divelab/Sys2Bench huggingface dataset:
Question: Rose is two years older than Bruce who is twice as old as Chris. If the total of the ages of Rose, B and Chris be 27, the how old is Bruce ?
Options:
A)8 years
B)10 years
C)12 years
D)13 years
E)14 years
Rationale: Let Chris's age be x years. Then, Bruce's age = 2x years.Rose's age = (2x + 2) years.
(2x + 2) + 2x + x = 27
5x = 25
x = 5.
Hence, Bruce's age = 2x = 10 years.
B
Answer: B
Question: The true discount on a bill due 9 months hence at 16% per annum is Rs. 180.The amount of the bill is
Options:
A)Rs. 1386
B)Rs. 1764
C)Rs. 1500
D)Rs. 2268
E)None of these
Rationale: Solution
32.5
Let P.W. be Rs. x.Then,S.I. on Rs.x at 16% for 9 months =Rs.180.
∴ x 16 x9/ 12x 1/100}= 180 or x = 1500.
∴ P.W. = Rs. 1500.
Answer C
Answer: C
Question: A store raised the price of an item by exactly 25 percent. Which of the following could NOT be the resulting price of the item?
Options:
A)130
B)250
C)375
D)500
E)62.5
Rationale: If it raised the value by 25% then the resulting price is 1.25 times the original price. So the final answer has to be a multiple of 1.25. Check the answers.
130 - No
250 - Yes
375 - Yes
500 - Yes
62.50 - Yes
Answer : A
Answer: A
Question: How many shots of 1cm radius can be prepared from a sphere of 7cm radius?
Options:
A)343
B)388
C)327
D)88
E)99
Rationale: 4/3 Ï€ * 7 * 7 * 7 = 4/3 Ï€ * 1 * 1 * 1 * x
x =343
Answer: A
Answer: A
Question: Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 % fescue. If a mixture of X and Y contains 30% ryegrass, what percent of the weight of the mixture is X?
Options:
A)10%
B)33 1/3%
C)40%
D)50%
E)66 2/3%
Rationale: Assuming the weight of the mixture to be 100g**, then the weight of ryegrass in the mixture would be 30g.
Also, assume the weight mixture X used in the mixture is Xg, then the weight of mixture Y used in the mixture would be (100-X)g.
So we can now equate the parts of the ryegrass in the mixture as:
0.4X + 0.25(100-X) = 30
0.4X + 25 - 0.25X = 30
0.15X = 5
X = 5/0.15 = 500/15 = 100/3
So the weight of mixture X as a percentage of the weight of the mixture
= (weight of X/weight of mixture) * 100%
= (100/3)/100 * 100%
= 33%
ANSWER:B
Answer: B
Question: In 1970 there were 8,902 women stockbrokers in the United States. By 1978 the number had increased to 19,947. Approximately what was the percent increase?
Options:
A)45%
B)125%
C)145%
D)150%
E)225%
Rationale: We can use approximation to get the answer quickly.
Like,
19,947 is approx 20000
8902 is approx 8900
So, Total increase = 20000 - 8900 = 11100
Hence, % increase = 11100 / 8900 = approx. 11/9 = 1.22 = approx. 1.25
So, Choice is B.
Answer: B
Question: The area of a square is equal to five times the area of a rectangle of dimensions 125 cm * 64 cm. What is the perimeter of the square?
Options:
A)700
B)800
C)900
D)600
E)500
Rationale: Area of the square = s * s = 5(125 * 64)=> s = 25 * 8 = 200 cm
Perimeter of the square = 4 * 200 = 800 cm.
Answer: Option B
Answer: B
Question: The expression (11.98 x 11.98 + 11.98 x x + 0.02 x 0.02) will be a perfect square for x equal to:
Options:
A)0.02
B)0.2
C)0.04
D)0.4
E)None
Rationale: EXPLANATION
Given expression = (11.98)2 + (0.02)2 + 11.98 x x.
For the given expression to be a perfect square, we must have
11.98 x x = 2 x 11.98 x 0.02 or x = 0.04
Answer C
Answer: C
Question: In digging a grave 200 m * 100 m * 3 m the volumes of the dirt to be extracted will be?
Options:
A)600 cu.m
B)12000 cu.m
C)6000 cu.m
D)60000 cu.m
E)4000 cu.m
Rationale: 200 * 100 * 3 = 60000
ANSWER:D
Answer: D
Question: Sreenivas sells a table to Shiva at 10% profit and Shiva sells it to Mahesh at 10% loss. At what price did Sreenivas purchase the table if Mahesh paid Rs. 2178?
Options:
A)2887
B)2665
C)2200
D)2665
E)2771
Rationale: Let the cost price of table for Sreenivas be Rs. x and given that, cost price of table for Mahesh = Rs. 2178.
=> (90%) of (110%) of x = Rs. 2178.
=> (90/100)(110/100)x = 2178
=> x = (2178 * 100)/(9 * 11)
=> x = Rs. 2200
Answer:C
Answer: C
Question: A company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R does a certain job in 20 hrs and a machine of type S does the same job in 60 hours. If the company used the same number of each type of machine to do the job in 3 hours, how many machines of type R were used?
Options:
A)3
B)4
C)5
D)6
E)8
Rationale: Type R completes 1/20 of the job each hour.
Type S completes 1/60 of the job each hour.
Together R and S complete 1/20 +1/60 = 1/15 of the job each hour.
Let n be the number of each type of machine.
3*n*(1/15) = 1 job completed
n = 5
The answer is C.
Answer: C
Question: The ages of X and Y are in the proportion of 6:5 and total of their ages is 44 years. The proportion of their ages after 8 years will be
Options:
A)3:6
B)6:3
C)8:7
D)9:5
E)None of these
Rationale: Explanation:
Let current ages of X and Y correspondingly, is 6A & 5A
Given: 6A + 5A = 44
=> A = 4
Proportion of ages after 0.8 decades will be
6A + 8 : 5A + 8 32:28 (or) 8:7 . Answer: C
Answer: C
Question: The sum of the present ages of A, B, C is 33 years. Three years ago their ages were in the ratio 1:2:3. What is the present age of A?
Options:
A)5
B)7
C)9
D)11
E)13
Rationale: Three years ago:
a:b:c = 1:2:3
Let a = 1x, b = 2x, and c = 3x.
Today:
(x+3) + (2x+3) + (3x+3) = 33
x = 4
So the present age of A is x+3 = 7
The answer is B.
Answer: B
Question: A shop owner has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 years. How much sale must she have in the 6th year so that she gets an average sale of Rs. 6500?
Options:
A)5000
B)5500
C)4991
D)5600
E)5700
Rationale: Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.
Required sale = Rs. [ (6500 x 6) - 34009 ]
= Rs. (39000 - 34009)
= Rs. 4991
C
Answer: C
Question: The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 7 years ago was 20 years. The present age of the husband is:
Options:
A)20 years
B)30 years
C)60 years
D)36 years
E)35 years
Rationale: D
36 years
Sum of the present ages of husband, wife and child = (27 x 3 + 3 x 3) years = 90 years.
Sum of the present ages of wife and child = (20 x 2 + 7 x 2) years = 54 years.
Husband's present age = (90 - 54) years = 36 years.
Answer: D
Question: At the end of the month, a certain ocean desalination plant’s reservoir contained 30 million gallons of water. This amount is twice the normal level. If this amount represents 75% of the reservoir’s total capacity, how many million gallons short of total capacity is the normal level?
Options:
A)10
B)15
C)20
D)25
E)30
Rationale: the Q talks of total capacity, normal level, present level, shortage etc..
so it is all about not going wrong in these terms
30 MG= 75% of Total..
total = 30/.75=40MG..
Normal level= 1/2 of 40=20MG..
shortage of normal level=40-20=20MG..
C
Answer: C
Question: An elementary school houses primarily kindergartners, first graders, second graders, and third graders. If an assembly included 30 kindergartners, 20 first graders, and 80 other children, and the ratio of first graders to third graders is tripled, what is the ratio of the first graders to the rest of the children?
Options:
A)40: 20
B)42: 21
C)50: 100
D)15: 25
E)It cannot be determined by the information given.
Rationale: While you can figure out most of this question, there is not enough information to completely tell. the answer is E.
Answer: E
Question: The number 42 can be written as the sum of the squares of 3 different positive integers. What is the sum of these 3 integers?
Options:
A)17
B)10
C)15
D)14
E)13
Rationale: I think brute force with some common sense should be used to solve this problem.
Write down all perfect squares less than 42: 1, 4, 9, 16, 25, 36.
Now, 42 should be the sum of 3 of those 8 numbers. Also to simplify a little bit trial and error, we can notice that as 42 is an odd numbers then either all three numbers must be odd (odd+odd+odd=odd) OR two must be even and one odd (even+even+odd=odd).
We can find that 42 equals to 1+16+25=1^2+4^2+5^2=42 --> 1+5+4=10.
Answer: B.
Answer: B
Question: In one hour, a boat goes 15 km/hr along the stream and 9 km/hr against the stream. The speed of the boat in still water (in km/hr) is:
Options:
A)12 kmph
B)13 kmph
C)14 kmph
D)15 kmph
E)16 kmph
Rationale: Explanation:
Let the speed downstream be a km/hr and the speed upstream be b km/hr, then
Speed in still water =1/2(a+b) km/hr
Rate of stream = 1/2(a−b) km/hr
Speed in still water = 1/2(15+9) kmph = 12 kmph.
Answer: Option A
Answer: A
Question: Nitrous oxide (laughing gas) worth Rs 126 per kg and Rs 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs 153 per kg, then the price of the third variety per kg will be
Options:
A)Rs 169.50
B)Rs 170
C)Rs 175.50
D)Rs 180
E)Rs 190
Rationale: Suppose the quantities of Nitrous oxide worth Rs 126 per kg, Rs 135 per kg and Rs.x per kg purchased are Y, Y and 2Y kg respectively:
(126 Y +135 Y + 2XY )/ 4Y = 153
=> 261 + 2X = 612
=> X= 351/2 = 175.50
Nitrous oxide of the third varity is purchased at the rate of Rs 175.50 per Kg
ANSWER:C
Answer: C
Question: Walking at 25% of his usual speed a man takes 24 minutes more to cover a distance. What is his usual time to cover this distance?
Options:
A)8
B)16
C)42
D)48
E)54
Rationale: speed is inversly proprtional to time
walking at 25% of speed meand 1/4s takes 4t.
It takes 24 minutes extra to cover the distance. then
4t=t+24
3t=24
t=8.
Option A is correct
Answer: A
Question: Population of a city in 20004 was 1000000. If in 2005 there isan increment of 15% , in 2006 there is a decrements of 35 % and in 2007 there is an increment of 45%, then find the population of city atthe end of the year 2007
Options:
A)346656
B)466577
C)854343
D)354567
E)1083875
Rationale: Required population = P (1 + R1/100)(1 - R2/100)(1 + R3/100)
= P (1 + 15/100)(1 - 35/100)(1 + 45/100)
= 1083875
E
Answer: E
Question: What is the least integer greater than –4+0.5?
Options:
A)–3
B)–1
C)0
D)1
E)2
Rationale: This question is just about doing careful Arithmetic and remembering what makes a numberbiggerorsmallercompared to another number.
First, let's take care of the Arithmetic:
(-4) + (0.5) = -3.5
On a number line, since we're ADDING +.5 to a number, the total moves to the RIGHT (so we're moving from -4 to -3.5).
Next, the question asks for the LEAST integer that is GREATER than -3.5
Again, we can use a number line. Numbers become greater as you move to the RIGHT. The first INTEGER to the right of -3.5 is -3.
Final Answer:
A
Answer: A
Question: A train passes a station platform in 28 seconds and a man standing on the platform in 10 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?
Options:
A)270
B)277
C)240
D)88
E)112
Rationale: Speed = [54 * 5/18] m/sec = 15 m/sec.
Length of the train = (15 * 10) m = 150 m.
Let the length of the platform be x meters.
Then, x + 150 / 28 = 15
x + 150 = 420
x = 270 m.
Answer:A
Answer: A
Question: If x and y are negative integers, which of the following statements must be true?
I. (-x)^y is positive.
II. (-x)^(-y) is positive.
III. x^(-y) is positive.
Options:
A)None
B)II only
C)I and II
D)I and III
E)I, II and III
Rationale: Let's x = -2 and y = -3
I. (-x)^y = (2)^(-3) = 1/(2^3) --> Positive
II. (-x)^(-y) is positive. = 2^3 --> Positive
III. x^(-y) is positive. = -2^3 --> Negative
Answer: C
Answer: C
Question: The average of first 10 odd numbers is?
Options:
A)11
B)10
C)18
D)12
E)19
Rationale: Explanation:
Sum of 10 odd no. = 100
Average = 100/10 = 10
Answer:B
Answer: B
Question: If Re.1 amounts to Rs.9 over a period of 20 years. What is the rate of simple interest?
Options:
A)27
B)29
C)26
D)40
E)12
Rationale: 8 = (1*20*R)/100
R = 40%.Answer: D
Answer: D
Question: David completes a piece of work in 5 days, John completes the same work in 9 days. If both of them work together, then the number of days required to complete the work is?
Options:
A)3.2 days
B)8 days
C)10 days
D)12 days
E)14 days
Rationale: IfA can complete a work in x days and B can complete the same work in y days, then, both
of them together can complete the work in x y/ x+ y days.
That is, the required No. of days = 5 × 9/14= 3.2 days
Answer : A
Answer: A
Question: A vessel contains 20 liters of a mixture of milk and water in the ratio 3:2. 10 liters of the mixture are removed and replaced with an equal quantity of pure milk. If the process is repeated once more, find the ratio of milk and water in the final mixture obtained?
Options:
A)4:5
B)9:1
C)2:5
D)3:4
E)6:7
Rationale: Milk = 3/5 * 20 = 12 liters, water = 8 liters
If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.
Remaining milk = 12 - 6 = 6 liters
Remaining water = 8 - 4 = 4 liters
10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.
The ratio of milk and water in the new mixture = 16:4 = 4:1
If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = 4/5 * 10 = 8 liters.
Amount of water removed = 2 liters.
Remaining milk = (16 - 8) = 8 liters.
Remaining water = (4 -2) = 2 liters.
final mixture obtained = (8 + 10):2 = 18:2 = 9:1.
ANSWER B
Answer: B
Question: Car X began traveling at an average speed of 35 miles per hour. After 72 minutes, car Y began traveling at an average speed of 49 miles per hour. When both cars had traveled the same distance, both cars stopped. How many miles did car X travel from the time car Y began traveling until both cars stopped?
Options:
A)105
B)120
C)140
D)147
E)168
Rationale: Car Y began travelling after 72 minutes or 1.2 hours.
Let t be the time for which car Y travelled before it stopped.
Both cars stop when they have travelled the same distance.
So,
35(t+1.2) = 49t
or 5t+6=7t
or t=3
Distance travelled by car X from the time car Y began traveling until both cars stopped is
35 x 3
= 105 miles
Answer:- A
Answer: A
Question: The cost of fencing a square field @ Rs. 20 per metre is Rs.10.080.How much will it cost to lay a three meter wide pavement along the fencing inside the field @ Rs. 50 per sq m
Options:
A)73889
B)27788
C)73879
D)73800
E)2799
Rationale: Explanation:
perimeter = total cost / cost per m = 10080 /20 = 504m
side of the square = 504/4 = 126m
breadth of the pavement = 3m
side of inner square = 126 - 6 = 120m
area of the pavement = (126 x126) - (120 x 120) = 246 x 6 sq m
cost of pavement = 246*6*50 = Rs. 73800
Answer: D)
Answer: D
Question: If n is an odd integer, which of the following must also be an odd integer?
Options:
A)2n
B)2n^2
C)n^2 + 2n
D)n^2 + 3n
E)n^2 + 2n + 3
Rationale: (A)	2n = even*odd = even.
(B)	2n^2 = even*odd = even.
(C)	n^2 + 2n = odd + even*odd = odd + even = odd.
(D)	n^2 + 3n = odd + odd*odd = odd + odd = even.
(E)	n^2 + 2n + 3 = odd +even + odd = even.
Answer: C.
Answer: C
Question: An oratorical society consists of six members, and at an upcoming meeting, the members will present a total of four speeches. If no member presents more than two of the four speeches, in how many different orders could the members give speeches?
Options:
A)720
B)1080
C)1170
D)1470
E)1560
Rationale: You need to remember that of the 3 people, 1 is going twice and two are going once. Since one of the 3 is going once, there are 3C1 possibilities for the double speaker.
So, to convert your work to the correct answer, you need to multiply:
240 * 3C1 = 240 * 3 = 720. ANS:A
Answer: A
Question: Andrew investe 12 percent of his March earnings. In April he earned 20 percent more than he did in March, and he invested 9 percent of his April earnings. The amount he invested in April was what percent of the amount he invested in March?
Options:
A)80%
B)90%
C)100%
D)110%
E)120%
Rationale: Let Andrew's March earnings be 100 --> Amount invested = 12% of 100 = 12
Andrews April earning = 1.2 * 100 = 120 --> Amount invested = 9% of 120 = 10.8
Amount invested in April = (10.8/12)*100 = (108/120)*100 = 90% of the amount invested in March
Answer: B
Answer: B
Question: Kelly has had 3 pay cuts in her salary in the past 6 months. If the first pay cut was 8%, the second pay cut was 14% and the third was 18%. What will be the percentage decrease, if the salary is decreased in a single shot?
Options:
A)35.9%
B)34.12%
C)32.12%
D)35.12%
E)31.12%
Rationale: Let Rs.100 be initial salary.
Salary After 1st decrease, 8% = 92
Salary After 2nd decrease, 14% = 79.12 i.e. reduced by 14 percent of 92
Salary After 3rd decrease, 18% = 64.88 i.e. reduced by 18 percent of 79.12
So if its decreased in single shot =
i = ( ( b - a ) /b ) * 100
=35.12%
ANSWER:D
Answer: D
Question: An amount of money is to be distributed among P, Q and R in the ratio of 5 : 8 : 12 respectively. If the total share of Q and R is four times that of P, what is definitely P’s share?
Options:
A)3,000
B)5,000
C)8,000
D)Data inadequate
E)None of these
Rationale: P : Q : R = 5 : 8 : 12
Total share of QandR/Share of P=8+12/5=20/5=4
So, we see that no new information has been given in the question and P’s share can’t be determined.
Answer D
Answer: D
Question: Find the 12th term of an arithmetic progression whose first term is 2 and the common difference is 8.
Options:
A)45
B)38
C)44
D)90
E)96
Rationale: n th term of A.P = a +(n-1) *d
= 2+(12-1)*8 , = 2 + 88 = 90.
ANSWER:D
Answer: D
Question: A boat having a length 3 m and breadth 2 m is floating on a lake. The boat sinks by 2 cm when a man gets on it. The mass of man is
Options:
A)12 kg
B)60 kg
C)72 kg
D)120 kg
E)None
Rationale: Solution
Volume of water displaced	= (3 x 2 x 0.02)m3
= 0.12m3.
Mass of man	=Volume of water displaced ×Density of water
= (0.12×1000) kg
= 120 kg.
Answer D
Answer: D
Question: The numerator of a certain fraction is 8 less than the denominator. If 3 is added to the numerator and 3 is subtracted from the denominator, the fraction becomes 3/4. Find the original fraction?
Options:
A)A)24.5
B)B)28.5
C)C)30
D)inadequate
E)of these
Rationale: Explanation:
Let C.P. be Rs. 100.
Then, S.P. = Rs. 123.50
Let marked price be Rs. x. Then, 95/100 x = 123.50
x = 12350/95 = Rs. 130
Now, S.P. = Rs. 130, C.P. = Rs. 100
Profit % = 30%.
Answer: Option C
Answer: C
Question: At a certain university, the ratio of the number of teaching assistants to the number of students in any course must always be greater than 3:85. At this university, what is the maximum number of students possible in a course that has 5 teaching assistants?
Options:
A)140
B)141
C)142
D)143
E)144
Rationale: (5/3)*85=141 plus remainder
The answer is B.
Answer: B
Question: A cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 9 hours. If both the taps are opened simultaneously, then after how much time will the cistern get filled?
Options:
A)7.5
B)7.4
C)7.9
D)7.2
E)7.1
Rationale: Net part filled in 1 hour = (1/4 - 1/9) = 5/36
The cistern will be filled in 36/5 hrs i.e., 7.2 hrs.Answer: D
Answer: D
Question: If z is a multiple of 9702, what is the remainder when z^2 is divided by 4?
Options:
A)0
B)1
C)2
D)3
E)It depends on z.
Rationale: 2 is a factor of 9702, so 2 is a factor of z.
Then 2^2 = 4 is a factor of z^2.
Then the remainder when z^2 is divided by 4 is 0.
The answer is A.
Answer: A
Question: Company P had 15 percent more employees in December than it had in January. If Company P had 460 employees in December, how many employees did it have in January?
Options:
A)391
B)400
C)410
D)423
E)445
Rationale: Given: January*1.15=December --> January*1.15=460 --> January=460/1.15=400.
Answer: B.
Answer: B
Question: One robot builds a robot in 1 hour, another one builds a robot in 2 hours. The 2 robots work together and when a new robot is complete, it joins the others, working at a constant rate of one robot in 2 hours. How much time will it take until there are 6 robots altogether, if the robots build only one robot at a time?
Options:
A)70 min
B)94 min
C)109 min
D)114 min
E)120 min
Rationale: First the 2 robots work at the rate of
1 + 1/2 = 3/2
so they complete one robot in 2/3 rd of an hour = 40 minutes - (1)
Now the 3 robots work together at the rate of
1 + 1/2 + 1/2 = 4/2 = 2/1
so they complete one robot in 1/2 an hour,i.e 30 minutes - (2)
Now the 4 robots work together at the rate of
1 + 1/2 + 1/2 + 1/2 = 5/2
so they complete one robot in 2/5th of an hour, i.e. 24 minutes - (3)
Now the 4 robots work together at the rate of
1 + 1/2 + 1/2 + 1/2+1/2 = 3
so they complete one robot in 1/3th of an hour, i.e. 20 minutes - (4)
And now we have 5 robots
So Total = (1) + (2) + (3) +(4)= 114minutes
Answer - D
Answer: D
Question: In a party there are 5 couples. Out of them 5 people are chosen at random. Find the probability that there are at the least two couples?
Options:
A)5/21
B)5/28
C)5/22
D)5/25
E)5/27
Rationale: Number of ways of (selecting at least two couples among five people selected) = (⁵C₂ * ⁶C₁)
As remaining person can be any one among three couples left.
Required probability = (⁵C₂ * ⁶C₁)/¹⁰C₅
= (10 * 6)/252 = 5/21
Answer:A
Answer: A
Question: An army’s recruitment process included N rounds of selection tasks. For the first A rounds, the rejection percentage was 60 percent per round. For the next B rounds, the rejection percentage was 50 percent per round and for the remaining rounds, the selection percentage was 70 percent per round. If there were 10000 people who applied for the army and 350 were finally selected, what was the value of N?
Options:
A)4
B)5
C)6
D)9
E)8
Rationale: fastly i reduce 60% till it gets closer to our required 2000 candidates
step(1) 40000 accepted.
step(2) another 40%of 40000=16000 accepted.
here it is quiet observable that if we further deduct candidate by 60% it would change our probablity of easy going 2000 candidate.
so i would get to second stage of recruitment where 50 % is accepted
step(3) 50% of 16000=8000
step(4) 50% of 8000=4000
Step(5) 50% of 4000=2000
Step(6) 50% of 2000=1000
Step(7) 50% of 1000=500
last step of accepting 70% of 500=350(our target)
total 8 steps required.
Ans D
Answer: D
Question: A class consists of 15 girls and 10 boys. Three students are to be randomly selected. Find the probability that one boy and two girls are picked.
Options:
A)21/46
B)21/49
C)21/42
D)26/46
E)21/43
Rationale: ANSWER: A
Answer: A
Question: In a division, when a student mistook the divisor as 12 instead of 21 and obtained 35 as quotient the remainder is 0 . What is the correct quotient?
Options:
A)14
B)18
C)20
D)24
E)29
Rationale: Let x be the number
x÷12=35, remainder = 0
=> x=35×12
correct quotient
=x21=35×1221=5×123=5×4=20
C
Answer: C
Question: What are the next two numbers in this sequence?
7, 14, 17, 21, 27, 28, 35, 37, ?
Options:
A)40
B)41
C)42
D)43
E)44
Rationale: these are digits which are either multiples of 7 or containing the digit 7.
s0 next numbers are 42
ANSWER:C
Answer: C
Question: How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
Options:
A)5
B)10
C)15
D)20
E)25
Rationale: Explanation:
Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 x 5 x 4) = 20.
Answer: D
Answer: D
Question: Which of the following is a possible length for side AB of triangle ABC if AC = 3 and BC = 9?
I. 6
II. 9 √3
III. 11.5
Options:
A)I only
B)II only
C)III only
D)II and III
E)I, II, and III
Rationale: (9-3) < AB < (9+3)
The answer is C.
Answer: C
Question: At a certain pizza parlor, the diameter of a large pizza is 40% larger than the diameter of a small pizza. What is the percent increase in total amount of pizza, from a small to a large?
Options:
A)20%
B)40%
C)64%
D)80%
E)96%
Rationale: A real world problem —- yum! —- but again, the specifics of the shape don’t matter. Diameter increases 40%, so k = 1.4 in this case. “Amount of pizza” just means the area of the circle, so k^2 = (1.4)^2 = 1.96, which is the multiplier for a 96% increase. That’s the percent increase in area. Answer = E
Answer: E