1.Form the pair of linear equations in the following problems. (1 mark each)
a) The cost of 2 kg of apples and 1kg of grapes on a day was found to be
Rs.160. After a month, the cost of 4 kg of apples and 2 kg of grapes is
Rs.300.
b) Half the perimeter of a rectangular garden, whose length is 4 m more
than its width, is 36 m.
c) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will
be twice as old as Sonu.
d) A fraction becomes 9/11 if 2 is added to both the numerator and the
denominator. If, 3 is added to both the numerator and the denominator
it becomes 5/6.
e) The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later,
she buys 3 bats and 5 balls for Rs.1750.
f) The sum of a two-digit number and the number formed by
interchanging its digits is 110. If 10 is subtracted from the first number,
the new number is 4 more than 5 times the sum of the digits in the first
number.
Answers
Answer:
Correct equations :-
\sf{\:\dfrac{4}{16x+24z}+\dfrac{12}{21x-14z}=\dfrac{1}{2}}
16x+24z
4
+
21x−14z
12
=
2
1
\sf{\:\dfrac{14}{4{\bf{x}}+6z}+\dfrac{4}{3x-2z}=2}
4x+6z
14
+
3x−2z
4
=2
Solution :-
Given first equation:
\sf\implies\dfrac{4}{16x+24z}+\dfrac{12}{21x-14z}=1/2⟹
16x+24z
4
+
21x−14z
12
=1/2
\sf\implies\dfrac{4}{4(x+6z)}+\dfrac{12}{7(3x-2z)}=1/2⟹
4(x+6z)
4
+
7(3x−2z)
12
=1/2
\sf\implies\dfrac{1}{4x+6z}+\dfrac{12}{7(3x-2z)}=1/2⟹
4x+6z
1
+
7(3x−2z)
12
=1/2
Assume that :-
4x + 6z = u
3x - 2z = v
\sf\implies\dfrac{1}{u}+\dfrac{12}{7v}=1/2⟹
u
1
+
7v
12
=1/2
\sf\implies\dfrac{7v+12u}{7vu}=1/2⟹
7vu
7v+12u
=1/2
\sf\implies 14v + 24 u = 7vu---(1.)⟹14v+24u=7vu−−−(1.)
Given second equation:
\sf\implies\dfrac{14}{4x+6z}+\dfrac{4}{3x-2z}=2⟹
4x+6z
14
+
3x−2z
4
=2
\sf\implies\dfrac{14}{u}+\dfrac{4}{v}=2⟹
u
14
+
v
4
=2
\sf\implies\dfrac{14v+4u}{vu}=2⟹
vu
14v+4u
=2
\sf\implies{14v+4u}=2vu---(2.)⟹14v+4u=2vu−−−(2.)
Eq(1) - Eq(2)
\implies⟹ 14v+24u - (14 v + 4 u) =7vu-2vu
\implies⟹ 24u - 4u = 5vu
\implies⟹ 20 u = 5vu
\implies⟹ 4 = v
Put v in eq(2)
\implies⟹ 14(4) + 4u = 2(4)u
\implies⟹ 56 + 4u = 8u
\implies⟹ 56 = 8u - 4u
\implies⟹ 56 = 4u
\implies⟹ 56/4 = u
\implies⟹ 14 = u
Now we get :-
u = 14 = 4x + 6z
v = 4 = 3x - 2z ---(3)
Multiply equation (3) with 3
\implies⟹ 3(4 = 3x -2z)
\implies⟹ 12 = 9x - 6z ---(4)
Add u and eq(4)
\implies⟹ 14+12 = 4x + 6z + 9x - 6z
\implies⟹ 26 = 13x
\implies⟹ 26/13 = x
\implies⟹ 2 = x
So the value of x is 2.
Put this value of x in u
\implies⟹ 14 = 4(2) + 6z
\implies⟹ 14 = 8 + 6z
\implies⟹ 14 - 8 = 6z
\implies⟹ 6 = 6z
\implies⟹ 1 = z
So the value of z is 1.