In a division sum, the divisor is ten times the quotient and five times the remainder. If the remainder is 46, then find the dividend?

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

Number of trees that can be planted on one side of road = (Length of road/gap between two trees ) + 1
Number of trees that can be planted on one side of road = ( 1760/20 )+ 1
Number of trees that can be planted on one side of road = 88 + 1 = 89
? Threes on the both sides
= 2 x 89 = 178

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

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

Given expression = 42060 / 15 + 5
= 2804 +5
=2809.

x = 13200 - 9873 = 3327