SChand Composite Mathematics Class 7 Chapter 3 Decimals Exercise 3D

 Exercise 3D

Question 1

Convert 

(i) 8 km to m

=8×1000 m

=8000 m

(ii) 4.8 hg to cg

=4.8×10000 cg

=48000 cg

(iii) 18.5 L to ml

=18.5×1000

=18500 ml

(iv) 23.35 dl to cl

=28.35×10 cl

=283.5 cl

(v) 93 dag to dg

=93×100 dg

=9300 dg

(vi) 29.83 hm to m

=29.83×100 m

=2983 m

(vii) 9600 mm to m

$=9600 \times \frac{1}{1000}$ m

=9.6 m

(viii) 32.48 dag to kg

$=32.48 \times \frac{1}{100}$

=0.3248 kg

(ix) 125 cm to dam

$=125 \times \frac{1}{1000}$

=0.125 dam

(x) 292 ml to cl

$=292 \times \frac{1}{10}$

=29.2 cl

(xi) 179.6 dg to hg

$=179.6 \times \frac{1}{1000}$

=0.1796 hg

(xii) 0.85 dal to cl

=0.85×1000 cl

=850 cl

(xiii) 750 m to km 

$=750 \times \frac{1}{1000}$

=0.75 km

(xiv) 1600 ml to l

$=1600 \times \frac{1}{1000}$

=1.6 l

(xv) 8.75 kg to g

=8.75×1000 g

=8750 g

Question 2

Convert :

(i) 46 m 18 cm to cm

=46×100 cm + 18 cm 

=4600 cm + 18 cm

=4618 cm

(ii) 1 kg 2g to g

=1×1000 g + 2 g

=1000 g + 2g

=1002 g

(iii) 6 km 350 m to m

=6×1000 m + 350 m

=6000 m + 350 m

=6350 m

(iv) 3749 m to km and m

=3000 m + 749  m

$=3000 \times \frac{1}{1000}$ km + 749  m

=3 km 749 m

(v) 349 cl to l and cl

= 300 cl + 49 cl

$=300 \times \frac{1}{100}$ l + 49 cl

=3 l 49 cl

(vi)  5003 g to kg and g

= 5000 g + 3 g

$=5000 \times \frac{1}{1000}$ kg + 3 g

=5 kg 3 g

(vii) 6095 mg to g and mg

= 6000 mg +  9 5   mg 

$=6000 \times \frac{1}{1000}$g+95 mg

=6g 95 mg

(viii) 2302 cm to m and cm

=2300 cm + 2 cm

$=2300 \times \frac{1}{100}$ m + 2 cm 

=23 m 2 cm

(ix) 18 km 5 m to m

=18×1000 m + 5 m

=18000 m + 5 m 

=18005 m

 (x) 5 kl 250 l to l 

=5000 l + 250 l 

=5250 l

Question 3

The Mariana Trench is about 10,900 m deep in the pacific Ocean . Is this greater or less than 8.85 km height of Mt.Everest.

Sol :

Mariana Trench is greater because height of Mt.Everest 8.85×100 m=8850 m

=10900-8850

=2050 m greater

Question 4

The thickness of 54 sheets of paper is 15.12 mm.What is the thickness of each sheet in cm?

Sol :

15.12 mm into cm $=15.12 \times \frac{1}{10}$ cm

=1.512 cm

Thickness of 54 sheets=1.512 cm

Thickness of 1 sheet$=\frac{1.512}{54}$

=0.028 cm 

Question 5

An inch contains 2.54 cm. How many cm will be in 12.5 inches?

Sol :

If 1 inch contains 2.54 cm

∵cm in 12.5 inches=2.54×12.5

=31.75 cm

Question 6

Gauri spends 0.75 of her salary every month. If she earns 12000 per month in how many month will she save 39000 ?

Sol :

Gauri Save=1-0.75

=0.25 of the salary

=0.25×1200=3000

Gauri save 3000 per month

∴She will save 3900 in $=\frac{39000}{3000}$=13 month

Question 7

How many jars each holding 1.75 kg of jam can be filled from a vessel containing 100 kg of jam and how much remains?

Sol :

Each jars contains=1.75 kg jam

So from 100 kg vessel fill $=\frac{100}{1.75}=\frac{100 \times 100}{1.75}$

$=\frac{400}{7}$

=57 jars will filled

and $\frac{1}{7}$  of 1.75 is left remains

SChand New Learning Composite Mathematics Class 6 Chapter 3 Playing With Numbers Exercise 3D

Exercise 3D

Question 1

(a)12= 5+7

(b) 30= 11+19

(c) 68= 31+37

(d) 72= 31+41

Question 2

(a) 12= 5+7

(b) 30= 13+17

(c) 78= 37+41

(d) 90= 47+53

Question 3

(a) 15 = 3+5+7

(b) 23= 5+7+11

(c) 35= 5+7+23

(d) 55= 13+19+23

Question 4

The five pairs of prime numbers less than 20, whose sum is divisible by 5 —

(2,3) , (3,7), (2,13), (3,17) and (13,17).

Question 5

(a) 151 is divisible by 1 and 151 , so it is a prime number.

(b) 69 is divisible by 1, 3, 2, 69 ,so, it is not a prime number.

(c) 241 is divisible by 1 and 241 so, it is a prime number.

(d) 265 is divisible by 1, 5, 53, 265 so it is not a prime number.

Question 6

The tree prime numbers are 5, 7 and 11

5x7x11= 385

Question 7

F

S.chand publication New Learning Composite mathematics solution of class 7 Chapter 3 Rational numbers Exercise 3D

 Exercise 3D

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Q1 | Ex-3D | Class 7 | Rational Numbers | SChand | New Learning Composite Maths | Chapter 4 | myhelper

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Question 1

Find Each Sum

(a) $\frac{1}{2}+\left(\frac{7}{-8}\right)$

Sol :

$=\frac{1}{2} -\frac{7}{8}$

$=\frac{4-7}{8}=\frac{-3}{8}$

(b) $-\frac{1}{4}+\left(\frac{7}{-12}\right)$

Sol :

$=\frac{-1}{4}-\frac{7}{12}$

$=\frac{-3-7}{12}=\frac{-10}{12}$

$=\frac{-5}{6}$

(c) $\frac{43}{5}+\left(-\frac{71}{10}\right)$

Sol :

$=\frac{43}{5}-\frac{71}{70}$

$=\frac{86-71}{10}=\frac{15}{10}=\frac{3}{2}$

(d) $-1\frac{7}{15}+\left(-\frac{19}{30}\right)$

Sol :

$=-\frac{22}{15}-\frac{19}{30}$

$=\frac{-44-19}{30}=\frac{-63}{30}$

$=-2\frac{1}{10}$

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Question 2

Subtract:

(a) $\frac{1}{13} \text{ from } -\frac{5}{26}$

Sol :

$=\frac{-5}{26}-\frac{1}{13}$

$=\frac{-5\times 1-1\times 2}{26}$

$=\frac{-5-2}{26}=-\frac{7}{26}$

(b) $\frac{8}{-15} \text{from } \frac{-11}{30}$

Sol :

$=\frac{-11}{30}-\left(\frac{8}{-15}\right)$

$=\frac{-11}{30}+\frac{8}{15}$

$=\frac{-11\times 1+8\times 2}{30}=\frac{-11+16}{30}$

$=\frac{+5}{30}=\frac{1}{6}$

(c) $1\frac{4}{25} \text{ from } -3\frac{7}{50}$

Sol :

$=-3\frac{7}{50}-1\frac{4}{25}$

$=-\frac{157}{50}-\frac{29}{25}$

$=-\frac{215}{50}=-\frac{43}{10}$

(d) $-1\frac{3}{11} \text{ from } 3\frac{5}{22}$

Sol :

$=3\frac{5}{22}-\left(-1\frac{3}{11}\right)$

$=\frac{71}{22}+\frac{14}{11}$

$=\frac{99}{22}=4\frac{1}{2}$

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Question 3

Add.

(a) $\frac{7}{-8} \text{ and } \frac{5}{11}$

Sol :

$=\frac{7}{-8}+\frac{5}{11}=\frac{-77+40}{88}$

$=-\frac{37}{88}$

(b) $1\frac{2}{3} \text{ and } -2\frac{1}{5}$

Sol :

$=\frac{5}{3}-\frac{11}{5}$

$=\frac{25-33}{15}=-\frac{8}{15}$

(c) $-1\frac{3}{7} \text{ and } 1\frac{3}{8}$

Sol :

$=-1\frac{3}{7}+1\frac{3}{8}$

$=\frac{10}{7}+\frac{11}{8}$

$=\frac{-80-77}{56}$

$=-\frac{3}{56}$

(d) $-14 \frac{7}{11} \text{ and } -35\frac{9}{10}$

Sol :
$=-14\frac{7}{11}+\left(-35\frac{9}{10}\right)$

$=\frac{-161}{11}-\frac{359}{10}$

$=\frac{-5559}{110}=-50\frac{59}{110}$

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Question 4

(a) $\frac{3}{7} \text{ from } \frac{5}{-9}$

Sol :

$=\frac{5}{-9}-\frac{3}{7}$

$=-\frac{5}{9}-\frac{3}{7}$

$=\frac{-35-27}{63}=\frac{-62}{63}$

(b) $\frac{-9}{22} \text{ from } \frac{-5}{21}$

Sol :

$=-\frac{5}{21}-\left(-\frac{9}{22}\right)$

$=-\frac{5}{21}+\frac{9}{22}$

$=\frac{-110+189}{462}$

$=\frac{79}{462}$

(c) $-1\frac{1}{7} \text{ and } -2\frac{3}{10}$

Sol :

$=-2\frac{3}{10}-\left(-1\frac{1}{7}\right)$

$=-\frac{23}{10}+\frac{8}{7}$

$=\frac{-161+80}{70}=-\frac{81}{70}$

$=-1\frac{11}{70}$

(d) $-6\frac{5}{8} \text{ from } 1\frac{4}{11}$

Sol :

$=\frac{15}{11}+\frac{53}{8}$

$=\frac{120+583}{88}=\frac{703}{88}$

$=7\frac{87}{88}$

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Question 5

Find the sum.

(a) $\frac{-17}{20}+\left(-\frac{19}{25}\right)$

Sol :

$=-\frac{17}{20}-\frac{19}{25}$

$=\frac{-85-76}{100}=\frac{-161}{100}$

$=-1\frac{61}{100}$

(b) $1\frac{5}{8}+\frac{9}{-12}$

Sol :

$=1\frac{5}{8}+\frac{9}{-12}$

$=\frac{13}{8}-\frac{9}{12}$

$=\frac{39-18}{24}=\frac{21}{24}$

(c) $-2\frac{5}{6}+2\frac{9}{14}$

Sol :

$=-\frac{17}{6}+\frac{37}{14}$

$=\frac{-119+111}{42}$

$=-\frac{8}{42}$

$=-\frac{4}{21}$

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Question 6

Subtract:

(a) $-1\frac{3}{8} \text{ from } 5\frac{7}{12}$

Sol :

$=\frac{67}{12}+\frac{11}{8}$

$=\frac{134+33}{24}$

$=\frac{167}{24}=6\frac{23}{24}$

(b) $-1\frac{4}{15} \text{ from } \frac{11}{-20}$

Sol :

$=\frac{11}{-20} -\left(-1\frac{4}{15}\right)$

$=-\frac{11}{20}+\frac{19}{15}$

$=\frac{-33+76}{60}=\frac{43}{60}$

(c) $4\frac{7}{10} \text{ from } \frac{18}{-25}$

Sol :

$=\frac{18}{-25}-4\frac{7}{10}$

$=\frac{-8}{25}-\frac{47}{10}$

$=\frac{-36-235}{50}=\frac{-271}{50}$

$=-5\frac{21}{50}$

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Question 7

Simplify:

(a) $-10+ \left(-6\frac{1}{5}\right)$

Sol :

$=-16\frac{1}{5}$

(b) $-\frac{19}{80}-(-3)$

Sol :

$=2\frac{61}{80}$

(c) $\frac{18}{-19}+1\frac{1}{38}-2\frac{7}{57}$

Sol :

$=-2\frac{5}{114}$

(d) $-\frac{5}{17}+\left(-\frac{2}{5}\right)$

Sol :

$=\frac{-59}{85}$

(e) $6 + \frac{9}{-11}$

Sol :

$=5\frac{2}{11}$

(f) $-\frac{5}{18}+\frac{17}{-24}+1\frac{29}{36}– 2$

Sol :

$=\frac{-85}{72}$

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Question 8

$\frac{5}{-9}-\frac{7}{-12}-2-\left(-\frac{-5}{-6}\right)$

Sol :

$=-\frac{5}{9}+\frac{7}{12}-2+\frac{5}{6}$

$=\frac{-20-21-72+30}{36}=-\frac{41}{36}=-1\frac{5}{36}$

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Question 9

Subtract $\left(\frac{3}{7}+\frac{5}{-9}\right)$ from the negative of $\frac{8}{-63}$

Sol :

$\frac{8}{-63}-\left(\frac{3}{7}+\frac{5}{-9}\right)$

$=-\frac{8}{63}-\left(\frac{3}{7}-\frac{5}{9}\right)$

$=\frac{-8}{63}-\frac{3}{7}+\frac{5}{9}$

$=\frac{-8-27+35}{63}$

=0

S.chand publication New Learning Composite mathematics solution of class 8 Chapter 3 Squares and Square roots,Cube and Cube roots Exercise 3D

 Exercise 3D

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Question 1

Find the least number which must be subtracted from each of the following numbers to make it perfect square.

(a) 740

Sol :

$\begin{array}{r|l}&27\\ \hline 2&740\\&4\\\hline 47&340 \\ &329\\ \hline &11\\\end{array}$

∴Remainder=11

∴Now subtract 11 from 740. Then we get a perfect square.

∴740-11=729

∴√729=27

(b) 1535

Sol :

$\begin{array}{r|l}&39\\ \hline 3&1535\\&9\\\hline 69&635 \\ &621\\ \hline &14\\\end{array}$

∴Remainder=14

∴Now subtract 14 from 1535. Then we get a perfect square.

∴1535-14=1521

(c) 7926

Sol :

$\begin{array}{r|l}&89\\ \hline 8&7926\\&64\\\hline 169&1526 \\ &1521\\ \hline &5\\\end{array}$

∴Remainder=5

∴Perfect square=7926-5=7921

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Question 2

Find the least number which must be added to each of the following numbers to make it a perfect square.

(a) 708

Sol :
$\begin{array}{r|l}&26\\ \hline 2&708\\&4\\\hline 46&308 \\ &276\\ \hline &32\\\end{array}$

∴We observe that

26²<708<27²

∴The number to be added

=27²-708

=729-708

=21

Then, 708 would become=708+21=729
∴√729=27


(b) 1840

Sol :

$\begin{array}{r|l}&42\\ \hline 4&1840\\&16\\\hline 82&240 \\ &164\\ \hline &76\\\end{array}$

∴We observe that

42²<1840<43²

∴The number to be added

=43²-1840

=1849-1840

=9

(c) 3219

Sol :

$\begin{array}{r|l}&56\\ \hline 5&3219\\&25\\\hline 106&719 \\ &636\\ \hline &83\\\end{array}$

∴We observe that

56²<3219<57²

∴The number to be added

=57²-3219

=3249-3219

=30

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Question 3

4225 soldiers are arranged in different rows so that there are as many rows as the number of soldiers in a row. Find the number of rows.

Sol :

The number of rows=4225

=√4225=65

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Question 4

A gardener plants 3364 trees in such a way that there are as many rows as there are trees in a row. Find the number of trees in a row.

Sol :

The number of trees in a row=√3364

=58

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Question 5

A club collected Rs. 7225 for a charity. Each member contributed as many rupees as there were number of members of the club. How many members does the club have.

Sol :

Members of the club=√7225

=85

RS Aggarwal solution class 8 chapter 3 Squares and Square roots Exercise 3D

Exercise 3D

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Q1 | Ex-3D | Squares and Square roots | Chapter 3 | Class 8 | RS AGGARWAL  | myhelper

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Question 1:

Find the square root of number by using the method of prime factorisation:
225

Answer 1:

By prime factorisation method:

$$\begin{gathered} 225=3\times 3\times 5\times 5 \\ \sqrt{225}=3\times 5=15 \end{gathered}$$

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Question 2:

Find the square root of number by using the method of prime factorisation:
441

Answer 2:

By prime factorisation:

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Question 3:

Find the square root of number by using the method of prime factorisation:
729

Answer 3:

Resolving into prime factors:

$729=3\times 3\times 3\times 3\times 3\times 3$

∴ $\sqrt{729}=3\times 3\times 3=27$

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Question 4:

Find the square root of number by using the method of prime factorisation:
1296

Answer 4:

Resolving into prime factors:

$1296=2\times 2\times 2\times 2\times 3\times 3\times 3\times 3$

∴$\sqrt{1296}=2\times 2\times 3\times 3=36$

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Question 5:

Find the square root of number by using the method of prime factorisation:
2025

Answer 5:

Resolving into prime factors:

$2025=3\times 3\times 3\times 3\times 5\times 5$

∴$\sqrt{2025}=3\times 3\times 5=45$

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Question 6:

Find the square root of number by using the method of prime factorisation:
4096

Answer 6:

Resolving into prime factors:
$4096=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2$

∴ $\sqrt{4096}=2\times 2\times 2\times 2\times 2\times 2=64$

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Question 7:

Find the square root of number by using the method of prime factorisation:
7056

Answer 7:

Resolving into prime factors:

$7056=2\times 2\times 2\times 2\times 3\times 3\times 7\times 7$

∴$\sqrt{7056}=2\times 2\times 3\times 7=84$

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Question 8:

Find the square root of number by using the method of prime factorisation:
8100

Answer 8:

Resolving into prime factors:

$8100=2\times 2\times 3\times 3\times 3\times 3\times 5\times 5$

∴$\sqrt{8100}=2\times 3\times 3\times 5=90$

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Question 9:

Find the square root of number by using the method of prime factorisation:
9216

Answer 9:

Resolving into prime factors:

$9216=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3$

∴ $\sqrt{9216}=2\times 2\times 2\times 2\times 2\times 3=96$

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Question 10:

Find the square root of number by using the method of prime factorisation:
11025

Answer 10:

Resolving into prime factors:

$11025=3\times 3\times 5\times 5\times 7\times 7$

∴$\sqrt{11025}=3\times 5\times 7=105$

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Question 11:

Find the square root of number by using the method of prime factorisation:
15876

Answer 11:

Resolving into prime factors:

$15876=2\times 2\times 3\times 3\times 3\times 3\times 7\times 7$

∴$\sqrt{15876}=2\times 3\times 3\times 7=126$

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Question 12:

Find the square root of number by using the method of prime factorisation:
17424

Answer 12:

Resolving into prime factors:

$17424=2\times 2\times 2\times 2\times 3\times 3\times 11\times 11$

∴ $\sqrt{17424}=2\times 2\times 3\times 11=132$

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Question 13:

Find the smallest number by which 252 must be multiplied to get a perfect square. Also, find the square root of the perfect square so obtained.

Answer 13:

Resolving into prime factors:



Thus, the given number must be multiplied by 7 to get a perfect square.

New number = $252\times 7=1764$

∴$\sqrt{1764}=2\times 3\times 7=42$

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Question 14:

Find the smallest number by which 2925 must be divided to obtain a perfect square. Also, find the square root of the perfect square so obtained.

Answer 14:

Resolving into prime factors:

$2925=3\times 3\times 5\times 5\times 13$

13 is the smallest number by which the given number must be divided to make it a perfect square.

New number = $2925÷13=225$

$\sqrt{225}=3\times 5=15$

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Question 15:

1225 Plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

Answer 15:

Let the number of rows be x.
Therefore, the number of plants in each row is also x.
Total number of plants $=\left(x \times x\right)=x^2=1225$
$$\begin{gathered} x^2=1225=5\times 5\times 7\times 7 \\ x=\sqrt{1225}=5\times 7=35 \end{gathered}$$

Thus, the number of rows is 35 and the number of plants in each row is 35.

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Question 16:

The students of a class arranged a picnic. Each student contributed as many rupees as the number of students in the class. If the total contribution is Rs 1156, find the strength of the class.

Answer 16:

Let the number of students be $x$.
Hence, the amount contributed by each student is Rs x.

Total amount contributed $=x\times x=x^2=1156$

$$\begin{gathered} 1156=2\times 2\times 17\times 17 \\ x=\sqrt{1156}=2\times 17=34 \end{gathered}$$

Thus, the strength of the class is 34.

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Question 17:

Find the least square number which is exactly divisible by each of the numbers 6, 9, 15 and 20.

Answer 17:

The smallest number divisible by each of these numbers is their L.C.M.
L.C.M. of 6, 9, 15, 20 = 180

Resolving into prime factors: 
$180=2\times 2\times 3\times 3\times 5$
To make it a perfect square, we multiply it with 5.

Required number = $180\times 5=900$

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Question 18:

Find the least square number which is exactly divisible by each of the numbers 8, 12, 15 and 20.

Answer 18:

The smallest number divisible by each of these numbers is their L.C.M.
L.C.M. of 8, 12, 15, 20 = 120

Resolving into prime factors: 
$120=2\times 2\times 2\times 3\times 5$

To make this into a perfect square, we need to multiply the number with $2\times 3\times 5=30$.

Required number = $120\times 30=3600$