Problem Solving Methods (Arithmetic)

Front 1

Avg. 50units for n days. Adding 90units increases avg. to 55units.
n = ?

Back 1

# added = sum2 - sum1
_________________________
1) 90 = 55(n+1) - 50n
2) 5n = 35
A) n = 7

Front 2

Avg. of 750 & 780

Back 2

Avg.= [(n2 - n1)/2] + n1
_________________________
1) 780-750 = 30
2) 30/2 = 15
A) 750+15 = 765 = Avg

Front 3

p students avg. is 70. q students avg. is 92. Combined avg. is 86.
p/q = ?

Back 3

Combined Avg
= (sum1+sum2)/(n1+n2)
_________________________
1) (70p+92q)/(p+q) = 86
2) 70p+92q = 86p+86q
3) 6q = 16p
4) 6q/16 = p
A) p/q = 6/16

Front 4

238*647 = ?
(a) 143,734 (b) 152,736
(c) 153,734 (d) 153,986
(e) 155,737

Back 4

Last Digit Method
_________________________
238
*647
--------
...66
...20
--------
...86 >> 153,986
A) (d)

Front 5

3.63*5.89 = ?
(a)14.387 (b) 17.8934
(c) 21.3807 (d) 22.57
(e) 27.684

Back 5

Last Digit Method & Decimal Spaces
_________________________
3.63
*5.89
---------
......7
......0
......0
---------
......7 >> which is going to be 4 places to the right of the decimal point because 3.63 & 5.89 each have 2 placeholders to the right of the decimal for a total of 4.
A) (c) is the only choice with a 7 in the ten thousands place.

Front 6

3.63*5.80 = ?
(a)21.0054 (b) 17.8934
(c) 21.054 (d) 22.57
(e) 27.684

Back 6

Last Digit Method & Decimal Spaces w/ zero as a place holder
_________________________
3.63
*5.80
---------
......0
....40
...500
---------
...540 >> which is going to have 3 decimal places because 3.63 has 2 placeholders to the right of the decimal and 5.8 only has 1 placeholder.
A) (c) is the only choice with a 5 in the hundreds place and a 4 in the ten thousands place; zero does not hold a place when it is on the far right of the decimal point.