0.3×1000)mm=300mm

(g) 7 dam = How many meter

Sol : (7×10)m=70m

(h) 5 hm = How many meter

Sol : (5×100)m=500m

(i) 2 m = How many decemeter.

Sol : (2×10)dm=20dm


Question 2

(a) 478 cm = How many meter

Sol : $=\frac{478}{100}$m=4.78m


(b) 3209 centimeter = How many meter

Sol : $=\frac{3209}{100}$ m


(c) 90 milimeter = How many how many centimeter.

Sol : $=\frac{9}{10}$cm=0.9cm


(d) 49 meter = How many kilometer.

Sol : $=\frac{49}{1000}$km==0.049km


(e) 2000 m = How many kilometer

Sol : $=\frac{2000}{1000}$km=2km


(f) 200 cm = How many meter.

Sol : $=\frac{200}{100}$m=2m


(g) 428 milimeter = How many meter

Sol : $=\frac{428}{1000}$m=0.428m


(h) 3 milimeter = how many meter

Sol : $=\frac{3}{1000}$m=0.003m


(i) 8 dacemeter = How many kilometer.

Sol : $=\frac{8}{100}km$=0.08km


Question 3

(a) 5 kg = How many gram

Sol : (5×1000)g=5000

(b) 3.7 kg = How many gram

Sol : (3.7×1000)g=3700

(c) 0.289 kg = How many gram

Sol : (0.289×1000)g=289

(d) 3.59 g = How many miligram

Sol : (3.59×1000)g=3590

(e) 9 g = How many miligram.

Sol : (9×1000)g=9000

(f) 0.05 g = How many miligram.

Sol : (0.05×1000)g=50

(g) 2 q = How many kilogram

Sol : (2×100)kg=200kg

(h) 2.9 t = how many kilogram

Sol : (2.9×1000)kg=2900kg

(i) 0.032 kg = How many gram

Sol : (0.032×1000)g=32g


Question 4

(a) 4000 g = How many kilogram

Sol : $=\frac{4000}{1000}$kg=4kg

(b) 1800 gram = How many kilogram.

Sol : $=\frac{1800}{1000}$kg=1.8kg

(c) 46 g = How many kilogram

Sol : $\frac{46}{1000}$kg

(d) 729 mg = How many gram.

Sol : $\frac{729}{1000}$kg=0.729g

(e) 700 mg = How many gram.

Sol : $=\frac{700}{1000}$g=0.7g

(f) 67 mg = How many gram.

Sol : $=\frac{67}{1000}$g=0.067g

(g) 8 mg = How many gram

Sol : $=\frac{8}{1000}$g=0.008g

(h) 9 g = How many kilogram

Sol : $=\frac{9}{1000}$kg=0.009kg

(i) 2 cg = How many miligram.

Sol : (2×10)mg=20


Question 5

(a) 4 hg = How many gram.

Sol : (4×100)g=400g

(b) 1.9 km = How many decemeter.

Sol : (1.9×10000)g=19000g

(c) 8.3 decagram = How many dg.

Sol : (8.3×100)dg=830dg

(d) 75 dam = How many kilometer

Sol : $=\frac{75}{100}$km=0.75km

(e) 4 dg = How many miligram

Sol : =(4×100)mg=400mg

(f) 300 kg = How many quintal.

Sol : $=\frac{300}{100}$q=3q


Question 6

Compare and put <,> or == in the given boxes.

(a) 5,000 mm ____ 5 m

Sol : 5,000 mm = 5 m

(b) 3.8 kg ____ 400 g

Sol : 3.8 kg > 400 g

(c) 5 mg ____ 1 g

Sol : 5 mg < 1 g

(d) 3 g ____ 30,000 mg

Sol : 3 g < 30,000 mg

(e) 0.0087 m ____ 8.7 mm

Sol : 0.0087 m = 8.7 mm

(f) 0.2 km ____ 2000 mm

Sol : 0.2 km > 2000 mm

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