1Class - VI Mathematics Question Bank
1.What fractions do the shaded parts in each of the following figures represent?
(i) (ii)
(iii) (iv)
Ans. (i)5
8(ii)
7
12(iii)
2
7(iv)
1
8
2. Write a fraction for each of the following :
(i) 4 parts out of 9 equal parts (ii) 5 parts out of 11 equal parts
(iii) two-fifths (iv) four-sevenths (v) seven-tenths
Ans. (i)4
9(ii)
5
11 (iii)
2
5(iv)
4
7 (v)
7
10
3. Point out the numerator and denominator in each of the following fractions:
(i)5
6(ii)
9
14(iii)
12
25
Ans. (i) Numerator = 5, Denominator = 6
(ii) Numerator = 9, Denominator = 14
(iii) Numerator = 12, Denominator = 25
4. Write five fractions equivalent to each of the following :
(i)2
3(ii)
7
9
Ans. (i)2 2 2 4 2 2 4 8 2 2 3 6
, , ,3 3 2 6 3 3 4 12 3 3 3 9
× × ×= = = = =
× × ×
4FRACTIONS
Class - VI Mathematics Question Bank2
2 2 5 10 2 2 6 12,
3 3 5 15 3 3 6 18
× ×= = = =
× ×
∴ Five equivalent fractions to 2
3are
4 6 8 10, , ,
6 9 12 15 and
12
18
(ii)7 7 2 14 7 3 21
,9 9 2 18 9 3 27
× ×= = = =
× ×
7 4 28 7 5 35 7 6 42
9 4 36 9 5 45 9 6 54
× × ×= = = = = =
× × ×
∴ Five equivalent fractions to 7
9are
14 21 28 35, , ,
18 27 36 45 and
42.
54
(iii)3 3 2 6 3 3 9
14 14 2 28 14 3 42
× ×= = = =
× ×
3 4 12 3 5 15 3 6 18
14 4 56 14 5 70 14 6 84
× × ×= = = = =
× × ×
∴ Five equivalent fractions to 3
14 are
6 9 12 15, , ,
28 42 56 70and
18.
84
5. Which of the following are the pairs of equivalent fractions?
(i)4
5 and
24
30(ii)
8
12 and
14
21
Ans. (i)4 24 24 6 4
,5 30 30 6 5
÷= =
÷ which are equivalent.
(ii)8 8 4 2
12 12 4 3
÷= =
÷ and
14 17 7 2.
21 21 7 3
÷= =
÷ So, both are equivalent.
6. Write an equivalent fraction of :
(i)4
5 with numerator 32 (ii)
7
9 with numerator 63
(iii)10
13 with denominator 91 (iv)
32
48with numerator 2
(v)10
13 with numerator 3 (vi)
24
56 with numerator 3
(vii)6
9 with numerator 10 (viii)
18
24 with denominator 28
3Class - VI Mathematics Question Bank
Ans. (i)4 4 8 32
5 5 8 40
×= =
×(ii)
7 7 9 63
9 9 9 81
×= =
×(iii)
10 10 7 70
3 13 7 91
×= =
×
(iv)32 32 16 2
84 48 16 3
÷= =
÷(v)
21 21 7 3
28 28 7 4
÷= =
÷(vi)
24 24 8 3
56 56 8 7
÷= =
÷
(vii)6 6 3 2 2 5 10
9 9 3 3 3 5 15
÷ ×= = = =
÷ ×(viii)
18 18 6 3 3 7 21
24 24 6 4 4 7 28
÷ ×= = = =
÷ ×
7. Write the missing numerals in place holders :
(i)4
9 63= (ii)
3 18
4= (iii)
42 3
70=
Ans. (i)4 4 7 28
9 9 7 63
×= =
× (ii)
3 3 6 18
4 4 6 24
×= =
× (iii)
42 42 14 3
70 70 14 5
÷= =
×
8. Which of the following fractions are in simplest form?
(i)21
40(ii)
35
49(iii)
64
81(iv)
25
36
(v)28
39(vi)
56
65
Ans. (i)21
40 is in the simplest form (ii)
35 5
49 7= is not in the simplest form.
(iii)64 8
81 9= is not in the simplest form. (iv)
25
36 is in the simplest form.
(v)28
39is in the simplest form (vi)
56
65 is in the simplest form
9. Reduce each of the following fractions to its lowest terms
(i)72
90(ii)
140
252
Ans. (i) Now, 72 = 2 × 2 × 2 × 3 × 3 and 90 = 2 × 3 × 3 × 5
So, HCF of 72 and 90 = 2 × 3 × 3 = 18.
∴72 72 18 4
.90 90 18 5
÷= =
÷
(ii) Now , 140 = 2 × 2 × 5 × 7 and 252 = 2 × 2 × 7 × 3 × 3.
∴ H C F of 140 and 252 = 2 × 2 × 7 = 28.
∴140 140 28 5
.252 252 28 9
÷= =
÷
Hence, 140
252in its lowest terms is
5.
9
Class - VI Mathematics Question Bank4
10. Point out the proper and improper fractions from the following :
(i)7
8(ii)
9
9(iii)
15
11(iv)
18
25
Ans. (i)7
,8
Proper fraction (ii)9
,9
Improper fraction
(iii)15
,11
Improper fraction (iv)18
,25
Proper fraction
11. Convert each of the following mixed numerals into an improper fraction :
(i)2
63
(ii)11
513
(iii)3
104
(iv)1
333
Ans. (i)2 (6 3) 2 18 2 20
63 3 3 3
× + += = =
(ii)11 (5 13) 11 65 11 76
513 13 13 13
× + += = =
(iii)3 (10 4) 3 40 3 43
104 4 4 4
× + += = =
(iv)1 (33 3) 1 99 1 100
333 3 3 3
× + += = =
12. Convert each of the following improper fractions into a mixed numeral :
(i)31
8(ii)
89
8(iii)
101
3(iv)
115
13(v)
208
19
Ans. (i)31
8 ∴
31 73
8 8= (ii)
89
8
111
8= . (iii)
101
3
233
3=
(iv)115
13
118
13= (v)
208
19
1810
19=
13. Convert each of the following sets of unlike fractions into like fractions :
(i)2 1 5 7 11
, , , ,3 4 6 8 12
(ii)2 1 5 7 13
, , , ,3 6 9 12 18
(iii)1 4 9 11 37
, , , ,2 7 14 21 42
Ans. (i)2 1 5 7 11
, , , ,3 4 6 8 12
LCM of 3, 4, 6, 8, 12, = 24
∴2 2 8 16 1 1 6 6
,3 3 8 24 4 4 6 24
× ×= = = =
× ×
5Class - VI Mathematics Question Bank
5 5 4 20 7 7 3 21 11 11 2 22, ,
6 6 4 24 8 8 3 24 12 12 2 24
× × ×= = = = = =
× × ×
∴ Fractions become 16 6 20 21 22
, , , ,24 24 24 24 24
(ii)2 1 5 7 13
, , , ,3 6 9 12 18
. L C M of 3, 6, 9, 12, 18 = 36
∴ 2 2 12 24 1 1 6 6 5 5 4 20
, ,3 3 12 36 6 6 6 36 9 9 9 36
× × ×= = = = = =
× × ×
7 7 3 21 13 13 2 26
,12 12 3 36 18 18 2 36
× ×= = = =
× ×
∴ Fractions become 24 6 20 21 26
, , , ,36 36 36 36 36
(iii)1 4 9 11 37
, , , ,2 7 14 21 42
L C M of 2, 7, 14, 21, 42 = 42
∴ 1 1 21 21 4 4 6 24 9 9 3 27
, , ,2 2 21 42 7 7 6 42 14 14 3 42
× × ×= = = =
× × ×
11 11 2 22 37 37 1 37
,21 21 2 42 42 42 1 42
× ×= = = =
× ×
∴ Fractions become 21 24 27 22 37
, , , ,42 42 42 42 42
14. Fill in the place holders with > or < :
(i)7 5
9 9(ii)
9 11
13 13(iii)
3 3
5 4(iv)
7 7
9 11
Ans. (i)7 5
9 9> (ii)
9 11
13 13< (iii)
3 3
15 4< (iv)
7 7
9 11>
15. Arrange the following fractions in ascending order :
(i)2 5 7 11 13
, , , ,3 6 9 12 18
(ii)2 1 5 5 7
, , , ,3 4 6 8 12
(iii)2 4 7 11 23
, , , , .3 5 15 20 30
Ans. (i) L C M of 3, 6, 9, 12 and 18 = 36
∴2 2 12 24 5 5 6 30
, ;3 3 12 36 6 6 6 36
× ×= = = =
× ×
7 7 4 28 11 11 3 33
: ;9 9 4 36 12 18 3 36
× ×= = = =
× ×
2
2
3
3
3 – 6 – 9 – 12 – 18
3 – 3 – 9 – 6 – 9
3 – 3 – 9 – 3 – 9
1 – 1 – 3 – 1 – 3
1 – 1 – 1 – 1 – 1
Class - VI Mathematics Question Bank6
13 13 2 26
18 18 2 36
×= =
× ∴ LCM = 2 × 2 × 3 × 3
Since 24 < 26 < 28 < 30 < 33, so,
24 26 28 30 33
36 36 36 36 36< < < < or
2 13 7 5 11
3 18 9 6 12< < < <
(ii) L C M of 3, 4, 6, 8 and 12 = 24
∴2 2 8 16 1 1 6 6 5 5 4 20
, ,3 3 3 24 4 4 6 24 6 6 4 24
× × ×= = = = = =
× × ×
5 5 3 15 7 7 2 14,
8 8 3 24 12 12 2 24
× ×= = = =
× ×
Since, 6 14 15 16 20
24 24 24 24 24< < < < , we have
1 7 5 2 5
4 12 8 3 6< < < <
(iii) L C M of 3, 5, 15, 20, 30, = 3 × 5 × 2 × 2 = 60.
∴ 2 2 20 40 4 4 12 48
; ;3 3 20 60 5 5 12 60
× ×= = = =
× ×
7 7 4 28 11 11 3 33
;15 15 4 60 20 20 3 60
× ×= = = =
× × and
23 23 2 46.
30 30 2 60
×= =
×
so, 28 33 40 46 48
.60 60 60 60 60
< < < < ∴ 7 11 2 23 4
.15 20 3 30 5
< < < <
16. Arrange the following fractions in descending order :
(i)4 4 4 4 4
, , , ,7 3 9 5 11
(ii) 2 5 7 9 1
, , , ,3 6 9 12 2
(iii) 17 7 19 13 9
, , , ,32 12 48 24 16
Ans. (i) Among two fractions with same numerator, the one with greater denomina-
tor is least of the two.
so, 4 4 4 4 4
.3 5 7 9 11
> > > >
(ii) L C M of 3, 6, 9, 12 and 2 = 36
∴2 2 12 24
3 3 12 36
×= =
×;
5 5 6 30 7 7 4 28; ;
6 6 6 36 9 9 4 36
× ×= = = =
× ×
9 9 3 1 1 8 18
12 12 3 2 2 18 36
× ×= = = =
× ×
Since, 30 > 28 > 27 > 24 > 18
5 7 9 2 1
6 9 12 3 2> > > >
2
2
2
3
3 – 4 – 6 – 8 – 12
3 – 2 – 3 – 4 – 6
3 – 1 – 3 – 2 – 3
3 – 1 – 3 – 1 – 3
1 – 1 – 1 – 1 – 1
2
2
3
3 – 6 – 9 – 12 – 2
3 – 3 – 9 – 6 – 1
3 – 3 – 3 – 3 – 1
1 – 1 – 3 – 1 – 1
7Class - VI Mathematics Question Bank
(iii) L C M of 32, 12, 48, 24 and 16 = 96.
∴17 17 3 51 7 7 8 56
; ;32 32 3 96 12 12 8 96
× ×= = = =
× ×
19 19 2 38 13 13 4 52; ;
48 48 2 96 24 24 4 96
× ×= = = =
× ×
9 9 6 54
16 16 6 96
×= =
×
Since , 56 > 54 > 52 > 51 > 38
∴7 9 13 17 19
12 16 24 32 48> > > >
17. Find the sum:
(i)3 5
5 6+ (ii)
7 18
12 18+ (iii)
3 8 7
10 15 20+ +
(iv)8 11 13 5
9 18 27 6+ + + (v)
1 5 74 2 3
6 8 12+ +
Ans. (i) L C M of 5 and 6 = 30
3 5 18 25 43 131
5 6 30 30 30
++ = = =
(ii)7 13
12 18+ =
7 13,
12 18+ L C M of 12 and 18 = 36
= 21 26 47 11
136 36 36
+= =
(iii)3 8 7
10 15 20+ + =
3 8 7
10 15 20+ + = [L C M of 10, 15, 20 = 60]
= 18 32 21 71 11
160 60 60
+ += =
(iv)8 11 13 5
9 18 27 6+ + + [LC M of 9, 18, 27 and 6 = 54]
= 8 33 26 45
54
4 + + + =
152 76 222
54 27 27= =
(v)1 5 7
4 2 36 8 12
+ + = 25 21 43
6 8 12+ + =
100 63 86 249 310
24 24 8
+= =
L C M of 6, 8, and 12 = 24
2
2
2
2
2
3
32 – 12 – 48 – 24 – 6
16 – 6 24 – 12 – 3
8 – 3 12 – 6 – 3
4 – 3 – 6 – 3 – 3
2 – 3 – 6 – 3 – 3
1 – 3 – 3 – 3 – 3
–1 – 1 –1 – 1 – 1
Class - VI Mathematics Question Bank8
18. Find the difference :
(i)2 3
5 – 33 4
(ii)1 1
5 – 48 12
(iii)1 7
6 – 36 10
Ans. (i)2 3
5 – 33 4
= 2 3 17 15
5 – 3 –3 4 3 4
= = 68 – 45 23 11
112 12 12
= = [L C M of 3 and 4 = 12]
(ii)1 1 41 49
5 – 4 –8 12 8 12
= = 123 – 98 25 1
124 24 24
= = [L C M of 8 and 12 = 24]
(iii) 1 7
6 – 36 10
37 37
–6 10
= = 185 –111 74 37 7
230 30 15 15
= = = [L C M of 6 and 10 = 30]
19. Simplify : (i)2 1 2
6 4 – 23 6 9
+ (ii)3 1 5
10 – 5 – 44 8 12
Ans. (i) 2 1 2
6 4 – 23 6 9
+ = 20 25 20
–3 6 9
+ [L C M of 3, 6, 9 = 18]
= 120 75 – 40 155 11
818 18 18
+= =
(ii)3 1 5
10 – 4 – 44 8 12
= 43 41 53
– –4 8 12
[L C M of 4, 8 and 12 = 24]
= 258 –123 –106 258 – 229
24 24= =
29 51
24 24=
20. Subtract the sum of 3
94
and 5
36
from 7
15 .12
Ans. Required number = 7 3 5
15 – 9 312 4 6
+
=
187 39 23–
12 4 6
+
= 187 39 23
– –12 4 6
[L C M of 12, 4 and 6 = 12]
= 187 –117 – 46 187 –163
12 12= =
242
12=
21. Subtract the sum of 5
212
and 3
34
from the sum of 1
73
and 1
46
Ans.1 1 5 3
7 4 – 2 33 6 12 4
+ +
=
22 25 29 15–
3 6 12 4
+ +
=
44 25 29 45–
6 12
+ +
= 69 74 138 – 74 64 16 1
– 56 12 12 12 3 3
= = = =
9Class - VI Mathematics Question Bank
22. Add the difference of 3
35
and 7
210
to the difference of 7
215
and 9
320
.
Ans.3 7 9 7
3 – 2 3 – 25 10 20 15
+
=
18 27 69 37– –
5 10 20 15
+
= 36 – 27 207 –148
10 60
+
=
9 59 54 59 113 531
10 60 60 60 60
++ = = =
23. From a rope of length 1
202
m, a piece of length 5
38
m is cut off. Find the length of
the remaining rope.
Ans. Total length of rope = 1
202
m. Length cut off = 5
38
m.
Remaining rope = 1 5
20 – 32 8
= 41 29
–2 8
= 41 4 – 29 1 164 – 29
8 8
× ×= =
135 716
8 8= m.
24. The weights of three packets are 3
22
kg,1
33
kg and 2
55
kg. Find total weight of all
three packets.
Ans. ∴ Total weight of three packets = 3 1 2 11 10 27
2 3 54 3 5 4 3 5
+ + = + +
= 11 15 10 20 27 12
60
× + × + × [∵ L.C.M. of 4, 3, 5 = 60]
= 165 200 324
60
+ + =
689 2911
60 60= kg.
25. I bought fruits worth Rs 3
274
and vegetables worth Rs 1
102
. If I gave a fifty- rupee
note to the shopkeeper, how much will I get back ?
Ans. Cost of fruits Rs 3
274
.
Cost of vegetables = Rs 1
102
Total cost = 3 1
27 104 2
+ = 111 21 111 42
4 2 4
++ = = Rs
153
4
Class - VI Mathematics Question Bank10
∴ Remaining money = Total – the cost of fruit and vegetables.
153 200 –153
50 –4 4
= = Rs47
4 = Rs
311
4
26. If the speed of a car is 1
505
km/ hr, find the distance covered by it in 3
35
hours.
Ans. Speed of car = 1
505
km/hours, Time = 3
35
hours
Distance = Speed × Time = 1 3
50 35 5
× = 251 18 4518
5 5 25× = =
18180
25km.
27. Find the product.
(i)11
2 1827
× Ans. 11 65 65 2 130 1
2 18 18 4327 27 3 3 3
×× = × = = =
(ii)1 1
8 121 13
× Ans. 1 1
8 121 13
× = 169 14 13 2 26 2
821 13 3 1 3 3
× = × = =
(iii)3 1 4
1 2 44 7 5
× × Ans.3 1 4 7 15 24
1 2 4 18.4 7 5 4 7 5
× × = × × =
(iv)1 3 4
3 2 25 4 11
× × Ans.16 11 26
5 4 11× × =
4 1 26
5 1 1× × =
104 420
5 5=
28. Simplify :
(i)13 2
21 7÷ Ans. (i) =
13 7 13 1 13 12
21 2 3 2 6 6× = × = =
(ii)1
3 58
÷ Ans. 1
3 58
÷ = 25 5 25 1 5
8 1 8 5 8÷ = × =
(iii)5
7 349
÷ Ans. 68 34 68 1 2
9 1 9 34 9÷ = × =
(iv)1 1
4 62 2
÷ = 9 13 9 2 9
2 2 2 13 13÷ = × =
(v)7 1
5 310 6
÷ = 57 6 3 3 9 4
110 19 5 1 5 5
× = × = =
(vi)5 11
10 17 14
÷ = 5 11 75 25 75 14
10 1 67 14 7 14 7 24
÷ = ÷ = × =
(vii)8 2
15 39 3
÷ = 8 2 143 11
15 39 3 9 3
÷ = ÷ = 143 3 13 1
49 11 3 3
× = =
11Class - VI Mathematics Question Bank
29. Simplify :
(i)1 5 1
1 1 36 9 3
÷ ×
Ans.7 14 10
6 9 3÷ × =
7 9 10 1 1 5 5 12
6 14 3 2 1 1 2 2× × = × × = =
(ii)1 3
13 7
÷ of 5 1
2 18 9
+
Ans.1 3
13 7
÷ of 5 1
2 18 9
+ = 4 3
3 7÷ of
21 10 4 3 21 10
8 9 3 7 8 9+ = ÷ × +
= 4 9 10 4 8 10
3 8 9 3 9 9÷ + = × + =
32 10 32 30 62 82
27 9 27 27 27
++ = = =
(iii)1 1
6 35 10
÷ of 1 1
22 4
÷
Ans.1 1
6 35 10
÷ of 1 1
22 4
÷ = 31 31 5 1
5 10 2 4÷ × ÷ =
31 31 1 31 4 4
5 4 4 5 31 1
÷ ÷ = × ×
= 4 16 1
4 35 5 5
× = =
(iv)2 3
3 –3 11
of 3 1 2 1
2 14 4 3 3
÷ × +
Ans.2 3
3 –3 11
of 3 1 2 1
2 1 14 4 3 3
÷ × + = 11 3 11 5 5 1
–3 11 4 4 3 3
× ÷ × +
= 11 3 5 5 1
–3 4 4 3 3
÷ × + = 11 3 4 5 1
–3 4 5 3 3
× × +
= 11 1 1 11 – 3 1
–3 1 3 3
++ = =
12 – 3 93
3 3= =
(v)2
27
of 3 1 4
15 24 4 7
× ÷ of 5
28
Ans.2
27
of 3 1 4
15 24 4 7
× ÷ of 5
28
= 16 63 9 4 21
7 4 4 7 8× × ÷ ×
= 9 3 9 2
36 36 3 9 2 54.4 2 4 3
× ÷ = × × = × × =
= 4 5 1
1 – 37 4 2
÷ + ÷ = 7 5 1 1
1 –4 4 2 3
× + ×
Class - VI Mathematics Question Bank12
= 7 5 1 7 1 5
– –4 4 6 4 6 4
+ = + = 21 2 –15 23 –15 8 2
12 12 12 3
+= = =
(vii)1 1 1 2 1 1
2 3 3 –13 2 3 9 8 12
+ ÷
Ans.1 1 1 2 1 1
2 3 3 –13 2 3 9 8 12
+ ÷
1 5 10 2 25 13
–3 2 3 9 8 12
+ ÷
= 1 15 20 2 75 – 26
3 6 9 24
+ ÷
=
1 35 2 49
3 6 9 24× ÷ ×
= 35 98
18 216÷ ⇒
35 216 12 30 25 4
18 98 14 7 7× = × = =
(viii)1 1 1 1 1
–4 9 2 4 3
÷ + ÷
Ans.1 1 1 1 1
–4 9 2 4 3
÷ + ÷
=
1 1 1 1 1–
4 9 2 4 3
÷ + ÷
= 9 – 4 1 3
36 2 4
÷ +
=
5 2 3 5 5 5 4 1
36 4 36 4 36 5 9
+ ÷ = ÷ = × =
30. A man earns Rs 12000 per month. He spends 7
12 on household items and
3
24 on
education of children. How much does he save ?
Ans. Total salary = Rs 12000, Amount spend on household = 7
12000 Rs 700012
× =
Amount pend on education = 3
12000 Rs 150024
× =
∴ Money saved = 12000 – 7000 – 1500 = Rs 3500
31. In a school, 25
54of the students are girls and the rest are boys. If the number of
boys is 2030, find the number of girls.
Ans. Let the total number of students be x.
∴ Fraction of girls = 25
54of x and Number of boys = 2030
As per conditions, 25
– 203054
x x = ⇒ 54 – 25
203054
x x=
⇒ 54x – 25x = 2030 × 54 ⇒ 29x = 109620
13Class - VI Mathematics Question Bank
⇒ x = 109620
378029
=
Hence, number of girls = 3780 – 2030 = 1750.
32. When Ajay travelled 18km, he found that one third of his journey was still left. What
was the length of his journey ?
Ans. Let the total length of journey = x km. Distance travelled = 18km.
Journey left to covered = 1
3x
As per condition, 1
– 183
x x =
⇒ 3 – 2
183 3
x xx= = ⇒
318
2x = × ∴ x = 27
Hence, total length of his journey is 27 km.
33. A sum of Rs 56000 is shared between Mohan, Sohan and Krishna. Mohan gets
1
4of it and Sohan gets
1
5of it. How much does Krishna get ?
Ans. Total sum = Rs 56,000, Mohan get = 1
56000 Rs 140004
× =
Sohan get = 1
56000 Rs 112005
× =
Krishna get = 56000 – 14000 – 11200 = 56000 – 25200 = Rs 30800.
34. From a rope 3
124
m long, a piece of length 5
36
has been cut off. What is the length
of remaining piece ?
Ans. Length of remaining piece = 3 5
12 – 34 6
m = 51 23
–4 6
m = 153 – 46
12 m
[L C M of 4 and 6 = 12]
= 107 11
m 512 12
= m
35. On one day a labourer earned Rs 1
38 .2
Out of this money, he spent Rs 3
164
on food,
Rs 1
82
on tea and Rs 1
44
on other eatables. How much is left with him ?
Class - VI Mathematics Question Bank14
Ans. One day a labourer earned= Rs 1
382
= Rs 77
2
Total expenditure of a labourer in one day = Rs 3
164
+ Rs 1
82
+Rs1
44
= Rs 67 17 17 67 34 17
4 2 4 4
+ + + + =
= Rs
118
4 = Rs
59
2
Money left with him = Rs 77 59
–2 2
= Rs 18
2= Rs 9
36. A man had Rs 15000. He gave 1
5of his money to his daughter,
1
3of the remainder
to his son and the rest to his wife. Find the share of each.
Ans. Total money a man had = Rs 15000
∴ Daughter’s share = 1
5of 15000 = Rs 3000
Son’s share = 1
3 of (15000 – 3000) =
1
3× 12000 = Rs 4000
∴ Wife’s share = Rs 15000 – (3000 + 4000) = Rs 8000
37.2
3of the students in a class are boys. If there are 17 girls in the class, how many are
boys ?
Ans. Let total students be x. ∴ Number of boys = 2
3x
As per condition, 2
– 173
xx = ⇒
3 – 217
3
x x= ⇒ 17
3
x=
⇒ x = 17 × 3 ⇒ x = 51
∴ Number of boy = 2 2 51
343 3
x×
= =
38. After reading 5
8of a book, 168 pages are left. How many pages are there in all in the
book ?
Ans. Let the number of pages in the book be x ∴ Pages read = 5
8 of x =
5
8x
As per condition, 5
– 1688
x x = ⇒ 8 – 5
1688
x x= ⇒
3168
8
x=
15Class - VI Mathematics Question Bank
∴168 8
56 83
x×
= = × = 448
∴ Total pages in the book = 448.
40. A man gave 1
4 of his money to his elder son,
1
4 of the remainder to his younger son
and the rest to his daughter. What part of the money did the daughter get?
Ans. Money a man had = 1 (let) Share of elder son = 1
4 of 1 =
1
4
Share of younger son = 1
4 of
1 1 3 31
4 4 4 16
− = × =
∴ Daughter’s part = 1 3
14 16
− +
= 4 3
116
+− =
16 7 9
16 16
−=
41. In an orchard, 1
3of the trees are guava trees,
1
8are banana trees and the rest are
mango trees how many trees in all are there?
Ans. Let the total number of trees be x. Guava trees = 1
3of x
3
x=
Banana trees = 8
x. ∴ Mango tree = x – –
3 8
x x =
24 – 8 – 3 13
24 24=
x x x x
But 13
11724
x= (given) ⇒
117 24
13x
×= ∴ x = 216
Thus, total number of trees = 216.
42. The area of a rectangle is 2
465
sq. cm. If its length is 1
75
cm find, its breadth.
Ans. Breadth of the rectangle = 2 1
46 75 4
÷
cm
= 232 29
5 4
÷
cm =
232 4
5 29× =
8 4
5 1× cm =
32
5 cm =
26
5 cm
43. Mr. Sekhar had some money. He spent 1
4of it on sweets and
2
3of remainder on
fruits. The rest of money he spent on refreshment. If the amount spent on therefreshment is Rs 60, how much money did he have in the beginning ?
Class - VI Mathematics Question Bank16
Ans. Let the total money in beginning be Rs x. Amount spent on sweets = 1
4of
Rs = 4
x
Remaining money = Rs –4
xx
= 4 –
4
x x = Rs
3
4
x
Amount spent on fruits = 2
3of Rs
3
4
x = Rs
2 3
3 4
x× = Rs
2
x
∴ Remaining money = Rs 3
–4 2
x x
According to statement, 604
x= ⇒ x = 60 × 4 = 240
∴ Total money in beginning = Rs 240.
44. A drum of kerosene oil is 3
4full. When 30 litres of oil are drawn from it, it is
7
12full. Find the capacity of the drum.
Ans. Let the total capacity of drum be x. Kerosene in drum = 3
4 full
After drawing 30 litres capacity = 7
12 full.
As per condition, 3 7
– 304 12
x
=
⇒ 9 – 7
3012
x
=
⇒ 2
3012
x = ⇒ 1
306
x = ⇒ x = 30 × 6 = 180 litres.
The capacity of the drum is 180 litre.
45. Two friends Albert and Robert decide to buy a car. Albert pays 3
7of the cost and
Robert pays the rest. (i) What fraction of the cost does Robert pay ?
(ii) Robert pays Rs 31540 more than Albert. Calculate the cost of the car.
Ans. (i) Let the cost of car = 1. Amount Albert pays = 3
7
∴ Robert pay = 3 7 – 3 4
1–7 7 7
= =
17Class - VI Mathematics Question Bank
(ii) Excess contribution by Robert is = 4 3 1
–7 7 7
=
Let, x be the cost of the car, then 1
7 of x = 31,540 ⇒ x = 31540 × 7 = 220780.
46. I bought wheat worth Rs 3
124
rice worth Rs 3
254
and vegetables worth Rs 1
10 .4
If I gave a hundred rupee note to the shopkeeper ; how much will he return to me ?
Ans. Money given to shopkeeper = Rs 1000
Total amount of goods bought = Rs 1 3 1
10 25 102 4 4
+ +
(Wheat, Rice and Vegetable)
= 25 103 41
2 4 4+ +
= 50 103 41
4
+ += Rs
194
4
∴ Money returned by shopkeeper = Rs 194 400 –194
100 –4 4
=
= Rs 206
4 = Rs
103
2 = Rs
151
2.
47. In a school, 4
5 of the children are boys, and the number of girls is 100. Find the
number of boys.
Ans. Let the total number of boys and girls = x
Total number of boys = 4 4
of5 5
xx =
As per condition
4100
5
xx − =
⇒ 1005
x=
∴ x = 500
Thus, number of boys = total strength – number of girls
= 500 – 100
= 400.
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