if the ten's digit of a two digit positive number is 2 more than three times the unit's digit.....?

Question

A collection of 43 coins consists of dimes and quarters. The total value is 7.60 How many dimes and how many quarters are there?

The ten’s digit of a two digit positive number is 2 more than three times the unit’s digit. If the digits are interchanged, the new number is 13 less than half the given number. Find the given integer. Hint= let x = the ten’s place digit and y = the unit’s place digit; then 10x + y is the number.)

Answer 1

d+q = 43
10d + 25q = 760

-10d - 10q = -430
10d + 25q = 760

.........15q = 330

...........q = 22 d = 21

Answer 2

#1 If all 43 were dimes,the total value would be $4.30
this is $3.30 less than the actual value
The reason is that there were some quarters,value of which is $0.15 more than the dime
Hence no. of quarters=3.30/0.15=22
and the no. of dimes=43-22=21.
#2 let the units digit be x
so,the tens digit is 3x+2
The number is 10(3x+2)+x
=31x+20
If the digits are interchanged,the new number is
10x+3x+2
=13x+2
By the problem.
13x+2=(31x+20)/2 -13
=>2(13x+2)=31x+20-26 [multiplying all the terms by 2]
=>26x+4=31x-6
=>-5x=-10
=> x=2
The unit's digit is 2 and the tens digit is 3*2+2 or 8
The number is 82

Answer 3

7.6-4.3
3.3
answer/.15
22
answer*.25
5.5

22 Quarters, 21 dimes