Question 1
Solve the following equation and check your solution: = 2Solution:
= 2
Multiplying both sides by 3x,
5x â€“ 7 = 2 Ã— 3x
â‡’ 5x â€“ 7 = 6x
â‡’ 5x â€“ 6x = 7 [Transposing â€“7 to the R.H.S. and 6x to the L.H.S.]
â‡’ x = 7
âˆ´ x = 7
Check:
Substituting the value of x =  7,
L.H.S. = =
= = = 2 = R.H.S.
Hence the solution is correct.
Solution:
=
Cross multiplying,
5(2 â€“ z) = 3 (z + 16)
â‡’ 10 â€“ 5z = 3z + 48
â‡’ 5z â€“ 3z = 48  10 [Transposing 3z to the L.H.S. and 10 to the R.H.S.]
â‡’ 8z = 38
â‡’ z =
âˆ´ z =
Check:
Substituting the value of z = ,
L.H.S. = =
Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â = Ã—
Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â = = R.H.S.
Hence the solution is correct.
Solution:
=
Cross multiplying,
Â Â Â 7(2y +3) = 2 (y  9)
â‡’ 14y + 21 = 2y  18
â‡’ 14y â€“ 2y = 18  21 [Transposing 2y to the L.H.S. and 21 to the R.H.S.]
Â Â Â Â Â Â â‡’ 12y =  39
Â Â Â Â Â Â Â â‡’Â Â Â y =
âˆ´Â Â Â Â Â Â Â Â Â Â y =
Check:
Substituting the value of y = ,
L.H.S. = =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = Ã—
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = R.H.S.
Hence the solution is correct.
Â
Solution:
=  1
Multiplying both sides by 3y + 4,
Â Â Â Â 2y  9 = 1 (3y + 4)
â‡’Â Â 2y  9 =  3y  4
â‡’ 2y + 3y =  4 + 9 [Transposing 3y to the L.H.S. and 9 to the R.H.S.]
â‡’ 5y = 5
â‡’ y =
âˆ´ y = 1
Check:
Substituting the value of y = 1,
L.H.S. = =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =  1
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = R.H.S.
Hence the solution is correct.
Solution:
Multiply both sides by (3z + 7)
â‡’5zÂ âˆ’11 =Â âˆ’ 2 (3z + 7)
â‡’5zÂ âˆ’11 =Â âˆ’ 6zÂ âˆ’14
â‡’5z + 6z = 11 âˆ’ 14 (Transposing 6z to the L.H.S. and 11 to the R.H.S.)
â‡’11z =Â âˆ’3
â‡’ z =
Check Substitute for the value of Z =
=Â
Â Â Â Â Â Â Â Â Â Â Â Â Â =Â Â
Â Â Â Â Â Â Â Â Â Â Â Â Â =Â Â Â
Â Â Â Â Â Â Â Â Â Â Â Â Â =Â Â Â Â
Â Â Â Â Â Â Â Â Â Â Â Â Â Â =Â Â Â Â Â Â Â =Â Â Â 2
Solution:
= 
Cross multiplying,
3 (2y â€“ 4) = 2 (3y + 2)
â‡’ 6y  12 =  6y  4
â‡’ 6y + 6y =  4 + 12 [Transposing â€“ 6y to the L.H.S. and  12 to the R.H.S.]
â‡’ 12y = 8â‡’ y =
âˆ´ y =
Check:
Substituting the value of y = ,
L.H.S. = =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = Ã—
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = R.H.S.
Hence the solution is correct.
Solution:
= â€“
Cross multiplying,
7 (5 â€“ 7y) = â€“ 8 (2 + 4y)
â‡’ 35 â€“ 49y = â€“ 16 â€“ 32y
â‡’ â€“ 49y + 32y = â€“ 16 â€“ 35 [Transposing â€“ 32y to the L.H.S. and 35 to the R.H.S.]
â‡’ â€“ 17y = â€“ 51â‡’ y =
âˆ´ y = 3
Check:
Substituting the value of y = 3,
L.H.S. = =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = R.H.S.
Hence the solution is correct.
Solution:
=
Cross multiplying,
Â 22 (2k â€“ 5) = 3 (5k + 2)
â‡’ 44k â€“ 110 = 15k + 6
â‡’ 44k â€“ 15k = 6 + 110 [Transposing 15k to the L.H.S. and 110 to the R.H.S.]
Â Â Â Â Â Â Â Â â‡’ 29k = 116Â Â Â Â Â Â Â Â Â Â Â â‡’ k = Â Â Â Â Â Â Â Â Â Â Â Â Â Â âˆ´ k = 4
Check:
Substituting the value of k = 4,
L.H.S. = =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = R.H.S.
Hence the solution is correct.
Solution:
= 
Cross multiplying,
Â Â Â 4 (8p â€“ 5) =  5 (7p + 1)
â‡’Â Â 32p â€“ 20 =  35p  5
â‡’ 32p + 35p =  5 + 20 [Transposing â€“35p to the L.H.S. and 20 to the R.H.S.]
â‡’ 67p = 15
âˆ´ p =
Check:
Substituting the value of p = ,
L.H.S. = =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = Ã—
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = R.H.S.
Hence the solution is correct.
Solution:
=
Cross multiplying,
3 Ã— [ x + 1] = 5 (x + )
â‡’ 2 x + 3 = 5x +
â‡’ 2x â€“ 5x =  3 [Transposing 5x to the L.H.S. and 3 to the R.H.S.]
â‡’ 3x =
â‡’ x =
âˆ´ x =
Check:
Substituting the value of x = ,
L.H.S. = =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = Ã—
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = Ã—
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = R.H.S.
Hence the solution is correct.
Solution:
Â Â Â Â Â = â‡’ =
â‡’ Ã— =
Â Â Â Â Â Â â‡’ =
Cross multiplying,
7 Ã— 4 Ã— (8x â€“ 3) = 4 (63x + 4)
Dividing both sides by 4,
â‡’ 7 (8x â€“ 3) = (63x + 4)
â‡’ 56x â€“ 21 = 63x + 4
â‡’ 56x â€“ 63x = 4 + 21 [Transposing 63x to the LHS and â€“ 21 to the RHS]
â‡’  7x = 25
âˆ´ x =  .
Check:
Substituting the value of x = ,
L.H.S. = =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = Ã—
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = R.H.S.
Hence the solution is correct.
Solution:
Â Â Â Â Â =
â‡’ =
â‡’ Ã— =
â‡’ =
Cross multiplying,
4 Ã— 5 Ã— (3y + 28) = 5 Ã— 4 (2y  20)
Dividing both sides by 5 Ã— 4,
â‡’ 3y + 28 = 2y â€“ 20â‡’ 3y â€“ 2y = â€“ 20 â€“ 28 [Transposing 2y to the LHS and 28 to the RHS]
âˆ´ y = â€“ 48
Check:
Substituting the value of y = â€“ 48,
L.H.S.Â Â Â = =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (29) Ã—
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = R.H.S.
Hence the solution is correct.
Solution:
Â = 
â‡’ = â€“
â‡’ Ã— Â =Â â€“
â‡’ Â = â€“
Cross multiplying,
20 Ã— 3 (5z â€“12) = (3 ) Ã— 20 (4 â€“ 21z)
Dividing both sides by 3 Ã— 20,
â‡’ 5z â€“ 12 = â€“ (4 â€“ 21z)
â‡’ 5z â€“ 12 = â€“ 4 + 21z
â‡’ 5z  21z = â€“ 4 + 12 [Transposing 21z to the LHS and 12 to the RHS]
Â Â Â \ â€“ 16 z = 8
Â Â Â Â Â Â Â Â Â Â Â Â z = 
Check:
Substituting the value of z = ,
L.H.S. = =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = Ã—
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = R.H.S.
Hence the solution is correct.
Solution:
= 
â‡’ = â€“
â‡’ = â€“
Cross multiplying,
3 (3x + 10) = 8 (6x + 11)
â‡’  9x + 30 = â€“ 48x  88
â‡’  9x + 48 x = â€“ 88 â€“ 30 [Transposing â€“ 48x to the LHS and 30 to the RHS]
â‡’ 39x = â€“ 118
âˆ´ x =
Check:
Substituting the value of x = ,
L.H.S. =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â = Ã—
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â = R.H.S.
Hence the solution is correct.
Solution:
=
â‡’ =
â‡’ =
Cross multiplying,
23 ( y  3) = 1 (3 â€“ 5y)
â‡’ â€“ 23y â€“ (3 Ã— 23) = 3 â€“ 5y
â‡’ â€“ 23y â€“ 69 = 3 â€“ 5y
â‡’ â€“ 23y + 5y = 3 + 69 [Transposing 5y to the LHS and 69 to the RHS]
â‡’ â€“ 18 y = 72
âˆ´ y =
y =  4
Check:
Substituting the value of y =  4,
L.H.S. =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â = R.H.S.
Hence the solution is correct.
Solution:
Â Â Â Â Â Â Â Â = 6
â‡’ = 6
â‡’ = 6
Â Â Â Â Â Â Â Â Â Â â‡’ = 6
Cross multiplying,
Â Â â€“ 3x â€“ 2 = 6 (5x + 1)
â‡’  3x â€“ 2Â = 30x + 6
â‡’ â€“ 3x â€“ 30x = 6 + 2
Â Â Â Â Â Â â‡’  33x = 8
x = 
Check:
Substituting the value of x =
L.H.S. =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â = Ã—
Â Â Â Â Â Â Â Â = 2 Ã— 3
Â Â Â Â Â Â Â Â = 6 = R.H.S.
Hence the solution is correct.
Solution:
Â Â Â Â Â Â = 2
â‡’ = 2
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â â‡’ = 2
Cross multiplying,
Â Â Â Â Â Â Â Â xÂ = 2 (x â€“ 5)
Â Â Â Â Â â‡’ x = 2x â€“ 10
â‡’ x â€“ 2xÂ =  10 [Transposing 2x to the LHS]
Â Â Â Â Â Ãž  x=  10
Â Â Â Â Â Â Â Â Â Â xÂ = 10
Check:
Substituting the value of x = 10,
L.H.S. =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â = 2
Â Â Â Â Â Â Â Â = R.H.S.
Solution:
=
Cross multiplying,
â‡’ 9 (x^{2} â€“ 9) = â€“ 5 (5 + x^{2})
â‡’ 9x^{2} â€“ 81 = â€“ 25 â€“ 5x^{2 }
Â Â 9x^{2} + 5x^{2} = â€“ 25 + 81 [Transposing â€“ 5x^{2} to the LHS, and â€“ 81 to the RHS]
Â Â Â Â Â â‡’ 14x^{2} = 56
Â Â Â Â Â Â Â Â Â â‡’ x^{2 }=
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 4âˆ´ The value of x = = 2
Eliminating the negative value of x, take the positive value of x.
x= 2
Check x = 2
L.H.S =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â Â = Â Â Â =Â Â R.H.S
Hence the answer.
Solution:
=
Cross multiplying,
â‡’ 2 (y^{2} + 4) = 1 (3y^{2}+ 7)
Â Â Â â‡’ 2y^{2} + 8 = 3y^{2} + 7
Â â‡’ 2y^{2} â€“ 3y^{2} = 7 â€“ 8 [Transposing 3y^{2} to the LHS, and 8 to the RHS]
Â Â Â Â Â Â Â Â Â Ãž â€“ y^{2} = â€“ 1
Â Â Â Â Â Â Â Â Â Â Â Â â‡’ y^{2 }= 1âˆ´ The value of y = = 1
Eliminating the negative value of y, take the positive value of y.
âˆ´ y = 1.
checkÂ In
put y = 1
Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â =
Â Â Â Â Â Â Â Â = = R.H.S
Question 23
The difference between two positive integers is 36. The quotient, when one integer is divided by the other is 4. Find the two integers.Solution:
Let one integer be x.
â‡’ The other integer = x + 36
The quotient, when one integer is divided by the other is 4.
= 4
Cross multiplying,
â‡’ x + 36 = 4x
â‡’ x â€“ 4x = â€“ 36 [Transposing 4x to the LHS and 36 to the RHS]
Â Â Â Â Â Â â€“ 3x = â€“ 36
Â Â Â Â Â Â Â âˆ´ x = 12
âˆ´ One integer = 12.
âˆ´ The other integer = 12 + 36 = 48.âˆ´ The two integers are 12 and 48.
Question 24
The sum of two positive integers is 98. The integers are in the ratio 3:4. Find the integers.Solution:
Let one integer be x.
The sum of two integers is 98.
â‡’ The second integer = 98 â€“ x.
The integers are in the ratio 3 : 4.
â‡’ x : 98 â€“ x = 3 : 4
â‡’ =
Cross multiplying,
4x = 3 (98 â€“ x)
â‡’ 4x = 3 Ã— 98 â€“ 3x
â‡’ 4x = 294 â€“ 3x
â‡’ 4x + 3x = 294 [Transposing â€“ 3x to the LHS]
â‡’ 7x = 294
âˆ´ x = 42
âˆ´ One of the integer is 42.
â‡’ The other integer is 98 â€“ x = 98 â€“ 42 = 56.
âˆ´ The two integers are 42 and 56.
Question 25
The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is Find the rational number.Solution:
Let the numerator of the rational number be x.
The denominator is greater than its numerator by 8.
â‡’ The denominator = x + 8
If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is .
=
=
Cross multiplying,
2 (x + 17) = 3 (x + 7)
â‡’ 2x + 34 = 3x + 21
â‡’ 2x â€“ 3x = 21 â€“ 34 [Transposing 3x to the LHS and 34 to the RHS]
â‡’ â€“ x =  13
âˆ´ x = 13
âˆ´ The numerator = 13
âˆ´ x + 8 = 13 + 8 = 21
âˆ´ The denominator = 21
âˆ´ The rational number is .
Question 26
One number is 3 times another number. If 15 is added to both the numbers, then one of the new numbers becomes twice that of the other new number. Find the numbers.Solution:
Let one number be x.
Then the other number = 3x
If 15 is added to both the numbers, then one of the new numbers becomes twice that of the other new number.
When 15 is added to the numbers, the new numbers are x + 15 and 3x + 15.
â‡’ 2 (x+ 15) = 3x + 15
â‡’ 2x + 30 = 3x + 15
â‡’ 2x â€“ 3x = 15 â€“ 30 [Transposing 3x to the LHS and 30 to the RHS]
â‡’ xÂ Â Â Â Â Â Â = â€“ 15
âˆ´ xÂ Â Â Â Â Â Â Â Â = 15
âˆ´ one of the number = 15.
â‡’ The other number = 3x
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 3 Ã— 15
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 45
âˆ´ The two numbers are 15 and 45.
Solution:
Let one multiple of 5 be x.
The next consecutive multiple of 5 = x + 5.
The sum of the two consecutive multiples of 5 = 55
â‡’ x + (x + 5) = 55
â‡’ 2x + 5 = 55
â‡’ 2x = 50
â‡’ x = = 25
â‡’ One multiple of 5 = 25.
â‡’ The other consecutive multiple of 5 = 25 + 5 = 30.
âˆ´ The two consecutive multiples are 25 and 30.
Solution:
Let one multiple of 6 be x.
â‡’ The second multiple of 6 = x + 6
â‡’ The third multiple of 6 = x + 12
The sum of the three consecutive multiples of 6 is 666.
â‡’ x + (x + 6) + (x + 12) = 666
â‡’ x + x + 6 + x + 12 = 666
â‡’ 3x + 18 = 666
â‡’ 3x = 666 â€“ 18 [Transposing 18 to the RHS]
âˆ´ x = = 216
âˆ´ The first multiple of 6 = 216.
âˆ´ The second multiple of 6 = x + 6 = 216 + 6 = 222
âˆ´ The third multiple of 6 = x + 12 = 216 + 12 = 228.
âˆ´ The three consecutive multiples of 6 are 216, 222 and 228.
Solution:
Let one multiple of 9 be x.
â‡’ The second multiple of 9 = x + 9
â‡’ The third multiple of 9 = x + 18
The sum of the three consecutive multiples of 9 is 999.
â‡’ x + (x + 9) + (x + 18) = 999
â‡’ x + x + 9 + x + 18 = 999
â‡’ 3x + 27 = 999
â‡’ 3x = 999 â€“ 27 [Transposing 27 to the RHS]
âˆ´ x = = 324
âˆ´ The first multiple of 9 = 324.
âˆ´ The second multiple of 9 = x + 9 = 324 + 9 = 333
âˆ´ The third multiple of 9 = x + 18 = 324 + 18 = 342.
âˆ´ The three consecutive multiples of 9 are 324, 333 and 342.
Question 30
The ages of Ruby and Reshma are in the ratio 5 : 7. Four years later, their ages will be in theÂ ratio 3:4. Find their ages.Solution:
Since the ages of Ruby and Reshma are in the ratio 5:7,we may take Let
the ages of Ruby and Reshma to be 5x and 7x respectively,.
The age of Ruby after four years = 5x + 4
The age of Reshma after four years = 7x + 4
The ratio of their ages after four years = 3: 4
â‡’ 5x + 4 : 7x + 4 = 3: 4
â‡’ =
Cross multiplying,
4 (5x + 4) = 3 (7x + 4)
â‡’ 20x + 16 = 21x + 12
â‡’ 20x â€“ 21x = 12 â€“ 16 [Transposing 21x to the LHS and 16 to the RHS]
â‡’  x =  4
âˆ´ x = 4
âˆ´ The age of Ruby = 5x = 5 Ã— 4 = 20 years.
âˆ´ The age of Reshma = 7x = 7 Ã— 4 = 28 years.
âˆ´ The ages of Ruby and Reshma are 20 years and 28 years respectively.
Question 31
Five years ago, Luckee was three times as old as Lovely. 10 years later, Luckee would be twice as old as Lovely. How old are they now? [Hint: First find how old they were five years ago.]Solution:
Let the age of Lovely five years ago be x years.
Then the age of Luckee = 3x years.
As the age of Lovely five years ago is x years then 10 years later (5 + 10),
3x + 15 = 2(x+ 15)Â Â Â Â Â Â Â Â Â Â Â Â Â (Given)
3x + 15 = 2x + 30
3x â€“ 2x = 30 â€“ 15
Â Â Â Â âˆ´ x = 15
â‡’ 3x = 3 Ã— 15 = 45
â‡’ Lovelyâ€™s present age = x + 5 = 20 years.
â‡’ Luckeeâ€™s present age = 3x + 5
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 3 Ã— 15 + 5
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 45 + 5
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 50 years.
âˆ´ The present age of Lovely = 20 years.
âˆ´ The present age of Luckee = 50.
Question 32
The perimeter of a rectangle is 240 cm. If its length is decreased by 10% and its breadth is increased by 20%, we get the same perimeter. Find the length and the breadth of theÂ rectangle. [Hint: If l and b denote respectively the length and the breadth, then b = 120l)].Solution:
Let l cm be the length of the rectangle and b cm be the breadth of the rectangle.
The perimeter of the rectangle = 240 cm
â‡’ 2(l + b) = 240 cm
l + b = 120
â‡’ b = 120 â€“ l
The length of the rectangle when it is decreased by 10% = 0.9l
The breadth of the rectangle when it is increased by 20% = 1.2b
When the length is decreased by 10% and its breadth is increased by 20%, the perimeter remains the same.
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Ãž 2(0.9l + 1.2b) = 240
Â Â Â Â Â Â Â Â Â Â Â Â Â Â (Dividing throughout by 2)
Â Â Â Â Â Â Ãž 0.9l + 1.2 (120  l) = 120 [Substituting the value b = 120 â€“ l]
â‡’ 0.9 l + 1.2 Ã— 120 â€“ 1.2 l = 120
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â â‡’  0.3 l = 120 â€“ 144
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â â‡’  0.3 l =  24
â‡’ l = =
â‡’ b = 120 â€“ 80 = 40
âˆ´ The length of the rectangle is 80 cm and the breadth of the rectangle is 40 cm.
Question 33
Sum of the digits of a twodigit number is 9. The number obtained by interchanging the digitsÂ exceeds the given number by 27. Find the given number.Solution:

Let the tens digit be x.
Sum of the digits of a two digit number = 9.
â‡’ The unitâ€™s digit = 9 â€“ x
â‡’ The original number = 10x + 9 â€“ x = 9x + 9
The number obtained when the digits are interchanged = 10 (9 â€“ x) + x
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 90 â€“ 10x + x
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 90 â€“ 9x
Digits interchanged 

9  x 
x 
10 (9 â€“ x) + x = 90 â€“ 10x + x Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 90 â€“ 9x 
When the digits are interchanged, the number got exceeds the original number by 27.
â‡’ 90  9x = (9x + 9) + 27
â‡’  9x â€“ 9x = 9 + 27 â€“ 90 [Transposing 9x to the LHS and 90 to the RHS]
â‡’ â€“ 18x =  54
Â Â Â Â Â â‡’ x =
âˆ´ x = 3
âˆ´ The tens digit = 3.
âˆ´ The units digit = 9 â€“ 3 = 6
âˆ´ The given number = 36.
Question 34
Sum of the digits of a twodigit number is 12. The given number exceeds the numberÂ obtained by interchanging the digits by 36. Find the given number.Solution:

Let the tens digit be x.
Sum of the digits of a two digit number = 12.
â‡’ The unitâ€™s digit = 12 â€“ x
â‡’ The number = 10x + 12 â€“ x = 9x + 12
The number obtained when the digits are interchanged = 10 (12 â€“ x) + x
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 120 â€“ 10x + x
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 120 â€“ 9x
When the digits are interchanged, the given number exceeds the number obtained by 36.
â‡’ (120 â€“ 9x) + 36 = 9x + 12
â‡’ â€“ 9x â€“ 9x = 12 â€“ 120 â€“ 36 [Transposing 9x to the LHS and (+120 + 36) to the RHS]
â‡’ â€“ 18x = 12 â€“ 156
â‡’ â€“ 18x = â€“ 144
â‡’ x =
âˆ´ x = 8
âˆ´ The tens digit = 8.
âˆ´ The units digit = 12 â€“ 8 = 4
âˆ´ The given number = 84.
Question 35
Each side of a triangle is increased by 10 cm. If the ratio of the perimeters of the newÂ triangle and the given triangle is 5 : 4, find the perimeter of the given triangle.Â Â Â Â Â Â Â Â Â [Hint: If the sides are a, b, c then the perimeter is a + b + c = x say].
Solution:
Let the perimeter of a triangle with sides a cm, b cm and c cm be x cm.
When the sides are increased by 10 cm each, then the perimeter will also increase.
The perimeterÂ = (a+10) cm + (b + 10) cm + (c + 10) cm
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (a + b + c) + 30 cm
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = x + 30 cm [Since a + b + c = x cm]
The ratio of the perimeters of the new triangle to the given triangle = x + 30 : x
It is given that the ratio of the perimeters of the new triangle and the given triangle is 5 : 4.â‡’ x + 30 : x = 5 : 4
â‡’ =
Cross multiplying,
4(x + 30) = 5x
â‡’ 4x + 120 = 5x
â‡’ 120 = 5x â€“ 4x (Transposing 4x to the R.H.S.)
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â âˆ´ x = 120
âˆ´ The perimeter of the given triangle is 120 cm.
Question 36
Kanchan receives a certain amount of money on her retirement from her employer. She gives half of this money and an additional sum of Rs. 10000 to her daughter. She also gives onethird of the money received and an additional sum of Rs. 3000 to her son. If the daughter gets twice as much as the son, find the amount of money Kanchan received on her retirement.
Solution:
Let the amount received by Kanchan be Rs. x.

Amount given to her daughter by Kanchan = Rs.
Additional amount given to her daughter = Rs. 10 000
âˆ´ Total amount received by daughter = Rs. [ + 10 000]
Amount given to her son by Kanchan = Rs.
Additional amount given to her son = Rs. 3 000
\ Total amount received by daughter = Rs. [ + 3 000]
It is given that the amount daughter gets is twice as much as the son gets.
2 [ + 3 000] = [ + 10 000]
2 Ã— =
=
2 (2x + 18000) = 3 (x + 20000)
4x + 36000 = 3x + 60000
â‡’ 4x â€“ 3x = 60000 â€“ 36000(Transposing 3x to the L.H.S. And 36000 to the R.H.S.)
â‡’ x = 24000
âˆ´ The amount Kanchan received on her retirement is Rs. 24000.