MATHS Solutions For Class 10 MBOSE II Maths Ex 1(a)


 

Chapter-1                                            REAL NUMBER

EXERCISE – 1(a)


SHORT ANSWER TYPE QUESTION (SLAB-I)


1.      USE EUCLID’S DIVISION ALGORITHM TO FIND HCF OF EACH OF THE FOLLOWING PAIR OF NUMBERS.


(i)    18,24

SOLUTION: THE GIVEN NUMBERS ARE 18 AND 24.

SINCE 24>18.

WE TAKE  a=24 AND b=18.

APPLYING EUCLID’S DIVISION LEMMA TO

‘a’ AND ‘b’, we get:

24=18*1+6

18=6*3+0                                                                                     REMAINDER =0 (STOP HERE!)

∴HCF (24, 18)=6.

       (ii)  70, 30

SOLUTION: THE GIVEN NUMBERS ARE 70 AND 30.

SINCE 70>30.

WE TAKE a=70 AND b=30.

APPLYING EUCLID’S DIVISION LEMMA TO

‘a’ AND ‘b’, we get:

70=30*2+10

30=10*3+0


∴ HCF (70, 30)=10.

 

(iii) 714, 924.

SOLUTION: THE GIVEN NUMBERS ARE 714 AND 924.

SINCE  924>714.

∴ WE TAKE a=924 AND b=714.

APPLYING EUCLID’S DIVISION LEMMA TO

‘a’ AND ‘b’, we get:

924=714*1+210

714=210*3+84

210=84*2+42

84=42*2+0

       

     HCF(714,924)=42.

 

(iv) 100, 190

SOLUTION: THE GIVEN NUMBERS ARE 100 AND 190.

SINCE 190>100.

∴ WE TAKE a=190 AND b=714.

APPLYING EUCLID’S DIVISON LEMMA TO

‘a’  AND “b’, we get:

190=100*1+90

100=90*1=10

90=10*9=0.


∴ HCF (100,190)=10.

 

(v) 105,120

SOLUTION: THE GIVEN NUMBERS ARE 105 AND 120.

SINCE 120>105.

∴ WE TAKE a=120 AND b=105.

APPLYING EUCLID’S DIVISION LEMMA TO

‘a’  AND ‘b’, we get:

120=105*1+15

105=15*7+0.


∴ HCF (105,120)=15.

 

(vi) 155, 1385

SOLUTION:  THE GIVEN NUMBERS ARE 155 AND 1385.

SINCE 1385>155.

∴ WE TAKE a=1385 AND b=155.

APPLYING EUCLID’S DIVISON LEMMA TO

‘a’ AND ‘b’, we get:

1385=155*8+145

155=145*1+10

145=10*14+5

10=5*2+0.


HCF (155,1385)=5.

 

 

 

(vii) 4641, 4095.

SOLUTION: THE GIVEN NUMBERS ARE 4641 AND 4095.

SINCE 4641>4095.

WE TAKE a=4641 AND b=4095.

APPLYING EUCLID’S DIVSION LEMMA TO

‘a’ AND ‘b’, we get:

4641=4095*1+546

4095=546*7+273

546=273*2+0

∴ HCF(4641,4095)=273.

(viii) 9367, 3451.

SOLUTION: THE GIVEN NUMBER ARE 9367 AND 3451.


SINCE 9367>3451.


∴ WE TAKE a=9367 AND b=3451.


APPLYING EUCLID’S DIVISION LEMMA TO


‘a’ AND ‘b’, we get:

9367=3451*2+2465

3451=2465*1+986

2465=986*2+493

986=493*2+0


HCF(9367,3451)=493.

 

       2.  USING EUCLID’S DIVISION ALGORITHM FIND THE HCF OF THE FOLLOWING NUMBERS.

 

(i) 296,999,925.

SOLUTION: THE GIVEN NUMBERS ARE 296, 999, 925.


SINCE 999>296

WE TAKE a=999 AND b=296


APPLYING EUCLID’S DIVISION LEMMA TO

‘a’ AND ‘b’, we get:

999=296*3+111

296=111*2+74

111=74*1+37

74=37*2+0

HCF(999,296)=37.

AGAIN,

  SINCE 925>37


∴ WE TAKE a=925 AND b=37


APPLYING EUCLID’S DIVISION LEMMA TO

‘a’ AND ‘b’ we get:

SO,

925=37*25+0


∴ HCF (296,999,925) =37.



(ii) 480, 704, 3680)


SOLUTION: THE GIVEN NUMBERS ARE 480, 704, 3680.


SINCE 704>480

∴ WE TAKE a=704 AND b= 480

APPLYING EUCLID’S DIVISION LEMMA TO

‘a’ AND ‘b’, we get:

704=480*1+224

480=224*2+32

224=32*7+0


∴ HCF (704,480)=32.


AGAIN,


SINCE 3680>32

∴ WE TAKE a= 3680 AND b=32.

APPLYING EUCLID’S DIVISON LEMMA TO

‘a’ AND ‘b’, we get:

3680=32*115+0


∴ HCF (3680,704,480)=32.

 

(iii) 1215, 513, 1134.

SOLUTION: THE GIVEN NUMBERS ARE 1215, 513 AND 1134.

SINCE 1215>513

∴ WE TAKE a=1215 AND b= 513.

APPLYING EUCLID’S LEMMA TO

‘a’ AND ‘b’, we get:

1215=513*2+189

513=189*2+135

189=135*1+54

135=54*2+27

54=27*2+0

HCF (1215, 513) = 27.

  AGAIN,

SINCE 1134>27

∴ WE TAKE a=1134   b=27

APPLYING EUCLID’S LEMMA TO

‘a’ AND ‘b’, we get:

1134=27*42+0


∴ HCF (1215, 513, 1134) = 27.

 

3. FIND THE LARGEST POSITIVE INTEGER WHICH DIVIDES 615 AND 963 LEAVING REMAINDER 6 IN EACH           CASE.

SOLUTION:  HERE, 6 IS THE REMAINDER IN BOTH CASE.

615-6=609   ,    963-6= 957                             (SUBSTRACT THE GIVEN NUMBERS WITH THE REMAINDER)

SINCE 957>609


HERE a=957 AND b=609.

      

    APPLYING EUCLID’S DIVISION LEMMA TO

       ‘a’ AND ‘b’, we get:

        957=609*1+384

        609=384*1+261

        384=261*1+87

        261=87**3+0

        HCF (615 , 963) = 87.

 

4. DETERMINE THE GREATEST NUMBER BETWEEN WHICH WILL DIVIDE 445,572,699 LEAVING        THE   REMAINDER 4,5,6 REPECTIVELY.


SOLUTION: HERE,

   445-4=441      572-5=567      699-6=693        (SUBSTRACT THE NUMBERS WITH THE GIVEN REMAINDER RESPECTEVELY)


SINCE 567>441

HERE a=567  AND b=441.

APPLYING DIVISION LEMMA TO

‘a’ AND ‘b’, we get:

567=441*1=126

441=126*3+63

126=63*2+0.


∴ HCF(441,567) = 63.

AGAIN,

HERE 693>63

APPLYING DIVISION LEMMA TO, WE GET;

693=63*11+0.


∴ HCF(445,567,699) = 63.

 

5. USING EUCLID’S DIVISION ALGORITHM STATE WHETHER THE NUMBERS 47 AND 149 ARE COPRIMES OR NOT.

SOLUTION: SINCE 149>47

HERE a=149 AND b=47

∴APPLYING EUCLID’S DIVISION LEMMA TO

‘a’ AND ‘b’, we get:

149=47*3+8

47=8*5+7

8=7*1+1

7=1*7+0 (zero Remainder)


∴ THE GIVEN NUMBERS ARE COPRIME.


6. USING EUCLID’S DIVISION ALGORITHM FIND WHICH OF THE FOLLOWING PAIRS OF NUMBERS ARE CO-PRIMES:

(i). 272,1032.

SOLUTIION: SINCE 1032>272.

HERE a=1032 AND b=272

∴ APPLYING EUCLID’S DIVISION LEMMA TO

‘a’ AND ‘B’, we get:

1032=272*3+216

272=216*1+56

216=56*3+48

56=48*1+8

48=8*5+8

8=8*1+0


∴ THE NUMBERS ARE NOT PRIME.

 

(ii). 2160,847

SOLUTION:  SINCE 2160>847

HERE a=2160 AND b=847

∴ APPLYING EUCLID’S DIVISION LEMMA TO

‘a’ AND ‘b’, we get:

2160=847*1+466

847=466*1+381

466=381*1+85

381=85*4+41

85=41*2+3

41=3*13+1

3=1*3+0

THE NUMBERS ARE CO-PRIME.


(iii). 867,255

SOLUTION: SINCE 867>255

HERE a=867 AND b=255

APPLYING EUCLID’S LEMMA TO

‘a’ AND ‘b’, we get:

867=255*3+102

255=102*2+51

102=51*2+0


∴ THE NUMBERS ARE NOT CO-PRIME.

 

(iv). 616,309

SOLUTION: SINCE 616>309

HERE a=616 AND b=309

APPLYING EUCLID’S LEMMA TO

‘a’ AND ‘b’, we get:

616=309*1+307

309=307*1+2

307=2*153+1

2=1*2+0


∴ THE NUMBERS ARE CO-PRIME.


7. FIND THE HCF OF 135 AND 225 AND EXPRESS IT AS A LINEAR COMBINATION OF 135 AND 225.

SOLUTION:  SINCE 225>135.

HERE a=225 AND b=135

APPLYING EUCLID’S LEMMA TO

‘a’ AND ‘b’, we get:

225=135*1+90                                                           ⟶ (i)

135=90*1+45                                                              ⟶ (ii)

90=45*2+0                                                                   ⟶ (iii)


∴ HCF OF 135 AND 225=45.

45=135-90*1                                                              ⟶FROM EQUATION (ii).

     =135-(225-135*1)*1                                           ⟶ FROM EQUATION (i).

     =135*2-225*1                                        


HENCE, 45=135x + 225y; WHERE x=2 AND y=-1.

 


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