mathewhe/aqua_rat huggingface dataset:
Question: The sum of the 1st and 3rd of 3consecutive odd integers is 131 less than three times the second integer. Find the three integers?
(A) 121
(B) 133
(C) 145
(D) 150
(E) 165
Rationale: Let x, x + 2 and x + 4 be three integers. The sum of the first x and third x + 4 is given by
x + (x + 4)
131 less than three times the second 3(x + 2) is given by
3(x + 2) - 131
"The sum of the first and third is 131 less than three times the second" gives
x + (x + 4) = 3(x + 2) - 131
Solve for x and find all three numbers
x = 129 , x + 2 = 131 , x + 4 = 133
As an exercise, check that the sum of the first and third is 131 less than three times
B
Answer: (B) 133
Question: Some of the people in Town F are left-handed, some are tall, some are both, and some are neither. In Town Y, three times as many people are left-handed > as are left-handed in Town F, three times as many people are tall as are tall in Town F, three times as many people are both as are both in Town F, but no one is neither. If the total number of people in Town F is four times greater than the total number of people in Town Y, which of the following could be the number of people in Town F who are neither left-handed nor tall?
(A)  23
(B)  39
(C)  72
(D)  143
(E)  199
Rationale: Yes, correct answer is indeed D.
{F}={Left} + {Tall} - {Both} + {Neither};
{Y} = 3*{Left} + 3*{Tall} - 3*{Both};
Since the total number of people in Town F is four times greater than the total number of people in Town Y, then:
{Left} + {Tall} - {Both} + {Neither}=4*(3*{Left} + 3*{Tall} - 3*{Both});
{Neither}=11*({Left} + {Tall} - {Both}), which means that # of people in Town F who are neither left-handed nor tall must be a multiple of 11.
Only answer choice D, is a multiple of 11: 143=11*13.
Answer: D.
Answer: (D)  143
Question: What are the number of ways to select 3 men and 2 women such that one man and one woman are always selected?
(A) 12
(B) 10
(C) 30
(D) 99
(E) 123
Rationale: The number of ways to select three men and two women such that one man and one woman are always selected = Number of ways selecting two men and one woman from men and five women
= ⁴C₂ * ⁵C₁ = (4 * 3)/(2 * 1) * 5
= 30 ways.
Answer:C
Answer: (C) 30
Question: If n is the product of three consecutive integers, and n is divisible by 4, which of the following is not necessarily a divisor of n ?
(A) 26
(B) 24
(C) 28
(D) 30
(E) 32
Rationale: IF ONE NO. IS 4 THEN OTHER CONSEQUTIVE NO. ARE 2*3*4 =24 , 3*4*5=60 , 4*5*6 =120
24 IS NOT A DIVISOR OF N
B
Answer: (B) 24
Question: The output of a factory is increased by 10% to keep up with rising demand. To handle the holiday rush, this new output is increased by 10%. By approximately what percent would the output of the factory now have to be decreased in order to restore the original output?
(A) 17%
(B) 24%
(C) 30%
(D) 32%
(E) 79%
Rationale: take it as original output =100.
to meet demand increase by 10%, then output=110.
to meet holiday demand , new output increase by 10% then output equals 121
To restore new holidy demand output to original 100.
final -initial/final*100
=21/121*100=8/33*100=17% approxiamately.
option A is correct.
Answer: (A) 17%
Question: If 34 men completed 2/5th of a work in 8 days working 9 hours a day. How many more man should be engaged to finish the rest of the work in 6 days working 9 hours a day?
(A) 89
(B) 98
(C) 102
(D) 142
(E) 152
Rationale: From the above formula i.e (m1*t1/w1) = (m2*t2/w2)
so, (34*8*9/(2/5)) = (x*6*9/(3/5))
so, x = 136 men
number of men to be added to finish the work = 136-34 = 102 men
Answer : C.
Answer: (C) 102
Question: If Z is a negative integer, which of the following must be a negative integer?
(A) Z + 1
(B) Z/2+3
(C) Z^4
(D) Z – y
(E) Z^11
Rationale: We need to check each option here:
A. Z + 1
This will depend on the value of Z. Nothing can be said
B. Z/2+3
This will depend on the value of Z. Nothing can be said
C. Z^4
This will always be positive. Even powers of negative numbers are positive and odd powers are negative
D. Z – y
This will depend on the value of Z. Nothing can be said
E. Z^11
This will always be negative. Even powers of negative numbers are positive and odd powers are negative
Correct Option: E
Answer: (E) Z^11
Question: A circular box kept around a circular place. If the difference between the circumference of the outer circle and the inner circle is 44m, find the width of the box?
(A) 4 m
(B) 5 m
(C) 6 m
(D) 7 m
(E) 8 m
Rationale: Let radius of the outer circle be R and radius of the inner circle be r.
Circumference of the outer circle =
Circumference of the inner circle =
But,
or, R - r =
Thus, width of the road = 7 m
D
Answer: (D) 7 m
Question: If Jan 1, 2006 was a Tuesday, What was the day of the week Jan 1, 2010?
(A) Friday
(B) Thursday
(C) Tuesday
(D) Saturday
(E) Sunday
Rationale: Explanation :
Given that Jan 1 2006 was a Tuesday
Number of odd days in the period 2006-2009
= 3 normal years + 1 leap year
= 3 x 1 + 1 x 2 = 5 (note that we have taken the complete year 2006 because the period in 2006 is from 01-Jan-2006 to 31-Dec-2006, which is the whole year 2006. Then the complete years 2007, 2008 and 2009 are also involved)
Hence Jan 1 2010 = (Tuesday + 5 odd days) = Sunday.. Answer : Option E
Answer: (E) Sunday
Question: Three persons invested Rs.9000 in a joint business. The second person invested Rs.1000 more than the first and the third Rs.1000 more than second. After two years, they gained Rs.5400. How much third person will get?
(A) 2400
(B) 2878
(C) 2699
(D) 2662
(E) 1821
Rationale: First persons investment = x
Second persons investment = x + 1000
Third persons investments = x + 2000
x + x + 1000 + x + 2000 = 9000
3x = 6000
x = 2000
Ratio = 2000 : 3000 : 4000
2:3:4
4/9 * 54000 = 2400
Answer: A
Answer: (A) 2400
Question: What is the sum of odd integers from 35 to 85, inclusive?
(A) 1,560
(B) 1,500
(C) 1,240
(D) 1,120
(E) 1,100
Rationale: Number of odd integers = (85-35)/2 + 1
= 50/2 + 1
= 26
Sum of odd integers = (35+85)/2 * 26
= 60 * 26
= 1560
Answer A
Answer: (A) 1,560
Question: After successive discounts of 20%, 10% and 5% a certain good is sold for Rs. 6800. Find the actual price of the good.
(A) s. 6000
(B) s. 9000
(C) s. 10800
(D) s. 9942
(E) s. 9980
Rationale: Let actual price was 100.
After three successive discount this will become,
100 ==20% discount => 80 ==10% discount => 72 ==5% discount = 68.4
Now Compare,
68.4 = 6800
1 = 6800/68.4
100 = (6800*100)/68.4 = Rs. 9942.
Answer: Option D
Answer: (D) s. 9942
Question: From his monthly salary a man spends 10% of his salary for house rent, 20% on food, 25% for all other expenses and saves Rs 2520 per month. His monthly salary is?
(A) Rs 4000
(B) Rs 4200
(C) Rs 5600
(D) Rs 5880
(E) Rs 5900
Rationale: Explanation: Total salary = 100%
His spending are on house rent 10%, on food 20% and for other expenses 25%.
Total expenditure = 10 % + 20% + 25% = 55%
His savings are => Income – Expenses
=> 100% - 55% => 45% = Rs 2520
Hi salary is 1005 => 2520 45
x 100
= Rs 5600
C
Answer: (C) Rs 5600
Question: By selling 9 pencils for a rupee a man loses 20%. How many for a rupee should he sell in order to gain 20%?
(A) 8
(B) 9
(C) 7
(D) 4
(E) 6
Rationale: 80% --- 9
120% --- ?
80/120 * 9 = 6
Answer: E
Answer: (E) 6
Question: Let A = {3, 5, 7}, B = {2, 3, 4, 6}.Find A ∩ B
(A) {12}
(B) {5}
(C) {1}
(D) {3}
(E) {7}
Rationale: A = {3, 5, 7}
B = {2, 3, 4, 6}
correct answer :A ∩ B = {3}
D
Answer: (D) {3}
Question: If 3 women can color 180 m long cloth in 2 days, then 5 women can color 300 m long cloth in?
(A) 2 days
(B) 3 days
(C) 5 days
(D) 1 day
(E) 4 days
Rationale: The length of cloth painted by one woman in one day = 180/ 3× 2 = 30 m
No. of days required to paint 300 m cloth by 5 women = 300/ 5 × 30 = 2 days
Answer : A
Answer: (A) 2 days
Question: At what rate of interest is an amount doubled in two years, when compounded annually?
(A) 41.8%
(B) 41.4%
(C) 81.4%
(D) 48.4%
(E) 11.4%
Rationale: Let the rate of interest be r. The amount gets doubled in two years => P(1 + r/100)2 = 2p => 1 + r/100 = √2 => r/100 = √2 - 1 = 0.414 => r = 100(0.414) = 41.4%.Answer:B
Answer: (B) 41.4%
Question: The compound interest earned by Sunil on a certain amount at the end of two years at the rate of 8% p.a. was Rs.2828.80. Find the total amount that Sunil got back at the end of two years in the form of principal plus interest earned?
(A) 19828.88
(B) 19828.8
(C) 19828.87
(D) 19828.84
(E) 19828.81
Rationale: Let the sum be Rs.P
P{ [ 1 + 8/100]2 - 1 } = 2828.80
P(8/100)(2 + 8/100) = 2828.80 [a2 - b2 = (a - b) ( a + b)] P = 2828.80 / (0.08)(2.08) = 1360/0.08 = 17000 Principal + Interest = Rs.19828.80
Answer:B
Answer: (B) 19828.8
Question: A crate of mangoes contains one bruised mango for every thirty mango in the crate. If three out of every four bruised mango are considerably unsaleble and there are 12 unsaleable mangoes in the crate then how msny mango are there in the crate?
(A) 100 mangoes
(B) 110 mangoes
(C) 120 mangoes
(D) 130 mangoes
(E) 125 mangoes
Rationale: 120 mangoes
ANSWER:C
Answer: (C) 120 mangoes
Question: 63 small identical cubes are used to form a large cube. How many more cubes are needed to add one top layer of small cube all over the surface of the large cube ?
(A) 64
(B) 128
(C) 152
(D) 216
(E) 153
Rationale: 63 small cube will make a large cube with 4 cubes in each line i.e.
Adding one layer will require one cube at each end and hence new cube will have 6 cubes in each line.
Total number of small cubes in new cube = 6^3 = 216
Extra cube required = 216 - 63 = 153
Hence, E is the answer.
Answer: (E) 153
Question: How many positive integers less than 50 have a reminder 01 when divided by 5?
(A) 10
(B) 14
(C) 15
(D) 16
(E) 17
Rationale: 1 also gives the remainder of 1 when divided by 5. So, there are total of 10 numbers.
Answer: A.
Answer: (A) 10
Question: A flagstaff 17.5 m high casts a shadow of length 40.25 m. The height of the building, which casts a shadow of length 28.75 m under similar conditions will be?
(A) 10 m
(B) 12.5 m
(C) 12m
(D) 13 m
(E) 14 m
Rationale: Let the height of the building x metres.
Less lengthy shadow, Less in the height (Direct Proportion)
40.25 : 28.75 :: 17.5 : x 40.25 x x = 28.75 x 17.5
x =	28.75 x 17.5/40.25
x = 12.5
Answer:B
Answer: (B) 12.5 m
Question: Calculate the circumference of a circular field whose radius is 8 centimeters.
(A) 17 π cm
(B) 18 π cm
(C) 1 π cm
(D) 16 π cm
(E) 6 π cm
Rationale: Circumference C is given by
C = 2πr = 2π*8 = 16 π cm
Answer: D
Answer: (D) 16 π cm
Question: If C and D are non-zero integers, and C/D > 1, then which of the following must be true?
(A) CD > D
(B) 2C > D
(C) C^2< D^2
(D) C > D
(E) C^3 < D^3
Rationale: MULTIPLICATION OF TWO INTEGER IS OBIVIOUSLY GREATER THAN EACH ONE OF THEM.
A
Answer: (A) CD > D
Question: Let the function Q(n) represent the product of the first n prime numbers, where n > 0. If x = Q(n) + 1, which of the following must be true?
(i) x is always odd
(ii) x is always prime
(iii) x is never the square of an integer
(A) ii only
(B) iii only
(C) i and ii only
(D) i and iii only
(E) ii and iii only
Rationale: Q(n) is always even, because the first prime is 2 and no matter what n is, 2 always will be a divisor of Q(n). Thus, Q(n) + 1 = even + 1 = odd. So, (i) is always true.
Now, use logic:
If (ii) is true (so if x is always prime), then (iii) must automatically be true: no prime is the square of an integer. So, the correct answer must be i only; i, ii, and iii only; or i and iii only. since onlyi and iii onlyis among the options, then it must be true.
Or, since (i) is always true, then from options the answer must be either C or D. C cannot be correct because if (ii) is true, then so must be (iii). Thus only D remains.
Answer: D.
Answer: (D) i and iii only
Question: At Joes Steakhouse the hourly wage for a chef is 20% greater than that of a dishwasher, and the hourly wage of a dishwasher is half as much as the hourly wage of a manager. If a managers wage is $7.50 per hour, how much less than a manager does a chef earn each hour?
(A) $1.40
(B) $3
(C) $3.40
(D) $4.40
(E) $5.40
Rationale: Manager wages per hour = $7.50
Dishwasher wages per hour = half of manager's wages. = 1/2($7.50) ==> $3.75
Chef wages per hour = 20% greater than Dishwasher wages
--> 20% of $3.75 = (20*($3.75)) /100
--> ($75)/100
--> $0.75
Therefore,
Chef wages per hour = $3.75 + $0.75 ==> $4.5
Difference of wages between manager and chef = $7.50 - $4.5 ==> $3
ANSWER:B
Answer: (B) $3
Question: Which is greatest 33 1/3%, 4/15 and 0.35 ?
(A) 33 1/3%
(B) 4/15
(C) 0.35
(D) Cannot be compared
(E) None of these
Rationale: Answer
331/3% = (100/3) /100 = 1/3 =0.33;
4/15 = 0.26
Clearly 0.35>0.33>0.26
∴ 0.35 is greatest
Correct Option: C
Answer: (C) 0.35
Question: A man was engaged on a job for 30 days on the condition that he would get a wage of Rs. 10 for the day he works, but he have to pay a fine of Rs. 2 for each day of his absence. If he gets Rs. 216 at the end, he was absent for work for ... days.
(A) 7 days
(B) 5 days
(C) 3 days
(D) 4 days
(E) 8 days
Rationale: The equation portraying the given problem is:
10 * x – 2 * (30 – x) = 216 where x is the number of working days.
Solving this we get x = 23
Number of days he was absent was 7 (30-23) days.
Answer:A
Answer: (A) 7 days
Question: What is the tenth digit to the right of the decimal point, in the decimal expansion of (1/5)^10
(A) 0
(B) 2
(C) 4
(D) 6
(E) 8
Rationale: Actually 1/5 = 0.2 = 2*10^-1
so 2^10 * 10^-10
2 has sequence 2,4,8,6.
Hence digit will be 4
Answer is C
Answer: (C) 4
Question: In a school 50% of the students are younger than 10, 1/20 are 10 years old and 1/10 are older than 10 but younger than 12, the remaining 70 students are 12 years or older. How many students are 10 years old?
(A) 5
(B) 6
(C) 9
(D) 8
(E) 10
Rationale: Let us write the fraction for each group of students
Group A: younger than 10: 50% = 50/100 = 1/2
Group B: 10 years old: 1/20
Group C: older that 10 but younger than 12: 1/10
Group D: 12 years or older: 70 students
The fraction for group A, B and C together is given by
1 / 2 + 1 / 20 + 1 / 10 = 10 / 20 + 1 / 20 + 2 / 20 , common denominator
= 13 / 20 , add numerators
The fraction for group D is given by
20 / 20 - 13 / 20 = 7 / 20 and corresponds to 70 students
If X is the total number of students, then
7 / 20 of X = 70
or (7 / 20) ? X = 70
Solve for X
X = 70 ? (20 / 7) = 200
Students who are 10 years old have a fraction of 1 / 20 of the total X and their number is equal to
(1 / 20) of 200 = 10 students
correct answer E
Answer: (E) 10
Question: A is 1.5 times as fast as B. A alone can do the work in 30 days. If A and B working together in how many days will the work be completed?
(A) 23
(B) 18
(C) 21
(D) 24
(E) 25
Rationale: A can finish 1 work in 30 days
B can finish 1/1.5 work in 30 days - since A is 1.5 faster than B
This means
B can finish 1 work in 30*1.5 days = 45 days
Now using the awesome GMAT formula when two machines work together they can finish the job in = AB/ (A+B) = 45*30/(45+30) = 20*30/50 = 18 days
So answer is B
Answer: (B) 18
Question: Which of the following numbers is between 1⁄2 and 2⁄5?
(A) .45
(B) .35
(C) .29
(D) .22
(E) .20
Rationale: 1/2 = .5
2/5 = .4
The only number between these two is 0.45.
The answer is A.
Answer: (A) .45
Question: When a number is divided by 6 &then multiply by 12 the answer is 10 what is the no.?
(A) 4.5
(B) 5
(C) 5.5
(D) 5.8
(E) 6
Rationale: if $x$ is the number, x/6 * 12 = 10
=> 2x = 10
=> x = 5
B
Answer: (B) 5
Question: If x and y are different integers and x^2 = xy, which of the following must be true ?
I. x = 2
II. y = 0
III. x = y
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) I, II, and III
Rationale: Answer choice A is correct.
The question gives you x^2=xy
you can simplify that as x^2-xy=0
x(x-y)=0 the solution will be either x=0 or x=y, since x and y are different integers x cannot equal y, that leaves you x=0 to be the only answer.
Answer C
Answer: (C) III only
Question: A wildlife preserve is being planned for 7,000 rhinoceroses. The preserve is to contain a total of 10,000 acres of watering area, plus 100 acres of grazing area for each rhinoceros. If the number of rhinoceroses is expected to increase by 10 percent, how many thousand acres should the preserve have in order to provide for the increased population?
(A)  340
(B)  330
(C)  320
(D)  780
(E)  300
Rationale: Number of rhinos = 7000
Watering area = 10,000 acres
Number of rhino to increase by 10 percent ,then number of rhino = 7700
Grazing area for a rhino = 100
Total grazing area for 7700 rhinos= 7700 * 100 = 7,70,000
Total area required for the wildlife preserve = 7,70,000 + 10,000 = 7,80,000 = 780 * 1000
Answer D
Answer: (D)  780
Question: Pipe P can fill a tank in 6 hours, pipe Q in 12 hours and pipe R in 24 hours. If all the pipes are open, in how many hours will the tank be filled?
(A) 2 hours
(B) 2.4 hours
(C) 3 hours
(D) 3.4 hours
(E) 4.5 hours
Rationale: Explanation:
Part filled by (P + Q + R) in 1 hour = (1/6 + 1/12 + 1/24) = 7/24
All the three pipes together will fill the tank = 24/7 = 3.4 hours
ANSWER D
Answer: (D) 3.4 hours
Question: In a shower, 7 cm of rain falls. The volume of water that falls on 1.5 hectares of ground is:
(A) 75 cu. m
(B) 1050 cu. m
(C) 7500 cu. m
(D) 75000 cu. m
(E) 7.50 cu. m
Rationale: 1 hectare = 10000 sq mtr
The volume of water that falls on 1.5 hectares of ground = 0.07*1.5*10000 = 1050 cub mtr .
ANSWER:B
Answer: (B) 1050 cu. m
Question: What will come in place of the x in the following Number series? 5,28,57,88,125,166,209,256
(A) 309
(B) 159
(C) 166
(D) 170
(E) 180
Rationale: 28=23+5
57=29+28
88=31+57
125=37+88
166=41+125
209=43+166
256=47+ 209
309 =53+ 256
=309
A
Answer: (A) 309
Question: The averge score of a cricketer for ten matches is 64 runs. If the average for the first six matches is 78. Then find the average for the last four matches?
(A) 48
(B) 49
(C) 40
(D) 43
(E) 36
Rationale: sum of last 4 matches = (10 × 64) – (6 × 78)= 172
Average =172/4
= 43
ANSWER:D
Answer: (D) 43
Question: The sum of three numbers is 98. The ratio of the first to the second is 2/3, and the ratio of the second to the third is 5/8. Find the second number.
(A) 10
(B) 20
(C) 30
(D) 40
(E) 50
Rationale: Let the three numbers be x, y and z.
Sum of the numbers is 98.
x + y + z = 98………………(i)
The ratio of the first to the second is 2/3.
x/y = 2/3.
x = 2/3 × y.
x = 2y/3.
The ratio of the second to the third is 5/8.
y/z = 5/8.
z/y = 8/5.
z = 8/5 × y.
z = 8y/5.
Put the value of x = 2y/3 and z = 8y/5 in (i).
2y/3 + y + 8y/5 = 98
49y/15 = 98.
49y = 98 × 15.
49y = 1470.
y = 1470/49.
y = 30 .
Therefore, the second number is 30.
Answer is C.
Answer: (C) 30
Question: The speed of a motor boat is that of the current of water as 30 : 5. The boat goes along with the current in 5 h 10 min. It will come back in
(A) 5 hour 15 minutes
(B) 6 hour
(C) 6 hour 50 minutes
(D) 12 hour 10 minutes
(E) 8 hour
Rationale: Let speed of motor boat is 30x and Current is 5X.
Speed of the boat along with current = 30X + 5X = 35X.
Given Time = 5h 10 min = 310/60 = 31/6 h.
Distance = (32/6) * 35X = 180.83X . . . . . (1)
Speed of the Boat in coming back = 30X - 5X = 25X.
Let time taken is T.
Now,
Distance = 25X *T
180.83X = 25X * T
T = 7.21 hours.
Note: If given ratio is 36:5 then answer would be 6 hours 50 minutes.
answer :C
Answer: (C) 6 hour 50 minutes
Question: The perimeter of one face a of cube is 20 cm. Its volume must be
(A) 125 cm3
(B) 400 cm3
(C) 1000 cm3
(D) 8000 cm3
(E) None
Rationale: Solution
Edge of the cube	= (20 / 4) cm
‹=›5 cm.
Volume	= (5×5×5)cm3
‹=›125 cm3.
Answer A
Answer: (A) 125 cm3
Question: During a trip on an expressway, Son drove a total of x miles. His average speed on a certain 5-mile section of the expressway was 30 miles per hour, and his average speed for the remainder of the trip was 60 miles per hour. His travel time for the x-mile trip was what percent greater than it would have been if he had traveled at a constant rate of 60 miles per hour for the entire trip?
(A) 8.5%
(B) 50%
(C) x/12%
(D) 60/x%
(E) 500/x%
Rationale: so Son drove (x-5) miles at 60mph and 5 miles at 30mph:
(x-5)/60 + (10/60) = (X+5)/60 - current time to drive the whole distance
if the entire trip @60mph than time = x/60
so to get percent [(X+5)/60 ]/[x/60] x100
=(x+5)/x *100
=100+500/x
so increase of 500/x%[/u]=E
Answer: (E) 500/x%
Question: Before being simplified, the instructions for computing income tax in country R were to add 6 percent of one's annual income to the average(arithmetic mean) of 100 units of Country R's currency and 1 percent of one's annual income. Which of the following represents the simplified formula for computing the income tax, in country R's currency, for a person in that country whose annual income is I?
(A) 50 + 13I/200
(B) 50 + 3I/100
(C) 50+ I/40
(D) 100+ I/50
(E) 100+ 3I/100
Rationale: Income of a person is I
6 percent of ones annual income = 6I/100
100 units of Country R's currency = 100
1 percent of one's annual income = I/100
Average of 100 units and 1 percent annual income = 50 + I/200
Sum of average and 6 percent = 6I/100 + 50 + I/200
= 50 + 13I/200
So answer is A
Answer: (A) 50 + 13I/200
Question: George's drawer has 10 loose black socks, 12 loose blue socks, and 8 loose white socks. If George takes socks out of the drawer at random, how many would he need to take out to be sure that he had at least two matching pairs of socks in two different colors?
(A) 11
(B) 13
(C) 15
(D) 17
(E) 19
Rationale: The worst case scenario is choosing 12 blue socks. Then he would need 3 more socks to be sure of a matching pair of black socks or white socks.
The answer is C.
Answer: (C) 15
Question: The avgof marks obtained by 120candidates was 35. If the avg of marks of passed candidates was 39 & that of failed candidates was 39 and that of failed candidates was 15, the no.of candidateswho passed the examination is?
(A) 70
(B) 80
(C) 100
(D) 120
(E) 140
Rationale: Let the number of candidate who passed = y
Then, 39y + 15(120 - y) =120 x 35
⇒ 24y= 4200 -1800
∴ y= 2400 / 24 = 100
C
Answer: (C) 100
Question: Two assembly line inspectors, Lauren and Steven, inspect widgets as they come off the assembly line. If Lauren inspects every fifth widget, starting with the fifth, and Steven inspects every third, starting with the third, how many of the 98 widgets produced in the first hour of operation are not inspected by either inspector?
(A) 91
(B) 59
(C) 53
(D) 47
(E) 45
Rationale: Widgets inspected by Lauren: ((95-5)/5)+1=18+1=19
Widgets inspected by Steven: ((96-3)/3)+1 =31+1 =32
Widgets inspected by both: ((90-15)+1 =6
Total : 19+32 -6=45
hence, widgets not inspected: 98-45=53
ANSWER:C
Answer: (C) 53
Question: A deer is standing 50 meters in from the west end of a tunnel. The deer sees a train approaching from the west at a constant speed ten times the speed the deer can run. The deer reacts by running toward the train and clears the exit when the train is 35 meters from the tunnel. If the deer ran in the opposite direction it would barely escape out the eastern entrance just as the train came out of the eastern entrance. How long is the tunnel in meters?
(A) 91
(B) 97
(C) 103
(D) 109
(E) 115
Rationale: Let x be the length of the tunnel.
When the deer runs 50 meters west, the train goes 500 meters to a point 35 meters from the west entrance of the tunnel.
When the deer runs east, the deer runs x-50 meters while the train goes x + 500 + 35 meters.
x + 500 + 35 = 10(x-50)
9x = 1035
x = 115 meters
The answer is E.
Answer: (E) 115
Question: In the biology lab of "Jefferson" High School there are 0.037*10^5 germs, equally divided among 74000*10^(-3) Petri dishes. How many germs live happily in a single dish?
(A) 10
(B) 20
(C) 30
(D) 40
(E) 50
Rationale: 0.037*10^5 can be written as 3700
74000*10^(-3) can be written as 74
Required= 3700/74= 50
ANSWER:E
Answer: (E) 50
Question: A question paper has 2 parts, A & B, each containing 6 questions. If a student has to choose 4 from part A &3 from part B, in how many ways can he choose the questions?
(A) 100
(B) 200
(C) 350
(D) 150
(E) 140
Rationale: There 6 questions in part A out of which 4 question can be chosen as = 6C4.
Similarly, 3 questions can be chosen from 6 questions of Part B as = 6C3.
Hence, total number of ways,
=6C4*6C3
=[6!/(2!4!)]*[6!/(3!*3!)]
= {15}*{6*5*4*3/(3*2*1)}
= 300.
D
Answer: (D) 150