mathewhe/aqua_rat huggingface dataset:
Question: If b + xy = b and x is not equal to 0, which of the following must be true?
(A) y=b
(B) y=0
(C) x=y
(D) x<y
(E) x+y=0
Rationale: b + xy = b --> xy=0. Since x is not equal to 0, thus y = 0.
Answer:
B
Answer: (B) y=0
Question: The ratio of three numbers is 3 : 4 : 7 and their product is 18144. The numbers are
(A) 9, 12, 21
(B) 15, 20, 25
(C) 18, 24, 42
(D) 22, 36. 48
(E) None
Rationale: Sol.
Let the numbers be 3x, 4x and 7x. Then
3x × 4x × 7x = 18144
⇔x3 = 18144/84= 216
⇔x3 = 63
⇔x = 6.
∴ The numbers are 18, 24 and 42.
Answer C
Answer: (C) 18, 24, 42
Question: Two trains of equal are running on parallel lines in the same direction at 42 km/hr and 36 km/hr. The faster train passes the slower train in 36 sec. The length of each train is?
(A) 30 m
(B) 72 m
(C) 80 m
(D) 82 m
(E) 84 m
Rationale: Let the length of each train be x m.
Then, distance covered = 2x m.
Relative speed = 42 - 36 = 6 km/hr.
= 6 * 5/18 = 5/3 m/sec.
2x/36 = 5/3=> x = 30.
ANSWER:A
Answer: (A) 30 m
Question: What is the lowest possible common multiple D of 2 distinct integers, each greater than 329?
(A) 330·331
(B) 330^2
(C) 33·331
(D) 660
(E) 330
Rationale: Lowest possible common multiple D of 2 distinct integers will have to be multiple of the number greater than 329...which 330..
so one number will be 330...another one will be 660..
Ans is D.
Answer: (D) 660
Question: A train moves fast a telegraph post and a bridge 220 m long in 8 sec and 20 sec respectively. What is the speed of the train?
(A) 79.5
(B) 39
(C) 66
(D) 79.2
(E) 79.1
Rationale: Let the length of the train be x m and its speed be y m/sec.
Then, x/y = 8 => x = 8y
(x + 220)/20 = y
y = 18.33
Speed = 18.33 m/sec = 18.33 * 18/5 = 66 km/hr.
Answer:C
Answer: (C) 66
Question: A 270 m long train running at the speed of 120 km/hr crosses another train running in opposite direction at the speed of 80 km/hr in 9 sec. What is the length of the other train?
(A) 230
(B) 783
(C) 291
(D) 112
(E) 221
Rationale: Relative speed = 120 + 80 = 200 km/hr.
= 200 * 5/18 = 500/9 m/sec.
Let the length of the other train be x m.
Then, (x + 270)/9 = 500/9 => x = 230.
Answer:A
Answer: (A) 230
Question: If a box contains 10 bulbs,of which just three are defective. If a random sample of five bulbs is drawn, find the probability that the sample contains no defective bulbs.
(A) 1/17
(B) 1/99
(C) 1/11
(D) 1/12
(E) 1/10
Rationale: Explanation:
Total number of elementary events = \inline {\color{Black}10C_{5}}
Number of ways of selecting no defective bulbs i.e., 5 non-defective bulbs out of 7 is \inline {\color{Black}7C_{5}}.
So,required probability =\inline {\color{Black}7C_{5}/10C_{5}} = 1/12.
Answer: D) 1/12
Answer: (D) 1/12
Question: If a:b = 4:8 and b:c = 3:2 find a:b:c?
(A) 3:8:4
(B) 3:4:8
(C) 3 : 6 : 4
(D) 4:6:3
(E) None of these
Rationale: Explanation:
a:b = 4:8, b:c = 3:2
4:8
3:2
(a = 4 × 3 = 12, b = 8 × 3 = 24 and c = 8 × 2 = 16)
(a = a × b, b = b × b and c = b × c)
a:b:c = 12 : 24 : 16
a : b : c = 3 : 6 : 4
Answer: Option C
Answer: (C) 3 : 6 : 4
Question: A wheel that has 12 cogs is meshed with a larger wheel of 14 cogs. When the smaller wheel has made 21 revolutions, then the number of revolutions made by the larger wheel is:
(A) 9
(B) 10
(C) 6
(D) 18
(E) 7
Rationale: Explanation :
Let the required number of revolutions made by larger wheel be x.
Then, More cogs, Less revolutions (Indirect Proportion)
14:12::21:x => 14 * x = 21 * 12
=> x = (21 * 12)/14
=> x = 18
Answer : D
Answer: (D) 18
Question: A person walking at 4 Kmph reaches his office 8 minutes late. If he walks at 6 Kmph, he reaches there 8 minutes earlier. How far is the office from his house?
(A) 3 1/5 Km
(B) 2 1/5 Km
(C) 4 1/5 Km
(D) 5 1/5 Km
(E) None of these
Rationale: Formula = S1Ã—S2/S2-S1 Ã— T1+T2/60
= 4Ã—6/2 Ã— 16/6
= 24/2 Ã— 16/60
= 8 Ã— 2/5
= 16/5 = 3 1/5 Km
ANSWER IS A
Answer: (A) 3 1/5 Km
Question: In a apartment, 30% of the people speak English, 20% speak Hindi and 10% speak both. If a people is selected at random, what is the probability that he has speak English or Hindi?
(A) 1/5
(B) 1/3
(C) 2/5
(D) 2/7
(E) 3/5
Rationale: P (E) = 30 / 100 = 3 / 10 , P (H) = 20 / 100 = 1 / 5 and P (E ∩ H) = 10 / 100 = 1 / 10
P (E or H) = P (E U H)
= P (E) + P (H) - P (E ∩ H)
= (3 / 10) + (1 / 5) - (1 / 10) = 4 / 10 = 2/5
C
Answer: (C) 2/5
Question: A sum of money at simple interest amounts to Rs. 815 in 2 years and to Rs. 854 in 4 years. The sum is:
(A) s. 650
(B) s. 690
(C) s. 698
(D) s. 700
(E) s. 737
Rationale: S.I. for 1 year = Rs. (854 - 815) = Rs. 39.
S.I. for 2 years = Rs.(39 x 2) = Rs. 78.
Principal = Rs. (815 - 78) = Rs. 737.
Answer: Option E
Answer: (E) s. 737
Question: A tourist purchased a total of $1,000 worth of traveler’s checks in $10 and $50 denominations, During the trip the tourist cashed 7 checks and then lost all of the rest. If the number of $10 checks cashed was one more or one less than the number of $50 checks cashed, what is the minimum possible value of the checks that were lost?
(A) $1,430
(B) $1,310
(C) $1,290
(D) $1,270
(E) $770
Rationale: since total cashed checks = 7 ans $10 check cashed = one less or more than $50 checks
Therefore either $10 checks = 3 and $50 checks = 4 OR $10 checks = 4 and $50 checks = 3
using this information 1st option gives us the maximum cashed value therefore if we negate this value from total value we will get minimum value of which checks have lost
Therfore 1000 - 230 = 770 hence answer is E
Answer: (E) $770
Question: When tossed, a certain coin has an equal probability of landing on either side. If the coin is tossed 4 times, what is the probability that it will land on the same side each time?
(A) 1/4
(B) 1/6
(C) 1/8
(D) 1/12
(E) 1/16
Rationale: On the first toss, the coin will land on one side or the other.
On the second toss, the probability of landing on the same side is 1/2.
On the third toss, the probability of landing on the same side is 1/2.
On the fourth toss, the probability of landing on the same side is 1/2.
P(same side all four times)=1/2*1/2*1/2=1/8.
The answer is C.
Answer: (C) 1/8
Question: $4,000 is deposited in a savings account that pays 8% annual interest compounded semiannually. To the nearest dollar, how much is in the account at the end of the year?
(A) $4326.40
(B) $4116.20
(C) $4256.21
(D) $4316.54
(E) $4312.45
Rationale: this is the case of semi-annual compoundingso, multiply time period by 2 and divide rate by 2
So,
new time=2periods and new rate=8/2=4%
Now,
CA=4000(1+4/100)^2=$4326.40
Hence answer is A
Answer: (A) $4326.40
Question: A salesman’s terms were changed from a flat commission of 5% on all his sales to a fixed salary of Rs.1100 plus 2.5% commission on all sales exceeding Rs. 4,000. If his remuneration as per new scheme was Rs. 600 more than that by the previous schema, his sales were worth?
(A) 14000
(B) 12000
(C) 12019
(D) 12197
(E) 12012
Rationale: [1100 + (X-4000) * (2.5/100)] - X * (5/100)
= 600
X = 14000
Answer: A
Answer: (A) 14000
Question: Suresh started a business with Rs.20,000. Kiran joined him after 4 months with Rs.30,000. After 2 more months, Suresh withdrew Rs.5,000 of his capital and 2 more months later, Kiran brought in Rs.20,000 more. What should be the ratio in which they should share their profits at the end of the year?
(A) 17:12
(B) 12:17
(C) 21:32
(D) 32:21
(E) 17:32
Rationale: Here capital is not the same.
Suresh invested 20000 for initial 6 months and 15000 for the next 6 months. Hence his term of ratio
=(20000×6+15000×6)
Kiran invested Rs.30000 for 4 months and Rs.50000 for next 4 months. Hence his term of ratio
=(30000×4+50000×4)
Suresh : Kiran
=(20000×6+15000×6) :(30000×4+50000×4)
=(20×6+15×6):(30×4+50×4)=(20×3+15×3):(30×2:50×2)
=105:160=21:32
Answer is C.
Answer: (C) 21:32
Question: The radius of a circular wheel is 1.75 m, how many revolutions will it make in traveling 1 km?
(A) 1000
(B) 2991
(C) 8216
(D) 2991
(E) 2881
Rationale: 2 * 22/7 * 1.75 * x
= 11000
x = 1000
Answer: A
Answer: (A) 1000
Question: Harold and Millicent are getting married and need to combine their already-full libraries. If Harold, who has 1/2 as many books as Millicent, brings 1/3 of his books to their new home, then Millicent will have enough room to bring 1/3 of her books to their new home. What fraction of Millicent's old library capacity is the new home's library capacity?
(A) 1/2
(B) 2/3
(C) 3/4
(D) 4/5
(E) 5/6
Rationale: Because we see H willbring 1/3 of his booksto the new home --> try to pick a number that isdivisible by 3.
Before:
Assume H = 30 books
H = 1/2 M --> M = 60 books
After:
H' = 1/3H = 10 books
M' = 1/3M = 20 books
Total = 30 books
M' = 30 = 1/2*60
Ratio: 1/2
Ans: A
Answer: (A) 1/2
Question: 39 persons can repair a road in 12 days, working 5 hours a day. In how many days will 30 persons, working 6 hours a day, complete the work?
(A) 11
(B) 12
(C) 13
(D) 15
(E) 14
Rationale: First codition=total hours to complete the work=39 persons * 12 days * 5 hr
Second codition=total hours to complete the work=30 persons * x days * 6 hr
since the work is still the same so
39*12*5=30*x*6
x=13
answer :C
Answer: (C) 13
Question: Find the area of a cuboid of length 15cm , breadth 10 cm . and height 16cm.
(A) 2400 cm cube
(B) 2410 cm cube
(C) 2420 cm cube
(D) 2430 cm cube
(E) 2480 cm cube
Rationale: Area of a cuboid = lxbxh =15cm x 10cm x 16cm = 2400 cm cube
answer :A
Answer: (A) 2400 cm cube
Question: By purchasing an article at 20 % discount on the original price and then selling it at a price of 25% above the original price, a trader earns Rs. 200 as the profit. What was the original price of the article?
(A) 444.44
(B) 444.409
(C) 444.42
(D) 444.429
(E) 444.428
Rationale: Explanation:
Let the original price of the article be Rs. 100. Hence the purchase price for the trader would be Rs. 80 and his selling price would be Rs. 125. Thus, he would earn a profit of Rs. 45 (125 – 80). Therefore,
Profit is Rs. 45 if the original price is Rs. 100
Hence, if profit is Rs. 200, then original price will be..
= 100x 200/45
= Rs. 444.44
ANSWER: A
Answer: (A) 444.44
Question: If A lends Rs.3500 to B at 10% per annum and B lends the same sum to C at 15% per annum then the gain of B in a period of 3 years is?
(A) 525
(B) 500
(C) 510
(D) 530
(E) 550
Rationale: (3500*5*3)/100 => 525
Answer: A
Answer: (A) 525
Question: The ratio between the speeds of two trains is 7 : 8. If the second train runs 400 km in 4 hours, then the speed of the first train is:
(A) 87 km/hr
(B) 88 km/hr
(C) 87.5 km/hr
(D) 67 km/hr
(E) 77 km/hr
Rationale: Let the speed of two trains be 7x and 8x km/hr.
Then, 8x =(400/4)= 100
x =100/8= 12.5
Speed of first train = (7 x 12.5) km/hr = 87.5 km/hr.
ANSWER C
Answer: (C) 87.5 km/hr
Question: If d and t are integers and d^2 – t^2 is an odd integer, which of the following must be an even integer?
I. d + t + 2
II. d^2 + 2dt + t^2
III. d^2 + t^2
(A) None
(B) I only
(C) II only
(D) III only
(E) I, II, and III
Rationale: d^2–t^2 is an odd integer means that either d is even and t is odd or d is odd and t is even.
Check all options:
I. d + t + 2 --> even+odd+even=odd or odd+even+even=odd. Discard;
II. d^2 + 2dt + t^2 --> even+even+odd=odd or odd+even+even=odd. Discard;
III. d^2 + t^2 --> even+odd=odd or odd+even=odd. Discard.
Answer: A.
Answer: (A) None
Question: The difference between a two-digit number and the number after interchanging the position of the two digits is 72. What is the difference between the two digits of the number?
(A) 4
(B) 6
(C) 3
(D) Cannot be determined
(E) None of these
Rationale: Let the two-digit no. be l0x + y.
Then, (10x + y) â€“ (10y + x) = 72
or, 9(x â€“ y) = 72
Answer E
Answer: (E) None of these
Question: There are three different hoses used to fill a pool: hose U, hose y, and hose z. Hose U can fill the pool in a days, hose y in b days, and hose z in c days, where a > b > c. When all three hoses are used together to fill a pool, it takes d days to fill the pool. Which of the following must be true?
I. d<c
II. d>b
III. c/3<d<a/3
(A) I
(B) II
(C) III
(D) II &III
(E) I &II
Rationale: Name T = full pool
U fills a pool in a days ==> 1 day U fills: T/a
Y fills a pool in b days ==> 1 day Y fills: T/b
Z fills a pool in c days ==> 1 day Z fills: T/c
1 day (U+Y+Z) together fill: T(1/a + 1/b + 1/c)
d days (U+Y+Z) together fill: T
==> d = T1 / T(1/a+1/b+1/c) = abc/(ab+bc+ca)
==>d = abc/(ab+bc+ca)
d cannot be greater than c ; d is greater than b ; d is greater than c/3 but not less than a/3
B
Answer: (B) II
Question: Billy read 2 books. He read the first one in one week with 25 pages everyday. He read the second book in 12 days with 23 pages everyday. What is the total number of pages that Billy read?
(A) 251 pages
(B) 151 pages
(C) 751 pages
(D) 451 pages
(E) 351 pages
Rationale: Pages read in the first book in one week which is 7 days with 25 pages everyday.
25 ? 7 = 175 pages
Pages read in the second book in 12 days with 23 pages everyday.
23 ? 12 = 276 pages
Total number of pages read
175 + 276 = 451 pages
correct answer D
Answer: (D) 451 pages
Question: If 2 men can color 50 m long cloth in 2 days, then 4 men can color 100 m long cloth in?
(A) 3 days
(B) 4 days
(C) 5 days
(D) 2 days
(E) 6 days
Rationale: The length of cloth painted by one man in one day = 50/ 2× 2 = 12.5 m
No. of days required to paint 100 m cloth by 4 men = 100/ 4 × 12.5 = 2 days
Answer : D
Answer: (D) 2 days
Question: Marla starts running around a circular track at the same time Nick starts walking around the same circular track. Marla completes 28 laps around the track per hour and Nick completes 13 laps around the track per hour. How many minutes after Marla and Nick begin moving will Marla have completed 4 more laps around the track than Nick?
(A) 5
(B) 8
(C) 12
(D) 16
(E) 20
Rationale: Maria's rate - 28 laps per hour --> 28/60 laps/min
Nick's rate - 13 laps per hour --> 13/60 laps/min
lets set equations:
28/60*t=4 (since Maria had to run 4 laps before Nick would start)
13/60*t=0 (Hick has just started and hasn't run any lap yet)
(28/60-13/60)*t=4-0 (since Nick was chasing Maria)
t=16 min needed Maria to run 4 laps
Answer : D
Answer: (D) 16
Question: By selling 32 carrots in a dollar, a man loses 40%. How many carrots must he sell in a dollar to gain 20% profit?
(A) 16
(B) 18
(C) 20
(D) 25
(E) 28
Rationale: This is a question that is much easier than it first appears. If you work with the percents directly, you can solve it very quickly. If the man sells 32 carrots for a dollar, he loses 40%. That means he is at 60% of his cost (100% - 40% = 60%). We are trying to figure out how many carrots he has to sell in order to make a 20% profit, or be at 120% (100% + 20% = 120%). 120% is double 60%, meaning that we simply cut the number of carrots in half to double our returns (from 60% to 120%), yielding 32/2 = 16 carrots, or answer choice A.
Answer: (A) 16
Question: A farmer spent $38 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?
(A) $37.80
(B) $38.50
(C) $39.20
(D) $39.50
(E) $41.80
Rationale: A farmer spent 40% money on chicken feed, so he spent 0.4*$38=$15.2 on chicken feed, thus he spent the remaining 38-15.2=$22.8 on goat feed.
Now, since he bought chicken feed at a 20% discount then the original price of it was x*0.8=$15.2 --> x=$19.
Therefore if the farmer had paid full price for both the chicken feed and the goat feed, then he would he have spent 19+22.8=$41.8.
Answer: E.
Answer: (E) $41.80
Question: A train 260 m long passed a pole in 40 sec. How long will it take to pass a platform 940 m long?
(A) 180
(B) 184
(C) 178
(D) 167
(E) 156
Rationale: Speed = 260/40= 6.5 m/sec.
Required time = (260 + 940)/6.5 = 184 sec.
Answer: Option B
Answer: (B) 184
Question: The ratio of the two natural numbers is 6 : 5. If a certain number is subtracted to both the numbers, the ratio becomes 5 : 4. If the larger number exceeds the smaller number by 5, find the number subtracted?
(A) 8
(B) 5
(C) 15
(D) 22
(E) 6
Rationale: Let the two numbers be 6x and 5x.
Let the numbers subtracted to both so that their ratio becomes 5 : 4 be k.
(6x - k) / (5x - k) = 5/4
=> 24x - 4k = 25x - 5k => k = x.
6x - 5x = 5 => x = 5
k = x = 5.
Answer:B
Answer: (B) 5
Question: On a playground, there is a number of children, out of which 1/4 wear red t-shirts, 1/20 wear blue t-shirts, and the remaining 14 children wear white t-shirts. How many children wear blue t-shirts?
(A) 6
(B) 20
(C) 1
(D) 3
(E) 5
Rationale: Let x be the number of children on the playground.
x/4+x/20=6x/20=3x/10 is the number of children who wear red and blue t-shirts
The number of the children who wear white t-shirts is x-3x/10=7x/10. Thus,
7x/10=14 implies x=20.
1/20*20=1. Hence, only 1 child wears a white t-shirt.
Answer C 1
Answer: (C) 1
Question: At a certain laboratory, chemical substance are identified by an unordered combination of three different colors. If no chemical may be assigned the same three colors as any other, what is the maximum number of substances that can be identified using five colors?
(A) 21
(B) 35
(C) 105
(D) 135
(E) 10
Rationale: The prompt tells us that substance is identified by an UNORDERED combination of 3 colors and that no chemical may be assigned the same 3 colors as any other chemical.
As an example, a chemical with the colors A/B/C means that the following combinations CANNOT be used by any other chemical:
ABC
ACB
BAC
BCA
CAB
CBA
By extension, the first part of your calculation would be the number of possible substances that could be identified with 5 colors: 5c3 = 10 possible substances
E
Answer: (E) 10
Question: The number of cars present at a particular time was measured at 3,999 different locations on Tuesday and on Wednesday. The number of locations that had more cars on Wednesday was 20% higher than the number of locations that had more cars on Tuesday. How many of the locations had more cars on Tuesday?
(A) 1,159
(B) 1,333
(C) 1,817
(D) 2,460
(E) 2,829
Rationale: The question states that all these 3,999 locations were measured twice, once on each day.
It also breaks the 3,999 into two groups, the higher-on-Tuesday locations and the higher-on-Wednesday locations, which we can call T and W.
Therefore, we can write two equations:
For each of the first and second day count of locations, T + W = 3999
But, for the second day count, W = 1.20 T
i.e. for Second day count, T + 1.20T = 3999
i.e. 2.20 T = 3999
i.e. T = 1817
C
Answer: (C) 1,817
Question: The average (arithmetic mean) of y numbers is x. If 30 is added to the set of numbers, then the average will be x - 6. What is the value of y in terms of x ?
(A) x/6 - 6
(B) x/6 - 5
(C) x/7 - 5
(D) x/5 - 7
(E) x/5 - 6
Rationale: (A1 + A2 + .. + Ay)/y = x
(A1 + A2 + .. + Ay + 30)/(y+1) = x - 6
=> (xy + 30)/(y+1) = x - 6
=> xy + 30 = yx -6y + x - 6
=> 36 = x - 6y
=> 6y = x - 36
=> y = x/6 - 6
Answer - A
Answer: (A) x/6 - 6
Question: 1: 3 = 1 2/3: x. The value of x is?
(A) 1
(B) 3
(C) 4 1/6
(D) 5
(E) 6
Rationale: Explanation:
x * 1 = 3 * 5/3
x = 5
ANSWER IS D
Answer: (D) 5
Question: Two trains are traveling in the same direction at speeds of 50 kmph and 20 kmph respectively If the faster train passes the driver in the slower train in 20 seconds, What is the length of the faster train?
(A) ANSWER: 169 m
(B) ANSWER: 162 m
(C) ANSWER: 136 m
(D) ANSWER: 166 m
(E) ANSWER: 366 m
Rationale: Explanation:
Given :
Speed of fast train = 50 kmph.
Speed of slow train = 20 kmph.
The driver in the slower train is moving with the speed of the slower train which is 20 kmph.
The faster train will pass the driver when it has gained a distance. Distance gained by faster train
in 1 hour = (50 – 20) = 30 km
= 8.33 m/sec
Distance gained by faster train in 20 sec = 20 * 8.33 = 166.6 m.
Therefore, Length of faster train = 166.6 m.
ANSWER:D
Answer: (D) ANSWER: 166 m
Question: For a particular American football game, the probability of a team's quarterback throwing an intercepted pass on each throw is 1/10. What is the least number of times that the quarterback will throw the ball that will increase the probability of throwing an intercepted pass at least once to more than 50%.
(A) 10
(B) 3
(C) 9
(D) 7
(E) 5
Rationale: Rule of Subtraction: P(A) = 1 - P(A')
Rule of Multiplication: P(A ∩ B) = P(A) P(B)
Probability quarterback throws an intercepted pass at least once in 2 throws is 1- (9/10)^2 = 1 - 81/100 = 11/100 < 50%
Probability quarterback throws an intercepted pass at least once in 3 throws is 1- (9/10)^3 = 1 - 729/1,000 = 271/1,000 < 50%
Probability quarterback throws an intercepted pass at least once in 4 throws is 1- (9/10)^4 = 1 - 6,561/10,000 = 4,439/10,000 < 50%
Probability quarterback throws an intercepted pass at least once in 5 throws is 1- (9/10)^5 = 40,951/100,000 < 50%
Probability quarterback throws an intercepted pass at least once in 5 throws is 1- (9/10)^6 = 468,559/1,000,000 < 50%
Probability quarterback throws an intercepted pass at least once in 5 throws is 1- (9/10)^7 = 5,217,031/10,000,000 > 50%
Answer: D
Answer: (D) 7
Question: An investor purchased 100 shares of stock X at 8 1/8 dollars per share and sold them all a year later at 32 dollars per share. If the investor paid a 2 percent brokerage fee on both the total purchase price and the total selling price, which of the following is closest to the investor's percent gain on this investment?
(A) 92%
(B) 240%
(C) 380%
(D) 300%
(E) 290%
Rationale: If the purchase price was 8$ per share then the investor would have made a profit of 300%.
Since the purchase price is slightly more than 8$ the profit would be slightly less than 300%.
Also a 2% brokerage is negligible and it brings down the profit percentage only by a small value. Approximation is very useful to solve these kind of problems as the answer choices are far apart.
Answer: E
Answer: (E) 290%
Question: A certain sum of money is invested for one year at a certain rate of simple interest. If the rate of interest is 3% higher, then the invest earned will be 25% more than the interest earned earlier. What is the earlier rate of interest?
(A) 28
(B) 65
(C) 89
(D) 12
(E) 14
Rationale: If the interest earned is 25% more than the earlier interest then the rate of interest also should be 25% higher than the earlier rate.
Let the earlier rate of interest be x%.
Now it will be (x + 3)%
% increase = (x + 3) - x/x * 100 = 25
=> x = 12.Answer: D
Answer: (D) 12
Question: In a new housing development, trees are to be planted along the sidewalk of a certain street. Each tree takes up one square foot of sidewalk space, and there are to be 14 feet between each tree. How many trees can be planted if the road is 151 feet long?
(A) 8
(B) 9
(C) 10
(D) 11
(E) 12
Rationale: Let T be the number of trees. Then the length required for trees on the sidewalk will be 1*T= T
To maximize the number of trees, the number of 14 feet spaces between trees should be 1 less than total number of trees.
For example, If there are 3 trees, then there should be 2 spaces between them.
So the number of 14 feet spaces will be T-1. Then, the length of sidewalk required for 14 feet spaces will be 14*(T-1)
It is given that total lenght of sidewalk is 166 feet.
or 14(T-1)+T = 151
or 14T-14+T = 151
or 15T = 165
or T=11
Answer:- D
Answer: (D) 11
Question: Month	Average Price
per Dozen
April $1.0
May $1.20
June $1.5
The table above shows the average (arithmetic mean) price per dozen of the large grade A eggs sold in a certain store during three successive months. If 2/3 as many dozen were sold in April as in May, and twice as many were sold in June as in April, what was the average price per dozen of the eggs sold over the three-month period?
(A) 1.1
(B) 1.29
(C) 1.32
(D) 1.42
(E) 1.5
Rationale: Say x dozen eggs were sold in May.
Then 2x/3 dozen eggs were sold in April and 2*2x/3=4x/3 dozen eggs were sold in May.
Total x+2x/3+4x/3=3x dozen eggs were sold in three months.
Total sales: 2x/3*1.+x*1.2+4x/3*1.5= $3.86x. The average price per dozen 3.86x/3x=$1.29.
B
Answer: (B) 1.29
Question: Walking at 5/6th of its usual speed a cab is 12mnts late. Find its usual time to cover the journey?
(A) 25m
(B) 45m
(C) 32m
(D) 50m
(E) 60m
Rationale: New speed = 5/6th of usual speed
New time = 6/5th of usual time
6/5 UT - UT = 12m
UT/5 = 12m
UT = 60m
Answer is E
Answer: (E) 60m