Find the maximum number of trees which can be planted 20 m apart on the two sides of a straight road 1760 m long.

Let ten's place digit be a and unit's place digit be a²
Original number = 10 x a + 1 x a² = 10a + a²
The number formed by interchanging the digits,
New number = 10 x a ² + 1 x a = 10a ² + a
According to the question
(10a² + a) - ( 10a + a²) = 54
? 10a² + a - 10a - a ² = 54
? 9a² - 9a = 54
? 9( a² - a) = 54
? ( a² - a) = 54/9
? ( a² - a) = 6
? a² - a - 6 = 0
? a² - 3a + 2a - 6 = 0
? a (a - 3) + 2 (a - 3) = 0
? (a - 3) (a + 2) = 0
? a = 3, - 2
? Ten,s digit = a = 3
Unit's digit = a² = 3² = 9
Original number = 39
? Required number = 39 x 40/100 = 15.6

Let hundred's place digit = p
Then according to question,
unit's digit = 4 x hundred's place digit = 4p
and Ten's place digit = 3 x hundred's place digit = 3p
So Number = 100 x p + 10 x 3p + 1 x 4p= 134p

If the digit in the unit's place and the ten's places are interchanged according to question,
unit's digit = 3p
and Ten's place digit = 4p
Now Number = 100 x p + 10 x 4p + 1 x 3p = 143p
According to the question, after interchanging,
143p - 134p = 18
? 9p = 18
? p = 2
Original number = 134p = 134 x 2 = 268
25% of original number = 268 x 25/100 = 67

Let total number of candidates in Exam = a
Number of candidates answered 5 questions = a x 5% = a x 5/100 = 5a/100 = a/20
Number of candidates answered not any questions = a x 5% = a x 5/100 = 5a/100 = a/20
? Remaining students = a - ( a/20 + a/20) = a - ( 2a/20 ) = a - ( a/10 ) = (10a - a)/10 = 9a/10
Number of candidates answered only 1 question = ( 9a/10 ) x 25% = ( 9a/10 ) x 25/100 = 9a/40
Number of candidates answered 4 questions = ( 9a/10 ) x 20% = ( 9a/10 ) x 20/100 = 9a/50
Given number of candidates awarded either 2 questions or 3 questions = 396
? a - ( a/20+ a/20 + 9a/40 + 9a/50 ) = 396
? a - ( a/10 + 9a/40 + 9a/50 ) = 396
? a - ( ( a x 20 + 9a x 5 + 9a x 4 )/200) = 396
? a - ( ( 20a + 45a + 36a )/200) = 396
? a - ( 101a/200) = 396
? ( 200a - 101a)/200 = 396
? ( 99a)/200 = 396
? a = 396 x 200/99
? a = 4 x 200 = 800
? a = 800
Hence, number of candidates = 800

Given that,
The cost of meal of Y = ? 100
Now, according to the question,
The cost of the meal of Z = 20% more than that of Y
The cost of the meal of Z = (100 + 100 x 20 %) = (100 + 100 x 20/100 ) = (100 + 20) = ? 120
and the cost of the meal of X = 5/6 as much as the cost of the meal of Z = 5/6 x 120 = ? 100
? Total amount that all the three has to be paid = 100 + 120 + 100
Total amount that all the three has to be paid = ? 320

let the fixed charges = ? p for first 5 km
and the additional charges = ? q per km
according to the question
p + 5q = 350........... (1)
p + 20q = 800...........(2)
on subtracting Eq. 1 from Eq. 2 we get
15q = 450
q = 30
On putting the value of q in Eq. (1) we get
p = 200
? charge for a distance of 30 km = p + 25q
= 200 + 30 x 25 = ? 950

Let the number be P.
According to the question,
P - P/7 = 180
? 6P/7 = 180
? P = 180 x 7/6 = 210

Divisor = 5 x 46 = 230
Also, 10 x Quotient = 230
? Quotient = 23
We know that
Dividend = (Divisor x Quotient ) + Remainder
? Dividend = (230 x 23) + 46
? Dividend = 5290 + 46 = 5336

Let two consecutive odd number be (A + 1) and (A + 3).
According to the question.
(A + 1) (A + 3) = 6723
? A² + 3A + A + 3 = 6723
? A² + 4A + 3 - 6723 = 0
? A² + 4A - 6720 = 0
? A² + 84A - 80A - 6720 = 0
? A(A + 84) - 80 (A + 84) = 0
? (A - 80) (A + 84) = 0
? A = 80, (A = ? - 84)
Hence, the greater number = 80 + 3 = 83

Let four consecutive even numbers are P, P + 2, P + 4 and P + 6
According to the question,
P + P + 2 + P + 4 + P + 6 = 284
? 4P + 12 = 284
? 4P = 284 - 12 = 272
? P = 272/4 = 68

Let be the ten's digit be P and unit's digit be Q.
The two-digit number = 10P + Q
(where, P > Q)
According to the question,
         P + Q = 14 ......(i)
and   P - Q = 2 .......(ii)
solving Eqs. (i) and (ii), we get
P = 8 and Q = 6
? Required product = 8 x 6 = 48