X mathematics solutions chapter 1 exercise 1.2

EXERCISE 1.2

1. Express each number as a product of its prime factors:

(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

(i) 140             = 2 x 2 x 5 x 7            =2^2 x 5 x 7

(ii)156             =2 x 2 x 3 x 13           =2^2 x 3 x 13

(iii)3825          =3 x 3 x 5 x 5 x 17    =3^2 x 5^2 x 17

(iv)5005          =5 x 7 x 11 x 13        =5 x 7 x 11 x 13

(v)7429           =17 x 19  x 23           =17 x 19 x 23

                                                              

                                                                                                                                           

2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF =product of the two numbers. (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54

(i)                     26        = 2 x 13

91        =7 x 13

HCF    = 13

LCM    =2 x 7 x 13     =182

Product of two value           26 x 91           = 2366

Product of HCF and LCM   13 x 182         = 2366

Hence, product of two numbers = product of HCF × LCM

(ii)                    510     = 2 x 3 x 5 x 17

92        =2 x 2 x 23

HCF    =2

LCM    =2 x 2 x3 x 5 x 17 x 23         = 23460

Product of both values        510x92           = 46920

Product of HCF and LCM   2x23460         =46920

Hence, product of two numbers = product of HCF × LCM

(iii)                   336     = 2 x 2 x 2 x 2 x 3 x 7

                                   54        = 2 x 3 x 3 x 3

HCF    = 2 x 3                                                = 6

LCM    = 2 x 2 x 2 x 2 x 3 x 3 x 3 x 7           =3024

Product of number              336 x 54         =18144

Product of HCF and LCM   6 x 3024         = 18144

Hence, product of two numbers = product of HCF × LCM

3. Find the LCM and HCF of the following integers by applying the prime factorization method.(i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25

(i)          12        = 2 x 2 x 3

15        =3 x 5

 21       =3 x 7

HCF    = 3

LCM = 2 x 2 x 3 x 5 x 7        = 420

(ii)        17        = 1 x 17

23        = 1 x 23

29        = 1 x 29

HCF    = 1

LCM    = 1 x 17 x 19 x 23     = 11339

(iii)

8          =1 x 2 x 2 x 2

9          =1 x 3 x 3

25        =1 x 5 x 5

HCF    =1

LCM    = 1 x 2 x 2 x 2 x 3 x 3 x 5 x 5           = 1800

4. Given that HCF (306, 657) = 9, find LCM (306, 657).

We have the formula that

Product of LCM and HCF   =          product of number

LCM x 9         =          306 x 657

Divide both side by 9 we get

LCM                =          (306 x 657)/ 9   = 22338

5. Check whether 6n can end with the digit 0 for any natural number n.

 If any digit has last digit 10 that means It is divisible by 10 And the factors of

 10 = 2x5 So value 6n should be divisible by 2 and 5 both 6n is divisible by 2 but not divisible by 5 So it can not end with 0.

6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

      7 × 11 × 13 + 13

      Take 13 common there we get

      13  (7 x 11 +1 )

      13(77 + 1 )

      13  (78)

      It is product of two numbers and both numbers are more than 1 so it is a composite number

      7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

      Take 5 common there we get

      5(7 x 6 x 4 x 3 x 2 x 1 +1)

      5(1008 +1)

      5(1009)

      It is product of two numbers and both numbers are more than 1 so it is a composite number

7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

      They will be meet again after LCM of both values at the starting point.

      To get the LCM we have to factorize the number.

            18        = 2 × 3 × 3

            12        = 2 × 2 × 3

            LCM    = 2 × 2 × 3 × 3 = 36 

            Therefore, they will meet together at the starting point after 36 minutes.

X class ncert solutions chpater -1 exercise1_1

EXERCISE 1.1

1. Use Euclid’s division algorithm to find the HCF of :

(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255

(i)

Here 225 > 135 we always divide greater number with smaller one.

Divide 225 by 135 we get 1 quotient and 90 as remainder so that

225= 135*1 + 90

Divide 135 by 90 we get 1 quotient and 45 as remainder so that

135= 90*1 + 45

Divide 90 by 45 we get 2 quotient and no remainder so we can write it as

90 = 2*45+ 0

As there are no remainder so deviser 45 is our HCF

(ii)

38220>196 we always divide greater number with smaller one.

Divide 38220 by 196 then we get quotient 195 and no remainder so we can write it as

38220 = 196 * 195 + 0

As there is no remainder so deviser 196 is our HCF

(iii)

           
867>255  we always divide greater number with smaller one.

divide 867 by 255 then we get quotient 3 and remainder is 102

so we can write it as

867 = 255 * 3 + 102

Divide 255 by 102 then we get quotient 2 and remainder is 51

So we can write it as

255 = 102 * 2 + 51

Divide 102 by 51 we get quotient 2 and no remainder

So we can write it as

102 = 51*2+ 0

As there is no remainder so deviser 51 is our answer

2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Let take a as any positive integer and b = 6.

Then using Euclid’s algorithm we get  a = 6q + r  here r is remainder and value of q is more 

than or equal to 0  and r = 0, 1, 2, 3, 4, 5 because 0 ≤r < b  and the value of b is 6 

So total possible forms  will  6q+0 , 6q+1 , 6q+2,6q+3,6q+4,6q+5 

6q+0   6 is divisible by 2 so it is a even number

6q+1   6 is divisible by 2 but 1 is not divisible by 2 so it is a odd number

6q+2  6 is divisible by 2   and 2 is also divisible by 2 so it is a even number

6q+3  6 is divisible by 2 but 3 is not divisible by 2 so it is a odd number

6q+4  6 is divisible by 2 and 4 is also divisible by 2 it is a even number

6q+5  6 is divisible by 2 but 5 is not divisible by 2 so it is a odd number

So odd numbers will in form of 6q + 1, or 6q + 3, or 6q + 5

3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

To get the maximum number column here we always find HCF and for minimum number we find LCM

So can use Euclid’s algorithm to find the HCF.

Here 616> 32 so always divide greater number with smaller one

When we divide 616 by 32 we get quotient 19 and remainder 8

So we can write it as

616 = 32 * 19 + 8

Now divide 32 by 8 we get quotient 4 and no remainder

So we can write it as

32 = 8 * 4 + 0

As there are no remainder so our HCF will 8

So that maximum number of columns in which they can march is 8.

4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Let take a as any positive integer and b = 3.

Then using Euclid’s algorithm we get  a = 3q + r  here r is remainder and value of q is more 

than or equal to 0  and r = 0, 1, 2 because 0 ≤r < b  and the value of b is 3 So our possible 

values will 3q+0 , 3q+1 and 3q+2

Now find the square of values 

Use the formula (a+b)^2 = a^2 + 2ab +b^2 to open the square bracket 

(3q)^2             = 9q^2   if we divide by 3 we get no remainder  

we can write it as 3*(3q^2)  so it is in form of 3m  here m = 3q^2

(3q+1)^2         = (3q)^2 + 2*3q*1  + 1^2       

                        =9q^2 + 6q +1 now divide by 3 we get 1 remainder

so we can write it as 3(3q^2 + 2q) +1 so we can write it in form of 

3m+1 and value of m is 3q^2 + 2q  here

(3q+2)^2         = (3q)^2 + 2*3q*2  + 2^2 

                        =9q^2 + 12q +4  now divide by 3 we get 1 remainder

so we can write it as 3(3q^2 + 4q +1) +1 so we can write it in form 

of 3m +1 and value of m will 3q^2 + 4q +1

Square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

According to Euclid’s Division Lemma

Let take a as any positive integer and b = 9. 

Then using Euclid’s algorithm we get  a = 9q + r  here r is remainder and value of q is more 

than or equal to 0  and r = 0, 1, 2, 3, 4, 5 , 6 , 7 , 8, because 0 ≤r < b  and the value of b is 9

Sp possible forms will  9q, 9q+1, 9q+2,9q+3,9q+4,9q+5,9q+6,9q+7 and 9q+8

to get the cube of these values use the formula

(a+b)^3           = a^3 + 3a^2b+ 3ab^2 + b^3

      In this formula value of a is always 9q

So plug the value we get

(9q+b)^3          = 729q^3 + 243q^2b + 27qb^2 + b^3

Now divide by 9 we get quotient = 81q^3 + 27q^2b + 3qb^2  and remainder is b^3

So we have to consider the value of b^3

b = 0                     we get 9m+0  = 9m

b = 1                     then  1^3  = 1 so we get  9m +1

b = 2                     then 2^3 = 8 so we get 9m + 8

b = 3                     then 3^3 = 27  and it is divisible by 9  so we get 9m

b = 4                     then 4^3 =64 divide by 9 we get 1 as remainder so we get 9m +1

b=5                       then 5^3=125 divide by 9 we get 8 as remainder so we get 9m+8

b=6                       then 6^3=216 divide by 9 no remainder there so we get 9m

b=7                       then 7^3 = 343 divide by 9 we get 1 as remainder so we get 9m+1

b=8                       then 8^3 = 512 divide by 9 we get 8 as remainder so we get 9m+8

So all values are in form of 9m , 9m+1 or 9m+8 

Define the term . How many types of solutions are formed? Write briefly about each type with an example.

Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

A solution is a homogenous mixture of two or more than two substances whose composition can change within a certain limits.  A solution of two substances is called binary solution.

In binary solution, the component which is present in smaller amount is called solute and other one is called solvent.

Types of solutions and there example:-

         first component is solute and second one solvent

1.    Solid in solid              -stainless steel (C and Fe )

2.    Liquid in solid             - water and sodium carbonate (10H2O+Na2CO3  )

3.    Gas in solid                 -H2 in Pd

4.    Solid in liquid             -salt water

5.    Liquid in liquid           -alcohol in water

6.    Gas in liquid               -O2 in water

7.    Solid in gas               -Camphor in N2 gas
   
8.    Liquid in gas              - moisture

9.    Gas in gas                 - O2 and N2 gas in air

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An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL–1, then what shall be the molarity of the solution?

Molecular mass of glycol C2H4(OH)2 =62

Number of moles of C2H4(OH)2 = mass of glycol / molar mass of glycol

= 222.6/62=3.59  moles

Molality of solution = number of moles of solute / mass of solvent in Kg  

Mass of solvent = 200g = 200/1000 Kg = 0.2 kg

Plug the values we get

Molality = 3.59/0.2

Molality = 17.95

Formula of molarity of solution = number of moles of solute / volume of solution in Kg

Formula of volume = mass / density

Volume = 422.6g/(1.072 g/ml)

Volume = 394.21 ml

Convert in liter   

Volume in liter = 394.21 ml /1000 liter  =0.394 liter

Molarity =9.1M

How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both?

In our problem both Na2Co3 and NaHCO3 both are equimolar that means molairy of both substance are equal

So number of moles of Na2CO3  = number of moles of NaHCO3  --------------(1)

Let find moles of both substances

Let take mass of Na2CO3 = x g

Then mass of NaHCO3  = 1-x g

Molar mass of Na2CO3 = 2*23+ 12+ 3*16 =  106 g/ mol

Molar mass of NaHCO3 = 1*23 + 1 + 12 + 3*16 = 84 g/mol

So number of moles of Na2CO3 = mass/ molar mass = x/106 moles     ……….(2)

Number of moles of NaHCO3 is = (1-x)/ 84 gram

Plug the value in equation (1)

  We get

x/106 = 1-x/84

cross multiply them we get

84x = 106  - 106 x  solve it we get

X = 106/190 = 0.558 gram

Plug the value back in equation (2) we get = 0.558/106    = 0.00526 moles

Our reactions will

Na2CO3 + 2HCl ----> 2NaCl  +  H2CO3

NaHCO3+ HCl ------> NaCl +  H2CO3

Here in our reaction 2 moles of  HCl is required each for of Na2CO3 so that

 Moles of HCl required to react with Na2CO3 = 2* 0.00526   = 0.01052

And moles of HCl required for NaHCO3 will also 0.00526

Total number of HCl moles required = 0.01052 + 0.00525 = 0.01578 moles 

Molarity of HCl  = 0.1M   given in our problem

Molarity = number of moles of solute / volume of solution in liter

0.1  = 0.01578/ volume in liter  

Cross multiply now we get

0.1 * volume in liter  = 0.01578

Divide by 0.1 we get

Volume in liter  = 0.01578/0.1   = 0.1578 liter

Volume in ml = 0.01578 liter * 1000 ml / 1liter  = 157.8 ml 

 

A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.

Mass of component in first solution = 25% of 300 gram  = 25*300/100 = 75 gram

Mass of component in second solution   = 40 % of 400 gram = 40*400/100 = 160 gram

Total mass of solution = 300+400 = 700 gram

Total mass of component  = 75+ 160  = 235 grams

Mass % of final solution     = mass of component * 100 / total mass

                                         =235*100/700  = 33.5%

Total of % is always 100

So % of water in solution = 100 – 33.5  = 66.5 %