Read RD Sharma Solutions Class 7 Chapter 8 Linear Equations in One Variable below, students should study RD Sharma class 7 Mathematics available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 7 Mathematics have been prepared by teacher of Grade 7. These RD Sharma class 7 Solutions have been designed as per the latest NCERT syllabus for class 7 and if practiced thoroughly can help you to score good marks in standard 7 Mathematics class tests and examinations
Exercise 8.1
Question 1: Verify by substitution that:
(i) x = 4 is the root of 3 x – 5 = 7
(ii) x = 3 is the root of 5 + 3 x = 14
(iii) x = 2 is the root of 3 x – 2 = 8 x – 12
(iv) x = 4 is the root of (3x/2) = 6
(v) y = 2 is the root of y – 3 = 2y – 5
(vi) x = 8 is the root of (1/2) x + 7 = 11
Solution 1:
(i) x = 4 is the root of 3 x – 5 = 7
Replacing x = 4 at the place of ‘x’ in the equation,
= 3(4) – 5 = 7
= 12 – 5 = 7
7 = 7
LHS = RHS
So, x = 4 is the root of 3 x – 5 = 7.
(ii) x = 3 is the root of 5 + 3 x = 14.
Replacing x = 3 at the place of ‘x’ in the equation,
= 5 + 3(3) = 14
= 5 + 9 = 14
14 = 14
LHS = RHS
So, x = 3 is the root of 5 + 3 x = 14.
(iii) x = 2 is the root of 3 x – 2 = 8 x – 12.
Replacing x = 2 at the place of ‘x’ in the equation,
= 3(2) – 2 = 8(2) – 12
= 6 – 2 = 16 – 12
4 = 4
LHS = RHS
So, x = 2 is the root of 3 x – 2 = 8 x – 12.
(iv) x = 4 is the root of 3x/2 = 6.
Replacing x = 4 at the place of ‘x’ in the equation,
= ((3×4))/2 = 6
= (12/2) = 6
6 = 6
LHS = RHS
So, x = 4 is the root of (3x/2) = 6.
(v) y = 2 is the root of y – 3 = 2y – 5.
Replacing y = 2 at the place of ‘y’ in the equation,
= 2 – 3 = 2(2) – 5
= -1 = 4 – 5
-1 = -1
LHS = RHS
So, y = 2 is the root of y – 3 = 2y – 5.
(vi) x = 8 is the root of (1/2) x + 7 = 11.
Replacing x = 8at the place of ‘x’ in the equation,
= (1/2)(8) + 7 =11
= 4 + 7 = 11
= 11 = 11
LHS = RHS
So, x = 8 is the root of 12 x + 7 = 11.
Question 2: Solve each of the following equations by trial – and – error method:
(i) x + 3 =12
(ii) x -7 = 10
(iii) 4x = 28
(iv) (x/2) + 7 = 11
(v) 2x + 4 = 3x
(vi) (x/4) = 12
(vii) (15/x) = 3
(vii) (x/18) = 20
Solution 2:
(i) x + 3 =12
LHS = x +3 and RHS = 12
Exercise 8.2
Solve each of the following equations and check your answers:
Question 1: x – 3 = 5
Solution 1:
x – 3 = 5
By Adding 3 to both side,
x – 3 + 3 = 5 + 3
x = 8
Verification:
Replacing x = 8 in LHS,
LHS = x – 3 and RHS = 5
LHS = 8 – 3 = 5 and RHS = 5
Verified LHS = RHS
Question :2. x + 9 = 13
Solution 2:
x + 9 = 13
By Subtracting 9 from both sides
x + 9 – 9 = 13 – 9
x = 4
Verification:
Replacing x = 4 on LHS,
LHS = 4 + 9 = 13 = RHS
Verified LHS = RHS
Question 3: x – (3/5) = (7/5)
Solution 3:
x + 9 = 13
By Subtracting 9 from both sides
x + 9 – 9 = 13 – 9
x = 4
Verification:
Substitute x = 2 in LHS
2 – (3/5) = (7/5)
((10-3))/5 = (7/5)
(7/5) = (7/5)
Verified LHS = RHS
Question 4: 3 x = 0
Solution 4:
3 x = 0
By dividing both sides by 3,
(3x/3) = (0/3)
x = 0
Verification:
Replacing x = 0 in LHS
3 (0) = 0
And RHS = 0
Verified LHS = RHS
Question :5. (x/2) = 0
Solution 5:
(x/2) = 0
By multiplying both sides by 2,
(x/2) × 2 = 0 × 2
x = 0
Verification:
Replacing x = 0 in LHS,
LHS = 0/2= 0 and RHS = 0
LHS = 0 and RHS = 0
Verified LHS = RHS
Question 6: x – (1/3) = (2/3)
Solution 6:
x – (1/3) = (2/3)
By Adding (1/3) to both sides,
x – (1/3) + (1/3) = (2/3) + (1/3)
x = ((2+1))/3
x = (3/3)
x =1
Verification:
Replacing x = 1 in LHS,
1 – (1/3) = (2/3)
((3-1))/3 = (2/3)
(2/3) = (2/3)
Verified LHS = RHS
Question 7: x + (1/2) = (7/2)
Solution 7:
x + (1/2) = (7/2)
By Subtracting (1/2) from both sides,
x + (1/2) – (1/2) = (7/2) – (1/2)
x = ((7-1))/2
x = (6/2)
x = 3
Verification:
Replacing x = 3 in LHS
3 + (1/2) = (7/2)
((6+1))/2 =(7/2)
(7/2) = (7/2)
Verified LHS = RHS
Question 8: 10 – y = 6
Solution 8:
10 – y = 6
By subtracting 10 from both sides,
10 – y – 10 = 6 – 10
-y = -4
By Multiplying both sides by -1,
-y × -1 = – 4 × – 1
y = 4
Verification:
Replacing y = 4 in LHS,
10 – y = 10 – 4 = 6 and RHS = 6
Verified LHS = RHS
Question 9: 7 + 4y = -5
Solution 9:
7 + 4y = -5
By Subtracting 7 from both sides,
7 + 4y – 7 = -5 -7
4y = -12
By Dividing both sides by 4,
y = (-12)/4
y = -3
Verification:
Replacing y = -3 in LHS,
7 + 4y = 7 + 4(-3) = 7 – 12 = -5, and RHS = -5
verified LHS = RHS
Question 10: (4/5) – x = (3/5)
Solution 10:
(4/5) – x = (3/5)
By Subtracting (4/5) from both sides,
(4/5) – x – (4/5) = (3/5) – (4/5)
– x = ((3-4))/5
– x = ((-1)/5)
x = (1/5)
Verification:
Replacing x = (1/5) in LHS
(4/5) – (1/5) = (3/5)
((4-1))/5 = (3/5)
(3/5) = (3/5)
Verified LHS =RHS
Question 11: 2y – (1/2) = ((-1)/3)
Solution 11:
x2y – (1/2) = ((-1)/3)
By Adding (1/2) from both the sides,
2y – (1/2) + (1/2) = ((-1)/3) + (1/2)
2y = ((-1)/3) + (1/2)
2y = ((-2+3))/6 [LCM of 3 and 2 is 6]
2y = (1/6)
Now divide both the side by 2,
y = (1/12) + 9 = 13
By Subtracting 9 from both sides
y + 9 – 9 = 13 – 9
y = 4
Verification:
Replacing y = (1/12) in LHS
2 (1/12) – (1/2) = ((-1)/3)
(1/6) – (1/2) = ((-1)/3)
((2-6))/12 = ((-1)/3) [LCM of 6 and 2 is 12]
((-4)/12) = ((-1)/3)
((-1)/3) = ((-1)/3)
Verified LHS = RHS
Question 12: 14 = (7x/10) – 8
Solution 12:
14 = (7x/10) – 8
By Adding 8 to both sides,
14 + 8 = (7x/10) – 8 + 8
22 = (7x/10)
By Multiply both sides by 10 ,
220 = 7 x
x = (220/7)
Verification:
Replacing x = (220/7) in RHS,
14 = (7/10) × (220/7) – 8
14 = 22 -8
14 = 14
Verified LHS = RHS.
Question 13: 3 (x + 2) = 15
Solution 13:
3 (x + 2) = 15
By Dividing both sides by 3,
3 ((x+2))/3 = (15/3)
(x + 2) = 5
Now subtracting 2 by both sides,
x + 2 -2 = 5 -2
x = 3
Verification:
Replacing x =3 in LHS ,
3 (3 + 2) = 15
3 (5) = 15
15 = 15
Verified LHS = RHS
Question 14: (x/4) = (7/8)
Solution 14:
(x/4) = (7/8)
By Multiply both sides by 4,
(x/4) × 4 = (7/8) × 4
x = (7/2) Verification:
Replacing x = (7/2) in LHS,
((7/2))/4 = (7/8)
(7/8) = (7/8)
Verified LHS = RHS
Question 15: (1/3) – 2 x = 0
Solution 15:
(1/3) – 2 x = 0
By Subtract (1/3) from both sides,
(1/3) – 2 x – (1/3) = 0 – (1/3)
– 2 x = – (1/3)
2 x = (1/3)
By Divide both side by 2,
2x/2 = ((1/3))/2
x = (1/6)
Verification:
Replacing x = (1/6) in LHS,
(1/3) – 2 (1/6) = 0
(1/3) – (1/3) = 0
0 = 0
Verified LHS = RHS
Question 16: 3 (x + 6) = 24
Solution 16:
3 (x + 6) = 24
By Divide both the sides by 3,
3 ((x+6))/3 = (24/3)
(x + 6) = 8
By Subtract 6 from both sides,
x + 6 – 6 = 8 – 6
x = 2
Verification:
Replacing x = 2 in LHS,
3 (2 + 6) = 24
3 (8) =24
24 = 24
Verified LHS =RHS
Question 17: 3 (x + 2) – (2 x – 1) = 7
Solution 17:
3 (x + 2) – 2 (x – 1) = 7
By simplifying the brackets,
3 × x + 3 × 2 – 2 × x + 2 × 1 = 7
3 x + 6 – 2 x + 2 = 7
3 x – 2x + 6 + 2 = 7
x + 8 = 7
By Subtracting 8 from both sides,
x + 8 – 8 = 7 – 8
x = -1
Verification:
Replacing x = -1 in LHS,
3 (x + 2) -2(x -1) = 7
3 (-1 + 2) -2(-1-1) = 7
(3×1) – (2×-2) = 7
3 + 4 = 7
Verified LHS = RHS
Question 18: 8 (2 x – 5) – 6(3 x – 7) = 1
Solution 18:
8 (2 x – 5) – 6(3 x – 7) = 1
By simplifying the brackets,
(8 × 2 x) – (8 × 5) – (6 × 3 x) + (-6) × (-7) = 1
16 x – 40 – 18 x + 42 = 1
16 x – 18 x + 42 – 40 = 1
-2 x + 2 = 1
By Subtracting 2 from both sides,
-2 x + 2 – 2 = 1 -2
-2 x = -1
By Multiplying both sides by -1,
-2 x × (-1) = -1× (-1)
2 x = 1
By Dividing both sides by 2,
2x/2 = (1/2)
x = (1/2)
Verification:
Replacing x = (1/2) in LHS,
(8 × (2 × (1/2)–5) – (6 × (3 × (1/2)-7) = 1
8(1 – 5) – 6( 3/2 – 7) = 1
8× (-4) – (6 × 3/2 ) + (6 × 7) = 1
– 32 – 9 + 42 = 1
– 41 + 42 = 1
1 = 1
Verified LHS = RHS
Question 19: 6 (1 – 4 x) + 7 (2 + 5 x) = 53
Solution 19:
6 (1 – 4 x) + 7 (2 + 5 x) = 53
By simplifying the brackets,
(6 ×1) – (6 × 4 x) + (7 × 2) + (7 × 5 x) = 53
6 – 24 x + 14 + 35 x = 53
6 + 14 + 35 x – 24 x = 53
20 + 11 x = 53
By Subtracting 20 from both sides,
20 + 11 x – 20 = 53 – 20
11 x = 33
By Dividing both sides by 11,
11x/11 = 33/11
x = 3
Verification:
Replacing x = 3 in LHS,
6(1 – 4 × 3) + 7(2 + 5 × 3) = 53
6(1 – 12) + 7(2 + 15) = 53
6(-11) + 7(17) = 53
– 66 + 119 = 53
53 = 53
Verified LHS = RHS
Question 20: 5 (2 – 3 x) -17 (2 x -5) = 16
Solution 20:
5 (2 – 3 x) -17 (2 x – 5) = 16
By expanding the brackets,
(5 × 2) – (5 × 3 x) – (17 × 2 x) + (17 × 5) = 16
10 – 15 x – 34 x + 85 = 16
10 + 85 – 34 x – 15 x = 16
95 – 49 x = 16
By Subtracting 95 from both sides,
– 49 x + 95 – 95 = 16 – 95
– 49 x = -79
Dividing both sides by – 49,
(-49x)/(-49) = (-79)/(-49)
x = 79/49
Verification:
Replacing x = (79/49) in LHS ,
5 (2 – 3 × (79/49) – 17 (2 × (79/49) – 5) = 16
(5 × 2) – (5 × 3 × (79/49)) – (17 × 2 × (79/49)) + (17 × 5) = 16
10 – (1185/49) – (2686/49) + 85 = 16
((490-1185-2686+4165))/49 = 16
784/49 = 16
16 = 16
Verified LHS = RHS
Question 21: ((x-3))/5 -2 = -1
Solution 21:
(((x-3))/5) -2 = -1
Adding 2 to both sides,
(((x-3))/5) – 2 + 2 = -1 + 2
((x-3))/5 = 1
Multiply both sides by 5,
((x-3))/5 × 5 = 1 × 5
x– 3 = 5
Add 3 to both sides,
x – 3 + 3 = 5 + 3
x = 8
Verification:
Replacing x = 8 in LHS,
(((8-3))/5) -2 = -1
(5/5) – 2 = -1
1 -2 = -1
-1 = -1
Verified LHS = RHS
Question 22: 5 (x – 2) + 3 (x +1) = 25
Solution 22:
5 (x – 2) + 3 (x +1) = 25
By simplifying the brackets,
(5 × x) – (5 × 2) +3 × x + 3× 1 = 25
5 x – 10 + 3 x + 3 = 25
5 x + 3 x – 10 + 3 = 25
8 x – 7 = 25
By Adding 7 to both sides,
8 x – 7 + 7 = 25 + 7
8 x = 32
By Dividing both sides by 8,
8x/8 = 32/8
x = 4
Verification:
Replacing x = 4 in LHS,
5(4 – 2) + 3(4 + 1) = 25
5(2) + 3(5) = 25
10 + 15 = 25
25 = 25
Verified LHS = RHS
Exercise 8.3
Solve each of the following equations. Also, verify the result in each case.
Question 1: 6 x + 5 = 2 x + 17
Solution 1:
6 x + 5 = 2 x + 17
Rearranging 2x to LHS & 5 to RHS,
6 x – 2 x = 17 – 5
4 x = 12
Now, Dividing 4 both sides,
4x/4 = 12/4
x = 3 Verification:
Replacing x = 3 in the given equation,
6 × 3 + 5 = 2 × 3 + 17
18 + 5 = 6 + 17
23 = 23
Verified LHS = RHS
Question 2: 2 (5x – 3) – 3 (2x – 1) = 9
Solution 2:
2 (5 x – 3) – 3 (2 x – 1) = 9
Simplifying the brackets,
2 × 5 x – 2 × 3 – 3 × 2 x + 3 × 1 = 9
10 x – 6 – 6 x + 3 = 9
10 x – 6 x – 6 + 3 = 9
4 x – 3 = 9
By adding 3 to both sides,
4 x – 3 + 3 = 9 + 3
4 x = 12
By dividing both sides by 4,
4x/4 = 12/4
x = 3. Verification:
Replacing x = 3 in LHS,
2(5 × 3 – 3) – 3(2 × 3 – 1) = 9
2 × 12 – 3 × 5 = 9
24 – 15 = 9
9 = 9
Verified LHS = RHS
Question 3: (x/2) = (x/3) + 1
Solution 3:
(x/2) = (x/3) + 1
Rearranging (x/3) to LHS
(x/2) – (x/3) = 1
((3x-2x))/6 = 1 [LCM of 3 and 2 is 6]
x/6 = 1
By multiplying 6 to both sides
x = 6 Verification:
Replacing x = 6 in given equation
(6/2) = (6/3) + 1
3 = 2 + 1
3 = 3
Verified LHS = RHS
Question 4: (x/2) + (3/2) = (2x/5) – 1
Solution 4:
(x/2) + (3/2) = (2x/5) – 1
Rearranging (2x/5) to LHS and (3/2) to RHS,
(x/2) – (2x/5) = – 1 – (3/2)
((5x-4x))/10 =((-2-3))/2 [LCM of 5 and 2 is 10]
x/10 = (-5)/2
By multiplying both sides by 10 ,
(x/10) × 10 = ((-5)/2) × 10
x = ((-50)/2)
x = -25
Verification:
Replacing x = -25 in given equation
((-25)/2) + (3/2) = ((-50)/5) – 1
((-25+3))/2 = -10 – 1
((-22)/2) = -11
-11 = -11
Verified LHS = RHS
Question 5: (3/4) (x -1) = (x – 3)
Solution 5:
(3/4) (x -1) = (x – 3)
On simplifying the brackets both sides ,
(3/4) x – (3/4) = (x – 3)
Now Rearranging (3/4) to RHS and (x – 3) to LHs
(3/4) x – x = (3/4) – 3
((3x-4x))/4 = ((3-12))/4
(-x)/4 = ((-9)/4)
Multiply both sides by -4
(-x)/4 × -4 = ((-9)/4) × -4
x = 9 Verification:
Replacing x = 9 in the given equation:
(3/4) (9 – 1) = (9 -3)
(3/4) (8) = 6
3 × 2 = 6
6 = 6
Verified LHS = RHS.
Question 6: 3 (x – 3) = 5 (2 x + 1)
Solution 6:
3 (x – 3) = 5 (2 x + 1)
By simplifying the brackets,
3 x – 9 = 10 x + 5
Rearranging 10 x to LHS and 9 to RHS
3 x – 10 x = 5 + 9
-7 x = 14
Now dividing both sides by -7
(-7x)/(-7) = 14/(-7)
x = -2 Verification:
Replacing x = -2 in the given equation
3 (-2 – 3) = 5 (-4 + 1)
3 (-5) = 5 (-3)
-15 = – 15
Verified LHS = RHS
Question 7: 3 x – 2 (2 x -5) = 2 (x + 3) – 8
Solution 7:
3 x – 2 (2 x -5) = 2 (x + 3) – 8
By simplifying the brackets on both sides,
3 x – 2 × 2 x + 2 × 5 = 2 × x + 2 × 3 – 8
3 x – 4x + 10 = 2 x + 6 – 8
- x + 10 = 2 x – 2
Rearranging x to RHS and 2 to LHS,
10 + 2 = 2 x + x
3 x = 12
By dividing both sides by 3,
3x/3 = 12/3
x = 4 Verification:
Replacing x = 4 on both sides,
3(4) – 2{2(4) – 5} = 2(4 + 3) – 8
12 – 2(8 – 5) = 14 – 8
12 – 6 = 6
6 = 6
Verified LHS = RHS
Question 8: x – (x/4) – (1/2) = 3 + (x/4)
Solution 8:
x – (x/4) – (1/2) = 3 + (x/4)
Rearranging (x/4) to LHS and (1/2) to RHS
x – (x/4) – (x/4) = 3 + (1/2)
((4x-x-x))/4 = ((6+1))/2
2x/4 = 7/2
(x/2) = 7/2
x = 7 Verification:
Replacing x = 7 in the given equation
7 – (7/4) – (1/2) = 3 + (7/4)
((28-7-2))/4 = ((12+7))/4
19/4 = 19/4
Verified LHS = RHS
Question 9: ((6x-2))/9 + ((3x+5))/18 = (1/3)
Solution 9:
((6x-2))/9 + ((3x+5))/18 = (1/3)
((6x (2) – 2 (2) + 3x + 5))/18 = (1/3)
((12x – 4 + 3x + 5))/18 = (1/3)
((15x + 1))/18 = (1/3)
By multiplying both sides by 18
((15x + 1))/18 × 18 = (1/3) × 18
15 x + 1 = 6
Rearranging 1 to RHS,
= 15 x = 6 – 1
= 15 x = 5
By dividing both sides by 15,
= 15x/15 = 5/15
= x = (1/3)
Verification:
Replacing x = (1/3)both sides,
((6 (1/3) – 2)/9 + (3 (1/3) + 5))/18 = (1/3)
((2 – 2)/9 + (1 + 5))/18 = (1/3)
(6/18) = (1/3)
(1/3) = (1/3)
Verified LHS = RHS
Question 10: m – ((m-1))/2 = 1 – ((m-2))/3
Solution 10:
m – ((m-1))/2 = 1 – ((m-2))/3
((2m – m + 1))/2 = ((3 – m + 2))/3
((m + 1))/2 = ((5 – m))/3
((m + 1))/2 = (5/3) – (m/3)
(m/2) + (1/2) = (5/3) – (m/3)
Rearranging (m/3) to LHS and (1/2) to RHS
(m/2) + (m/3) = (5/3) – (1/2)
((3m+2m))/6 = ((10-3))/6
5m/6 = (7/6)
5m = 7
By dividing both sides by 5,
5m/5= 7/5
m = 7/5 Verification:
Replacing m = 7/5 on both sides,
(7/5) – ((7-5))/10 = 1 – ((7-10))/15
(7/5) – (2/10) = ((15+3))/15
((14-2))/10 = ((15+3))/15
12/10 = 18/15
(6/5) = (6/5)
Verified LHS = RHS
Question 11: ((5x – 1))/3 – ((2x-2))/3 = 1
Solution 11:
((5x – 1))/3 – ((2x-2))/3 = 1
((5x – 1 – 2x + 2))/3 = 1
((3x+ 1))/3 = 1
By multiplying both sides by 3
((3x+ 1))/3 × 3 = 1 × 3
(3 x + 1) = 3
By subtracting 1 from both sides
3 x + 1 – 1 = 3 – 1
3 x = 2
By Dividing both sides by 3,
3x/3 = 2/3
x =2/3 Verification:
Replacing x = 2/3 in LHS,
((5 (2/3) – 1))/3 – ((2 (2/3) – 2))/3 = 1
(((10/3) – 1))/3 – (((4/3) – 2))/3 = 1
((7/3) )/3 – (((-2)/3) )/3 = 1
(7/9) + (2/9) = 1
(9/9) = 1
1 = 1
Verified LHS = RHS
Question 12: 0.6 x + (4/5) = 0.28 x + 1.16
Solution 12:
0.6 x + (4/5) = 0.28 x + 1.16
Rearranging 0.28 x to LHS and 45 to RHS,
0.6 x – 0.28 x = 1.16 – 45
0.32 x = 1.16 – 0.8
0.32 x = 0.36
By dividing both sides by 0.32,
0.32 x 0.32 = 0.360.32
x = (9/8) Verification:
Replacing x = (9/8) on both sides,
0.6(9/8) + 45 = 0.28(9/8) + 1.16
(5.4/8) + (4/5) = (2.52/8) + 1.16
0.675 + 0.8 = 0.315 + 1.16
1.475 = 1.475
Verified LHS = RHS
Question 13: 0.5 x + (x/3) = 0.25 x + 7
Solution 13:
0.5x + (x/3) = 0.25x + 7
(5/10) x + (x/3) = (25x/100) + 7
(x/2) + (x/3) = (x/4) + 7
Rearranging(x/4) to LHS
(x/2) + (x/3) – (x/4) = 7
((6x + 4x – 3x))/12 = 7
(7x/12) = 7
By multiplying both sides by 12
(7x/12) × 12 = 7 × 12
7 x = 84
By dividing both sides by 7
(7x/7) = (84/7)
x = 12 Verification:
Replacing x = 12 in given equation
0.5 (12) + (12/3) = 0.25 (12) + 7
6 + 4 = 3 + 7
10 = 10
Verified LHS = RHS
Exercise 8.4
Question 1: If 5 is subtracted from three times a number, the result is 16. Find the number.
Solution 1:
Let the x be the number
Then, given that 5 subtracted from 3 times x
= 3 x – 5 = 16
By adding 5 to both sides,
= 3 x – 5 + 5 = 16 + 5
= 3 x = 21
By dividing both sides by 3,
= (3x/3)= (21/3)
= x = 7
Therefore, required value of x is 7.
Question 2: Find the number which when multiplied by 7 is increased by 78.
Solution 2:
Let x be the number
Multiplied by 7, it gives 7 x, and x increases by 78.
According to the given information we can write as 7 x = x + 78
Rearranging x to LHS,
= 7 x – x = 78
= 6 x = 78
Now, Divide by 6 both sides,
= (6x/6) = (78/6)
= x x =13
Therefore, the required value of x is 13.
Question 3: Find three consecutive natural numbers such that the sum of the first and second is 15 more than the third.
Solution 3:
Let x be the first number.
According to the given information second number is x + 1 and the third is x + 2
Addition of first and second numbers = (x) + (x + 1).
According to question:
= (x) + (x + 1) = 15 + (x + 2)
= 2 x + 1 = 17 + x
Rearranging x to LHS and 1 to RHS,
=2 x – x = 17 – 1
= x = 16
So, first number = x = 16,
Second number = x + 1 = 16 + 1 = 17
And third number = x + 2 = 16 + 2 = 18
Therefore, the consecutive natural numbers are 16, 17 and 18.
Question 4: The difference between two numbers is 7. Six times the smaller plus the larger is 77. Find the numbers.
Solution 4:
Let x be the smaller number
So, the larger number = x + 7.
According to given information:
= 6 x + (x + 7) = 77
= 6 x + x + 7 = 77
On simplifying
= 7 x + 7 = 77
By Subtracting 7 from both sides,
= 7 x + 7 – 7 = 77 – 7
= 7 x = 70
By Dividing both sides by 7,
= (7x/7) = (70/7)
= x = 10
Therefore, the smaller number = x = 10
And the larger number = x + 7 = 10 + 7 = 17.
The two required numbers are 10 and 17.
Question 5: A man says, “I am thinking of a number. When I divide it by 3 and then add 5, my answer is twice the number I thought of”. Find the number.
Solution 5:
Let x be the required number
So, according to the given information:
= (x/3)+ 5 = 2 x
Rearranging x/3 to RHS,
= 5 = 2x x – (x/3)
= 5 = (((6x-x))/3)
= 5 = (5x/3)
By Multiplying both sides by 3
= 5 × 3 = (5x/3) × 3
= 15 = 5 x
By Dividing both sides by 5
= (15/5) = (5x/5)
= 3 = x
So, the number thought of by the man is 3.
Question 6: If a number is tripled and the result is increased by 5, 50. Find the number.
Solution 6:
Let x be the required number
According to given information:
= 3 x + 5 = 50
By Subtracting 5 from both sides,
= 3 x + 5 – 5 = 50 – 5
= 3 x = 45
By Dividing both sides by 3,
= (3x/3) = (45/3)
= x = 15
So, the required number is 15.
Question 7: Shikha is 3 years younger to her brother Ravish. If the sum of their ages 37 years, what are their present age?
Solution 7:
Let x year be the present age of Shikha.
The present age of Shikha’s brother Ravish is (x + 3) years.
So, sum of their ages = x + (x + 3)
= x +( x + 3) = 37
= 2 x + 3 = 37
By Subtracting 3 from both sides,
= 2 x + 3 – 3 = 37 – 3
= 2 x = 34
By Dividing both sides by 2,
= (2x/2) = (34/2)
= x = 17
So, the Shikha’s present age is 17 years, and
The Ravish’s present age is
= x + 3 = 17 + 3 = 20 years.
Question 8: Mrs Jain is 27 years older than her daughter Nilu. After 8 years she will be twice as old as Nilu. Find their present ages?
Solution 8:
Let x be the present age of Nilu
The present age of Nilu’s mother is (x + 27) years
And after 8 years, Nilu’s age = (x + 8),
And Mrs Jain’s age = (x + 27 + 8) = (x + 35) years
= x + 35 = 2(x + 8)
Expanding the brackets,
= x + 35 = 2 x + 16
Rearranging x to RHS and 16 to LHS,
= 35 – 16 = 2 x – x
= x = 19
So, the Nilu’s present age is x = 19 years,
And the present age of Nilu’s mother that is Mrs Jain
= x + 27 = 19 + 27 = 46 years.
Question 9: A man 4 times as old as his son. After 16 years, he will be only twice as old as his son. Find their present ages.
Solution 9:
Let x be the present age of the son.
The present age of his father = 4 x years.
So, after the 16 years,
Son’s age is (x + 16) and father’s age is (4 x + 16) years
According to the given information:
= 4 x + 16 = 2(x + 16)
= 4 x + 16 = 2x + 32
Rearranging 2x to LHS and 16 to RHS,
= 4 x – 2x = 32 – 16
= 2 x = 16
By dividing both sides by 2,
= (2x/2) = (16/2)
= x = 8
So, the present age of the son is 8 years,
And the present age of the father = 4 x = 4(8) = 32 years.
Question 10: The difference in age between a girl and her younger sister is 4 years. The younger sister in turn is 4 years older than her brother. The sum of the ages of the younger sister and her brother is 16. How old are the three children?
Solution 10:
Let X be the age of the girl.
The age of her younger sister is (x – 4) years.
Thus, the age of the brother is (x – 4 – 4) years = (x – 8) years.
According to the given information:
= (x – 4) + (x – 8) = 16
= x + x – 4 – 8 = 16
= 2 x – 12 = 16
By adding 12 to both sides,
= 2 x – 12 + 12 = 16 + 12
= 2 x = 28
By dividing both sides by 2,
= (2x/2) = (28/2)
= x = 14
So, the girl’s age is 14 years,
The age of the younger sister
= x – 4 = 14 – 4 = 10 years,
The age of the younger brother
= x – 8 = 14 – 8 = 6 years.
Question 11: One day, during their vacation at beach resort, Shella found twice as many sea shells as Anita and Anita found 5 shells more than sandy. Together sandy and Shella found 16 sea shells. How many did each of them find?
Solution 11:
Let x be the number of sea shells found by sandy
So, the number of sea shells found by Anita = (x + 5).
The number of sea shells found by Shella = 2 (x + 5).
According to the question,
= x + 2(x + 5) = 16
= x + 2 x +10 = 16
= 3 x + 10 = 16
Subtracting 10 from both sides,
= 3 x + 10 – 10 = 16 – 10
= 3 x = 6
Dividing both sides by 3,
= (3x/3) = (6/3)
= x = 2
Thus, the number of sea shells found by Sandy = 2,
The number of sea shells found by Anita
= x + 5 = 2 + 5 = 7,
The number of sea shells found by Shelia
= 2 (x + 5) = 2 (2 + 5) = 2(7) = 14.
Question 12: Andy has twice as many marbles as Pandy, and Sandy has half as many has Andy and Pandy put together. If Andy has 75 marbles more than Sandy. How many does each of them have?
Solution 12:
Let x be the number of marbles with Pandy
The 2 x be the number of marbles with Andy
Thus, the number of marbles with Sandy = (x/2) + (2x/2) = (3x/2)
According to the given information:
= x = (3x/2) + 75
By Rearranging
=2 x – (3x/2) = 75
= (((4x-3x))/2) = 75
= (x/2) = 75
= x = 150
Here, Pandy has 150 marbles,
Andy has 2 x = 2 (150) = 300 marbles,
Sandy has 3x/2 = 225 marbles.
Question 13: A bag contains 25 paise and 50 paise coins whose total value is Rs 30. If the number of 25 paise coins is four times that of 50 paise coins, find the number of each type of coins.
Solution 13:
Let x be the number of 50 paise coins
the money value contribution of 50 paise coins in bag = 0.5 x.
The number of 25 paise coins in bag = 4 x
The contribution of 25 paise coins in bag = 0.25 (4 x) = x.
According to the given information
= 0.5 x + x = 30
= 1.5 x = 30
By dividing both sides by 1.5,
= 1.5x/1.5 = 30/1.5
= x = 20
Thus, the number of 50 paise coins x = 20,
The number of 25 paise coins
= 4 x = 4 (20) = 80.
Question 14: The length of a rectangular field is twice its breadth. If the perimeter of the field is 228 meters, find the dimensions of the field.
Solution 14:
Let x meter be the breadth of the rectangle.
According to the given information,
Length of the rectangle = 2 x metres
Perimeter of a rectangle = 2 (length + breadth)
So, 2 (2 x + x) = 228
= 2 (3 x) = 228
= 6 x = 228
By dividing both sides by 6,
= 6x/6= 228/6
= x = 38
So, the breadth of the rectangle is 38 metres,
The length of the rectangle
= 2 x = 2(38) = 76 metres.
Question 15: There are only 25 paise coins in a purse. The value of money in the purse is Rs 17.50. Find the number of coins in the purse.
Solution 15:
Let x be the number of coins in the purse
So, the value of money in the purse = 0.25 x.
0.25 x = 17.50 Rs.
By dividing both sides by 0.25,
= 0.25x/0.25 = 17.5/0.25
= x = 70
Thus, the number of 25 paise coins in the purse = 70.
Question 16: In a hostel mess, 50kg rice are consumed every day. If each student gets 400gm of rice per day, find the number of students who take meals in the hostel mess.
Solution 16:
Let x be the number of students in the hostel
Quantity of rice consumed by each student = 400 gm.
So, daily rice consumption in the hostel mess = 400 (x).
But, daily rice consumption = 50 kg = 50 × 1000 = 50000gm [since 1 kg = 1000gm].
According to the given information
= 400 x = 50000
By dividing both sides by 400,
= 400x/400 = 50000/400
= x = 125
So, 125 students have their meals in the hostel mess.