Class 8: Chapter 23 – Simultaneous
Linear Equations -Exercise 23-B
Q.1 The sum of two numbers is 60 and their difference is 14. Find the numbers.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑏𝑒 𝑥 𝑎𝑛𝑑 𝑦
𝑥 + 𝑦 = 60 … … … … … … 𝑖
𝑥 − 𝑦 = 14 … … … … … … 𝑖𝑖
𝐴𝑑𝑑 (𝑖) 𝑎𝑛𝑑 (𝑖𝑖)
2𝑥 = 74 𝑜𝑟 𝑥 = 37
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑦 = 37 − 14 = 23
𝐻𝑒𝑛𝑐𝑒 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑎𝑟𝑒 23 𝑎𝑛𝑑 37
Q.2 Twice a number is equal to thrice the other number. If the sum of the numbers is 85,
find the numbers.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑎𝑟𝑒 𝑥 𝑎𝑛𝑑 𝑦
2𝑥 = 3𝑦 … … … … … … 𝑖
𝑥 + 𝑦 = 85 … … … … … … 𝑖𝑖
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑥 𝑎𝑛𝑑 𝑦
2𝑥 − 3𝑦 = 0
(−)2𝑥 + 2𝑦 = 170
−5𝑦 = −170
𝑜𝑟
𝑦 = 34
3
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑥 = × 34 = 51
2
𝐻𝑒𝑛𝑐𝑒 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑎𝑟𝑒 51 𝑎𝑛𝑑 34.
Q.3 Find two numbers such that twice the first added to thrice the second gives 70 and
twice the second added to thrice the first gives 75.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑏𝑒 𝑥 𝑎𝑛𝑑 𝑦
2𝑥 + 3𝑦 = 70 … … … … … … 𝑖
3𝑥 + 2𝑦 = 75 … … … … … … 𝑖𝑖
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑥 𝑎𝑛𝑑 𝑦
𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑖) 𝑏𝑦 2 𝑎𝑛𝑑 𝑖𝑖) 𝑏𝑦 3 𝑎𝑛𝑑 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡 𝑖𝑖) 𝑓𝑟𝑜𝑚 𝑖)
6𝑥 + 9𝑦 = 210
(−) 6𝑥 + 4𝑦 = 150
5𝑦 = 60
⇒ 𝑦 = 12
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑖𝑛 𝑖)
70 − 3𝑦 70 − 36
⇒ 𝑥=
=
= 17
2
2
𝐻𝑒𝑛𝑐𝑒 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑎𝑟𝑒 17 & 12
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Q.4 Find two numbers which differ by 9 and are such that four times the larger added to
three times the smaller gives 92.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑏𝑒 𝑥 𝑎𝑛𝑑 𝑦
𝑥 − 𝑦 = 9 … … … … … … 𝑖)
4𝑥 + 3𝑦 = 92 … … … … … … 𝑖𝑖)
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑏𝑦 𝑥 𝑎𝑛𝑑 𝑦
𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑖) 𝑏𝑦 4 𝑎𝑛𝑑 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡 𝑖𝑖) 𝑓𝑟𝑜𝑚 𝑖)
4𝑥 − 4𝑦 = 36
(−)4𝑥 + 3𝑦 = 92
−7𝑦 = −56
⇒ 𝑦=8
𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑛𝑔 𝑓𝑜𝑟 𝑥
⇒ 𝑥 = 𝑦 + 9 = 8 + 9 = 17
𝐻𝑒𝑛𝑐𝑒 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑎𝑟𝑒 8 & 17
Q.5 The sum of two numbers is 30 and the difference of their squares is 180. Find the
numbers.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑏𝑒 𝑥 𝑎𝑛𝑑 𝑦
𝑥 + 𝑦 = 30 … … … … … … 𝑖)
𝑥 2 − 𝑦 2 = 180 … … … … … … 𝑖𝑖)
𝑆𝑖𝑚𝑝𝑙𝑖𝑓𝑦𝑖𝑛𝑔 𝑖𝑖)
(𝑥 − 𝑦)(𝑥 + 𝑦) = 180
180
⇒𝑥−𝑦=
=6
30
𝑥 − 𝑦 = 6 … … … … … … 𝑖𝑖𝑖)
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑥 𝑎𝑛𝑑 𝑦
𝐴𝑑𝑑 𝑖) 𝑎𝑛𝑑 𝑖𝑖𝑖)
𝑥 + 𝑦 = 30
(+) 𝑥 − 𝑦 = 6
2𝑥 = 36
⇒ 𝑥 = 18
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑖𝑛 𝑖) 𝑦 = 30 − 18 = 12
𝐻𝑒𝑛𝑐𝑒 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑎𝑟𝑒 18 & 12
Q.6 The sum of the digits of a two-digit number is 8. On adding 18 to the number, its digits
are reversed. Find the number.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑏𝑒 𝑥𝑦
𝑥 + 𝑦 = 8 … … … … … … 𝑖)
10𝑥 + 𝑦 + 18 = 10𝑦 + 𝑥 … … … … … … 𝑖𝑖)
𝑆𝑖𝑚𝑝𝑙𝑖𝑓𝑦𝑖𝑛𝑔 𝑖𝑖)
9𝑥 − 9𝑦 = −18
⇒ 𝑥 − 𝑦 = −2 … … … … … … 𝑖𝑖𝑖)
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑥 𝑎𝑛𝑑 𝑦
𝐴𝑑𝑑 𝑖) 𝑎𝑛𝑑 𝑖𝑖𝑖)
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𝑥+𝑦 =8
(+) 𝑥 − 𝑦 = −2
2𝑥 = 6
⇒ 𝑥=3
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑖𝑛 𝑖) ⇒ 𝑦 = 8 − 3 = 5
𝐻𝑒𝑛𝑐𝑒 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑖𝑠 35
Q.7 Two digit number is three times the sum of its digits. lf 45 is added to the number, its
digits are reversed. Find the original number.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑏𝑒 𝑥𝑦
3(𝑥 + 𝑦) = 10𝑥 + 𝑦 … … … … … … 𝑖)
10𝑥 + 𝑦 + 45 = 10𝑦 + 𝑥 … … … … … … 𝑖𝑖)
𝑆𝑖𝑚𝑝𝑙𝑖𝑓𝑦𝑖𝑛𝑔 𝑖) 𝑎𝑛𝑑 𝑖𝑖)
7𝑥 − 2𝑦 = 0 … … … … … … 𝑖𝑖𝑖)
9𝑥 − 9𝑦 = −45
⇒ 𝑥 − 𝑦 = −5 … … … … … … 𝑖𝑣)
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑥 𝑎𝑛𝑑 𝑦
𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑦𝑖𝑛𝑔 𝑖𝑣) 𝑏𝑦 7 𝑎𝑛𝑑 𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡 𝑖𝑣) 𝑓𝑟𝑜𝑚 𝑖𝑖)
7𝑥 − 2𝑦 = 0
(−)7𝑥 − 7𝑦 = −35
5𝑦 = 35
⇒ 𝑦=7
𝐻𝑒𝑛𝑐𝑒 𝑥 = 𝑦 − 5 = 7 − 5 = 2
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑖𝑠 27
Q.8 A two-digit number is seven times the sum of its digits. lf 27 is subtracted from the
number, its digits get interchanged. Find the number.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 𝑏𝑒 𝑥𝑦
10𝑥 + 𝑦 = 7(𝑥 + 𝑦)
⇒ 3𝑥 − 6𝑦 = 0 … … … … … … 𝑖)
10𝑥 + 𝑦 − 27 = 10𝑦 + 𝑥
⇒ 9𝑥 − 9𝑦 = 27
⇒ 𝑥 − 𝑦 = 3 … … … … … … 𝑖𝑖)
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑥 𝑎𝑛𝑑 𝑦
𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑦𝑖𝑛𝑔 𝑖𝑖) 𝑏𝑦 3 𝑎𝑛𝑑 𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡 𝑖𝑖) 𝑓𝑟𝑜𝑚 𝑖)
3𝑥 − 6𝑦 = 0
(−) 3𝑥 − 3𝑦 = 9
−3𝑦 = −9
⇒ 𝑦=3
𝐻𝑒𝑛𝑐𝑒 𝑥 = 𝑦 + 3 = 6
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑖𝑠 63
3
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Q.9 Find a fraction which reduces to 2/3 when 3 is added to both its numerator and
denominator; and reduces to 3/5 when 1 is added to both its numerator and denominator.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑏𝑒
𝑥
𝑦
𝑥+3 2
=
𝑦+3 3
⇒ 3𝑥 + 9 = 2𝑦 + 6
⇒ 3𝑥 − 2𝑦 = −3 … … … … … … 𝑖)
𝑥+1 3
=
𝑦+1 5
⇒ 5𝑥 + 5 = 3𝑦 + 3
⇒ 5𝑥 − 3𝑦 = 2 … … … … … … 𝑖𝑖)
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑥 𝑎𝑛𝑑 𝑦
𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑦𝑖𝑛𝑔 𝑖) 𝑏𝑦 5 𝑎𝑛𝑑 𝑖𝑖) 𝑏𝑦 3 𝑎𝑛𝑑 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡 𝑖𝑖) 𝑓𝑟𝑜𝑚 𝑖)
15𝑥 − 10𝑦 = −15
(−)15𝑥 − 9𝑦 = −6
=
−𝑦 = −9
⇒ 𝑦=9
2𝑦 − 3 18 − 3
𝐻𝑒𝑛𝑐𝑒 𝑥 =
=
=5
3
3
5
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑖𝑠
9
Q.10 On adding 1 to the numerator of a fraction, it becomes ½. Also, on adding 1 to the
denominator of the original fraction, it becomes 1/3. Find the original fraction.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑏𝑒
𝑥
𝑦
𝑥+1 1
=
𝑦
2
⇒ 2𝑥 − 𝑦 = −2 … … … … … … 𝑖)
𝑥
1
=
𝑦+1 3
⇒ 3𝑥 − 𝑦 = 1 … … … … … … 𝑖𝑖)
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑥 𝑎𝑛𝑑 𝑦
𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑦𝑖𝑛𝑔 𝑖) 𝑏𝑦 3 𝑎𝑛𝑑 𝑖𝑖) 𝑏𝑦 2 𝑎𝑛𝑑 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡 𝑖𝑖) 𝑓𝑟𝑜𝑚 𝑖)
6𝑥 − 3𝑦 = −6
(−) 8𝑥 − 2𝑦 = 2
−𝑦 = −8
⇒ 𝑦=8
1+8
𝐻𝑒𝑛𝑐𝑒 𝑥 =
=3
3
3
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑖𝑠
8
4
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Q.11 In a given fraction, if the numerator is multiplied by 2 and the denominator is reduced
by 5, we get 6/5. But, if the numerator of the given fraction is increased by 8 and the
denominator is doubled, we get 2/5. Find the fraction.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑏𝑒
𝑥
𝑦
2𝑥
6
=
𝑦−5 5
⇒ 10𝑥 − 6𝑦 = −30 … … … … … … 𝑖)
𝑥+8 2
=
2𝑦
5
⇒ 5𝑥 − 4𝑦 = −40 … … … … … … 𝑖𝑖)
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑥 𝑎𝑛𝑑 𝑦
𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑦𝑖𝑛𝑔 𝑖𝑖) 𝑏𝑦 2 𝑎𝑛𝑑 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡 𝑖𝑖) 𝑓𝑟𝑜𝑚 𝑖
10𝑥 − 6𝑦 = −3
(−)10𝑥 − 8𝑦 = −80
2𝑦 = 50
⇒ 𝑦 = 25
4𝑦 − 40 100 − 40
𝐻𝑒𝑛𝑐𝑒 𝑥 =
=
= 12
5
5
12
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑖𝑛
25
Q.12 5 years ago, a lady was thrice as old as her daughter. 10 years hence, the lady would
be twice as old as her daughter. What are their present ages?
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑎𝑔𝑒 𝑜𝑓 𝑙𝑎𝑑𝑦 = 𝑥
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑃𝑟𝑒𝑠𝑒𝑛𝑡 𝑎𝑔𝑒 𝑜𝑓 𝐷𝑎𝑢𝑔ℎ𝑡𝑒𝑟 = 𝑦
𝑥 − 5 = 3(𝑦 − 5)
⇒ 𝑥 − 3𝑦 = −10 … … … … … … 𝑖)
𝑥 + 10 = 2(𝑦 + 10)
⇒ 𝑥 − 2𝑦 = 10 … … … … … … 𝑖𝑖)
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑥 𝑎𝑛𝑑 𝑦
𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡 𝑖𝑖) 𝑓𝑟𝑜𝑚 𝑖)
𝑥 − 3𝑦 = −10
(−) 𝑥 − 2𝑦 = 10
−𝑦 = −20
⇒ 𝑦 = 20
𝐻𝑒𝑛𝑐𝑒 𝑥 = 2𝑦 + 10 = 50
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 ∶
𝐿𝑎𝑑𝑦’𝑠 𝐴𝑔𝑒 = 50 𝑦𝑒𝑎𝑟𝑠
𝐷𝑎𝑢𝑔ℎ𝑡𝑒𝑟’𝑠 𝐴𝑔𝑒 = 20 𝑦𝑒𝑎𝑟𝑠
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Q.13 The sum of the ages of A and B is 39 years. In 15 years’ time, the age of A will be twice
the age of B. Find their present ages.
𝐿𝑒𝑡 𝐴’𝑠 𝐴𝑔𝑒 = 𝑥
𝐿𝑒𝑡 𝐵’𝑠 𝐴𝑔𝑒 = 𝑦
𝑥 + 𝑦 = 39 … … … … … … 𝑖)
𝑥 + 15 = 2(𝑦 + 15)
⇒ 𝑥 − 2𝑦 = 15 … … … … … … 𝑖𝑖)
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑥 𝑎𝑛𝑑 𝑦
𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡 𝑖𝑖) 𝑓𝑟𝑜𝑚 𝑖)
𝑥 + 𝑦 = 39
(−) 𝑥 − 2𝑦 = 15
𝑜𝑟 𝑦 = 8
3𝑦 = 24
⇒ 𝑥 = 39 − 𝑦 = 39 − 8 = 31
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒
𝐴’𝑠 𝐴𝑔𝑒 = 31 𝑦𝑒𝑎𝑟𝑠
𝐵’𝑠 𝐴𝑔𝑒 = 8 𝑦𝑒𝑎𝑟𝑠
Q.14 A is 15 years elder than B. 5 years ago A was four times as old as B. Find their present
ages.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑎𝑔𝑒 𝑜𝑓 𝐵 = 𝑥,
⇒ 𝐴𝑔𝑒 𝑜𝑓 𝐴 = 𝑥 + 15
𝑥 + 15 − 5 = 4(𝑥 − 5)
⇒ 𝑥 + 10 = 4𝑥 − 20
3𝑥 = 30 𝑜𝑟 𝑥 = 10
𝐻𝑒𝑛𝑐𝑒 𝐴’𝑠 𝐴𝑔𝑒 = 10 + 15 = 25
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒
𝐵’𝑠 𝐴𝑔𝑒 = 10 𝑦𝑒𝑎𝑟𝑠
𝐴’𝑠 𝐴𝑔𝑒 = 25 𝑦𝑒𝑎𝑟𝑠
Q.15 Six years ago, the ages of Geeta and Seema were in the ratio 3 : 4. Nine years hence,
their ages will be in the ratio 6 : 7. Find their present ages.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑎𝑔𝑒 𝑜𝑓 𝐺𝑒𝑒𝑡𝑎 = 𝑥
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑎𝑔𝑒 𝑜𝑓 𝑆𝑒𝑒𝑚𝑎 = 𝑦
𝑥−6 3
=
𝑦−6 4
⇒ 4𝑥 − 24 = 3𝑦 − 18
⇒ 4𝑥 − 3𝑦 = 6 … … … … … … 𝑖)
𝑥+9 6
=
𝑦+9 7
⇒ 7𝑥 + 63 = 6𝑦 + 54
7𝑥 − 6𝑦 = −9 … … … … … … 𝑖𝑖)
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑥 𝑎𝑛𝑑 𝑦
𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑦𝑖𝑛𝑔 𝑖) 𝑏𝑦 7 𝑎𝑛𝑑 𝑖𝑖) 𝑏𝑦 4 𝑎𝑛𝑑 𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡 𝑖𝑖) 𝑓𝑟𝑜𝑚 𝑖)
6
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28𝑥 − 21𝑦 = 42
(−)28𝑥 − 24𝑦 = −36
3𝑦 = 78
⇒ 𝑦 = 26
6 + 3 × 26
𝐻𝑒𝑛𝑐𝑒 𝑥 =
= 21
4
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒
𝐺𝑒𝑒𝑡𝑎’𝑠 𝐴𝑔𝑒 = 21 𝑦𝑒𝑎𝑟𝑠
𝑆𝑒𝑒𝑚𝑎’𝑠 𝐴𝑔𝑒 = 26 𝑦𝑒𝑎𝑟𝑠
Q. 16 4 knives and 6 forks cost Rs. 200, while 6 knives and,7 forks together cost Rs. 264.
Find the cost of a knife and that of a fork.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑐𝑜𝑠𝑡 𝑜𝑓 𝑘𝑛𝑖𝑣𝑒𝑠 = 𝑥
𝐿𝑒𝑡 𝑡ℎ𝑒 𝐶𝑜𝑠𝑡 𝑜𝑓 𝑓𝑜𝑟𝑘 = 𝑦
4𝑥 + 6𝑦 = 200 … … … … … … 𝑖)
6𝑥 + 7𝑦 = 264 … … … … … … 𝑖𝑖)
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑥 𝑎𝑛𝑑 𝑦
𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑦𝑖𝑛𝑔 𝑖) 𝑏𝑦 3 𝑎𝑛𝑑 𝑖𝑖) 2
12𝑥 + 18𝑦 = 600
(−)12𝑥 + 14𝑦 = 528
4𝑦 = 72
⇒ 𝑦 = 18
200 − 6 × 18
𝐻𝑒𝑛𝑐𝑒 𝑥 =
= 23
4
𝐻𝑒𝑛𝑐𝑒 𝐶𝑜𝑠𝑡 𝑜𝑓 ∶
𝐾𝑛𝑖𝑣𝑒 = 23 𝑅𝑠.
𝐹𝑜𝑟𝑘 = 18 𝑅𝑠.
Q.17 The cost of 13 cups and 16 spoons is Rs. 296, while the cost of 16 cups and 13 spoons
is Rs. 284. Find the cost of2 cups and 5 spoons.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑐𝑜𝑠𝑡 𝑜𝑓 𝐶𝑢𝑝 = 𝑥
𝐿𝑒𝑡 𝑡ℎ𝑒 𝐶𝑜𝑠𝑡 𝑜𝑓 𝑆𝑝𝑜𝑜𝑛 = 𝑦
13𝑥 + 16𝑦 = 296 … … … … … … 𝑖)
16𝑥 + 13𝑦 = 284 … … … … … … 𝑖𝑖)
𝑆𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑥 𝑎𝑛𝑑 𝑦
𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑖) 𝑏𝑦 16 𝑎𝑛𝑑 𝑖𝑖) 𝑏𝑦 13 𝑎𝑛𝑑 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡 𝑖𝑖) 𝑓𝑟𝑜𝑚 𝑖)
208𝑥 + 256𝑦 = 4736
(−) 208𝑥 + 169𝑦 = 3692
87𝑦 = 1044
⇒ 𝑦 = 12
296 − 16 × 12
𝐻𝑒𝑛𝑐𝑒 𝑥 =
=8
13
𝐻𝑒𝑛𝑐𝑒 𝐶𝑜𝑠𝑡 𝑜𝑓:
𝐶𝑢𝑝 = 𝑅𝑠. 8
𝑆𝑝𝑜𝑜𝑛 = 𝑅𝑠. 12
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Q.18 Rahul covered a distance of 128 km in 5 hours, partly on bicycle at 16 kmph and
partly on moped at 32 kmph. How much distance did he cover on moped?
𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 128 𝑘𝑚
𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 = 5 𝐻𝑟𝑠.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑐𝑜𝑣𝑒𝑟𝑒𝑑 𝑏𝑦 𝑐𝑦𝑐𝑙𝑒 = 𝑥 𝑘𝑚
𝐿𝑒𝑡 𝑡ℎ𝑒 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑐𝑜𝑣𝑒𝑟𝑒𝑑 𝑏𝑦 𝑚𝑜𝑝𝑒𝑑 = (120 − 𝑥) 𝐾𝑚.
𝑥 120𝑥
⇒
+
=5
16
32
𝑆𝑖𝑚𝑝𝑙𝑖𝑓𝑦 ⇒ 2𝑥 + 128 − 𝑥 = 160
⇒ 𝑥 = 160 − 128 = 32
𝐻𝑒𝑛𝑐𝑒 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝐶𝑜𝑣𝑒𝑟𝑒𝑑 𝑏𝑦:
𝐶𝑦𝑐𝑙𝑒 = 32𝐾𝑚.
𝑀𝑜𝑝𝑒𝑑 = 96𝐾𝑚.
Q.19 A boat can go 75 km downstream in 5 hours and, 44 km upstream in 4 hours. Find i)
the speed of the boat in still water (ii) the rate of the current.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑏𝑜𝑎𝑡 𝑖𝑛 𝑠𝑡𝑖𝑙𝑙 𝑤𝑎𝑡𝑒𝑟 = 𝑥
𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑆𝑡𝑟𝑒𝑎𝑚 = 𝑦
75
=5
𝑥+𝑦
⇒ 𝑥 + 𝑦 = 17𝑐𝑖)
4𝑥
=4
𝑥−𝑦
⇒ 𝑥 − 𝑦 = 11 … … … … … … 𝑖𝑖)
𝑆𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑥 𝑎𝑛𝑑 𝑦,
𝑎𝑑𝑑 𝑖) 𝑎𝑛𝑑 𝑖𝑖)
2𝑥 = 28 𝑜𝑟 𝑥 = 14 𝐾𝑚 /ℎ𝑟
𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑆𝑡𝑟𝑒𝑎𝑚 = 3 𝐾𝑚/ℎ𝑟
Q.20 The monthly incomes of A and B are in the ratio 4 : 3 and their monthly savings are in
the ratio 9 : 5. If each spends Rs. 3500 per month, find the monthly income of each.
𝐿𝑒𝑡 𝐴’𝑠 𝐼𝑛𝑐𝑜𝑚𝑒 = 𝑥 𝑅𝑠.
𝐿𝑒𝑡 𝐵’𝑠 𝐼𝑛𝑐𝑜𝑚𝑒 = 𝑦 𝑅𝑠.
𝑥 4
=
𝑦 3
⇒ 3𝑥 = 4𝑦 … … … … … … 𝑖)
𝑥 − 3500 9
=
𝑦 − 3500 5
⇒ 5𝑥 − 17500 = 9𝑦 − 31500
⇒ 5𝑥 − 9𝑦 = −14000 … … … … … … 𝑖𝑖)
4
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑥 = 𝑦 𝑖𝑛 𝑖𝑖)
3
4
⇒ 5( 𝑦) − 9𝑦 = −14000
3
8
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−7𝑦 = −14000 × 3 𝑜𝑟 𝑦 = 6000 𝑅𝑠.
4
𝐻𝑒𝑛𝑐𝑒 𝑥 = × 6000 = 8000 𝑅𝑠.
3
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 ∶
𝐴’𝑠 𝐼𝑛𝑐𝑜𝑚𝑒 = 𝑅𝑠. 8000
𝐵’𝑠 𝐼𝑛𝑐𝑜𝑚𝑒 = 𝑅𝑠. 6000
Q.21 6 nuts and 5 bolts weigh 278 grams, while 8 nuts and 3 bolts weigh 268 grams. Find
the weight of each nut and that of each bolt. How much do 3 nuts and 3 bolts weigh
together?
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑁𝑢𝑡 = 𝑥 𝑔𝑚.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝐻𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝐵𝑜𝑙𝑡 = 𝑦 𝑔𝑚.
6𝑥 + 5𝑦 = 278 … … … … … … 𝑖)
8𝑥 + 3𝑦 = 268 … … … … … … 𝑖𝑖)
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑥 𝑎𝑛𝑑 𝑦
𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑖) 𝑏𝑦 8 𝑎𝑛𝑑 𝑖𝑖) 𝑏𝑦 6 𝑎𝑛𝑑 𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡 𝑖𝑖) 𝑓𝑟𝑜𝑚 𝑖)
48𝑥 + 40𝑦 = 2224
(−)48𝑥 + 18𝑦 = 1608
22𝑦 = 616
⇒ 𝑦 = 28
278 − 5 × 28
𝐻𝑒𝑛𝑐𝑒 𝑥 =
= 23
6
𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓:
𝑁𝑢𝑡 = 23 𝑔𝑚.
𝐵𝑜𝑙𝑡 = 28𝑔𝑚.
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 3 𝑁𝑢𝑡𝑠 𝑎𝑛𝑑 3 𝐵𝑜𝑙𝑡𝑠 𝑤𝑖𝑙𝑙 𝑤𝑒𝑖𝑔ℎ𝑡 = 3 × 23 + 3 × 28 = 153 𝑔𝑚.
Q.22 There are some girls in two classrooms, A and B. If 12 girls are sent from room A to
room B, the number of girls in both the rooms will become equal. If 11 girls are sent from
room B to room A, then the number of girls in room A would be double the number of girls
in room B. How many girls are there in each class room?
𝐿𝑒𝑡 𝑁𝑜. 𝑜𝑓 𝑔𝑖𝑟𝑙𝑠 𝑖𝑛 𝑐𝑙𝑎𝑠𝑠𝑟𝑜𝑜𝑚 𝐴 = 𝑥 𝑎𝑛𝑑 𝑖𝑛 𝐵 = 𝑦
𝑥 − 12 = 𝑦 + 12
⇒ 𝑥 − 𝑦 = 24 … … … … … … 𝑖)
2(𝑦 − 11) = 𝑥 + 11
⇒ 𝑥 − 2𝑦 = −33 … … … … … … 𝑖𝑖)
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑥 𝑎𝑛𝑑 𝑦,
𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡 𝑖𝑖) 𝑓𝑟𝑜𝑚 𝑖)
𝑥 − 𝑦 = 24
(−) 𝑥 − 2𝑦 = −33
𝑦 = 57
𝐻𝑒𝑛𝑐𝑒 𝑥 = 24 + 57 = 81
𝑁𝑜. 𝑜𝑓 𝐺𝑖𝑟𝑙𝑠 𝑖𝑛 𝐶𝑙𝑎𝑠𝑠 𝑅𝑜𝑜𝑚:
𝐴 = 81
𝐵 = 57
9
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Q.23 4 men and 4 boys can do a piece of work in 3 days, while 2 men and 5 boys can finish
it in 4 days. How long would it take 1 man alone to do it?
𝑆𝑢𝑝𝑝𝑜𝑠𝑒 1 𝑚𝑎𝑛 𝑓𝑖𝑛𝑖𝑠ℎ𝑒𝑑 𝑡ℎ𝑒 𝑤𝑜𝑟𝑘 𝑖𝑛 𝑥 𝑑𝑎𝑦𝑠 𝑎𝑛𝑑 1 𝐵𝑜𝑦 𝑓𝑖𝑛𝑖𝑠ℎ𝑒𝑑 𝑡ℎ𝑒 𝑤𝑜𝑟𝑘 𝑖𝑛 𝑦 𝑑𝑎𝑦𝑠
1
𝑇ℎ𝑒𝑟𝑒 1 𝑚𝑎𝑛’𝑠 1 𝑑𝑎𝑦’𝑠 =
𝑥
1
𝐴𝑛𝑑, 1 𝐵𝑜𝑦’𝑠 1 𝐷𝑎𝑦’𝑠 𝑤𝑜𝑟𝑘 =
𝑦
𝑁𝑜𝑤 4 𝑚𝑒𝑛 𝑎𝑛𝑑 4 𝐵𝑜𝑦𝑠 𝑓𝑖𝑛𝑖𝑠ℎ𝑒𝑑 𝑡ℎ𝑒 𝑤𝑜𝑟𝑘 𝑖𝑛 3 𝑑𝑎𝑦𝑠
4 5 1
⇒ + = … … … … … … 𝑖)
𝑥 𝑦 3
𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦 2 𝑚𝑒𝑛 𝑎𝑛𝑑 5 𝑏𝑜𝑦𝑠 𝑓𝑖𝑛𝑖𝑠ℎ𝑒𝑑 𝑖𝑛 4 𝑑𝑎𝑦𝑠
2 5 1
⇒ + = … … … … … … 𝑖𝑖)
𝑥 𝑦 4
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑥 𝑎𝑛𝑑 𝑦
𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑖𝑖) 𝑏𝑦 2 𝑎𝑛𝑑 𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡 𝑖𝑖) 𝑓𝑟𝑜𝑚 2
4 4 1
⇒ + =
𝑥 𝑦 3
4 10 1
(−) +
=
𝑥 𝑦
2
⇒
𝑜𝑟 𝑦 = 36
−6 −1
=
𝑦
6
𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦 𝑥 = (
4
)=
4 × 36
= 18
8
1
4
–
3 36
𝑆𝑜 𝑜𝑛𝑒 𝑚𝑒𝑛 𝑐𝑎𝑛 𝑓𝑖𝑛𝑖𝑠ℎ𝑒𝑑 𝑡ℎ𝑒 𝑤𝑜𝑟𝑘 𝑖𝑛 18 𝑑𝑎𝑦𝑠.
Q.24 A takes 3 hours longer than B to walk 30 km. But, if A doubles his pace, he is ahead of
B by 1 hour 30 minutes. Find the speeds of A and B.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝐴 = 𝑥 𝑘𝑚/𝐻𝑟.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝐵 = 𝑦 𝑘𝑚/𝐻𝑟.
30
𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 𝐵𝑦 𝐴 =
𝑥
30
𝑇𝑖𝑚𝑒 𝑇𝑎𝑘𝑒𝑛 𝐵𝑦 𝐵 =
𝑦
30 30
⇒
=
+ 3___________________𝑖)
𝑥
𝑦
30
𝐼𝑓 𝐴 𝐷𝑜𝑢𝑏𝑙𝑒 ℎ𝑖𝑚 𝑅𝑎𝑐𝑒 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 𝑏𝑦 𝐴 =
2𝑥
30 3 30
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒
+ =
2𝑥 2
𝑦
30 30
(−)
=
+3
𝑥
𝑦
⇒
−30 3
+ = −3
2𝑥
2
10
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10
1
⇒ 𝑥=
= 3 𝐾𝑚/𝐻𝑟.
3
3
30 30
𝐻𝑒𝑛𝑐𝑒
=
×3−3 =6
𝑦
10
30
⇒ 𝑦=
= 5 𝑘𝑚/𝐻𝑟.
6
Q.25 If the length of a rectangle is reduced by 1 m and breadth increased by 2 m, its area
increases by 32 m2. If however, the length is increased by 2 m and breadth reduced by 3 m,
then the area is reduced by 49 m2. Find the length and breadth of the rectangle.
𝐿𝑒𝑡 𝑡ℎ𝑒 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒 = 𝑥
𝐿𝑒𝑡 𝑡ℎ𝑒 𝐵𝑟𝑒𝑎𝑑𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒 = 𝑦
⇒ (𝑥 − 1)(𝑦 + 2) = 𝑥𝑦 + 32
⇒ 𝑥𝑦 − 𝑦 + 2𝑥 − 2 = 𝑥𝑦 + 32
⇒ 2𝑥 − 𝑦 = 34_____________𝑖)
⇒ (𝑥 + 2)(𝑦 − 3) = 𝑥𝑦 − 49
⇒ 3𝑥 − 2𝑦 − 3𝑥 − 6 = 𝑥𝑦 − 49
⇒ 3𝑥 − 2𝑦 = 43____________𝑖𝑖)
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑥 & 𝑦
𝐹𝑟𝑜𝑚 𝑖) 𝑦 = 2𝑥 − 34
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑖𝑛 𝑖𝑖) ⇒ 3𝑥 − 2(2𝑥 − 34) = 43
⇒ 3𝑥 − 4𝑥 + 68 = 43
⇒ 𝑥 = 25𝑚
𝐻𝑒𝑛𝑐𝑒 𝑦 = 2(25) − 34 = 16𝑚
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒:
𝐿𝑒𝑛𝑔𝑡ℎ = 25𝑚
𝐵𝑟𝑒𝑎𝑑𝑡ℎ = 16𝑚.
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