Let the distance be ‘ x ’ km. (Recall that time = \(\frac { Distance }{ Speed } \))
Time taken to cover ‘ x ’ km at 35 km/hr: T₁ = \(\frac { x }{ 35 } hr\)
Time taken to cover ‘ x ’ km at 50 km/hr: T₂ = \(\frac { x }{ 50 } hr\)
According to the problem, the difference between two timings
= 4–(–5)
= 4+5 = 9 minutes
= \(\frac { 9 }{ 60 } \)hour (changing minutes to hour)
Given, T₁ – T₂ =\(\frac { 9 }{ 60 } \)
\(\frac { x }{ 35 } \)-\(\frac { x }{ 50 } \) = \(\frac { 9 }{ 60 } \)
\(\frac { 10x-7x }{ 350 } =\frac { 9 }{ 60 } \)
\(\frac { 3x }{ 350 } =\frac { 9 }{ 60 } \)
x \(=\frac { 9 }{ 60 } \times \frac { 350 }{ 3 } \)
The distance to his office x = 17\(\frac { 1 }{ 2 } \) km.

Let the two digit number be xy (i.e., ten’s digit is x, ones digit is y)
Its value can be expressed as 10 x + y.
Given, x + y = 8 which gives y = 8 − x
Therefore its value is 10 x + y
= 10x + 8 − x
= 9x + 8.
The new number is yx with value is 10y + x
= 10(8 − x) + x
= 80 – 9x
Given, when 18 is added to the given number (xy) gives new number (yx)
(9x + 8) + 18 = 80 – 9x
This simplifies to 9x + 9x = 80 – 8 – 18
18x = 54
x = 3 ⇒ y = 8 – 3 = 5
The two digit number is xy = 35

Let the original fraction be \(\frac { x }{ y } \)
Given that y = x + 3. (Denominator = Numerator + 3).
Therefore, the fraction can be written as \(\frac { x }{ x+3 } \).As per the given condition, \(\\ \frac { x+2 }{ (x+3)+9 } =\frac { 5 }{ 6 } \)
By cross multiplication, 6( x +2) = 5 ( x +3+9)
6 x +12 = 5( x +12)
6 x +12 = 5 x +60
6 x − 5 x = 60 − 12
x = 60 − 12
x = 48.
Therefore, the original fraction is \(\frac { x }{ x+3 } =\frac { 48 }{ 48+3 } =\frac { 48 }{ 51 } \).

Given condition: After two years, Mother’s age = 4 times of Daughter's age
5 x +2 = 4 ( x +2)
5 x +2 = 4 x +8
5 x − 4x = 8 − 2
x = 6
Hence daughter’s present age = 6 years;
and mother’s present age = 5 x = 5 × 6 = 30 years 

Let us assume that the length of the first piece is x cm.
Th en the length of the second piece is (200cm – x cm) i.e., (200 − x) cm.
According to the given statement (change m to cm),
First piece = 40 less than twice the second piece.
x = 2× (200 − x) − 40
x = 400 − 2x − 40
x + 2x = 360
3x = 360
x = \(\frac { 360 }{ 3 } \)
x = 120
Thus the length of the first piece is 120cm and
the length of second piece is 200cm − 120cm = 80cm, which happens to be the smaller.

Let the breadth of the field be ‘ x ’ metres; then its length (x + 9) metres.
Perimeter of the P = 2(length + breadth) = 2(x + 9 + x) = 2(2x + 9)
Given that, 2(2x + 9) = 154.
4x + 18 = 154
4x = 154 − 18
4x = 136
x = 34
Th us, Breadth of the rectangular fi eld = 34m
Length of the rectangular fi eld = x + 9 = 34 + 9 = 43m

Let the number of passengers having Rs.8 tickets be y. Then, the number of passengers with Rs.10 tickets is (56−y).
Total money received from the passengers = Rs.500
That is, y × Rs.8 + (56 − y) × Rs.10 = 500
8y +560 −10y = 500
8y−10y = 500 − 560
− 2y = − 60
y = \(\frac { 60 }{ 2 } \)
y = 30
Hence, the number of passengers having,
(i) Rs.8 tickets = 30
(ii) Rs.10 tickets = 56−30 = 26

Let the smaller number be x and the greater number be x + 8
Given: the sum of two numbers = 36
x + (x+8) = 36
2 x + 8 = 36
2 x = 36 − 8
2 x = 28
x = \(\frac { 28 }{ 2 } \)
x = 14
The smaller number, x = 14
The greater number, x + 8 = 14 + 8 = 22