Maths World
(Teacher Manual)
Class-6
CLASS-6
Review
1.
Numbers
321000
their different = 321000 –
100023 = 220977
7. Place value of 3 in 1790 3 2 456 = 3
10000 = 30000
Face value if 4 in 1790 3 2456 = 4
(digit itself )
8. (a) ascending order :
29326732, 423258016,
491785931, 968253241
(b) do same as above manner.
9. (a) descending order :
278123512, 248468210,
100529582, 41835194
(b) do same as above manner.
10. (a) Sum = 100000073 + 3267924 +
42532 = 103310529
(b) do same as above or by usual
method.
11.(a) 90000000 – 32408232 =
57591768
(b) do by usual method
12. (a) 42326 101 = 4274926
(b)-(d) : do as above i.e by usual
method.
13. (a) 999989 + 1 = 999990
(b)-(d) : do same as (a).
14. (a) 1000000 – 1 = 999999
(b)-(d) : do same as part (a).
15. (a)
522 1624 3
–1566
HCF 58 522 9
–522
0
InternatioIndian Numnal Number
ber System
System
(a)
7175129
71,75,129
7,175,129
(b)
823659
8,23,659
823,659
(c)
1538361
15,38,361
1,538,361
(d)
2595689
25,95,689
2,595,689
2. (a) 1,53,26,897= One crore fifty
three lakhs twenty six thousand
eight hundred and ninety seven.
(b) 792,685= Seven lakhs ninety
two thousand six hundred eight
five.
(c) 9,57,211= Nine lakh fifty seven
thousand two hundred and eleven.
(d) 51,731,217= Fifty one million
seven hundred thirty one thousand
two hundred and seventeen.
3. 7654369 + 1000 = 7655369,
7656369, 7657369.
4. (a) 96, 89, 32, 000 = 900000000 +
60000000+8000000+900000+
30000+2000
(b) 9, 89, 00, 001 = 90000000+
8000000+900000+1
(c)-(f) : do same as above manner.
5.(a) 7000000+200000+20000+
2000+ 20+2 = 72, 22, 022
(b)-(d) : do same as above manner.
6. The smallest 6-digit number=
100023
The greatest 6- digit number =
1
2
3
5
7
<
<
<
15
12 19 18
(b) do same as above.
21. (a) In descending order :
15
7
6
9
<
<
<
8
8
7
12
required HCF of 1624, 522,
(d) do same as above.
1276 = 58.
1
100
(b)-(h) : do same manner as above 22. (a) 33 3 % = 3 %
part (a).
100
1
16. (a) LCM of 1572, 1344
=
=
3
100
3
required LCM = 2 2 3 2
2 2 2 7 131
(b)-(f) do same manner as above
= 12 16 7 131
part (a).
= 176064 Ans
1 100
23. (a) 1 =
=4%
=
4
2 1344, 1572
25
Now, 58 1276
–116
116
–116
0
22
2
3
2
2
2
2
7
131
672, 786
336, 393
112, 131
56, 131
28, 131
14, 131
7, 131
1, 131
1, 1
(b)-(h) : do same manner as above
part (a).
17.(a) MDLXXX1 = M+D+L+X+X
+X+1
= 1000 + 500 + 50 + 30 + 30 + 30 +
1 = 1581
(b)-(I) : do same as above manner.
18. (a) 55 = 50 + 5 = L + V = LV Ans
(b)-(L) : do same as above manner
i.e, part (a).
19. (a)
9
135
= 8
120
(b)-(f) : do same as above ie, cancel
common factors.
20. (a) In Ascending order :
(b)-(f) do same as part (a).
24.
7
20
= 0.35
(b)-(f) : do as usual method.
1
1
25. (a) 12 – 4 3 – 6 4
144 – 52 – 75
12 13
25
–
–
=
12
1
3
4
17
5
=
=1
12
12
=
(b)-(f) : do same as above manner.
26.(a) 319.12–51.789+219.332–
91.219=(319.12+219.332)–(51.78
9+ 91.219)
= 538.452 – 143.008
= 395.444 Ans
(b) do same as above.
27. (a) 4 days 6 hr = 4 days + 6hr
= 4 24hr + 6hr
[... 1 day =
24hr]
= 96hr + 6hr = 102hr Ans
(b)-(d) : do same as above manner.
2
28. (a) elapsed time = 8.00– 2.30 =
(b) Sum of angles = 60 + 60 + 70 =
5.30 = 5hr 30 min
190° 180° No
9
(b)-(d) : do same as above.
100 % = 75%
=
12
29. (a) 10:00 AM = 1000hr
42. required %
(b) 12:15 PM= 1215hrs
(c)-(d) : do same as above manner.
43.Passing marks = 18 + 36 = 54.
HCF LCM
net max. mark = y
=
one number
30. Other number
3636% of Maximum marks = 54
36
y = 54
= 720
= 144
100
54
y=
100 = 150
36
31.Least number will be LCM of
(150, 180).
Max. Mark = 150
LCM of 150, 180 = 900 Ans
44. 1 hour = 60 min
12min
32.The greatest number will be HCF
100 %
of 2016 & 1056 and HCF of (1056,
required % = 60min
2016) = 96
1
= 20 %
91
33.Prime number less than 50 = 2, 3,
2
5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 45. required No. of bottles =
1
1
41, 43, 47.
183
2
34.Sum = 3.003+2.05=5.053,
2
183
= 61
=
= 3
difference=3.003–2.05=0.953
3
required result=5.053–0.953 = 4.1
2
35. do same as Q.N. 34.
46. do yourself
36. (a) 3 (7.4–2.1) – 1.2 (1.8–0.6) = 3
47.Man's annual income (ie, 12
5.2 – 1.2 1.2 = 156 –1.44 =
months income) = `2,50,000
250000
14.16
his
monthly
income
=
12
(b) do yourself by BODMAS Rule.
= 1`20833.33
37. (a) Yes, since 5+3 > 4, 3+4 > 5, 5+
48. 2 Milk weight = 241.5gm
4>3
3
3
(b) do similar as above part (a).
(241.5gm)
4 Milk weight = 2
38. (a) Scalene, isosceles, equilateral
724.5
(b) acute, obtuse, right
=
2
= 362.25gm
39.(a) Complement angle of 89 =
90°–89° = 1°
(b)-(c) : do same as above ie, part 49. Amount of money left with Reena
(a).
= 250 – (12 15.50 + 2 1.25)
40.(a) supplement angle of 100° =
= 250 – (186 + 2.5)
180°–100° = 80° Ans
= 250 – 188.5 = `61.5
(b)-(c) : do same as above part (a).
50. 25m of cloth cost = `1062.50
41.(a) Sum of angles = 20 + 60 + 100
1062.50
12m of cloth cost
12
=
= 180° Yes
25
3
= `510
51.The Minimum distance should
each walk
= LCM
33, 36, 42
3 33,
36, of42,
11 11, 12, 14
2 1, 12, 14
6 1,
6,
7
7 1,
1,
7
1,
1,
1
= 3 11 2 6 7 = 2772
52.Total distance between two cities
=1692 km
Time taken by train = Thursday at
8PM – Saturday at 8A.M
distance
= 24 + 12 = 36hr
=
time
Avg
1692speed of the train
= 47 km/h
=
36
C.P (100 + P%)
S.P =
100
53.
650
= (100+20)
650
= 120
d
r=
2
54. (a) diameter = 9.2
9.2
r
o
A
=
= 4.6cm
4.6
2
(b) do same as above part (a).
55. (a) radius = 5cm
Taking 5cm measurement in
r
o
A
Ruler, draw a circle as
5cm
(b) do same as above.
Knowing Numbers
Exercise - 2A
1. (a) The smallest natural number =1
(b) The first and tenth natural
4
number = 1 and 10.
(c) The smallest and first whole
number = 0.
(d) 4 digit numbers start from 1000
to 9999.
Required No. of 4-digit
numbers = 9999 – 999 = 9000
(e) 7- digit numbers are 1000000 to
9999999
Required No. of 7-digit number
= 9999999 – 999999 = 9000000
2. (a) Eight thousand twenty-two =
8022
(b) One billion fifty thousand six =
1,00,00,50,006
(c) Four billion five million =
4,005,000,000
(d) fifty lakh nine thousand eightyfive = 50,09,085.
(e) Six arab eight lakh seven
thousand three hundred ninetynine = 6,00,08,07,399
3. (a) 596851
In Indian system : 5,96,851
Five lakh, ninety-six thousand,
eight hundred fifty-one;
(b) 906538
In Indian System : 9,06,538
Nine lakh, six thousand, eight
hundred fifty-one;
(c) 3480706
In Indian System : 34,80,706.
thirty-four lakh, eighty thousand,
seven hundred and Six;
(d) 80950252
In Indian system : 8,09,50,252
Eight Crore, nine lakh, fifty
thousand, two hundred fifty-two;
4. (a) 79312735
In International System : 79, 312,
735
Seventy-nine million, three
Units
Tens
Hundreds
Thousand
Ten
thousand
Thousands
Hundreds
Tens
Units
4
Lakhs
5
0
2
4
5
8
1
6
5
5
0
8
Ans (a)
Ans (b)
Ones
Thousands
9
Millions
Hundred
Millions
Ten
Millions
Millions
Hundred
Thousands
Ten
Thousands
Ten Lakhs
Crores
Ten crores
hundred twelve thousand, seven 7. (a) Not Possible (i.e, there is no
hundred thirty-five;
greatest and last whole number)
(b) Not Possible (i.e, there is no
(b) 97142160
greatest and last natural number)
In International System : 97, 142,
8. In Indian Word System :
160
(a) 65,392 = sixty-five thousand
N i n e t y - s eve n m i l l i o n , o n e
three hundred ninety-two
hundred forty-two thousand, one
(b) 56,789 = Fifty-Six thousand
hundred sixty;
seven
hundred eight-nine
(c) 8574635
(c) 3,21,53,215 = Three crore
In International System : 8, 574,
twenty-one lakh fifty-three
635
thousand two hundred fifteen
Eight million five hundred
(d) 9,54,327 = Nine lakh fifty-four
seventy-four thousand, six
thousand three hundred twentyhundred thirty five;
seven
(d) 235120
9. In International Word System :
In International System : 2, 35, 120
(a) 136,889 = One hundred thirty
Two hundred thirty five thousand,
six thousand eight hundred eightyone hundred twenty.
nine
5. Numbers in Indian place value
(b) 56,205,306,760 = Fifty-six
chart :
billion two hundred five million
three hundred six thousand seven
Periods
hundred
sixty
Crores Lakhs ThousOnes
ands
(c) 433,972,522,827 = Four
hundred thirty-three billion nine
hundred seventy-two million five
hundred twenty-two thousand
Ans (a)
4 1 8
eight hundred twenty-seven
(d) 5,032,801 = five million thirtyAns (b)
7 2 6 8 9
two
thousand eight hundred one
Ans (c)
3 4 6 7 0 9 9
(e) 1, 000, 660 = One million six
Ans (d) 4
1 1 5 0 7 9 7 8
hundred sixty.
(f) 5,110,600,600 = Five billion
6. Numbers in the International place
one hundred ten million six
value chart :
hundred thousand six hundred
10. 8783957
Periods
Ans (c)
7
0
1
0
4
0
0
0
7
Ans (d)
6
0
3
0
5
0
4
0
7
Place Value of 8 = 8
10000 = 80000
Place Value of 8 = 8
1000000 = 8000000
Required difference = 8000000
– 80000 = 7920000
5
11. 9654321587
million
21.1 billion = 100 crore = 1000
Place Value of 5 =
million = 1000 10 lakh = 10,000
5 100 = 500
lakh
Face value of 5 = 5
required difference = 22. Reverse number of 579 = 975
Sum of 579 and its reverse
500 – 5 = 495
number = 579 + 975 = 1554
Place Value of 5 = 5
10000000 = 50000000 23. 1 Million = 10 lakh = 1,000,000
Reverse of 1,000,000 = 0000001
and face value of 5 = 5
required difference = 999999
required difference = 50000000
24. 3-digit number using the digits (0,
– 5 = 49999995
6, 7) = 607, 706, 760, 670
12.(a) The greatest 5-digit number =
25.Smallest 4-digit natural number
99999
having three different digits = 1002
The smallest 5-digit = 10000
required total number of 5digit 26. The smallest 3-digit number = 101
= 99999 – 9999 = 90000
(reversible also)
(b) : do as above.
The greatest 3-digit number = 999
13.Required whole numbers between
(reversible also)
55 and 155 = 99
27.From 100 to 1000, 2 occur in ten's
14.No, all whole number are not
and unit's place = 180 times
natural numbers but all natural 28.159060708 : Fifteen crore ninety
number are whole numbers.
lakh sixty thousand seven hundred
15. In expanded form :
eight.
(a) 58376 = 5 10000 + 8 1000
and smallest number =100056789.
+ 3 100 + 7 10 + 6
Exercise - 2B
= 50000 + 8000 + 300 + 70 + 6
1. (a) Predecessor of 2659 = 2659 – 1
(b), (c), (d) : do as above manner
= 2658
16.In short form
(b), (c), (d) up to (i) - do same as
(a) 1100000 + 80000 + 9000 + 600
above
+ 10 + 5 or 11 100000 + 8
2. (a) Successor of 2589700 =
10000 + 9 1000 + 6 100 + 1
2589700 +1=2589701
10 + 5 1 = 1189615
(b)- (i) - do same as above.
(b) Similarly do as above manner
3. Representation of given number on
17. Since 1 lakh = 100 thousand
number line :
10 lakhs = 10 100 = 1000
thousand
A
B
D
E
F
C
18.Since 1 arab = 100 crore = 100
4. (a) 0> (b)
<
(c)
<
(d)
<
(e)
< (f)40> (g) >
10000000
8
16 20 22 25 30
(h) >
= 1000000 1000 = 100000
5. In descending order : 1650 > 1507
10,000 = 100000 ten thousands
> 1408 > 1379 > 672 > 542 > 291 >
19. 10 Millions = 1 crore = 100 lakhs =
26 > 5.
10 ten lakhs
20.1 billion = 100 crore = 1000 6. In ascending order : 0 < 12 < 15 <
6
their difference = 612
16.All Possible 3-digit number = 631,
613, 316, 361, 136, 163
17.3 Consecutive whole numbers just
preceding 3769001 are 3768998,
3768999, 3769000
18. (a) F (b) F (c) T (d) T (e) F
Exercise - 2C
1. Shikha bought cloth of one kind =
72m 25cm
Shikha bought cloth of another
kind = 7m 50cm
Total length of the cloth bought
by Shikha = 79m 75cm
2. Total weight of wheat being carried
by the truck = 459kg + 437kg +
456kg + 695kg + 724kg = 2771 kg
3. Shopkeeper has total sugar =
1500kg
... Sugar sold 1 day = 90kg
sugar sold 14 days = 90 14kg =
1260kg
now, required quantity of sugar is
left = 1500kg – 1260kg = 240kg
4. Total Population in the village =
73,25,568
Number of males in the village =
34,19,439
Number of Females in the
village = 7325568 – 3419439 =
3906129
5. Price of a Scooter = `49,283
Discount on it = `5,328
Amount paid by latika = `49283
– `5328 = `43955
6. Distance covered by sagarika in 1
day = 3km 125m
Distance covered by sagarika in
26 days = 26 (3km + 125m)
= 26 (3125m) = 81250m
122 < 1358 < 15976
7. Four continuous Predecessor of
526 526 – 1 = 525
525 – 1 = 524
524 – 1 = 523
523 – 1 = 522
Similarly, four predecessors of 1749
= 1748, 1747, 1746, 1745
8. 5 Continuous Successors of 1341 =
1341 + 1 = 1342
1342 + 1 = 1343
1343 + 1 = 1344
1344 + 1 = 1345
Similarly 5 continuous successors
of 1509 = 1510, 1511, 1512, 1513.
9. (a) < (b) < (c) < (d) > (e) < (f) > (g) =
(h) >
10. (a) The Smallest & 8 digit number
= 10000000
(b) The Smallest 9-digit number =
100000000
(c) The greatest 5-digit number =
99999
(d) the greatest 8-digit number =
99999999
11.The greatest 4-digit number = 4320
The Smallest 4-digit number =
2034
required difference = 2286
12.(a) All Possible number without
repeating any digit are 809, 890,
908, 980
(b) 358, 385, 835, 853, 583, 538
(c) 476, 467, 746, 764, 647, 674
13.(a) Possible 4-digit number = 1000
(b) Possible 4-digit number = 5002,
5020, 5200, 2005, 2050, 2500
14. All Possible 2-digit number
(a) 80 (b) 68,86
(c) 36, 35, 63, 53,65,56
15.The greatest 3-digit number = 820
The smallest 3-digit number = 208
7
= 81.250km = 81km 250m
to the nearest 10 = 30
(d) to (j) - Similarly, do as above.
2. Estimate to the nearest 100 :
(a) 129 Since ten place digit 2<5,
therefore
replacing this digit and unit digit by
‘0’ and keeping the remaining digit
same we have Estimated to the
nearest 100 = 100
(b) 445 4 45
81250
7. ... 98km
workers earned in 1 day =
1000
`3120
98 workers earned in 11 day = 11
(` 3120) = `34320
8. Correct answer when multiplied
7290 by 48 = 7290 48 = 349920
Wrong answer when multiplied by
7290 by 84 = 7290 84 = 612360
Difference in both answers =
Estimated to nearest 100 = 400
612360 – 349920 = 262440
(c)
256 2 56
Hence, Aditya's answer was greater
by = 262440
Estimated
to nearest 100 = 300
9. Total length of cloth Mohan had =
4 00
(d)
959
9
59
28m 14cm
= 2800cm + 14cm = 2814cm
Estimated to nearest 100 = 1000
Shirts Stitched by tailor from this
3 00
cloth = 14
(e)-(j) : Similarly, do as above.
cloth was used to make 1 shirt = 3. Estimate to the nearest 1000 :
2814cm 14
(a) 7128, since hundred place digit
10 00
1<5, therefore
replacing this digit,
tens digit and unit digit by ‘0’ and
keeping the remaining digit same,
10.Rohini sells milk in 1 day =
2814
we
have
129litres
cm = 201cm or 2m1cm.
=
Estimated to the nearest 1000 =
There14are 30 days in the month of
7000
June.
i.e, 7128 7 128
Rohini sells milk in 30 days =
129l 30 = 3870 litres.
Estimated to nearest 1000 =
Exercise - 2D
7000
1. Estimate to the nearest 10 :
(b) 9823, since hundred place digit
(a) 91, since 91 is more close to 90.
8>5,
therefore replacing this digit,
Estimated to the nearest 10 = 90
tens
digit
and
unit digit by ‘0’ and
7 000
(b) 85, since the unit place digit is 5,
increasing
thousands
digit by 1 and
therefore replacing 5 by ‘0’ and
keeping
the
remaining
digit same,
increasing 1 to the tens place digit,
we
have
we get
Estimated to the nearest 1000 =
Estimated to the nearest 10 = 90
10000
(c) 32, Since the unit digit is 2 which
i.e,
9823 9 823
is less than 5, therefore replacing 2
by ‘0’ and keep the remaining digit
Estimated to nearest 1000 =
(i.e here 3) same, we get Estimated
8
10000
(c) 6189 6 189
598 estimated to the nearest 100 =
600
236 estimated to the nearest 100 =
200
required estimation = 600 – 200
= 400
(b)-(c) do as above.
9. (a) 63681 – 27650
63681 estimated to the nearest 1000
= 64000
27650 estimated to the nearest 1000
= 28000
required estimation = 64000 –
28000 = 36000
10.Rounding off the nearest 100 :
(a) 267 132
267 is rounding off the nearest 100
= 300
132 is rounding off the nearest 100
= 100
required estimation = 300 100
= 30000
(b) (c) similarly do as above.
Exercise - 2E
1. In Roman Numerals :
(a) 84 = 50+10+10+10+4 =
LXXXIV
(b) 95 = 90+5 = XCV
(c) 30 = 10+10+10 = XXX
(d) 43 = 40+3 = XLIII
2. In Hindu-Arabic Numerals :
(a) LXIV = 50+10+4 = 64
(b) XXI = 10+10+1 = 21
(c) LXII = 50+10+2 = 62
(d) LXXIX = 50+10+10+9 = 79
3. In Ascending order : XIII < XX <
XLIX < XC or XIII, XX, XLIX,
XC.
4. In Descending order : M, XC, LX,
XXXIX, X.
CCE DRILL-1
Tick () the correct answer :
Estimated
to nearest 1000 =
10 000
6000
(d)-(J) : Similarly, do as above.
4. (a) 58+61
58 estimated to the nearest 10 = 60
6 to
000
61 estimated
the nearest 10 = 60
required estimation = 60+60 =
120
(b)-(d) : Similarly as above.
5. (a) 1466 + 2314
1466 estimated to the nearest 100 =
1500
2314 estimated to the nearest 100 =
2300
required estimation =
1500+2300 = 3800
(b)-(d) : similarly as above.
6. (a) 63682+27642
63682 estimated to the nearest 1000
= 64000
27642 estimated to the nearest 1000
= 28000
required estimation = 64000 +
28000 = 92000
(b) 4832 + 1394
4832 estimated to the nearest 1000
= 5000
1394 estimated to the nearest 1000
= 1000
required estimation = 5000 +
1000 = 6000
(c) do as above.
7. (a) 62 – 26
62 estimated to the nearest 10 = 60
26 estimated to the nearest 10 = 30
required estimation = 60 – 30 =
30
(b)-(d) do as above.
8. (a)598 – 236
9
(a) +1 (b) 0 (c) 3009 (d) 7921 (e) 1 (f)
ten (g) hundred (h) 99 (I) XLIX (j)
1000gm (k) 50 (l) 8900
CCE DRILL - 2
1. Fill in the blanks :
(a) Natural (b) 1 (c) 9 (d) 6000 (e)
100 (f) 1 (g) 900000 (h) 679239 (i)
362481 (j) 100000
2. Write T for true of F for false for
the following :
(a) T (b) F (c) F (d) F (e) F (f) T (g) T
(h) F (i) T (j) F
HOTS QUESTIONS
1. Ramya spent on a DVD player =
`4892
Ramya spent on a television =
`26884
Now, the sum of 4892+26884 to
the nearest hundreds.
`4892 estimated to the nearest
hundred = `5000
`26884 estimated to the nearest
hundred = `27000
required estimation = `27000 +
`5000 = `32000
Hence, Ramya's estimation is
correct.
2. In a magic basket, the number of
oranges get double after every
hours.
Let, initially the no. of oranges
were = x
In 1 hour it becomes = 2x
Basket becomes full (20x) in = 10
hours
Basket becomes half full (10x) in
= 9 hours.
3. We consider this problem as
follows:
Hence, they have aRaghav
day off together
Works
Off
on 13 Jan.Off Off
Operations
Whole
27,
28, 29, 30, 31, 1, on
2, 3, 4,
5, 6, 7, 8, Numbers
9, 10, 11, 12, 13
Exercise -3A
Raghav
Raj Both
Both
Off Raj
1.
Works
Works have
Works have
off
days
off
days
2. (a) 0,8
1 of
2 38. 4 5 6
0 is on the0left
Number line
(b) +9, +2
0 1
2
3 4
5
7
8
6 7
8
Left of 8
+2 is on the left of +9.
(c)-(d) : do same as above manner.
3. (a) 4567+6812 = 11379
(b)-(f) : do
same as part 3 (a) and
0 1 2 3 4 5 6 7 8 9
6812+4567 = 11379
4. (a) (456+320) + 472
Left of 9
(a+b) + c = a + (b+c) (Associative
law)
(456+320) + 472 = 776+472 =
1248
and 456 + (320+472) = 456+792 =
1248
(b)-(f) : do same manner as above
4(a).
5. (a) 663 + (922+834) = 663+ (1756)
= 2419
(b)-(f) : do same manner as above 5
(a).
6. (a) 2468 + 1437 + 9563 + 3532 =
17000
(b) 12592 + 7663 + 408 + 2337 =
10
23000
7. (a) 5+2
Start from 2, jump 5 times to the
right. The last number on which
jump completes is 7. so 2+5 = 7
(b)-(a) : do same manner as above.
8. (a) 6 3 start from 0, more 3 unit a
1 right.
2 3 4 now
5 6 make
7 8 9 6 such
time to 0the
move. we reach the pint 18. so = 3
6 = 18
(b)-(d) : do same manner as above
in part (a).
9. largest number of 6-digit = 999999
Smallest number of 7-digit =
1000000
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
difference = 1000000 – 999999 =
1
10. do same as above (Q.No. 9 )
11. (a) 621 – (422 – 121) = 621 – 301 =
320
(621 – 422) – 121 = 199 – 121 = 78
(b)-(d) : do same as above part (a).
12. (a) 427 – 126 = 301
Check : Result + smaller number =
larger whole number = 301+126 =
427
(b)-(f) : do same as above manner is
part (a)
13. (a) 22 27 = 27 22
Commutative property.
(b) 32 35 + 32 5 = 32 (35 + 5)
distributive property of multiplication over addition
(c) 5672 = 0 closure property
(d) 34 40 – 34 10 =34 (40 –
10)
distributive property of multiplication over subtraction.
14. largest 4-digit number = 9999
largest 6-digit number = 999999
Their product = 999999 9999
= 9998990001
15.Yes, the product of any two odd
whole number is odd. true.
16. (a) 203245 = 91440
17. (a) 999 99 = 99 (900 + 99)
= 99 900 + 99 99
= 89100 + 9801 = 98901
(b)-(d) : do same manner as above.
18. (a) 124 15 + 124 45 = 124 (15
+ 45) = 124 60 = 7440
(b)-(d) : do same manner as above
part
19. (a) 87540 = 87200 = 17400
(b)-(f) : do same manner as above
part (a).
20. (a) 0 443 = 0
(b) 19732 + (459565 5) = 19732 +
91913 = 111645
(c) 495 – (625 25) = 495 – 25 =
470
(d) 972 1 = 972
(e) (2122 2122) – (2121 2121)=
1–1=0
(f) (15000 150) + 200 (15000
15) = 100 + 200 1000 = 100 +
200000 = 200100
21.Smallest 7-digit number = 1000000
required 7-digit number
which is exactly divisible by 79
= 1000000 + (79 – 18)
= 1000000+61 = 1000061
11
12658
79 1000000
–79
210
–158
520 - 3B
Exercise
–474
1. Total number
of people in the
village = 7564460
+ 2122 + 436 =
–395
10122
2. Amita spent in all650
= `144 + `1227 +
–632
`356 = `1727
18
3. Number of cars manufactured by
the company in 3 months = 10909
+ 23082 + 24091 = 58082
4. Total marks get by Ishan = 28 + 54
+ 92 + 46 = 220
5. Total voters in the city = 44000
women voters = 15000
no of man voters = 44000 –
15000 = 29000
6. Total population of village = 1785
no. of men = 785
no. of women = 500
no. of children = 1785 – (785 +
500)
= 1785 – 1285 = 500
no. of old men = 200
no. of old women= 126
no. of young men = 785 – 200 =
585
no. of young women = 500 – 126 =
374
7. Speed of the train = 45km/hr,
Time taken in traveling = 87hr
Total distance traveled = speed
time
= 45 87 = 3915km
8. No of sections in class VI = 4
Total no. of students in 4 sections =
40 + 40 + 40 + 40 = 160
The fee taken by each student = `75
Total collection of fee from class
VI = 160 75 = `12000
9. ... Cost of 55 coats =` 308055
Cost of 1 coats =
10. Let the second number be y.
Then. y 28 = 18816
308055
11.Divided = 98928321,
quotient =
`
55
304394
Remainder
= 271, divisor (i.e, the
=
`5601
number) = ?
We know that, dividend = (Divisor
quaint) + remainder
18816
= d 304394 + 271
98928321
y=
28 = 672
98928321 – 271 = d304394
98928050 = d3043943
Exercise - 3C
1. (a) 1, 10,100, 1000, 10000, 100000
(b) 2, 8, 32, 128, 512, 2048
(c) 5, 8, 11, 14, 17, 20
(d) 11, 33, 99, 297, 891, 2673
2. 1 + 3+5+7+9 = 25, 1 + 3 + 5 + 7 + 9
+ 11 98928050
= 36
= 325
d
=
3. 1 10000
304394+ 1111 = 11111, 1
100000 + 11111 = 111111
4. 20000000 2 = 10000000,
2000000000 2 = 1000000000.
C.C.E. DRILL - 1
Tick () the correct answer :
(a) 372 (b) 0 (c) Associative law of
Addition (d) Commutative property (e) 8 0 (f) 0 1 (g) Yes (h) (a
b) c = a (b c) (I) 9792 (j)
(125 4) (20 5)
C.C.E. DRILL -2
1. Fill in the blanks :
12
(a) 360, 200 (b) 0 (c) 0 (d) 10 (e) 0 (f)
0 (g) face (h) a (I) whole number (j)
commutative
2. Write T for true of F for false for
the following :
(a) T (b) T (c) F (d) F (e) F (f) F (g) F
(h) T (I) T
HOTS QUESTIONS
1. In a bouquet, flowers are = 8+9 =
17
In 7 bouquets, flowers will be =
177 = 119 flowers
2. Number between 20 and 30 which
divide 92 and 71 gives the
remainder as 2 is 23.
e.g
= 36 – [18{14 – 11}] = 36 – [18 – 3]=
36 – 15 = 21
7. 120 40 20 + 15 – 3 = 120 2 +
15 – 3 = 240 + 15 – 3 = 240 +12 =
252
8. 81 [73 – {8 6 + (15 – 3 2)}] = 81
[73 – {8 6 + 9}]
=81 [73 – {48 + 9}] = 81 [73 – 57] =
81 [16] = 81 16 = 1296
9. 19 – [4 + {16 – (12 – 2)}] = 19 – [4 +
{16 – 10}] = 19 – [4+ 6] = 19 – 10 = 9
10. 27– [18 – {16 – (5 – 4 – 1)}] = 27 –
[18 – {16 – (5 – 3)}] = 27– [18 – {16
– 2}]
= 27 – [18 – 14] = 27 – 4 = 23
Exercise - 4B
1. (a) Factor : An exact divisor of a
number is a factor of that number.
e.g. 1, 6, 36 are factors of 36.
(b) multiple : A number is a
multiple of any of it factors.
e.g; 36 is the multiple of 1, 6 and
36.
3. Yes, there is only one such whole
number which when multiplied by 2. (a) 26
itself gives the product equal to the
23 94
Factor of 26 = 1, 2, 13, 26
number
itself4which is 1.
92
(b)-(d) do as the same manner as in
e.g 1 1 = 1 (number itself)
2 reminder
part (a).
Playing with Numbers
3. Prime numbers; A number which
Exercise
4A
23 71 3
has only two factors, namely 1 and
1. 500 –69
5 + 200 4 = 100 + 50 =150
the
number itself, is called a prime
2. 12 8 –2 4+16
= 12 4 +16 = 3
reminder
number
+16 = 19
e.g; 2, 3, 5, 7, 11 etc.
3. 24 8 + 9 2 = 3 +18 = 21
1
26
4.
Composite
number;
the numbers
4. 30 – [6+{32 – (40 – 10)}] = 30 – [6
2
13
having
more
than
two
factors are
+{32 – 30}]
called composite number.
= 30 –[6 + 2] = 30 – 8 = 22
e.g, 4, 6, 8, 9, 10, 12, 14 etc.
5. 60 – [40 – {20 – (10 – 8)}] = 60 – [40
5.
(a) 12
– {20 – 2}]
12
1 = 12, 12 2 = 24, 12 3 =
= 60 – [40 – 18] = 60 – 22 = 38
36, 12 4 = 48, 12 5 = 60
6. 36 – [18 – {14 – (15 – 4 2 2)}] =
First five multiples of 12 = 12,
36 – [18 – {14 – (15 – 8 2)}]
24, 36, 48, 60
13
(b)-(e) : do same as above par (a)
6. even numbers : (b), (d), (e), (g)
odd numbers (a), (c), (f), (h)
7. (a) 40 and 8 41, 43, 47, 53, 59, 61,
67, 71, 73, 79
(b)-(c) do same as above rule :
8. (a) smallest prime number = 2
(b) all even prime number = 2 only
(c) smallest odd prime number = 3
9. 5 consecutive odd numbers : (1, 3,
5, 7, 9)
10. greatest multiple of 5 less than 60
= 55
11. smallest multiple of 90 less than
190 = 180
12. multiple of 8 : 24, 72
13. prime number : 97, 7, 89
14. Yes, A composite number can be
odd. e.g, 9.
15.co-primes : If the two numbers
have no common factor other than
1, they are called 6 primes.
e.g; (a) 2, 3 (b) 3, 4
16.Twin-Primes : two consecutive
odd prime numbers are called
twin-primes.
e.g (a) 3, 5 (b) 5,7 (c) 11, 13 (d) 17,
19 (e) 29,31, (f)41,43 (g) 71, 73
17.(a) 36 = 17+18 (b) 42 = 5+37 (c)
108 = 19+89 (d) 86 = 13+73
18. Yes
19.(a) 24 = 11+13 (b) 36 = 17+ 19
(c) 120 = 59+61 (d) 84 = 41 + 43 (e)
144 = 71+73
20.(a) False (b) False (c) False (d)
False (e) True
21. 43 = 7+17+19
22. 19 = 3+5+11
23.Prime numbers between 2 – 50
which are the sum of two equal
number = 2
24.There are only one even prime
number i.e, 2.
Exercise- 4C
1. (a) 6573 = 3, 7 (b) 5368 = 2, 4, 8, 11
(c) 8135 = 5 (d) 4895 = 5 (e) 7168 =
2, 4, 8 (f) 110010 = 2, 5, 7, 10, 11 (g)
76285 = 5 (h) 9092 = 2, 4 (i) 396 = 2,
3, 4, 6, 9, 11 (j) 3178965 = 3, 5 (k)
17908 = 2, 4, 11(l) 568762 = 2
2. (a) 2570 Yes, since unit digit is 0
(b) 23075 No, since unit digit is 5.
(c) 594321 No, since unit digit is
odd i.e, 1.
3. (a) 45678
Sum of its digit = 4+5+6+7+8 = 30
30 is divisible by 3. So, 45678 is
divisible by 3.
30 is not completely divisible by 9.
So, it is not divisible by 9.
(b)-(c) : do same as above manner.
4. (a) 6252
last two digit of it =
so, 6253 is divisible by 4.
(b)-(c) : do same as manner.
5. (a) 3265
(i) Since last digit i.e unit place digit
is 5 so 3265 is divisible by 5.
(ii) It's unit digit is 5.
Number formed by remaining
digits is 326.
326–25 = 326–10 = 316, which
is not exactly divisible by 7.
(b)-(c) : do same as above manner
52
(i.e, Rule).
= 13
4
6. (a) 26910
It's unit digit is 0 It is divisible by
2.
The sum of its digits = 2 + 6 + 9 + 1
+ 0 = 18, Which is also divisible by
3.
So, 26910 is divisible by 6.
(b)-(c) : do same as above manner
14
(i.e Rule).
7. (a) 10206
It's Unit digit is 6.
Number formed by remaining
digit is 1020.
1020 –26 = 1020 –12 = 1008,
exactly divisible by 7.
(b)-(c) : do same as above manner
(i.e Rule).
8. (a) 756840
last three digit =
exactly divisible by 8.
so, 756840 is divisible by 8.
(b)-(c) : do same as above manner
i.e Part (a).
9. (a) 5687
Sum of digit in even places = 6 + 7
= 13
Sum of digits in odd places = 5 + 8
=13
difference of these sum = 13 – 13 =
0
840 is divisible by
So, the number 5687
= 105, which is
8
11.
(b)-(c) : do same as above part (a).
10. (a) 459734 (b) 587686 © 258456
11. (a) 37088 (b) 6504 © 52144
12. (a) False (b) True © True
13. (a) do same as Q.No. 9 above.
(b) do same as Q.No. 4 above.
(c) do same as Q.No. 3 above.
14. Prime numbers (b), (d), (e), (f), (g),
(h).
Exercise - 4D
1. (a)
7
part (a).
2. Prime Factorization are (b), (c).
3. (a) 4900
56
2 28
2 14
4900 = 2 2 5
5
2 7 77
(b)-(j) : do same as part (a).
4. The largest 5-digit number = 99999
Prime factorize of 99999 is
99999
= 3 3 41 271
2 4900
2 2450
5 1225
5
245
7
49
7
7
5. The smallest 3-digit number = 100
1
Prime factorize of 100
100 = 2 2 5 5
3 99999
Exercise - 34E33333
1. (a) 12, 15
41 11111
12 = 2 2 3
271
271
15 = 5 3
1
HCF of 12, 15 = 3
(b)-(i) : do same as above part (a).
2 100
2. (a) 80, 246
2 50
So, 56 = 2 2 2
5 25
5
5
(b)-(e) : do same as
1
15
2 12
3 15
2 6
5 5
3 3
1
1
required HCF = 2
(b)-(c) : do same as part (a).
(d) 70, 100, 165
385 621 1
3
First,80we246
find H.C.F
of 70 & 100.
–385
–240
226 385 1
6 80 13
–226
–6
159 226 1
20
–159 of 385 and 621 is 1.
Since, H.C.F
–18
159
This shows67that
385 2and 621 are coHCF
2 6 3
–134
prime.
6
25as67
2 manner
(b)-(f)
: do same
above
–50
i.e, part 3 (a).
17 25 1
The HCF of 70 & 100 is 10.
4. Two consecutive even numbers = 2,
–17
Now, we find the H.C.F of 10 and
4
8 17 2
165
2= 2 1
16
4= 2 2
1 8
HCF = 2
8
70 100 1
5.
The
greatest
number
will
be
HCF
0
–70
of (300-6) and (185- 3).
30 70 2
i.e, H.C.F of 294 and 182.
–60
The greatest number = 14
10 30 3
6.
Three
consecutive number = 2, 3,
the H.C.F. of 10 &–30
165 = 5
4
the H.C.F of (70,
100, 165) = 5
2 = 2 1, 3 = 3 1, 4 = 2 2 1
(e)-(f) ; do same as above part (d).
HCF of (2,3, 4) = 1
3. (a) 385, 621
7. The greatest number will be the
If H.C.F of 385 and 621 is 1, so
H.C.F of (1354-10), (1866-10) and
10 165 will
16 be co-prime
these numbers
(2762-10) i.e, 1344, 1856, 2752.
–10
otherwise not.
Now do same process as in Q.No
65
2(d).
–60
The greatest number will be = 64
5 10 2
–10
8. (a)
0
16
8
182 294 1
–182
112 182 1
–112
70 112 1
–70
42 70 1
–42
28 42 1
–28
14 28
–28
0
2. (a) 84, 126, 288
2We24
2 72
2 60
have
12 = 2 22
362 2 2
2L.C.M
2 303
23 6 7 = 2016 2 18
3 15
3 3
3 9
5 5
1
1
3 3
1
2
(b)-(i) : do same manner as part 2
(b)-(d) : do same as above.
(a).
9. Te greatest number will be H.C.F
3.
since 35 – 18 = 17
of (2011-9) and (2623-5) i.e HCF of
45
– 28 = 17
2002 and 2618.
55
–
38 =21784, 126, 288
Now, Find H.C.F of 2002 and 2616
63, 144= (L.C.M of
The least
2 42,number
by using division method same as
35, 45, 55)
17 63, 72
Q.No. 2 above.
2 –21,
The greatest number = 154
2 21, 63, 36
10. do same as Q.No. 2 above
2 21, 63, 18
(The greatest number = 27)
3 21, 63, 9
Exercise - 4F
3 7, 21, 3
108 54 27
21672, 60
1. (a) 24,
7 7, 7, 1
=
=
=
256 128 64 32
We have
1, 1, 1
= (3465) – 17 = 3448
4. The least number = (L.C.M of 6, 7,
8, 9, 10, 12) + 1
= 2520 + 1 = 2521
5. The least number = L.C.M of (74,
104, 108, 206) = 10701288
6. do same as Q.No. 4.
7. do same as Q.No. 5.
24 = 2 2 2 3
45, 55 must by
5 35,
8.The
least
number
72 = 2 2 2 3 3
subtracted
from
319 will be the
60 = 2 2 3 5
7 7, 9, 11
L.C.M of 5, 7, 9.
L.C.M = (2 2 3) 2 3 5
11 1, 9, 11
= 360
9 1, 9, 1
(b)-(i) : do same as part 1 (a).
1, 1, 1
17
L.C.M of 5, 7, 9 = 5 7 9 = 315
required least number to be
L.C.M = 131323 = 1014
subtracted = 319 – 315 = 4
(b)-(f) : do same as above part (a).
9. do yourself
13 1014 78
10. do same as Q.No. 8 above.
3. H.C.F =
–91
Exercise - 4G
104
1. (a) 87, 145
–104
Product of two numbers = 87 4. do same as Q.No.
0 3 above (Ans :
145 = 12615
450)
H.C.F of 87,
5. a b = H.C.F L.C.M
5, 7, 9
5 145
115 b = 5 1955
7 1, 7, 9
13 13, 1014
9 1, 1, 9
1, 78
13
b
1, 1, 1
1, 56above (ans :
2 Q.No.
6. do same as
207)
1,
3
3
H.C.F = 29
7. given, product 1,
of two
1 numbers =
8732
L.C.M of 87, 145
H.C.F of these numbers = 2
Product of two numbers =
L.C.MProduct
H.C.F of Two numbers
8732 = LCM2L.C.M
2200
LCM
= 440 = 5
L.C.M = 2935 = 435
Exercise - 4H
and H.C.F L.C.M = 29 435 =
1. The greatest possible length of each
12615
87 145 1
cloth
L.H.S = R.H.S Proved
–87
=
H.C.F of (54m, 69m, 78m) = 3m
(b)-(i) : do same
58 as
87above
1 part 1 (a).
2.
The
minimum capacity of bag =
2. (a) 13, 1014
–58
5 1955 204kg, 120kg)
L.C.M
= 85
= of (144kg,
H.C.F of 13, 1014
29 58 2
115
–59
0
29 87, 145
3, 5
3
1, =513
5
HCF of 13, 1014
1, 1
L.C.M of 13, 1014
8732
= 4326175
= 12240kg
= 4366
=
2
3. L.C.M of 8, 12, 6, 20 minutes = 2
18
2 2 3 5 = 120min = 2hour
They will ring together = 11am +
2h = 1P.M
= 7min + 12 sec = 7min 12sec
They will change again
simultaneously after = 7min 12sec.
C.C.E DRILL - 1
Tick () the correct answer :
(a) Yes (b)3No9,(c)12,
1715
(d) 333 (e) 16 (f)
3,
11,
2 (g) even 3
(h) factor (i)5 1 (j) 16.15
C.C.E
4, 5 - 2
4 1,DRILL
1. Fill in the 5blanks
: 5
1, 1,
(a) even (b) even
1, 1,(c) 12 (d) factor (e)
equal (f) H.C.F (g) 1 (h) 4 (i)
composite (j) twin-primes.
2. Write T for true or F for false for
the following :
(a) F (b) F (c) F (d) T (e) T (f) F (g) T
(h) T (i) T (j) F
HOTS QUESTIONS
1. Do Yourself
2. If a number is divisible by both 2
and 3 then that number is divisible
by 6 also.
since 2 and 3 both are factors of 6.
3. do yourself.
4 144, 204, 120
3 36, 51, 30
2 12, 17, 10
6, 17
, 5 of (84cm,
6 tape
4. The longest
= HCF
1, =
17,
138cm,171200cm)
6cm5
1,
1,
5 of the
5. The maximum
capacity
5
container which
can
measure
the
1, 1,
1
kerosene of both the tanks exactly
= H.C.F of (102l, 1054l) = 34l
6. Find the L.C.M of 90cm, 85cm,
80cm. (Ans : 12240cm)
7. L.C.M of 6, 8, 12 seconds
= 2 2 2 3 = 24 seconds
2 6, 8, 12, 20
2 3, 4, 6, 10
2 3, 2, 3, 5
3 3, 1, 3, 5
Basic Concepts of Geometry
They flash
1, 1, 5 again = 24
5 1, together
Exercise- 5A
sec.
1, 1, 1, 1
8. L.C.M of 9, 12, 15 = 3 3 4 5 1. (a) Name of the lines are m, n and
o.
= 180min = 3 hours
(b) Lines are DA, AB, AC, CD and
BC.
2. (a) Examples of lines : edge of
ruler, meeting place of two walls.
(b) Examples of planes : Surface of
table, surface of carrom board.
3. Yes, we can draw unlimited lines
They will next ring together after
through point P.
= 3 hours
4. (a) Line segments are AM, AN, AB,
9. required least number = LCM of
6, 8, 12
MN, MB, NB, MO, OQ, NO, OP,
(25, 40, 60)2+ 9 = 600+9 = 609
CP,
PQ, QD, CQ, CD, PD.
3,
4,
6
3 48, 72, 108 = 432
10. L.C.M of
(b)
Rays
are OM, ON, NB, MB,
seconds 2 1, 4, 2
MA, NA, OP, OQ, PC, QC, PD,
= 420 seconds
2, seconds
4
2 1,+ 12
QD.
1, 1, 1
19
10.(a) Location of countries in a map
represent plane.
(b) The edge of diary represent
lines.
(c) A fine drop of shower represent
curve.
(d) Wire stretched between two
poles represent lines.
11.(a) A line contains infinite number
of points.
(b) There can be drawn infinite
number of lines through a given
point.
(c) Only one line can be drawn
through two given points.
12. Concurrent lines : When three or
more lines pass through a same
X
Y
point in a plane is called concurrent
lines. For example.
(c) Non-interscting line segments
are NO, AM, MN, NB, MO, OQ,
OP, PQ, CP, QD.
5. Do yourself.
6. From the given figure, all possible
parallel edges are AB||CD,
AB||HG, AC||BD, AC||HF,
CF||DE, AH||CF, BD||GH,
FE||HG, GE||HF
7. Total number of lines segments= 6,
which are as AB, BC, CD, AD, BD,
AC.
(b) Total number of lines segments
A
= 6, which
are as NQO,B OM, QM,
M
PR, PN, RN.
(c) In it, Total no. of line segments =
O
12, which are ED,
DC, EG, CG,
C
D
P QBC, HD,
AB, BH, HF, AF,
FE, AG
(d) From fig, Total no. of line
segments = 48, which as DJ, DL, 13.No, They can intersect on
DC, JL, LC, JC, EM, EN, EF, MN,
producing line segment which
MF, NF, GO, OP, GH, GP, OH,
meet at same point of intersection.
PH, AI, AK, AB, IK, IB, KB, DE,
DG, DA, EG, EA, GA, JM, JO, JI,
MO, MI, OI, LN, LP, LK, NP, NK,
PK, CF, CH, CB, FH, FB, HB.
8. (a) Parallel line segments are
AF||EB, EB||CD, AF||CD,
14. (a) Yes (b) No
EF||AB, BC||ED.
15.(a)
Four parts of intersecting lines
(b) Parallel line segments are
are
cd,
dc, ch, bc.
AB||EF, BC||GF, CD||GH,
(b)
Four
collinear points are f, a, b,
ED||AH.
g.
(c) From the fig. Parallel line
(c) Three non-collinear points - e, f,
segments are AC||FD, AE||BD,
g.
EC||BF
(d)
ThreeA concurrent lines - ad, fg,
9. Only one line can be drawn through
ac.
there two points.
B
(e) Three lines whoseF points of
C
20
D
E
intersection are a and c are fg, ac,
ad.
20. One.
16.(a) Pairs of parallel lines : lm, mn,
P
nl.
(b) Points on intersecting lines : p, q,
r,Cs,1 t, u, v. O
D
Exercise- 5B
C
1
(c)
O lines whose point of intersection
1. (a) closed curves - (b), (c)
is P : x, y.
D
(b) Open curves(a), (d), (e)
(d) collinear points : p, q, s, u and p, 2. Polygons : A polygon is a simple
e
r, t, v.
closed are made up of three or
more line segments
e.g. triangle,
A
B
quadrilateral.
3. (a) Open curve (b) Closed curve
A
4. Septagon
(e) lines whose point of intersection
is S : X, m.
(f) lines whose point of intersection
is V : l, y
(g) The points of intersection of
line y : v, t, r, p.
(h) the points that intersect x are u,
s, q, p.
(i) lines that intersect x and y both :
l, m, n.
17. Yes, they all are concurrent.
P
18.(a) Collinear points
- b, c, d; ab; ac
and ad.
Q
R
l
m
B
n
C
n
l
m
5. (a) Yes
(b) It is closed curve
Exercise- 5C
1. Three examples of angles : Door,
Window, Table.
2. Vertex O; arms of POQ are OP,
OQ.
3.
(a)
There are 4 angles in the fig (a)
n
namely
ABC, BCD , CDA,
T
S
m
DAB.
V
U
(b)There are 3 angles in this fig(b)
l
namely CAB, ABC, BCA
X
Y
(c) There are 10 Eangles in this fig (c)
a
(b) concurrent line and concurrent
namely F POQ,
a QOR,
ROS,
D
Points - AB, AC, AD at Point A.
SOT, POS,
POT,
QOS,
QOT,
a
a
ROT, POR
AB, DB at Point B.
4. exteriorGpoint of AXB C= R, M
AD, BD at point D.
a
a
interior points ofa AXB = P, O
19. Three
A
21
B
points which lie on AXB =Qn
triangular region.
Exercise - 5E
1. quadrilateral are (a), (b), (d) since
quadrilateral have four sides.
5. From the given figure, we see
2. (a) sides : AB, BC, CD, DA;
that
Vertices : A, B, C, D; Angles : A,
(a) 1 = interior; (b) 2 = exterior
B, C, D.
(c) 3 = interior; (d) 12 = exterior
(b)
sides : PQ, QR, RS, SP; vertices
(e) 15 = exterior; (f) 8 = exterior
:
P,
Q, R, S Angles : P, Q, R,
6. (a) False (b) False (c) True (d) False
S.
(e) True (f) True.
3. (a) opposite sides = (AB, DC); (BC,
7. (a) No (b) No
AD)
Exercise- 5D
(b)
Adjacent sides = (AB, BC);
1. No, Three collinear point do not
(AD,
DC) & (AB, AD), (BC, DC)
form a triangle.
(c)
Adjacent
angles = ( A D, C),
P
2. We get triangle as
( A,
B)
. .. .
B
Triangle
...... ....
Q
..... .... .. . . region
R
(d) opposite angle = ( ...... ..A,
( B,
.............. .......C);
.
.
.
.
.
.
.
.
.
.
. .. . .. .. . .
R
P
B .
C
Q D) Triangle
O
A
X
n
m
3. In triangle ABC
4. (a) side adjacent to NO = LO
(a) side opposite vertex C = AB
(b) angle opposite to M = O
(b) Side opposite vertex A = BC
(c) side adjacent to LM = MN
(c) Side opposite vertex B = AC
(d) side opposite to OL = MN
(d) the angle opposite BC = A
(e) the angle opposite AC = B
(f) the angle opposite AB = C
(g) vertex opposite AB = C
(h) vertex opposite AC = B
(i) vertex opposite BC = A
4. Fill in the blanks :
Exercise- 5F
(a) 3 (b) 3 (c) 3 (d) 6 (e) 180°
5. Triangle : A triangle is C a plane
1.
P
figure that is closed by 3 straight
B
A
lines segment.
Triangular region : The interior
2. Which statement are true and
Q part of triangle
Ais known Bas
R
which areCfalse?
D
22
(a) False (b) True (c) True (d) True
(e) True (f) True (g) True
3. Fill in the blanks :
6.
(a) Passes (b) Two equal (c) (The
center, The circle)
(d) arc (d)
L
Diameter
4. (a) Secant of a circle : A line 7.
segmentO intersectingM any two
points of a circle is called secant of
the circle for example
:
N
Major Sector
OADB
O0
Major
Segment
C.C.E DRILL - 1
Tick
()
the
correct answer :
C
D
Chord
(a) Tip of your pencil (b) Infinite
O
points (c) Infinite (d) On the angle
Chord
A
B
P
(e) Infinite (f) 6 (g) same
center (h)
D
P
(b) Sector of a circle : The region
diameter (i) triangle (j) 4 sides
A
bounded by an arc of a circle and
O
C.C.E
DRILL
-B2
two radii is called a sector of the 1. Fill in theMblanks :
N
C
circle, Here OACB is a sector of
(a) point (b) plane (c) collinear (d)
circle.
6.2 cm
simple
curve (e)O5.3cm
3 (f) vertices
(g) 3
4cm
O
O
(h) 180 (i) 2 (j) diameter.
2. Write
T for true
of F for false
for
A
B
C
the following :
(a) T (b) F (c) FO(d) F (e) T (f) F (g) F
(h) F (i) T (j) T
65
(c) Segment of a circle : A chord
A
B
divides the circle into two parts.
Sector OAB
Each part is called the segment of a
circle.
0
Chard
Radius
diameter
3.
5
secant
5. (a) chord = PQ
(b) Arc = BD
(c) Diameter = AB
(d) Radius = OA
D = OB = OD
Major Sector
(e) Semi-circle = ACBA
OADB
O
(f) Secant = MN
A
C
B
Major Sector
OACB
23
cm
O0
3.5
cm
secant
m
4c
24
m
6c
HOTS QUESTION
4. Do it yourself by measuring length
between two points.
1. Since AB and BA are just opposite
rays, because a ray starts from its 5. Let we draw a triangle assuming
initial point A and passes through
AB = 4Cm, BC = 5cm AC = 6cm,
B extending indefinitely, that is
then
A
why Aarti and Aakash are
AB+BC = 4+5 = 9cm
confused.
AC = 6cm
AB +BC > AC
A
B
B
A
(Initial point)
i.el 9cm > 6cm. B
(Initial point)
C
5cm
2. Yes, Arnav can draw a circle as
Exercise - 6B
follows :
1. (a) ONM> CBA (b) XYZ>
since diameter is the longest chord
TSR (c) CED> ZXY (d)
radius = diameter = 6cm
ABC> MNO
= 3cm
2
2
2. (a) 1 = 2 (b) a = b (c) m = n
(d) a = b
3. (a) 30° = Acute (b) 45° = Acute (c)
Radius
O
3cm
90° = Right (d) 75° = Acute (e) 120°
= obtuse (f) 195° = Reflex (g) 180°
= straight (h) 135° = obtuse (i)
3. No, It is not possible to make a
77.6° = Acute (j) 0° = No angle (k)
triangle with three collinear points.
360° = complete (l) 90.5° = obtuse
e.g,
l
o
m
n
4.
(a)
(b)
Since all the points are in a straight
line, while to make a triangle at
least one point must be nonObtuse
collinear. So, Amit can't make a
(d) Right
(c)
triangle with these three collinear
points.
Understanding Elementary Shapes
Acute
Reflex
Exercise - 6A
(e)
(f)
1. (a) Using Ruler and Divider AB =
O
A
2.5cm
Complete
Acute
(b) Using Ruler and Divider, CD =
2.5cm
5. (a) two straight angles = 180° +
2. No, It is not reliable comparing line
180° = 360°
segments by merely observation.
(b) one right angle and a half right
3.
angle
= 90° + 90 = 135°
A 2cm B
C
D
2
5cm
(c) 2 right angle = 2 90° = 60°
3
3
8cm
(d) Three straight angle = 3(180°)
It is clear from the figure, B lies
= 540°
between A and C, C lies between B
6. (a) 12'o clock = 0° (b) 9 o'clock =
and D.
Exercise - 6C
90° (b) 6 o'clock = 180° (d) 2
o'clock = 60° (e) 10 past 10 = 110° 1. (a)
(b) Y
B
(approx) (d) 9 to 12 o'clock = 90°
= 75°
= 60°
8. (a) East and North = 90° (b) South
O
X
O
and East = 90° (c) West and South
A
= 90° (d) South and West = 90° (e)
West and North = 90° (f) North
(c) Q
(d) O
N
and West = 90° (g) North and
=125°
=60°
South = 90°+90° = 180° (h) West
P
and East = 90+90 = 180° (i) North
O
M
and South-West = 90+45° = 135°
(e)
(f)
O
(j) East and North-West =
O
R
90+45°=135° (k) South-East and
60°
60°
North-East = 45°+45° = 90° (l)
South to South = 90° + 90° + 90° +
C
D
T
90° = 360°
N
2. (a) 30°; Draw a ray OA. Place the
mid-point of protractor along the
90° 90°
raj
OA. Start from O° on the side of
W
E
90°
90°
A go to 30° on the protractor and
mark point B. Now, remove the
protractor
and draw line OB, We
S
find
AOB
measured 30° as the
9. Straight angles : NOS, WOE,
required
angle.
B
N-WO S-E, S-EO N-E
Right angles : NOE, WON,
WOS, SOE, N-W O N-E, NE O S-E, S-E O S-W, S-W O NW
N
N-E
N-W
O
W
E
30°
O
A
(b)-(h) do as above.
3. do it yourself with the help of
protractor and same as above.
P
C
4. (a)
(b)
120°
S-E
S-W
30°
B
S
Q
A
(c)
M
Acute angles : E O N-E, N-E
ON, W O N-W, N-W O N,
90°
WO S-W, S O S-E, S-E O E
O
N
Obtuse angles : E O N-W, N-E
O W, N O S-W, N-W O S, W O
Exercise - 6D
S-E, S-W O E, S O N-E, S-E O 1. (a) 64°+26° = 90° compleN
mentary (b) 22 + 68 = 90°
complementary (c) 77 + 102 =
25
2.
3.
4.
5.
6.
179° Not complementary (d) 45
+ 46 = 91° Not complementary
(e) 65° + 26° = 91° Not
complementary (f) 1° + 89° = 90°
complementary
(a) 102° + 178° = 280° Not
Supplementary (b) 44° + 136° =
180° Supplementary (c) 90° +
90° = 180° Supplementary (d)
101° + 79° = 180°
Supplementary (e) 75° + 105° =
180° Supplementary (f) 5° + 85°
= 90° Not Supplementary
(a) 180° – 130° = 50° (b) 180° – 55°
= 125° (c) 180° – 65° = 115° (d)
180° –0° = 180°
(a) 90° – 45° = 45° (b) 90 – 33 = 51°
(c) 90° – 17° = 73° (d) 90 – 48° = 42°
(a) 90° – 45° = 45° (b) 180° – 90° =
90° (c) 150° (d) 72° (e) 60°
Pairs of adjacent angles are
( POQ, QOR), ( QOR, ROS),
( ROS, SOT), ( POR, ROS),
( POS, SOT), ( QOS, SOT),
( POR, ROT)
NOC); ( NOC, COP); ( COP,
DOP); ( APO, OPB); ( BPM,
APM); ( OPB, BPM)
8. No, BPO is not at point O.
9. (a) No (b) No (c) 180–90° (Right
angle)
(d) Obtuse (... obtuse + acute =
180°)
10.Linear Pair : ( 1, 2); ( 1, 3);
( 3, 4); ( 4, 2); ( 5, 6); ( 5,
7); ( 7, 8); ( 6. 8); ( 9, 10);
( 9, 11); ( 10, 12), ( 11, 12)
Vertically opposite angles : ( 1,
4); ( 2, 3); ( 5, 8); ( 6, 7);
( 9, 2); ( 11, 10)
2
5 6
7 8
1
4
3
9 10
11 12
11. (a) BOC + AOC = 180°
75° + x° = 180°
C
x° = 180 – 75 = 105°
(b) do same as above.
x°
75
B
A
O opp.
(c) BOD = AOC (vertically
T
S
angle) = 45°
R
(d) 3x + 4x +9x + 7x + 2x = 360°
Q
(Sum of all the angles at a point is
P
O
7. (a) linear pair
: ( DON,
NOC);
360°)
D
( NOC, COP); ( COP, DOP);
B
25x = 360°
( APO, OPB); ( BPM, APM);
4
( OPB, BPM); ( APO, APM);
1o 3
x = 360 = 14.4°
2
( DON, DOP)
12. given25
1 = 46°
A
3 = 1 (vertically opp.
angle ) C
N
3 = 46°
4 + 1 = 180° (linear Pair)
D
C
O
4 + 46° = 180°
4 = 180 – 46° = 134°
A
B
2 = 4 (vertically opp angle)
P
2 = 134°
(b) Adjacent
angles : ( DON, 13. (a) 55 (90° – 55°) = 35°
M
26
(b) 75° (90 – 75°) = 15°
(c) 20° (90° – 20°) = 70°
14. (a) 180 – 110 = 70°
(b) 180° – 89° = 91°
(c) 180 – 65° = 115°
Exercise - 6E
1. (a) No (b) Yes (c) Yes (d) No (e) Yes
(each 60°) (f) No (g) Yes (h) Yes
2. equilateral triangle has each angle
60°, i.e,
6.
P
Then 4x + 3x + 2x + x = 360°
10x = 360°
360
x =
= 36°
10
D
C
x
2x
3x
D = xA=4x360°, C =
2xB = 72°
B = 3x = 108° , A = 4x = 144°
Let other two angle be x° each
Then 60° + 100° + x° + x° = 360°
2x = 360– 160°
60
60
60
R
3. IsoscelesQ
4. Scalene
5. equilateral
6. (a) Yes (b) Yes (c) No 45°
7. (a) equilateral (a=b=c),90(b)
45°scalene
(abc), (c) isosceles (a=bc),
(d) equilateral (a=b=c)
8. (a) equilateral (each angle = 60°),
(b) acute (each angle <90°), (c)
acute (each angle <90°).
Exercise - 6F
1. (a) trapezium (b) rectangle (c) nonparallel (d) 360° (e) 90°
2. (a) Square (b) rectangle (c)
trapezium (d) parallelogram (e)
rhombus
3. P + Q + R + S = 360°
75° + 100° + 120° + x° = 360°
S
295° + x = 360°
x = 360 – 295° = 65°
P
4. In quadrilateral, we know
R
120° + x + x + x = 360°
Q
3x = 360 – 120 = 240°
240
each
angles are 80°.
x = three
= 80°
3
5. Let angles are 4x, 3x, 2x, x.
200
x
= 100°
=other
2 two angle are 100° each.
7. (a) True (b) False (c) False (d) True
(e) False (f) True (g) False (h) False.
Exercise - 6G
1. Tick () the correct answer :
(a) length, breadth and height (b)
cube (c) cylinder (d) sphere (e)
cuboid (f) none of these
2. Fill in the blanks :
(a) solid (b) 6, 12, 8 (c) opposite (d)
sphere (e) cube (f) 4, 8 (g) 3, 6 (h) 6,
3, 2, 9,
3. (a) examples of a cone : ice cream,
clown's cap, a conical tent, a
conical vessel
(b) examples of a cuboid : brick,
wooden box, book, almirah
(c) a cylinder : circular pipe,
measuring jar, gas cylinder, test
tube.
C.C.E DRILL- 1
Tick () the correct answer :
(a) 1000m (b) 90° (c) 6 o’ clock (d) 4 (e)
obtuse (f) 90° (g) scalene (h) a straight
angle (i) a square (j) 1
C.C.E DRILL - 2
1. Fill in the blanks :
(a) acute (b) equal (c) straight (d)
27
isosceles (e) protactor (f) quadrilateral
(g) 360 (h) rectangle (i) equal (j) reflex
2. Write T for true of F for false for
the following :
(a) T (b) T (c) F (d) F (e) F (f) T (g) T (h)
F (i) F (j) F
HOTS QUESTIONS
1. Ye s, c o m p a r i s o n b y m e r e
observation we can say easily
which segment is shorter and
which is longer. so, Maya's
comparison can be reliable.
2. No. Raghu draw this triangle
because In any triangle, sum of all
the angles is 180°. But, Here
A+ B+ C = 60° + 70° + 60° =
190°180°.
therefore, he can't make a triangle
with these angles.
3. No, He is not correct because 170°
is an obtuse angle i.e, 90 < 170° <
180°.
Integers
Exercise - 7A
1. (a) Decrease of water — Increase
of water, (b) Increase in population
— Decrease in population, (c)
Spending money — Earing money,
(d) Going to north-east — coming
from north-east, (e) 400BC —
400AD, (f) Fall of temperature —
rise in temperature,(g) goup — Go
up down.
2. (a) + (b) + (c) – (d) – (e) + (f) – (g) +
(h) + (i) – (j)+ (k) + (l) –
3. (a) +26°C (b) –32°C (c) –`600 (d) +
`25 (e) +`1892 (f) +`2400 (g)
–1124m (h) –24°C (i) –26 (j) +`1818
(k) –`292 (l) +`5000
4. (a) –3 (b) –5 (c) 10 (d) 14 (e) 0 (f) 1
(g) –472 (h) –77 (i) 476
5. (a) –16 (b) –10 (c) –1472 (d) 33 (e)
–1 (f) –10 (g) –14752356897 (h) –15
(i) –16
6. Given numbers form (a) to (h) are
on number line is as follow :
F1
E1 D1 C1
B1
A1
–15 –10 –8 –5 –3 –2
O
A
B
C
D
E
0
2
5
10
14
15
7. (a) integers between (0 and 15) = 1,
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14
(b) integers between (0 and 7) = 1,
2, 3, 4, 5, 6
(c) integers between (299 and 298)
= None
(d) integers between (5 and –6) = 4,
3, 2, 1, 0, –1, –2, –3, –4, –5
(e) integers between (–241 and
–246) = –242, –243, –244, –245
(f) integers between (–21 and –15) =
–13, –14
8. (a) 10 > –14 (b) –99 > –120 (c) 40
> –140 (d) 4444 > –5555 (e) –111
< –110 (f) 0 > –50
9. Increasing order is
(a) –25, –10, –5, 15, 20, 105
(b) –112, –16, –15, –11, 0, 1, 100
(c) –400, –300, –200, 200, 300, 400
(d) –5000, –4000, –3000, 1000,
2000
10. decreasing order is
(a) 100, 10, 8, 5, 0, –5, –6, –100
(b) 15, 4, –2, –8, –20, –25
(c) 700, 600, 549, –549, –600, –700
(d) 9, 5, 1, –13, –17, –22, –26
11.The absolute value of
(a) |–1005| = 1005 (b) |–771| =
771 (c) |–500| = 500 (d) |–5061|
= 506 (e) |–450| = 450 (f) |–375|
= 375 (g) |500–673| = |–173| =
173 (h) |300–506| = |–206| = 206
12.(a) 4 more than –8 – 8 + (+4) =
–4
28
1. Representation of given number
on number line is as follows :
–8–7 –6–5 –4 –3–2–1 0 0 1 2 3 4 5 6 7 8
We count from –8 and go 4 steps to
the eight.
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
So, we get 4 more than –8 is –4.
2. do it your self by marking given
number and add or subtract as
(b) 6 more than 10 = 10+(+6) = 16
given in the questions.
e.g, (a) 121+(–22) =121–22 = 99
–10 –9 –8 –7 –6 –5 –4 –3 –2 –1
H
G
F E
A
B
C
D
–3–2 –1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
We count from 6 and go 10 steps to
the right and obtain 16 as shown
0
25
50
75
99
100
3. (a) –24 + 24 + 18 = 42
– 24
=From
18 o
above on the number line.
Move
Right
(b) 15 + 11 = 26
So, 6 more than 10 is 16.
(c) 9 + (–6) = 9 – 6 = 3
(c) (d) (f) do same as part (a).
(d) 5 + (–11) + 5 – 11 = –6
(e) 2 less than –5 = –5+(–2) = –5–2
(e) 40 + 100 + (–140) = 140 – 140 =
= –7
0
(f) (–77) + (–22) + 10 = –99 + 10 =
–89
–8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5
4.
(a)
–10000 + 9999 = –1
Here, we go two steps left from –5.
(b)
–333 + 400 = 67
So, we get-7 as answer.
(c) 400 + (–460) = –60
(e), (h) do same as above part (e).
(d) –4283 + (–3616) = –4283 – 3616
(i) do as part (b).
= –7899
13.(a) +1 (b) 7 (c) 3 (d) –5 (e) 19 (f) 25
(e) –426 + 400 + 419 = – 426 + 819
14.–19, –18, –17, –16, –15, –14, —13,
= 393
–12 all are greater negative integers
(f) 262 + (–212) = 262 – 212 = 50
than (–20).
(g) 3006 + (–7000) = 3006 – 7000 =
15.–5, –4, –3, –2, –1 > (–6).
–3994
16 –24, –23, –22, –21, –20, –19, –18 >
(h) (–613) + (–322) + 192 = –613 –
(–25).
322 + 192 = –935 + 192 = –743
17.Successor of —121 = –120
(i) (–51) + 203 + 38 + (–48) = (–51 –
Predecessor of –121 = –122
48) + (203 + 38) = –99 + 261 = 142
18. Left
5. (a) (–4) + (–8) = –12
(b) (–4) + 9 = 5
(c) (–18) + (–200) = –218
–6 –5 –4 –3 –2 –1 0 1 2 3 4 5
19. Right
(d) (–100) + (–99) = –199
(e) (–8) + (–9) + (–10) = –27
(f) 119 + (–19) + (–100) = 0
–3 –2 –1 0 1 2 3 4 5 6 7 8 9
20. –19, –18, –17, –16.
(g) 0 + (–9) = –9 (h) 0 + (–3) + 3 = 0
Exercise- 7B
6. (a) –91 +91 (b) 0 0 (c) –199213
29
+199213 (d) 1183 –1183 (e)
2312 –2312 (f) –63333 63333
(g) 4206 –4206 (h) 3065 –3065
7. (a) Successor = (–499) + 1 = –498
Predecessor = (–499) –1 = –499 – 1
= –500
(b)-(f) : do same as part (a).
8. (a) 1 + x = 0
(b) x + 9 = 0
x = –1
x = –9
(c) x + (–10) = 0 (d) –20 + x = 0
x = 10
x = 20
9. required distance to the south =
190km – 160km = 30km
10.(a) (–4) + (–5) + (–6) + (–8) + 6 = –
4 – 5 – 6 – 8 + 6 = – 17
(b) 87 + (–43) + (–92) + (–10) = 87 –
43 – 92 – 10 = –58
(c) (–432) + (–265) + (–49) + 125 =
– 432 – 265 – 49 + 125 = –621
(d) 5622 + (–5623) + (–38) + 100 =
5622 – 5623 – 38 + 100 = 61
11.His total balance = `2218 (–`1218)
+ `2000
= `4218 – `1218 = `3000
12.Total lost money due to a hole =
`90 + `30 + `140 + `78 = `338
but he found money on the road =
`300
his net loss = `338 – `300 = `38
Exercise - 7C
1. (a) –40 – (20) = – 40 – 20 = – 60 (b) –
81 – (36) = – 18 – 36 = – 117 (e)-(l) :
do same as above (a), (b).
2. (a) [47 – (–9)] + [49 – (–38)] = [47 +
9] + [49 + 38] = 56 + 87 = 143 (b)
[–19 – (–18)] + [–57 – (–88) ] = [–19
+ 18] + [–57 + 88] = [–1] + [31] =
31 – 1 = 30 (c) [–122 – (–225)] – [–
322 – (–428)] = [– 122 + 225] – [–
322 + 428] = [225 – 122] – [428
–322] = 103 – 106 = – 3
3. The other integer = – 45 – (426) = –
45 – 426 = – 471
4. (a) – 8 + – 17 = –25
(b) – 12 – 8 = –20
(c) – 20 + – 48 = –68
(d) – 81 – 28 = –109
(e) – 254 + 211 = –43
(f) – 5662 – 4372 = –10034
(g) – 5677 – 1000 = –6677
(h) 322 + (– 677) = –355
5. (a) (–6) + (–12) = (–12) + (–6) (b)
–47 – (–88) > (–88) + 47 (c) –25 –
(+25) = – 50 (d) – 428 – 525 > – 628
– 733
6. (a) 44 – 82 = –38 and (–82) –44 =
–126
they are not equal.
(b) do same as above.
7. The other integer = (sum of two
integers) – (one of them)
= (840) – (–1426) = 840 + 1426 =
2266
8. Sum = (–99) + (–426) = –99 –426 =
–525
142 – (–525) = 142 + 525 = 667
9. do same as above Q. No 8
10.do same as above Q. NO 8.
11.do same as above Q. No 8, 9, 10.
12.difference = (–89) – (–95) = – 89 +
95 = 6
Sum of both = 6 + (–46) = 6 – 46
= –40
13.do same as above Q. No 8, 9, 10.
14.Let b = x, then a = x – 1. Now a – b
=x–1–x=–1
15.Let a, m, b are three integers such
that a = x, m = x+1, b = x +2.
Now, a – m = x – (x + 1) = x – x – 1 =
–1
b – m = x + 2 – (x + 1) = x + 2 – x – 1
=1
... a – m b – m, so It is False.
30
16.do same as Q. No 14.
17.(a) + 2 – 2 + 2 – 2 + 2 – 2 + - - - - - - =
0
(b) + 2 – 2 + 2 – 2 + - - - - - + up to
90th Term = 0
18. – 1 + 2 – 3 + 4 – 5 + 6 - - - - - – 19 +
20 = – (1 + 3 + 5 + 7 + 9 + 11 + 13 +
15 + 17 + 19) + (2 + 4 + 6 + 8 + 10+
12 + 14 + 16 + 18 + 20)
= –100 + 110 = 110 – 100 = 10
19.Temperature on monday = 20°c
Reduced temperature on same day
= 14°c
required fall in temperature =
20°c – 14°c = 6°c
20.The required distance between
peak of minar &
Point under the earth = 1800m +
456m = 2256m
C.C.E DRILL - 1
Tick () the correct answer :
(a) less than (b) greater (c) closure
property (d) 7 (e) 5 (f) 5 (g) –11 (h)
+700 (i) +2 (j) 1
C.C.E DRILL - 2
1. Fill in the blanks :
(a) 0 (b) –y (c) positive (d) greater (e)
left (f) (a – b) (g) –28 (h) –355
2. Write T for true or F for false for
the following :
(a) T (b) F (c) T (d) F (e) F (f) T (g) F
(h) T (i) F (j) F
HOTS QUESTIONS
1. Total money spent of Reshu on all
items = `60 + `180 + `50 = `290
Total sold items = `40 + `140 = `180
her net loss = `290 – `180 = `110
2. It is dear form the number line
Amit shall move to right to reach at
5 of the number line.
A (Amit)
B
5
6 and 5
4 right
–2
–1 0move
1 2to 3the
3. It should
steps to reach at 3. Right
A
–2
–1
dice
3
0
1
2
Fractions
Exercise - 8A
1. (a) shaded region =
4 2
=
6 3
11
(c) shaded region =
20
2 1
(d) shaded region = =
4 2
3
(e) shaded region =
7
2. (a) unshaded region =
6
7 2 1
(b) unshaded region = =
6 3
9
(c) unshaded region = 20
2 1
(d) unshaded region = 4 = 2
3
(e) unshaded region =
7
1
3. (a)
7
(b) shaded region =
(b) 1 =
8
2
(c) =
6
(d) 1 =
2
1
4. (a) Fraction
=
8
31
4
Nr. 5
(b)-(d) : do same
as =
(a).
= Dr.
27
of 24 flowers == 7
18
7
7
24
8
8 has book = 128
15. Som
14.
5. (a) seven-ninth
7
(b) five-eights = 9
5
=
(c) Nine-twentients
8
(d) Five-twelfths
6. (a)
Ravi get toffees
=
flowers
he21
gave
his brother Ram =
=
5
=
Three-eights
12
3
(b) 8 = Nine-twelfths
9
20
he gave to his brother = 1 of 128
4
= 32 Books
1
128
4
16.(a)
(b) 2 of 25 = 2 25 = 10
5
5
6
6
(c)
of 95 = 95 = 30
19
19
3
(d) 3
of 60 = 60 = 9
20
20
17. (a) 7 of 625 = 7 625 = 175
25
25
29
R
(b)
; Nr. = 29, Dr. = 22
1
(b)
22
O 1
1
4
4 R
17 : do same as (a), (b).
(c)-(d)
; Nr. = 17, Dr. = 25
(c) 2
25
2
1
8. 1 hour = 60 minutes.
9 O
9
R
required fraction
(d) 3
1
9. do same as Q. No 8. = 20min = 1
3
6 O
6
60min
R
10.Total milk = 25l, sold milk = 19l 3
11 the given representation of
18.From
O
11
1
14 line
remaining unsold milk = 25l – 19l =
number
14
6l
(a)
(b)
required fraction
11.Renu has frocks = 3
(c) L = 2
(d) M =–2
3
Blue Dye on Frocks = 2 6
3
=
25
required fraction
(e) N = 6
(f) O = 5
3
3
12. In a day = 24 hours
required fraction
(g) P = 8
(h) Q = –6
= 2
3
3
3
–1
–4
R=
=
= 6 hours
3Exercise -S8B
3
13.Laxman's father had toffees
= 18
24 hours
6
1
he gave
1. Proper fractions (Nr. < Dr.)
=
= to him = 7 toffees
24 4
9
(c) 12 = Six-fifteenths
6
(d) 15 = Five-ninths
5
7. (a) 9 =
32
8. do same
Q. 9No 36
7 in part (a)-(d)
4 as4
=
=
above.7 7 9 63
9. (a)
(b)
5 , 5 , fractions
18 , 1 (Nr Dr.)
2.
Improper
6 8 20 2
14, 7 , 17 , 20
3. (a) 11 4 16 19
+ 3 35part
3 (75)
+ 3 (a).38
(b)-(f)
= same as above
7 : do
=
=
5
5
5
5
4. Mixed fraction are
2 22
(c) 20 = 2 (d)
=
3 33
9
90
(e) 14 = 7 (f) 6 = 30
8
16
5 25
9 63
1 27
10.(a) No
=
=
10
70
3 81
(b)
(c) 2 , 42 = 2 ; Yes
5. (a)7 , 5 , 5 , 6
3 63 3
1
2
4
3
6
12
3
7
8
4
4
11.(a) ,
(b)
= ; Yes
(b) 338 = 3 38
14
7
7
100
100
52 13
24 1
2
12
(c) 72 =
(d) 76 = 19
=2
3
5
5
21 7
64
1
5 12 2
(e) 72 = 24
(f) 128 = 2
–10
12 = 2 2
(c)
(d) 2
100 1
144
9
5
5
(g) 1600 =
(h) 500 = 5
100
2 (f) 62
8
(e) 27
10 1
=5
=6
5
9
5
9
=Exercise - 8C 35 = 5
50 5
49 7
1
89
6. (a) 33 = 4 5
=8
7
11
7
11
1. Like Fractions
6 6 2 12
1 , 2 i.e same Dr.
=
=
7 7 2 14
3 3
2. Unlike
6 6 3 18
4 , 5 Fractions
i.e
same
Dr.
=
=
7 7 3 21
3 3
3, 2 , 7, 1
3. (a)
(b)
6
4 as24part (a).
6 : do
(b)-(d)
same
=
=
8 3
12 4
7 4 28
7 = 49
7. (a) Dr
(c) 9 < 11 (d) 4 < 5
12
7
12
7
6
7
2
13
(e)
(f)
(b) Nr4= 28 4 7 = 28
>
>
=
15
9
15
18
7
7 7 49
7
16
12
< 5
4. (a)
>
16
19
8
57
(c) Nr 4= 48 4 7 28
=
=
2 1
7
7 7 49
=
= 0.25
8 : do4same as part (a).
(b)-(f)
(d) Dr = 63
4 2
5. (a) 4Comparing
we
4 4 12= 48
>
= 1 = 0.5thenumerator,
=
8 8
8
get 3 < 42< 5 < 7
7 7 12 84
33
is the required
ascending order.
(b) do same as above in part (a).
3,4 , 5 , 7
(c) 8converting
8 8 8the given fractions
into like fractions.
L.C.M of (2, 5, 6, 8) = 120
Thus, 1050> 700 > 420 > 300
is the
210 210
210 order.
required210
descending
(c)-(d) :10
do same
10part
10(b) above.
10 as
> > ->8D
Exercise
2 3 5 7
1. (a)
9 9 15 135 ,
=
=
8 8 15 120
9 9 20 180 ,
=
=
6 6 20 120
9 9 24 216 ,
=
=
5 5 24 120
Since 135<180<216<540
9 9 60 540
=
=
2 2 60 120
Hence,
thus,
135 180 216 540
<
<
<
120 120 120 120
(d) do same as part (c) above.
9 9 9
6. (a) Since97>4>2>1
< < <
8 6 5 2
(b)-(f) do same as part (a) above.
2. (a) 7 + 11 = 7 + 11 =18
15
15
15 15
13 5 + 1 13 + 10 + 8
=
+
16
(b) 16 8 2
15
31
=1
16 [LCM of 2, 8, 16 = 16]
16
1 5 + 1 6 + 10 + 1
=
+
24
4 12 24
(c)-(f) ; do same as part (a), (b)
17 [LCM of 4, 12, 24 = 24]
above.
=
3. (a) 24
2 45
8 [LCM of 8,
5
+
5
+2 =
3
8
8
3=
24]
(b)-(d) : do same as part 33(a)
above.
7
135 + 64 199 =
is the required
=
8
24
24
descending order.
4. (a)24
(b) Making the Dr. of all fractions
7 i.e,4 > 2 > 1
same
>
6 (2, 63, 5,67) = 210
6 of
1
3 5 [LCM of 4,
3
L.C.M
+1 = +
4
8 4
8
= 8]4 (a)
(b)-(d) : do same as 8part
13
5
3
+
10
above.
=
= =1
8
8 Exercise
-88E
10 = 10 105= 1050
2 2 105
1. (a)
[LCM (17, 17) = 17]
210
10 10 30
=
= 300
7
7 30
210
13 5
–
10 10 42
420
17 17
=1050 > 700=> 420 > 300
since,
(b)
[LCM (19, 19) = 19]
5
5 42
210
13 – 5 13 – 5 8
=
=
10 10 70
17
17
17 17
=
= 700
Hence,
3
3 70
210
34
10 7
–
19 19
(e)-(f) : do same as (a), (b) above.
10 – 7 7
=
=
9
19
2. (a)
the following :
(a) F (b) F (c) F (d) F (e) T (f) F (g) T
(h) F (i) T (j) T
HOTS QUESTIONS
1. Let me = A, Now A
[LCM of 8, 16 = 16]
11 5 11 – 10 1
–
=
=
16
16
16 8
(b)
Since Dr. is Prime80 number
4
= be a=prime
therefore, Dr .should
100 5
factor of 100.
[LCM of 20, 35 = 140]
63 (b) above.
1
16: do9same64as–(a),
(c)-(f)
–
=
=
140
140
35 20
2. Let there are total x passengers at
80 at the
4 first station, It drop
first.
=
100 5
3. (a)
1
3
15 9
–2
–
4 = 4
4
4
(b)
15 – 9
6 3
1
=
= = =1
4
4 2
2
2
5
17 5
2 –1
–
3 =
3
(c)-(d)6: do same as6(a), (b) above.
17 – 10 7
1
=
= =1
6
6
6
4. (a)
3
remainder passengers
1
x
of x =
3
3
at the first station, It takes
x
= x – = 2x
3
3
2 1 13 8 – 1
1
+
+2 – =
3 4
3 4
2
2
13 6 + 8 4 – 1 3[LCM (2, 3,
=
4) = 12]
12
78 + 32 – 3
110 – 3
=
(b)-(d)
:12
do same =
as (a)12
above.
107 C.C.E
11 - DRILL- 1
=
=8
Tick ()
answer :
12 the correct
12
(a) 13 (b) 6 (c) Numerator (d) 1/2 (e)
6/4 (f) 31/50 (g) Like (h)
(i) 6/24
(j) 20/30
C.C.E - DRILL - 2
2
1. Fill in the blanks :
3
3
(a) Fraction (b) denominator
(c)
equivalent (d) like (e) proper (f) 1/3
(g) 38/7 (h) 3/4 (i) 8 (j) 45
2. Write T for true or F for false for
6
35
= 1 × 2x + 96
3 (i.e second) station, it
at 2the next
drops
= x + 96 Passengers
3
remainder passengers at second
station
x
1 x=
=
+ 96 = 6 + 48 Passengers
2 3
x
x
–
+96
+48 It takes
at3the
second6station,
x x
x
–
+ (96 – 48) = + 48
=
3 6
6
at the next (i.e; third) station,
passengers
left = 248
x
+ 48 + 12 = x + 60
=
6
6
D.1. 25483 > 25400
x
2. LXIV = 50+10+4 (=64)
+ 60 = 248
6
3. Successor of 3457269 3457270
x
There
= 248
– 60
= 188
4. 3265 to the nearest hundred 3300
was
1128
Passengers in
6
the train initially.
E. Correct multiply = 5290 48 =
x = 188 6 = 1128
253920
3. Let the purse contain at first = `x
wrong multiply = 5290 84 =
Taking out of its
444360
difference = 444360 – 253920 =
remainder
x
190440
1
5
F. (1) 2633 + 8262 + 1427 = 12322
5
given,
of remainder = `7.40
x 4x
(2) 3627 + 1437 + 2486 + 3523 =
x –
5
5
11073
G.
(1)
233 25 + 233 5 – 233 20
1
=
233
(25 + 5 – 20)
12
= 233 (30 – 20) = 233 10 =
1
4x
2330
12
there was5 `111 in the pure at first.
(2) 36 686 + 40 686 + 14 686
Formative
Assessment-I
7.40 60
= 686 (36 + 40 + 14) = 686 90 =
x=
= 7.40 15 = `111
4 correct option :
A. Tick () the
61740
1. on the angle 2. triangle 3. 90° 4. H. Amita spent in all = `144 + `1227 +
obtuse 5. –11 6. 6/4
`356 = `1727
B. Fill in the blanks :
I. 1. 1, 10, 100, 1000, 10000, 100000
1. 3 2. 180° 3. straight 4. 360° 5. –y
2. 11, 33, 99, 297, 891 2673
6. 38/7
j. (1) 81 [73 – {8 6 + (15 – 3
C. Write T for true of F for false for
2)}] = 81 [73 – {48 + (15 – 6)}]
the following :
= 81 [73 – {48 + 9 }] = 81 [73 –
1. F 2. T 3. T 4. F 5.T 6. T 7. T 8. F
57]
Summative Assessment-I
=81 16 = 1296
A.1. Four billion five million = 4,005,
(2) 60 – [40 – {20 – (10 – 8)}] = 60 –
000, 000.
[40 – {20 – 2}] = 60 – [40 – 18] = 60
2. Seven hundred one million forty
– [22] = 38
= 701,000040.
K.(1) 43290
3. Fifty lakh nine thousand = 50,
Test by T; It's unit digit = 0
09000.
Number formed by reaming
4. Eight thousand twenty two =
digit = 4329
8,022
4329 – 2 0 = 4329 – 0 = 4329,
B. 10 lakhs = 1000 thousands
Which is not divisible by 7.
C. 190560807 = Nineteen crore five
Test by 9 :
lakh sixty thousand eight hundred
Sum of its digits = 4 + 3 + 2 + 9 + 0
seven.
= 18, Which is divisible by 9.
Smallest number = 100056789.
43290 is divisible by 9 not 7.
36
(2), (3), (4) : do same manner as
above.
L. (1) 216, 252
HCF of 216, 252
(b)
2.007
Ones
Decimal
Tenths
Hundreds
5
3
2
.
6
6
2
.
0
0
Thousands
Tens
(a) 532.46
Hundreds
Numbers
216
252 252
1 = 36
H.C.F
of 216,
–216
L.C.M of 216,
252
36 216 6
–216
0
Thousand
1
180
Remaining articles
= 5ofof yellow
color = 180 – 36 – 96 =1 48
= 5 180 = 36
Decimals
8 of 180
Exercise - 9A=
15
8
= 15 180 = 96
7
2 216, 252
.
(c)
0.8
0
8
108, 126
2 22976
L.C.M =
= 1512
.
(d)
5
7 8
0
40.578
4
54, as
63above.
(2), (3) : do9same
.
8
7 6, 7 (b) 2 = exterior (e)
2. (a)4.8
(b) 4
M.(a) 1 = exterior
.
(f)
5 0 8
0.508
0
(c) 3 = interior
6 6, 1 (d) 12 = exterior
.
(e) 15 = interior
1,1 (f) B = exterior (g) (c)16.3
3
1 6
(d)
N. (a) 120° (b) 30° (c) 90°
.
(h) 120.7
7
1 2 0
O. (1) (–4) + (–5) + (–6) + (–8) + 6 = – 4 (i) (e)
(f)
.
4392.86 4 3 9 2
8 6
– 5– 6 – 8 + 6 = – 17
(2), (3), (4) : do same as above and 3. (a) 4
12 = 1.5
= 0.16
usual method.
25
8
P. The distance between the peak of
7 = 0.875
(b) 8 = 0.8
minar and the point under the earth
10
8
= 1800m + 456 = 2256m
2
3
(c)-(f) : =do0.4same as part
(a), (b)
= 0.75
5
4
above.
Q. (1)
4. (a) 5 1 = 11 = 5.5(b)
2
2
(2), (3), (4) : do same as above.
2
77
(c) 3
(d)
=
= 3.08
25 25
R. Articles of red color
(e)
(f)
13 5
1
+ +
16 8
2
25 5
5. (a) Four
2.5 =tenths
0.4= 4 = 2
=
31
+ 8 =color
Articles
of green
10 2
10 5
= 13 + 10
16
16
37
4
3
(b) Six
and8 five
like because of containing same
0.8=
= tenths 0.3=
10 5
10
number of decimal places.
Exercise - 9C
0.6 = 6 = 3
7.6 = 76 = 38
10 5
10 5 1. (a) 0 < 0.8 < 1 (b) 1 < 1.2 < 2 (c) 7 <
(c) Seven and Four tenths = 7.4
7.5 < 8 (d) 4 < 4.8 < 5
= 4and
= 0.4
(d) Fifty-seven
Thirty-five
10
2. (a) 1.8
hundredths = 57.35
(e) Twenty-nine and six tenths =
(b) 0.4
29.6 5
=
+
= 6 +and
0.5 =three
6.5 tenths =
(f)6Seventeen
10
17.3
(c)-(d) : do same as (a), (b) above.
P
6. (a) 9.3 = Nine and three tenths.
Exercise - 9D
O
1
1.8 2
(b) 4.9 = Four and nine tenths
1. (a) 319.092
P
(c) 13.7 = thirteen and seven tenths
= 300 + 10 + 9
1
O
0.4
(d) 39.3 = thirty nine and three tenths
(e) 7.9 = Seven and nine tenths
= 300 + 10 + 9 + 0.009 + 0.002
(f) 215.7 = Two hundred fifteen
and seven tenths
(b) 0.98 = 0 +
0.9 + 0.08
0
9
2
Exercise - 9B
+
+
+
10 100 1000
(c) 6.32 =6 +
= 6 + 0. 3 +
1. (a)
(b)
0.02
(e)-(h) : do same as part (a), (b), (c)a
9
8
(c)
above.
+
=
10 100
2. (a) 50 + 6 +
3
2
(d)-(j) : do same as (a), (b), (c)
+
10
10
above.
= 50+ 6 + 0.7 + 0.09 = 56.79
36 = 0.36
2. (a) 92 = 9.2
(b)
10
100
(b)
7
9
4243
+
(c)
(d)
= 4.243
10
100
1000
= 7 + 0.4 + 0.08 + 0.007 = 7.487
(e)
(f)
(c)-(f) : do same as part (a), (b)
above.
17
4
8 -79E
1.7 =
(g) 0.6 = 6
7 + Exercise
+
+
10
10
10 100 100
1. (a) In 25.78, the whole number part
214
is 25.
(h) 3.9 = 39
2.14 =
100
10
In 27.21, the whole number part is
2843
27.
3. (a) 7.8
= 7.800
(b) =
35.683 =
28.43
=
0.83
100
100
Since 27 > 25
35.60
3
27.21 > 25.78
3.99
2.91 = 2.91
0.03==3.990
100
(b)-(f) : do same as part (a) above.
1.242 = 1.242
3.46 = 3.46
10532
2. (a) 0.9 > 0.09 (b) 0.783 < 8.3 (c)
(c)-(f)
: do same
10.532
= as (a), (b) above.
1000of decimals are
62.9 < 69.2 (d) 6.100 = 6.1 (e)
4. (a), (c), (d) groups
38
28
7.05 < 7.50 (f) 2.95 > 2.59
(c) 2683g
= 2.683kg
=
kg
100
1
.
.
3. (a) comparing the whole number
[ . 1g = 1000 kg ]
365
parts :
=
kg
1000
0 < 4 < 5 and 0.3 < 0.37
(d) 2kg 750g = 2kg + 750g
So, the required order is : 0.3 < 0.37
2683
< 4.32 < 5.27
= 2kg
=
kg
1000
1
(b) Comparing the whole number
kg ]
[... 1g =
parts :
= 2kg + 0.75kg = 2.75kg 1000
2<3<4<5
(e)-(f) : do same as above.
so, the required order is : 2.14 <
750
3.68 < 4.27 < 5.66
3. (a) 8mm
+
kg
1000
(c)-(d) : do same as above in part
(a), (b).
(b) 5cm 25mm = 5cm +25mm
4. (a) In descending order : 3.8 > 3 >
2.625 > 0.6 > 0.25
= 5cm +
8
(b) 5.270 > 4.320 > 3.471
=
cm = 0.8cm
10
... 1mm = 1
(c) 0.342 > 0.032 > 0.023 > 0.0032
cm ]
=5cm + 2.5cm =[(5+2.5)cm
10
(d) 13.5 > 8.464 > 7.5 > 2.73
= 7.5cm
Exercise - 9F
(b), (c), (e) and (f) : do same as part
25
(a) & (d) above.
cm . .
1
10
[ . 1mm = cm ]
1. (a) 8Paise
= 0.08 rupees
10
4. (a) 298cm
= 2.98m
(b) 40Paise
= 0.40 rupee
(b) 23m80cm = 23m + 80cm
8
= 23m
=
rupee
100
298
=
(c) 465Paise . .
m
1
100 [... 1cm = 1 m ]
[
.
1Paise
]
=
rupee
= 23m + 0.80m = 23.80m
= 4.65rupee
100
100
40
(b), (c), (e), (f) : do same as part (a)
rupee= 3rupees +
(d) 3rupees= 60paise
100
1 m
& (d) above.
60paise
[... 1cm =
]
...1Paise = 1 rupee]
80
[
100
= 3rupees
+
cm
100
465
5. (a) 5ml 100
= 0.005l
=
rupee
100
=3rupees + 0.60rupees
(d) 2l 28l = 2l + 28ml
= 3.60rupees
(e)-(f) : do same as above.
60
= 2l
= 2l + 0.028l = 2.028l
+
rupee
5
100
=
(b),
(c),
(e),
(f)
:ldo same as part (a)
2. (a) 28g
= 0.028kg
1000
1 l
& (d) above.
[... 1ml =
]
1000
Exercise
9G
(b) 365g
1 l
[... 1ml =
]
1. (a) 7 tenths + 9 tenths
= 0.365kg
1000
28
= 0.7++ 0.9 =l 1.6
1000
39
(b) do same as above part (a).
(c) 3 hundredths + 8 hundredth
= 7 + 9
10
10
(d) do same as above part (c).
2. (a)
(b)
= 8 + 18 + 0.25 = 26.25km
6. Shikha bought cloth = 2m 50cm
250
+
Shalini bought
1000cloth = 10m 20cm
They bought in all = 2m 50cm +
10m 20cm
= 3 + 8 = 11 = 0.11
(c) 100 100(d)100
(e)
(g)
3. (a)
(c)
(e)
1.3
+ 0.6
1.9
0.8
+ 0.5
1.3
37.00
+ 28.29
65.29
108.3
+68.09
176.39
(f)
(h)
(b)
(d)
2.3
– 0.5
(f)
1.8
2.5
+ 0.8
3.3
52.31
+ 27.96
80.27
54.23
+ 25.50
79.73
18.67
+2.97
21.64
20.65
– 3.87
16.78
0.8
– 0.1
0.7
20 =m70.25g
7. Rakhi's50
weight
25g
m += 70kg
10 +
= 2+
100weight = 78kg
100 890g =
Raghu's
78.890g70 = 12 + 0.7 = 12.7m
= 12 +
Clearly100
78.890 > 70.25
So, Raghu's weight is heavier than
Rakhi's weight and the required
difference = 78.890 – 70.250 =
8.640g
8. Total distance traveled by Geeta =
12km 125m
distance traveled by bus = 5km
225m
distance traveled by auto =
12.125 – 5.225 = 6.900km or
6km900m.
Exercise - 9H
1. (a) 3.27 = 22.4
(b) 2.91000 = 2900
(c) 19.8916
0.6
–
0.3
(g)
(h)
0.3
13.730
0.9
– 6.324
– 0.3
4. Aman had money = `20.25
7.406
0.6
his friend gave him = `9.75
13.286
59.35
Now, Aman have total
amount =
– 6.900
– 40.07
20125 +
9.75
=
`30.00
6.386
19.28
5. Payal walked on Monday = 8km
19.89
Payal walked on Tuesday = 18km
(d)-(i) : do same
as16
part (c).
250m
Exercise
11934- 9I
She walked in two days = 8km + 1. (a) 283.214 1989
18km 250m
318.24
= (8 + 18
) km
40
283.2
2832
2832
=
=
14
14 10 140
140 2830 20.228
–280
320
= 20.228 –280
(b) 35.5 35 =400
1.014
–280
200
(c) 25.5 5
–120
(d)-(f) : do similar80
as above part (a).
=
The length of the other piece =
12m – 4.29m = 7.71m
4. Sum of the two numbers = 30.64
One number = –21.28
Other number = 9.36
5. Total amount of wheat bought by
them altogether = 1.5kg + 7.25kg +
18kg = 26.75kg
6. Let x should be added.
x + 729.28 = 1000 x = 1000 –
729.28
x = 270.72
7. Length of the rectangle
2. (a) 0.213 0.3
25.5
=
= 5.1
5
(b) 76.363 0.07
= 5.625cm
8. ... Cost of 1l of milk = `28.25
Cost of 17l of milk = 17 28.25
2
= `480.25
(c) 6.25 0.025
Area 10.125 cm
=
=
0.213
9. The
Price of 1m
cm = `45.50
breath
1.8 of cloth
=
= 0.71
0.3
The Price of 12.75 of cloth = 12.75
(d) 7.25 100
76.363
45.50 = `580.125
=
= 1090.9
0.07
10. The Price of a Fancy dress =
`50.50
6.25
= above.
(e)-(i) : do same as
0.025
The Price of 1.5 such dresses
625 1000
Exercise
=15 50.50 = `757.50
=
= 250- 9J
25 100
1. Sunita Purchased a725
Pen = `20.25
11.The Price of 25.3kg mangoes =
=
100a Pencil
100 = `5.36
`243.81
Sunita Purchased
725
The Price of 1kg mangoes
=
0.0725
=
Sunita Purchased ruler = `10.31
10000
Total amount spent = `35.92
2. Anita gave money to her sister =
12. Milkman carries total milk with
`2099.99
himself = 11.50l
Anita gave money to her brother =
he
supplies milk to two customers
`1991.11
243.81
= 3.5l
+ 7.85l
11.35l
==
`9.63
Total amount she gave to both =
=`
25.3quantity of milk left with
Total
`4091.10
himself = 11.50 – 11.35 = 0.15l
3. Total length of a piece of wire =
13. Aditya has money = `410.50
12m
Ravi has money = `316.25
one piece = 4.29m
Aditya has more money by =
41
410.50 –316.25 = `94.25
`11.15
14. Weight of Amit = 79.35kg
2. Since 3<3.798 <4. therefore 3.798
lie between 3 and 4. the difference
Weight of Sumit = 89.9kg
between the two whole numbers is
Clearly, 89.9kg > 79.35kg
1, yes the difference of two
Sumit's weight more and
consecutive whole numbers is
required more weight = 89.90kg
always same.
–79.35kg = 10.56kg
3.
Total
weight purchased by soniya =
15. Product of two numbers = 108.9
6kg925g
+ 8kg625g = 14kg1550g
One numbers = 60.5
Now 14kg 1550g = 14kg + 1550g
The other number = 108.9
60.5
=14kg
= 14kg + 1.55kg = 15.55kg
C.C.E. DRILL- 1
Total 1550
weight .Purchased
Tick () the correct answer :
1 by
[ .. the
1g =
kg]
+= 15.55kg
kgBut
Soniya
weight
1000 of
(a)108.9
2.36, 2.360,
1000
1089 2.3600 (b) 2.30,
the
bag
after
putting
the
vegetable
=
1.8
=
=
2.36,
2.98 (c)
0.041 (d) 0.71 (e) 9.8
60.5
605
and fruits = 16kg
(f) 9.380kg (g) 4.5 (h) 5.5 (i) 0.318l
Weight of the emptied bag =
(j) 0.9
16kg
– 15.55kg = 0.45g
C.C.E. DRILL -2
Introduction
to Algebra
1. Fill in the blanks :
Exercise
- 10A
(a) 9324 (b) 3000.07 (c) 43.6 (d)
1.
(a)
In
a
row
=
5dot
greater (e) 2.36 (f) 0.001 (g) like (h)
In ‘r’ rows = 5r dots
decimus (i) point (j) thousandth.
(b) In a box = 100ladoos
2. Write T for true or F for false for
In ‘l’ box = 100ladoos
the following :
(c) Perimeter of rectangle = l + b + l
(a) T (b) F (c) T (d) T (e) T (f )F (g) T
+ b = 2 (l + b)
(h) F (i) F (j) T
(d) Perimeter of septagon = 7
HOTS QUESTIONS
side = 7Scm
1. Sanya bought apples = `23.65
(e)
Raj's age = x years.
Sanya bought oranges = `165.20
Rahuls's
age = (x – 8) years.
Sanya bought Total items worth
Exercise - 10B
= `23.65 + `165.20 – `188.85
1. (a) Let the number be x. then, x + 4
She Paid to the shopkeeper = `200
= 23
Shopkeeper should return =
(b) Let the number be x, then, x – 7
`200 – `188.85 = `11.15
= 13
But shopkeeper returned to her =
(c) Let the number be x. then,
`18.885
No, Sanya is not correct, she has
(d) Let the number be x. then 3x + 8
made wrong calculation.
= 12
Shopkeeper should return to her =
(e)
42
x
5=2
(f) x + 9 = 37 (g) x x = 100 x2 =
100
(h) 4x – 3 = 89 (i) 5x = x + 4
x
3 = 21
(j)
2. (a) x – 9 = 12 a number decreased
by 9 equals 12.
(b) 3p = 18 Three times a number
x to 18.
equals
3 = x – 42
Solution
x
L.H.S.
R.H.S.
0
15 0 – 45 = –45
60
1
15 1 – 45 = –30
60
2
15 2 – 45 = –15
60
15the
3solution
– 45 = 0
3 x = 7 is
60
(c)
x when 5 is added
(b)-(l)
:
do
same
manner
as
above
15 4 – 45 = 15
4
60
to one-fifth of a number equals the
part 8(a).
15 5 – 45 = 30
60
5
number.
9.6(b), (c),15
(e),6 – (f),
(h), (g) are60 all
45 = 45
(d) 15y – 45 = 15 fifteen times a
equations because of having (=)
number
15 7 – 45 = 60
7
60
1 decreased by 45 equals 15.
sign.
5x+5=
Exercise - 10C
(e)
two-thirds of a number
1.
Let
two
consecutive even numbers
equals 8.
be x, x+2.
(f) 5x – 6 = 19 five times a
x + (x+2) = 42
number decreased by 6 equal 19.
2x + 2 = 42 2x = 42 – 2
3. (a) x+6 = 14 (8, 20)
2a
2x
= 40 x =
= 20
x ==14–6
=8
3 8
8 is the solution.
numbers are 20, 22
(b) x–1 = 15 (2, 16, 8)
2. According to questions,
x = 1+15
x + x+ 3 + x + 4 = 27
x = 16
3x + 6 = 27 3x = 27 – 6 3x =
16 is the solution
40
21
(a)-(f) : do same as above part (a) &
2
(b).
numbers are 7, 9, 11
4. 2x – 4 = 8 2x = 8 + 4 2x = 12
3. According to questions,
x +x +1 + x + 2 = 99
3x + 3 = 99 3x = 99 – 3 = 96
x = 6 is the solution of the given
equation.
5. 6. 7. do same as Q.No. 4 above.
21 are 32, 33, 34
numbers
x=
=7
8. (a) 15x – 45 = 60
4. P = 2 (l + 3b)
Here, L.H.S
72 = 2 (x + x – 8)
12 is 15x – 45 and R.H.S
= Now,
= 6evaluate the L.H.S of
is x60.
72 = 2 (2x – 8)
2
equation for same values of y till
36 = 2x – 8 36 + 8 = 2x
the L.H.S of equation equal to
2x = 44
R.H.S
96 = 22m
x
length
= ==x32
3
43
breadth = (x – 8) = 22–8 = 14m
5. Let his Present age = x years
x + 10 = 3x 10 = 2x x = 5
years
6. Let Puja’s age = x year, then
Father’s age = 3xyears
44
= 22
3x + 15 = 2 x
(x=+ 15)
2
3x + 15 = 2x + 30 3x –2x = 30
– 15 x = 15
Puja’s Present age = 15years
Father Present age = 3 15 =
45years
7. Let First number = x
Then, Second number = 4x
4x – x = 90 (given)
3x = 90 x = 30
Number are = 30, 120
8. do same as Q.No. 6 above.
9. Let one number = x
other number = x + 15
x (x + 15) = 75 2x = 75 – 15
2x = 60
x = 30 numbers are 30&45
10.Let Manu’s age = x, then Anu’s age
= 2x years
2x – x = 2 x =2
Hence after 10 years manu’s age = 2
+ 10 = 12 years
and after 10 years, Anu’s age = 2
2 + 10 = 14years
11.6x = 95 + x 6x – x = 95 5x = 95
13. do same
95 as above Q.No 8.
x=
= 19
5
14.Let Navin’s
Present age = x years
8 years ago, his age = (x – 8)years
x + 32 = 3 (x –8)
x + 32 = 3x – 24 32 + 24 =
3x – x
2x = 56
his present age = 28 years
15.Let his sister’s age = x years
then Umesh’s age = (x – 5) years
x + (x – 5) = 45 2x = 45 + 5
Hence, Umesh’s age = 20 years
and his sister age = 25 years
16.Let the number be x
then 2x + 5 = 45 56
x=
= 28
2
number is 20.
17.Let 50 paise coins = x
than 25 paise coins = 6x
value of5050 paise coin in rupees
x =
x = 25
2
value of 25 paise coin in rupees
x=
45 – 5 40
= = 20
2
2
number is 19
2x = 50 x = 25
12.Let breadth = x m
50 paise coins = 25
than length = 3x m
50
=
25` paise
100 coins = 25 6 = 150
Area of rectangle = l b
18.Let girl age = x year
48 = (3x) (x) 3x2 = 48 x2 =
young sister = y years
16
25
= ` her brother = z years
and
x = 16 = 4m
100
x – y = 4 ...........(1)
breadth = 4m, length = 12m
44
y = z + 450 ...........(2)
25
+ `
= `50
y+xz =` 16
100...........(3)
100
from
and (3), we 200x
get
50x +(2)
150x
= 50
= 50
z + 4100
+ z = 16 2z =
16 – 4
100
From (2), we get y = z + 4 = 6
+ 4 = 10
From (1), we get x – 10 = 4 x =
10 + 4 = 14
girls age = x year = 14 years
young sister = y year = 10 years
and her brother = z year = 6
years
19. Let Larger number = y
and smaller number = x
then
– x==6 7 ..........(1)
z =y 12
2 = 77
y + 6x
–7x = –70
– x = 8 (h) 8 (i) 7 (j) 33
2. (a) true (b) false (c) false (d) true (e)
true (f) false (g) false (h) true (I)
false (j) true
HOTS QUESTIONS
1. Let the number be x
x + 5x = 20
But she answered the number (x) =
15
She is wrong and we are not
agree with her.
2. do same as above in Q.No 20 of
exercise- 10C.
3. Let Swati’s Present age = x years
then after 10 years his age = (x + 10)
years
x + 10 = 3x
3x – x = 10 2x = 10
x = 5
20
Thus, Swati’s Present
x = age = 5 years
6
Ratio and Proportion
Exercise - 11A
From (1), y – 10 = 7 y = 17
numbers are (x, y) = (10, 17)
1. (a) 80 : 30
20.Let Nidhi Present age = x years
and father Mr. Saxena’s Present
(b) 19 : 38
age = (x + 25) years
(c)-(f) : do same as above part (a)
(x + 25) + 5 = 2 (x + 5)
& (b)
x + 30 = 2x + 10
30 – 10 = 2x – x 20 = x
or, x = 20
Nidhi’s Present age = x = 20
x = 10
2
years
–70
or,
x=
=
10
Mr. Saxens’s
–7Present age = (x + 25)
= 45 yers
C.C.E. DRILL - 1
Tick () the correct answer :
(a) 12, (b) 2 (c) 4 (d) x/3 (e) 2x – 3 =
= 80 = 8 = 8 : 3
30 36
3 (f) –31 (g) 48 (h) 3 (i) 33 (j) 3
C.C.E. DRILL - 2
= 19 = 1 = 1 : 2
38 2
1. (a) 3y (b) –1 (c) solution (d)
equation (e) 1 (f) Trail, error (g) 36
45
1 =1 : 20
20
`800
2. (a) `40 : `800 = `40 =
(b)-(e) : do same as above part (a).
3. (a) 45min : 1 hour =
= 3=3:4
4
45min 45min
=
1hour 60min
(b)-(e) : do same as above part (a).
4. (a) required ratio = 1 : 7 (b) required
ratio = 1 : 10 (c) required ratio = 3 : 4
(d) required ratio = 2 : 1
(e) required ratio = 3 : 4
5. Time taken by Rahim from Delhi to
Ghaziabad is 3 times that of Ram.
2
20
(b) The no. of villages in India is
3
times that of cities.
(c) The no. of good bats Produced in a
factory is 5 times that of bad bats
Their total earnings = 45000 + 65000 =
`110000
(a) Avneesh’s earning : his wife earning
= 45000 : 65000 = 45 : 65 = 9 : 13
(b) Avnessh’s earning : Their total
earnings = 45000 : 110000 = 45 : 110 =
9 : 22
(c) Avnessh’s wife's earning : Avnessh
earning = 65000 : 45000 = 65 : 45 = 13 :
9
(d) Wife’s earnings : difference of both
earnings = 65000 : (65000 – 45000) =
65000 : 20000 = 13 : 4
9. do same as above Q.No. 8.
10. do same as above Q.No. 8.
distance 45
= Time = 5
11. speed of a cycle
= 9km/h
distance 120
= Time = 6
speed of a car
= 20km/h
2
9 3
4
=
required ratio = 9km/h : 20km/h = 9
6. (a) 9 : 15 15 = 5 and 4 : 5 = 5
: 20
4 3
> 4 : 5 > 9 : 15 12. Boys : girls = 5 : 4, Total girls = 80
Since 4 > 3
5 5
Let Boys = 5x, girls = 4x
80
24 4
4
= 20
x=
4 : 16 =
=
(b) 24 : 30 =
4
30 5 and
16
4x = 80
1
=
4
Boys = 5x = 5 20 = 100
L.C.M of (4, 5 ) = 20
Hence Boys = 100 and Total students =
100 + 80 = 180
13. p : q : r = 6 : 9 : 10
Sum of ratio
6 = 6 + 9 + 10 = 25
4 4 4 16
=
=
5 5 4 20
1 45 5
and =
=
4 4 5 20
Since 16 > 5 16 > 5
20 20
4
1
> 24 : 30 > 4 : 16
5 4
(c)-(d) : do same as above in Part (a), (b)
7. Raju earned = `2,50,000
Raju expend = `50,000
(a) Income : expenditure = 2,50,000 :
50,000 = 5 : 1
(b) Expenditure : income = 50,000 :
2,50,000 = 1 : 5
8. Avneesh earns per month = `45,000
His wife earns per month = `65,000
P = `
25
9
`
q=
25
10
`
r=
25
2
= =2:7
14. required ration = 200 : 700 7
15. do same as above Q.No. 13.
16. Factory Produces bolts per day = 435
In every 15 bolts defective are = 2
46
2
15
In every 1 bolts defective are =
and2In 435 bolds defective are
15
435 = 58
(b)-(f) : do same as above part (a).
17. Expenditure : Saving =220
11 : 4
11x = 2200 x = 11 = 200
Saving = 4x
= 4 200 = `800
18. Price of 12 Maths Books = 240
`240
`
12
Price of 1 Math Books =
Price of 15 English Book =`135
135
`
Price of 1 English Book =
15
2. (a) 15 : 25 : : 9 : 15
Product of mean = Product of
extremes
25 9 = 15 5
225 = 225
LHS = RHS
It is true
(b)-(f) : do same as above Part 2(a)
3. (a) 40 : 10 = 80 : 20
= `20
= `9
40
4
= =4
10
1
RHS = 80 : 20 80
4
=
= =4
20
1
LHS = 40 : 10
required ratio = Math : English =
20 : 9
19. Let l = 5x, b = 3x
P = 2 (l + b) 48 = 2(5x + 3x)
48 = 248
8x 48 = 16x
x=
16
=3
length (l) = 5x = 5 3 = 15cm
breadth (b) = 3x = 3 3 = 9cm
20. Let first part of money = 3x
and second part of money = 4x
3x + 4x = 7777
7777
7x = 7777 x = 7 = 1111
150m
1
=
= 1 : 300
300 150m2 300
=
3
4
5
=
and 5 : 6 6
3 5
4 6
3:4 5:6
= 16
(b)-(d) : do6same as above part 4(a).
5. Let First Term be x. then x : 92 : : 87 :
116
x
87 92
87
x=
=
92
116 = 69
116
6. Let Third Term be x.
Then 25 : 5 : : x : 4
23. do same as above Q.No. 7 and Q.NO.
8.
Exercise - 11B
1. (a) 3, 4, 5, 6
3:4
LHS = RHS, Hence verified.
(b)-(f) : do same as above part 3(a).
4. (a) 8 : 6 : : x : 12
Product of means = Product of
extremes
6 x = 8 12
x = 8 12
First Part of Money = 3x = `3333
Second Part of Money = 4x = `4444
21. do same as above Q.No. 13.
22. (a) length : breadth = 300m : 150m = 2 : 1
(b) breadth : Area = 150m : (300
150m2)
=
=
25
=
x
x = 20
5
4 be x.
7. Let Fourth
term
Then 117 : 13 : : 81 : x
117 81
13 81
x=
13 = x
117
x
=
9
2
8. b = ac
b2 = 144 169
b = 144 169 = 12 13 = 156
9. given 25 : x : : x : 5
x x = 25 5 x2 = 125
x = 125
47
10. given fourth term = 60
8 : 15 : : x : 60
1. Cost of 28m cloth = `504
Cost of 1m cloth =
8
60 8
x
x=
=
15
60
x = 415
8 = 32
504
28
504
6 = `108
2. Cost of 5 oranges = ``48
28
Cost of 6m cloth = `
11. (a) 121 : x : : x : 289
x
121
x2 = 121 289
x = 289
x = 121 289 = 11 17 = 187
(b)-(d) : do same as above part 11(a).
12. (a) Let x be the mean proportion
between 64 and 81.
i.e; 64 : x : : x : 81
x2 = 64 81
[... b2
= ac]
x = 64 81 = 8 8 9 9
x = 8 9 = 72
(b)-(d) : do same as above part 12(a).
13. (a) 18, 36, 19, 38
We have, 18 38 = 36 19 .........(1)
Thus, First term = 5 and fourth term =
38
second term = 36 and third term = 19
5 : 36 : : 19 : 38 ......... (2)
We can write equation (1) as 18 38 =
19 36
First term = 18, Second term = 36
Third term = 19, Fourth term = 36
18 : 19 = 36 : 38
Now clearly, two more proportion are
36 : 5 : : 38 : 19 and 19 : 18 : : 38 : 19
Hence, the required proportions are :
18 : 36 : : 19 : 38 and 18 : 19 : : 36 : 38 and
19 : 18 : : 38 : 36
(b)-(d) : do same as above manner.
Cost of 1 oranges =
48
5
3. Cost of 120 articles = 48
`64000
23 = `220.8
5 120
Cost of 1 articles = `64000
Cost of 23 oranges =`
Cost of 20 articles = `
64000
= `10666.7
20
4. do same as above Q. No.120
1, 2, 3.
5. 5 dozen (= 5 12 = 60) soap-cakes cost
= `80
1 soap-cakes cost = `
80
60
and 5 soap-cakes cost = `
80
3.
5 = `6.66
6. 7. do same as above Q. NO.601, 2,
8. 155.75 kg weight is of = 15m
1kg weight is of =
15 m
155.75
250 kg weight is of =
15
= 24.07m
250
155.75
9. 800 men need rice in 1 day
= 200kg
1 men need rice in 1 day =
14. (a) 25 : 10 : : 10 : 4
800
575 men need rice in 3 days
= kg
200
25 10
2.5 = 2.5, Hence
verified=
10
4
(b)-(d) : do are as above manner.
15. b2 =ac where a : b : : b : c and b is called
mean proportion
16. (a) 25, 10, 4
2.5 = 2.5
25 10
25, 10, 4 arein proportional.
=
4
(b) 16, 84, 441 10
200
573 3 kg
800
1725 = 431.25
kg
10.=Car Covers
the distance
in 21 hr 39 min
4
= 1299km
given numbers are in proportional.
= 3.
11. do same as above Q. No. 1, 2,
2
=
649.5km
12. motorbike covers the distance in per
16
4 - 11C
84 Exercise
4
84 = 441
21 = 21
or car covers the distance in (21 60 +
39 = 1299min)
car coves the half time in
1299
48
litre i.e, 1l = 60km
motorbike covers the distance 20l = 20
60 = 1200km
the required distance between Delhi
& Pune = 1200km
13. In 8.5 gm of an allow weight of copper
= 4.7kg
In 1gm of an allow weight of copper
30 8.5
km = 63.75km
19.=For 1254men provisions is sufficient for
= 30days
For 1 men provision is sufficient for =
30 125days
For 300 men provisions is sufficient for
30 125
days = 12.5 days
=
20. Cost300
of 10m of a cloth = `3320
4.7
25.5kg (= 2500gm) of an allow
g
=
weight
8.5 of copper
cost of 1m of a cloth =
`
3320
(a) The cost of 35m of the same
10cloth
3320
14. In4.7
`1225
of =
chairs
purchased = 35
no.
2500
14.10kg
=
35 = ` 11620
=
10`3320 cloth comes = 10m
(b) In
8.5
In `1 no. of chairs purchased =
In `1 cloth will come = 10
In `1400 no. of chairs purchased 35
3320
m
and in `6640 cloth comes
1225
10 66m = 20m
3320 C.C.E DRILL - 1
=
35 travels
15. A cyclist
12km in =
1400 = 40
=
1225
2
hr
3
A cyclist travels 144km =
in 2
hr
3 12
(a) 15 : 25, (b) 1 : 8, (c) 15/17 (d) 1 : 20,
(e) 4 : 5, (f) ad = bc, (g) 5 : 8 : : 90 : 144,
(h) 25, (i) 104, (j) 2, 5, 2, 5.
C.C.E DRILL - 2
1. Fill in the blanks :
(a) Fraction (b) equal (c) continued (d)
2 taken
16.=Time
4 workers
proportion (e) less (f) number (g)
by
144
= 8hr = 88days
3
12 taken by 1 worker = 88
simplest (h) product of extremes (i) no
So,
time
(j) ratio
4days
2. Write T for true F for false fo the
Time taken by 11 workers
following :
88 4 days
= 32days
=
(a) false (b) true (c) false (d) true (e) true
11
17. In 16 days a piece of work finished
by
(f) true (g) false (h) true (I) false (j) true
= 25men
HOTS QUESTIONS
In 1 day a piece of work finished by = 1. In `96 Savitri bought wheat = 12kg
25 16men
In 8 days a piece of work finished by =
In `1 Savitri bought wheat = 12 kg
A cyclist travels 1km in
96
25 16men = 50men
18.=engineer
8 uses 4cm to represent = 30km
engineer used 1cm to represent =
engineer uses 8.5cm to represent 30 km
In `160 Savitri bought wheat
12
160kg = 20kg
=
96
2. Ragini’s earning’s
4
49
=
2
2
81000 =
81000
(2+7)
9
2 9000 = `18000
and Sangini’s earnings
=
7
7
81000= 81000
(2+7)
9
= 7 9000 = `63000
Clearly, Sangini earns higher.
Sangini earn by = `63000 – `18000 =
`45000more
3. do yourself
Formative Assessment - III
A. 1. 4.5, 2. 0.318l, 3. 2x – 3 = 3, 4. 33,
5.15/17 , 6. 104
B. 1. 43.6, 2. like, 3. equation, 4. 33, 5.
fraction, 6. equal.
C. 1. true, 2. false 3. true 4. true 5. false 6.
true 7. true 8. true
Exercise - 12A
1. (a) Perimeter of the given figure will be
the sum of all its sides
P = 20cm+ 5cm + 20cm + 5cm =
50cm
(b) P = 30 + 30 + 30 + 30 = 120cm
(c) P = 4cm + 3cm + 3cm + 4cm + 2cm +
4cm + 3cm + 3cm = 26cm
(d) P = 6cm + 1cm + 3cm + 4cm + 3cm +
2cm + 3cm = 2cm + 3cm + 3cm +1cm
= 35cm
2. (a) l = 2.5cm, b = 1.5cm
P = 2 (l+b)
= 2(2.5+1.5) = 2 4 = 8cm
(b)-(d) : do same as above part (a).
3. (a) Perimeter of square = 4 side
= 4 4cm = 16cm
(b) do same as above part (a)
(c) 2m 70cm = 2m + 70cm = 2
100cm + 70cm = 270cm
P = 4 size = 4 270cm = 1080cm
or 10.8m
4. distance covered by girl in 1 round = 26
+ 14 + 26 + 14 = 80cm
distance covered by girl in 5 round =
80 5 = 400cm
5. (a) Perimeter of pentagon = 5 side =
5 5 = 25cm
(b) Perimeter of square = 4 side = 4
5 = 20cm
(c) Perimeter of septagon = 7 side =
7 5 = 35cm
6. Perimeter of the floor of a rectangle = 2
(7 + 4) = 22m
7. P = 2 (l + b) 28 = 2 (l + 4) 14 = l + 4
l = 14 – 4 = 10m
8. The required length of wire to fence the
field = Perimeter of the field
= 20m + 16m + 20m + 16m = 72m
Cost of 1m fencing the wire = `3
Cost of 72m fencing the wire = 3
72 = `216
Exercise - 12B
1. (a) It contain 12 complete squares.
So, the area of one square = 1cm
1cm = 1cm2
then, area of 12 complete square = 12
1cm2 = 12cm2
(b) It contains 7 complete and 4 more
than half squares, we neglect less than
half parts of square.
Total area = 7 complete square + 4
more than half square
= 7cm2 + 4cm2 = 11cm2
(c) Total area = 6 complete square + 2
more than half square + 2 (1/2 half
square)
= 6cm2 + 2cm2 + 1 full square =- 9cm2
(d) Total area = 3 complete square + 3
(1/2 square)
= 3cm2 + 3/2cm2 = 3cm2 + 1.5cm2 =
4.5cm2
(e) Total area = 4 complete square + (4
more than half squares)
= 4cm2 + 4cm2 = 8cm2
(f) Total area = 3 complete square + (2
more than half square) + 2 (1/2
squares)
= 3cm2 + 2cm2 + 1cm2 = 6cm2
(g) Total area = 2 complete square + 4
(1/2 squares)
= 2cm2 + (2 full square)
= 2cm2 + 2cm2 = 4cm2
(h) Total area = 13 complete square +
(2 more than half square)
= 13cm2 + 2cm2 = 15cm2
(i) it contains 8 complete squares
So, area of 8 complete squares = 8
1cm2 = 8cm2
(j) It contains 12 complete square. we
neglect less than half parts two of
50
square.
Area of 1 square = 1cm × 1cm = 1cm2
So, area of 12 complete squares = 12
1cm2 = 12cm2
Exercise - 12C
1. (a) l = 8cm, b = 9cm
Area = l b = 8 9 = 72cm2
(b) do same as above in part (a).
(c) l = 5dm = 5 10cm = 50cm
b = 2.5cm
[...1dm = 10cm]
Area = l b = 50 2.5 = 125cm2
(d) l = 28cm
b = 100mm =
= 10cm
Area = l b100
= 28 10 = 280cm2
cm
2. (a) Area of square
10 = side side
= 225 225 = 50625cm2
(b) Area of square = side side
= 19 19 = 361m2
3. Area = 240cm2, b = 16cm, l = ?
Area of rectangle = l b
240 = l 16 l =
4. P = 80m, l = 25m
(a) P = 2 (l + b)
80 = 2 (25 + b)
240
= 15m
80 = 50 + 2b
16
80 – 50 = 2b
30 = 2b
breadth = b = 15m
(b) Area = l b
30 = 375m2
=25m
b = 15m
= 15
5. Perimeter
2 of square = 48cm
i.e P = 48cm
P = 4a
48 = 4a
a = 12
Area of square = a a = 12 12 =
144cm2
6. Area of square48
= 225cm2
one side
= a = Area
(a) length of
4
= 225 = 15cm
(b) P = 4 one side
= 4 15 = 60cm
7. Area of square = 121cm2
(a) Area = (side)2 = a2
121 = a2
a = 121 = 11cm
(b) a = 11 10mm
10mm]
= 110mm
[...1cm =
(e)
(d) a = 11
a=
11
m = 0.11m
100
= 1.1dm
8. l = 48dm = 480cm
1
b =11
3.6m
dm=[.360cm
..1cm =
dm]
=
10
Area = l b 10
= 480 × 360 =
172800cm2
P = 2 (l +b) =2 × (480 + 360) = 1680cm
9. l = 12dam
b = 1km = 100dam
Area = l b = 12 100 = 1200dam2
10. do same as above Q. No. 2.
2
[...1km
= 100dam]
11. Area of square = (side)
2
2
121cm = (side)
Side = 121 = 11cm = 11 10mm
= 110mm
12. l = 425dam, b = 300dm
[...1cm = 10mm]
A = l b = 425 300 = 127500dm2
2
... 1dm
2
2
A = 127500
100
100mm
=10cm
22
2
= 1275000000mm
1cm = 100mm
2
2
P = 2 (l + b)=1dm
2 (425
300)dm
=2
=+
100
100mm
725dm
= 1450dm = 1450 100mm
= 145000mm
13. Total cost of painting a wall = `336
Rate of painting = `2per m2
Then, area of the wall[...1dm = 10mm]
Now, the area = 168m2
total cost
2
=
breadth
= 120cmm =
rate per m2
Then, length
336
= 168m2
2
14. l = 225m, b = 20m= 120 m = 1.2m
100
Area of carpet = l
b = 225 20
51
area
168
= 140m
= 4500m2 =
=
1.2
...Cost of 1m2 =breadth
`25
Cost of 4500m2 = 25 4500 =
`112500
15. Side of a tile = 1cm
Area of rectangular field
one side
22. =
breadth
7 hectares
50m
70,000m2
50m
23.=required no. of=note-books
Area of a tile = (side)2
= 1400m
Area 27m2
= 5.4m
= = `45000
24. Total cost=oflength
producing
1
5m2
dm
=
rate
of
producing
=
`2.25m
10 2
1
Then, Area of the rectangular field
required no. of tiles needed
= 10 to cover the
105 22
floor
= 15 2 = 77
1
dm2
=
100
=
Area of floor
Area of one square type tile
total cost
2
=
length
rate per m2 m
2
16. Let
the side
of square (a) = x cm
960dm
= 96000
=(a) side is doubled
(a) = 2x cm
1
2
2
Areadm
= (side)
= (2x)2 = 4x2 = 4 time
100
the previous area.
(b)-(c) : do same as above part 16 (a).
17. Let length = x can, breadth = y cm
(a) Area = l b = (2x) (2y) = 4xy =
4times
(b) Area = l b = (2x) (y) = 2xy =
2times
(c) Area = l b = (x) (2y) = 2xy =
2times
(d) Area = l b =
(e) Area = l b =x(3x)
= 9xy =
y (3y)xy
=
9times
2
2
4
2
2
2
18. 5cm
1 = 5 100mm = 500mm
.
.
2
2
times
[=. 41cm
= 100mm ]
19. The area of two rectangles having
same perimeter will be equal.
20.Square contains larger area
21. other side
45000
2
20000m field = 2 (l + b)
=
Perimeter
of=rectangular
2.25
m
= 2 (4000 Area
+ 5) = (8000
20000+ 10) m =
= 4000
=
8010m =
5
... Cost ofBreadth
fencing 1m = `7.50
Cost of fencing 8010 = 8010 7.50
= `60075
25. Perimeter of rectangle = Perimeter of
square.
2 (l + b) = 4 side
2 (20 + b) = 4 15
40 + 2b = 60
2b = 60 – 40 = 20
Now, Area of rectangle = l b
= 20 10 = 200m2
and, area of square = (side)2 = (15)2 =
20
225m2
= 10m
b=
2
room of square type has greater
area.
required difference = (225 – 200) =
25m2
26. Let when side of square = y cm, area
= y2 cm2
but side = 2y area = (2y)2 = 4y2
Area increased by 4 times
27. l = 132cm
52
Perimeter of rectangle = Perimeter of
square
2 (l + b) = 4 side
2 (132 + b) = 4 100
b = 200 – 132 = 68cm
Area of Rectangle = l b = 132
68
= 8976cm2
28. Area of rectangle room = 26 13 =
338cm2
Area of square type room = (side)2 =
(23)2 = 529
square room has larger area.
Exercise - 12D
1. Area of inner rectangle PQRS
= (l b) = (24 18) = 432m2
Area of outer rectangle ABCD
= (l b) = (24 + 4 + 4) (18 + 4 + 4)
= 32 26 = 832m2
A
Area
of the path (i.e, shadedDportion)
= AreaP of outer rectangle –S Area of
inner rectangle
18m
26m 4m2
= 832m
– 432m2 24m
= 400m2
2. Let ABCD
is a square
Q
4mfield. R
32m
AreaBof square ABCD
= (side)2C
2
2
= (20) = 400m
A
20m
D
length of PS = (20 – 2 – 2) = 16m
P
length of PQ = RS = (20
– 2 – 2) = S16m
Area of inner square PQRS = (side)2
2
16m
20m 2m
= (16 16) = 256m
Area of path = Ara of ABCD – Area
16m
of PQRS
R
Q
2
2
2 2m
= 400m – 256m =B144m
C
3. Area of rectangular room = l b = 30
15 = 450m2
C 2
D
Area of square
carpet = (side)
2
2
R
S
= (21) = 21 21 = 441m
Area left uncovered = 450m2 – 441m2
21m 15m
= 9m2
4. Area of rectangular playing ground = l
P
21m
Q
b
230m
A
B
= 25 15 = 375m
Area of inner rectangular ground = l
b
C
D
= (25 – 3 – 3 ) 3m
(15 – 3 – 3) =R(25 – 6)
S
2
(15 – 6) = 19 9 = 171m
2
2
18m
Area
of path = 375m
– 171m
=
15m
3m
3m
204m2
P 3m
5. Area of hall
ABCD = l b Q
B
25m
A
= (10.5 8.5)m2 = 89.25m2
Area of path PQRS (i.e, total part
R
S
enclosed
with road) = l b
3.5m
D
C
= (10.5 + 3.5 + 3.5) (8.5 + 3.5 + 3.5)
2
= 17.5 15.5 = 271.25m
hall
8.5m
15.5m 3.5m
Area of the road
= Area3.5m
of path
10.5m
PQRS – Area
ABCD B
A of hall 3.5m
= 271. 25
Q
P – 89.25 17.5m
= 182.00 = 182m2
6. do same as above.
(hints : area of path (shaded region) =
90m2)
20m
53
S 5m 2R
= (210D + 250 – 15)m
= 445m2 C
Area pf whole rectangular field = l b
X 5m 2 W
K
N
= 70
50 = 3500m
3m
3m
50m
AreaL of remaining
parts of
M the filed
Y
Z
2
= 3500 – 445 = P3055m
Q
70m PQRS = l B
9. Area of
b
A grafty lawn
= 45 25 = 1125m2
7. Let KLMN and PQRS are two cross
roadsD parallel to length & breadth
of
C
rectangular
S field respectively.
R
6m
10m
10m
Q
P
A
15m
B
The Area of road KLMN = l b = (80
10) = 800m2
S 10m R
D
The Area
of road
PQRS = l bC = (60
2
10) =K600m X
W
N
Now, It is clear that path XYZW are
60m
10m
common to both roads KLMN and
L
M
Y
Z
PQRS.
P
Q
2
So. theAarea of XYZW
= (10 10)m
=
B
60m
2
100m
Area of path to be gravelled = (800 +
600 – 100)m2 = 1300m2
Cost of gravelling the path = `(5
1300) = `6500
8. Area of path KLMN = l b
70 3 = 210m2
Area of path PQRS = l b
= 50 25 = 250m2
Since, Path XYZW is common to both
sides (Paths) KLMN and PQRS.
So, the area of XYZW = (l b)m2 = (5
3)m2 = 15m2
Area of ABCD = l b
= (45 + 2 + 2) (25 + 2 + 2) =49 29 =
1421m2
Area
of Path = Area of CABCD –
D
2
Area ofSgrafty lawn
PQRS R
= 1421 – 1125 = 296m2
2
Cost of gravelling the25path29= `(4.30
45
296) = `1272.80
Q
2
10. (a) AreaP of rectangle
ABCD =B l b
A
2 49
= 5 2 = 10cm
Similarly, Area of rectangle PQRS
= (l b) = 5 2 = 10cm2
Area of rectangle KBPM = l b
= 2 2 = 4cm2
Area of three shaded region = Area
of ABCD + Area of KBPM + Area of
PQRS
= (10 + 4 + 10)
= 24cm2
(b) Area of rectangle ABCD = l b
= 20 5 = 100cm2D2cm C
S 2cm R
Area of APQR = l b
= 17 4 = 68cm2
Area of shaded region
3cm = 100 – 68 =
5cm
32cm2
5cm
Area of paths = (Area of path
KMLN) + (Area of path PQRS) – Area
of common path XYZW
K
A
2cm
B
P
6cm
54
M
Q
(c) Area of square = side side
= 11 11 = 121m2
20cm
D
C
given, area of four region in side the
1cm
square
R PQRS = A + B + C +
Q D
= 5m2 + 2m2 + 5m2 + 2m2 = 14m2
Area of shaded region = Area of
4cm
4cm
PQRs
– Area of 4 region
inside the
square
2
= 121A– 14 = 107m
P
17cm
3cm B
A
4m
C
M
A
2
B
D
2
C
(g) Area of shaded region = Area of (1)
2
+ Area of (2) + Area
of (3) + Area of
(4)
T
U
P 2 Q
2
= (7 1)m52 + (5 1)m2 + (3 1)m
+
5
2
(1 1)m
S 2 R2
= 7m2 + 5m2 W+ 3mV2 + 1m2 =
16m
2
2
Q
(e) Area of shaded region = Area of
ABCD + Area of PQ RS + Area of
P
C
KLMC
2
= (8 2)2m2 + (2
2)m2 + (2 2)m2
= 16m2 + 4m2 + 4m22 = 24m2
2
10cm
2
2
A
Q
2m
(d) Area of square ABCD = (side)
= (10)2 = 100m2 S
R
Area of 3 inside square =A = (side)2 +
2m
A = 5m
(side)2 + (side)2
2
2
2
= (2) + (2) + (2) = 4 + 4 + 4 = 12m2
Area of shaded region = Area of
ABCD – Area of inside
A = 5m square 2m
A=
= 100m2 – 12m2 = 88m2
P
2m
P
2
2
B
(f) Area of shaded region = Area of
TUVW + Area of ABCD + Area of
PQRS
8m
S
R
= (5 2)m2 + (9 2)m2 + (5 2)m2
2
2
2
2
L
= 10m + 18m + K10m 2m
= 38m
D
2
2m
10cm
2
B
C.C.E. DRILL - 1
1m
Tick () the correct
:
G answer
F
(a) 4 side (b)1m2 (l + b)
(c) 7 side (d) 7
1m
E
H (g) four times
(e) 35 (f) 112
(h) 100cm2
1m
1m
2
D
I
(i) 15cm (j) 10000m
1m C.C.E. DRILL - 2 1m
C
J
1. Fill1min the blanks :
1m
2
(a) length
(b) m (c)
A
7m 100 (d) 2 (l +
B b) (e) 8
(f) volume (g) Area (h) 100 (i) Perimeter
(j) length
2. Write t for true or F for false of the
following :
(a) T (b) F (c) T (d) T (e) F (f) T (g) F (h)
55
F (i) T (j) T
HOTS QUESTIONS
1. Distance covered by Ishan in 1 round =
2 (l + b)m
= 2 (98 + 55) = 2 153 = 360m
distance covered by Ishan in 10
rounds = 306 10 = 3060m
2. Area of flour of her room = (side)2 =
(10)2 = 100m2
in 1m2 of the area she need flowers =
100
In 100m2 of the area, she will need
flowers = 100 100 = 10000
3. Vivek’s friends are correct since 16
cubes of this dimension cannot be
placed in cuboidal box because of
cuboidal box does not have height.
Exercise - 13A
1.
Numbers
Tally marks
14
17
2. do same as above Q. NO. 1
19
3.
21
Exercise - 13B
1. (a) We see from given figure,
Students play hockey = 5 pictures
But 1 Picture of student = 8 students
5 pictures = 5 8 = 40 students
No. of students paly hockey = 40
students.
(b) Students play cricket = 8 pictures
But 1 picture = 8 students
8 pictures = 8 8 = 64 students
No. of students play cricket = 64
students
(c) Since No. of picture are more in
basketball row.
Basketball is played by most of the
students.
(d) No. of students play cricket than
basketball = 8 9 – 8 8 = 72 – 64 = 8
2. From the given pictograph, we can
conclude that
(a) The sale of books was more on
thursday.
(b) The sale of books was less on
wednesday.
3. Let = 10 students. Then, the
pictograph will be
Frequency
5
1
6
4
25
5
27
4
28
1
30
1
(a) highest
obtained
10
Tally
marks =Frequency
Marks marks
(b) No. of students got the highest
5
4
marks = 3
6
4 less
(c) No. of students got the marks
than 7 =
4
+
4
=8
7
2
(d) No 8of students got the lowest4marks
= 4 (i.e, 5 marks)
(e) No.9of students got the marks3 more
3
than 810
=3+3=6
56
(a) In section D, there are 4 pictures,
and 1 picture of = 10 students
Section
of=Students
4 pictures = 4No.
10
40 students
A In section E, students are = 6 10 =
(b)
60
B
In section D, students are = 4 10 = 40
C
difference = 60 – 40 = 20
D
there are 20 more students in sec E
E that of sec D.
with
(c) since section A more picture than
others
section A has the highest no. of
students.
(d) Section D has the lowest no. of
y
90
90
70
(i)
80
80
80
Marks
student
In sec D, 4 pictures 4 10 = 40
students.
4. do as above.
Exercise - 13C
1. (a) The maximum number of
restaurants are in = Delhi
(b) The minimum number of
restaurants are in = Chennai
(c) No. of restaurants in Delhi = 28
(d) required more restaurants = 26 – 24
=2
2.
75
72
David secure maximum65marks in =
Mathematics (90)
60
(ii)
50
David secure least marks in =
English (65).
40
5. 30
do same as above Q. No. 3, 4.
C.C.E. DRILL - 1
20
Tick () the correct answer :
10
(a) pictograph (b)
(c) raw (d) equal
x
o
Math Sci S.St Hindi Eng Sans
(e)
Subject
C.C.E. DRILL - 2
1. Fill in the blanks :
(a)
(b) data (c) pictograph (d)
pictograph (e) scale
2. Write T for true of F for false for the
following :
(a) F (b) F (c) F (d) T (e) F
HOTS QUESTIONS
1.
y
40
40
35
3. 35
10
5
o
15
Thursday
15
20
Wednesday
Monday
20
Friday
25
25
Tuesday
terms
30
y
900
y
700
Population
2. 600
Let 1 object = 600
17
than
85
object
=
500
3. 400
500
4.
400
speed per min
800
800
x
days
950
450
355
300
300
200
100
400
350
200
o
250
150
100
o
Nitin
NiteeshManeesh Rishi
A
B
C
Villages
200
Varmit
57
D
E
x
8. do yourself.
4 line
1 line
C.C.E DRILL - 1
1. Fill in the blanks :
(a) symmetry (b) line of symmetry (c)
Pentagon(d) symmetrical (e) line
not symmetry
2. Write
false for the
noneT for true or F for
3 line
a
following
: b
7
7
(a) T (b) T (c) F (d) T (e) T
c
HOTS
QUESTIONS 6
Scalene
triangle
equilateral
1. Yes, a equilateral triangle
can betriangle
drawn
having 3 lines of symmetry. e.g.
Frequency
Numbers Tally marks
Exercise - 14
3
1
1. (a),
5
2 (b), (c), (d)
2. (a),
3 (d).
6
3
4
3. (a)
(b)
5
5
6
4
7
1
8
2
(c)
(d)
9
4
(e)
(f)
3
4
7
(g)
(h)
6
(i)
2. Priyanka can draw many lines of
symmetry in a circle.
4
7
3
(j)
a
a
a
4. All the letter of English alphabet
having one line symmetry : A, B, C D,
E, M, T, U, V, W, X, Y
5. All the English alphabet having two
lines of symmetry : H, I
6. All the English alphabet having two lies
of symmetry : F, G, J, L, N, P, Q, R, S, Z
7. (a)
(b)
3. English alphabet having one line of
symmetry are A,B, C, D, E, M, T, U, V,
W, X, Y.
No, she is not correct. She has not
found all. The remaining is above.
Exercise - 15A
1. radius = 3.6cm
circle
2. radius = 2cm
(c)
(e)
(d)
(f)
1 line
a
Many line
3.
o
3.6cm
A
r
a
Circle
isoceles triangle
circle
58
4.5cm on the ruler.
Now, we take the compass in the same
position and put its metal side on A and
withAhelp of pencil mark on B.
l
3.6cm B
So, the segment AB has the required
length of 4.5cm.
5cm
D
o
2 A
C
B
4cm
3cm
4.
Circle
radius =
5.
diameter 60
cm = 3cm
=
2
2
B
3cm O 3cm
A
d
D
(b)-(c) : do same as above part 2(a).
l
3. do yourself.
A
4. (a) Let PQ and QR be given line
segments. we draw any line l and mark
a point P to find a point Q on l such that
PQ + QR = PR
Open the compass and put its metal
point on A and pencil on B.
(a) No, A is not exterior of the circle
r
(b) Yes, B is on the
C circle.B
3cm
6. Triangle is obtained.
A
A
r
B
7. Square is obtained, when we join the
two diameter of a circle which we
perpendicular to each other.
8. Yes, PQ RS and RS PQ
ROQ = ROP = 90°
POS = QOS = 90°
C
90º
A
B
o
D - 15B
Exercise
1. (a) 3.6cm; First draw a line and take
3.6cm fro it.
R
(b)-(e) : do same as abvoe
2. (a) 4.5cm; Draw any
90º line l and mark a
point A on Pit. Now,
place the metal
Q
o
point of the compass at zero mark of
the ruler and the pencil point indicating
S
A
B
Now take the compass in same position
and put the metal point on P and with
pencil mark a small point Q on line l.
Now again put the metal point on Q
and the pencil point on R.QTransfer the
compass on l in the same position and
put the Pmetal point on Q and make a
R
Q by pencil point at R.
small cut
he, PR = PQ + QR is required segment.
l
P same as above.
R
Q
(b) do
5. given PQ = 4.5cm, RS = 3.6cm
(a) PQ + RS : Draw any line l and mark
a point P on it. Now, place the metal
point of the compass at zero mark of
the ruler and the pencil point indicating
4.2cm on the ruler.
Now, we take the compass in the same
position and put its metal side on p and
with the help of pencil mark on Q.
Again take the compass in the same
position and put its metal side on Q and
with the help of pencil mark on R.
59
Again, take the compass in the same
position and put its metal side on R and
with the help of pencil mark on S.
So, the segment PQ and RS has the
required length of 4.2cm and 3.6cm
respectively.
square.
(e) PR is required perpendicular i.e, PR
OQ.
(b)-(f) : do as above manner.
6. Name of rays : MP, MN, MO, MQ,
NO, NQ, NM, NP, OQ, ON, OM, OP.
Q
Q
R
(a) No, both start with the same initial
point M.
(b) Yes, both rays start from different
l
initial
and O respectively.
P point
R 3.6cm S
4.2cm N Q
(c) Yes, but it not rays because it have no
end points.
7. Draw
any
line
l
and
mark
a point
A
on
it. Now, place the metal point of the
compass at zero mark of the ruler and
Q
P
Mpoint Nindicating
O
the pencil
2.5cm
on
the ruler.
Now, we take the compass in the
position and put its metal side on A and
with the help of pencil mark on B.
So, the segment AB has the required
length of 2.5cm.
Similarly, putting the metal side of
compass on B and with the help of
pencil mark on C. Proceed similar
manner we get the segment CD.
l
2.5cm=C3
2.5cm
D
A 2.5cm
thus, AD
= 3 B AB
2.5cm
=
7.5cm
Exercise - 15C
1. (A) using ruler and set squares :
steps :
(a) Draw a line OQ and give the point P
on it.
(b) Place the ruler such that one of its
edges along OQ.
(c) Holding the ruler firmly place a set
square along the edge of the ruler such
that the right angled corner coincides
with point P.
(d) Draw PR along the edges of the set
S
OP
l S
P (O)
l
(B) Using ruler and compass :
steps :
(a) Draw a line OQ and give the point P
on it.
(b) Draw a semi-circle of any radius
with center P on OQ
at at A and B.
R
(c) with point A, any radius more than
PA draw an arc.
P
(d) with point
B and the Qsame radius
O
draw another are that cut the are
previously draw at R.
Join PR, then we find PR OQ.
2. (A) Using set square and ruler
steps :
(a) Using a line OQ and give the Point
Pout side it.
R
(b) Place a set square on OQ Firmly.
(c) Place a ruler along the longest side
of the set square opposite the right
angle.
(d) Side theAset square
along the ruler
B Q
P
O
till the point P coincides with the other
arm of the set square.
(e) Draw a line PR along the edge
meeting OQ at R.
60
(f) PR is the required perpendicular to
OQ i.e, PR OQ.
5.2 as Q. No. 3 above.
4. do same
2.6cm Pmeans bisector of
5. =
axis 2of =symmetry
line segment AB i.e, axis of symmetry
A
P
m
(B) Using ruler and compass :
steps : (a) Draw a line OQ and give the
point P outside it.
(b) with P as a center, draw an arc AB
on OQ.
(c) with Point A as center and radius
greater than half of AB draw on arc.
(d) with point B as center and the same
radius draw the other arc that cuts the
previously drawn arc at R.
(e) Join PR cutting OQ at S.
Then, PS is the required Perpendicular
on OQ.
B
O
now, do same above
Q. No. 3, 4.
6. (a) Draw a line segment AB = 8cm
(b) with A as a center and taking radius
2cm, mark on lineQsegment AB three
times as P, Q and R.
P
P
m
m
ABAP 8cm
Thus,
= PQ = QR = RB = 2cm
= 2 = 2 = 4cm
7. do same
as Q. No. 3 and 4 above.
8. do same as Q. No. 3 and 4 above.
9. Steps : (a) Draw a line m and Q is a
point on it. take T as a point outside it.
(b) Join t Q.
(c) Draw angle O T Q such that T Q
8cm
B
= O T Q.
B
A
2cm
P
2cm
2cmboth
R sides
(d)2cm
Produce
O TQ on
and
make a line.
thus, line O T passes through T and O T
|| AB.
P
3. (a) Draw a line segment AB = 5.2cm.
(b) with A as a center and radius more
than half of AB, draw arcs
each side of
B
A
S
AB.
Q
O
(c) Now with center B and same radius
draw the arcs each side of AB that cuts
the arcs at Q and P, Previously drawn.
R
(d) Join OP cutting
AB at O.
Thus O is the midpoint of AB and OB
10. do same manner as Q. No. 9 above.
11. Steps : (a) draw a line segment OP =
3cm
(b) with O as center
andTtaking radius
O
3cm, draw circle with the help of
compass as given below.
A
61
Q
B
(c) take two points R and S on the
circumference of the circle and join
them.
(d) Now with R as a center and radius
more than half of RS, draw arcs each
O
P
3cm
side of RS.
M
(e) similarly, with center S and same
radius draw the arcs each side of RS
S
R
that cuts the Qarcs, at M and N,
Previously drawn. N
(f) Join MN cutting RS at Q. Thus MN
is the Perpendicular bisector
(g) on producing MN up side, we see
that QM passes through the center O of
the circle.
12. Do same as above Q. No. 11.
Exercise - 15D
1. To Draw an angle (by using protractor)
: given angle POQ = 80º.
Steps : (a) Draw a ray OP with O as
initial Point (b) Place the mid point of
protractor at O and base line of
protractor along OP.
(c) Mark the point Q against 80º
starting from Oº on the side of P.
(d) Remove protractor and draw OQ.
3. To construct a square of side 5cm.
steps : (a) First of all, Draw a ray of any
length with A as initial point.
(b) Place the metal point of the
compass at O mark of the ruler and the
pencil point indicating 5cm on the
ruler, mark on this ray point B.
(c) Now taking A & B as center
respectively draw two arcs of radius
5cm on the above of AB.
(d) Now, mark them arc C and D.
(e) Join CB, DC, DA.
Thus, we get ABCD as required square.
4. steps : (a) Draw ray OA.
D
5cm
A
2. (a) 60º
steps : (a) Draw a ray OA.
(b) Draw on arc CD that cuts OA at D
Q
with center O and any
suitable radius.
(c) with D as a center and OD as radius
draw on arc cut at C.
(d) Join OC and Produce to OB, then
AOB = 60º
80º
(d)-(i) : do as Osimilar manner Pas above
B
part (a).
C
60º
O
D
A
5cm
C
5cm
5cm
B
(b) with center O and any suitable
radius, draw on arc cutting OA at B.
F
(c) Now with
B as center and same
radius drawE on arc cutting at C, then
with C and same radius cut the arc at D.
D
(d) Take radius more
than
half CD and
C
P
draw an arc with c as center and again
with D as a center.
(e) Join OOE and Produce
B
A to F, then
AOF = 90º
(f) Draw the bisector of AOF, then
AOP = 45º
5. do yourself as above.
6. do yourself by measuring angles of
both figures with the help of protractor.
then constructs required angles.
We will find (a) ABC = 60º
(b) ABC = 120º
62
C.C.E DRILL - 1
Tick () the correct answer :
(a) 60º (b) 2 (c) 3cm (d) 45º (e)
C.C.E DRILL - 2
1. Fill in the blanks :
(a) Pencil, Compass (b) Perpendicular
(c) 90º (d) 45º (e) 180º
2. Write T for true or F for false for the
following :
(a) F (b) T (c) T (d) F (e) F
HOTS QUESTIONS 11°
1. No, he can’t draw the circle22because
2
radius of the circle is more than the
dimensions of the paper.
2. Yes, he is correct. He can make an
angle of 90º because while making a
perpendicular bisector of the given line
segment we always get an angle of 90º
as given in the figure.
Q
62º
O
A.1.
P
Q. 2, 3, 4 do same as Q. 1
B. given 2x – 4 = 8
L.H.S = 2x – 4 (Put x = 6)
= 2 6 – 4 = 12 – 4 = 8
thus, putting x = 6 in L.H.S of the given
equation we get R.H.S of the equation.
so x = 6 is the solution of the equation
2x – 4 = 8
46of a girl, her younger sister
C. Let 92
the ages
and10
her=brother
are x, y, z respectively
5
Then, x – y = 4 ..........(i)
y = z + 4 ..........(ii)
y + z = 16 ..........(iii)
on solving these equations, we get x =
14, y = 10, z = 6
Hence, the age of a girl = x = 14years
The age of her sister = y = 10years
and the age of her brother = z = 6years
D. 1. required ratio
3. steps : (a) Draw a ray OP with O as
initial point.
(b) Place the mid point of protractor at
O and base line of protractor along OP.
90º point
90º Q against 62º
(c) Mark the
starting from Oº on the side ofl P.
(d) Remove Protractor and draw OQ.
Formative Assessment - IV
A. Tick () the correct option :
1. 7 side 2. four times 3. raw 4.
pictograph 5. 45º 6. 2
B. Fill in the blanks :
1. 90º 2. 180º 3. a symmetrical 4.
symmetry 5. pictograph 6. m2
C. 1. T 2. T 3. F 4. F 5. T 6. T 7. T 8. F
Summative Assessment - II
2, 3, 4 do as above.
E. Let fourth term be x. then 117 : 13 : : 81 :
x
117 x = 13 81
Hence, Fourth term is 9.
1 in part 1
F. 1. Perimeter of the given
`40 figure
1 : 20
= ==
=
= 20cm + 5cm + 20cm
+ 5cm
50cm
20
`800
2. P = 30 + 30 + 30+ 30 = 120cm
3. P = 6 + 1 + 3 + 4 + 3 + 2 + 3 + 2 + 3 +
4 + 3 + 1 = 35cm
G. Area of carpet = length breadth =
2
225 20 = 4500m
[... Product of extremes2 =
... Cost of per
squareofmeter
i.e,] 1m =
Product
means
63
x=
13 81
=9
117
`25
Cost of per 4500m2 = 25 4500 =
`112500
H. Area of ABCD = l b
= 45 25 = 1125m2
Area of PQRS = l b
= (45 + 2 + 2) (25 + 2 + 2)
= 49 49 = 1421m2
Area of the path (i.e, shaded
Portion) = Area of PQRS – Area of
ABCD
R
S – 1125 = 296m2
= 1421
2M
C
... Cost ofD gravelling
1m2 path = `4.30
2
29M
Cost2Mof gravelling 296m25
path29= 296
4.30 = `1272.80
I.
A
P
Numbers
J.
14
45M
49M
Tally marks
y
90
90
80
72
75
80
80
70
K. In given figures, the given line is
60 symmetrical
line in (a), (d)
65
50 Triangle
L.
M.
40 steps : (a) Draw a circle of radius 5cm
30 with center as O.
20 (b) take any point B on the
circumference of the circle and join O
10
to B.
x
o Produce
Eng Hindi
Science
(c)
OB
to Asans
i.e,S.Scien
PointMath
A on the
krit
circumference fo the circle.
(d) Now, AB is the first and longest
chord of the circle. This is called
diameter also.
B
Q
(c) Now taking A and B as center and
more than half radius draw two arcs K
and L above and below
chord AB. Join
K
KL. They meet at O. M
(f) takeA twoO point C
and D on the
B longest
Q and join
chordthem. It is
circumference
C
D
chord
chord CD.
(g) Now, taking CN and D as center and
more than half radius
draw two arcs M
L
and N above and below chord CD. join
MN. They meet O and P.
Thus the perpendicular bisector of two
chords meet at the center.
Frequency
5
17
1
19
6
21
4
25
5
27
4
28
1
30
1
64
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