Daily Ohio statesman. [volume] (Columbus, Ohio) 1855-1870, 1867-06-15
EHEPUzzLEs's OeLUmN.

/ lE ilHGNT f3tt5siit u aG
Our correspondents are upon us in vast numbers
this week. So numerous. SO pleasant and SO in
structive they that we hardly know who to give
the frstplace. We think, however, all viii concede
it to Delta after several little ones of the State
Street School have been heard. 70
Annie. Louis. Charley. Belle and Martha send an
| swer tp No. IL :, inch.sI aa -raaoil s1g 81, .
| @a wAsHINaToN. GUERNsEY Co., o., june 10. ?
MIR. EDrToEs-Some of yeur contributors are 80
given to testing problems that they become quite
tssty tnsmsssyss am goins to differ h n WILD some
doire, to sow no Seeds Of unkindness, nor Root Of
discoustssy which woul be in. danger of Robin Us
would rather at all times. Minister to the good that
s 8 rsS sd 81 81 st Es $ 1's nd i $ tBs
ldty of spherical segment. Divide this by base of
They must have @ regarded the ssmsphse sem as cone.
! About the round and the square daughter-
one sse 0i E ssnpy snd RoSn
answers; ;NI will take the Square. This is my reaz
square. and the circumference and area of circle,
Square divided by Perimeter=2o -ag Hei7-sS :
| Square divided by Side of $50 raie .5zs.1l scr
| Circle divided by Circumference=2O
Hence y. Diameter=2o and Diameter=so. If Di-
smeter=sO, Area of Circle=5O26 Sc. a.sle
equal the ratio before stated is smaller in square
than in circle If the square is less this ratio is
greater its perineter compared to its area. If them
when the ratio between the perimeter and area of
Of circle the ear not be equal to nor 5ess
than circle; 1t must be greater. 80 - falslg
We haze seen that tie si.ppstsr Of circle is equal
exceeds the other as any square exceeds the Do
XIV SoSt ssss sse%f ns S SssS his Problem No
namely S;l3 inches. Arewe all wrong that had 1t
t 50; oSs non "a W. not hv ons Coo,
Woodnan spare that tree. ifH
answer to PrSbIs@ 75.-IS rods d b by 90 rods 70 On
Answer to Problem 79.-One inch.l ~ .-,-r...,L-
Answer 70 Problem so.-..Distsnce from base Of
distsnee from base Of Sd, niso5 K length Of cord.
12O, ,20X. intended giving Demonstration of this
Il Answer to Problem s1.--9-52 Of sn hour. ,
GSds Is and
rods each daughters, So acres roods and "A rods.
p share = dsSs
ters share. 81 500; value of widows share ss 500
Answer to Problem S4--A's 3S 2 rods; BIS 6l 3 rods
Impossible By the letter. but solved according to
"a 1nsWerto ) Problem ss-By applying diagonal Of
angles stsrding on same base Attitudes are each
other as to 4; and the base may be any length.
think this problem is indeterminate. #-s.r.
Answer to Problem e8-Dimensions are abstract
Of bushels is650S S-lO. 'a.9lsl -I-'r
Answer to Problem so-A's share of capital. % :
capital 81 03s So AS share Of gain $780, BIS share
Of gain. ss2s Cs share Of gain $ls5 But BY BIS de-
parture, leaving behind S465 So of his money, A
gains $366 o.: gains s1s9 "a. DELTA.
Answer to Problem No. 4s-Here are the number
and age of the required animals to fill the conditions
of my problem. No: 49: cows; two year old heif-
or; yearling do. Your two cows and two year old
heifer have calves keep the best heifer calf and
sell the other two sell one of the old COWS he hef-
ossl and ss the end Of the year you have the same
stock yon begun with viz; head. as marked above.
4 Answer Isl 70 Problem No. ss-Distance between the
decimal places CAS- the hight of the statue is not
given, L make no allowance for it.) -. =.
this problem all that is required is to ascertain
which had the largest estate. not the amount
The square piece contained the most in a ratio of
Answer to Problem No 75-A's distance from
bird 1863 Y; Bs distance from bird 3d. C's distance
from bird l27 <; C's distance from 81 s. Cs dis
'AlGs Answer 1 to Problem No. 7s-Remainder inch.
Answer to Problem No. sI-lO 6-Is minutes.
sI'rFff ''iTa COLUMBUS, o.. May II, 1857
Mr. EDIroR-The answer. to Problem No. "A is I:
Answer ss sT to Pfobiem No. 4sHs distance from the
bird is lSs 50 feet nearly; Bs feet Cs 7 feet
9ssTy cs DISTANCE zee sI 1888 feet nearly C'
Answer to Problem No. 7s-A. s. Simpsons log
will be foot too Short unless he allows only foot
Answer @ to Problem No. 5s-0ns inch. = 60,
Answer to Problem No. S3-lfl understand Prob
lem No. SO, the answers The whole fortune is
SI2 500; the sons share $7200 the mothers share
S3 600 the daughters share 81 S00 this Problem can
be solved very easy bv mental arithmetic, but as
George likes proof. will give it to him: By the
first condition the son was to receive twice much
as the mother. and by the second CON-
dition the mother was to receive twice much
as the daughter therefore if the sons share be rep
resented by Al the mothers share would be 27. and
the daughters share would be 1t of the whole
estate; then by the third condition 2-3 minus 27=
S-2l of the estate what the mother lost, which is
a. SOO therefore if s-21=S4 200. 1-21 would equal
$500, and 2I-2l or 7-7=S12 500; then Al would
equal o. 200, 81 S, 800. and 1T 81 500. .sf
ygreen horn would know at glance that the last
Problem mo. S3 could not receive correct answer
Proof-A by digging 10 rods at 81 25 rod would
ceive $50 and would have only 60 rods to dig at
rs cents per rod, which would be s45 and $50 plus
| $45=95, which is not 800.
| Answer to No. s4-The area of the parallelogram
is 800 square feet; the of the triangle 1000
So uare feet That's right, brother Sammy" send
such problems as that. and you viii have them an
| Win L. W. Seeds tell me what measure he takes
in finding the snlid contents Of his bin Gin Problem
No. 66O Please be more explicit.
Answer to Problem No. 81 The boy had 10 mar
| bles. ll..r r-1 o.. fz-1-.- -7-. ROBIN,
Answer to Problem No. 5s-The mark on the hoop
went ll3.0878 feet. / just agreo with Guernsey in
i regard to Matte aud her problem.
i s777-' -.'1E = -'Iti r. L. RoDEHAvER,
| a'f.7-f - -5i -a -74 Walhonding. Ohio.
| GUERNsEY's ANSWERS TO PROBLEMS.