mathewhe/aqua_rat huggingface dataset:
Question: A car drove from Town A to Town B without stopping. The car traveled the first 80 miles of its journey at an average speed of 25 miles per hour. What was the car’s average speed, in miles per hour, for the remaining 80 miles if the car’s average speed for the entire trip was 40 miles per hour?
(A) 28
(B) 48
(C) 50
(D) 55
(E) 100
Rationale: KAPLANOFFICIAL SOLUTION:
To solve this problem, you must remember that average speed means total distance divided by total time over an entire journey. This should not be confused with the average of the speeds. The total distance in this problem is 160 miles, but we will need to express the total time in a more complex way.
For the first 80 miles, the car traveled at 25 miles per hour. Therefore, we can say that the time this took was 80/25, because distance divided by rate equals time.
For the last 80 miles, we are trying to solve for the rate, so we can call this speed R. Thus, the time for the final 80 miles is 80/R.
If we put all of this together, knowing that the average speed over the entire journey is 40 miles per hour, we get the equation 160/(80/25 + 80/R) = 40. We can now solve for R and reach our answer.
160/(80/25 + 80/R) = 40
R = 100
100 is choice (E) and that’s our correct answer.
Answer: (E) 100
Question: If x and y are positive real numbers such that x + y = 1, which of the following could be the value of 100x + 500y?
I. 90
II. 250
III. 499
(A) II only
(B) III only
(C) I and II
(D) I and III
(E) II and III
Rationale: We know that x and y are both between 0 and 1.
The minimum limit of 100x + 500y is 100, when x is close to 1 and y is close to 0.
The maximum limit of 100x + 500y is 500, when x is close to 0 and y is close to 1.
The expression can take on any value between 100 and 500.
The answer is E.
Answer: (E) II and III
Question: What is the units digit of 2222^(333)*4444^(222)?
(A) 0
(B) 2
(C) 4
(D) 6
(E) 10
Rationale: Each of the other explanations to this question has properly explained that you need to break down the calculation into pieces and figure out the repeatingpatternof the units digits.
Here's another way to organize the information.
We're given [(2222)^333][(4444)^222]
We can 'combine' some of the pieces and rewrite this product as....
([(2222)(3333)]^222) [(2222)^111]
(2222)(3333) = a big number that ends in a 6
Taking a number that ends in a 6 and raising it to a power creates a nice pattern:
6^1 = 6
6^2 = 36
6^3 = 216
Etc.
Thus, we know that ([(2222)(3333)]^222) will be a gigantic number that ends in a 6.
2^111 requires us to figure out thecycleof the units digit...
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256
So, every 4powers, the pattern of the units digits repeats (2, 4, 8, 6.....2, 4, 8, 6....).
111 = 27 sets of 4 with a remainder of 3....
This means that 2^111 = a big number that ends in an 8
So we have to multiply a big number that ends in a 6 and a big number that ends in an 8.
(6)(8) = 48, so the final product will be a gigantic number that ends in an 8.
Final Answer:
E
Answer: (E) 10
Question: The dimensions of a certain machine are 48*32*52,if the size of the machine is increased proportionally until sum of dimensions equals 156.
(A) 56*40*50
(B) 56*40*40
(C) 56*40*30
(D) 56*40*20
(E) 56*40*60
Rationale: Given dimensions of a machine are 48*32*52
if the size is raised proportionally untill the sum of dimensions
sum of given dimensions is 48+32+52=132
difference of actual and increased=156-132=24
24/3=8
this means each dimension has to raise for 8 units
thus the new dimensions are 56*40*60
ANSWER:E
Answer: (E) 56*40*60
Question: A multiple choice test consists of 4 questions, and each question has 5 answer choices. In how many R ways can the test be completed if every question is unanswered?
(A) 24
(B) 120
(C) 625
(D) 720
(E) 1024
Rationale: 5 choices for each of the 4 questions, thus total R of 5*5*5*5 = 5^4 = 625 ways to answer all of them.
Answer: C.
Answer: (C) 625
Question: x and y are positive integers. If xy + x is odd, then which of the following must be even?
.
(A) x
(B) y
(C) x + y
(D) xy − x
(E) x^2 − y
Rationale: Two possibilities:
(1) even + odd = odd
(2) odd + even= odd
In this case, x can not be even as it will make xy also even .
Therefore, (1) possibility present => y must be even.
Answer: B
Answer: (B) y
Question: A sum amount to Rs.1344 in two years at simple interest. What will be the compound interest on the same sum at the same rate of interest for the same period?
(A) 10.75%
(B) 10.28%
(C) 10.25%
(D) 16.25%
(E) 20.25%
Rationale: 100
5 ---- I
5
0.25 --- II
-----------
10.25%.Answer:C
Answer: (C) 10.25%
Question: The prices of tea and coffee per kg were the same in june. In july the price of coffee shot up by 20% and that of tea dropped by 20%. if in july , a mixture containing equal quantities of tea and coffee costs 60/kg. how much did a kg of coffee cost in june ?
(A) 40
(B) 45
(C) 50
(D) 55
(E) 60
Rationale: Let the price of tea and coffee be x per kg in June.
Price of tea in July = 1.2x
Price of coffee in July = 0.8x .
In July the price of 1/2 kg (500gm) of tea and 1/2 kg (500gm) of coffee (equal quantities) = 50
1.2x(1/2) + 0.8x(1/2) = 60
=> x =60
E
Answer: (E) 60
Question: X does a work in 10 days. Y does the same work in 15 days. In how many days they together will do the same work?
(A) 10
(B) 15
(C) 3
(D) 6
(E) 12
Rationale: X's 1day's work= 1/10
Y's 1 day's work=1/15
(X+Y)'s 1day's work= (1/10+1/15)=1/6
both together will finish the work in 6 days.
correct option is D
Answer: (D) 6
Question: Bob finishes the first half of an exam in one-thirds the time it takes him to finish the second half. If the whole exam takes him an hour, how many minutes does he spend on the first half of the exam?
(A) 20
(B) 24
(C) 27
(D) 30
(E) 40
Rationale: Many times, it is easiest to think of problems like this conceptually (as opposed to formulaically). Conceptually, if the first half of the exam takes 1/3's the time of the second half (3/3's), we can see that the hour the entire exam took to finish can be broken down into fifth's. (Another way to look at this problem is via ratio's - first half is 1:5 and the second half is 3:5). With each fifth of an hour being 15 minutes (60/4), the first half of the exam would have taken 30 minutes. Correct answer is D.
Answer: (D) 30
Question: In a store, the profit is 320% of the cost. If the cost increases by 25% but the SP remains constant, approximately what %of the sp is the profit?
(A) 50%
(B) 70%
(C) 80%
(D) 100%
(E) 120%
Rationale: Let C.P. = Rs. 100. Then, profit = Rs. 320, S.P. = Rs. 420
New C.P. = 125% of Rs. 100 = Rs. 125.
New S.P. = Rs. 420
Profit = 420 - 125 = Rs. 295
Required percentage = 295/420 * 100 = 1475/21 = 70%
B
Answer: (B) 70%
Question: The ratio of ages of Aman, Bren, and Charlie are in the ratio 5:8:7 respectively. If 8 years ago, the sum of their ages was 76, what will be the age of Charlie 9 years from now?
(A) 44
(B) 52
(C) 57
(D) 65
(E) 80
Rationale: Let the present ages of Aman, Bren, and Charlie be 5x, 8x and 7x respectively.
5x-8 + 8x-8 + 7x-8 = 76
x = 5
Present age of Charlie = 7*5 = 35
Charlie's age 9 years hence = 35+9 = 44
Answer = A
Answer: (A) 44
Question: A shopkeeper sold an article for Rs 2591.36. Approximately what was his profit percent if the cost price of the article was Rs 2400
(A) 4%
(B) 5%
(C) 6%
(D) 8%
(E) 10%
Rationale: Explanation:
Gain % = (191.36*100/2400) = 7.97 % = 8% approx
Option D
Answer: (D) 8%
Question: A team has to design a flag. The team has two yellow strips of cloth and five green strips of cloth that they must use as is to make the flag. How many different flags can the team design with the materials at hand?
(A) 21
(B) 121
(C) 240
(D) 720
(E) 1040
Rationale: The # of permutations of total 2+5=7 strips where 2 are identical yellow and 5 are identical green is 7!/(2!*5!)=21 (note that they must use all the materials at hand, also I guess that the strips must be either only vertical or only horizontal).
Answer: A.
Answer: (A) 21
Question: How many numbers between 0 and 1670 have a prime tens digit and a prime units digit?
(A) 268
(B) 272
(C) 202
(D) 112
(E) 262
Rationale: There are 4 single digit prime numbers: 2, 3, 5 and 7. Hence, last two digits (tens and units) can take 4*4=16 different values: 22, 23, ..., 77.
So, in each hundred there are 16 such numbers. In 17 hundreds there will be 17*16=272 such numbers, but 4 out of them will be more than 1670, namely: 1672, 1673, 1675 and 1677. Which means that there are 272-4=268 numbers between 0 and 1670 which have a prime tens digit and a prime units digit.
Answer: A.
Answer: (A) 268
Question: If 'b' is a positive integer exactly divisible by 5 or 7 but not divisible by 11.What could possibly be the value of 'b' ?
(A) 55
(B) 35
(C) 77
(D) 110
(E) 165
Rationale: 55 and 77 are both divisible by 5 and 7 but also by 11. so they are not the right answer.
110 and 165 are both clearly not divisible by 7(not correct)
35 is both divisible by 5 and 7 but not by 11.
answer : (B) 35
Answer: (B) 35
Question: If x and y are non-zero integers, and 16x^4 – 4y^4 = 4x^2 + 2y^2, which of the following could be the value of x2 in terms of y?
(A) 2y^2
(B) 2y^2+3/4
(C) 2y^2+1/4
(D) 2y^2+1/2
(E) 2y^2+1/6
Rationale: 16x^4 – 4y^4 = 4x^2 + 2y^2
(4x^2 + 2y^2) (4x^2 - 2y^2) = 4x^2 + 2y^2
4x^2 - 2y^2 = 1
4x^2 = 2y^2+1
x^2 = 2y^2+1/4
Answer : C
Answer: (C) 2y^2+1/4
Question: What sum of money will produce Rs.70 as simple interest in 4 years at 3 1/2 percent?
(A) 339
(B) 500
(C) 277
(D) 266
(E) 2812
Rationale: 70 = (P*4*7/2)/100
P = 500
Answer:B
Answer: (B) 500
Question: The average of first five prime numbers greater than 20 is?
(A) 32.5
(B) 32.0
(C) 32.4
(D) 32.2
(E) 32.8
Rationale: 23 + 29 + 31 + 37 + 41 = 161/5
= 32.2
Answer: D
Answer: (D) 32.2
Question: How long does a train 130 m long running at the speed of 80 km/hr takes to cross a bridge 140 m length?
(A) 13.9 sec
(B) 12.3 sec
(C) 17.9 sec
(D) 61.9 sec
(E) 47.98 sec
Rationale: Speed = 80 * 5/18 = 22 m/sec
Total distance covered = 130 + 140 = 270 m.
Required time = 270/22 = 12.3 sec.
Answer:B
Answer: (B) 12.3 sec
Question: A case of 12 rolls of paper towels sells for $9. The cost of one roll sold individually is $1.What is the percent E of savings per roll for the 12-roll package over the cost of 12 rolls purchased individually?
(A) 9%
(B) 11%
(C) 15%
(D) 25%
(E) 90%
Rationale: Cost of 12 paper towels individually = 1*12=12
Cost of a set of 12 paper towels = 9
Cost of one roll = 9/12 = 3/4 = 0.75
Savings per roll = 1-.75 = 0.25
% of savings is E= .25/ 1 *100 = 25% D is the answer.
Answer: (D) 25%
Question: On a sum of money, the S.I. for 2 years is $600, while the C.I. is $609, the rate of interest being the same in both the cases. The rate of interest is?
(A) 2%
(B) 3%
(C) 4%
(D) 5%
(E) 6%
Rationale: Difference in C.I. and S.I for 2 years = $609 - $600 = $9
S.I for one year = $300
S.I. on $300 for 1 year = $9
Rate = (100 * 9)/(300) = 3%
The answer is B.
Answer: (B) 3%
Question: Positive integers indicated by x and y satisfy 3 1/x * y 2/5 = 13 3/4, the fractions being in their lowest terms, then x = ? and y = ?
(A) x = 4, y = 8
(B) x = 4, y = 4
(C) x = 2, y = 4
(D) x = 8, y = 4
(E) None of these
Rationale: 3 1/x * y 2/5 = 13¾
=> (3x + 1) / x + (5y + 1) / 2 += 55 / 4
only choice (4) i.e x = 8 and y = 4 satisfies the given equation.
3 (1/8) * 4 (2/5) = (25 / 8 ) * (22 / 5) = 55/4 = 13¾
ANSWER:D
Answer: (D) x = 8, y = 4
Question: There is a train and car. The ratio btw the speed of a train & a car is 16:15 respectively. Also, a bus covered a distance of 480km in 8h. The speed of the bus is 3/4th of the speed of the train. How much distance will the car cover in 6h?
(A) 410 km
(B) 420 km
(C) 460 km
(D) 430 km
(E) 450 km
Rationale: Sol. Speed of bus = 480/8 = 60km/ h
Speed of train = (60*4)/3=80 km / h
Speed of train : Speed of car = 16 : 15
Speed of car =80/16 * 15 = 75 km/ h
Distance covered by car in 6 hr = 75 × 6 = 450 km
E
Answer: (E) 450 km
Question: Paul purchased a house for Rs.46000 and a site for Rs.14000 respectively, if he sold the house for Rs.44000 and the site for Rs.20000, then find the resultant percentage of gain?
(A) 3.33%
(B) 4.44%
(C) 5.55%
(D) 6.66%
(E) 7.77%
Rationale: Sol.
Total CP = 60000 & total SP =64000 ⇒ gain = SP/CP = 64000/60000 = 1.0666 = 6.66%.
Answer D
Answer: (D) 6.66%
Question: In an examination series, which consists of 5 exams, Arvind scored an average of 60 marks in
the first 3 exams and an average of 70 in the last 4 exams, what is the average score of
Arvind for the first and last two exams if his average scores of all the five exams are 65?
(A) 63.33
(B) 6
(C) 64
(D) 12
(E) 67.33
Rationale: let a,b,c,d,e be the five exams
(a+b+c+d+e)/5=65
a+b+c+d+e=325
first 3 exams avg is (a+b+c)/3=60
a+b+c=180
d+e=325-180=145
avg of last 4 exams is (b+c+d+e)/4=70
b+c+d+e=280
a=325-280=45
avg of first and last two exams are(a+d+e)/3=(45+145)/3=(190)/3=63.33
ANSWER:A
Answer: (A) 63.33
Question: A reduction of 25% in the price of oil enables a house wife to obtain 5kgs more for Rs.800, what is the reduced price for kg?
(A) 33
(B) 77
(C) 40
(D) 88
(E) 112
Rationale: 800*(25/100) = 200 ---- 5
? ---- 1 => Rs.40
Answer: C
Answer: (C) 40
Question: In how many ways can a committee of 7 people comprising at least 3 boys be formed using a group of 5 boys and 6 girls?
(A) 69
(B) 64
(C) 65
(D) 66
(E) None of these
Rationale: There are 2 options: the selected groups have no girls or one girls.
-First case: 1 girl + 3 boys: 6C1 * 5C3 = 6*10 = 60
-Second case: 4 boys: 5C4 = 5
So there are 60 + 5 = 69 ways in total.
A
Answer: (A) 69
Question: A man can row with a speed of 10 kmph in still water. If the stream flows at 25 kmph, then the speed in downstream is?
(A) 10 kmph
(B) 35 kmph
(C) 20 kmph
(D) 22 kmph
(E) 25 kmph
Rationale: M = 10
S = 25
DS = 10 + 25 = 35
ANSWER:B
Answer: (B) 35 kmph
Question: A total of 3000 chocolates were distributed among 120 boys and girls such that each boy received 2 chocolates and each girl received 3 chocolates. Find the respective number of boys and girls?
(A) 32
(B) 60
(C) 27
(D) 29
(E) 21
Rationale: Let the number of boys be x.
Number of girls is 120 - x.
Total number of chocolates received by boys and girls = 2x + 3(120 - x) = 300
=> 360 - x = 300 => x = 60.
So, the number of boys or girls is 60.
Answer: B
Answer: (B) 60
Question: The time it took car P to travel 750 miles was 2 hours less than the time it took car R to travel the same distance. If car P’s average speed was 10 miles per hour greater than that of car R, what was car R’s average speed, in miles per hour?
(A) 40
(B) 56
(C) 60
(D) 70
(E) 80
Rationale: Let speed of car R be=x
Then speed of car P= x+10
A/Q,
(750/x)-(750/(x+10))=2
Solving for x=56 miles\hr.
B
Answer: (B) 56
Question: How many 2's are there between the terms 112 to 375?
(A) 156
(B) 157
(C) 158
(D) 159
(E) 160
Rationale: Let us calculate total 2's in the units place. (122, 132, 142 ... 192), (201, 212, 222, ... 292), (302, 312, ... 372) = 8 + 10 + 8 = 26
Total 2's in tenth's place, (120, 121, 122, ..., 129) + (220, 221, ..., 229) + (320, 321, ..., 329) = 30
Total 2's in hundred's place = (200, 201, ... 299) = 100.
Total 2's between 112 and 375 = 26 + 30 + 100 = 156
Answer:A
Answer: (A) 156
Question: If 12 men and 16 boys can do a piece of work in 5 days; 13 men and 24 boys can do it in 4 days, then the ratio of the daily work done by a man to that of a boy is?
(A) 2:1
(B) 2:98
(C) 2:2
(D) 2:9
(E) 2:4
Rationale: Let 1 man's 1 day work = x and 1 boy's 1 day work = y.
Then, 12x + 16y = 1/5 and 13x + 24y = 1/4
Solving these two equations, we get:
x = 1/100 and y = 1/200
Required ratio = x:y = 1/100 : 1/200 = 2:1.
Answer: A
Answer: (A) 2:1
Question: Plane leaves from Chelsea every 50 minutes from West Ham airport. Air hostess told Frank Lampard that the plane had left 25 minutes ago and the next plane for Chelsea will leave at 7:00 p.m.
At what time air hostess convey this information to Frank Lampard?
(A) 5:35 p.m
(B) 6:35 p.m
(C) 7:35 p.m
(D) 8:35 p.m
(E) 9:35 p.m
Rationale: Solution:
6:35 pm
Explanation:
Time of Frank Lampard last missed plane = 7:00 p.m - 50 min = 6:10 p.m
As this information has been given to Frank Lampard 15 minutes after the plane has already left, the air hostess gave the information at 6:10 p.m + 25 min = 6:35 p.m
Answer B
Answer: (B) 6:35 p.m
Question: Circular gears P and Q start rotating at the same time at constant speeds. Gear P makes 10 revolutions per minute and Gear Q makes 40 revolutions per minute. How many seconds after the gears start rotating will gear Q have made exactly 15 more revolutions than gear P?
(A) 12
(B) 15
(C) 20
(D) 24
(E) 30
Rationale: Gear Q makes 30 more revolutions each 60 seconds.
The time to make 15 = 30/2 more revolutions is 60/2 = 30 seconds.
The answer is E.
Answer: (E) 30
Question: A certain population of bacteria doubles every 10 minutes. If the number of bacteria in the population initially was 10^5, then what was the number in the population 1 hour later?
(A) 2(10^5)
(B) 6(10^5)
(C) (2^6)(10^5)
(D) (10^6)(10^5)
(E) (10^5)^6
Rationale: x bacteria would become 2x in 10 mins, which would become 4x in the next 10 (total 20 mins)...
So in 60 mins x becomes 2^6 times into x
Therefore Answer is C, (2^6)(10^5)
Answer: (C) (2^6)(10^5)
Question: How many paying stones, each measuring 2 1/2 m * 2 m are required to pave a rectangular court yard 30 m long and 16 1/2 m board?
(A) 99
(B) 88
(C) 77
(D) 65
(E) 44
Rationale: 30 * 33/2 = 5/2 * 2 * x => x = 99
Answer: A
Answer: (A) 99
Question: Anil spends 40% of his income on rent, 30% of the remaining on medicines and 20% of the remaining on education. If he saves Rs. 840 every month, then find his monthly salary?
(A) Rs. 1800
(B) Rs. 2000
(C) Rs. 2200
(D) Rs. 2500
(E) None of these
Rationale: Let's Anil's salary be Rs. 100.
Money spent on Rent = 40% of 100 = Rs. 40.
Money spent on medical grounds = 30% of (100 - 40) = 3/10 * 60 = Rs. 18.
Money spent on education = 20% of (60 - 18) = 1/5 * 42 = Rs. 8.40
Anil saves 100 - (40 + 18 + 8.40) i.e., Rs. 33.60
for 33.6 ---> 100 ; 840 ---> ?
Required salary = 840/33.6 * 100 = Rs. 2500
ANSWER:D
Answer: (D) Rs. 2500
Question: The banker's discount on Rs.1800 at 4% per annum is equal to the true discount on Rs.1872 for the same time at the same rate. Find the time?
(A) 3 months
(B) 4 months
(C) 5 months
(D) 6 months
(E) 12 months
Rationale: Solution
S.I on Rs.1800 = T.D on Rs.1872.
P.W on Rs.1872 is Rs.1800.
Rs.72 is S.I on Rs. 1800 at 4%.
Time	=(100x72 / 4x1800)
= 1 year
= 12 months.
Answer E
Answer: (E) 12 months
Question: The ratio of men to women in a class is 4 to 3. If 4 men drop out of the class and the ratio of men to women after the drop out is 10 to 9, then how many women are currently in the class?
(A) 14
(B) 16
(C) 18
(D) 20
(E) 22
Rationale: Equation on the basis of question
4x-4/3x=10/9
x=6
Original number of men = 4*6=24
Original Number of Women = 3*6=18 (Answer)
Current number of men = 24-4=20
Current number of women = 18
Ratio = 20/18= 10/9 ( Just to verify)
Ans C
Answer: (C) 18
Question: A can do a piece of work in 10 hours; B and C together can do it in 9 hours, while A and C together can do it in 18 hours. How long will B alone take to do it?
(A) 5 5/4 hours
(B) 6 5/8 hours
(C) 15 8/6 hours
(D) 6 8/9 hours
(E) 6 6/14 hours
Rationale: A's 1 hour's work = 1/10 ;
(B + C)'s 1 hour's work = 1/9 ;
(B + C)'s 1 hour's work = 1/18
(A + B + C)'s 1 hour's work = (1/10 + 1/9) = 19/90
B's 1 hour's work = (19/90 - 1/18) = 14/90
B alone will take 90/14 hours to do the work.
=6 6/14
Answer = E
Answer: (E) 6 6/14 hours
Question: Working together, Tim and Tom can type 12 pages in one hour. If they would be able to type 14 pages in one hour if Tom increases his typing speed by 25%, what is the ratio of Tom's normal typing speed to that of Tim?
(A) 2/1
(B) 1/2
(C) 1/3
(D) 4/1
(E) 1/4
Rationale: Lets say Tim types x pages an hour and Tom types y pages an hour.
We know that x+y=12
Tom increase speed by 25% means he will type 1.25y pages an hour.
So we get x+1.25y=14
We need to know the ratio of Tom's speed to Tim's speed. This is going to be proportional to the number of pages each can type in an hour, hence (y/x).
Subtracting both : 0.25y=2 so y=8 ... so x=4
(y/x)=8/4 =2/1
Answer is (A)
Answer: (A) 2/1
Question: A person got Rs.52 more when he invested a certain sum at compound interest instead of simple interest for two years at 7% p.a. Find the sum?
(A) 10612
(B) 11234
(C) 10678
(D) 2667
(E) 2671
Rationale: P = (d * 1002) / R2
=> (52 * 100 * 100) / 7 * 7 = Rs.10612
Answer:A
Answer: (A) 10612
Question: Find the average of first 4 multiples of 11?
(A) 10
(B) 12.6
(C) 27.5
(D) 31.3
(E) 40.8
Rationale: average = (11+22+33+44)/4 = 27.5
Answer is C
Answer: (C) 27.5
Question: Of the science books in a certain supply room, 30 are on botany, 37 are on zoology, 90 are on physics, 50 are on geology, and 110 are on chemistry. If science books are removed randomly from the supply room, how many must be removed to ensure that 80 of the books removed are on the same area science?
(A) 81
(B) 121
(C) 164
(D) 225
(E) 276
Rationale: The worst case scenario is to remove all the books on botany, zoology, and geology.
Also we can remove 79 books each from physics and chemistry.
The total is 30 + 37 + 50 + 79 + 79 = 275.
The next book removed must either be physics or chemistry, ensuring 80 books in that subject.
The answer is E.
Answer: (E) 276
Question: A box contains 4 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If two marbles are picked at random, what is the probability that they are either blue or red?
(A) 8/10
(B) 5/10
(C) 9/10
(D) 6/10
(E) 7/10
Rationale: Given that there are 4 blue marbles, 3 red marbles, 6 green marbles and 2 yellow marbles. Probability that both marbles are blue = 4Câ‚‚/Â¹âµCâ‚‚ = (3 * 2)/(15 * 14) = 2/35
Probability that both are red = 3Câ‚‚/Â¹âµCâ‚‚ = (1)/(15 * 14) = 1/35
Probability that one blue and other is red = (4Câ‚ * 3Câ‚)/Â¹âµCâ‚‚ = (4*2)/(15 * 14) = 4/35
Required probability = 2/35 + 1/35+ 4/35
= 7/10
Answer:E
Answer: (E) 7/10
Question: From a square piece of a paper having each side equal to 10 cm, the largest possible circle is being cut out. The ratio of the area of the circle to the area of the original square is nearly :
(A) 4⁄5
(B) 3⁄5
(C) 5⁄6
(D) 6⁄7
(E) None of these
Rationale: Area of the square = (10)2 = 100 cm2
Area of the circle = 22⁄7 × (5)2 = 22×25/7
Required ratio = 22×25/7×100=22/28=11/14
= 0.785 ≈ 0.8 = 4⁄5
Answer A
Answer: (A) 4⁄5
Question: A shop owner sells 50mtr of cloth and gains SP of 10 mtrs. Find the gain %?
(A) 25%
(B) 40%
(C) 50%
(D) 60%
(E) 70%
Rationale: Here, selling price of 10 m cloth is obtained as profit.
Profit of 10 m cloth = (S.P. of 50 m cloth) – (C.P. of 50 m cloth)
Selling price of 40 m cloth = Selling Price of 50 m of cloth
Let cost of each metre be Rs. 100.
Therefore, cost price of 40 m cloth = Rs. 4000 and S.P. of 40 m cloth = Rs. Rs. 5000
Profit% = 10/40 × 100 = 25%
Profit of 25% was made by the merchant.
A
Answer: (A) 25%
Question: If |m+5|−5 = m, which of the following must be true?
(A) m = 0
(B) m = −10
(C) m is a multiple of 5.
(D) m ≥ −5
(E) |m+5| > 0
Rationale: Case 1 : |m+5| = 0, then m= -5
|-5+5|−5 = -5 --> true
Case 2; |m+5| < 0, then m< -5
|m+5|−5 = -m - 5 -5 = m , then m = -5 --> false
Case 3: Case 2; |m+5| > 0, then m> -5
|m+5|−5 = m+5 -5 = m , then m = m --> always true
Only case 1 and 3 satisfy the equation, so D is the answer.
Answer: (D) m ≥ −5
Question: Each digit 1 through 5 is used exactly once to create a 5-digit integer. If the 1 and the 5 cannot be adjacent digits in the integer, how many 5-digit integers are possible?
(A)  48
(B)  66
(C)  76
(D)  78
(E)  None of these
Rationale: Number of arrangements using 5 distinct digits = 5!
Number of arrangements in which 1 and 5 are adjacent - consider 1 and 5 together as one group. Now you have 4 numbers/groups to arrange which can be done in 4! ways. In each of these arrangements, 1 and 5 can be arranged as 15 or 51.
Number of arrangements in which 3 and 4 are not adjacent = 5! - 2*4! = 72
Answer : E
Answer: (E)  None of these
Question: If 0 < m < 1 < n< 7, which of the following must be less than 1? I.m/n II. mn III. n - m
(A) II,III only
(B) I,III only
(C) I only
(D) III only
(E) II only
Rationale: n > 1. So, option II, III is not possible.
So, Option I is correct.
Answer : C
Answer: (C) I only