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
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 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 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
Given expression = (7 + 7 + 7 ÷ 7) / (5 + 5 + 5 ÷ 5)
= (14 + 1) / (10 + 1)
=15 / 11
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