2 TH—Easy Maths—5
Chapter 1: Number and Numeration
Exercise 1.1 Refer answers at the end of book.
Exercise 1.2
3.Greatest 6 digit number= 999999
Greatest 5 digit number= 99999
Total 6 digit numbers = Greatest 6 digit number – Greatest 5 digit number
= 999999 – 99999
= 900000
4. Refer answers at the end of book.
5.-7. Refer answers at the end of book.
Exercise 1.3-1.4
Refer answers at the end of book.
Exercise 1.5
1.–3. Refer answers at the end of book.
4. To write the greatest 7 digit number we select the greater 7 digits from
given digits they are 3, 2, 7, 8, 4, 6 and 5. Now place them in such a
way that the greatest digit occupy the greatest place value i.e., 8765432
is the greatest seven digit number.
Similarly to form smallest seven digit number, we place them in such
a way that the smallest come at the highest place value i.e., 1234567
is the smallest seven digit number.
Similarly for eight digit the greatest numeral is 87654321 and the
smallest numeral is 12345678.
(b) 0, 2, 4, 1, 3, 6, 8, 9.
Greatest seven digits: Seven greater digits are 9, 8, 6, 4, 3, 2 and 1
Hence greatest seven digit number = 9864321 and
Smallest seven digit number is = 1234689
Greatest eight digit number using 0, 2, 4, 1, 3, 6, 8, 9 are 98643210
Smallest eight digit number = 10234689.
Remember that ‘0’ can not be placed at crore’s place as it will become
an seven digit number.
5.–6. Refer answers at the end of book.
7. Smallest 8 digit number = 10000000
Greatest number using digit 3, 8, 6, 5, 1, 2, 0 is 8653210
Difference = 10000000
8653210
1346790
Smallest number using digits 3, 8, 6, 5, 1, 2, 0 is 1023568.
TH—Easy Maths—5
Price : `60.00
3
Difference = 10000000
–1023568
8976432
8. Greatest 8 digit number= 99999999
Greatest 7 digit number = –9999999
90000000
Greatest 8 digit number= 99999999
Smallest 7 digit number = –1000000
98999999
Exercise 1.6
1.–8. Refer answers at the end of book.
Chapter 2: Roman Numerals
Exercise 2.1
1.–5. Refer answers at the end of book.
6. (a)XXXVIII = 38
(b)LXXXVI = 86
XXV = 63
XI = 11
Sum = 63
Sum = 97
63 = LXIII
97 = XCVII
(c)
XC = 90
(d) XXIII = 23
CX = 110
XXVII = 27
Sum = 200
200 = CC
7. Refer answers at the end of book.
Sum = 50
50 = L
Chapter 3: Addition and Subtraction
Exercise 3.2
2. (a) 35,27,644
+ 72,04166
10731810
10731810 – One crore seven
lakh thirty one thousand
eight hundred ten.
4 (b) 1060408
+ 1227391
2287799
2287799 – Twenty two lakh
eighty seven thousand seven
hundred ninety nine.
TH—Easy Maths—5
(c) 25842
0741287
+ 4499087
5266216
5266216 – Fifty two lakh
sixty six thousand two
hundred sixteen.
(d) 1924477
39042
+ 5234659
7198178
7198178 – Seventy one lakh
ninety eight thousand one
hundred seventy eight.
Exercise 3.3
1.–2. Refer answers at the end of book.
3.(a) 49531956
+ 99813335
149345291
(b) 24365741
+ 55848511
80214252
149345291 – Fourteen crore
ninety three lakh fourty
five thousand two hundred
ninety one.
80214252 – Eight crore two
lakh fourteen thousand two
hundred fifty two.
8763566
(d) 87087771
2136009
01178518
+ 10177835
+ 59900866
21077410
148167155
(c)
21077410 – Two crore
ten lakh seventy seven
thousand four hundred ten.
148167155 – Fourteen crore
eighty one lakh sixty seven
thousand one hundred fifty
five.
Exercise 3.4
1. (a) Production of toys in 2009–2010 = 5369831
Production of toys in 2010–2011 = + 8567842
Total 13937673
(b) Production of toys in 2011–2012 = 19678395
Production of toys in 2012–2013 = + 23698743
43377138
Total
Total production of toys in 2009–2013 13937673
= Answer of (a) + Answers of (b) = + 43377138
57314811
2. Cost of house = `5632154
Cost of office = + `8362158
`13994312
Total
Total money spent by Mohan Lal is `13994312.
TH—Easy Maths—5
5
3. Books sold in 1st week = 42621273
Books sold in 2nd week = + 85321540
Total books sold
127942813
4.
Total books sold at the end of two weeks is 127942813.
7456231
Number of copies sold in January =
7714098
Number of copies sold in February =
+ 27849125
Number of copies sold in March =
43019454
Total copies sold
5.
Total copies sold in three months is 43019454.
Money allocated in 2010
=
`38741981
Money allocated in 2011
=
`33141928
Money allocated in 2012
= + `37425627
`109309536
Total money allocated
Total money allocated is `109309536.
Exercise 3.5
1. (a) Actual sum
Estimated sum
269
270
1792
1790
+ 731
+ 730
2792
2790
Difference between the actual sum and estimated sum is 2792 –
2790 = 2.
(b) Actual sum
Estimated sum
444
440
241
210
+ 879
+ 880
1537
1530
Difference between the actual sum and estimated sum is 1537 –
1530 = 7.
(c) Actual sum
Estimated sum
342
340
879
880
+ 1895
+ 1900
3116
3120
6 Difference between the actual sum and estimated sum is 3120 –
3116 = 4.
TH—Easy Maths—5
(d) Actual sum
343
486
+ 7284
8113
Estimated sum
340
490
+ 7280
8110
Difference between the actual sum and estimated sum is 8113 –
8110 = 3.
2. (a) Actual sum
Estimated sum
4865
4900
349
300
+ 7426
+ 7400
12640
12600
Difference between the actual sum and estimated sum is 12640 –
12600 = 40.
(b) Actual sum
Estimated sum
14600
14562
829
800
+ 4862
+ 4900
20253
20300
Difference between the actual sum and estimated sum is 20300 –
20253 = 47.
(c) Actual sum
Estimated sum
8462
8500
1491
1500
+ 2942
+ 2900
12895
12900
Difference between the actual sum and estimated sum is 12900 –
12895 = 5.
(d) Actual sum
Estimated sum
3421
3400
6882
6900
+ 3149
+ 3100
13452
13400
Difference between the actual sum and estimated sum is 13452 –
13400 = 52.
3. (a) Actual sum
Estimated sum
6348
6000
1720
2000
+ 8592
+ 9000
16660
17000
Difference between the actual sum and estimated sum is 17000 –
16660 = 340.
TH—Easy Maths—5
7
(b) Actual sum
3948
2736
+ 5481
12165
(c) Actual sum
3333
4811
+ 36245
44389
(d) Actual sum
74926
83440
+ 26609
184975
Difference between the actual sum and estimated sum is 12165–
12000 = 165.
Estimated sum
3000
5000
+ 36000
44000
Difference between the actual sum and estimated sum is 44389–
44000 = 389.
Estimated sum
75000
83000
+ 27000
185000
Difference between the actual sum and estimated sum is 185000–
184975 = 25.
4. (a) Actual cost
`378
`1549
`352
+ `772
`3051
(b) Actual cost sum
`378
`1549
`352
+ `772
`3051
8 Estimated sum
4000
3000
+ 5000
12000
Estimated cost
`380
`1550
`350
+ `770
`3050
Difference between actual cost and estimated cost is `(3051–3050)
= `1.
Estimation is accurate.
Estimated cost sum
`400
`1500
`400
+ `800
`3100
Difference between estimated cost sum and actual cost sum is
`(3100–3051) = `49.
Estimation is accurate.
TH—Easy Maths—5
Exercise 3.6
1. Refer to the answer given
2. Refer to the answer given
at the end of the book. at the end of the book.
3. (a) One of the number 97262987
Difference
– 88066143
Other number
9196844
Other number is 9196844.
78432178
(b)Sum
– 58543218
One number
19888960
19888960 should be added to 58543218 to get 78432178.
(c) Greatest number of 8 digits 99999999
– 63547458
Number
36452541
36452541 should be added to 63547458 to get greatest 8 digit
number.
(d) Money earned
`5362187
Money spent
– `3267898
Money saved
`2094289
He saved `2094289.
`9876540
(e) Greatest number
– `4056789
Smallest number
`5819751
Difference between greatest and smallest number firmed using
the digits 8, 5, 7, 0, 6, 4, 9 is 5819751.
(f) Production after 2 years
`8632167
Production initially
– `5346247
`3285920
3285920 was increase in the production.
Exercise 3.7 Refer answers at the end of book.
TH—Easy Maths—5
9
Exercise 3.8
1. First find the sum of 1362521 2. First find the sum of 25678232
and 2658324.
and 33784701.
1362521
+ 2658324
4020845
Now find the sum of 4321624 and 5672158.
4321624
+ 5672158
9993782
Now subtract 4020845 from
9993782.
25678232
+ 33784701
59462933
Now subtract 59462933 from
79836260.
79836260
– 59462933
20373327
9993782
– 4020845
5972937
3. First find the sum of 74583267 Now subtract 47139156 from
and 2874032.
77457299
77457299
74583267
– 47139156
+ 2874032
30318143
77457299
Ans:
Sum of 74583267 and
Now find the sum of
2874032
is greater than the
43874280 and 3264876.
sum of 43874280 and 3264876
43874280
by 30318143.
+ 3264876
47139156
4. First find the sum of 5672672 5. First find the sum of the children
and 3821426
less than 10 years old and their
find the number of people in
5672672
the age group of 10–70 years.
+ 3821426
9494098
2173269
Now subtract 9494098 from
+ 2378621
10000000.
4551890
10000000
Now subtract 4551890 from
5834678.
– 9494098
505902
5834678
– 4551890
1282788
There are 1282788 whose age
is more than 70 years.
10 TH—Easy Maths—5
6. First add the number of7. First add the amount allocated
students who voted for winner
in the first year and second
and who voted for loser.
year.
`20844000
1st year
9447932
+
`11582400
2nd
year
+ 7949268
`32426400
Total
17397200
Now subtract this from the total
Now subtract this sum from
annual allocation.
the total number of student
`37814000
voters.
– `32426400
17849600
`5387600
– 17397200
452400
The remaining amount due to
be allocated in the third year is
 452400 students did not
`5387600.
vote in the election.
8. To find the number of magazine sold in July, subtract 4550000 from
21739298.
21739298
– 4550000
17189298
17189298 were sold in July. To find the number of magazine sold in
August add 7524000 to 17189298.
17189298
+ 7524000
24713298
In the month of August 24713298.
9. Refer answers at the end of book.
Chapter 4: Multiplication
Exercise 4.1
1.(a) 5621
×7
39347
(c) 6254
× 32
12508
18762 ×
200128
TH—Easy Maths—5
(b) 8625
×9
77625
(d) 5678
× 63
17034
34068 ×
357714
11
(f) 5544
8765
× 42
× 33
17530
16632
35060 ×
16632 ×
368130
182952
(g)
(h)
6789
7878
× 761
× 820
6789
157560
40734 ×
63024 × ×
47523 × ×
6459960
5166429
2. (a) Cost of 1 toy car = `4621
Cost of 7 toy cars = `4621 × 7 = `32347.
Cost of 7 toy cars = `32347.
(b) Cost of 1 book = `46
(c) Cost of 1 bicycle = `1562
Cost of 6214 books
Cost of 322 bicycles
= `46 × 6214
= `1562 × 322
1562
6214
× 322
× 46
3124
37284
3124×
24856×
4686××
285844
502964
Thus cost of 6214 books is
Thus cost of 322 bicycles is
`285844.
`502964.
(d) To find the bill paid by Mr. Mohan add answers obtained in (i),
(ii) and (iii).
(i) Cost of 1 dress = `536
Cost of 8 dresses = `536 × 8 = `4288
(ii)Cost of 1 book = `326 (iii) Cost of 1 pizza = `232
Cost of 63 books =
Cost of 83 pizzas = `232 × 83
`326 × 63
232
× 83
326
696
× 63
1856×
978
`19256
1956×
`20538
(e)
Total bill =
He paid `44082 in total.
12 4288
20538
+ 19256
44082
TH—Easy Maths—5
(e)Cost of one chair =
Cost of 764 chairs =
`1486
× 764
5944
8916×
10402××
1135304
Cost of 764 such chairs is `1135304.
Exercise 4.2 Refer answers at the end of book.
Exercise 4.3
1.(a)
55555 × 626
333330
111110×
333330××
34777430
(d) 470896 × 813
1412688
470896×
3767168××
382838448
2.(a) 26732 × 562
53464
160392×
133660××
15023384
(d)
254002 × 246
1524012
1016008×
508004××
62484492
(b)
60121 × 585
300605
480968×
300605××
35170785
(e) 764897 × 252
1529794
3824485×
1529794××
192754044
(b) 400634 (c)
18357 × 369
165213
110142×
55071××
6773733
(f) 385462 × 518
3083696
385462×
1927310××
199669316
(c)
83945 × 323
1201902
801268×
1201902××
129404782
× 151
83945
419725×
83945××
12675695
(e) 605035
× 86
3630210
4840280×
52033010
3.a.-b. Refer answers at the end of book.
Exercise 4.4
1. Product of 3254 and 74
= 3254 × 74 = 240796
Greatest number formed by the
digits 4, 7, 5, 3, 6, 8 is 876543.
TH—Easy Maths—5
Now
876543
– 240796
635747
13
2.
3.
Smallest three digit number using 5, 0, 7 is 507.
Greatest five digit number 99999
Product of 507 × 99999 = 50699493.
Number of toys in one
326406
day = 326406
× 24
Number of toys in 24
1305624
days = 326406 × 24
652812×
7833744
7833744 toys are produced in 24 days.
4. Number of soap cakes
5. Cost of 1 shirt = `156
in a carton = 450
Cost of 3624 shirts = `3624 × 156
Number of cartons = 1585 3624
To find the number of
× 156
soap cakes in 1585 cartons,
21744
multiply = 1585 × 450
18120×
3624××
1585
565344
× 450
Cost
of 1 trouser = `867
0000
Cost of 4567 trouser = `867× 4567
7925×
4567
6340××
× 867
713250
31969
There are 713250 soap
27402×
cakes in 450 cartons.
36536××
3959589
Total money he made =
`565344 + `3959589 = `4524933.
6. Cost of 175 bags of urea
= `356 × 175
Cost of 214 bags of phosphorus= `415 × 214
= `88810
Total amount spent
= `(62300 + 88810)
= `151110
7. Weight of 1 bag
= `62300
= 150 kg
Number of bags a truck can carry = 883
Weight a truck can carry
= 150 × 883 kg
= 132450 kg
Weight carried by 65 trucks
= 132450 × 65 kg
= 8609250 kg
14 TH—Easy Maths—5
8. Number of bags
= 214
Weight of 1 bag
= 95 kg
Total weight of 214 bags = 214 × 95 kg
= 20330 kg
Cost of 1 kg rice
= `38
Cost of 20330 kg rice
= 38 × 20330 = `772540
9. One day earning
= `957
One month or 26 days earning = `957 × 26 = `24882
One year (or 12 months) earning = `24882 × 12 = `298584
Exercise 4.6
1. Savings of 8 months = `48000
Savings of 1 months = `48000 ÷ 8 = `6000
Mohan saves `6000 in a month.
2. Number of pens fitted in a box = 250
Number of pens fitted in 40 boxes = 250 × 40 = 10,000
10,000 pens can be fitted in 40 such boxes.
3. Money spent in 12 months (or 1 year) = `10800
`10800
Money spent in 1 month =
= `900
12
Money spent in 6 months = `900 × 6 = `5400
Rohit spend `5400 in 6 months for maintaining the garden.
4. Cost of 16 books = `640
Cost of 1 book = `640 ÷ 16 = `40
Cost of 9 books = `40 × 9 = `360
Cost of 9 books is `360.
5. Distance covered in 8 minutes = 480 m.
Distance covered in 1 minute = (480 ÷ 8) m = 60 m.
Distance covered in 20 minutes = 60 × 20 = 1200 m.
Athlete will cover 1200 m in 20 minutes.
6. 1 dozen = 12
Cost of 12 bananas = `48
Cost of 1 banana = `48 ÷ 12 = `4
Cost of 1 banana is `4.
7. 1 dozen = 12
Cost of 12 oranges = `24
Cost of 1 orange = `24 ÷ 12 = `2
Cost of 15 oranges = `2 × 15 = `30
8. Toys produced in 30 days = 82440
Toys produced in 1 day = 82440 ÷ 30 = 2748
Toys produced in 5 days = 2748 × 5 = 13740
13740 toys are produced in 5 days.
TH—Easy Maths—5
15
Chapter 5: Division
Exercise 5.1
1.(a) 5
56
544 9
430
86 (b)
– 504
– 40
40
30
×
– 30
×
Q = 86 R = 0
Q = 9 R = 40
(c) 164
752 4 (d)
9
2016
224
656
– 18
96
21
– 18
36
– 36
×
Q = 4 R = 96
Q = 224 R = 0
(e) 6075 ÷ 81
(f) 8928 ÷ 93
81
6075
75 93
8928
96
– 567
– 837
405
558
– 405
– 558
×
×
Q = 75 R = 0
Q = 96 R = 0
(g) 3780 ÷ 254
(f) 2016 ÷ 329
254 3780
329 2016
6
14 1974
– 254
42
1240
×
– 1016
224
Q = 14 R = 224
Q = 6 R = 42
2. (a) 12 objects = 1 dozen
3912 objects = 3912 ÷ 12 = 326
There are 326 dozen if number of objects is 3912.
(b) Number of trees
= 3256
Number of gardens
= 37
Number of trees in each garden= 3256 ÷ 37
= 88
Number of trees in each garden 88.
(c) Number to be multiplied
= 4032 ÷ 72
= 56
 56 should be multiplied by 72 to give 4032.
16 TH—Easy Maths—5
(d) Cost of 1 book= `26
Number of books that can be bought for `8424
= 8424 ÷ 26 = 324.
 324 books can be bought for `8424.
(e) Cost of 154 kg rice= `5236
Cost of 1 kg rice = `5236 ÷ 154
= `34.
Cost of 1 kg rice is `34.
Exercise 5.2
1. Refer answers at the end of book. 2. Refer answers at the end of book.
Exercise 5.3
1. (a) 67892 ÷ 74
74
67892 917 – 666
129
– 74
552
– 518
34
Q = 917 R = 34
(b) 54386 ÷ 86
86
54386 632
– 516
278
– 258
206
– 172
34
Q = 632 R = 34
(c) 286752 ÷ 51
51
286752 5622
– 255
317
– 306
115
– 102
132
– 102
30
Verification
= Quotient × Divisor + Remainder
= 917 × 74 + 34
= 67892 = Dividend
Verification
= Quotient × Divisor + Remainder
= 632 × 86 + 34 = 54352 + 34
= 54386 = Dividend
Verification
= Divisor × Quotient + Remainder
= 51 × 5622 + 30 = 286722 + 30
= 286752
Q = 5622 R = 30
TH—Easy Maths—5
17
(d) 458432 ÷ 68
68
458432 6741
– 408
504
– 476
283
– 272
112
– 68
44
(e) 726908 ÷ 58
726908
12532
58
– 58
146
– 116
309
– 290
190
– 174
168
– 116
52
Q = 12532 R = 52
Verification
= Divisor × Quotient + Remainder
= 68 × 6741 + 44 = 458388 + 44
= 458432
Q = 6741 R = 44
(f) 2947349 ÷ 36
36
2947349
– 288
67
– 36
313
– 288
254
– 252
29
18 8187
Verification
= Quotient × Divisor + Remainder
= 12532 × 58 + 52
= 726908
= Dividend
Verification
= Quotient × Divisor + Remainder
= 81870 × 36 + 29
= 2947320 + 29
= 2947349 = Dividend
Q = 81870 R = 29
TH—Easy Maths—5
(g) 64836921 ÷ 59
64836921
59
– 59
583
– 531
526
– 472
549
– 531
182
– 177
51
1098930
Q = 1098930 R = 51
(h) 1075602 ÷ 84
84
1075602
12804
– 84
235
– 168
676
– 672
402
– 336
66
Verification
= Quotient × Divisor + Remainder
= 1098930 × 59 + 51
= 64836870 + 51
= 64836921 = Dividend
Verification
= Divisor × Quotient + Remainder
= 84 × 12804 + 66
= 1075536 + 66
= 1075602 = Dividend
Q = 12804 R = 66
2. (a) Dividend = Divisor × Quotient + Remainder
= 47 × 321 + 13
= 15087 + 13
= 15100
 Dividend = 15100
(b) Quotient = (Dividend – Remainder) ÷ Divisor
= (637428 – 12) ÷ 24
= 637416 ÷ 24
= 26559
 Quotient = 26559
(c) Dividend = Quotient × Divisor + Remainder
= 52643 × 38 + 58
= 2000434 + 58
= 2000492
 Dividend = 2000492
TH—Easy Maths—5
19
(d) Dividend = 789615
Divisor = 26
26
789615
30369 – 78
96
– 78
181
– 156
255
– 234
21
Q = 30369 R = 21
(f) Dividend = 8765419
Divisor = 69
69
8765419
127035
– 69
186
– 138
485
– 483
241
– 207
349
– 345
4
Q = 127035 R = 4
Exercise 5.4
1. (a) 86732 ÷ 371
371
86732
– 742
1253
– 1113
1402
– 1113
289
Q = 233 R = 289
20 233
(e) Dividend = 56783
Divisor = 62
62
56783
915
– 558
98
– 62
363
– 310
53
Q = 915 R = 53
Checking
Dividend = Quotient × Divisor +
Remainder
= 233 × 371 + 289
= 86443 + 289
= 86732
TH—Easy Maths—5
(b) 96753 ÷ 632
632
96753
– 632
3355
– 3160
1953
– 1896
57
Q = 153 R = 57
(c) 876543 ÷ 423
876543
423
– 846
3054
– 2961
0933
– 846
87
Q = 2072 R = 87
(d) 554433 ÷ 675
554433
675
– 5400
01443
– 1350
933
– 675
258
Q = 821 R = 258
153
Checking
Dividend = Quotient × Divisor +
Remainder
= 153 × 632 + 57
= 96696 + 57
= 96753
2072 Checking
Dividend = Quotient × Divisor +
Remainder
= 2072 × 423 + 87
= 876456 + 87
= 876543
821
(e) 5028672 ÷ 252
252 5028672
19955
– 252
2508
– 2268
2406
– 2268
1387
– 1260
1272
– 1260
12
Q = 19955 R = 12
TH—Easy Maths—5
Checking
Dividend = Divisor × Quotient +
Remainder
= 252 × 19955 + 12
= 5028660 + 12
= 5028672
21
(f) 4167620 ÷ 515
(g) 23021405 ÷ 148
Checking
Dividend = Divisor × Quotient +
Remainder
= 515 × 8092 + 240
= 4167380 + 240
= 4167620
515
4167620
8092
– 4120
4762
– 4635
1270
– 1030
240
Q = 8092 R = 240
148
23021405
– 148
822
– 740
821
– 740
814
– 740
740
– 740
5
155550
Verification
Dividend= Divisor × Quotient +
Remainder
23021405 = 148 × 155550 + 5
= 23021400 + 5
= 23021405
Q = 155550 R = 5
(h) 35607520 ÷ 227
227
35607520
– 227
1290
– 1135
1557
– 1362
1955
– 1816
1392
– 1362
300
– 227
73
156861
 Dividend = Divisor ×
Quotient +
Remainder
= 227 × 156861 + 73
= 35607447 + 73
= 35607520
Q = 156861 R = 73
2. (a) Dividend = Divisor × Quotient + Remainder
= 321 × 1234 + 46
= 396114 + 46
= 396160
 Dividend = 396160
22 TH—Easy Maths—5
(b) Dividend = Quotient × Divisor + Remainder
425
5632471
13252
– 425
1382
– 1275
1074
– 850
2247
– 2125
1221
– 850
371
Q = 13252 R = 371
(c) Dividend = Divisor × Quotient + Remainder
= (Dividend – Remainder) ÷ Quotient = Divisor
= (4823597 – 893) ÷ 5008 = Divisor
= 4822704 ÷ 5008 = Divisor
963 = Divisor
(d) Dividend – Divisor × Quotient = Remainder
= 52156789 – 235 × 221943 = Remainder
= 52156789 – 52156605 = Remainder
= 184 = Remainder
(e) Dividend = Quotient × Divisor + Remainder
= 58841 × 835 + 530
= 49132235 + 530
= 49132765
= Dividend
Yes, the given sum is correct.
Exercise 5.5
1. Divisor = 435 Quotient = 5643 Remainder = 46
 Dividend= Divisor × Quotient + Remainder
= 435 × 5643 + 46
= 2454705 + 46
= 2454751
 The number was 2454751.
TH—Easy Maths—5
23
2. Apples in one carton = 563
Cartons required to pack 18241275 apples
= 18241275 ÷ 563
563
18241275
32400
– 1689
1351
– 1126
2252
– 2252
75
 32400 cartons are required to pack 18241275 apples and 75 apples
will be left.
3. Donation by 782 person = `3406392
Donation by 1 person = `3406392 ÷ 782
= `4356
Donation by 987 person = `4356 × 987
= `4299372
 987 person will donate `4299372 in a month for CRY foundation.
4. To find the number that must be subtracted from 186039 so that the
number is exactly divisible by 328 we need to find the remainder
when 186039 is divided by 328.
328
186039
567
– 1640
2203
– 1968
2359
– 2296
63
 63 must be subtracted from 186039 to make it exactly divisible
by 328.
5. Greatest 6 digit number = 999999
Greatest 3 digit number using 7, 8 and 9 = 987
Now divide 999999 by 987 and subtract the remainder so obtained
from 999999 to get the largest six digit number divisible by 987.
987
999999
1013
– 987
1299
– 987
3129
– 2961
168
 999999 – 168 = 999831
 999831 is the largest six digit number exactly divisible by 987.
24 TH—Easy Maths—5
6. 3427895 + 8261439 = 11689334
11689334
45483
257
– 1028
1409
– 1285
1243
– 1028
2153
– 2056
974
– 771
203
Sum of the given number is not exactly divisible by 257 and remainder
is 203.
7. Number of students = 856
Total amount collected= 342400 paise
= `3424 (Divide 342400 by 100)
Amount donated by each student= `3424 ÷ 856
= `4
 Each student donated `4.
8. Selling price or cost of 275 bicycles = `479875
Selling cost of 1 bicycle
= `479875 ÷ 275
= `1745
Selling cost of 135 bicycles
= `1745 × 135
= `235575
9. Total students
= 67 + 73 = 140
Total mess fees paid
= `1571080
Mess fees paid by one student= `1571080 ÷ 140
= `11222
 Each student paid `11222.
10. Cost of 364 microchips
= `1257984
Cost of 1 microchip
= `1257984 ÷ 364
= `3456
Cost of 136 defective microchips = `3456 × 136
= `470016
Cost of remaining microchips = `1257984 – `470016
= `787968
TH—Easy Maths—5
25
11. Greatest 7 digit number = 9999999
Greatest number using 3, 7, 4
= 743
9999999
13458
743
– 743
2569
– 2229
3409
– 2972
4379
– 3715
6649
– 5944
705
Divisor = 743
Quotient = 13458
Dividend = 9999999
Remainder = 705
Dividend = Divisor × Quotient + Remainder
= 743 × 13458 + 705
12. Total consumption = 47963324
Total days= August + September + October + November
= 31 + 30 + 31 + 30
= 122
Consumption per day = 47963324 ÷ 122
= 393142 kg
 393142 kg vegetable is consumed per day.
13. To let students try by themselves.
Exercise 5.6
1. (a) 396 is rounded to 400
(b)
44 is rounded to 40
 400 ÷ 40 = 10
(c) 719 is rounded to 720
(d)
84 is rounded to 80
 720 ÷ 80 = 9
(e) 2877 is rounded to 2880
(f)
77 is rounded to 80
 2880 ÷ 80 = 36
(g) 4863 is rounded to 4860
(d)
63 is rounded to 60
 4860 ÷ 60 = 81
26 184 is rounded to 180
63 is rounded to 60
 180 ÷ 60 = 3
149 is rounded to 150
45 is rounded to 50
 150 ÷ 50 = 3
2974 is rounded to 2970
31 is rounded to 30
 2970 ÷ 30 = 99
5849 is rounded to 5850
49 is rounded to 50
 5850 ÷ 50 = 117
TH—Easy Maths—5
Chapter 6: Multiples and Factors
Exercise 6.1
1.–4. Refer answers at the end of book.
5.(a) 9
36 4 (b)
13
216 16
– 36
– 13
0
86
– 78
6
Since remainder is zero Remainder is 6 so 13 is not a
hence 9 is a factor of 36. factor of 216.
(c) 25
125 5
– 125
×
 25 is a factor of 125.
6. (a)24
(b)36
Factors of 24 are 1, 2, 3, 4, Factors of 36 are 1, 2, 3, 4,
6, 8, 12 and 24
6, 9, 12, 18 and 36
24 = 1 × 24
2 × 12
3×8
4×6
6×4
(c)108
Factors of 108 are 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54 and
108
108 = 1 × 108
2 × 54
3 × 36
4 × 27
6 × 18
9 × 12
36 = 1 × 36
2 × 18
3 × 12
4×9
6×6
(d)225
Factors of 225 are 1, 5, 15,
45, 225, 3, 75, 9, 25
225 = 1 × 225
3 × 75
5 × 45
15 × 15
9 × 25
7. (a) Factors of 14 are 1 , 2, 7 , 14
Factors of 35 are 1 , 5, 7 , 35
Common factors of 14 and 35 are 1 and 7.
(b) Factors of 27 are 1 , 3 , 9 , 27
Factors of 36 are 1 , 2, 3 , 4, 6, 9 , 12, 18, 36
Common factors of 27 and 36 are 1, 3 and 9.
TH—Easy Maths—5
27
(c) Factors of 48 are 1 , 2 , 3, 4 , 6, 8 , 16, 12, 16, 24 and 48
Factors of 32 are 1 , 2 , 4 , 8 , 16, 32
Common factors of 48 and 32 are 1, 2, 4, 8, 16.
(d) Factors of 36 are 1 , 2, 3 , 4, 6, 9 , 12, 18, and 36
Factors of 45 are 1 , 3 , 5, 9 , 15 and 45
Common factors of 36 and 45 are 1, 3 and 9.
8. Refer answers at the end of book.
Exercise 6.2
1. (a) 15 and 33
Factors of 15 are 1, 3, 5, 15
Factors of 33 are 1, 3, 11, 33
Common factors of 15 and 33 are 1 and 3
 15 and 33 are not coprime as they have 3 as common factor
other than 1.
(b) 24 and 29
Factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24
Factors of 29 are 1, 29
Common factors of 24 and 29 is 1 only. Hence they are coprime.
(c) 34 and 57
Common factors of 34 are 1, 2, 17, 34
Common factors of 57 are 1, 3, 19, 57
Common factors of 34 and 57 is 1. Hence 34 and 57 are coprime.
(d) 18 and 67
Factors of 18 are 1, 2, 3, 6, 9, 18
Factors of 67 are 1, 67
Common factors of 18 and 67 is 1. Hence 18 and 67 are coprime.
(e) 14 and 52
Factors of 14 are 1, 2, 7, 14
Factors of 52 are 1, 2, 4, 13, 26, 52
Common factors of 14 and 52 is 1 and 2. Hence they are not
coprime.
2. (a)36 is a composite number as it has more than 2 factors i.e., 1, 2, 3,
4, 6, 9, 12, 18, 36 are its factors.
(b) 37 is prime as it has only two factors and they are 1 and 37.
(c) 31 is a prime as its has only two factors and they are 1 and 31.
(d) 59 is a prime as it has only two factors and they are 1 and 59.
(e)29 is a prime number as it has only two factors and they are 1 and
29.
3. Refer answers at the end of book.
Exercise 6.3
1. Refer answers at the end of book.
28 TH—Easy Maths—5
2. (a)4623
(b)9000
(c) 1438
(d)7596
(e) 8875 unit digits are 3 and 5 so it is not divisible by 2.
(f)20014 has units digits as 0, 8, 6 and 4 so they all are divisible
by 2.
3. (a) 3123 = 3 + 1 + 2 + 3 = 9
(b) 4826 = 4 + 8 + 2 + 6 = 20
(c) 52596 = 5 + 2 + 5 + 9 + 6 = 27
(d) 84463 = 8 + 4 + 4 + 6 + 3 = 25
(e) 76321 = 7 + 6 + 3 + 2 + 1 = 19
Sum of digits of (a) and (d) is divisible by 3 hence 31, 23 and 52596 is
divisible by 3.
4. (a)486
(b)512
(c)4920
(d)58862
(e)9636
(f)7773
The number formed by last two digits of b, c and e i.e., 12, 20, and 36
are divisible by 4. Hence b, c and e are divisible by 4.
5. (a)5912
(b)83224
(c)46246
(d)7732
Since 912, 224 are divisible by 8, So the number 5912 and 83224 are
divisible by 8.
6. (a) 68978 = 6 + 8 + 9 + 7 + 8 = 38
(b) 6987 = 6 + 9 + 8 + 7 = 30
(c) 8424 = 8 + 4 + 2 + 4 = 18
(d) 9405 = 9 + 4 + 0 + 5 = 18
Sum of digits of 8424 and 9405 is divisible by 9, So 8424 and 9405 are
divisible by 9.
7.(a) + 5 + 9 + 2 + 1 = 17 +
A number greater than 17 divisible by 3 is 18 so 18 – 17 = 1 should
be there in the box as the smallest digit. The number so formed is
divisible by 3.
(b) 8 + + 2 + 2 = 12 +
12 is divisible by 3 so the smallest digit that can be filled in the
blank is 0.
(c) 59 – 6 = 5 + 9 + 6 +
Smallest = 20 + but the smallest number greater than 20 but
divisible by 3 is 21 so 21–20 = 1 should be the digit at the blank
space.
TH—Easy Maths—5
29
(d)52 51
5 + 2 + 5 + 1 + = 13 +
The smallest number greater than 13 divisible by 3 is 15 so 15 – 13
= should be the digit at the blank space.
8. (a)495 8
9. (a)365
0 as 08 or 8 is divisible by 6 as 656 is divisible by 8.
4.
(b)3726
4 as 264 is divisible by 8.
(b)340 2
(c)637 4
1 as 12 is divisible by 4.
0 as 704 is divisible by 8.
(c)5425
(d)83 2
2 as 52 is divisible by 4.
1 as 312 is divisible by 8.
(d)3927
2 as 72 is divisible by 4.
10. (a)572 = 5 + 7 + 2 + = 14 +
A number greater than 14, divisible by 9 is 18. Hence 18 – 14 = 4
is the digit to be fitted in the blank space.
(b)47 58 = 4 + 7 + 5 + 8 + = 24 +
A number greater than 24 but divisible by 9 is 27. So 27 – 24 = 3 is
the digit to be fitted in the blank space.
(c)197 5 = 1 + 9 + 7 + 5 + = 22 +
A number greater than 22, but divisible by 9 is 27 hence 27 – 22
= 5 is the digit to be fitted in the blank space.
(d)4 12 = 4 + 1 + 2 + = 7 +
A number greater than 7, but divisible by 9 is 9. Hence 9 – 7 = 2
so 2 is to be fitted in the blank space.
Exercise 6.4
1. (a) 42, 84
2 42 2
3 21
2
7 7
3
1
7
84
42
21
7
1
(b) 36, 63
2 36 3
2 18
3
3 9
7
3 3
1
63
21
7
1
42 = 2 × 3 × 7 × 1
84 = 2 × 3 × 7 × 2 × 1
36 = 2 × 2 × 3 × 3 × 1
63 = 1 × 7 × 3 × 3
HCF (42, 84) = 2 × 3 × 7 = 42.
HCF (36, 63) = 3 × 3 = 9.
30 TH—Easy Maths—5
(c) 12, 18, 27
2 12 2 18
2 6
3 9
3 3
3 3
1
1
12 = 2 × 2 × 3 × 1
18 = 2 × 3 × 3 × 1
27 = 3 × 3 × 3 × 1
HCF (12, 18, 27) = 3
(e) 25, 65, 95
5 25 5 65 5 95
5 5 13 13 19 19
1
1
1
25 = 5 × 5 × 1
65 = 5 × 13 × 1
95 = 5 × 19 × 1
HCF 25, 65, 95 = 5
(f) 42, 63
3 42
2 14
7 7
1
42 = 2 × 3 × 7 × 1
63 = 3 × 3 × 7 × 1
HCF (42, 63) = 3 × 7 = 21
(g) 25, 90
5 25
5 5
1
(h) 18, 24, 32
2 18
3 9
3 3
1
18 = 2 × 3 × 3 × 1
24 = 2 × 2 × 2 × 3 × 1
32 = 2 × 2 × 2 × 2 × 2 × 1
HCF (18, 24, 32) = 2
2
5
3
3
3 27
3 9
3 3
1
90
45
9
3
1
25 = 5 × 5 × 1
90 = 2 × 5 × 3 × 3 × 1
HCF (25, 90) = 5
(i) 64, 80, 120
2 64 2 80 2 120
2 32 2 40 2 60
2 16 2 20 2 30
2 8 2 10 3 15
5
2 4 5 5 5
1
1
2 2
1
64 = 2 × 2 × 2 × 2 × 2 × 2 × 1
80 = 2 × 2 × 2 × 2 × 5 × 1 120 = 2 × 2 × 2 × 3 × 5 × 1 HCF (64, 80, 120) = 2 × 2 × 2 = 8 TH—Easy Maths—5
(d) 22, 66, 121
2 22 2 66 11 121
11 11 3 33 11 11
1 11 11
1
1
22 = 2 × 11 × 1
66 = 2 × 3 × 11 × 1
121 =
11 × 11 × 1
HCF (22, 66, 121) = 11
3 63
3 21
7 7
1
2
2
2
3
24
12
6
3
1
2
2
2
2
2
32
16
8
4
2
1
(j) 180, 136, 152
2 108 2 136 2 152
2 54 2 68 2 76
3 27 2 34 2 38
3
9 17 17 19 19
3
3
1
1
1
108 = 2 × 2 × 3 × 3 × 3 × 1
136 = 2 × 2 × 2 × 17 × 1
152 = 2 × 2 × 2 × 19 × 1
HCF (108, 135, 152)
= 2 × 2 = 4
31
2. (a) 12, 28
12 28 2
– 24
4 12 3
– 12
×
∴ H.C.F. (12, 28) = 4.
(b) 60
420 7
– 420
×
60 924 15
– 60
324
– 300
24 60 2
– 48
12 24 2
– 24
×
∴ H.C.F. (60, 420, 924) = 12.
(c) 42, 330
42 330 7
– 294
36 42 1
– 36
6 36 6
– 36
×
∴ H.C.F. (42, 330) = 6.
(e) 154, 770, 910
3. (a) 24, 35
154 770 5
24 35 1
– 770
– 24
×
11 24 2
– 22
154 910 5
2 11 5
– 770
– 10
140 154 1
1 2 2
– 140
–2
14 140 10
×
– 140
×
∴ H.C.F. (24, 35) = 1.
∴ H.C.F. (154, 770, 910) = 14.
32 (d)78, 210
78 210 2
– 156
54 78 1
– 54
24 54 2
– 48
6 24 4
– 24
×
∴ H.C.F. (78, 210) = 6.
TH—Easy Maths—5
(b) 36, 252
36 252 7
– 252
×
H.C.F. (36, 252) = 36.
(d) 210, 420, 540
(e) 90, 140, 210
210 420 2
90 140 1
– 420
– 90
×
50 90 1
– 50
210 540 2
40 50 1
– 420
– 40
120 210 1
10 40 1
– 120
– 40
90 120 1
×
– 90
30 90 3
10 210 21
– 90
– 20
×
10
H.C.F. (210, 420, 540) = 30.
10
×
H.C.F. (90, 140, 210) = 10.
(f) 45, 25, 65
25 45 1
– 25
20 25 1
– 20
5 20 4
– 20
×
5 65
65
(c) 18, 36, 45
18 36 2
– 36
×
18 45 2
– 36
9 18 2
– 18
×
H.C.F. (18, 36, 45) = 9.
13
H.C.F. (45, 25, 65) = 5.
TH—Easy Maths—5
33
Exercise 6.5
1. (a) 28, 35
2 28
5 35
2 14
7 7
7 7
1
1
28 = 2 × 2 × 7
35 = 5 × 7
LCM = 2 × 2 × 5 × 7
= 140
∴ LCM (28, 35) = 140
(c) 22, 66
2 22
11 11
1
34 2 66
3 33
11 11
1
22 = 2 × 11
66 = 2 × 11 × 3
LCM (22, 66) = 2 × 11 × 3
= 66
(e) 125, 180, 210
5 125
2
5 25
2
5
5
3
1
3
5
180
2 210
90
5 105
45
3 21
15
7
7
5
1
1
125 = 5 × 5 × 5
180 = 2 × 2 × 3 × 3 × 5
210 = 2 × 5 × 3 × 7
LCM (125, 18, 210)
= 5 × 5 × 5 × 3 × 3 × 2 × 2 × 7
= 31500
(b) 48, 72
2 48
2 72
2 24
2 36
2 12
2 18
2 6
3 9
3 3
3 3
1
1
48 = 2 × 2 × 2 × 2 × 3
72 = 2 × 2 × 2 × × 3 × 3
HCM = 2 × 2 × 2 × 2 × 3 × 3
= 144
(d) 36, 48, 96
2 36
2 18
3 9
3 3
1
2
2
2
2
3
48
24
12
6
3
1
2
2
2
2
2
3
96
48
24
12
6
3
1
36 = 2 × 2 × 3 × 3
48 = 2 × 2 × 3 × 2 × 2
96 = 2 × 2 × 3 × 2 × 2 × 2
LCM (36, 48, 96)
= 2 × 2 × 2 × 2 × 2 × 3 × 3
= 288
(f) 198, 216, 360
2 198
2
3 99
2
3 33
2
11 11
3
1
3
3
216
2 360
108
2 180
54
2 90
27
3 45
9
3 15
3
5
5
1
1
198 = 2 × 3 × 3 × 11
216 = 2 × 2 × 2 × 3 × 3 × 3
360 = 2 × 2 × 2 × 3 × 3 × 5
LCM (198, 216, 360)
= 2 × 2 × 2 × 3 × 3 × 3 × 11 × 5
= 11880
TH—Easy Maths—5
2. (a) 20, 35, 45
5 20, 35, 45
2 4, 7, 9
2 2, 7, 9
7 1, 7, 9
3 1, 1, 9
3 1, 1, 3
1, 1, 1
LCM (20, 35, 45)
= 5 × 2 × 2 × 7 × 3 × 3
= 1260
(c) 36, 64, 72, 96, 120
2 36, 64, 72, 96, 120
2 18, 32, 36, 48, 60
2 9, 16, 18, 24, 30
2 9, 8, 9, 12, 15
2 9, 4, 9, 6, 15
3 9, 2, 9, 3, 15
2 3, 2, 3, 1, 5
3 3, 1, 3, 1, 5
5 1, 1, 1, 1, 5
1, 1, 1, 1, 1
LCM (36, 64, 72, 96, 120) = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5
= 2880
(b) 10, 25, 65
5 10, 25, 65
2 2, 5, 13
5 1, 5, 13
13 1, 1, 13
1, 1, 1
LCM (10, 25, 65)
= 5 × 2 × 5 × 13
= 650
(d) 42, 60, 84, 108
2 42, 60, 84, 108
2 21, 30, 42, 54
3 21, 15, 21, 27
3 7, 5, 7, 9
3 7, 5, 7, 3
7 7, 5, 7, 1
5 1, 5, 1, 1
1, 1, 1, 1
LCM (42, 60, 84, 108)
= 2 × 2 × 3 × 3 × 3 × 5 × 7
= 3780
(e) 27, 45, 60, 72, 96
3. (a) 30, 75
3 27, 45, 60, 72, 96
5 30, 75
3 9, 15, 20, 24, 32
3 6, 15
3 3, 5, 20, 8, 32
2
2, 5
5 1, 5, 20, 8, 32
5
1, 5
2 1, 1, 4, 8, 32
1, 1
2 1, 1, 2, 4, 16
LCM (30, 75)
2 1, 1, 1, 2, 8
= 5 × 3 × 2 × 5
2 1, 1, 1, 1, 4
= 150
2 1, 1, 1, 1, 2
1, 1, 1, 1, 1
LCM (27, 45, 60, 72, 96)
= 3 × 3 × 3 × 5 × 2 × 2 × 2 × 2 × 2
= 4320
TH—Easy Maths—5
35
36 (b) 45, 66
3 45, 66
2 15, 22
3 15, 11
5 5, 11
11 1, 11
1, 1
LCM (45, 66)
= 3 × 2 × 3 × 5 × 11
= 990
(d) 144, 120
2 144, 120
2 72, 60
2 36, 30
2 18, 15
3 9, 15
3 3, 5
5 1, 5
1, 1
LCM (144, 120)
= 2 × 2 × 2 × 2 × 3 × 3 × 5
= 720
(f) 36, 63, 81
3 36, 63, 81
3 12, 21, 27
3 4, 7, 9
3 4, 7, 3
7 4, 7, 1
2 4, 1, 1
2 2, 1, 1
1, 1, 1
LCM (36, 63, 81)
= 3 × 3 × 3 × 3 × 7 × 2 × 2
= 2268
(h) 162, 270, 108
2 162, 270, 108
3 81, 135, 54
3 27, 45, 18
3 9, 15, 6
3 3, 5, 2
5 1, 5, 2
2 1, 1, 2
1, 1, 1
(c) 135, 175
5 135, 175,
3 27, 35
3 9, 35
3 3, 35
5 1, 35
7 1, 7
1, 1
LCM (135, 175)
= 5 × 3 × 3 × 3 × 5 × 7
= 4725
(e) 20, 25, 60
5 20, 25, 60
2 4, 5, 12
2 2, 5, 6
5 1, 5, 3
3 1, 1, 3
1, 1, 1
LCM (20, 25, 60)
= 5 × 2 × 2 × 5 × 3
= 300
(g) 21, 42, 105
3 21, 42, 105
7 7, 14, 35
2 1, 2, 5
5 1, 1, 5
1, 1, 1
LCM (21, 42, 105)
= 3 × 7 × 2 × 5
= 210
LCM (162, 270, 108)
= 2 × 3 × 3 × 3 × 3 × 5 × 2
= 1620
TH—Easy Maths—5
Exercise 6.6
1. LCM of 8, 16, and 24 is 48
Now divide greatest number
of 6 digit by 48.
48
999999 20833
– 96
399
– 384
159
– 144
159
– 144
15
∴ The greatest six digit
number divisible by 8, 16
and 24 is 999999 – 15
= 999984
3. LCM of 18, 24, and 36 = 72
Greatest 5 digit number
= 99999
Now divide 99999 by 72
72
99999
1388
– 72
279
– 216
639
– 576
639
– 576
63
∴ (9999 – 63) + 7 = 99943 is the greatest 5 digit number
which leaves a remainder 7
when divided by 18, 24 and
36.
TH—Easy Maths—5
2.LCM of 48, 60 and 64 is 960.
Greatest 4 digit number
= 9999
960 9999 10
– 960
399
∴ 9999 – 399 = 9600 is the
largest 4 digit number
divisible by 960.
4. LCM of 12, 16, 24 and 36 = 144
Greatest 4 digit number
= 9999
Now divide 9999 by 144
144
9999 69
– 864
1359
– 1296
63
∴ (9999 – 63) + 5 = 9941
9941 is the greatest 4 digit
number which when divided
by 12, 16, 24 and 36 leaves a
remainder 5.
37
5.
1036 – 1 = 1035
347 – 2 = 345
633 – 3 = 630
Now find the HCF of 345, 630
and 1035.
345 630 1
– 345
285 345 1
– 285
60 285 4
– 240
45 60 1
– 45
15 45 3
– 45
×
6.
373 – 5 = 368
484 – 4 = 480
542 – 6 = 536
Now find the HCF of 368, 480 and 536.
368 480 1
– 368
112 368 30
– 336
32 112 3
– 96
16 32 2
32
×
7. LCM of 8, 15 and 24 = 120
2 8, 15, 24
2 4, 15, 12
3 2, 15, 6
2 2, 5, 2
5 1, 5, 1
1, 1, 1
38 15
1035 69
– 90
135
– 135
×
∴ 15 is the greatest number
that divides 1036, 347 and 633
leaving remainders 1, 2, and 3
respectively.
16 536
528
×
33
8 16 2
16
×
HCF of 368, 480, 536 = 8
∴ 8 is the greatest number
that divides 373, 484 and 542
leaving remainders 5, 4, and 6
respectively.
= 2 × 2 × 3 × 2 × 5 = 120
Smallest 5-digit number
= 10000
Now, divide 10000 by 120.
TH—Easy Maths—5
120 10000
–960
400
–360
40
83
∴ 10000 + (120 – 40)
⇒ 10000 + 80 = 10080 is
the smallest 5-digit number
which is exactly divisible by
8, 15 and 24.
8. 14 m 25 cm = 1425 cm
5 m 50 cm = 550 cm
6 m = 600 cm
1425, 550 and 600
600 1425 2
– 1200
225 600 2
– 450
150 225 1
– 150
75 150 2
– 150
×
550 7
– 525
25 75 3
– 75
×
Hence 25 cm is the length of
the longest tape that can be
used to measure exactly the
dimensions of the given hall.
9. 108 and 180 is
108 180 1
108
72 108 1
72
36 72 2
72
×
Hence 36 is the largest
number of students among
whom 108 chocolates and 180
cookies can be distributed
equally.
Now to find the number of
chocolates and cookies each
students will get, divide
the number of cookies and
chocolates by 36.
i.e.,
Chocolates
36 108 3
– 108
×
Cookies
36 180 5
– 180
×
Hence each students will get
3 chocolates and 5 cookies.
10. LCM of 120, 180 and 240 is
180 240 1
– 180
60 180 3
– 180
×
60 120 2
–120
×
∴ 60 litres is the capacity of
the greatest container that can
be used to measure 120 litres,
180 litres and 240 litres of oil
exactly.
TH—Easy Maths—5
75
39
11. `1 = 100 paise
LCM of 125,5000
∴ `1.25 = 125 paise and
= 5 × 5 × 5 × 5 × 8 = 5,000
`50 = 5000 paise.
Amount in paise
Now find the LCM of 125 and 5000. = 5,000
Amount in rupees
5 125, 5000
= 5,000 ÷ 100 = `50
5 25, 1000
The number of days
5 5, 200
= `50 ÷ 1.25 = 40 days
5 1, 40
8 1, 8
1, 1
Exercise 6.7
1. We know that
HCF × LCM = 1st number × 2nd number
∴ 5 × 280 = 35 × 2nd number
5 × 280
⇒ 2nd number =
= 40
35
2. HCF × LCM = 1st number × 2nd number
∴ 3 × 120 = 15 × 2nd number
3 × 120
⇒ 2nd number =
= 24
15
3. HCF × LCM = 1st number × 2nd number
∴ 6 × 180 = 36 × 2nd number
6 × 180
⇒ 2nd number =
= 30
36
4. HCF × LCM = 1st number × 2nd number
∴ 27 × 2079 = 189 × 2nd number
27 × 2079
⇒ 2nd number =
= 297
189
5. LCM × HCF = 12 × 64
∴ 192 × HCF = 12 × 64
12 × 64
HCF =
=4
192
6. HCF × LCM = Product of the numbers
∴ LCM × 6 = 432
432
⇒ LCM =
= 72
6
7. HCF × LCM = Product of the numbers
∴ LCM × 7 = 735
735
⇒ LCM =
= 105
7
40 TH—Easy Maths—5
Chapter 7: Fractions
Exercise 7.1
1. Equivalent fractions are obtained by multiplying the numerator and
denominator of the given fraction by the same number.
1×4 4
1×5 5
1×6 6
(a)
= ; = ; =
4 × 4 16 4 × 5 20 4 × 6 24
1×4 4
1×5 5
1×6 6
(b)
= ; = ; =
5 × 4 20 5 × 5 25 5 × 6 30
1×4 4
1×5 5
1×6 6
(c)
= ; = ; =
8 × 4 32 8 × 5 40 8 × 6 30
1×4 4
1×5 5
1×6 6
(d)
= ; = ; =
9 × 4 36 9 × 5 45 9 × 6 54
2. (a)In like fractions, the fraction with greater numerator in greater
11 > 9
∴
14
14
3
15
3
15 = 3
= ∴
55
11
55 11 11
(c) Convert unlike fractions into like fractions by finding LCM of
9 and 12.
3 9, 12
3 3, 4 LCM = 3 × 3 × 4
4 1, 4
= 36
1 1, 1
5 5 × 4 20
5
5 × 3 15
Now =
= and =
=
9 9 × 4 36
12 12 × 3 36
(b) Since =
20 > 15
5 > 5
or
36
36
9
12
In a given pair of fractions whose numerators are same the fraction
with greater denominator is smaller.
(d) Find the LCM of 35 and 15
LCM = 5 × 7 × 3 = 105
5 35, 15 7, 3
12 12 × 3 36
10 10 × 7 70
Now =
=
and
=
=
35 35 × 3 105
15 15 × 7 105
36 < 70
12
10
∴
or <
105
105
35
15
31 = 31
(e)
45
45
∴
TH—Easy Maths—5
41
1
16
1
= 48 3 3
(f)
∴ 1
15
1
=
45 3 3
and 1 = 1
or
3
3
16 = 15
48
45
3. (a) Divide 18 by 10
10
18 1
1
=
10
8
8
10
=
1
4
5
(b) Divide 37 by 6, 26
(c) 5
⇒
5
⇒
5 1
5
11
(c) 7
26 5 –25
1
⇒
7
⇒
1 4
7
27
(e) 2
11 1 –7
4
⇒
⇒
13 49
(f) 8
27 13 –2
×7
–6
1
⇒
49 6 –48
1
⇒
6 37 6
6
–36
1
2
8
=
6
1
6
1
2
1
8
4. (a) Find the LCM of 5, 20, 10 LCM = 5 × 2 × 2 = 20
5 5, 20, 10
2 1, 4, 2
1, 2, 1
3 3 × 4 12
5
5
2
2×2
4
Now =
= ; = ; =
=
5 5 × 4 20
20 20
10 10 × 2 20
Arranging in ascending order:
42 4
5 12
2
5
3
<
< or <
<
20 20 20
10 20 5
TH—Easy Maths—5
(b) LCM of 5, 10, 20 is 20.
2 2×4 8
2×2
4
6
6×1
6
Now =
= ; = ; =
=
5 5 × 4 20
10 × 2 20
20 20 × 1 20
4 6
8
2
6
2
∴ < < or <
<
20 20 20
10 20 5
(c) LCM of 2, 4, 8 = 8
1 1×4 4
1×2 2
1 1×1 1
Now =
= ; = ; =
=
2 2×4 8
4×2 8
8 8×1 8
1 2 4
1 1 1
∴ < < or < <
8 8 8
8 4 2
5. (a) As per the reason given in Answer 2(a).
3 2 1
> >
4 4 4
(b) As per the reason given in Answer 2(c).
1 1 1
> >
4 6 9
(c) LCM of 4, 8, 16 = 16
2 2×4 8
5 × 2 10
11 11 × 1 11
∴ =
= ; = ; =
=
4 4 × 4 16
8 × 2 16
16 16 × 1 16
11 10 8
11 5 2
Hence >
> or > >
16 16 16
16 8 4
6. (a) Number of players allowed to play at one time = 11
Total number of players = 16
11
Hence the fraction of the team that played the match =
16
(b) Dividing 32 students in 4 equal groups i.e. 32 ÷ 4 = 8 students
3 groups of 8 students each includes = 3 × 8 = 24 students
Hence, 24 students are less than 10 years old.
(c) Improper fraction
3 (3 × 4) + 3 12 + 3 15
(d) 3 =
=
=
4
4
4
4
7
(4 × 12) + 7 55
(e) Tina jumped = 4 ft or
=
ft
12
12
12
1
(3 × 6) + 1 18 + 1 19
Meena jumped = 4 ft or
=
= ft
6
6
6
6
Changing both fractions into like fractions, we get
19
=
6
19 × 2
38
55
= and 6×2
12
12
55 38
7
1
Since >
or 4 > 3 12 12
12
6
TH—Easy Maths—5
43
Since the whole number 4 is greater than 3.
7
1
∴ 4 > 3 12
6
Hence Tina jumped further.
Exercise 7.2
1. To reduce the fraction in lowest fraction we divide the numerator
and denominator by common factor. The process is continued till the
common factor of numerator and denominator is 1.
10 10 ÷ 5 2
26 26 ÷ 13 2
(a) =
= (b)
=
=
15 15 ÷ 5 3
39 39 ÷ 13 3
60 60 ÷ 6 10 ÷ 2 5
42 42 ÷ 7
6÷2 3
(c) =
=
= (d)
=
=
=
96 96 ÷ 6 16 ÷ 2 8
56 56 ÷ 7
8÷2 4
66 66 ÷ 11 6 ÷ 3 2
49 49 ÷ 7 7
(e) =
=
= (f)
=
=
99 99 ÷ 11 9 ÷ 3 3
63 63 ÷ 7 9
34 34 ÷ 17 2
30 30 ÷ 6 5
(g) =
= (h)
=
=
85 85 ÷ 17 5
48 48 ÷ 6 8
45 45 ÷ 9 5
21 21 ÷ 7 3
(i) =
= (j)
=
=
72 72 ÷ 9 8
28 28 ÷ 7 4
2 2÷2 1
6
6÷2
3
2.(a) =
= (b)
=
=
4 4÷2 2
20 20 ÷ 2 10
24 24 ÷ 8 3 ÷ 3 1
12 12 ÷ 4
3
(c) =
=
= (d)
=
=
72 72 ÷ 8 9 ÷ 3 3
16 16 ÷ 4
4
20 20 ÷ 5 4
4
4÷4
1
(e) =
= (f)
=
=
25 25 ÷ 5 5
16 16 ÷ 4
4
6 6÷3
9
9÷3
3.(a) =
= 2
(b)
=
=3
3 3÷3
3
3÷3
8
8÷2
14 14 ÷ 7
(c) =
= 4
(d)
=
=2
2
2÷2
7
7÷7
25 25 ÷ 5
10 10 ÷ 10
(e) =
= 5
(f)
=
=1
5
5÷5
10 10 ÷ 10
16 16 ÷ 4
30 30 ÷ 6
(g) =
= 4
(h)
=
=5
4
4÷4
6
6÷6
49 49 ÷ 7
(i) =
=7
7
7÷7
4. Dividing numerator and denominator by common factors.
9
9÷9
1
8
8÷8
1
(a) =
= (b)
=
=
27 27 ÷ 9 3
24 24 ÷ 8
3
10 10 ÷ 10 1
24 24 ÷ 24
(c) =
= (d)
=
=1
30 30 ÷ 10 3
24 24 ÷ 24
44 TH—Easy Maths—5
10 10 ÷ 10 1
3
3÷3
1
(e) =
= (f)
=
=
60 60 ÷ 10 6
15 15 ÷ 3
5
4 (2 × 7) + 4 14 + 4 18
5. (a) 2 =
=
= 7
7
7
7
9 (5 × 10) + 9 50 + 9 59
=
=
= 10
10
10
10
(c) 5 5 (1 × 8) + 5
8 + 5 13
(e) 1 =
=
= 8
8
8
8
7 (3 × 9) + 7 27 + 7 34
(b) 3 =
=
=
9
9
9
9
1 (5 × 9) + 1 45 + 1 46
(d) 5 =
=
=
9
9
9
9
3 (3 × 4) + 3 12 + 3 15
(f) 3 =
=
=
4
4
4
4
Exercise 7.3
3
3
8
8
1. (a) 7 + = 7 (b) 15 +
= 15
5
5
9
9
5
5
(a) 9 + = 9
11
11
When we have like fractions, we simply add numerators keeping the
denominator same.
5
3 5+3 8
8
7 8 + 7 15
(d) +
=
= (e)
+
=
=
or 1
11 11
11
11
15 15
15
15
5 4 + 2 + 5 11
2 + 5 + 4 = 2 + 5 + 4 = 11 or 1 4
(f) 4 + 2 +
=
= (g)
13
7
13 13 13
13
7 7 7
7
7
3
5 7 + 3 + 5 15
7
(h) +
+
=
=
16 16 16
16
16
1 3
1 3
1+3
4
4
(i)5 + 5 = 5 +
+
=5+
=5+ =5
5
5 5
5
5
5
7
5
5
7
5+7
12
(j)4 + 11 = 4 +
+
=4+
=4+
11
11 11
11
11
=4+1
1
1
1
1
= (4 + 1) +
=5+
=5
11
11
11
11
(k)7
(l)3
2 1
2+1
3
+ =3+
=3+ =3+1=4
3 3
3
3
(m)2
3
3
3
3
+ 7 = (2 + 7) + = 9 + = 9
5
5
5
5
(n)4
2
2
2
2
+ 6 = (4 + 6) + = 10 + = 10
9
9
9
9
When denominators are not same we change them into like fraction
by LCM method.
11 9
11 + 9
20
5
+
=7+
=7+
=7+1
15 15
15
15
15
= (7 + 1) +
TH—Easy Maths—5
5
5
1
=8
or 8
15
15
3
45
5
3
5 3
+7 =8+7+ +
8
4
8 4
(o)8
LCM of 8 and 4 is 8.
= 15 +
(p)13
Since LCM of 3 and 9 is 9 ∴
Now 21 +
(q)5
(r)4
Since LCM of 16, 8, 2 is 16 ∴
Now, 15 +
or 15 + 1
3
∴ 20 +
∴
5 6
5+6
11
3
3
+ = 15 +
= 15 +
or 15 + 1 = 16
8 8
8
8
8
8
2
1
2 1
+ 8 = (13 + 8) + +
9
3
9 3
1 1×3 3
=
=
3 3×3 9
2 3
5
5
+ = 21 + = 21
9 9
9
9
3
4
2
3
4 2
+2 +7
= (5 + 2 + 7) +
+
+
11
11
11
11 11 11
= 14 +
3+4+2
9
9
= 14 + = 14
11
11
11
7
5
1
7 5 1
+ 3 + 8 = (4 + 3 + 8) +
+ +
16
8
2
16 8 2
5 5 × 2 10
1 1×8 8
=
=
and =
=
8 8 × 2 16
2 2 × 8 16
7 10 8
7 + 10 + 8
25
+
+
= 15 +
= 15 +
16 16 16
16
16
9
9
= 16
16
16
7
3
9
7 3 9
+ 12 + 5
= 3 + 12 + 5 +
+ +
10
5
10
10 5 10
LCM of 5, 10 is 1 ∴
3 3×2 6
=
=
5 5 × 2 10
7
6
9
22
2
2
1
+
+
= 20 +
or 20 + 2 = 22
or 22
10 10 10
10
10
10
5
2. (a) Fraction of money spent on books =
46 3 3×2 6
=
=
4 4×2 8
1
2
Fraction of money spent on snacks and drinks =
1 1
+
2 4
1 1×2 2
LCM of 2 and 4 is 4 ∴ =
=
2 2×2 4
2 1 3
Now, + =
4 4 4
1
4
Total fraction of money spent =
TH—Easy Maths—5
1
kg
2
(b) Apples purchased = 2
Oranges purchased = 1
Grapes purchased = 3
Total weight of fruits purchased = 2
1
kg
4
1
kg
4
1
1
1
+1 +3
2
4
4
LCM of 2 and 4 is 4
1 1×2 2
∴ =
=
2 2×2 4
2
1
1
2 1 1
Now 2 + 1 + 3 = (2 + 1 + 3) +
+ +
4
4
4
4 4 4
4
=6+
= 6 + 1 = 7 kg.
4
2
(c) Distance jogged on Monday = 4 km
3
1
km
3
Distance jogged on Tuesday = 5
Distance jogged on Wednesday = 3
Total distance jogged = 4
= (4 + 5 + 3) +
(d) Part of cake eaten by friends =
Part of cake eaten by family members =
Total part of the cake eaten =
Changing
6÷3 2
= is the total part of the cake eaten.
9÷3 3
= 12 + 1
TH—Easy Maths—5
2
km
3
2
1
2
+5 +3
km
3
3
3
2 1 2
2+1+2
5
+ +
= 12 +
= 12 +
3 3 3
3
3
2
2
or 13 km.
3
3
4
9
2
9
4 2 4+2 6
+ =
=
9 9
9
9
2
into lowest form.
3
47
(e) Part of book read on Monday =
4
9
Part of book read on Tuesday =
2
9
Part of book read on Wednesday =
1
9
Part of the book finished reading =
4 2 1 4 + 2 +1 7
+ + =
=
9 9 9
9
9
Exercise 7.4
1. (a) 8 –
48 5
9
8 8 × 9 72
Meera =
=
1 1×9 9
72 5 72 – 5 67
4
∴ – =
= or 7
9 9
9
9
9
11 11 × 8 88
5 88 5
(b) 11 – =
– as =
=
1
1×8
8
8 8 8
83
3
= or 10
8
8
7
47
(c) 11 – 5 or 11 –
8
8
11 11 × 8 88
88 47 88 – 47 41
1
Here, =
= ∴ –
=
= or 5
1
1×8
8
8
8
8
8
8
9
75
(d) 14 – 6 or 14 –
11
11
14 14 × 11 154
Here, =
=
1
1 × 11
11
154 75 154 – 75 79
2
∴
–
=
= or 7
11 11
11
11
11
3
31
(e)7 – 3 or – 3
4
4
3 3 × 4 12
Here, =
=
1 1×4 4
31 12 31 – 12 19
3
∴ –
=
= or 4
4
4
4
4
4
3
3
3
3
(f)9 – 6 = (9 – 6) +
= 3 + or 3
11
11
11
11
13
388
(g)25 – 21 or – 21
15
15
21 21 × 15 315
Here, =
=
1
1 × 15
15
388 315 388 – 315 73
13
Now,
–
=
= or 4
15
15
15
15
15
TH—Easy Maths—5
7 4
(h) –
9 9
Since fractions are like fractions we simply subtract the numerators
and write the answer over the common denominator.
7–4 3
3÷3 1
i.e., = or
=
9
9
9÷3 3
15 9 15 – 9 6
(i) –
=
=
17 17
17
17
2 2×2 4
9 2
9
4 9–4 5
– or –
=
=  =
=
7 7 × 2 14
14 7
14 14
14
14
3 1
3–1
3 1
2
2
1
(k)2 – = 2 +
–
=2+
= 2 + = 2 or 2
4 4
4
4 4
4
4
2
12
5
(l)9 – 6
17
17
(j)
= (9 – 6) +
LCM of 9 and 5 is 45
∴
We subtract the whole number and fractions separately.
12 5
–
17 17
12 – 5
7
7
= 3 +
= 3 + or 3
17
17
17
3
4
(m) 8 – 6
9
5
Changing mixed numbers into improper fractions,
3 75
4 34
= 8 = and 6 =
9 9
5 5
75 × 5 375
34 × 9 306
=
and =
9×5
45
5×9
45
375 306 375 – 306 69
24
8
Now,
–
=
= or 1 or 1
45
45
45
45
45
15
4
7
(n)12 – 3
9
12
Changing into improper fractions,
12
LCM of 9 and 12 is 36.
∴
7 43
4 112
=
and 3 =
9
12 12
9
112 112 × 4 448
43 43 × 3 129
=
=
and =
=
9
9×4
36
12 12 × 3 36
448 129 448 – 129 319
31
Now,
–
=
=
or 8
36
36
36
36
36
TH—Easy Maths—5
49
1
7
3
+1 –7
12
9
18
(o)9
9
Now,
=
Changing into improper fractions,
7 16
3
1 43
1 109
=
; 1 = and 7 or 7 =
9 9
18
6 6
12 12
LCM of 12, 9 and 6 is 36.
109 109 × 3 327
16 16 × 4 64
=
=
=
=
12
12 × 3
36
9
9 × 4 36
43 43 × 6 258
327 64 258
and =
=
So,
+
–
6
6×6
36
36 36
36
(327 + 64) – 258 391 – 258 133
25
=
=
or 3
36
36
36
36
3
7
1
(p)29 + 28
– 11
15
12
6
Changing into improper fractions
29
3
1 146
or 29 =
5
15
5
28
7 343
1 67
=
and 11 =
12 12
6 6
Now, LCM of 5, 12 and 6 = 60
∴ 146 146 × 12 1752
343 343 × 5 1715
=
=
=
=
5
5 × 12
60
12
12 × 5
60
67 67 × 10 670
1752 1715 670
=
=
Now,
+
–
6 6 × 10
60
60
60
60
=
146 343 67
+
–
5
12
6
(1752 + 1715) – 670 3467 – 670 2797
37
=
=
or 46
60
60
60
60
1
of cake
5
1
Mohan ate = of cake
6
Comparing the two fractions
1
1
1 1
= 1 × 6 > 1 × 5 Hence, > . ∴ Sohan ate more.
5
6
5 6
2. (a) Sohan ate =
50 TH—Easy Maths—5
(b) Fraction of pizza Rahul ate =
1
2
Changing into like fractions
LCM of 3 and 2 = 6
∴
∴
1 1×6 6
1 1×5 5
=
= and=
=
5 5 × 6 30
6 6 × 5 30
6
5 6–5 1
–
=
=
30 30
30
30
1
Sohan ate more by of the cake.
30
Now
2
3
Fraction of pizza Abhay =
2 2×2 4
1 1×3 3
=
= and=
=
3 3×2 6
2 2×3 6
Since 4 > 3.
4 3
> ∴ Rahul has more pizza.
6 6
(i) True (ii) False (iii) False
3
108
(c) Total length of wire = 21 or
m.
5
5
1
151
Length of wire cut off from it = 15 or m.
10
10
108 151
Remaining length of the wire =
–
m.
5
10
Changing fractions into equivalent fractions
Changing into like fractions
108 108 × 2 216
=
=
5
5×2
10
216 151
216 – 151
65
5
∴
–
m=
m = m or 6 m
10
10
10
10
10
1
or 6 m
2
1
1 36 9
(d) 7 – 4 =
–
5
2 5 2
LCM of 5 and 2 = 10
or
36 36 × 2 72
9 9 × 5 45
=
= =
=
5
5 × 2 10
2 2 × 5 10
TH—Easy Maths—5
51
72 45 72 – 45 27
7
–
=
= or 2
10 10
10
10
10
4
1
1
– 3 17 – 9
5
2
3
29 19
35 28
– = –
4
5
2
3
(29 × 5) – (19 × 4)
(35 × 3) – (28 × 2)
=
4×5
2×3
Now,
(e)7
1
4
=
=
=
105 – 56
=
6
49
=
=
6
1
=
=8
6
1
9
Since 8 > 3 ∴8 > 3
6
20
1
1
∴ 17 – 9 is greater.
2
3
3
(f) Other number = 16 – 2
4
11
= 16 –
4
(16 × 4) – 11 64 – 11 53
1
=
=
= or 13
4
4
4
4
1
(g) Fraction of black marbles =
3
1
Fraction of white marbles =
2
1 1 (1 × 2) + (1 × 3) 2 + 3 5
Fraction of marbles not green = + =
=
=
3 2
3×2
6
6
3
(h) Total length of cloth
=9 m
16
1
1
Pieces cut off from cloth
=2 m+4 m
2
4
3
1
1
Cloth left = 9
– 2 +4
m
16
2
4
3
5 17
3
10 17
3 27
=9
– +
=9
–
+
=9
–
16 2 4
16
4
4
16 4
∴ 52 145 – 76
20
69
20
9
3 20
LCM of 16 and 4 = 16
27 × 4 108
=
4×4
16
147 108
147 – 108 39
7
Now, –
m=
=
m or 2
m cloth is left
16
16
16
16
16
TH—Easy Maths—5
(i) Total flour used to make both cakes
= 3
1
1
7 9
+ 2 kg = + kg
2
4
2 4
LCM of 2 and 4 is 4.
7 7 × 2 14
∴ =
=
2 2×2 4
14 9
14 + 9 23
3
+
kg =
=
kg or 5 kg
4 4
4
4
4
Flour bought by the baker = 15 kg
23
Flour left after making cakes = 15 –
kg
4
15 × 4 – 23
(60 – 23)
=
kg =
kg
4
4
37
1
= kg or 9 kg
4
4
Now, Exercise 7.5
1. Refer answers at the end of book.
1
2. (a) Pizzas eaten by girls= 3
2
1
Pizzas eaten by boys= 7
2
1
1
Total pizzas eaten = 3 + 7
2
2
= (3 + 7) +
= 11 pizzas
(b) Paint bought= 2
Paint used
Paint left
TH—Easy Maths—5
1 1
2
+
= 10 + = 10 + 1
2 2
2
1
5
litres or litres
2
2
3
litres
2
5 3
5–3 2
=
–
litres =
= litres or 1 litre
2 2
2
2
=
53
2
cups
3
(c) White flour needed for recipe =
Whole wheat flour needed for recipe= 2
Total flour needed
= 3 cups
(d) Age of dog
=5
Age of cat
= (5 – 2) +
(f) Number of cups of water needed to drink= 8
3
4
Number of cups of milk needed to drink = 2
1
4
Total fluid intake= 8
54 Cat is younger than dog by
1
cups
3
1 2
= 2 +
cups
3 3
1 2
=2+
+
3 3
1+2
3
=2+
=2+ =2+1
3
3
1
years
2
1
= 4 years
3
1
1
–4
years
2
3
1 1
3–2
1
1
= (5 – 4) +
–
=1+
= 1+
years or 1 years
2 3
6
6
6
2
1
(e) Since 5 > 2 hence, 5 > 2
3
4
2
1
∴ Dad ate more by = 5 – 2
buns
3
4
= 5
2 1
2×4–1×3
–
=3+
3 4
3×4
8–3
5
5
=3+
=3+
buns or 3
buns
12
12
12
3
1
3 1
+2
cups = (8 + 2)
+
4
4
4 4
3+1
4
= 10 +
= 10 +
4
4
= (10 + 1) = 11 cups
TH—Easy Maths—5
Exercise 7.6
1
5
5 1
1. (a)Subtracting from , i.e., – =
9
9
9 9
1
4
Again subtracting from ,
9
9
4 1 3
1
i.e. – = or 9 9 9
3
5
Hence degree of closeness of to
9
1
5
(b)Subtracting from ,
6
6
5 1 4
i.e. – =
6 6 6
1
4
Again subtracting from .
6
6
4 1 3
i.e. – =
6 6 6
5
Hence degree of closeness of to
6
1
3
(c)Subtracting from ,
5
5
3 1 3–1 2
⇒ – =
=
5 5
5
5
1
2
Again subtracting from ,
5
5
2 1 1
i.e., – =
5 5 5
(d)Subtracting
Hence, degree of closeness of
1
1
7
from ,
8
8
2
7 1
6 1
5
– ⇒ – ⇒
8 8
8 8
8
3
1
4
8

→
−
1
is 2.
3
3
is 2.
6
3
1
to is 2.
5
5
4
1
3
8

→
−
8
7
3
Hence, degree of closeness of to is 4.
8
8
3
1
2. (a) To find degree of closeness of to .
8
2
1
3
3 1 4
1
Add to i.e., + = or 8
8
8 8 8
2
3
1
Hence, degree of closeness of to is 1.
8
2
TH—Easy Maths—5
8
4
9
55
Subtract
To find degree of closeness of
1
3
from
8
8
3 1 2
1
i.e., – = or 8 8 8
4
Hence, degree of closeness of
3
1
to .
8
4
3
1
to is 1.
8
4
3
As degree of closeness is same in both cases ∴ is equally closer
8
1
1
to and .
2
4
7
1
1
7
(b) To find degree of closeness of to , we add to .
16 2
16 16
7
1
8
1
i.e., +
= or . Hence it is 1
16 16 16
2
7
1
To find degree of closeness of (DOC) of to , we
16 4
1
7
subtract from
16
16
1
1
−
−
7 1
6
1
16 → 5 
16 → 4

i.e., + ⇒
or 16 16 16
16
16
4
∴ We can say
Hence DOC is 3.
4
1
1
is closer to than to .
16
2
4
5
1
1
5
(c) To find DOC of to , we subtract from .
12 3
3
12
5
1
4
1
i.e., –
= or 12 12 12
3
Hence, DOC = 1
To find the DOC of
+
+
5
1
6
2
12 → 7 
12 → 8

i.e., +
⇒
or 12 12 12
12
12
3
Hence, DOC is 3.
So, we can say that
5
2
1
5
to , we add to
.
12 3
12 12
1
56 1
5
1
2
is closer to than to .
12
3
3
TH—Easy Maths—5
1
1
1
1
to we subtract from .
6
2
6
6
(d) To find DOC of
+
1 1 2
3
1
6
→ or i.e., + = 
6 6 6
6
2
Hence, degree of closeness is 2.
To find DOC of
Hence, DOC is 1.
So we can say that
1
1
1
1
1
to , we add from .
6
3
6
6
1 1 2
1
i.e., + = or 6 6 6
3
1
1
1
is closer to than to .
6
3
2
Exercise 7.7
1 1 1 1 1 1+1+1+1+1 5
1
1. + + + + =
= or 1
4 4 4 4 4
4
4
4
2.
1 1 1 1 1 1+1+1+1+1 5
2
+ + + + =
= or 1
3 3 3 3 3
3
3
3
3.
3 3 3 3 3 3 + 3 + 3 + 3 + 3 15
+ + + + =
=
=3
5 5 5 5 5
5
5
4.
5 5 5 5 5 + 5 + 5 + 5 20
4
1
+ + + =
=
= 2 or 2
8 8 8 8
8
8
8
2
Exercise 7.8
5
1.(a) × 30 = 5 × 5 = 25
6
2
(c) × 18 = 2 × 2 = 4
9
3
3 × 7 21
(e) × 7 =
= or 2
8
8
8
4
4 × 9 36
(g) × 9 =
= or 5
7
7
7
6
× 32 = 6 × 4 = 24
8
1
(d) × 40 = 5
8
5
2
16
2
(f)
× 8 = or 2
8
7
7
7
1
6
6 × 2 12
2
(h)
×2=
= or 2
7
5
5
5
5
(b)
2. (a) 1 week = 7 days
1
1 1
∴ of a week =  ×  days = 1 day
7
7 7
TH—Easy Maths—5
57
(b) 1 day = 24 hours, 1 hour = 60 minute, 1 minute = 60 second
Now,
1
1
of a day =
× 24 hours = 1 hour
24
24
1
1
of 1 hour =
× 60 min = 1 min.
60
60
1
1
of a min =
× 60 s = 1 s
60
60
1
1
1
Hence, of of of a day = 1 s
60 60 24
(c) Denominator = 14 Numerator = 1
1
Hence, fraction =
14
(d) Numerator = 5
Denominator = ?
Since the fractions is equivalent to
Hence, denominator = 70
(e) Numerator = 1 Denominator = 4 × 1 = 4
1
Hence, fraction is
4
1
1×5
5
∴
=
14 14 × 5 70
Exercise 7.9
2
3
2
3
1.(a) of 15 = × 15 = 2 × 5 = 10 (b) of 12 =
3
4
3
4
2
4
6
(c) of 10 = × 10 = 4 × 2 = 8 (d) of 21 =
5
7
5
1
2
2 1 2×1 2
1
3
4
2.(a) × =
= or (b) ×
5 6 5 × 6 30
15
6
9
3
(c)
(e)
3
(d)
1
6
3 12
×
11 14 7
=
3 × 6 18
=
11 × 7 77
11 × 1 11
2
11 5
×
=
= or 3
1×3
3
3
51 3
1
26 7 × 26 182
2
(f) 7 × 5 = 7 ×
=
=
or 36
5
5
5
5
5
23 5 × 23 115
5
5
1
(g) 15 × 3 = 15 ×
=
=
or 57
6
2
2
2
62
(h) 12 × 3
(i)2
58 2
3×2
6
10
×
=
= 6
1×1
51
21
3
3
3
× 12 = 3 × 3 = 9
4
6
3
× 21 = 6 × 3 = 18
7
1×2 2
=
=
3×3 9
3 15
3
= 12 ×
= 3 × 15 = 45
4
4
1 7 7 7 7 × 7 49
4
× = × =
= or 5
3 3 3 3 3×3 9
9
TH—Easy Maths—5
3
2 7
3 × 7 21
1
27
7
× =
×
=
= or 4
5 9
5
×
1
5
5
5
91
(j)5
(k)4
(l)10
(m)4
13 21
2
5
1 21
×
×2 =4 ×
⇒
6
8
3 8
31
8
(n)3
5
3
8 × 7 56
2
32
7
×
×1 ⇒
=
= or 6
9
4
9
9
9
9
4
(o)6
2
3
8 × 7 56
1
32
7
×
×1 ⇒
=
= or 11
5
4
5
5
5
41 5 × 1
3 1 23 1 23 × 1 23
3
× =
× =
= or 2
5 2 5 2 5 × 2 10
10
4
32
7
2 7
4 × 7 28
1
×
× =
=
= or 9
3 8
3×1 3
3
3
81
=
13 × 7 91
3
= or 11
1×8
8
8
8
8
3. (a) Perimeter of park = 2
7
1
7
km or km.
3
3
Number of rounds taken = 3
7
1
Total distance run by the boy =  × 3  = 7 km
3


(b) Earning in a month = `10,000
1
Fraction spent on house rent =
5
1
∴ Money spent on house rent= of `10,000
5
1
× 10 , 000 2000
=
5
= `2,000
Fraction of earning spent on personal expenses =
Money spent on personal expenses =
=
(c) Total students = 50
1
Fraction of girls =
5
1
of `10,000
2
1
2
1
× 10 , 000 5000 = `5,000
2
1
of 50
5
1
10
× 50 = 10 girls students
=
5
Number of girl students=
TH—Easy Maths—5
59
(d) Number of bags received = 112
Fraction of bags containing vegetables =
1
2
1
∴
Number of vegetables bags= of 112
2
1
56
× 112
=
2
= 56 bags
1
Fraction of potato bags =
7
1
1
8
× 56 = 8 bags
Number of potato bags = of 56 =
7
7
(e) Total number of pages = 250
∴ Number of pages read =
=
= 50 pages
Fraction of book read =
1
5
1
of 250
5
1
50
× 250
5
Number of pages left = (250 – 50) pages = 200 pages
8
8
of a kg= kg
9
9
1
Fraction of chocolates Jessica ate=
3
1
8
1 8 8
∴
Amount of ate= of kg = × =
3
9
3 9 27
(f) Chocolates brought =
8 8
–
kg
9 27
24 8
=
–
kg
27 27
Amount of chocolates left=
(Making denominator same)
16
kg of chocolate in left with Jessica.
27
Exercise 7.10
=
9
15
5
6
29
13
1.(a) (b)
or (c)
(d)
(e)
4
12
4
5
13
9
3
5
5
5
(f)14
(g) (h)
(i)
(j)
25
49
22
36
60 TH—Easy Maths—5
1 7
7 3
= Hence Multiplicative Inverse (MI) of is .
3 3
3 7
1
1 3
3 2
(b) 6 MI of 6 is (c)1 = Hence, MI of is .
6
2 2
2 3
1
9
4
(d) 100 MI of 100 is
.(e)
MI is .
100
4
9
2
8
14 28
2
2
28 3
×
=
(f)2 × 3 =
∴ MI of is .
3
4
3 28
3
4
3
2
1 29 13 377
377
9
(g)9 × 4 =
×
=
Hence, MI of
is
.
3
3 3
3
9
9
377
3
9
11 33
1
2
33 2
× =
(h)4 × 3 =
Hence, MI of is .
2
3
2 33
2
3
2
2. (a)2
Exercise 7.11
1. (a) 1 –
1 1
– =0
2 2
1
We have successively subtracted from 1 and 2 times
2
1
∴ 1 ÷ = 2 ⇒ 1 × 2 = 2
2
1 2
2 1 1
1 1
(b) 1 – = …1
– = …2
– = 0 …3
3 3
3 3 3
3 3
1
∴ 1 ÷ = 3 or 1 × 3 = 3
3
1
1
1 1
(c) 2 – = 1 …1
1 – = 1 …2
2
2
2 2
1 1
1 1
1 – = …3
– = 0 …4
2 2
2 2
1
Hence 2 ÷ = 4 or 2 × 2 = 4
2
(d),(e) and (f) are done in the same manner.
2. (a) 4 ÷
1 1
= 2 2
1
as done is Question 1
4
Rest of the parts are done in the same manner.
Exercise 7.12
1
1 1 1×1 1
1
1.(a) ÷ 3 = × =
= ( reciprocal of 3 is )
4
2 3 2×3 6
3
2
4
1
4
×
=
(b) ÷ 6 =
8
8
63
9
9
1
9×1
9
(c) ÷ 19 = ×
=
=
10
10 19 10 ×19 190
TH—Easy Maths—5
1
2 ×1
1
=
12
×
8
3
4
61
5
= 3 × 5 = 15
4
10 12 24 × 13
12 × 13 156
1
(e) 24 ÷ =
=
=
or 31
13
5
5
5
10 5
(d) 12 ÷
4
=
5
(f) 36 ÷
8
=
15
27 81
(g) ÷
=
55 85
(h)5
(j)4
3
12 ×
9
36 ×
15
9 × 15 135
1
=
=
or 67
2
2
2
82
17
3
27
85
3 × 17 51
17
×
=
= or 11 × 9 99
33
81 9
11 55
1
51 1
51
÷ 16 =
×
=
10
10 16 160
(i)3
2
11 1 11
÷6=
× =
3
3 6 18
3
24 15
4 7
72
2
×
÷ =
=
= 10
5 15
7
7
5
7
3
9
7
1 6
3 × 7 21
5
×
(k)2 ÷ =
=
= or 2
4 7
4×2 8
8
4
62
3 3
÷
=
4 12
3
17
51 12
×
= 17 × 3 = 51
31
14
(l)12
(m)4
2
3 18 8
36
1
×
÷4 =
= or 1
4
8
35
4 35 35
(n)2
3
5
÷5 =
16
8
(o)17
2
3
5
÷4
=
7
14
2. (a) Cost of 6
1
7
7
35
8
=
×
18
45 9
2 16
2
2
122 14
=2×2=4
×
61 1
17
2
2
374
kg of rice = `124 or
3
3
3
 187 374
3 
374
2
×

÷6
= ` 
3
3
3
20
10 

187
7
= ` or 18
10
10
(b) 1 = 100 paise ∴ `3.50 = 3.50 paise `10 = 1000 paise
62 Cost of 1 kg rice= `
Hence, the fraction is
7
7
350
= .
1000 20 20
TH—Easy Maths—5
4
(c) of the distance covered in = 8 min.
9
4
min
9
9
= 2 8 × = 18 min
4
3

9 
 × 14 = 3 × 14 = 14 or 4 2
3
9
9  3 × 108
(d) ÷
÷
=
3
9 3  9
93
3
12 108 14  12


24
(e) Sweets distributed to 3 children =
9
8
24
8
24 1
Each child will get =
÷ 3 sweets =
= sweets
×
9
9
9
3
Total time taken to reach the house= 8 ÷
(f) Total sugar = 105 kg
Sugar in 1 packet =
Number of packets = 105 ÷ 15 = 7 packets
1
15
1
× 105 = 15kg
of 105 =
7
7
Chapter 8: Decimals
Exercise 8.1
1.–8. Refer answers at the end of book.
Exercise 8.2
1. (a) 3.5 = 3 +
1
5
1
1
= 3 + or 3
2
2
10 2
17
17
34
= 65 50
100 50
21
21
or 3
100
100
(b) 65.34 = 65
(d) 0.4 = 0 +
(f) Rest of the parts can be done in the same manner.
2
2
4
= 10 5 5
(c) 3.21 = 3 +
(e) 7.101 = 7 +
101
101
or 7
1000
1000
3
3
3
=7+
= 7 + 0.3 = 7.3 (b) 12
= 12 + 0.003 = 12.003
10
10
1000
2
7
(c)7
= 7 + 0.02 = 7.02
(d)
= 0.07
100
100
4
(e)2
= 2 + 0.0004 = 2.0004
10000
1
1×5
5
(f)17 = 17 +
= 17 +
= 17 + 0.5 = 17.5
2
2×5
10
2. (a)7
TH—Easy Maths—5
63
4 4×2 8
(g) =
=
= 0.8
5 5 × 2 10
3
3 × 25
75
(h)2 = 2 +
=2+
= 2.75
4
4 × 25
100
8
8×8
64
13 13 × 5 65
(i)
=
=
= 0.064 (j)
=
=
= 6.5
125 125 × 8 1000
2
2 × 5 10
21 21 × 4 84
17 17 × 4 68
(k) =
=
= 0.84
(l)
=
=
= .68
25 25 × 4 100
25 25 × 4 100
1
1×4
4
(m)3 = 3 +
=3+
= 3 + 0.04 = 3.04
25
25 × 4
100
99
99 × 8
792
1
1 × 25
(n)
=
=
= 0.792 (o) 11 = 11 +
= 11.25
125 125 × 8 1000
4
4 × 25
14 14 × 4 56
2
2×2
4
(p) =
=
= 0.56
(q) 7 = 7 +
=7+
= 7.4
25 25 × 4 100
5
5×2
10
Exercise 8.3
1. 10 mm = 1 cm
To convert ‘mm’ into ‘cm’ we divide the given mm by 10.
148
49
(a) 148 mm =
cm or 14.8 cm (b) 49 mm =
cm or 4.9 cm
10
10
4
(c) 6 cm 4 mm = 6 cm + cm = (6 + 0.4) cm or 6.4 cm
10
2
(d) 92 cm 2 mm = 92 +
cm = (92 + 0.2) cm = 92.2 cm
10
224
(e) 224 mm =
cm = 22.4 cm
10
2. 100 cm = 1 m
To convert ‘cm’ into ‘m’ we divide the number by 100.
845
(a) 845 cm =
m = 8.45 m
100
36
(b) 755 m 36 cm = 755 +
m = (755 + 0.36) m = 755.36 m
100
1468
(c) 1468 cm =
m = 14.68 m
100
12
(d) 983 m 12 cm = 983 +
m = 983 + 0.12 = 983.12 m
100
(e) 1245 cm =
1245
m = 12.45 m
100
3. 1000 gm = 1kg
To convert ‘gm’ into ‘kg’ we divide it by 1000.
64 (a) 25 kg 15 g = 25 +
15
kg = 25.015 kg
1000
TH—Easy Maths—5
(b) 7 kg 125 g = 7 +
(c) 8145 gm =
125
kg = 7 + 0.125 = 7.125 kg
1000
8145
kg = 8.145 kg
1000
15005
(d) 15005 gm =
kg = 15.005 kg
1000
350
(e) 8 kg 350 g = 8 +
kg = 8.350 kg
1000
4. 100 paise = 1 rupee
To convert paise into rupees we divide by 100.
25
= `(5 + 0.25) = `5.25
100
(a) 5 rupees 25 paise = ` 5 +
(b) 20 rupees 80 paise = ` 20 +
(c) 7575 paise = `
80
= `(20 + 0.80) = `20.80
100
7575
9003
= `75.75 (d) 9003 paise = `
= `90.03
100
100
5
(e) 65 rupees 5 paise = ` 65 +
= `(65 + 0.50) = `65.05
100
5. 1000 m = 1 km
2750
km = 2.750 km
1000
5222
(b) 5222 m =
km = 5.222 km
1000
92
(c) 84 km 92 m = 84 +
km = (84 + 0.092) km = 84.092 km
1000
795
(d) 795 m =
km = 0.795 km
1000
328
(e) 7 km 328 m = 7 +
km = (7 + 0.328) km = 7.328 km
1000
(a) 2750 m =
6. 1000 mL = 1 L
To convert mL in L we divide it by 1000.
(a) 8L 200 mL = 8 +
(b) 10500 mL =
200
L = (8 + 0.200) mL = 8.200 mL
1000
10500
L = 10.500 L
1000
275
(c) 12 L 275 mL = 12 +
L = (12 + 0.275) L = 12.275 L
1000
50
(d) 15 L 50 mL = 15 +
mL = (15 + 0.050) mL = 15.050 mL
1000
TH—Easy Maths—5
65
(e) 17030 mL =
17030
L = 17.030 L
1000
7.(a)`17.25 = 17 rupees 25 paise (b) 9.050 L = 9L 50 mL
(c) 15.375 km = 15 km 375 m
(d) `56.35 = 56 rupees 35 paise
(e) 21.300 L = 21 L 300 ml
(f) 12.950 km = 12 km 950 m
(g) 17.5 cm = 17 cm 5 mm
(h) 25.6 cm = 25 cm 6 mm
(i) 14.5 m = 14 m 50 cm
(j) 8.15 m = 8 m 15 cm
(k) 35.08 m = 35 m 8 cm
(l) 11.04 kg = 11 kg 40 g
(m)7.100 kg = 7 kg 100 g
Exercise 8.4
1. Decimals having equal number of decimal places are called like
decimals.
(a) (3.45, 141.01); (18.3, 39.9)
(b) (12.15, 3.75); (144.632, 0.149)
(c) (8.43, 119.87); (16.009, 8.114) (d) (115.125, 5.689); (8.5, 195.8)
2. In order to convert a group of decimals into like decimals we make the
number of digits to the right of decimal point in all of them same by
adding required number of zeros. This is done as adding any number
of zeros after the extreme right digit in the decimal number does not
make any change to its value.
(a) Since 11.738 has maximum 3 decimal places
∴ the required like decimals are: 11.500, 11.738, 512.510
(b) Since 39.118 has maximum 3 decimal places
∴ the required like decimals are: 745.090, 39.118, 12.800
(c) Since 0.489 has maximum 3 decimal places
∴ the required like decimals are: 81.500, 394.260, 0.489
(d) The required like decimals are: 346.620, 439.100, 30.623
3. Since adding any number of zeros after the extreme right digit in the
decimal number does not make any change to its value
∴ the equivalent fractions are:
(a) 6.7 = 6.700
(b) 10.01 = 10.010
(c) 144.89 = 144.890
4. (a)6.5 → The equivalent fractions are 6.50, 6.500
(b)11.80 → The equivalent fractions are 11.800, 11.8000
(c)91.700→ The equivalent fractions are 91.7000, 91.70000
(d)51.4 → The equivalent fractions are 51.40, 51.400
Exercise 8.5
1. (a) 0.76
(lowest digit in ones place)
(b) 1.468
(1 being the lowest whole number)
(c) 189.26
(lowest digit in tens place)
(d) 49.8723 (49 being the lowest whole number)
2. (a) Lowest 4.350 (4 being the lowest whole number)
Highest 453.0 (greatest digit in tens place)
(b) Lowest 146.267
Highest 1764.26
66 TH—Easy Maths—5
(c) Lowest 271.414
(minimum digit on left side of decimal)
Highest 271414.4
(maximum digits on right side of decimal)
(d) Lowest 46.24781
(same reason as above)
Highest 4781.426
3. (a) 37.17 < 37.7
(digit at tenth place is greater)
(b) 144.2 > 87.489 (number of digit more on right side of decimal)
(c) 75.5 = 75.500
(equivalent fraction)
(d) 8.345 < 80.45
(reason same as in (b))
(e) 78.190 < 87.91 (reason same as in (a))
(f) 197.1 > 97.95
(reason same as in (b))
4. By comparing the digits one by one, the ascending order will be
(a) 44.76 < 46.47 < 46.74 < 47.64 (b) 8.06 < 8.60 < 80.6 < 86.0
(c) 9.067 < 90.67 < 96.07 < 97.60 (d) 52.67 < 52.76 < 56.72 < 57.62
5. By comparing the digits one by one the descending order will be:
(a) 590.35 > 559.03 > 553.09 > 550.93 (b) 60.68 > 60.08 > 6.80 > 6.008
(c) 461.23 > 46.123 > 42.163 > 41.623
(d) 554.04 > 54.40 > 50.44 > 5.440
Exercise 8.6
1. If the digit in hundredths place is 5 or more than 5 then the digit in the
tenth place is increased by 1 and all the digits to its right are replaced
by ‘0’ or removed.
If the digits in hundredths place is less than 5, then the digit in tenth
place remain unchanged and all the digits after it are replaced by ‘0’
or removed.
(a) 17.2377 = 17.2000 (3 is less than 5)
(b) 5.645 = 5.600
(4 is less than 5)
(c) 14.5112 = 14.5000 (1 is less than 5)
(d) 13.61 = 13.60
(1 is less than 5)
(e) 3.663 = 3.700
(6 is greater than 5)
(f) 19.80 = 19.80
(0 is greater than 5)
(g) 0.667 = 0.700
(6 is greater than 5)
(h) 18.0695 = 18.1000 (6 is greater than 5)
2. (a)11.786 = 11.790(The digit in thousandths place, 6 is greater
than 5)
(b) 15.382 = 15.380
(2 is less than 5)
(c) 2.9968 = 3.00
(6 is greater than 5)
(d) 13.1234 = 13.1200 (3 is less than 5)
(e) 4.1773 = 4.180
(7 is greater than 5)
(f) 5.716 = 5.720
(6 is greater than 5)
(g) 17.8946 = 17.8900 (4 is less than 5)
(h) 19.490 = 19.49
(0 in thousandths place)
3. (a)2.5168 = 2.52
(6 is greater than 5)
TH—Easy Maths—5
67
(b)617.906286 = 617.90
(0 is less than 5)
(c)84.15 = 80
(4 is less than 5)
(d)98.842 = 98.8
(2 is less than 5)
(e)741.45340 = 741.453
(4 is less than 5)
(f)6.570104 = 7
(5 at tenth place)
(g)134123 = 1.3412
(3 is less than 5)
(h)211.26 = 211.3
(as 6 > 5)
4. Rounding off 7.6178 m to
(a) 1 decimal place = 7.6 m
(1 is less than 5)
(b) two decimal places = 7.62 m (7 is greater than 5)
(c) three decimal places = 7.618 m (8 is greater than 5)
(d) the nearest metres = 8 m (6 is greater than 5)
5. Rounding of 8.9463 kg to
(a) one decimal place = 8.9 kg
(b)two decimal places = 8.95 kg
(c) three decimal places = 8.946 kg (d)the nearest kg = 9 kg
Exercise 8.7
1. Write or arrange the numbers with decimal points below one
another
(a) 4 2 . 8 6 (b)
1 5 7 . 2 4 1 (c)
8 1 2 . 1 6 0
+ 2 3.4 2 0
+ 2 4 7.2 7 1
+ 2 0.1 3
1 8 0.6 6 1
1 0 5 9.4 3 1
6 2.9 9
(d) 3 2 2 . 1 2 0 (e)2 1 6 4 . 1 2 (f)
2 1 6.6 0 0
+ 4 6.4 0 1
1 4.0 0
0.0 0 4
3 6 8.5 2 1
+ 1 2 3.7 1
+
4.6 8 1
2 3 0 1.8 3
2 2 1.2 8 5
2. Numbers are put in columns with the larger number on top and the
decimal point.
Underneath are another empty place values are filled with zeroes so
that all the number have same number of decimal places.
1 2 . 9 9 3
(a) 2 7 . 5 8 3 (b)
3 7 . 0101 01(c)
–
2.2 8 0
–
0
.
0
3
1
– 0.2 2 0
1 0.7 1 3
3 6.9 6 9
2 7.3 6 3
1 1 5 . 5 5
(d) 1 2 . 1 7 0 (e)
1 . 2 5 (f)
– 1 1 1.6 2
– 0.1 0
– 1.3 8 7
0 0 3.9 3
1.1 5
1 0.7 8 3
Exercise 8.8
1. When a decimal is multiplied by 10, the decimal point is shifted by 1
place to the right.
When a decimal is multiplied by 100, the decimal point is shifted by
2 places to the right.
(a) 39.145 × 10 = 391.45
(b) 82.483 × 100 = 8248.3
68 TH—Easy Maths—5
(c) 436.385 × 1000 = 436385.0
(d) 114.01 × 100 = 11401.0
(e) 345.19 × 1000 = 345190(since the number of decimal place is
only 3 we add a zero)
(f) 17.5 × 400 = 17.5 × 4 × 100 = 70 × 100 = 7000
(g) 8.814 × 500 = 8.814 × 5 × 100 = 44.07 × 100 = 4407
(h) 19.365 × 2000 = 19.365 × 2 × 1000 = 38.730 × 1000 = 38730
2.(a) 2 1 1 2
×
4
8448
The number of digits after the decimal point in
the multiplicand is 2
∴ 21.12 × 4 = 84.48
(b)
×
The number of digits after decimal point in
the multiplicand = 1
The multiplier = 1
Total number of digits after decimal point = 2
∴ 46.7 × 43.8 = 2045.46
(c)
4 6 .7
43 8
373 6
1401 0
+18680 0
20454 6
×
+
(d)
×
+
1 2 .5 6
41
12 56
502 4×
514 96
∴ 12.56 × 41 = 514.96
26 72
12
53 44
267 20
320 64
∴ 26.72 × 1.2 = 32.064
(e) 6.789 × 0.41 ⇒
67 89
41
67 89
+2715 60
∴ 6.789
× 0.41 = 2.78349
2783 49
×
(f) 7.26 × 4
⇒
7 26
×
4
∴ 7.26
× 4 = 29.04
29 04
Exercise 8.9
1. • When a decimal is divided by 10, the decimal point move towards
the left by 1 place.
• When a decimal is divided by 100, the decimal point moves towards
the left by 2 places
• When a decimal is divided by 1000, the decimal point moves towards
the left by 3 places.
(a) 45.5 ÷ 10 = 4.55
(b) 395.6 ÷ 100 = 3.956
TH—Easy Maths—5
69
(c) 482.16 ÷ 1000 = 0.48216
(d) 39.145 ÷ 1000 = 0.039145
(A zero is placed before first digit if the number of digits is less than
the required number of decimal places.)
(e) 2.46 ÷ 30 = 2.46 ×
(f) 38.148 ÷ 50 =
(g) 4.86 ÷ 200 =
(h) 214.16 ÷ 500 =
2. (a) 18.0 ÷ 3 =
6
82
246
1
1
82
×
or =
= 0.082
30
100
30 10 1000
38148 1 2 38148 × 2
76296
× × =
=
= 0.76296
1000 50 2 1000 × 100 100000
243
486
1
243
×
=
= 0.0243
100
200 100 10000
21416
1
2 42832
×
× =
= 0.42832
100
500 2 1000
18 .0 ×
1
=6
3
2832
5664 1 2832
× = 100 = 28.32
100
2
(b) 56.64 ÷ 2 =
(c) 40.96 ÷ 16 =
(d) 188.28 ÷ 12 =
(e) 564 ÷ 4.8 = 564 ÷
(f) 10.15 ÷ 1.45 =
(g) 6.300 ÷ 2.10 =
63 0 0 1 0 0
63
×
=
=3
1 0 0 0 21 0
21 1
(h) 3375 ÷ 0.125 =
27
3375 1000
3375
125
27
×
÷
=
= or 2.7
10000 1000
10000
125 1 10
256
256
4096
1
=
or 2.56
×
100
16 1 100
1569
1569
18828
1
=
= 15.69
×
100
12 1 100
7
48
=
10
141
1015 ×
564 ×
10
141 × 10 1410
=
=
= 117.5
12
12
48 12
1
=7
145 1
3
70 TH—Easy Maths—5
3. (a) 0.85 ÷ 0.09 or 9 85 9.44
81
40
–36
40
36
4
(c) 6.9 ÷ 0.07 =
0.85
85
or 0.09
9
6.9 × 100 690
=
0.07 × 100
7
Now 7 690 98.571
–63
60
–56
40
–35
50
–49
10
–7
3
Rounding off the quotient
= 98.57
5.25
525
(e) 5.25 ÷ 1.78 =
or 1.78
178
(b) 0.99 ÷ 14
14 0.99 0.0707
–98
100
–98
2
Rounding off 0.0707 to two decimal
places = 0.07
4.62 × 10 46.2
=
2.5 × 10
25
Now 25 46.2 1.848
–25
212
–200
120
–100
200
–200
×
(d)4.62 ÷ 2.5 =
ounding off the quotient
R
= 1.85
39.44
3944
or
0.23
23
Now 23 3944 171.478
Now 178 525 2.949
–23
–356
164
1690
161
–1602
34
880
23
–712
110
1680
–92
–1602
180
78
–161
Rounding off the quotient
190
= 2.95
–184
6
Rounding off the quotient
= 171.48
(g) and (h) can be done in the same manner.
TH—Easy Maths—5
(f) 39.44 ÷ 0.23 =
71
Exercise 8.10
1. Total length of 3 pieces of wood is
2.4 6 m
1.3 5 m
+ 0.9 2 m
4.7 3 m
2. Total length of ribbon = 16.70 m
Length of the piece cut off from it = 8.45 m.
Length of the ribbon left = (16.70 – 8.45) m.
or 1 6.7 0
– 8.4 5
8.2 5 m
3.
4.
5.
6.
7.
8.
9.
10.
72 Length of each ribbon piece = 1.50 m.
Number of pieces = 27
Total length = (27 × 1.50) m = 40.50 m.
The rate at which man worked = `110.50 per hour
Number of hours worked = 30
Money earned = `(110.50 × 30) = `3315
Initial weight = 97.5 kg
Weight lost = 13.8 kg
Present weight = (97.5 – 13.8) kg = 83.7 kg
Distance covered in 16 litres = 157.6 km
Distance covered in 1 litre = (15.76 ÷ 16) km = 9.85 km
Cost of 12.5 kg of sugar = `291.25
Cost of 1 kg of sugar = `(291.25 ÷ 12.5) = `23.30
Cost of 3.5 kg of sugar = `(23.30 × 3.5) = `81.55
Money spent on ice cream = `12.06
Money spent on snacks = `63.24
Money spent on movie ticket = `90.90
Total money spent = `(12.06 + 63.24 + 90.90) = `166.2
Rate of cloth = `78.35 per metres
Number of metres purchased = 23
Total amount spent = `(78.35 × 23) = `1802.05
Total cloth = 18 m
Cloth required for 1 suit = 2.5 m
Number of suit that can be made from 18 m cloth
18 × 10
= 18 ÷ 2.5 or Now 25 180 7
2.5 × 10
–175
180
=
5
25
Number of suits that can be made is 7; cloth left will be 0.50 m.
TH—Easy Maths—5
Chapter 9: Simplification of Numbers
1. To simplify we use the DMAS rule
(a) 24 ÷ 2 + 5 = 12 + 5 = 17
(b) 6 + 15 ÷ 3 = 6 + 5 = 11
(c) 6 + 2 × 3 = 6 + 6 = 12
(d) 12 – 4 × 3 = 12 – 12 = 0
(e) 2 × 6 ÷ 3 = 12 ÷ 3 = 4
(f) 6 ÷ 3 × 5 = 2 × 5 = 10
(g) 33 ÷ 3 – 11 = 11 – 11 = 0 (h) 39 ÷ 13 + 10 = 3 + 10 = 13
(i) 42 × 6 ÷ 3 = 252 ÷ 3 = 84
(j) 63 – 3 × 21 = 63 – 63 = 0
(k) 100 – 40 ÷ 5 = 100 – 8 = 92
(l) 11 – 6 ÷ 2 × 3 = 11 – 3 × 3 = 11 – 9 = 2
2. We follow DMAS rule
Here multiplication and division rank equally and addition and
subtraction also rank equally.
(a) 30 ÷ 3 + 6 × 3 – 10 = 10 + 18 – 10 = 28 – 10 = 18
(b) 78 – 28 ÷ 7 + 4 = 78 – 4 + 4 = 70
(c) 18 × 20 ÷ 4 – 55 = 18 × 5 – 55 = 90 – 55 = 35
(d) 14 – 21 ÷ 7 + 4 × 2 = 14 – 3 + 8 = 11 + 8 = 19
(e) 82 – 42 ÷ 21 × 2 = 82 – 2 × 2 = 82 – 4 = 78
(f) 20 × 5 + 15 × 4 – 130 = 100 + 60 – 130 = 160 – 130 = 30
3. (a) (9.7 × 3) + (3 × 4.1 × 1.5) ⇒ (29.1) + 18.45 = 47.55
(b) (7 + 3.2 + 5.8) + (3 × 2.8) ⇒ 16 + 8.4 = 24.4
(c) (19.2 – 4.6) + (2.1 × 1) + 6.8 ⇒ 14.6 + 2.1 + 6.8 = 23.5
(d) 8.7 + 4 + 5.1 = 12.7 + 5.1 = 17.8
(e) (25.4 – 3.3) + (2.9 × 4.7) ⇒ 22.1 + 13.63 = 35.73
(f) 17.2 – 2.9 + 1 × 4.1 + 2.1 = 14.3 + 4.1 + 2.1 = 20.5
2
10 3 1
1
2 1
× ×
4. (a)3 ÷ 1 × =
=1
3
3 2
3
5 2
2
1
÷ 1 ⇒ 10 +
3
6
2
5
35 6
⇒ 10 + 10 = 20
×
3
7
3
7
14
3
7
45 31 1
(c) 1 + × 1
×4 ÷3⇒1+
×
×
×
=1+1=2
15
31
7
15
31
7
3
2
2
2
10 14
16 4
11
3
3
7
3 11
×
−
×
×
(d)1 ÷ 3
– 1 ÷ 2 × ⇒
7
14
9
4 32
7
45 9
9
11 32 2
(b) 10 + 11
(e)4
⇒
4 2 4–2 2
– =
=
9 9
9
9
1
1
1
9
1 3
9
2+9
÷
+1
⇒ ÷
+
= ÷
2
3
2
2
3 2
2
6
3
9 11
27
5
9 6
÷ or
×
=
or 2
2 6
11
2 11 11
TH—Easy Maths—5
73
4
(f)9
1
3
7
2
2
×2 ×1 ÷2
–1 ⇒
3
7
9
27
7
28 17 16 56 9
×
÷
–
×
9 27 7
3
7
1
9
4
16 9 27 9
8
−
× 17 × ×
− = 17 × 16 ×
=
14
7
3
9 14 56 7
7
=
136 9 136 – 9 127
1
– =
=
= 18
7
7
7
7
7
Exercise 9.2
1. To simplify we follow BODMAS rule.
(a) {57 – (11 + 2) } × 2 = {57 – 13 } × 2 = 44 × 2 = 88
(b) [100 – {90 – (20 × 3) ÷ 2 }]= [100 – {90 – 60 ÷ 2}]
= [100 – {90 – 30}] = 100 – 60 = 40
(c) (35 × 2) ÷ 7 – 3 = 70 ÷ 7 – 3 = 10 – 3 = 7
(d) (40 + 5 × 3) ÷ 11 × 5 = (40 + 15) ÷ 11 × 5 = 55 ÷ 11 × 5 = 5 × 5 = 25
(e) 20 – {3 + (5 × 7 – 5) ÷ 6}= 20 – {3 + (35 – 5) ÷ 6}
= 20 – {3 + 30 ÷ 6} = 20 – {3 + 5} = 20 – 8 = 12
(f) 200 – [7 {8 + (10 × 7 – 25) ÷ 9}] ⇒
200 – [7 {8 + (70 – 25) ÷ 9}]
= 200 – [7 {8 + 45 ÷ 9}] ⇒
200 – [7 {8 + 5}]
= 200 – [7 {13}] = 200 – 91 = 109
(g) 100 + {25 × {40 – (20 ÷ 5)}] ⇒
100 + [25 × {40 – 4 }]
= 100 + [25 × 36] = 100 + 900 = 1000
(h) 3 × 2 + 6 – 4 – (4 – 0) = 6 + 6 – 4 – 4 = 12 – 8 = 4
(i) 57.5 + 1 – (36.4 – 12.52) = 58.5 – 23.88 = 34.62
(j) 5.9 + {(18.42 + 26.8) – 50.1}]
= 5.9 + {45.22 – 50.1} = 5.9 – 4.88 = 1.02
(k) 75 – {31 + 1.2 + 4.12)} = 75 – {31 + 5.32 } = 75 – 36.32 = 38.68
(l) (8.6 + 26 + 34.3) – 53.31 ⇒ 68.9 – 53.31 = 15.59
7
2 1
3 9
2  3 28  2 7 2 + 28
(m)
×
+
÷
⇒ +  ×
= + =
3 4
4 28
12  4
12
9 3  12 3
30
6
1
= or 2 or 2
12
12
2
74 2  4 5   2  4
2  4 3


(n) ×  ÷  + 5  = ×  ×  + 5  = ×  + 5 
3  5 3   3  3

3  5 5  
2 19 38
2
= ×
=
=4
3 3
9
9
1
1  1  21 7  
1  16  3 21
11  

×
(o) + 5 +  ÷   = +  + 
4  3  66 11   4  3  66 6 2
7 1 



1  16 1 
1  32 + 3  1 35 3 + 70 73
1
=
=
=6
= +  +  = + 
= +
4  3 2
4  6  4 6
12
12
12
TH—Easy Maths—5
2 
3
15  38 
43  63 
(p)12 +  16 − 10  ÷ 1  =
+  16 −  ÷ 
3 
4
48 
3 
4  48 
12

38  64 − 43  48  38  1 21 48 
+
×
+ 
=

÷ =
3  4  63  3  1 4
63 3 


38 12 50
2
+
=
or 16
=
3
3
3
3
2. (a) 72 ÷ (3 × 4) = 72 ÷ 12 = 6
(b) 72 – (7 × 9) = 72 – 63 = 9
(c) (12 ÷ 4) + 13 = 3 + 13 = 16
(d) (93 – 73) × (3 + 2) = 20 × 5 = 100
(e) 0.5 × [84 – (12 × 6)] ⇒ 0.5 × [84 – 72] = 0.5 × 12 = 6.0
(f) [5 {(12 × 6) – (35 + 25)}] + 13 ⇒ [5{72 – 60 }] +13
⇒ [5 × 12] + 13 = 60 + 13 = 73
Chapter 10: Percentage
Exercise 10.1
1. Refer answers at the end of book.
2. (a) Shade 25 boxes.
(b) Shade 44 boxes.
(c) Shade 70 boxes.
(d) Shade 66 boxes.
Exercise 10.2
1. To convert a fraction into percentage, we multiply it by 100 and put
the symbol ‘%’.
(a)1
(c)2
3
3
20
3
23
5
× 100 = 60%
=
× 100 = 115% (b) =
5 51
20 20 1
5 21
2100
1
= × 100 =
= 262 %
8 8
8
2
5
5
5 × 25 125
2
25
× 100 =
=
or 41 %
(d) =
12 12 3
3
3
3
1
2 17
20
7 =
=
× 100 = 340% (f)
4
5 5
13
2 12
20
(g)2 = × 100 = 240% (h) =
25
5 5
(e)3
(i)
18 18
5
=
× 100 = 90% 20 20
TH—Easy Maths—5
(j) 29
25
× 100 = 725%
41
13
4
× 100 = 52%
25 1
9
9
2
=
× 100 = 18%
50 50 1
75
Exercise 10.3
1. To convert a decimal into percentage, move the decimal point two
places to the right and attach the per cent sign.
(a) 5.36 = 536%
(b) 4.6 = 460%
(c) 2.14 = 214%
(d) 0.158 = 15.8%
(e) 0.7 = 70%
(f) 43.51 = 4351%
(g) 0.3712 = 37.12%
(h) 5.021 = 502.1%
(i) 0.037 = 3.7%
(j) 0.43 = 43%
11 × 5 55
=
= 0.55
20 × 5 100
Converting into percentage = 0.55 × 100 % = 55%
3
(b) = .75 (as decimal)
4
9
9
(c)Converting
as decimal we have
= 0.375
24
24
2. (a) Converting into decimal =
Converting 0.375 into percentage = 0.375 × 100% = 37.5%
37
37 × 4 148
(d) as decimal =
=
= 1.48
25
25 × 4 100
1.48 as percentage = 1.48 × 100% = 148%
7
(e) as decimal = 0.7
10
0.7 as percentage = 0.7 × 100% = 70%
8
8×8
64
(f)
as decimal =
=
= 0.064
125
125 × 8 1000
0.064 as percentage = 0.064 × 100% = 6.4%
57
(g) as decimal = 1.1875
48
1.1875 as percentage = 1.1875 × 100% = 118.75%
1
(h)
as decimal = 0.01
100
0.01 as percentage = 0.01 × 100 = 1%
Exercise 10.4
1. To convert percentage into fraction, we write 100 under the number.
(a)82% =
82
82 ÷ 2 41
=
⇒
100
100 ÷ 2 50
7
.6
56
7 (b)5.6% 5=
=
or
100
1000 125 125
76 TH—Easy Maths—5
9
100
(c) 9% =
(d)132% =
(e) 33% =
33
100
3
1
15
15
1
3
%=
×
=
(f) 7 % or
2
2
2
100 20 40
2
117
117
1
117
(g)23 % or
%=
×
=
5
5
5 100 500
2
50
50
1
1
×
=
%=
(h) 16 % or
3
3
3
100 2 6
0.5
5
1
=
or
100
1000 200
=
(i) 0.5%
11
2.2
22
11
=
or
100
1000 500 500
=
(j) 2.2%
132 132 ÷ 4 33
8
=
=
or 1
100 100 ÷ 4 25
25
2. To convert percentage into decimals we move the decimal point two
places to the left .
35
= 0.35
100
125
(c) 125% =
= 1.25
100
4.7
(e) 4.7% =
= 0.047
100
130
(g) 130% =
= 1.30
100
0.5
(i) 0.5% =
= 0.005
100
85
= 0.85
100
4.5
4.5% =
= 0.045
100
32.6
32.6% =
= 0.326
100
230.5
230.5% =
= 2.305
100
2.2
2.2% =
= 0.022
100
(a) 35% =
(b) 85% =
(d)
(f)
(h)
(j)
Exercise 10.5
1. (a) 0.8 = 0.80 =
80
1
= 80 ×
= 80%
100
100
(b) 0.34 =
34
1
= 34 ×
= 34%
100
100
(d) 2.54 =
254
1
= 254 ×
= 254%
100
100
(e) 3.745 =
(c) 0.79 =
79
1
= 79 ×
= 79%
100
100
374.5
1
= 374.5 ×
= 374.5%
100
100
TH—Easy Maths—5
77
kg =
(f) 7.001 =
700.1
1
= 700.1 ×
= 700.1%
.100
100
2. (a) 5% =
5
5÷5
21
=
= 100 100 ÷ 5 20
(b) 2% =
2
2÷2
1
=
=
100 100 ÷ 2 50
(c) 3% =
3
100
(d) 8% =
8
8÷4
2
=
=
100 100 ÷ 4 25
(e) 42% =
42
42 ÷ 2 21
=
= 100 100 ÷ 2 50
Exercise 10.6
(f) 36% =
36
36 ÷ 4
9
=
=
100 100 ÷ 4 25
1. (a) 35% of 100 =
35
35
× 100 = 35
of 100 =
100
100
(b) 50% of 800 =
50
8
50
× 800 = 400
of 800 =
100
100
(c) 90% of 450 =
90
90
× 450 = 405
of 450 =
100
100
(d) 20% of 800 =
20
8
20
× 800 = 160
of 800 =
100
100
(e) 15% of 750 =
15
15
× 750
of 750 =
100
100 4 2
(f) 75 % of 250 =
3
30
15
=
225
= 112.5
2
75
750
75
× 250 =
= 187.5
of 250 =
100
100 4
4
2. (a) 1 kg = 1000 gm
25
25
× 1000 gm = 250 gm
of 1000 g =
100
1 00
∴ 25% of 1kg =
(b) 55% of 70 m =
55
55
77
of 70 =
×70 =
m = 38.5 m
100
100 2
2
(c) 40% of 500 L =
40
40
of 500 L =
× 5 00 = 200 L
100
1 00
(d) 60% of `1200 =
60
60
of `1200 =
× 12 00 = ` 720
100
1 00
33
1
11 1
11
1
3
1
×
of 75 kg = ×
× 75 kg =
kg = 4 kg
(e)5 % of 75 kg =
2
2 100
2 1 00 4
8
8
11
33
11
1
1
3
kg = 4 kg
×
× 75 kg ==
8
8
2 1 00 4
1
6
1
(f)1 % of 200 g = ×
of 200 gm
5
5 100
78 TH—Easy Maths—5
=
6
1
12
2
×
× 200 g =
g = 2 gm
5 100
5
5
3. (a) 1 kg = 1000 gm ∴ 5 kg = 5000 gm
50
× 100 = 1%
5
0
00
1
(b) 1L = 1000 mL
∴ Converting 250 mL of 1L as a percentage
Now 1
250
× 10 0 = 25%
10 0 0
1
(c) Converting 12 kg of 250 kg as a percentage
2
1
25
1
5
×
× 100 = 5%
=
2
250 10
=
(d)`1 = 100 paise ∴ `2 = 200 paise
Now percentage of 10 paise of 200 paise
(e) 1 km = 1000 m ∴ 6 km = 6000 m
Now percentage of 300 m of 6000 m is
(f)`125 of `600
(g)
24
36
2
× 100 = 0.8% (h)
× 1 00 = 72%
50
30 00 5
(i)
420
350
× 1 00 = 70% (j)
× 100 = 70%
600
500
(k) 1 h = 60 min ∴ 2 h = 120 min
(l)
=
=
5
10
× 100 = 5%
200
300
60 00
1
20
5
× 100 = 5%
Percentage is
125
× 100
600
=
5
20 %
6
4
7
Now
24
36
4
2
63
70
10
25
1
25
= 8 %
× 100 =
12 0 3
3
3
× 100 =
TH—Easy Maths—5
200
2
= 66 %
3
3
79
Exercise 10.7
1. (a)In `1 there are 100 paise
55
100
55 paise out of 100 paise =
In % it is equal to =
(b)
(c) 1 m =100 cm
∴ 25 cm out of 100 cm =
(d) 1 m = 100 cm ∴ 2 m = 200 cm
and 10 m = 1000 cm
250 cm out of 1000 cm =
In percentage, it is equal to =
(f)`12.50 out of `100 =
3
12
× 100 = 30%
40
55
× 100 = 55%
100
25 1
=
100 4
1
25
In percentage, it is equal to × 100 = 25%
4
Hence, 25 cm is 25 % of 1 m.
1
250
1
=
100 0 4 4
1
In percentage it is equal to = × 100 = 25%
4
(e) 1 kg = 1000 gm ∴ 20 kg = 20,000 gm
and 7 kg 500 gm = 7500 gm
3
750
3
=
Now 7500 gm out of 20,000 gm is =
20 , 00 0 80 80
12.50
100
In percentage, it is equal to =
10
2. (a) 20% of 150 =
12.50
× 100 = 12.50%
100
20
3
× 15 0 = 30
10 0 2
30
× 180 = 54
10 0
Since 54 > 30
∴ 30% of 180 is greater than 20% by 150.
80 30% of 180 =
3
5 15
× 100 =
= 37.5%
80 4
4
TH—Easy Maths—5
15
× 300 = 45
100
12
× 200 = 24
12% of 200 =
100
Since 45 > 24 ∴ 15% of 300 is greater than 12% of 200.
(b) 15% of 300 =
3. Kartik’s pocket money = `500
 80
× 500
Amount he spends = 80% of `500 = ` 
 100
4. Max. marks in the examination = 540
Raghav scored = 75 % of 540
5
 = `400

3
=
75
135
× 540
= 405
100 4 1
5. Ajeet’s weight = 75 kg
His son’s weight = 60% of 75 =
6. Total fruits sold = 80 kg
15
60
3
× 75 = 45 kg
100 4

 30
× 80  kg = 24 kg
Apples sold = 30% of 80 kg = 


 100
7. Milk sold in a day = 2.5 L
Water mixed in the milk = 20% of 2.5 L
=
20
× 2.5 = 0.5 L or 500 mL
100 5
Chapter 11: Perimeter and Area
Exercise 11.1
1. (a) Perimeter= Sum of all sides
= (18 + 30 + 22 + 17) m = 87 m
(b) Perimeter of rectangle = 2 (L + B)
= 2 (1.2 + 0.8) m ( 80 cm = 0.8 m)
= 2 × 2 = 4 cm
(c) Perimeter of rectangle = 2 (L + B)
= 2 (15 + 6) cm = 2 × 21 cm = 42 cm
(d) Perimeter= 2 (L + B)
= 2 (8.5 + 7) cm = 2 × 15.5 cm = 31 cm
(e) Perimeter of square = 4 × side
= 4 × 29 m = 116 m
TH—Easy Maths—5
81
(f) Perimeter of rectangle = 2 (L+ B)
= 2 (75 + 85) m = 2 × 160 m = 320 m
(g) Perimeter of triangle = Sum of all three sides
= (16 + 18 + 15) cm = 49 cm
2. (a) Perimeter of regular hexagon = 6 × side
= (6 × 25) cm
= 150 cm
(b) Perimeter of flower bed= 2 (L + B)
= Length = 5 + 1 + 1 = 7 m
= Breadth = 4 + 1 + 1 = 6 m
∴ Perimeter
= 2 (7 + 6) m = 2 × 13 m = 26 m
(c) Perimeter of equilateral triangle = 3 × side
= 3 × 4 m = 12 m
(d) Perimeter of regular pentagon = 5 × side
= 5 × 3 inches = 15 inches
(e)Perimeter of square= 4 × side
Now 20 feet= 4 × side
or (20 ÷ 4) feet= side
5 feet= side
(f)Perimeter of hexagon= 6 × side
Side of hexagon= Perimeter ÷ 6
= (42 ÷ 6) cm = 7 cm
3. Let us first find out of the length of x × y
6 cm
y
Length of y = (13 – 6) m= 7 m Length of x = (8 – 5) m = 3 m
6 cm
Now, perimeter of plot
8m
= [8 + 13 + (5 + 3) + (6 + 7)] m
6m
= [8 + 13 + 8 + 13] m = 42 cm
Exercise 11.2
1. (a) No. of square in length= 5
No. of square in breadth= 5
Hence area= 5 × 5 = 25 sq cm
(b) Length = 9 cm
Breadth = 9 cm
∴ Area = 9 × 9 = 81 sq cm
(c) Length = 7 cm
Breadth = 7 cm
∴ Area = 7 × 7 = 49 sq cm
(d) Length = 3 cm
Breadth = 3 cm
∴ Area = 3 × 3 = 9 sq cm
82 x
13 m
TH—Easy Maths—5
Exercise 11.3
1. (a) The no. of full boxes= 6
The no. of half boxes= 4
1
× 4 cm2 = 6 + 2 = 8 cm2
2
∴ Area of the figure= 6 +
(b) The no. of full boxes= 20
The no. of more than half boxes = 3
The no. half boxes= 0
∴ Area of the figure= (20 + 3) cm2 = 23 cm2
(c) The no. of full boxes= 24
The no. of half boxes= 8
1
∴ Area of the figure= 24 + × 8 cm2 = 24 + 4 cm2
2
= 28 cm2
(d) The no. of full boxes
= 16
The no. of more than half boxes= 4
∴ Area of the figure
= (16 + 4) cm2
= 20 cm2
(e) The no. of full boxes
=8
The no. of more than half boxes= 7
∴ Area of the figure
= (8 + 7) cm2 = 15 cm2
(f) The no. of full boxes
= 12
The no. of more than half boxes= 6
The no. of half box
=1
∴ Area of the figure
= (12 + 6 + 0.5) cm2 = 18.5 cm2
Exercise 11.4
1. Area of square = side × side
(a) Area = 7 × 7 cm2 = 49 cm2( side = 7 cm)
(b) Area = 13 × 13 cm2 = 169 cm2 ( side = 13 cm)
(c) Area = 17 × 17 cm2 = 289 cm2 ( side = 17 cm)
(d) Side = 24 cm
∴ Area = (24 × 24) cm2 = 576 cm2
2. Area of the rectangle = Length × Breadth
(a) l = 7 cm, b = 4 cm
∴ Area = (7 × 4) cm2 = 28 cm2
(b) l = 12 cm, b = 9 cm
∴ Area = (12× 9) cm2 = 108 cm2
(c) l = 19 m, b = 14 m
∴ Area = (19 × 14) m2 = 266 m2
(d) l = 35 cm, b = 12 cm
∴ Area = (35 × 12) cm2 = 420 cm2
TH—Easy Maths—5
83
1
× base × height
2
(a) h = 14 cm, b = 8 cm
1
∴ Area =
× 14 × 8 cm2 = 56 cm2
2
(b) h = 16 cm, b = 9 cm
1
∴ Area =
× 16 × 9 cm = 72 cm2
2
(c) b = 23 m, h = 14 m
1
1
∴ Area = × b × h =
× 23 × 14 m2 = 161 m2
2
2
(d) b = 27 cm, h = 18 cm
1
∴ Area =
× 27 × 18 cm2 = 243 cm2
2
4. Area of a rectangle= Length × Breadth
Length of a rectangle= Area ÷ Breadth
Breadth of a rectangle= Area ÷ Length
Perimeter of a rectangle= 2 (L + B)
Now
(a)
Length= 18 cm2 ÷ 3 cm = 6 cm
Perimeter= 2 (3 + 6) cm = 2 × 9 cm = 18 cm
(b)
Area= (9 × 4) cm2 = 36 cm2
Perimeter= 2 (9 + 4) cm = 2 × 13 cm = 26 cm
(c) Breadth= (3600 ÷120) cm (as 1m 20 cm = 120 cm) = 30 cm
Perimeter= 2 (120 + 30) cm = 2 × 150 cm = 300 cm
(d)
Length= (75 ÷ 5) cm = 15 cm
Perimeter= 2 (15 + 5) cm = 2 × 20 cm = 40 cm
(e)
Area= (9 × 6) cm = 54 cm
Perimeter= 2 (9 + 6) cm = 2 ×15 cm = 30 cm
(f)
Area= (220 × 160) cm2 = (as 2m 20 cm = 220 cm;
1 m 60 cm = 160 cm )
= 352 cm2 = or 3.52 m2
Perimeter= 2 (220 + 160) cm = 2 × 380 cm = 760 cm or 7.6 m
1
5. (a) Area of triangle= × b × h
2
Base of a triangle= (2 × Area) ÷ Height
Height of a triangle= (2 × Area) ÷ Base
Now as Height= (2 × 36) ÷ 12 = (72 ÷ 12) cm = 6 cm
1
(b) Area=
× 80 × 140 cm2
(as 1m 40 cm = 140 cm)
2
2
= 5600 cm
(c) Base= [(2 × 48) ÷ 24] m = 4 m
3. Area of a triangle =
84 TH—Easy Maths—5
(d)Height = [(2 × 150) ÷ 25] cm = 12 cm
1
(e) Area =
× 40 × 25 cm2 = 500 cm2
2
(f) Base= [(2 × 12000) ÷ 600] cm
= 40 cm
1
6. (a) Area of triangle= × base × height
2
(b)
5 cm
(as 6 m = 600 cm)

1
=  × 1.5 × 2  cm2 = 1.5 cm2

2


I
15 cm
II
12 cm
8 cm
The given figure is divided into 2 rectangles
Area of I rectangle = length × breadth
Where length = 5 cm and breadth = (15 – 12) cm = 3 cm
So,
Area = (5 × 3) cm2= 15 cm2
Area of II rectangle with length = 12 cm and breadth = 8 cm
= (12 × 8) cm2 = 96 cm2
Total area = sum of areas of both rectangle
= (96 + 15) cm2 = 111 cm2
(c) Area of rectangle of length= 2 cm and breadth = 1.2 cm
= (2 × 1.2) cm2 = 2.4 cm2
(d) The given figure is divided into a square into a triangle
Area of the square of side 7 cm is
= (7 × 7) cm2 = 49 cm2
Area of the triangle of height = 7 cm and base = 6 cm
1
=
× 7 × 6 cm2 = 21 cm2
2
Total area = Area of rectangle + Area of the triangle
= (149 + 21) cm2 = 70 cm2
(e) The given figure is divided into a square into a square and a
triangle.
Area of the square of side 6 cm = (6 × 6) cm2 = 36 cm2
Area of a triangle of base= 6 cm and height = 4 cm
1
=
× 6 × 4 cm2 = 12 cm2
2
Total area = Area of square + Area of triangle
= (36 + 12) cm2 = 48 cm2
TH—Easy Maths—5
85
Exercise 11.5
1. Area of the rectangle = length × breadth
∴ Area of the room = (10 × 6) m2 = 60 m2
Cost of carpet to cover 1m2 = `35
Cost of carpet to cover 60 m2 = `(35 × 60) = `2100
∴ The cost of carpeting the room is `2100.
2. Area of the wall = (9 × 7)m2 = 63 m2
Cost of paint to cover = 1 m2 = `4.75
Cost of paint to cover 63 m2 = `(4.75 × 63) = `299.25
∴ The cost of painting the wall is `299.25
3. (a) Area of the path = Area of outer square – Area of inner square
Side of inner square = 13 m
Side of outer square = (13 + 1 + 1) m = 15 m
Now,
Area of path= (15 × 15) m2 – (13 × 13) m2
= (225 – 169) m2 = 56 m2
Cost of laying 1m2 of path = `15
Cost of laying 56 m2 of path = `(56 × 15) m2 = `840
4. Breadth of plot= Area ÷ Length
= (600 ÷ 40) km = 15 km
5. Area of rectangle of length 15 m and breadth = 7 m
= (15 × 7) m2 = 105 m2
2
Area of square of side 10 m = (10 × 10) m = 100 m2
Since 10 > 100; ∴ Area of rectangle is bigger.
6. (a) Area of path = Area of outer rectangle – Area of inner rectangle
Length of outer rectangle= (21 + 1 + 1) m =23 m
Breadth of outer rectangle= (19 + 1 + 1) m = 21 m
Area of outer rectangle= (23 × 21) m2 = 483 m2
Area of inner rectangle of length= 21 m = breadth = 19 m
= (21 × 19) m2 = 399 m2
Now area of path= (483 – 399) m2 = 84 m2
(b) Cost of laying 1 m2 = of path= `12
Cost of laying 84 m2 of path= `(84 × 12) = `1008
7. Perimeter of rectangle= 2 (L + B)
or length= (Perimeter ÷ 2) – Breadth
Now length of wildlife sanctuary of perimeter 450 km and breadth
90 km is
Length= [(450 ÷ 2) –90] km
= (225 –90) km = 135 km
Area of wild life sanctuary= (90 × 135) km2 = 12150 km2
86 TH—Easy Maths—5
Chapter 12: Volume
Exercise 12.1
1. (a) Top layer has 2 rows of 6 cubes
∴ volume = 2 × 6 × 1 = 12 cm3
Since there are only 1 such row
(b) Top layer has 2 rows of 6 cubes
∴ volume = 2 × 6 × 4 = 48 cm3
(c) Top layer has 2 row of 2 cubes
∴ volume = 2 × 2 × 2 = 8 cm3
(d) Top layer has 2 rows of 3 cubes
∴ volume = 2 × 3 × 2 = 12 cm3
(e) Top layer has 3 rows of 3 cubes
∴ volume = 3 × 3 × 2 = 18 cm3
(f) Top layer has 2 rows of 10 cubes
∴ volume = 2 × 10 × 3 = 60 cm3
(g) Top layer has 3 rows of 5 cubes
∴ volume = 3 × 5 × 1= 15 cm3
Since there are 4 such rows
Since there are 2 such rows
Since there are 2 such rows
Since there are 2 such rows
Since there are 3 such rows
Since there are only 1 such rows
Exercise 12.2
1. Volume of cuboid = Length × Breadth × Height
(a) Volume = (7 × 4 × 3) cm3 = 84 cm3
(b) Volume = (10 × 9 × 2) cm3 = 180 cm3
2. Volume of cube = side × side × side
(a) Volume = (5 × 5 × 5) cm3 = 125 cm3
(b) Volume = (7.5 × 7. 5 × 7. 5) cm3 = 421.85 cm3
(c) Volume = (11 × 11 ×11) cm3 = 1331 cm3
(d) Volume = (8.5 × 8.5 × 8.5) cm3 = 614.125 cm3
1
1
1
1
or 8 × 8 × 8
= 614 cm3
2
2
2
8
TH—Easy Maths—5
87
3. (a) Volume = (25 × 25 × 10) cm3 = 6250 cm3
(b) Volume = (20 × 15 × 10) cm3 = 3000 cm3
(c) Volume = (5.2 × 5.2 × 5.2) cm3 = 140.608 cm3 or 140.61 cm3
(d) Volume = (8 × 12 × 5) cm3 = 480 cm3
(e) Volume = (4 × 4 × 4) cm3 = 64 cm3
27
= 3 cm
3×3
(b) Volume = (8 × 5 × 2) cm3 = 80 cm3
1200
(c) Breadth =
= 8 cm
15 × 10
3872
(d) Length =
= 11 cm
22 × 16
756
(e) Length =
= 18 cm
14 × 3
(f) Volume = (28 × 15 × 18) cm3 = 7560 cm3
Refer answers at the end of book.
4. (a) Height =
5.
Exercise 12.3
2578
1.(a)
L = 2.578 L
1000
(b) 6525 cm3
2. (a) Volume = 8 m3; Length = 4 m; Height = 2 m
(b) Length = 25 m; Breadth = 15 m; Height = 10 m
(c) Length = 10 cm; Breadth = 6 cm; Height = 3 cm
Volume of 1 brick = (10 × 6 × 3) cm3 = 180 cm3
Volume of 250 bricks = (180 × 250) cm3 = 45000 cm3
(d) Base area = 125 cm2; Height = 25 cm
Volume = Base area × Height
Volume = (125 × 25) cm3 = 3125 cm 3
(e) Volume of cube of side 20 cm = (20 × 20 × 20) cm3 = 8000 cm3
(f) Depth or height =
88 Breadth =
Volume
8
=
= 1m
L×H
4×2
Volume = (25 × 15 × 10) m3 = 3750 m3
Capacity =
8000
L = 8L
1000
Volume
1176
=
= 14 cm
Length × Breadth 12 × 7
TH—Easy Maths—5
(g) Base area = 78 cm2; Height = 13 cm
(h) Volume of carbon having L = 30 cm, B = 15 cm and H = 20 cm
= (30 × 15 × 20) cm3 = 9000 cm3
Volume of book having L = 5 cm; B = 3 cm; H = 3 cm
= (3 × 3 × 5) cm3 = 45 cm3
Number of books that can fit the carbon
Volume = Base area × Height
= (78 × 13) cm3 = 1014 cm 3
(30 × 15 × 20) cm3
= 200
(3 × 3 × 5) cm3
(i) Volume of water in tank = (35× 30 × 15) cm3 = 15750 cm3
15750
Capacity
L = 15.750 L
1000
(j) Base area of container with side 8 cm
= (8 × 8) cm2 = 64 cm2
Volume
384 cm3
Height of container =
=
= 6 cm
Base area
64 cm2
=
Exercise 12.5
1.(a) Cube : Dice
(b) Cuboid : Matchbox, pencil box
(c) Cylinder = Pencil, Pillar
2. (a)Cube
—
(b)Cuboid
—
(c)Cylinder
—
TH—Easy Maths—5
89
Chapter 13: Bills
Exercise 13.1
1.(a)
Name of Customer: Chandan
Date
S. No
Item
Quantity Cost per unit (`) Total cost (`)
1 Sugar
5kg
15.00
75.00
2 Jam
6
36.00
216.00
3
Mustard oil
4
Rice
2. S. No.
1
2
3
4
5
Grand Total
1
2L
10 kg
28.00
14.00
30.00
Grand Total
300.00
605.00
Total cost (`)
200.00
1750.00
154.00
140.00
75.00
2319.00
3.(a)Rita’s bill
Banana 3 × 22= `66.00
Apples 5 × 40= `200.00
Mangoes 2 × 18= `36.00
Total = `302.00
(b) Kiran had to pay = 2 × 80 = `160.00
(c)Mona’s bill
Water melons = 4 ×20= `80.00
Litchies = 6 × 26= `156.00
Total= `236.00
Mona had to pay `236.00
(d)i. Ram’s bill
Name of the customer: Mr. Ram
Date
S. No Item
Qty
Cost per unit (`) Total cost (`)
1
Sugar 2 kg
20.00
40.00
2
Litchi 2.5 kg
26.00
65.00
3
Banana 3 dozens
22.00
66.00
Grand Total
171.00
90 TH—Easy Maths—5
ii.
Name of customer: Rohan
Date
S. No Item
Qty Cost per unit (`) Total cost (`)
1 Mango 6.5 kg
18.00
117.00
2 Apple 7.5 kg
40.00
300.00
3 Sugar 3.5 kg
20.00
70.00
Grand Total
487.00
iii.
(e)
Name of customer: Mr. Manmohan
Date
S. No
Item
Qty Cost per unit (`) Total cost (`)
1
Sugar
1 kg
20.00
20.00
2
Cherries
4.5 kg
80.00
360.00
3
Watermelon 0.5 kg
20.00
10.00
Grand total
390.00
Name of the customer: Ms Madhvi
Date
S. No.
Item
Quantity Rate per unit Total cost
1
Banana
2 dozens
22.00
44.00
2
Mangoes 2.5 kg
18.00
45.00
3
Chikoos 0.5 kg
35.00
17.50
4
Apples
1.5 kg
40.00
60.00
Grand Total
166.50
4.
Name of the customer: Mr. Raman
S. No.
Item
Date
Quantity Rate per unit Total cost
1
Rice
2 kg
34.20
68.40
2
3
Wheat 5.5 kg
22.90
125.95
Sugar
20.50
102.50
4
Pulses 3.5 kg
40.00
140.00
5 kg
Grand Total
436.85
Chapter 14: Profit and Loss
Exercise 14.1
1. Profit = S.P – C.P
Loss = C.P – S.P
(a) SP > CP ∴ Profit = `(900 – 800) = `100
(b) CP > SP ∴ Loss = `(18.00 –16.50) = `1.50
(c) CP > SP ∴ Loss = `(200 – 175) = `25
(d) SP > CP ∴ Profit = `(149.50 – 132.25) = `16.75
TH—Easy Maths—5
91
(e) SP > CP ∴ Profit = `(141.00 – 132.25) = `8.75
(f) CP > SP ∴ Loss = `(435.75 – 375.25) = `60.50
2. (a) Mona purchased scooty for = `30,257
Money spent on repairing it = `425
Hence CP of scooty = `(30,257 +425) = `30,682
SP of scooty = `30,500
Since CP > SP
∴ loss = ∴(30,682 – 30,500) = `182
(b) Total CP of 20 packets chocolate = `(20 × 35) = `700
SP of 20 packets = `1000
Since SP > CP ∴ Gain = `(1000 – 700) = `300
(c) Cost price of 20 dozen pastries at `160 per dozen
= `(20 × 160) = `3200
SP of pastries at `9.50 per pastry
= (9.50 × 12 × 20) = `2280
Since CP > SP ∴ loss = `(3200 – 2280)
= `920
(d) Cost of car = `5,40,000
Money spent on repairing = `20,000
∴ CP of car = `(5,40,000 + 20,000)
= `5,60,000
SP of car = `5,87,250
Since SP > CP ∴ Gain = `(5,87,250 – 5,60,000) = `27,250
(e) Cost prices of 10 tables = `(10 × 60) = `600
Cost price of 60 chairs = `(60 × 40) = `2400
Total CP of tables and chairs = `(2400 + 600) = `3000
Total chairs and tables = (10 + 60) = 70
SP of both tables and chairs (70) = `(70 × 50) = `3500
Since SP > CP ∴ Gain = `(3500 – 3000) = `500
Exercise 14.2
1. Profit = SP – CP
∴ CP = SP – Profit and SP = CP + Profit
As Loss = CP – SP
∴ CP = Loss + SP and SP = CP – Loss
(a) SP = CP + P ∴ SP = `(475.00 + 12.50) = `487.50
(b) SP = `(2235.25 + 30.75) = `2266
(c) SP = CP – Loss ∴ SP = `(1050 – 175) = `875
(d) SP = `(925.25 – 3.75) = `921.50
2. (a) CP = SP + Loss ∴ CP = `(475 + 35) = `510
(b) CP = `(785.25 + 2.75) = `788
(c) CP = `SP – Profit ∴ CP = `(7055.00 – 501.25) = `6553.75
(d) CP = `(435.65 – 12.75) = `422.90
92 TH—Easy Maths—5
3. (a) CP of table = `550
Profit = `120
∴ SP = CP + Profit
SP = `(550 + 120) = `670
(b) SP of bicycle = `4700 and P
rofit = `200
We know that CP = SP – Profit
∴ CP = `(4700 – 200) = `4500
(c) CP of TV = `9990 and Profit = `2450
Since SP = CP + Profit
∴ SP = `(9990 + 450) = `12440
(d) SP of car = `1,32,050 and Profit = 10,000
Since CP = SP – Profit
∴CP = `(1,32,050 – 10,000) = `1,22,050
(e) CP of watch = `2250 and Loss = `200
Since SP = CP – Loss
∴ SP = `(2250 – 200) = `2050
(f) SP of fan = `650 and Loss = `300
Since CP = SP + Loss
∴ CP = `(650 + 300) = `950
(g) SP of stereo = `5625 and Loss = `475
Since CP = SP + Loss
∴ CP = `(5625 + 475) = `6100
(h) CP of casio = `5000 and Loss = `1250
Since SP = CP – Loss
∴ SP = `(5000 – 1250) = `3750
Exercise 14.3
10
10
of `500 = ` × 500 = `50
100
100
20
20
(b) Profit = 20% of `750 or
of `750 = ` × 750 = `150
100
100
15
15
(c) Loss = 15% of `900 or
of `900 = ` × 900 = `135
100
100
30
30
(d) Loss = 30% of `1500 or
of `1500 = ` × 1500 = `450
100
100
1. (a) Profit = 10% of `500 or
2. (a) CP = `800; SP = `500; Loss = CP – SP
Loss = `(800 – 500) = `300
(b) CP = `800; SP = `850; Profit = `(850 – 800) = `50
Profit
50
Profit % =
× 100 =
× 100 = 6.25%
CP
800
Loss % =
Loss
300
× 100 =
× 100 = 37.5%
CP
800
TH—Easy Maths—5
93
(c) SP = `1050; CP = `1200 Loss = `(1200 – 1050) = `150
150
Loss% =
× 100 = 12.5%
1200
(d) CP = `1500; SP = `1700
(e) CP = `250.50; SP = `300.50
Profit = `(1700 – 1500) = `200 Profit = `(300.50 – 250.50) = `50
50
200
1
Profit % =
× 100 = 13 % Profit % = 250.50 × 100 = 19.96%
1500
3
(f) CP = `1250; SP = `1000
Loss = `(1250 – 1000) = `250
250
Loss % =
× 100 = 20%
1250
3. (a) CP = `250; SP = `325
Profit = `(325 – 250) = `75
75
Profit % =
× 100 = 30%
250
(b) CP = `60; SP = `80
Profit = `(80 – 60) = `20
20
1
Profit % = × 100 = 33 %
60
3
(d)CP = `2500, Loss = `125
125
Loss % =
× 100 = 5%
2500
(c) CP = `5000, SP = `4500
Loss = `(5000 – 4500) = `500
500
Loss % =
× 100 = 10%
5000
(e) CP = `800, Profit = `150
150
Profit % =
× 100 = 18.75%
800
(f) CP = `32,000, SP = `30,000
Loss = `(32,000 – 30,000) = `2000
(g) CP = `62,000, Interest paid = `3000
∴ Total CP = `(62,000 + 3000) = `65,000
SP = 70,000
Gain = `(70,000 – 65,000) = `5000
5000
9
Gain % =
× 100 = 7 %
65000
13
Loss % =
2000
1
× 100 = 6 %
32,000
4
Chapter 15: Simple Interest
Exercise 15.1
P×R×T
1. S.I =
100
3000 × 6 × 2
(a) SI = ` = `360
100
94 (b) SI = ` 5000 × 10 × 5
= `2500
100
TH—Easy Maths—5
2600 × 8 × 3
= `624 (d)
100
1500 × 5 × 3
(e) SI = ` = `112.5 (f)
100 × 2
8000 × 15 × 2
(g) SI = ` = `1200 (h)
2 × 100
2500 × 9 × 3
(i) SI = ` = `337.50 (j)
2 × 100
2. (a) P = `1800; R = 4%; T = 5 years
1800 × 5 × 4
∴ SI = ` = `360
100
(c) SI = ` (b) P = `4000; R = 10%; T = 3 years
4000 × 10 × 3
∴ SI = ` = `1200
100
1200 × 11 × 13
= `858
100 × 2
10,000 × 8 × 5
SI = ` = `2000
100 × 2
500 × 5 × 3
SI = ` = `37.50
2 × 100
6000 × 5 × 15
SI = ` = `1125
100 × 2 × 2
SI = ` 1
7
(c) P = `1000; R = 3 % or %; T = 4 years
2
2
∴ SI = ` (d) P = `8000; R = 8%; T = 3
∴ SI = ` 1
9
1
5
(e) P = `1000; R = 4 % or %; T = 2 % years or years
2
2
2
2
∴ SI = ` 1000 × 4 × 7
= `140
100 × 2
1
7
years years
2
2
8000 × 8 × 7
= `2240
100 × 2
1000 × 9 × 5
= `112.50
2 × 2 × 100
Exercise 15.2
1. (a) Amount = Principal + Interest
Principal = Amount – Interest
Interest = Amount – Principal
Now (a) I = `(2530 – 2350) = `180
(b) Interest = `(9325.75 – 8750) = `575.75
(c) Principal = `(7250 – 630) = `6620
(d) Amount = `(3500 + 300) = `3800
(e) Amount = `(1523 + 477) = `2000
(f) Interest = `(510 – 425) = `85
TH—Easy Maths—5
95
2. (a) P = `4000; R = 4%; T = 1 year
4000 × 4 × 1
SI = ` = `160
100
Amount= Principal + Interest
= `(4000 + 160) = `4160
1
13
(b) P = `5000; R = 6 % or %; T = 1 year
2
2
Amount= `(5000 + 325) = `5325
(c) P = `7250; R = 10%; T = 1 year
7250 × 10 × 1
SI = ` = `725
100
SI = ` 5000 × 13 × 1
= `325
2 × 100
Amount = `(7250 + 725) = `7975
1
15
(d) P = `8000; R = 7 % or %; T = 1 year
2
2
8000 × 15 × 1
= `600
2 × 100
SI = ` Amount = `(8000 + 600) = `8600
Exercise 15.3
P×R×T
SI × 100
⇒T=
100
P×R
Now substituting the values
250 × 100
180 × 100
(a) T =
= 1 year
(b) T =
= 2 years
5000 × 5
1800 × 5
1. We know that SI =
(c) T =
600 × 100 × 2
1 15
= 5 years as 7 =
1600 × 15
2 2
SI × 100
P×T
Now substituting the values
120 × 100
(a) R =
= 2% per annum
2000 × 3
2. R =
(b) R =
192 100
×
= 15% per annum
640
2
(c) R =
450 × 100 × 2
= 4% per annum
7500 × 3
3. (a) P = `750; SI = `45; T = 2 years; R = ?
96 R=
SI × 100 45 × 100
=
= 3% per annum
P×T
750 × 2
TH—Easy Maths—5
(b) P = `1800; SI = `675; R =15 % T = ?
(c) P = `1500; SI = `675; T = 5 years; R =?
(d) P = `7200; SI = `3600; R = 10%; T = ?
(e) P = `8000; A = `8800; SI = `(8800 – 8000) = `800
T = 2 years, R = ?
1
25
(f) P = `20,000; R = 12 % or %; A = `25,000; T = ?
2
2
R=
R=
T=
R=
SI × 100 675 × 100
1
=
= 2 years
P×R
1800 × 15
2
SI × 100 675 × 100
=
= 9% per annum
P×T
1500 × 15
SI × 100 3600 × 100
=
= 5 years
P×R
7200 × 10
SI × 100 800 × 100
=
= 5% per annum
P×T
8000 × 2
∴ SI = `(25,000 – 20,000) = `5000
SI × 100 5000 × 100 × 2
Now T =
=
= 2 years
P×R
20,000 × 25
Chapter 16: Average
Exercise 16.1
Sum of given quantities
No. of quantities
18 + 24 + 32 + 46 120
(a) Average =
=
= 30
4
4
36 + 50 + 54 + 77 + 83 300
(b) Average =
=
= 60
5
5
1.1 + 2.2 + 3.3 + 4.4 + 5.5 16.5
(c) Average =
=
= 3.3
5
5
2
3
7
9
(d) Sum of all quantities = +
+
+
10 10 20 25
LCM of 10, 20, 25 = 100
∴ Sum =
1. Average =
20 + 30 + 35 + 36 121
=
100
100
No of quantities = 4
121
121 1 121
∴ Average =
÷4=
× =
100
100 4 400
TH—Easy Maths—5
97
(e) Sum of all quantities =
5
7
8
+
+
9 18 27
67
= 30 + 21 + 16 =
( LCM of 9, 18, 27 = 54)
54
54
No of quantities = 3
67
67 1 67
∴ Average =
÷3=
× =
54
54 3 162
(f) Sum of all quantities = 1
=
3
4
7
+2
+3
5
15
25
8 34 82
+ +
(LCM of 5, 15, 25 = 75)
5 15 25
536
= 120 + 170 + 246 =
75
75
No of quantities = 3
536
536 1 536
86
∴ Average =
÷3=
× =
or 2
75
75 3 225
225
Sum of all 4 incomes
2. (a) Average income =
4
558
`(133.75 + 135.50 + 143.25 + 145.50)
=
= ` = `139.50
4
4
98 Sum of rainfall in 11 consecutive years
11
(200 + 150 + 240 + 160 + 77 + 225 + 161 + 240 + 160 + 85 +205)
=
cm
11
1903
=
cm = 173 cm
11
Sum of marks obtained by 5 student
(c) Average=
5
(98 + 96 + 84 + 72 + 85) 435
=
=
= 87
5
5
(d) In 1 kg = 6 apples
In 17 kg = (17 × 6) = 102 apples
(e) Average earning of 12 persons = `3400
So their total earning for the month = `(3400 × 12) = `40,800
(f) Average of 4 numbers = 48
∴ Sum of 4 numbers = 48 × 4 = `192
Sum of their numbers = (48 + 54 + 56) = `158
∴ The fourth number is = 192 – 158 = 34
(g) Average height of 12 students = 150 cm
Total height of 12 students = (150 × 12) cm = 1800 cm
Average height of 18 students = 180 cm
(b) Average =
TH—Easy Maths—5
(h) Average of 5 numbers = 72
Sum of 5 numbers = 5 × 72 = 360
Average of 5 other numbers = 70
Sum of 5 other numbers = 5 × 70 = 350
Total numbers = 5 + 5 = 10
Total sum of 10 numbers = 360 + 350 = 710
710
Average of 10 numbers =
= 71
10
Total height of 18 students = (180 × 18) cm = 3240 cm
Total number of students in a class = (18 +12) = 30
Total height of 30 students = (1800 + 3240) cm = 5040 cm
5040
Average height =
cm = 168 cm
30
Chapter 17: Speed and Distance
Exercise 17.1
Distance
Time
(a) D = 2500 km, T = 20 hours
(b) D = 600 km, T = 3 hours
2500 km
600 km
Speed =
= 125 km/h Speed =
= 200 km/h
20 hours
3 hours
1. Speed =
(c) D = 13525 km, T = 25 hours
13525 km
Speed =
= 541 km/h
25 hours
(d) D = 800 km, T = 50 min or =
50
5
hours or hours
60
6
800 × 6
km/h = 960 km/h
5
2
5
(e) D = 2500 km, T = 1 h 40 min. or 1 h = h
3
3
2500 × 3
Speed =
= 1500 km/h
5
(f) D = 100 m, T = 2 min. 5 s or 125 s
100
Speed =
m/s = 0.8 m/s
125
2. Total distance travelled= (2500 + 600 + 13525) km
= 16625 km
Total time taken = (20 + 3 + 25) h = 48 h
16625 km
∴ Average speed =
= 346.35 km/h
48 h
Speed =
TH—Easy Maths—5
99
3. Total distance covered by car = (150 + 200 + 82) km
= 432 km
Total time taken = (3 + 5 + 4) h = 12 h
432 km
∴ Average speed =
= 36 km/h
12 h
Exercise 17.2
Distance
and 1 km = 1000 m and 1 h = 60 × 60 s = 3600
Time
18 × 1000
18 km
18 × 1000 m
= 5m/s
(a) Speed =
= =
1h
1 × 60 × 60 s
3600
36 km
36 × 1000 m
(b) Speed =
= = 10 m/s
1h
3600 s
1. Speed =
(c) Speed =
144 m
144 m
= = 2.4 m/s
1 min
1 × 60 s
(d) Speed =
198 km
198 × 1000 m
= = 55 m/s
1h
3600 s
(e) Speed =
400 km
400 × 1000 m
= = 111.11 m/s
1h
3600 s
(f) Speed =
270 km
270 × 1000 m
= = 4500 m/s
1 min
1 × 60 s
(g) Speed =
600 m
600 m
= = 0.17 m/s
1h
3600 s
90 km
90 × 1000 m
= = 90000 m/s
1s
1s
180 km
2. (a) Speed =
1h
as 1 km = 1000 m ∴ 180 km = 180 × 1000 m = 18,0000 km
1 h = 60 × 60 s = 3600 s
(h) Speed =
∴ Speed in m/s =
(b) Speed =
100 180000 m
= 50 m/s
3600 s
54 km
1h
as 1 km = 1000 m and 1 h = 3600 s
54 × 1000 m
∴ Speed =
= 15 m/s
3600 s
(c) Distance = 36 km
Time taken = 8:15 am – 7:45 am = 30 min
36 × 1000 m
∴ Speed in m/s =
= 20 m/s
30 × 60 s
TH—Easy Maths—5
(d) Distance = 390 km, Time taken = 6 h
390 km
Speed =
= 65 km/h
6h
(e) Distance = 2400 km or (2400 × 1000) m
1
10
Time taken = 3 h 20 min = 3 h or h
3
3
2400 × 3
km/h = 720 km/h
10
(i) Speed in km/h =
(ii) Speed in m/min. =
2400000 m
= 12000 m/min
200 m
( 1 h = 60 min ∴ 3 h 20 min = 3 × 60 + 20 = 200 min)
2400000 m
(iii) Speed in m/s =
= 200 m/s
200 × 60 s
Exercise 17.3
Distance
Speed = 40 m/s T = 20 s
Time
or Distance = Speed × Time
Distance = 40 m/s × 20 s = 800 m
(b) Distance = Speed × Time
as speed = 40 km/h and time = 8 h
∴ Distance = (40 × 8) k = 320 km
1
(c) Speed = 70 km/h; Time = 2 h 10 min or 2 h
6
70 km 13
2
Distance =
×
h = 151 km
h
6
3
(d) Speed of boat = 20 m/s
Time taken to cross river = 20 min or (20 × 60) s = 1200 s
Distance width of river= 20 m × 1200 s
= 24000 m or 24 km
(e) Speed of aeroplane = 7200 km/h
7200 × 1000 m
Speed of m/s =
= 2000 m/s
3600 s
Time = 20 min 20 s = (20 × 60) + 20 s
= 1200 + 20 = 1220 s
Distance a travelled= Speed × Time
m
= 2000
× 1220 s
s
= 2440000 m or 2440 km
(f) Speed = 50 m/s, Distance = 3 km or 3000 m
Distance 3000 m
Time =
=
= 60 s or 1 min
Speed
50 m/s
(a) Speed =
TH—Easy Maths—5
101
(g) Speed = 5 km/h, Distance = 20 km
Distance 20 km
Time =
=
=4h
Speed
5 km/h
(h) Speed = 15 m/s, Distance = 750 m
Distance
750 m
Time =
=
= 50 s
Speed
15 m/s
(i) (a) Speed of aeroplane = 1350 km/h or
1350 km
1h
1350 × 1000 m
= 375 m/s
1 × 60 × 60 s
1
(b) Distance travelled in 2 h 15 min (or 2 h)
4
Distance = Speed × Time
9
= 1350 km/h × h = 3037.50 km
4
(j) Speed from Mumbai to Goa = 60 km/h
Time taken = 3 h
∴ Distance = 60 km/h × 3 h = 180 km
Speed from Goa to Mumbai = 45 km/h
Distance = 180 km
Distance 180 km
Time =
=
=4h
Speed
45 km/h
Speed in m/s =
Chapter 18: Time and Timetables
Exercise 18.1
1. (a)Time lapsed from 3:00 pm to (b) 5:00
7:00 pm
– 4:00
7:00
1:00
– 3:00
4:00 or 4 hours
102 (c) 9:00
– 4:30
4:30 or 4 hours 30 min
(d) 8:00
– 6:45
1:15
or 1 hour
or 1 hour 15 min
TH—Easy Maths—5
(e) 10:00am to 1:00 pm
(f) 8:00 pm and 6:30 am
Step 1: Subtracting 10:00 am Step 1: Subtracting 8:00 pm
from 12:00 (noon)
from 12:00 midnight
12:00
12:00
– 10:00
– 8:00
2:00
4:00
Step 2: Adding 2:00 hours to Step 2: Adding 4:00 to
1:00 pm
6:30 am
(12:00 noon to 1:00 pm)
(12:00 mid night to 6:30 am)
1:00
4:00
+ 2:00
+ 6:30
10:30 or 10 hours 30 min
3:00 or 3 hours
2. (a) Rate of heart beat = 72 beats per min.
Number of min in a day = (24 × 60) = 1440 min.
Heart beat taking place in a day = (72 × 1440) beats = 103680 beats
(b) Time taken by Mrs. Webb to walk her dog once = 15 min
Since she takes her dog for walk a day
∴ time spent in a day = (2 × 15) = 30 min.
Days in a week = 7
Days in 4 week = 7 × 4 = 28
∴ time spent by Mrs. Webb in four weeks
= (28 × 30) min = 840 min.
(c) Time taken to walk to the bus stop = 3 min.
Number of days required for travelling 60 min walk
= 60 ÷ 3 = 20 days
(d) Time spent on jogging in each day = 30 min
24 hours of jogging = (24 × 60) min
= 1440 min
Number of days required to jog = 1440 min
= 1440 ÷ 30 = 48 days
Exercise 18.2
1. (a), (b), (c) Refer answers at the end of book.
(d) Time taken to complete the journey
= Arrival time – Departure time
= (2000 – 0800) h = 12 h
2. (a) 631 km
(b) Difference is = `(700.00 – 324.00) = `376
(c) Volvo bus leave from Delhi 15:30 h on 03.30 pm
(d) Ordinary bus starts from Delhi 1315 h or 1:15 pm
TH—Easy Maths—5
103
Exercise 18.3
1.(a)
8:50 or 1 h 40 min
– 7:10
1:40
(b) 11:30
– 7:10
4:20
(c) 11:32 or 2 min
– 11:30
00:02
(d) 11:32
+ 00:40
12:12
(e)
(f)
8 60
9:40
– 8:53
00:47
Time between 9:40 to 9:00 = 40 min
Hence total time = (40 + 7) = 47 min
(g) Delhi (7:10 – 7:00 = 10 min)
or 4 h 20 min
pm
9:45 am
– 9:40 am
0:05 min
Train stops at Agra for
5 min
(h)
9:40 or 2 h 30 min
– 7:10
2:30
1
(i) h = 30 min
2
∴ Train will reach Agra at (9:40 + 0:30) am
9:40 am (70 min = 1 min 60 s)
+ 0:30 am
10:10 am
Chapter 19: Lines and Angles
Exercise 19.1
1.–5. Refer answers at the end of book.
6.–9. Let students try by themselves.
Exercise 19.2
1.–4. Refer answers at the end of book.
Exercise 19.3
1. Refer answers at the end of book.
2. Refer answers at the end of book.
3. Refer answers at the end of book.
Exercise 19.4
1.–3. Let students try by themselves. 4. Refer answers at the end of book.
104 TH—Easy Maths—5
Chapter 20: Triangles and Quadrilaterals
Exercise 20.1
1. (a) Isosceles (2 sides are equal)
(b) Scalene (all sides are different)
(c) Equilateral (all sides are equal)
2. (a) Acute angled triangle (all angles are acute)
(b) Obtuse angled triangle (has 1 obtuse angle)
(c) Right angled triangle (has 1 right angle)
3. Let students try by themselves. 4. Refer answers at the end of book.
Exercise 20.2
1. Refer answers at the end of book.
2. (a) No; since BC + AC < AB
(b) Yes; as (60° + 65° + 55° = 180°)
(c)Yes; as sum of lengths of two sides is greater than the length of
the third side.
(d) No; as (90° + 50° + 50° = 190° ≠ 180°)
(e) Yes; (reason same as in (c)).
(f) No; as (40° + 40° + 40° = 120° ≠ 180°)
(g) Yes; (reason same as in (c)
3. (a) Yes; since sum of two sides is greater than third side.
(b) No; since sum of two sides is greater than third side.
4.–5. Refer answers at the end of book.
6. Sums of angles of a triangle = 180°
(a)∠C= 180° – (∠A + ∠B)
(b)In PQR
= 180° – (55° + 72°)
∠R = 180° – (∠P + ∠Q)
= 180° – 127° = 53°
∠R = 180° – (37° + 82°)
Since side opposite to
∠R = 180° – 119° = 61°
bigger angle is longer
∠Q has longest side
∴ ∠B
opposite to it.
(c)In XYZ,
(d)In DEF
∠Z = 180° – (∠X + ∠Y)
∠F= 180° – (∠D + ∠E)
X = 180° – (105° + 30°) = 180° – (86° + 50°)
X = 180° – 135° = 45°
= 180° – 136° = 44°
∠X has longest side
∠D has longest side
opposite to it.
opposite to it.
Exercise 20.3
1. Yes if the points are non–collinear.2.–5. Let students try by themselves.
Exercise 20.4
1.–2. Refer answers at the end of book.
TH—Easy Maths—5
105
3. Sum of all angles of quadrilateral = 360°.
(a) In quadrilateral ABCD
∠D= 360° – (∠A + ∠B + ∠C)
= 360° – (100° + 65° + 80°)
= 360° – 245° = 115°
(b) In quadrilateral PQRS
∠S = 360° – (∠P + ∠Q + ∠R)
= 360° – (75° + 75° + 105°)
= 360° – 255° = 105°
(c) In quadrilateral WXYZ
∠Z = 360° – (∠W + ∠X + ∠Y)
= 360° – (64° + 138° + 110°)
= 360° – 312°
= 48°
(d) In quadrilateral LMNO
∠O= 360° – (∠M + ∠L + ∠N)
= 360° – (90° + 90° + 90°)
= 360° – 270°
= 90°
(e) In quadrilateral PQRS
∠S = 360° – (∠P + ∠Q + ∠R)
= 360° – (65° + 140° + 65°)
= 360° – 270° = 90°
4. Let students try by themselves.
Chapter 21: Circles and Spatial Geometry
Exercise 21.1
1.–4. Refer answers at the end of book.
Exercise 21.2
1. Diameter = 74 cm
1
1
Radius = × Diameter = × 74 cm = 37 cm
2
2
Hence, answer is (c).
Circumference= 3 × diameter (approx.)
= (3 × 74) cm = 222 cm (approx.)
106 TH—Easy Maths—5
2. Diameter = 9 units
Circumference= 3 × diameter (approx.)
= 3 × 9 units = 27 units (approx.)
3. Radius = 7 units
∴ Diameter = 2 × radius = 2 × 7 =14 units
Circumference= 3 × diameter (approx.)
= 3 × 14 units (approx.) = 42 units (approx.)
1
4. Radius = × diameter
2
(a) D = 7 cm
(b) D = 13 cm
1
1
∴ Radius = × 7 = 3.5 cm ∴ Radius = × 13 cm = 6.5 cm
2
2
(c) D = 16 cm
1
∴ Radius = × 16 cm = 8 cm
2
5. Diameter = 2 × radius
(a) R = 2.5 cm
(b) R = 4.5 cm
∴ D = 2 × 2.5 = 5 cm ∴ D = 2 × 4.5 cm = 9 cm
(c) R = 6 cm
∴ D = 2 × 6 = 12 cm
6. Circumference = 3 × diameter (approx.)
(a) D = 6.5 cm
∴ Circumference= 3 × 6.5 cm (approx.) = 19.5 cm (approx.)
(b) D = 7 cm
∴ Circumference = 3 × 7 cm (approx.) = 21 cm (approx.)
(c) D = 9.4 cm
∴ Circumference = 3 × 9.4 cm (approx.) = 28.2 cm (approx.)
7.–8. Let students try by themselves.
Exercise 21.3
Exercise 21.4
1.–3. Refer answers at the end of book. 1. Refer answers at the end of book.
Chapter 22: Number Patterns
Exercise 22.1
3. (a) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 82 = 64
(b) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 102 = 100
4. (a) 49 = (7)2 = 1 + 3 + 5 + 7 + 9 + 11 + 13
(b) 81 = (9)2 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17
5. None of them are square numbers as (a) has 3 (b) has 7 (c) has 8 and
(d) has 2 in the units place.
TH—Easy Maths—5
107
Exercise 22.2
1. (a) 21 (c) 10 and (d) 45 can make triangular pattern
To draw dot pattern → Let students try by themselves.
12 × 13
2. (a) 12th Triangular number =
= 6 × 13 = 78
2
25 × 26
(c)25th Triangular number =
= 25 × 13 = 325
2
3. 78 is 12th Triangular number.
13 × 14
= 13 × 7 = 91
2
14 × 15
14th Triangular number =
= 7 × 15 = 105
2
4. Let students try by themselves. 5. Since 28 = 21 + 7
∴ It can be arranged in a
triangular pattern.
∴ 13th Triangular number =
Chapter 23: Data Handling
Exercise 23.1
1.
Marks
10
11
12
13
14
15
16
17
18
19
20
2.
108 Tally Marks
No. of children
0
1
2
3
4
5
6
Total
Frequency
1
3
5
2
1
4
2
2
3
4
3
30
Tally Marks
Total
Frequency
3
5
8
5
4
3
2
30
TH—Easy Maths—5
3. Weight of student
35
36
37
38
39
40
41
42
43
44
45
Tally Marks
–
–
–
–
Total
Frequency
4
0
5
8
2
1
0
12
0
0
3
35
4. Let students try by themselves.
Exercise 23.2
1. (a) Wednesday; 20
(b) Friday; 90
(c)Tuesday;
(d) 60
(e) (80 + 30 + 20 + 60 + 90) = 280
2.–3. Refer answers at the end of book.
Exercise 23.3
1. (a) Black
(b) Blue
(c) Yellow and Black
(d) Black
(e) (8 + 6 + 11 + 13 + 17) = 55
(f) Blue
(g) Red
(h) Yellow
2. (a) School B
(b) School C
(c) Atheletics
(d) (25 + 35 + 28 + 32) = 120
(e) Let students try by themselves.
3. (a) 2nd week
(b) File sheets
(c) 1st and 3rd week
(d) In 3rd week (45 – 30) = 15 more pens were sold than erasers
In the whole month =
= (50 + 60 + 45 + 35) – (25 + 30 + 30 + 20)
= 190 – 105 = 85 more pen were sold than eraser.
(e) Let students try by themselves.
Exercise 23.4
1.(a)Burger
(b)Biscuits
(c)Chips
(d) (45 – 25)% = 20%
(e) (20 – 10)% = 10%
2. (a) Dogs are liked most by 30%.(b) Birds are liked least by 10%
(c) 20%
(d) Dogs and Cats
TH—Easy Maths—5
109
3. (a)11%
(b)Comedy
(c) Action
(d) Romance and Drama; Horror and Foreign
(e) Comedy and action
Revision Test -1
1.-2. Refer answers at the end of book.
3.
Step 1: Sum = 2145786 + 54866640 = 57012426
Step 2: Sum = 20014563 + 5410964 = 25425527
Step 3: Difference = 57012426 – 25425527 = 31586899
Actual product = 245 × 133 = 32585
4.
Estimated product = 200 × 100 = 20000
6.
LCM of 6 and 8 = 2 × 3 × 4 = 24 minutes
7. a.
8.
4
7
×
12
12
=
48
84
b.
54
72
÷
9
9
=
2 6, 8
3, 4
6
8
Distance covered by walking = 10
3
4
– 4
1
2
+5
2
3
=
9.
=
43
43
4
4
–
–
61
6
9
2
=
+
17
7
12
3
km
Lata’s savings = `(18000 – 18000 × 45%) = `(18000 – 8100) = `9900
10. Perimeter = 10 m + 5 m (including 2 m) + 1 m + 1 m + 10 m
(including 2 m) + 1 m + 1 m + 5 m = 34 m
11. Volume of box = length × breadth × height
3872 = length × 16 × 22 ⇒ length =
12. C.P. = `(1215 + 35) = `1250
110 Profit % =
16 =
Profit
Profit
C.P.
3872
16×22
= 11 cm
× 100
× 100 ⇒ Profit = 200
1250
S.P. = `(1250 + 200) = `1450
TH—Easy Maths—5
15. Data representing the number of students who scored more than
75% marks in different subjects.
English
Hindi
Science
Maths
Social Studies
1
= 5 Students
Revision Test -2
1.
2.a. Sum = LXXV + XXXVIII = 75 + 38 = 113 = CXIII
Refer answers at the end of book.
b. Difference = XCIV – LXVI = 94 – 66 = 28 = XXVIII
3.
Amount invested = `( 54622400 + 24456150 + 55255100) = `13,43,33,650
4.
Product = 99999 × 207 = 2,06,99,793
5.
62 students of a class collected `9672 for an annual function.
How much money was collected from each student?
6.
LCM of 4, 5, 7, 8 and 10 = 2 × 2 × 5 × 7 × 2 = 280
Cat will mew after 280 seconds = 4 minutes 40 seconds
2 4, 5, 7, 8, 10
2 2, 5, 7, 4, 5
5 5, 7, 2, 5
7, 2
7.
9.
1
5
2
= 750 × = 300
2
2
5
a. 19.2 – 3.4 + 2.6 × 3 – 7.2 = (19.2 + 7.8) – (3.4 + 7.2) = 27 – 10.6 = 16.4
2
2 9 52 28 56 5
1
7
1
b. 2 × 5 + 3 ÷ 2 – 1 = ×
+
÷
–
27
3 4
4
9
9
9 27 3
9
No. of students = 750 ÷ 2
= 750 ÷
= 13 +
28
×
27
–
5
= 13 +
3
–
5
9 56 3
3
2
13. a. A + B + C = 71° + 88° + 43° = 162° ≠ 180°(No)
=
77
6
= 12
5
6
b. Yes, since sum of two sides is greater than the third side.
14.a. R = 360° – (83° + 40° + 105°) = 360° – 228° = 132°
15. Refer answers at the end of book.
TH—Easy Maths—5
111
Revision Test -3
1.
Difference = 9999999 – 775522 = 9224477
3.
Amount spent = 550 × 32 + 165 × `138 = `(17600 + 22770) = `40,370
4.
Estimated Quotient = 4870 ÷ 20 = 243.5
Actual Quotient = 4872 ÷ 24 = 203
5.
Refer example under topic Finding LCM by Prime Factorization
Method on page 63.
7.
a. LCM of 7, 14, 10 and 6 = 2 × 7 × 5 × 3 = 210
5 15 75
8
4 4 30 120
8 21 168
= ×
=
,
×
=
,
=
×
=
,
14 15 210
10 10 21 210
7 7 30 210
2 35 70
2
=
×
=
6 35 210
6
5
70
75 120 168
2
4
8
Ascending order are
<
<
<
or
<
< <
210 210 210 210
6 14 7 10
b. Similar working as above.
2
3 2
8. Jay’s share = 1 – = of `25000 = × `25000 = `10000
5
5 5
9. a. 32% of x = 8 kg 256 g
100
x = 8.256 x
= 25.8%
32
25
b. 25% of 175 =
× 175 = 43.75
100
32
32% of 128 =
× 128 = 40.96
100
10. Perimeter = 2 × (12 + 9) m = 2 × 21 m = 42 m
11. Cost of fencing = 25 × 42 = 1050
12. a. Volume = l × b × h = 20 cm × 15 cm × 10 cm = 3000 cm3
b. Similar working as above.
13. C.P. = `2200
Profit
Profit % =
C.P.
Profit
20% =
S.P. = `(2200 + 440) = `2640
112 2200
or Profit =
20
100
× 2200 = `440
TH—Easy Maths—5
14. Refer chapter on Circles and Spatial Geometry.
15. Refer answers at the end of book.
Revision Test -4
2.
a. Refer topic Patterns with Consecutive Odd Numbers on page 234.
b. Refer topic Relation of Consecutive Odd Numbers to Cube on
page 234.
Income per day of Rajan = `23343 ÷ 31 = `753
3.
5.
Refer example under topic Finding HCF by Long Division Method
on page 62.
6.Refer example on page 65.
7.
8
Time taken by Satish =
16
Time taken by Sudha =
Satish took less time.
8.
Time taken =
9.
a.
1
+ 5
7
2
3
Distance
Speed
+
21
56
÷2
=
3
4
× 60 = 30 minutes
× 60 = 45 minutes
1.8 km 1.8 × 1000 m
=
= 60 seconds = 1 minute
30 m/s
30 m/s
1
=
4
1
7
=
b. 15
2
3
+
15 – 11
3
4
÷1
1
7
15
3
47
3
3
47
+
3
21
+
17
+
=
24
=
17
+
56
+
+
13
4
1
6
4
×
9
=
15 –
÷
39
24
1
7
+
47
4
=
47
3
10. Cost of ploughing = `20 × (100 x 80) = `1,60,000
11. SI =
P×R×T
100
TH—Easy Maths—5
⇒ 750 =
35
6
÷
=
251
42
=5
41
42
39
24
+2=
53
3
= 17
2
3
P × 8 13 × 2
100
113
⇒
75000 × 3
= P or P = `4500
12.-14. Refer answers at the end of book.
25 × 2
15. a.
b.
c.
Revision Test -5
2.
a. 1045 = 1000 + 40 + 5 = M + XL + V = MXLV
b. 299 = 200 + 90 + 9 = CC + XC + IX = CCXCIX
c. 723 = 700 + 20 + 3 = DC + XX + III = DCXXIII
Division = 999999 ÷ 999 = 1001
5.
6.
Refer example under topic Finding HCF by Long Division Method
on page 62.
Total pizza =
7.
1
3
+
1
4
+
1
8
=
8+6+3
24
8.
9.
17
24
4 3, 4, 8
3, 1, 2
LCM = 4 × 3 × 2 = 24
Distance covered = 2 × 5 × 125 = 1250 m = 1.25 km
Refer example under topic Calculating Time and Rate of Interest
using Formula on page 170.
10.a. Refer example on page 173.
=
b. Average =
2.36 + 3.14 + 5.65 + 1.28 + 7
=
19.43
= 3.886
5
5
12. Volume = Base area × Height = 98 m2 × 4 m = 392 m3
14. d. Number of persons purchasing shampoo T = 25 % of 100
25
=
× 100 = 25
100
15. Refer answers at the end of book.
114 TH—Easy Maths—5
TH—Easy Maths—5
115