Wonderlictest Questions- S
(Exam Elaborations)
S
The-correct-answer-is-(B).-The-eleventh-month-of-the-year-is-November.---✅✅-
eleventh
The -month-of-the-year-is:
A. January
B. November
C. October
D. May
The-correct-answer-is-(B).-The-total-profit-from-the-sale-is-the-selling-price-minus-
theScost-Sof-purchase,-which-is-$7300-$5500=-$1800.If-the-total-profit-is-$1800-and-
eachSshovel-Saccounts-for-$50-profit,-the-amount-of-shovels-is-1800/50=36.---✅✅-A-
shopSowner-Sbought-some-shovels-for-$5,500.-The-shovels-were-sold-for-$7,300,-with-a-
profitSof-$50-Sper-a-shovel.-How-many-shovels-were-involved?
A. 18
B. 36
C. 55
D. 73
E. 90
F. None-of-these
The-correct-answer-is-(D).-The-USA-has-50-states.---✅✅-How-many-states-are-
thereSin-Sthe-U.S.A?
A. 20
B. 30
C. 40
D. 50
E. 60
The-correct-answer-is-(E).-If-there-are-3-times-as-many-elephants-as-there-are-giraffes,-
Sgiraffes-can-be-defined-as-X,-and-elephants-as-3X,-and-build-the-following-equation:S3X-
+-X-=-88
4X-=-88-
SX-=-
88/4SX-
=-22
As-stated-above,-X-stands-for-the-number-of-giraffes.-Thus,-there-are-22-giraffes.-
AsSmentioned-in-the-problem,-there-are-three-times-as-many-elephants-as-there-are-
giraffes
,in-the-safari.-Thus,-to-find-how-many-elephants-there-are-in-the-safari,-multiply-the-number-
Sof-giraffes-by-3:
22-*-3-=-66
66-is-the-number-of-elephants-in-the-safari.---✅✅-There-are-three-times-
elephants-Scompared-to-giraffes-in-the-safari.-If-there-is-a-total-of-88-elephants-and-
giraffes,-howSmany-elephants-are-there-in-the-safari?
A. 22
B. 31
C. 43
D. 59
E. 66
F. None-of-these
The-correct-answer-is-(D).-This-series-is-made-up-of-two-separate-subseries-thatS alternate:
1. A-series-of-numbers-(the-odd-terms)-that-increase-by-5-(5,-10,-15).
2. A-series-of-numbers-(the-even-terms)-that-decrease-by-2-(15,-13,-11)
The-pattern-is:-5,-15,-10,-13,-15,-11,
As-you-can-see,-the-numbers-that-increase-by-5-are-bolded,-and-the-numbers-
thatSdecrease-by-2-are-underlined.-We-are-missing-the-seventh-number-which-is-an-
oddSterm;-Stherefore,-it-should-be-greater-than-the-last-odd-term-(the-fifth-number)-by-5.
The-fifth-number-is-15.-Thus,-the-answer-is-15-+-5-=-20---✅✅-What-is-the-next-number-in-
the-series?-5,-15,-10,-13,-15,-11,
A. 9
B. 13
C. 10
D. 20
E. 15
F. None-of-these
The-correct-answer-is-(D).-An-hour-consists-of-60-minutes.-A-quarter-of-an-hour-is-15-
minutes-as-15-*-4-=-60.-Thus,-divide-the-distance-traveled-per-hour-by-4,-which-is-
S85/4=21.25.---✅✅-A-car-travels-at-a-speed-of-85-miles-per-hour.-How-far-will-it-travel-
inS15-Sminutes?
A. 9-miles
, B. 11-miles
C. 13.25-miles
D. 21.25-miles
E. 22.75-miles
F. None-of-these
The-correct-answer-is-(F)---None-of-the-above.
First-of-all,-in-order-to-simplify-the-calculations-we-can-divide-all-the-numbers-by-
aScommon-denominator:-1000,-in-that-way-we-can-get-rid-of-all-those-annoying-zeroes.-
SNow-we-define-the-used-cars-as-U-(with-a-price-of-7-each)-and-the-new-cars-as-N-(with-a-
Sprice-of-16-each)-Now-we-can-build-two-equations,-the-first-representing-the-total-
priceSand-the-second-the-total-amount-of-cars:
a. 7U+16N=191
b. N+U=-17
So-according-to-the-second-equation-N-=-17-U,-and-we-can-now-substitute-N-in-the-first-
Sequation:-7U+16(17-U)=191-;-7U+272-16U=191-;-9U=81-;-U=9.-9-used-cars-were-sold.
Solving-Tip:-You-could-try-solving-this-question-by-elimination.-For-starters,-since-191-
isS an-odd-number-and-16-is-an-even-number,-then-the-number-of-used-cars-must-be-odd
(even+odd=odd)---which-means-we-can-eliminate-answers-A,-D-&-E.-Now-that-we-are-left-
Sonly-with-B-and-C,-let's-try-to-cal---✅✅-A-car-dealership-sells-used-cars-for-$7,000-and-
new-cars-for-$16,000.-If-a-total-of-17-cars-were-sold-for-$191,000,-how-many-of-the-cars-
Ssold-were-used?
A. 2
B. 3
C. 5
D. 8
E. 10
F. None-of-these
The-correct-answer-is-(B).-25%-equals-one-quarter-and-a-quarter-of-1200-is-1200/4=300.-
STherefore,-an-addition-of-a-quarter-to-1200-equals-1200+300=1500.--
✅✅-Joey-got-a-25%-raise-to-his-salary.-If-his-original-salary-was-$1,200,-how-
was-it-after-the-raise-was-implemented?
much
A. $1225
B. $1500
C. $1350
(Exam Elaborations)
S
The-correct-answer-is-(B).-The-eleventh-month-of-the-year-is-November.---✅✅-
eleventh
The -month-of-the-year-is:
A. January
B. November
C. October
D. May
The-correct-answer-is-(B).-The-total-profit-from-the-sale-is-the-selling-price-minus-
theScost-Sof-purchase,-which-is-$7300-$5500=-$1800.If-the-total-profit-is-$1800-and-
eachSshovel-Saccounts-for-$50-profit,-the-amount-of-shovels-is-1800/50=36.---✅✅-A-
shopSowner-Sbought-some-shovels-for-$5,500.-The-shovels-were-sold-for-$7,300,-with-a-
profitSof-$50-Sper-a-shovel.-How-many-shovels-were-involved?
A. 18
B. 36
C. 55
D. 73
E. 90
F. None-of-these
The-correct-answer-is-(D).-The-USA-has-50-states.---✅✅-How-many-states-are-
thereSin-Sthe-U.S.A?
A. 20
B. 30
C. 40
D. 50
E. 60
The-correct-answer-is-(E).-If-there-are-3-times-as-many-elephants-as-there-are-giraffes,-
Sgiraffes-can-be-defined-as-X,-and-elephants-as-3X,-and-build-the-following-equation:S3X-
+-X-=-88
4X-=-88-
SX-=-
88/4SX-
=-22
As-stated-above,-X-stands-for-the-number-of-giraffes.-Thus,-there-are-22-giraffes.-
AsSmentioned-in-the-problem,-there-are-three-times-as-many-elephants-as-there-are-
giraffes
,in-the-safari.-Thus,-to-find-how-many-elephants-there-are-in-the-safari,-multiply-the-number-
Sof-giraffes-by-3:
22-*-3-=-66
66-is-the-number-of-elephants-in-the-safari.---✅✅-There-are-three-times-
elephants-Scompared-to-giraffes-in-the-safari.-If-there-is-a-total-of-88-elephants-and-
giraffes,-howSmany-elephants-are-there-in-the-safari?
A. 22
B. 31
C. 43
D. 59
E. 66
F. None-of-these
The-correct-answer-is-(D).-This-series-is-made-up-of-two-separate-subseries-thatS alternate:
1. A-series-of-numbers-(the-odd-terms)-that-increase-by-5-(5,-10,-15).
2. A-series-of-numbers-(the-even-terms)-that-decrease-by-2-(15,-13,-11)
The-pattern-is:-5,-15,-10,-13,-15,-11,
As-you-can-see,-the-numbers-that-increase-by-5-are-bolded,-and-the-numbers-
thatSdecrease-by-2-are-underlined.-We-are-missing-the-seventh-number-which-is-an-
oddSterm;-Stherefore,-it-should-be-greater-than-the-last-odd-term-(the-fifth-number)-by-5.
The-fifth-number-is-15.-Thus,-the-answer-is-15-+-5-=-20---✅✅-What-is-the-next-number-in-
the-series?-5,-15,-10,-13,-15,-11,
A. 9
B. 13
C. 10
D. 20
E. 15
F. None-of-these
The-correct-answer-is-(D).-An-hour-consists-of-60-minutes.-A-quarter-of-an-hour-is-15-
minutes-as-15-*-4-=-60.-Thus,-divide-the-distance-traveled-per-hour-by-4,-which-is-
S85/4=21.25.---✅✅-A-car-travels-at-a-speed-of-85-miles-per-hour.-How-far-will-it-travel-
inS15-Sminutes?
A. 9-miles
, B. 11-miles
C. 13.25-miles
D. 21.25-miles
E. 22.75-miles
F. None-of-these
The-correct-answer-is-(F)---None-of-the-above.
First-of-all,-in-order-to-simplify-the-calculations-we-can-divide-all-the-numbers-by-
aScommon-denominator:-1000,-in-that-way-we-can-get-rid-of-all-those-annoying-zeroes.-
SNow-we-define-the-used-cars-as-U-(with-a-price-of-7-each)-and-the-new-cars-as-N-(with-a-
Sprice-of-16-each)-Now-we-can-build-two-equations,-the-first-representing-the-total-
priceSand-the-second-the-total-amount-of-cars:
a. 7U+16N=191
b. N+U=-17
So-according-to-the-second-equation-N-=-17-U,-and-we-can-now-substitute-N-in-the-first-
Sequation:-7U+16(17-U)=191-;-7U+272-16U=191-;-9U=81-;-U=9.-9-used-cars-were-sold.
Solving-Tip:-You-could-try-solving-this-question-by-elimination.-For-starters,-since-191-
isS an-odd-number-and-16-is-an-even-number,-then-the-number-of-used-cars-must-be-odd
(even+odd=odd)---which-means-we-can-eliminate-answers-A,-D-&-E.-Now-that-we-are-left-
Sonly-with-B-and-C,-let's-try-to-cal---✅✅-A-car-dealership-sells-used-cars-for-$7,000-and-
new-cars-for-$16,000.-If-a-total-of-17-cars-were-sold-for-$191,000,-how-many-of-the-cars-
Ssold-were-used?
A. 2
B. 3
C. 5
D. 8
E. 10
F. None-of-these
The-correct-answer-is-(B).-25%-equals-one-quarter-and-a-quarter-of-1200-is-1200/4=300.-
STherefore,-an-addition-of-a-quarter-to-1200-equals-1200+300=1500.--
✅✅-Joey-got-a-25%-raise-to-his-salary.-If-his-original-salary-was-$1,200,-how-
was-it-after-the-raise-was-implemented?
much
A. $1225
B. $1500
C. $1350
