Linear Equalities Practice Questions - DPP for JEE

 Page 2


Line which passes from (5, 0) and (0, 14) is
14 x + 5y = 70
? In the shaded region 14x + 5y = 70
Line which passes from (5, 0) and (19, 14) is
x – y – 5 = 0
? In the shaded region x – y = 5
Thus, inequations are 14 x + 5y = 70, x – y = 5, y = 14.
4. (c)
5. (d) We can add two inequalities of the same type.  So by adding above
two inequalities we get 
The range of x + y is given by, 2 < x + y < 38
6. (b)
7. (c)
and  or  or 
Taking the common values of x, we get 
8. (c) We have
IQ = × 100
? IQ = × 100 [ CA = 12 years]
= 
Page 3


Line which passes from (5, 0) and (0, 14) is
14 x + 5y = 70
? In the shaded region 14x + 5y = 70
Line which passes from (5, 0) and (19, 14) is
x – y – 5 = 0
? In the shaded region x – y = 5
Thus, inequations are 14 x + 5y = 70, x – y = 5, y = 14.
4. (c)
5. (d) We can add two inequalities of the same type.  So by adding above
two inequalities we get 
The range of x + y is given by, 2 < x + y < 38
6. (b)
7. (c)
and  or  or 
Taking the common values of x, we get 
8. (c) We have
IQ = × 100
? IQ = × 100 [ CA = 12 years]
= 
Given, 80 = IQ = 140
? 80 = = 140
? 240 = 25MA = 420
? = MA = 
? 9.6 = MA = 16.8
9. (b) Given in-equations are
–17 = 3x + 10 = –2
?  –27 = 3 x =–12
?  –9 = x = –4 …(i)
–22 = 5x + 13 = 3
?  –35 = 5x = –10
?  –7 = x  = –2 …(ii)
–19 = 2x – 9 = –3
?  –10 = 2x = 6
?  –5 = x = 3 …(iii)
The common range of values that satisfies all the equations is [– 5, –  4]
10. (b)
11. (c) Let Ankur got ‘x’  marks in fifth subject.
So, average marks = 
Given 70      75
350  300 + x  375
50  x  75
Page 4


Line which passes from (5, 0) and (0, 14) is
14 x + 5y = 70
? In the shaded region 14x + 5y = 70
Line which passes from (5, 0) and (19, 14) is
x – y – 5 = 0
? In the shaded region x – y = 5
Thus, inequations are 14 x + 5y = 70, x – y = 5, y = 14.
4. (c)
5. (d) We can add two inequalities of the same type.  So by adding above
two inequalities we get 
The range of x + y is given by, 2 < x + y < 38
6. (b)
7. (c)
and  or  or 
Taking the common values of x, we get 
8. (c) We have
IQ = × 100
? IQ = × 100 [ CA = 12 years]
= 
Given, 80 = IQ = 140
? 80 = = 140
? 240 = 25MA = 420
? = MA = 
? 9.6 = MA = 16.8
9. (b) Given in-equations are
–17 = 3x + 10 = –2
?  –27 = 3 x =–12
?  –9 = x = –4 …(i)
–22 = 5x + 13 = 3
?  –35 = 5x = –10
?  –7 = x  = –2 …(ii)
–19 = 2x – 9 = –3
?  –10 = 2x = 6
?  –5 = x = 3 …(iii)
The common range of values that satisfies all the equations is [– 5, –  4]
10. (b)
11. (c) Let Ankur got ‘x’  marks in fifth subject.
So, average marks = 
Given 70      75
350  300 + x  375
50  x  75
12. (a)
13. (a) Profit = Revenue – Cost
Revenue – cost > 0 for some profit
(5x + 20) – (2x + 80) > 0
3x – 60 > 0
x > 20
Thus, the manufacturer have to sale more than 20 items to earn some
profit.
14. (b) Let the shortest side be x cm.
Then, by given condition, second length = x + 3 cm
Third length = 2x cm
Also given, total length = 91
Hence, sum of all the three lengths should be less than or equal to 91
x + x + 3 + 2x = 91
? 4x + 3 = 91
Subtracting (–3) to each term,
–3 + 4x + 3 = 91 – 3
? 4x = 88
?  ?   x = 
? x = 22 cm ... (i)
Again, given that
Third length = second length + 5
? 2x = (x + 3) + 5
? 2x = x + (3 + 5)
Transferring the term x to L.H.S.,
2x – x = 8
? x = 8 ... (ii)
From equations (i) and (ii), length of shortest board should be greater
than or equal to 8 but less than or equal to 22, i.e., 8 = x = 22.
15. (a) The equations, corresponding to inequalities
Page 5


Line which passes from (5, 0) and (0, 14) is
14 x + 5y = 70
? In the shaded region 14x + 5y = 70
Line which passes from (5, 0) and (19, 14) is
x – y – 5 = 0
? In the shaded region x – y = 5
Thus, inequations are 14 x + 5y = 70, x – y = 5, y = 14.
4. (c)
5. (d) We can add two inequalities of the same type.  So by adding above
two inequalities we get 
The range of x + y is given by, 2 < x + y < 38
6. (b)
7. (c)
and  or  or 
Taking the common values of x, we get 
8. (c) We have
IQ = × 100
? IQ = × 100 [ CA = 12 years]
= 
Given, 80 = IQ = 140
? 80 = = 140
? 240 = 25MA = 420
? = MA = 
? 9.6 = MA = 16.8
9. (b) Given in-equations are
–17 = 3x + 10 = –2
?  –27 = 3 x =–12
?  –9 = x = –4 …(i)
–22 = 5x + 13 = 3
?  –35 = 5x = –10
?  –7 = x  = –2 …(ii)
–19 = 2x – 9 = –3
?  –10 = 2x = 6
?  –5 = x = 3 …(iii)
The common range of values that satisfies all the equations is [– 5, –  4]
10. (b)
11. (c) Let Ankur got ‘x’  marks in fifth subject.
So, average marks = 
Given 70      75
350  300 + x  375
50  x  75
12. (a)
13. (a) Profit = Revenue – Cost
Revenue – cost > 0 for some profit
(5x + 20) – (2x + 80) > 0
3x – 60 > 0
x > 20
Thus, the manufacturer have to sale more than 20 items to earn some
profit.
14. (b) Let the shortest side be x cm.
Then, by given condition, second length = x + 3 cm
Third length = 2x cm
Also given, total length = 91
Hence, sum of all the three lengths should be less than or equal to 91
x + x + 3 + 2x = 91
? 4x + 3 = 91
Subtracting (–3) to each term,
–3 + 4x + 3 = 91 – 3
? 4x = 88
?  ?   x = 
? x = 22 cm ... (i)
Again, given that
Third length = second length + 5
? 2x = (x + 3) + 5
? 2x = x + (3 + 5)
Transferring the term x to L.H.S.,
2x – x = 8
? x = 8 ... (ii)
From equations (i) and (ii), length of shortest board should be greater
than or equal to 8 but less than or equal to 22, i.e., 8 = x = 22.
15. (a) The equations, corresponding to inequalities
3x + 2y = 6 and 6x + 4y = 20, are 3x + 2y = 6 and 6x + 4y = 20. So the
lines represented by these equations are parallel. Hence the graphs
are disjoint.
16. (a) We have 
Now two cases arise :
Case I : When  i.e., x – 3. Then
{(x + 1) > 0 and x + 2 > 0}
or {x + 1 < 0 and x + 2 < 0}
{x > – 1 and x  > – 2}
or {x < – 1 and x < –2}
x > – 1 or x < – 2
 or 
[Since ] …(i)
Case II : When x + 3 < 0, i.e., x < – 3