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Secondary 1 MathematicsLesson 3: Numbers and the Four Operations I

Page 1 of 8 Blue Orange Pte Ltd

Name: ___________________________________ Date:______________________

Lesson 3: Revision of Numbers and Operations

Revision Question 1(a) Given that 26 x 112 = 7744, evaluate 7 .

(b) Write 1728 as a product of its prime factors. Hence find the value of 3 17

2

8

.

(c) (i) Express 3969 and 9261 as products of their prime factors in indexnotation.(ii) Hence, find the positive square root of 3969.(iii) Find the largest integer that is a factor of 3969 and 9261.

[5 marks]

Solution:

(a) 88118112)112(7744 326 ====

(b) 1728 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3

123432)32(1728236

====

(c) (i)

3969 = 34 x 72

9261 = 33 x 73

(ii) 637973)73(3969 224 ====

(iii) 3969 = 34 x 729261 = 33 x 73

H.C.F of 3969 and 9261 = 33 x 72 = 27 x 49 = 1323

2 17282 8642 4322 2162 1082 543 27

3 93 3

1

3 39693 1323

3 4413 1477 497 7

1

3 9261

3 3087

3 1029

7 343

7 49

7 7

1

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Revision Question 2

(a) Express 42 and 315 as products of their prime factors, giving your answer in indexnotation.(b) Find the smallest positive integer psuch that 315pis a perfect square.(c) Find the smallest possible integer qsuch that 42qis a multiple of 315.

[5 marks]Solution:

(a) 42 = 2 x 3 x 7315 = 32 x 5 x 7

(b) 315p= 32 x 5 x 7p= 32 x 52 x 72 (for it to be perfect square)

p= 5 x 7 = 35

(c)15

2q

5x3

2q

7x5x3

7qx3x2

315

42q2

===

For15

2q

315

42q= to be a smallest whole number, q= 15

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Revision Question 3(a) Given that 4752 = 22ax 3bx 114c, find the values of a, band c.(b) The H.C.F. of 56, x and 154 is 14 and their L.C.M. is 4312. Find the smallest possibleinteger, x.

[5 marks]Solution:

(a) 4752 = 22ax 3bx 114c= 24x 33x 11

22 = 4 = 222a = 2

a =

3b= 33b= 3

114c= 114c= 1

c=(b) 56 = 23 x 7

154 = 2 x 7 x 11

The H.C.F of 56 and 154 is 2 x 7 = 14The L.C.M of 56 and 154 is 23 x 7 x 11 = 6164312 616 = 7

The smallest possible integer x is 2 x 72 = 98

Check:

56 = 23x 798 = 2 x 72154 = 2 x 7 x 11

The H.C.F of 56 and 154 is 2 x 7 = 14The L.C.M of 56 and 154 is 23x 72x 11 = 4312

2 4752

2 2376

2 1188

2 594

2 297

3 99

3 33

11 11

1

2 4752

2 2376

2 1188

2 594

3 297

3 99

3 33

11 11

1

2 4312

2 2156

2 1078

7 539

7 77

11 11

1

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Revision Question 4Melanie has 3 pieces of string measuring 120 cm, 192 cm and 252 cm respectively. Shewants to cut the strings into smaller pieces of equal lengths with no remainders.(a) Find the greatest possible length of each of the smaller pieces of string.(b) How many of these smaller pieces of string can she get?

[5 marks]Solution:

120 = 2 x 2 x 2 x 3 x 5 = 23 x 3 x 5192 = 2 x 2 x 2 x 2 x 2 x 2 x 3 = 2 6 x 3252 = 2 x 2 x 3 x 3 x 7 =22 x 32 x 7

H.C.F of 120, 192, 252 = 22 x 3 = 12

(a) The greatest possible length of each of the smaller pieces of ribbon is 12 cm.

120 + 192 + 252 = 564 cm

564 12 = 47

(b) She can get 47 smaller pieces of ribbons of equal length.

Revision Question 5

During Youth Day celebrations, a school distributed 4125 files, 1485 pens and 7260writing pads equally to all the students of the school.(a) Find the largest possible number of students in the school that day.(b) Find the largest number of files, pens and writing pads each student received.

[5 marks]Solution:

4125 = 3 x 53 x 11

1485 = 33 x 5 x 11

7260 = 22 x 3 x 5 x 112

H.C.F of 4125, 1485 and 7260 = 3 x 5 x 11 = 165

(a) The largest possible number of students in the school that day is 165.

(b) Largest no of files = 4125 165 = 25

Largest no of pens = 1485 165 = 9

Largest no of pads = 7260 165 = 44

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Revision Question 6Ferry service Qleaves the jetty every 3 minutes. Ferry service Pleaves the jetty every 4minutes and Ferry Service Sleaves the jetty every 6 minutes. If all 3 ferry services haveferries leaving the jetty at 0600, when will the last time that 3 ferries leave the jettytogether, before 0730?

[5 marks]Solution:

4 = 2 x 26 = 2 x 3

L.C.M of 3, 4 and 6 is 3 x 4 = 12

The three ferry services have ferries leaving together every 12 min.

Multiples of 12:12, 24, 36, 48, 60, 72, 84, 96,

0600 + 0084 = 0724

3 ferries will leave the jetty together the last time before 0730 at 0724.

Revision Question 7(a) Evaluate 3}3)]18(12[21{ +

(b) Arrange the number in descending order.

10

1

,09.0,11

1

,10

1

,101.0,9

1

[5 marks]

Solution:

(a)213633)9021(3]}90[21{3}3]30[21{

3}3)]1812[21{3}3)]18(12[21{

==+===

=+

(b).

1.0......1111.09

1==

1.010

1=

..

090.0...0909090909.011

1==

..

090.0...0909090909.011

1==

In descending order:9

1,101.0,

10

1,09.0,

11

1,

10

1

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Revision Question 8

(a) The solution to a set of numbers is represented on the number line below. Describein words, the set of numbers.

(b) Express

(i)11

9

(ii)2723

(iii)54

53

[5 marks]

Solution:

(a) The set of numbers are real numbers greater than -4 but less than or equal to 5.

(b) (i)..

18.0...81818181.011

9==

(ii)..

158.0...851851851.027

23==

(iii)..

4189.0...9814814814.054

53==

0 1 2 3 4 5 6-6 -5 -4 -3 -2 -1

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Revision Question 9

(a) Express 23

)3

11(

8

1

36

3+ as a fraction in its lowest terms.

(b) Evaluate)2.3(8.2

)3.10(5

+

(i) as a fraction in its simplest form(ii) as a recurring decimal

[5 marks]

Solution:

(a) 9

7

9

161)9

16(2

1

2

1)3

4(2

1

6

3)3

11(

8

1

36

3 223 =+=+=+=+

(b) (i)12

78

12

103

60

515

6

5.51

)2.3(8.2

)3.10(5==

=

=

+

(ii).

358.8...58333333.812

78 ==

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Revision Question 10

(a) Given that 32 x and 15 y , find

(i) The smallest value ofyx

(ii) The greatest value of 32 yx

(iii) The value of y if 252 =y

(b) (i) Estimate correct to one significant figure, the value of02.4

197996.5

(ii) A number corrected to four significant figures is 3 980 000. Write down theminimum possible value of the number.

(iii) Express..

7020.2 to five significant figures.[5 marks]

Solution:

(a)

(i) The smallest value of 21

2=

=

y

x

(ii) The greatest value of 150)125(25)5()5( 3232 === yx

(iii) The value of y if 252 =y is 5=y

(b)

(i) 20214

146

4

1966

02.4

197996.5=

=

(correct to 1 sig. fig.)

(ii) 3 975 000 3 980 000 (correct to 4 sig. fig.)

Minimum possible value of the number is 3 975 000.

(iii) 0207.2...0207207.27020.2..

= (correct to 5 sig. fig)

~ END ~

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