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Class 8 Chapter 22 Exercise 22B DOWNLOAD

Class 8: Chapter 22- Linear Equations Exercise 22B
Q1. 17 less than four times a number is 11. Find the number.
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ = π‘₯
4π‘₯ βˆ’ 17 = 11
28
π‘₯=
=7
4
Q2. If 10 be added to four times a certain number, the result is 5 less than five times the
number. Find the number
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ=π‘₯
4π‘₯ + 10 = 5π‘₯ βˆ’ 5
π‘₯ = 15
Q3.
2
3
of a number is 20 less than the original number. Find the original number.
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘œπ‘Ÿπ‘–π‘”π‘–π‘›π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ = π‘₯
2
π‘₯ = π‘₯ βˆ’ 20
3
1
π‘₯ = 20
3
π‘₯ = 60
Q.4 A number is 25 more than its part. Find the number.
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ = π‘₯
5
π‘₯ = 25 + π‘₯
6
1
π‘₯ = 25
6
π‘₯ = 150
Q.5 A number is as much greater than 21 as is less than 71. Find the number.
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ = π‘₯
π‘₯ βˆ’ 21 = 71 βˆ’ π‘₯
2π‘₯ = 92
π‘₯ = 46
Q.6 6 more than one-fourth of the number is two-fifth of the number. Find the number.
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ = π‘₯
1
2
π‘₯+6 = π‘₯
4
5
2 1
30
6 = ( βˆ’ )π‘₯ =
π‘₯
5 4
20
π‘₯ = 40
1
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Q.7 One-third of a number exceeds one-fourth of the number by 15 . Find the number.
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ = π‘₯
1
1
π‘₯ βˆ’ π‘₯ = 15
3
4
1
π‘₯ = 15 π‘œπ‘Ÿ π‘₯ = 180
12
Q.8 If one-fifth of a number decreased by 5 is 16, find the number.
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ = π‘₯
1
π‘₯ βˆ’ 5 = 16
5
π‘₯ = 105
Q.9 A number when divided by 6 is diminished by 40. Find the number.
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ = π‘₯
π‘₯
= π‘₯ βˆ’ 40
6
5
π‘₯ = 40
6
π‘₯ = 48
Q.10 Four-fifths of a number exceeds two-third of the number by 10. Find the number.
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ = π‘₯
4
2
π‘₯ = π‘₯ + 10
3
3
2
π‘œπ‘Ÿ
π‘₯ = 10
15
π‘œπ‘Ÿ π‘₯ = 75
Q.11 Two numbers are in the ratio 3:4 and their sum is 84. Find the number
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘‘π‘€π‘œ π‘›π‘’π‘šπ‘π‘’π‘Ÿπ‘  𝑏𝑒 π‘₯ π‘Žπ‘›π‘‘ 𝑦
Therefore
3π‘₯ = 4𝑦
π‘₯ + 𝑦 = 84
Solving
4
𝑦 + 𝑦 = 84
3
7𝑦
= 84
3
𝑦 = 3 × 12 = 36
4
𝐻𝑒𝑛𝑐𝑒, π‘₯ = × 36 = 48
3
Q.12 Three numbers are in ratio 4:5:6 and their sum is 135. Find the numbers.
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘‘β„Žπ‘Ÿπ‘’π‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿπ‘  𝑏𝑒 π‘₯, 𝑦, 𝑧
Therefore,
4π‘₯: 5π‘₯: 6π‘₯
2
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4π‘₯ + 5π‘₯ + 6π‘₯ = 135
Solving,
15π‘₯ = 135
π‘₯=9
Therefore
π‘‡β„Žπ‘’ π‘‘β„Žπ‘Ÿπ‘’π‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿπ‘  π‘Žπ‘Ÿπ‘’ 36, 45, 54
Q.13 Two numbers are in the ratio 3:5. If each is increased by 10, then ratio between the
new numbers so formed is 5:7, Find the original numbers.
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘‘π‘€π‘œ π‘›π‘’π‘šπ‘π‘’π‘Ÿπ‘  𝑏𝑒 π‘₯ π‘Žπ‘›π‘‘ 𝑦
𝐺𝑖𝑣𝑒𝑛,
π‘₯
3
= … … … … … … … … 𝑖)
𝑦
5
π‘₯ + 10
5
= … … … … … … … 𝑖𝑖)
𝑦 + 10
7
π‘ π‘œπ‘™π‘£π‘–π‘›π‘”,
3
πΉπ‘Ÿπ‘œπ‘š 𝑖) π‘₯ = 𝑦
5
𝑆𝑒𝑏𝑠𝑑𝑖𝑑𝑒𝑑𝑖𝑛𝑔 𝑖𝑛 𝑖𝑖)
3
𝑦 + 10
5
5
=
𝑦 + 10
7
21
𝑦 + 70 = 5𝑦 + 50
5
4
20 = 𝑦
5
π‘œπ‘Ÿ 𝑦 = 25
3
π‘₯ = × 25 = 15
5
π‘‡π‘€π‘œ π‘›π‘’π‘šπ‘π‘’π‘Ÿπ‘  π‘Žπ‘Ÿπ‘’ 15 π‘Žπ‘›π‘‘ 25
Q.14 The sum of three consecutive odd numbers is 75. Find the numbers.
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘‘β„Žπ‘Ÿπ‘’π‘’ π‘π‘œπ‘›π‘ π‘’π‘π‘’π‘‘π‘–π‘£π‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿπ‘  𝑏𝑒
π‘₯, π‘₯ + 2, π‘₯ + 4
π‘‘β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’,
π‘₯ + π‘₯ + 2 + π‘₯ + 4 = 75
3π‘₯ + 6 = 75
3π‘₯ = 69
π‘₯ = 23
π‘‡β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’ π‘‘β„Žπ‘’ π‘‘β„Žπ‘Ÿπ‘’π‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿπ‘  π‘Žπ‘Ÿπ‘’ 23, 25 π‘Žπ‘›π‘‘ 27
Q.15 Divide 25 into two parts such that 7 times the first part added to 5 times the second
part makes 139.
Let π‘‘β„Žπ‘’ π‘‘π‘€π‘œ π‘π‘Žπ‘Ÿπ‘‘π‘  𝑏𝑒 π‘₯ π‘Žπ‘›π‘‘ 𝑦
Therefore
π‘₯ + 𝑦 = 25
7π‘₯ + 5𝑦 + 139
Solving we get
3
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π‘₯ = 25 βˆ’ 𝑦
7(25 βˆ’ 𝑦) + 5𝑦 = 139
175 βˆ’ 2𝑦 = 139
2𝑦 = 175 βˆ’ 139 = 36
π‘œπ‘Ÿ 𝑦 = 18
π‘‡β„Žπ‘’ π‘œπ‘‘β„Žπ‘’π‘Ÿ π‘π‘Žπ‘Ÿπ‘‘ = 7
Q.16 Divide 180 into two parts such that the first part is 12 less than twice the second part.
Let π‘‘β„Žπ‘’ π‘‘π‘€π‘œ π‘π‘Žπ‘Ÿπ‘‘π‘  𝑏𝑒 π‘₯ π‘Žπ‘›π‘‘ 2𝑦
Therefore
π‘₯ + 𝑦 = 180
π‘₯ + 12 = 2𝑦
Solving
𝑦 = 180 βˆ’ π‘₯
π‘₯ + 12 = 2(180 βˆ’ π‘₯)
3π‘₯ = 360 βˆ’ 12 = 348
π‘₯ = 116
π‘‡β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’ 𝑦 = 180 βˆ’ 116 = 64
Q.17 The denominator of the fraction is 4 more than its numerator. On subtracting 1 from
each numerator and denominator the fraction becomes . Find the original fraction.
π‘₯
Let the fraction be
𝑦
Given
𝑦 =π‘₯+4
π‘₯
π‘‡β„Žπ‘’π‘Ÿπ‘“π‘œπ‘Ÿπ‘’ π‘‘β„Žπ‘’ π‘“π‘Ÿπ‘Žπ‘π‘‘π‘–π‘œπ‘› =
π‘₯+4
Given,
π‘₯βˆ’1
1
=
π‘₯+4βˆ’1
2
2π‘₯ βˆ’ 2 = π‘₯ + 3
π‘₯ = 5 π‘Žπ‘›π‘‘ 𝑦 = 9
5
π‘‡β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’ π‘“π‘Ÿπ‘Žπ‘π‘‘π‘–π‘œπ‘› =
9
Q.18 The denominator of the fraction is 1 more than the double the numerator. On adding 2
to the numerator and subtracting 3 from denominator, we obtain 1. Find the original
fraction.
π‘₯
Let the fraction be
2π‘₯ + 1
Given
π‘₯+2
=1
2π‘₯ + 1 βˆ’ 3
π‘₯ + 2 = 2π‘₯ βˆ’ 2
π‘₯=4
4
Fraction =
9
4
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Q.19 The sum of the digits of a two-digit number is 5. On adding 27 to the number, its digits
are reversed. Find the original number.
Let π‘‘β„Žπ‘’ π‘‘π‘€π‘œ 𝑑𝑖𝑔𝑖𝑑 π‘›π‘’π‘šπ‘π‘’π‘Ÿ 𝑏𝑒 π‘₯𝑦
𝐺𝑖𝑣𝑒𝑛
π‘₯ + 𝑦 = 5 … … … … … … … . 𝑖)
π‘₯𝑦 + 27 = 𝑦π‘₯
10π‘₯ + 𝑦 + 27 = 10𝑦 + π‘₯
9π‘₯ + 27 = 9𝑦
π‘œπ‘Ÿ π‘₯ + 3 = 𝑦 … … … … … … … 𝑖𝑖)
π‘†π‘œπ‘™π‘£π‘–π‘›π‘” 𝑖) π‘Žπ‘›π‘‘ 𝑖𝑖) π‘‘π‘œπ‘”π‘’π‘‘β„Žπ‘’π‘Ÿ.
π‘₯ + 3 = (5 βˆ’ π‘₯ )
2π‘₯ = 2
π‘₯=1
𝑦=4
𝐻𝑒𝑛𝑐𝑒 π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ = 14
Q.20 What same numbers should be added to each one of the number 15,23,29,44 to obtain
numbers which are in proportion?
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘Žπ‘‘π‘‘π‘’π‘‘ π‘‘π‘œ π‘’π‘Žπ‘β„Ž π‘œπ‘›π‘’ π‘œπ‘“ 15, 23, 29, 44 𝑏𝑒 π‘₯
15 + π‘₯
29 + π‘₯
=
23 + π‘₯
44 + π‘₯
660 + 59π‘₯ + π‘₯ 2 = 667 + 52π‘₯ + π‘₯ 2
7π‘₯ = 7
π‘₯=1
Q.21 The sum of two numbers is 110. One-fifth of the larger number is 8 more than oneninth of the smaller number. Find the numbers.
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘‘π‘€π‘œ π‘›π‘’π‘šπ‘π‘’π‘Ÿπ‘  𝑏𝑒 π‘₯ π‘Žπ‘›π‘‘ 𝑦
Given
π‘₯ + 𝑦 110
1
1
π‘₯ = 𝑦+8
5
9
Solving
1
1
π‘₯ = (110 βˆ’ π‘₯ ) + 8
5
9
1 1
110
182
( + )π‘₯ =
+8 =
5 9
9
9
182 × 45
π‘₯=
= 65
9 × 14
𝑦 = 10 βˆ’ 65 = 45
π‘‡π‘€π‘œ π‘›π‘’π‘šπ‘π‘’π‘Ÿπ‘  π‘Žπ‘Ÿπ‘’ 45 π‘Žπ‘›π‘‘ 65
Q22 A number is subtracted from the numerator of the fraction
1
12
and six times that
13
number is added to the denominator. If the new fraction is then find the number.
11
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘ π‘’π‘π‘‘π‘Ÿπ‘Žπ‘π‘‘π‘’π‘‘ π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘’π‘Ÿπ‘Žπ‘‘π‘œπ‘Ÿ = π‘₯
5
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12 βˆ’ π‘₯
1
=
13 + 6π‘₯
11
132 βˆ’ 11π‘₯ = 13 + 6π‘₯
17π‘₯ = 119
π‘œπ‘Ÿ π‘₯ = 7
Q.23 A right angled triangle having perimeter 120cm has its two side perpendicular side in
the ratio 5:12. Find the lengths of its sides.
π‘ƒπ‘’π‘Ÿπ‘–π‘šπ‘’π‘‘π‘’π‘Ÿ π‘œπ‘“ π‘Ÿπ‘–π‘”β„Žπ‘‘ π‘Žπ‘›π‘”π‘™π‘’π‘‘ π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ = 120
π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘–π‘π‘’π‘™π‘Žπ‘Ÿ 𝑠𝑖𝑑𝑒𝑠 = 5π‘₯ π‘Žπ‘›π‘‘ 12π‘₯
π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ = √(5π‘₯)2 + (12π‘₯ )2 = 13π‘₯
Therefore
5π‘₯ + 12π‘₯ + 13π‘₯ = 120
30π‘₯ = 120
π‘₯=4
Therefore length of side = 20, 48, 52
Q.24 The sum of the digits of a two-digit number is 9. If 9 is added to the number formed by
reversing the digits, then the result is thrice the original number. Find the original number.
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘‘π‘€π‘œ 𝑑𝑖𝑔𝑖𝑑 π‘›π‘’π‘šπ‘π‘’π‘Ÿ = π‘₯𝑦
π‘₯ + 𝑦 = 9 … … … . . 𝑖)
𝑦π‘₯ + 9 = 3(π‘₯𝑦)
10𝑦 + π‘₯ + 9 = 3(10π‘₯ + 𝑦)
10𝑦 + π‘₯ + 9 = 30π‘₯ + 3𝑦
𝑦 + 9 = 29π‘₯ … … … … 𝑖𝑖)
π‘†π‘œπ‘™π‘£π‘–π‘›π‘” 𝑖) π‘Žπ‘›π‘‘ 𝑖𝑖)
7(9 βˆ’ π‘₯ ) + 9 = 29π‘₯
63 βˆ’ 7π‘₯ + 9 = 29π‘₯
72π‘₯ = 36
π‘‚π‘Ÿ π‘₯ = 2
𝑦=9βˆ’2=7
π‘‡β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’ π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ = 27
Q.25 The lengths of a rectangle plot of land exceeds its breadth by 23 m. if the length is
decreased by 15 m. and the breadth is increased by 7 m. the area is reduced by 360 m 2
𝐹𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘™π‘’π‘›π‘”π‘‘β„Ž π‘Žπ‘›π‘‘ π‘‘β„Žπ‘’ π‘π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘™π‘œπ‘‘.
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘™π‘’π‘›π‘”π‘‘β„Ž = 𝑙 π‘Žπ‘›π‘‘ π‘π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž = 𝑏
𝑙 = 23 + 𝑏
Given
(𝑙 βˆ’ 15)(𝑏 + 7) = 𝑙𝑏 βˆ’ 360
(23 + 𝑏 βˆ’ 15)(𝑏 + 7) = (23 + 𝑏 )𝑏 βˆ’ 360
(𝑏 + 8)(𝑏 + 7) = 23𝑏 + 𝑏 2 βˆ’ 360
𝑏 2 + 15𝑏 + 56 = 23𝑏 + 𝑏 2 βˆ’ 360
416 = 8𝑏
π‘œπ‘Ÿ 𝑏 = 52π‘š
Therefore
𝑙 = 𝑏 + 23 = 52 + 23 = 75π‘š
6
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Q.26 The length of the rectangular park is twice its breadth. If the perimeter of the park is
186 m, find its length and breadth.
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘™π‘’π‘›π‘”π‘‘β„Ž = 𝑙 π‘Žπ‘›π‘‘ π‘π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž = 𝑏
𝑙 = 2𝑏
2𝑙 + 2𝑏 = 186
4𝑏 + 2𝑏 = 186
6𝑏 = 186
π‘œπ‘Ÿ 𝑏 = 31
𝑙 = 62
Q.27 The length of the rectangle is 7 cm more than its breadth. If the perimeter of the
rectangle is 90cm, find its length and breadth.
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘™π‘’π‘›π‘”π‘‘β„Ž = 𝑙 π‘π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž = 𝑏
𝑙 = 𝑏+7
𝐺𝑖𝑣𝑒𝑛
21𝑙 + 2𝑏 = 90
2(𝑏 + 7) + 2𝑏 = 90
4𝑏 = 76
π‘‚π‘Ÿ 𝑏 = 19 π‘π‘š
𝑙 = 19 + 7 = 26 π‘π‘š
Q.28 The length of a rectangle is 7 cm less than twice its breadth. If the length is decreased
by 2cm and breadth increased by 3cm, the perimeter of the resulting rectangle is 66 cm.
find the length and the breadth of the original rectangle
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘™π‘’π‘›π‘”π‘‘β„Ž = 𝑙 π‘Žπ‘›π‘‘ π‘π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž = 𝑏
𝑙 + 7 = 2𝑏
Given,
2(𝑙 βˆ’ 2) + 2(𝑏 + 3) = 66
2𝑙 βˆ’ 4 + 2𝑏 + 6 = 66
2𝑙 + 2𝑏 = 64
Solving,
2(2𝑏 βˆ’ 7) + 2𝑏 = 64
6𝑏 = 78
𝑏 = 13,
𝑙 = 2 × 13 βˆ’ 7 = 19
π‘π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž = 13 π‘π‘š
π‘™π‘’π‘›π‘”π‘‘β„Ž = 19π‘π‘š
Q.29 A man is five times as old as his son. In two years’ time, he will be four times as old as
his son. Find their present ages.
Let the π‘šπ‘Žπ‘›β€™π‘  π‘Žπ‘”π‘’ = 5π‘₯
𝐼𝑓 π‘ π‘œπ‘›β€™π‘  π‘Žπ‘”π‘’ = π‘₯
π‘‡π‘€π‘œ π‘¦π‘’π‘Žπ‘Ÿπ‘  π‘™π‘’π‘‘π‘‘π‘’π‘Ÿ
π‘€π‘Žπ‘›β€™π‘  π‘Žπ‘”π‘’ = 5π‘₯ + 2
π‘†π‘œπ‘›β€™π‘  π‘Žπ‘”π‘’ = π‘₯ + 2
5π‘₯ + 2 = 4(π‘₯ + 2)
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π‘₯ + 6 π‘¦π‘’π‘Žπ‘Ÿπ‘  = π‘ π‘œπ‘›β€™π‘  π‘Žπ‘”π‘’
π‘€π‘Žπ‘›β€™π‘  π‘Žπ‘”π‘’ = 30 π‘¦π‘Ÿπ‘ .
Q.30 A man is twice as old as his son. Twelve years ago, the man was thrice as old as his
son. Find their present ages.
Let the π‘ π‘œπ‘›β€™π‘  π‘Žπ‘”π‘’ = π‘₯
π‘€π‘Žπ‘›β€™π‘  π‘Žπ‘”π‘’ = 2π‘₯
12 π‘¦π‘’π‘Žπ‘Ÿπ‘  π‘Žπ‘”π‘œ
π‘†π‘œπ‘›β€™π‘  π‘Žπ‘”π‘’ = π‘₯ – 12
π‘€π‘Žπ‘›β€™π‘  π‘Žπ‘”π‘’ = 2π‘₯ – 12
2π‘₯ – 12 = 3 (π‘₯ – 12)
2π‘₯ – 12 = 3π‘₯ – 36
π‘₯ = 24 = π‘ π‘œπ‘›β€™π‘  π‘Žπ‘”π‘’
π‘€π‘Žπ‘›β€™π‘  π‘Žπ‘”π‘’ = 48 π‘¦π‘’π‘Žπ‘Ÿπ‘ 
Q.31 Seema is 10 years elder than Rekha. The ratio of their ages is 5:3. Find their ages.
𝐿𝑒𝑑 π‘…π‘’π‘˜β„Žπ‘Žβ€™π‘  π‘Žπ‘”π‘’ = π‘₯
π‘†π‘’π‘’π‘šπ‘Žβ€™π‘  π‘Žπ‘”π‘’ = π‘₯ + 10
𝑔𝑖𝑣𝑒𝑛
π‘₯ + 10
5
=
π‘₯
3
3π‘₯ + 30 = 5π‘₯
2π‘₯ = 30
π‘œπ‘Ÿ π‘₯ 15
π‘…π‘’π‘˜β„Žπ‘Žβ€² π‘ π‘Žπ‘”π‘’ = 15 π‘¦π‘Ÿπ‘ .
π‘†π‘’π‘’π‘šπ‘Žβ€² π‘ π‘Žπ‘”π‘’ = 25 π‘¦π‘Ÿπ‘ .
Q.32 5 years ago, the age of Parvati was 4 times the age of her son. The sum of their present
ages is 55 years. Find Parvati’s age.
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘π‘Ÿπ‘’π‘ π‘’π‘›π‘‘ π‘Žπ‘”π‘’ π‘œπ‘“ π‘ƒπ‘Žπ‘Ÿπ‘£π‘Žπ‘‘π‘– = π‘₯ π‘¦π‘Ÿπ‘ 
π‘Žπ‘”π‘’ π‘œπ‘“ π‘ π‘œπ‘› = 𝑦 π‘¦π‘Ÿπ‘ .
π‘₯ + 𝑦 = 55 … … … … … … . . … … . . 𝑖)
Five years before
π‘π‘Žπ‘Ÿπ‘£π‘Žπ‘‘π‘– = π‘₯ βˆ’ 5 π‘¦π‘Ÿπ‘ .
π‘ π‘œπ‘› = 𝑦 βˆ’ 5 π‘¦π‘Ÿπ‘ .
Given,
(π‘₯ βˆ’ 5) = 4(𝑦 βˆ’ 5)
π‘₯ βˆ’ 4𝑦 = βˆ’15 … … … … … … … … 𝑖𝑖)
π‘ π‘œπ‘™π‘£π‘–π‘›π‘” 𝑖) π‘Žπ‘›π‘‘ 𝑖𝑖)
π‘₯ βˆ’ 4 (55 βˆ’ π‘₯ ) = βˆ’15
5π‘₯ = 205
π‘œπ‘Ÿ π‘₯ = 44 = π‘π‘Žπ‘Ÿπ‘£π‘Žπ‘‘π‘– β€² π‘ π‘Žπ‘”π‘’
π‘†π‘œπ‘›β€™π‘  π‘Žπ‘”π‘’ = 55 – 44 = 22 π‘¦π‘’π‘Žπ‘Ÿπ‘ 
Q.33 A man is 56 years old and his son is 24 years old. In how many years, the father will be
twice as old as his son at that time?
π‘€π‘Žπ‘›β€™π‘  π‘Žπ‘”π‘’ = 56 π‘¦π‘’π‘Žπ‘Ÿπ‘ 
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π‘†π‘œπ‘›β€™π‘  π‘Žπ‘”π‘’ = 24 π‘¦π‘’π‘Žπ‘Ÿπ‘ 
Let in x years, man would be twice the age of son
56 + π‘₯ = 2( 24 = π‘₯ )
56 + π‘₯ = 48 + 2π‘₯
π‘œπ‘Ÿ π‘₯ = 8 π‘¦π‘’π‘Žπ‘Ÿπ‘ 
Q.34 9 years hence, a girl will be 3 times as old as she was 9 years ago. How old is she now?
Let the current age of the π‘”π‘–π‘Ÿπ‘™ = π‘₯
𝐺𝑖𝑣𝑒𝑛,
π‘₯ + 9 = 3( π‘₯ βˆ’ 9)
π‘₯ + 9 = 3π‘₯ βˆ’ 27
2π‘₯ = 36
π‘₯ = 18 π‘¦π‘’π‘Žπ‘Ÿπ‘ 
= π‘Žπ‘”π‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘”π‘–π‘Ÿπ‘™
Q.35 A man made a trip of 480km in 9 hours. Some part of trip was covered at 45 km/hr
and the remaining at 60km/hr. find the part of the trip covered by him at 60km/hr.
Let the distance covered at 45 π‘˜π‘š/ β„Žπ‘Ÿ = π‘₯
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘π‘œπ‘£π‘’π‘Ÿπ‘’π‘‘ π‘Žπ‘‘ 60 π‘˜π‘š/ β„Žπ‘Ÿ = 𝑦
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ = 480 π‘˜π‘š.
π‘₯ + 𝑦 = 480
π‘₯
𝑦
+
=9
45 60
Solving
480 βˆ’ 𝑦
𝑦
+
=9
45
60
π‘œπ‘Ÿ 𝑦 = 300 π‘˜π‘š π‘Žπ‘›π‘‘ π‘₯ = 180 π‘˜π‘š
Q.36 A motorist travelled from town A to town B at an average speed of 54km/hr. on his
return journey, his average speed was 60km/hr. if the total time taken is 9 hours, find the
distance between the two towns.
Let the distance between town 𝐴 π‘Žπ‘›π‘‘ 𝐡 = π‘₯
π‘₯
π‘₯
π‘‡β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’
+
= 9.5
54 60
π‘œπ‘Ÿ π‘₯ = 270 π‘˜π‘š.
Q.37 The distance between two stations is 300km. two motor-cyclist start simultaneously
from these stations and move towards each other. The speed of one of them is 7km/hr
faster than that of other. If the distance between them after 2 hours is 34km, find the speed
of each motor-cycle
π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ = 300 π‘˜π‘š
𝐿𝑒𝑑 π‘‘β„Žπ‘’ 𝑠𝑝𝑒𝑒𝑑 π‘œπ‘“ 1𝑠𝑑 𝑐𝑦𝑐𝑙𝑖𝑠𝑑 = π‘₯
π‘‡β„Žπ‘’π‘› 𝑠𝑝𝑒𝑒𝑑 π‘œπ‘“ 2𝑛𝑑 𝑐𝑦𝑐𝑙𝑖𝑠𝑑 = π‘₯ + 7
π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘π‘œπ‘£π‘’π‘Ÿπ‘’π‘‘ 𝑏𝑦 1𝑠 𝑐𝑦𝑐𝑙𝑖𝑠𝑑 𝑖𝑛 2β„Žπ‘Ÿ = 2π‘₯
π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘π‘œπ‘£π‘’π‘Ÿπ‘’π‘‘ 𝑏𝑦 2𝑛𝑑 𝑐𝑦𝑐𝑙𝑖𝑠𝑑 𝑖𝑛 2 β„Žπ‘Ÿ = 2( π‘₯ + 7)
Therefore
2π‘₯ + 34 + 2(π‘₯ + 7) = 300
4π‘₯ + 48 = 300
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252
= 63 π‘˜π‘š/β„Žπ‘Ÿ
4
𝑆𝑝𝑒𝑒𝑑 π‘œπ‘“ 1𝑠𝑑 𝑐𝑦𝑐𝑙𝑖𝑠𝑑 = 63π‘˜π‘š/β„Žπ‘Ÿ
𝑆𝑝𝑒𝑒𝑑 π‘œπ‘“ 2𝑛𝑑 𝑐𝑦𝑐𝑙𝑖𝑠𝑑 = 63 + 7 = 70 π‘˜π‘š/β„Žπ‘Ÿ
π‘₯=
Q.38 A boat travels 30km upstream in river in the same period of time as it travels 50km
downstream. If the ratio of stream be 5 km/hr, find the speed of the boat in still water.
Let the speed of boat = π‘₯ π‘˜π‘š/β„Žπ‘Ÿ
𝑆𝑝𝑒𝑒𝑑 π‘œπ‘“ π‘ π‘‘π‘Ÿπ‘’π‘Žπ‘š = 5 π‘˜π‘š/β„Žπ‘Ÿ
𝑆𝑝𝑒𝑒𝑑 π‘œπ‘“ π‘π‘œπ‘Žπ‘‘ π‘’π‘π‘ π‘‘π‘Ÿπ‘’π‘Žπ‘š = π‘₯ – 5 π‘˜π‘š/β„Žπ‘Ÿ
π‘†π‘π‘’π‘Žπ‘Žπ‘‘ π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘œπ‘Žπ‘‘ π‘‘π‘œπ‘€π‘›π‘ π‘‘π‘Ÿπ‘’π‘Žπ‘š = π‘₯ + 5 π‘˜π‘š/ β„Žπ‘Ÿ
30
50
π‘‡β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’
=
π‘₯βˆ’5
π‘₯+5
30π‘₯ + 150 = 50π‘₯ – 250
400 = 20 π‘₯
π‘‚π‘Ÿ π‘₯ = 20 π‘˜π‘š/β„Žπ‘Ÿ
Q.39 The length of each of the equal sides of an isosceles triangle is 4cm longer than the
base. If the perimeter of the triangle is 62cm, find the lengths of the sides of the triangle.
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘π‘Žπ‘ π‘’ = π‘₯ π‘π‘š
𝑆𝑖𝑑𝑒𝑠 = (π‘₯ + 4)
π‘ƒπ‘’π‘Ÿπ‘–π‘šπ‘’π‘‘π‘’π‘Ÿ = 62 π‘π‘š
π‘‡β„Žπ‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’
(π‘₯ + 4) + ( π‘₯) + ( π‘₯ + 4) = 62
3π‘₯ + 8 = 62
3π‘₯ = 54
π‘œπ‘Ÿ π‘₯ = 18
π΅π‘Žπ‘ π‘’ = 18 π‘π‘š, 𝑆𝑖𝑑𝑒𝑠 = 22π‘π‘š
Q.40 A certain number of candidates appeared for an examination in which one-fifth of the
whole plus 16 secured first division, one-fourth plus 15 secured second division and onefourth minus 25 secured third division, if the remaining 60 candidates failed, find the total
number of candidates appeared.
Let the number of π‘π‘Žπ‘›π‘‘π‘–π‘‘π‘Žπ‘‘π‘’π‘  = π‘₯
1
1
1
( π‘₯ + 6) + ( π‘₯ + 15 ) + ( π‘₯ βˆ’ 25) + 60 = π‘₯
5
4
4
1 1 1
6 + 60 = π‘₯ βˆ’ ( + + ) π‘₯
5 4 4
14
π‘₯ (1 βˆ’
) = 66
20
20 × 66
π‘₯=
= 220
6
π‘π‘œ. π‘œπ‘“ π‘π‘Žπ‘›π‘‘π‘–π‘‘π‘Žπ‘‘π‘’π‘  = 220
Q.41 Raman has 3 times as much money as Kamal. If Raman gives Rs. 750 to Kamal, then
Kamal will have twice as much as left with Raman. How much had each originally?
Let money with πΎπ‘Žπ‘šπ‘Žπ‘™ = π‘₯
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π‘‡β„Žπ‘’π‘› π‘šπ‘œπ‘›π‘’π‘¦ π‘€π‘–π‘‘β„Ž π‘…π‘Žπ‘šπ‘Žπ‘› = 3π‘₯
2(3π‘₯ βˆ’ 750) = (π‘₯ + 750)
π‘‚π‘Ÿ π‘₯ = 450 𝑅𝑠.
πΎπ‘Žπ‘šπ‘Žπ‘™ β„Žπ‘Žπ‘  450 𝑅𝑠. 𝐴𝑛𝑑 π‘…π‘Žπ‘šπ‘Žπ‘› 1350 𝑅𝑠.
Q.42 The angles of triangle are in ratio 2:3:4. Find the angles.
π‘…π‘Žπ‘‘π‘–π‘œ π‘œπ‘“ π‘Žπ‘›π‘”π‘™π‘’ = 2 ∢ 3 ∢ 4
Therefore
2π‘₯ + 3π‘₯ + 4π‘₯ = 180
9π‘₯ = 180
π‘₯ = 20
π‘‡β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’ π‘Žπ‘›π‘”π‘™π‘’ π‘Žπ‘Ÿπ‘’ 40, 60, 80π‘‘π‘’π‘”π‘Ÿπ‘’π‘’π‘ .
Q.43 A certain number man can finish a piece of work in 50 days. If there are 7 more men,
the work can be completed 10 days earlier. How many men were originally there?
Let π‘₯ π‘šπ‘’π‘› π‘“π‘–π‘›π‘–π‘ β„Ž π‘€π‘œπ‘Ÿπ‘˜ 𝑖𝑛 50 π‘‘π‘Žπ‘¦π‘ 
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘€π‘œπ‘Ÿπ‘˜ = 50π‘₯ π‘šπ‘Žπ‘› π‘‘π‘Žπ‘¦π‘ 
π‘₯ + 7 π‘šπ‘’π‘› π‘“π‘–π‘›π‘–π‘ β„Ž π‘€π‘œπ‘Ÿπ‘˜ 𝑖𝑛 49 π‘‘π‘Žπ‘¦π‘ 
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘€π‘œπ‘Ÿπ‘˜ = (π‘₯ + 7) × 40
Therefore
50π‘₯ = 40 (π‘₯ + 7)
5π‘₯ = 4π‘₯ + 28)
π‘₯ = 28
π‘‚π‘Ÿπ‘–π‘”π‘–π‘›π‘Žπ‘™ π‘›π‘œ π‘œπ‘“ π‘šπ‘’π‘› = 28
Q.44 Divide 600 in two parts such that 495 of one exceeds 60 of the other by 120.
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘‘π‘€π‘œ π‘π‘Žπ‘Ÿπ‘‘π‘  = π‘₯ π‘Žπ‘›π‘‘ 𝑦
π‘₯ + 𝑦 = 600
0.4π‘₯ – 0.6𝑦 = 120
π‘†π‘œπ‘™π‘£π‘–π‘›π‘” 𝑀𝑒 𝑔𝑒𝑑
π‘₯ = 480
𝑦 = 120
Q.45 A workman is paid Rs. 150 for each day he works and is fined Rs. 50 for each day he is
absent. In a month of 30 days he earned Rs. 2100. For how many das did he remain absent?
π‘†π‘Žπ‘™π‘Žπ‘Ÿπ‘¦ = 150 𝑅𝑠./π‘‘π‘Žπ‘¦
𝐹𝑖𝑛𝑒 = 50 𝑅𝑠./π‘‘π‘Žπ‘¦
𝐿𝑒𝑑 π‘₯ 𝑏𝑒 π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘‘π‘Žπ‘¦π‘  π‘€π‘œπ‘Ÿπ‘˜π‘’π‘‘
π‘‡β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’
150π‘₯ – (30 βˆ’ π‘₯)50 = 2100
π‘‚π‘Ÿ π‘₯ = 18 π‘‘π‘Žπ‘¦π‘ .
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