divelab/Sys2Bench huggingface dataset:
Question: In the manufacture of a certain product, 6 percent of the units produced are defective and 5 percent of the defective units are shipped for sale. What percent of the units produced are defective units that are shipped for sale?
Options:
A)0.125%
B)0.3%
C)0.8%
D)1.25%
E)2.0%
Rationale: 0.06*0.05 = 0.003 = 0.3%
The answer is B.
Answer: B
Question: Two men A and B start from place X walking at 4 ½ kmph and 5 ¾ kmph respectively. How many km apart they are at the end of 3 ½ hours if they are walking in the same direction?
Options:
A)5 km
B)10 4/3 km
C)4 3/8 km
D)18 km
E)6 km
Rationale: Explanation:
RS = 5 ¾ - 4 ½ = 1 ¼
T = 3 ½ h.
D = 5/4 * 7/2 = 35/8 = 4 3/8 km
C)
Answer: C
Question: In a coconut grove, (x + 3) trees yield 60 nuts per year, x trees yield 120 nuts per year and (x – 3) trees yield 180 nuts per year. If the average yield per year per tree be 100, find x.
Options:
A)3
B)4
C)5
D)6
E)None of the above
Rationale: (x+3)×60+x×120+(x−3)×180/(x+3)+x+(x−3)=100
⇒ 360x−360/3x = 100
⇒ 60x = 360 ⇒ x = 6
Answer D
Answer: D
Question: If the probability of rain on any given day in Chicago during the summer is 50%, independent of what happens on any other day, what is the probability of having exactly 3 rainy days from July 4 through July 10 inclusive?
Options:
A)7/32
B)11/64
C)35/128
D)49/128
E)65/128
Rationale: One possible case is: rainy-rainy-rainy-not rainy-not rainy-not rainy-not rainy.
The probability of this case is (1/2)^7 = 1/128
The number of possible cases is 7C3 = 35.
P(exactly 3 rainy days) = 35*1/128 = 35/128
The answer is C.
Answer: C
Question: A certain bacteria colony doubles in size every day for 16 days, at which point it reaches the limit of its habitat and can no longer grow. If two bacteria colonies start growing simultaneously, how many days will it take them to reach the habitat’s limit?
Options:
A)6.33
B)7.5
C)10
D)15
E)19
Rationale: If there is one bacteria colony, then it will reach the limit of its habitat in 16 days.
If there are two bacteria colonies, then in order to reach the limit of habitat they would need to double one time less than in case with one colony. Thus colonies need to double 15 times.
Answer: D.
Similar questions to practice:
Hope it helps.
Answer: D
Question: Steve is laying floor tiles in 2 rooms of his house. Each room requires 30 tiles, all the same size. How many of the following combinations could Steve buy if: A Small box (1 box@$8.00 w/tax) contains 10 tiles, and a Large box (1 box@$20.00 w/tax) contains 25 tiles. Steve only has $50 on him when he gets to the store.
Options:
A)6 Small Boxes
B)3 Large Boxes
C)3 Small Boxes and 1 Large Box
D)A and C
E)A, B and C
Rationale: The Correct Answer must total at least 60 tiles (2 rooms of 30) AND cost $50 or less
A=Total of 60 tiles (6*10) and costs $48 (6*$8.00) *Correct Answer*
B=Total of 75 tiles (3*25) (while sufficient amount of tiles, cost would be $60 which is $10 over budget)
C=Total of 55 tiles (3*10 +25) (while sufficient cash is on hand of $49 [3*8+25] he is 5 tiles short.
D=A is correct however C is not
E=A is correct however B and C are not
The Correct Answer is A
Answer: A
Question: The compound interest accrued on an amount of Rs.44000 at the end of two years is Rs.1193.60. What would be the simple interest accrued on the same amount at the same rate in the same period?
Options:
A)10568
B)10569
C)10560
D)10561
E)10562
Rationale: Let the rate of interest be R% p.a.
4400{[1 + R/100]2 - 1} = 11193.60
[1 + R/100]2 = (44000 + 11193.60)/44000
[1 + R/100]2 = 1 + 2544/1000 = 1 + 159/625
[1 + R/100]2 = 784/625 = (28/25)2
1 + R/100 = 28/25
R/100 = 3/25
Therefore R = 12 SI on Rs.44000 at 12% p.a. for two years
= 44000(2)(12)/100
=Rs.10560
Answer:C
Answer: C
Question: In a recent election, Ms. Robbins received 7,000 votes cast by independent voters, that is, voters not registered with a specific political party. She also received 10 percent of the votes cast by those voters registered with a political party. If N is the total number of votes cast in the election and 40 percent of the votes cast were cast by independent voters, which of the following represents the number of votes that Ms. Robbins received?
Options:
A)0.06N + 3,200
B)0.1N + 7,200
C)0.4N + 7,200
D)0.1N + 8,000
E)0.06N + 7,000
Rationale: total vote = N
vote cast by independent voters = 0.4 N
vote cast by registered voters = 0.6 N
vote received by Ms. Robbins = 7000 + 10% of vote be registered voters = 7000 + 0.06 N
Answer choice E
Answer: E
Question: If a sum of money borrowed at 5 percent/annum SI amounts to Rs.1020 in 4 yrs, then the sum of money borrowed is?
Options:
A)Rs. 850
B)Rs. 980
C)Rs. 990
D)Rs. 1000
E)Rs. 1100
Rationale: Sol We have, A : Rs.1U2D
Let the pr xnpal be R5 x
Then mterest u — 12 : wzo — x
Therefore, by u xglurmula.
_umxx
’ Rx’: (
, _ma>< 1nzn—xv
\4Ve|\a\e,Jc— 5“
:> ~(:51DD—S*(or6x:51UU
or, J: : w 350
. The sum of money burrowed : Rs. 850
A)
Answer: A
Question: A container has a capacity of 30 gallons and is full of spirit. 4 gallons of spirit is drawn out and the container is again filled with water. This process is repeated 5 times. Find how much spirit is left in the resulting mixture finally?
Options:
A)6(257/525) gallons
B)6(346/625) gallons
C)6.5 gallons
D)6.25 gallons
E)None of these
Rationale: There is a direct formula to calculate such repetition:
Left QTY / Orig QTY = ( (Orig QTY - Rep QTY) / Orig QTY )^n
where n is number of times mixing done:
Left QTY / 30 = (30 -4)^5 / 20^5
Left QTY = 30 (26/20)^5
Left QTY = 30 ( 13/10 )^5
Left QTY = 74.2586
Answer is E
Answer: E
Question: The G.C.D. of 1.08, 0.36 and 0.45 is?
Options:
A)0.03
B)0.09
C)0.18
D)0.108
E)none of these
Rationale: Given numbers are 1.08,0.36 and 0.45.
H.C.F of 108, 36 and 45 is 9
H.C.F of given numbers=0.09.
Correct Options:B
Answer: B
Question: Rectangular tile each of size 80cm by 55cm must be laid horizontally on a rectangular floor of size 210cm by 330cm,such that the tiles do not overlap and they are placed with edges jutting against each other on all edges. A tile can be placed in any orientation so long as its edges are parallel to the edges of floor. No tile should overshoot any edge of the floor. The maximum number of tiles that can be accommodated on the floor is:
Options:
A)26
B)16
C)18
D)19
E)20
Rationale: Area of tile = 80*55 = 4400
Area of floor = 210*330 = 72600
No of tiles = 72600/4400 = 16.5
So, the no of tile = 16
ANSWER:B
Answer: B
Question: If 15,555 – n is divisible by 11, and 0 < n < 11, what is n?
Options:
A)1
B)3
C)5
D)7
E)9
Rationale: 15,555 / 11 = 1414 with a remainder of 1.
We need to subtract the remainder to get a multiple of 11.
The answer is A.
Answer: A
Question: A scientist used a unique two-color code to identify each of the test subjects involved in a certain study. If the scientist found that choosing from among seven colors produced enough color codes to identify all but 6 of the test subjects, how many test subjects were in the study? (Assume that the order of the colors in the codes does not matter.)
Options:
A)25
B)26
C)27
D)28
E)29
Rationale: 7C2 = 21
The number of test subjects is 21 + 6 = 27
The answer is C.
Answer: C
Question: A bag holds 3 red marbles and 3 green marbles. If you removed two randomly selected marbles from the bag, without replacement, what is the probability that both would be red?
Options:
A)1/10
B)1/5
C)3/10
D)2/5
E)1/2
Rationale: Given: 3R and 3 G marbles
Required: Probability that 2 marbles removed without replacement are both red
Initially we have to pick one red from a total of 3 red and 3 green marbles
After one red has been picked, we need to pick 1 red from a total of 2 red and 3 green marbles.
P(Both red) = (3/6)*(2/5) = 1/5
Option B
Answer: B
Question: A man covers a distance on scooter .had he moved 3kmph
faster he would have taken 40 min less. If he had moved
2kmph slower he would have taken 40min more.the distance is.
Options:
A)42m
B)53m
C)37m
D)40m
E)29m
Rationale: Let distance = x m
Usual rate = y kmph
x/y – x/y+3 = 40/60 hr
2y(y+3) = 9x ————–1
x/y-2 – x/y = 40/60 hr y(y-2) = 3x —————–2
divide 1 & 2 equations
we get 2(y+3)=3(y-2)
2y + 6 = 3y-6
y= 12
Put y=12 in eq.--2
12*10=3x
by solving we get x = 40
Answer is D
Answer: D
Question: In country Z, 12% of the people do not have a university diploma but have the job of their choice, and 25% of the people who do not have the job of their choice have a university diploma. If 40% of the people have the job of their choice, what percent of the people have a university diploma?
Options:
A)35%
B)43%
C)55%
D)65%
E)75%
Rationale: Setting up a matrix is how I solve this one.
Diploma No Diploma Totals
Job of Choice w/Diploma
Job of Choice w/o Diploma = 12%
Job of Choice Total = 40%
Not Job of Choice with Diploma =.25X
Not Job of Choice w/o Diploma= .75X
Total Not Job of Choice = X
Total with Diploma
Total without Diploma
Total citizen = 100
If 40% of people have their job of choice, then 60% of people do NOT have their job of choice. 25% of 60% = 15%. We can also see that 28% of the people have their job of choice and a diploma (40%-12%=28%). 28% + 15% = 43%. Therefore 43% of the people in Country Z have a diploma.
Ans B
Answer: B
Question: New age consultants have three consultants Gyani, Medha, and Buddhi. The sum of the number of projects handled by Gyani and Buddhi individually is equal to the number of projects in which Medha involved. All three consultants are involved together in 6 projects. Gyani works with Medha in 14 projects. Buddhi has 2 projects with with Medha but without Gyani and 3 projects with Gyani but without Medha. The total number of projects for New age consultants is one less than twice the number of projects in which more than one consultant is involved.
What is the number of projects in which Medha alone is involved?
Options:
A)1
B)2
C)3
D)4
E)5
Rationale: gyani works with medha but with out buddhi=14-6=8.
buddhi with medha but without gyani=2
buddhi with gyani but without medha=3
all three involved projects=6
so number projects in which more than one consultant is involved=8+2+3+6=19
total no.projects for consultants is 2*19-1=37
among 37 projects in 19 projects more than one consultant is involved
so no. of projects done by onle one consultant=37-19=18
so b+g=16+x
b+g=17
x=1
ANSWER:A
Answer: A
Question: Compound interest earned on a sum for the second and the third years are Rs.2400 and Rs.2640 respectively. Find the rate of interest?
Options:
A)10% p.a.
B)20% p.a.
C)15% p.a.
D)12% p.a.
E)None of these.
Rationale: Rs.2640 - 2400 = Rs.240 is the interest on Rs.2400 for one year.
Rate of interest = (240/2400) * 100 = 10% p.a
ANSWER:A
Answer: A
Question: What is the possible value of 2*((1+1/100)^100)
Options:
A)5.4
B)6.4
C)7.4
D)8.4
E)4.4
Rationale: 1+1/100=1.01
1.o1^100==2.70
2.70*2
5.40 ans
ANSWER:A
Answer: A
Question: Running at the same constant rate, 6 identical machines can produce a total of 300 bottles per minute. At this rate, how many bottles could 10 such machines produce in 4 minutes?
Options:
A) 648
B) 1,800
C) 2,000
D) 10,800
E) 64,800
Rationale: 6 machines produce 300 bottles per minute;
1 machine produces 300/6=50 bottles per minute;
10 machines produce 50*10=500 bottles per minute;
In 4 minutes 10 machines produce 500*4=2,000 bottles.
Answer: C.
Answer: C
Question: n 10 years, A will be twice as old as B was 10 years ago. If A is now 9 years older than B the present age of B is
Options:
A)22
B)897
C)268
D)20
E)48
Rationale: A +10=2(B-10) ........(1)
A =B + 9 ......... (2)
from equations. 1 & 2
we get B = 39 A will be 39+9=48 years old.
Answer:E
Answer: E
Question: Each student is given a five-character identification code. The first four characters are selected from the numbers 0 to 9 inclusive, and the last character is selected from the 26 letters of the alphabet. If characters may be repeated and the same characters used in a different order constitute a different code, how many different identification codes can be generated following these rules?
Options:
A)240,000
B)245,000
C)250,000
D)255,000
E)260,000
Rationale: The number of possible codes is 10*10*10*10*26=260,000.
The answer is E.
Answer: E
Question: The average of first 12 natural numbers is?
Options:
A)7.6
B)6.5
C)6.1
D)6.8
E)7.2
Rationale: Sum of 12 natural no.
= 156/2 = 78
Average = 78/12
= 6.5
Answer:B
Answer: B
Question: Two pie pans of the same size remain on the counter. One pan has 1/2 of a pie, and the other
has 3/4 of a pie left in it. Mom wishes to divide all the pie into pieces that are each 1/8 of a pie. How
many pieces of pie will she have when she is finished?
Options:
A)20 pieces
B)9 pieces
C)10 pieces
D)19 pieces
E)15 pieces
Rationale: 1/2 + 3/4 = 2/4 + 3/4 = 5/4 pie
5/4 ÷ 1/8 = 40/32 ÷ 4/32 =
40 ÷ 4 = 10 pieces of pie
correct answer C
Answer: C
Question: If 3 persons can do 3 times of a particular work in 3 days, then, 7 persons can do 7 times of that work in
Options:
A)2 days
B)3 days
C)5 days
D)7 days
E)9 days
Rationale: That is, 1 person can do one time of the work in 3 days.
Therefore, 7 persons can do 7 times work in the same 3 days itself.
B)
Answer: B
Question: A committee is to be formed by randomly choosing 4 people from 8 candidates. If Alice and Bob are candidates, what is the probability that Alice will be selected but Bob will not?
Options:
A)1/3
B)2/5
C)3/5
D)1/7
E)2/7
Rationale: The total number of ways to form a committee is 8C4=70.
The number of committees which include Alice but not Bob is 6C3=20.
The probability that Alice is selected but Bob is not selected is 20/70=2/7.
The answer is E.
Answer: E
Question: Liam is pulled over for speeding just as he is arriving at work.He explains to the police officer that he could not afford to be late today, and has arrived at work only four minutes before he is to start. The officer explains that if Liam had driven 5mph slower for his whole commute, he would have arrived at work exactly on time. If Liam's commute is 10 miles long,how fast was he actually driving?(Assume that Liam drove at a constant speed for the duration of his commute.)
Options:
A)50 mph
B)45 mph
C)48 mph
D)30 mph
E)60 mph
Rationale: Let t be the number of hours he would need to reach office on time.
when he is driving with over speed, he reached office 4 min earlier! so the equation for this is s(t - 4/60) = 30
where s is the speed and 30 is the distance.
if he decreases his speed by 5mph then he would have reached his office on time: (s-5)t = 30
if you solve above equations, you will arrive at t = 2/3 hr and s = 30mph
therefore answer is D
Answer: D
Question: If m = 3^n, what is the greatest value of n for which m is a factor of 25!
Options:
A)8
B)10
C)12
D)14
E)16
Rationale: We should find the highest power of 3 in 25!: 25/3+25/3^2=8+2=10 (take only the quotient into account).. So the highest power of 3 in 25! is 10.
Answer: B.
Answer: B
Question: A bookshop had science and art books in the ratio of 2:5. By the end of the week, 20% of both type of books were sold and 2240 books were unsold. How many art books were there at the starting?
Options:
A)800
B)1200
C)1405
D)1540
E)2000
Rationale: science books sold = 2x × 0.2 = 0.4x
science books unsold = 2x – 0.4x = 1.6x
art books sold = 5x × 0.2 = x
art books unsold = 5x – x = 4x
total books unsold = 1.6x + 4x = 5.6x
5.6x = 2240
x = 400
2x science = 800
and 5x art books = 2000
E
Answer: E
Question: Find the average of first 10 multiples of 7
Options:
A)35.5
B)37.5
C)38.5
D)40.5
E)None of these
Rationale: Explanation:
=7(1+2+3+...+10)10
=7(10(10+1))10×2
=7(110)10×2=38.5
Answer: Option C
Answer: C
Question: Calculate the L.C.M of 5/6, 4/ 7, 9/10, 8/9 is :
Options:
A)560
B)364
C)320
D)330
E)360
Rationale: Required L.C.M =L.C.M. of 5, 4, 9, 8/H.C.F. of 6, 7, 10, 9 =360/1=360
Answer is E
Answer: E
Question: The average age of a husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is:
Options:
A)19 years
B)17 years
C)25 years
D)15 years
E)12 years
Rationale: Sum of the present ages of husband, wife and child
= (23 * 2 + 5 * 2) = 57 years.
Required average
= 57/3 = 19 years.
Answer: A
Answer: A
Question: A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?
Options:
A)277
B)240
C)278
D)2667
E)122
Rationale: Speed = (54 * 5/18) m/sec = 15 m/sec. Length of the train = (15 x 20)m = 300 m. Let the length of the platform be x meters. Then, (x + 300)/36 = 15 ==> x + 300 = 540 ==> x = 240 m.Answer: B
Answer: B
Question: Three pipes of same capacity A,B,C can fill a tank in 18 hours. If there are only two pipes B and A of same capacity, the tank can be filled in?
Options:
A)16
B)19
C)27
D)24
E)29
Rationale: The part of the tank filled by three pipes in one hour
= 1/6
=> The part of the tank filled by two pipes in 1 hour
= 2/3 * 1/18= 1/27.
The tank can be filled in 27 hours.
Answer:C
Answer: C
Question: A, B and C start swimming in a pool simultaneously from the same end. To complete 10 laps, A takes 10 minutes, B takes 8 minutes and C takes 2 minutes. What is the ratio of speeds A:B:C?
Options:
A)3:4:5
B)4: 5:20
C)5:4:3
D)4:6:5
E)12:15:25
Rationale: The slowest one is A, then comes B and the fastest one is C. So, the the ratio of speeds A:B:C must be in ascending order. Eliminate options C, D, and E.
A's speed (distance)/(time) = 10/10 = 1 lap per minute = 4 laps in 4 minute;
B's speed (distance)/(time) = 10/8 = 5/4 lap per minute = 5 laps in 4 minutes;
C's speed (distance)/(time) = 10/2 = 5 lap per minute = 20 laps in 4 minutes/
Therefore, the the ratio of speeds A:B:C is 4: 5:20.
Answer: B.
Answer: B
Question: Solve(0.76 × 0.76 × 0.76 − 0.008)/(0.76 × 0.76 + 0.76 × 0.2 + 0.04)
Options:
A)0.56
B)0.57
C)0.6
D)0.76
E)0.87
Rationale: Option 'A'
Answer: A
Question: If x < y < z and y-x > 7, where x is an even integer and y and z are odd integers, what is the least possible value of z - x?
Options:
A)6
B)7
C)11
D)8
E)10
Rationale: We have:
1) x<y<z
2) y-x>5
3) x=2k (x is an even number)
4) y=2n+1 (y is an odd number)
5) z=2p+1 (z is an odd number)
6) z-x=? least value
z-x=2p+1-2k=2p-2k+1=2(p-k)+1 - that means that z-x must be an ODD number. We can eliminate answer choices A, C and E
we are asked to find the least value, so we have to pick the least numbers
since y is odd and x is even, y-x must be odd. since y-x>7 the least value for y-x must be 11, the least value for x must be 2, and, thus, the least possible value for y must be 11 (y-2=9, y=11)
2<11<z, since z is odd, the least possible value for z is 13
z-x=13-2=11
Answer C
Answer: C
Question: Find the rate % of C.I and S.I on Rs. 1000 for 2 years if the difference between C.I and S.I is given to be 22.50.
Options:
A)15%.
B)20%
C)25%
D)30%
E)35%
Rationale: Difference between C.I and S.I =P(R/100)^2 1000(R/100)^2=22.50. ie)( R/100)^2=22.50/1000. R^2 =225 R =15%. Answer : A
Answer: A
Question: If in a certain code "RANGE" is coded as 12345 and "RANDOM" is coded as 123678. Then the code for the word
"GONE" would be
Options:
A)1234
B)2367
C)3547
D)5382
E)4735
Rationale: G-4
O-7
N-3
E-5
so for mango the code is 4735
ANSWER:E
Answer: E
Question: A farmer spent $36 on feed for chickens and goats. He spent 40% money on chicken feed, which he bought at a 20% discount off the full price, and spent the rest on goat feed, which he bought at full price. If the farmer had paid full price for both the chicken feed and the goat feed, what amount would he have spent on the chicken feed and goat feed combined?
Options:
A)$37.80
B)$38.50
C)$39.20
D)$39.60
E)$40.60
Rationale: A farmer spent 40% money on chicken feed, so he spent 0.4*$36=$14.4 on chicken feed, thus he spent the remaining 36-14.4=$21.6 on goat feed.
Now, since he bought chicken feed at a 20% discount then the original price of it was x*0.8=$14.4 --> x=$18
Therefore if the farmer had paid full price for both the chicken feed and the goat feed, then he would he have spent 18+21.6=$39.6.
Answer: D.
Answer: D
Question: When R is divided by W, the quotient is Q and the remainder is E. Which of the following expressions is equal to E?
Options:
A)QW + R
B)RW - Q
C)Q - RW
D)-QW + R
E)Q/R
Rationale: R/W = Q + E/W
R -QW = E
Amswer: D
Answer: D
Question: A train running at the speed of 60 km/hr crosses a pole in 9 seconds. Find the length of the train?
Options:
A)150
B)8827
C)166
D)199
E)121
Rationale: Speed = 60*(5/18) m/sec = 50/3 m/sec
Length of Train (Distance) = Speed * Time
(50/3) * 9 = 150 meter
Answer: A
Answer: A
Question: For positive integers k and n, the k-power remainder of n is defined as r in the following equation:
n = k^w + r, where w is the largest integer such that r is not negative. For instance, the 3-power remainder of 13 is 4, since 13 = 3^2 + 4. In terms of k and w, what is the largest possible value of r that satisfies the given conditions?
Options:
A)(k – 1)k^w – 1
B)k^w – 1
C)(k + 1)k^w – 1
D)k^(w+1) – 1
E)(k + 1)k^(w+1) – 1
Rationale: n = k^w + r, where w is the largest integer such that r is not negative
This means k^w <= n, and k^(w+1) > n
So n lies between k^w and k^(w+1)
Now, r = n - k^w, which means r is the distance between n and k^w
This distance is maximised at the highest possible value of n, which can be just below k^(w+1)
So n = k^(w+1) - 1, as n is an integer
Therefore, highest value of r = n - k^w = k^(w+1) - 1 - k^w = (k - 1)k^w - 1
A is the correct
Answer: A
Question: If n = 2×3×5×7, then which of the following statements must be true?
I. n2 is divisible by 60
II. n + 19 is divisible by 19
III. is even n+4 / 2 = even
IV n is divisible by 35
Options:
A)I only
B)b- II only
C)III only
D)none and IV only
E)e None
Rationale: n = 2×3×5×7= 210
I. n is not divisible by 60
II. n + 19 is not divisible by 19
III. is even n+4 / 2 = odd 107
IV n is divisible by 35
D
Answer: D
Question: A bat is bought for Rs.400 and sold at a gain of 20% find its selling price
Options:
A)Rs.500/-
B)Rs.480/-
C)Rs.430/-
D)Rs.490/-
E)Rs.500/-
Rationale: 100 % ------> 400 (100 * 4 = 400)
120 % ------> 480 (120 * 4 = 480)
Selling price = Rs.480/-
B
Answer: B
Question: A candy company sells premium chocolate at $5 per pound and regular chocolate at $4 per pound in increment of whole pound only.If Barrett buys a pack of candies which costs him $40, how many pounds of premium chocolates are in the box?
Options:
A)[A] 1
B)[B] 2
C)[C] 3
D)[D] 4
E)[E] 5
Rationale: This question is a bit of Special equations with a single set of unique values.
the Equation stands : 5X + 4Y = 31
Subtract multiple of 5 from 31 and check whether the remaining total is divisible by 4
31 - 5 = 26 -> Not divisible by 4
31 - 10 = 21 -> Not divisible by 4
31 - 15 =16 -> Divisible by 4
31 - 20 =11 -> Not divisible by 4
31 - 25 =6 -> Not divisible by 4
31 - 30 =1 -> Not divisible by 4
Any multiple of 5 hereon will give a negative value for regular chocolates bought and hence, inadmissible
Thus Amount spent on regular chocolate= $16
Amount spent on Premium chocolate= $15
Price per premium chocolate = $5
No. Of boxes of premium chocolate bought = 20 / 5 = 4
Correct Option : D
Answer: D
Question: A juice manufacturer organized taste-testing sessions featuring four brands of orange juice, A, B, C and D. All customers who participated told the organizer which variety they thought was the most similar to freshly-squeezed orange juice. 61% preferred brand A and exactly half as many preferred brand B. Only 60 chose brand C. Which of the following could be the number of customers who preferred brand D?
Options:
A) 1
B) 8
C) 14
D) 20
E) 27
Rationale: I get D, but I'm not sure that my method is sound.
You have A + B + C + D = total testers. Stating the variety they think is most similar to freshly squeezed orange juice means each person should get 1 vote.
x = total number of testers
A + B + C + D
.61x + (.5)(.61x) + 65 + D = x
.61x + .305x + 65 + D = x
I took total testers = 1000 because that gives me nice numbers to deal with. Also, if I use 200 total testers, that gives me A = 122, B = 61, C = 65...that's over 200 without considering D.
x = 400
A = 244, B = 122, C = 65..also over 400
x = 600
A = 366, B = 183 C = 65....also over 600
x = 800
A = 488, B = 244, C = 65...D = 3...not an option on the list.
x = 1000
A = 610 B = 305 C = 65 D = 27....the first time I get an answer that is an option in the question.
E
Answer: E