divelab/Sys2Bench huggingface dataset:
Question: How many 4-digit numbers (ABCD) can be formed such that |A – D| = 2?
Options:
A)2,000
B)1,900
C)1,800
D)1,600
E)1,500
Rationale: Digit A can be filled in 9 ways (excluding 0)
Digit B can be filled in 10 ways
Digit C can be filled in 10 ways
Digit D can be filled in 2 ways (because of modulus A-D )
The answer will be 9*10*10*2 = 1800
Ans C
Answer: C
Question: The time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream. If the speed of the boat in still water is 45 kmph, find the speed of the stream?
Options:
A)15
B)88
C)14
D)12
E)17
Rationale: The ratio of the times taken is 2:1.
The ratio of the speed of the boat in still water to the speed of the stream = (2+1)/(2-1) = 3/1 = 3:1
Speed of the stream = 45/3 = 15 kmph.
Answer:A
Answer: A
Question: Village X has a population of 68000, which is decreasing at the rate of 1200 per year. Village Y has a population of 42000, which is increasing at the rate of 800 per year .in how many years will the population of the two villages be equal?
Options:
A)13
B)15
C)16
D)12
E)11
Rationale: Let the population of two villages be equal after p years
Then,68000-1200p=42000+800p
2000p=26000
p=13
Answer is A.
Answer: A
Question: Shweta buys a product at 25% discount rate. At what % over the cost price should she sell the product to make an overall 25% profit over the listed price?
Options:
A)66.67%
B)66.69%
C)66.63%
D)66.64%
E)66.61%
Rationale: Let the listed price of the product be x
Hence, at 25% discount rate the Cost Price will become: 3x/4
Now to make an overall profit of 25% over listed price:
SP = CP+(25/100)CP
=> SP = x+x/4
=>SP = 5x/4
=>Selling Price over Cost price: (5x/4)-(3x/4)
=>x/2
=>%Selling price over Cost price: (x/2)/(3x/4)*100
=>
ANSWER: A
Answer: A
Question: If each year the population of the country grows by 10%, how many years will elapse before the population of the country doubles?
Options:
A)3
B)4
C)5
D)7
E)8
Rationale: Till Year 2000, population is 100 .
Year 2001: Population becomes 110 ............. 1 Year elapsed
Year 2002: Population becomes 121 ............. 2 Year elapsed
Year 2003: Population becomes>133 ............. 3 Year elapsed
Year 2004: Population becomes> 146 .................. 4 year elapsed
Year 2005: Population becomes> 160 .................. 5 year elapsed
Year 2006: Population becomes> 176 .................. 6 year elapsed
Year 2007: Population becomes> 193 .................. 7 year elapsed
Year 2008: Population > 200 .................. 8 year elapsed
Answer:E
Answer: E
Question: The ratio of the present ages of P and Q is 3:4. 5 years ago, the ratio of their ages was 5:7. Find the their present ages?
Options:
A)30,40
B)25,30
C)50,60
D)20,40
E)10,20
Rationale: Answer: Option A
Their present ages be 3X and 4X.
5 years age, the ratio of their ages was 5:7, then (3X - 5):(4X - 5) = 5:7
X = 35 - 25 => X = 10.
Their present ages are: 30, 40.
Answer: A
Question: The number of cars present at a particular time was measured at 2,999 different locations on Tuesday and on Wednesday. The number of locations that had more cars on Wednesday was 10% higher than the number of locations that had more cars on Tuesday. How many of the locations had more cars on Tuesday?
Options:
A)1,428
B)1,633
C)1,860
D)2,460
E)2,829
Rationale: The question states that all these 2,999 locations were measured twice, once on each day.
It also breaks the 2,999 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 = 2999
But, for the second day count, W = 1.10 T
i.e. for Second day count, T + 1.10T = 2999
i.e. 2.10 T = 2999
i.e. T = 1428
A
Answer: A
Question: On dividing a certain number by 234, we get 43 as remainder. If the same number is divided by 13, what will be the remainder?
Options:
A)6
B)9
C)4
D)7
E)8
Rationale: Description :
Solution : suppose that on dividing the given number by 234,
we get quotient=x and remainder= 43
then, number= 234*x+43----> (1).
=> (13*18x)+(13*3)+4
=> 13*(18x+3)+4.
So, the number when divided by 13 gives remainder=4.
Answer :(C)
Answer: C
Question: The average mark of the students of a class in a particular exam is 80. If 5 students whose average mark in that exam is 50 are excluded, the average mark of the remaining will be 90. Find the number of students who wrote the exam.
Options:
A)20
B)27
C)25
D)99
E)21
Rationale: Let the number of students who wrote the exam be x.
Total marks of students = 80 x.
Total marks of (x - 5) students = 90(x - 5)
80x - (5 * 50) = 90(x - 5)
200 = 10x => x = 20.Answer: A
Answer: A
Question: Two pipes A and B can fill a tank in 4 and 5 hours respectively. If they are turned up alternately for one hour each, the time taken to fill the tank is?
Options:
A)4 hrs 27 min
B)4 hrs 24 min
C)4 hrs 74 min
D)4 hrs 54 min
E)4 hrs 21 min
Rationale: 1/4 + 1/5 = 9/20
20/9 = 2 2/9
9/20 * 2 = 9/10 ---- 4 hours
WR = 1 - 9/10 = 1/10
1 h ---- 1/4
? ----- 1/10
2/5 * 60 = 24 = 4 hrs 24 min
Answer:B
Answer: B
Question: The perimeters of two squares are 80 cm and 64 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares.
Options:
A)42cm
B)44cm
C)46cm
D)48cm
E)49cm
Rationale: Explanation:
Side of first square = (80/4) = 20 cm;
Side of second square = (64/4)cm = 16 cm.
Area of third square = [(20)2 - (16)2] cm2 = (400 - 256) cm2 = 144 cm2.
Side of third square = (144)(1/2) cm = 12 cm.
Required perimeter = (12 x 4) cm = 48 cm.
Answer: Option D
Answer: D
Question: The work done by a woman in 8 hours is equal to the work done by a man in 6 hours and by a boy in 12 hours. If working 6 hours per day, 9 men can complete a work in 6 days then in how many days can 12 men, 12 women and 12 boys together finish the same work by working 8 hours per day?
Options:
A)1 1⁄3 days
B)3 2⁄3 days
C)3 days
D)1 1⁄2 days
E)None of these
Rationale: 8W = 6M = 12B
12M + 12W + 12B Þ 12M + 9M + 6M = 27M
\9 men can complete the work by working 1 hour per day in 6 × 6 days
\27 men working 8 hours per day = 6×6×9/27×8=112days
Answer D
Answer: D
Question: What is the dividend? Divisor is 9, the quotient is 49, and the remainder is 52.
Options:
A)Undetermined
B)0
C)1
D)546
E)493
Rationale: Dividend= Divisor * Quotient + Remainder
?= 9 * 49 + 52
441 + 52
?= 493
(e.g. Answer: E)
Answer: E
Question: A, B and C started a business with investment of Rs. 8000. Rs. 20000 and Rs. 32000. What is the share of B in the total profit of Rs. 5400 at the end of 1 year?
Options:
A)Rs. 2880
B)Rs. 720
C)Rs. 1800
D)Rs. 1650
E)None of these
Rationale: Explanation:
Ratio of capitals of A, B and C
= 8000 : 20000 : 32000 = 2 : 5 : 8
.: B's share = Rs. [(5/15) × 5400] = Rs. 1800
Answer: Option C
Answer: C
Question: Integer m has 4 different prime factors and n has 3 different prime factors. If m and n has the greatest common factor of 15, how many different prime factors does mn have?
Options:
A) 4
B) 5
C) 6
D) 7
E) 8
Rationale: As m and n has GCF of 15, 3*5 should be common factor for both m and n.
m=3*5*p1*p2
n=3*5*p3
mn=3^2*5^2*p1*p2*p3
mn has prime numbers of 3,5,p1,p2p3
B
Answer: B
Question: A, Band C can do a piece of work in 11 days, 5 days and 55 days respectively, working alone. How soon can the work be done if A is assisted by Band C on alternate days?
Options:
A)5 days
B)8 days
C)9 days
D)10 days
E)11 days
Rationale: (A+B)'s 1 day's work = 1/11+1/5 = 16/55
(A+C)'s 1 day's work = 1/11+1/55 = 6/55
work done in 2 day's = 16/55+6/55 = 2/5
2/5th work done in 2 days
work done= 5/2*2 = 5 days
ANSWER:A
Answer: A
Question: Which one of the given numbers is divisible by 11?
Options:
A)4823718
B)8423718
C)8432718
D)4132788
E)4852718
Rationale: Explanation:
4132788 = (4+3+7+8) – (1+2+8) = 11
So, the given number is divisible by 11
Answer:D
Answer: D
Question: Cn = 2Cn-1 + 4 and Qn = 4Qn-1 + 8 for all n > 1. If C5 = Q4 and C7 = 316, what is the first value of n for which Qn is an integer?
Options:
A) 1
B) 2
C) 3
D) 4
E) 5
Rationale: Assuming I have understood the symbols used correctly answer is C- 3
C7 = 2S6 +4 = 316
C6 = 312/2 = 156 = 2C5 + 4
C5 = 152/2 = 76 = Q4
Q4 = 4Q3 + 8 = 76
Q3 = 68/4 = 17
Q3 = 4Q2 + 8 = 17
Q2 = 9/4 = 2.25
Q2 = 4Q1 + 8
Q1 will not be integer
Answer: C
Question: Find the value of y from (12)^3 x 6^4 ÷ 432 = y?
Options:
A)2134
B)2245
C)3450
D)4500
E)5184
Rationale: Given Exp. = (12)3 x 64 = (12)3 x 64 = (12)2 x 62 = (72)2 = 5184
432 12 x 62
E
Answer: E
Question: RAJA complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.
Options:
A)200 km
B)250 km
C)224 km
D)255 km
E)260 km
Rationale: consider x -- > (1/2)x/21 + (1/2)x/24 = 10
==> x/21+x/24 = 20
15x = 168 * 20
==> 224km
ANSWER C
Answer: C
Question: Find avrg speed if a man travels at speed of 18km/hr up and 24km/hr dawn at an altitude of 300m.
Options:
A)25.8
B)26.8
C)27.8
D)20.5
E)21.8
Rationale: avg speed=2*x*y/(x+y)
=2*18*24/(18+24)=20.5
ANSWER:D
Answer: D
Question: A train 800 m long is running at a speed of 78 km/hr. If it crosses a tunnel in 1 min, then the length of the tunnel is?
Options:
A)300
B)400
C)500
D)600
E)700
Rationale: Speed = 78 * 5/18 = 65/3 m/sec.
Time = 1 min = 60 sec.
Let the length of the train be x meters.
Then, (800 + x)/60 = 65/3
x = 500 m.
Answer: Option C
Answer: C
Question: If the sides of a triangle are given as 5*x, 5*(x+1), 5*(x+2), then what could be the possible value of the product of their sides.
Options:
A)1000
B)2000
C)3000
D)4000
E)5000
Rationale: Since the sides are consecutive multiples of 5, the product will be a multiple of 125*2*3(=750).
750*4=3000
Answer:C
Answer: C
Question: Convert 350 inches into centimeter?
Options:
A)812cm
B)820cm
C)850cm
D)889cm
E)854cm
Rationale: 1 inch = 2.54cm
350 inches = 100*2.54 = 889cm
Answer is D
Answer: D
Question: A certain box has 13 cards and each card has one of the integers from 1 to 13 inclusive. Each card has a different number. If 2 different cards are selected at random, what is the probability that the sum of the numbers written on the 2 cards is less than the average (arithmetic mean) of all the numbers written on the 13 cards?
Options:
A)1/7
B)1/13
C)1/15
D)1/22
E)1/26
Rationale: The average of the numbers is 7
The total number of ways to choose 2 cards from 13 cards is 13C2 = 78.
The ways to choose 2 cards with a sum less than the average are:
{1,2},{1,3},{1,4},{1,5},{2,3},{2,4}
The probability is 6/78 = 1/13
The answer is B.
Answer: B
Question: If a clock takes 7seconds to strike 7, how long will the same clock take to strike 10?
Options:
A)10.5
B)11.5
C)12.5
D)13.5
E)14.5
Rationale: If a clock takes 7seconds to strike 7 , there are 6 time intervals of duration 7/6 secs.
At 10, there are 9 time intervals of 7/6 secs having total duration of 9*7/6= 10.5 secs for 10 strikes
ANSWER:A
Answer: A
Question: If one cat eats 1.5 rats in 1.5 days, then in howmany days 3 cats will eat 3 rats??
Options:
A)1 Day
B)2 Days
C)3 Days
D)4 Days
E)5 Days
Rationale: 1Cat ------ 3/2 Rat --------- 3/2 Days
1 Cat ------(3/2)*2 Rat -----(3/2)*2 Days (Rat doubled, days doubled)
1*3 Cat----- 3 Rat -----------3/3=1 Day (Cat*3 , Same rat , Days/3 )
ANSWER:A
Answer: A
Question: A set of data consists of the following 4 numbers: 0, 2, 4, 6, and 8. Which two numbers, if added to create a set of 7 numbers, will result in a new standard deviation that is close to the standard deviation for the original 5 numbers?
Options:
A)-1 and 9
B)4 and 4
C)3 and 5
D)2 and 6
E)0 and 8
Rationale: If you are in for some calculations , this is how I got to it
mean = 4
sd = \sqrt{8} = 2.8
Expected values for the SD to not change are - One value below SD from mean is (4 - 2.8) = 1.2 , and one value above SD is (4 + 2.8) = 6.8
This would mean , adding 1.2 ans 6.8 would have no impact on the SD . SD remains the same when these two numbers are added. Now for SD to change the least , we need to add two values that are closest to these two values.
Hence any two values that are closest to 1.2 and 6.8 would change the SD , the least.
1. -1 , 9
distance between (1,9) and (1.2 and 6.8) is 2.2 and 2.2
2. 4 , 4
distance etween (4,4) and (1.2 , 6.8) is 2.8 and 2.8
3. 3 , 5
Distance is - 1.8 and 1.8
4. 2 , 6
Distance is - 0.8 and 0.8
5. 0 , 8
Distnace is - 1.2 and 1.2
Hence from above , we see that adding 0 and 8 , results in a value that would change the SD to the least. Hence E
Answer: E
Question: A contractor undertakes to built a walls in 50 days. He employs 20 peoples for the same. However after 25 days he finds that only 40% of the work is complete. How many more man need to be employed to complete the work in time?
Options:
A)25
B)30
C)35
D)30
E)None of these
Rationale: 20 men complete 0.4 work in 25 days.
Applying the work rule, m1 × d1 × w2 = m2 × d2 × w1
we have,
20 × 25 × 0.6 = m2 × 25 × 0.4
or m2 = 20×25×0.6/25×0.4 = 30 men
AnswerD
Answer: D
Question: Find the area of a rhombus whose side is 25 cm and one of the diagonals is 30 cm?
Options:
A)602 sq.cm
B)620 sq.cm
C)200 sq.cm
D)605 sq.cm
E)600 sq.cm
Rationale: Consider the rhombus ABCD. Let the diagonals intersect at E. Since diagonals bisect at right angles in a rhombus.
BE2 + AE2 = AB2
252 = 152 + AE2 AE = √(625 - 225) = √400 = 20,
AC = 20 + 20 = 40 cm.
Area of a rhombus = 1/2 * d1d2
= 1/2 * 40 * 30 = 600 sq.cm.
Answer: E
Answer: E
Question: Peter wants to get $200,000 for his house. An agent charges 20% of the selling price for selling the house for Peter.What will be the agent's commission?
Options:
A)$25,000
B)$35,000
C)$60,000
D)$80,000
E)$50,000
Rationale: Let x be the selling price: x - 20%x = 200,000
a) Solve for x to find x = $250,000 which is the seeling price of the house,then
20% *250,000 = $50,000 the agent's commission
correct answer E)$50,000
Answer: E
Question: A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If two marbles are drawn at random, what is the probability that at least one is green?
Options:
A)23/35
B)23/38
C)23/18
D)53/18
E)23/22
Rationale: Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
Probability that at least one green marble can be picked in the random draw of two marbles = Probability that one is green + Probability that both are green
= (⁶C₁ * ⁹C₁)/¹⁵C₂ + ⁶C₂/¹⁵C₂
= (6 * 9 * 2)/(15 * 14) + (6 * 5)/(15 * 14)
= 36/70 + 1/7
= 46/70 = 23/35
Answer:A
Answer: A
Question: A certain university will select 1 of 5 candidates eligible to fill a position in the mathematics department and 2 of 10 candidates eligible to fill 2 identical positions in the computer science department. If none of the candidates is eligible for a position in both departments, how many different sets of 3 candidates are there to fill the 3 positions?
Options:
A)42
B)70
C)140
D)165
E)225
Rationale: 1 of 5 will be chosen for the math
2 of 10 will be chosen for the computer
None of the 3 chosen people can be in more than one departments.
We can choose any of the 5 candidates for the math dep., which gives as 5 selections.
We can choose 2 of the 10 candidates for the computer dep., which gives us 2 selections and 8 rejections.
So, the way to find how many different selections of 2 candidates we can have for the computer dep., we do:
10! /2!*8!= (9*10) / 2 = 90 / 2 = 45.
We are multiplying our individual selections: 5*45 = 225
In the bolded part, we don't have to multiply all of the numbers, as those in 8! are included in 10!, so we simplify instead.
Ans E
Answer: E
Question: If p and q are positive integers and q ≠ 1, then pq(q −2) is
Options:
A)pq^2
B)pq^2+2pq
C)pq^2-2pq
D)pq^2+pq
E)pq^2-6pq
Rationale: pq(q-2) = pq^2-2pq
Answer : C
Answer: C
Question: As a treat for her two children, a mother runs to the freezer in which she has three cherry ice pops, three orange ice pops, and five lemon ice pops. She chooses two at random to bring outside to the children, but realizes as she runs out the door that she cannot bring them different flavors without one being jealous of the other and getting even more upset. What is the probability that she has to return to the freezer to make sure that they each receive the same flavor?
Options:
A)7/11
B)9/22
C)19/33
D)29/44
E)39/55
Rationale: P(2 cherry) = 3/11 * 2/10 = 6/110
P(2 orange) = 3/11 * 2/10 = 6/110
P(2 lemon) = 5/11 * 4/10 = 20/110
P(2 of the same flavor) = 32/110
P(2 different flavors) = 1 - 32/110 = 78/110 = 39/55
The answer is E.
Answer: E
Question: If a point (x,y) is randomly selected within the square shown in the figure above, (the vertices are on 1,1 1,-1, -1,1 -1,-1) what are
the W odds that x^2+y^2>1?
Options:
A)1. pi/4
B)2. W=pi/16
C)3. W=1-pi/4
D)4. W=1-pi/16
E)5.4-pi
Rationale: area of the square will be 4, and the area of circle with center (0,0) and radius 1 will be pi.
then succesful outcomes = 4 - pi (the area where x2+y2 will be greater than 1)
total outcomes = 4
therefore probability = (4-pi)/4
C
Answer: C
Question: It takes 60 identical printing machines 2 hours to print 200,000 cards. How long would it take 50 of these machines to print the same total?
Options:
A)2 hr 36 min
B)2 hr 40 min
C)2 hr 30 min
D)2 hr 25 min
E)2 hr
Rationale: 50% of 60 = 30% or 0.30
2 hr x 60 min = 120 min
120 min x 0.30 = 36 min
36 min + 2 hr = 2 hr and 36 min
Answer is A
Answer: A
Question: A, B and C invests Rs.2000, Rs.3000 and Rs.4000 in a business. After one year A removed his money; B and C continued the business for one more year. If the net profit after 2 years be Rs.3200, then A's share in the profit is?
Options:
A)288
B)266
C)299
D)400
E)827
Rationale: 2*12 : 3*12 : 4*24
1: 3: 4
1/8 * 3200 = 400
Answer:D
Answer: D
Question: A participated in cycling contest and he drove the lap at the rate of 6kmph, 12kmph, 18kmph, 24kmph..What is his average speed?
Options:
A)15kmph
B)18kmph
C)14.25 kmph
D)11.52 kmph
E)16kmph
Rationale: =4x/(x/6 +x/12 + x/18 +x/24)
=11.52
ANSWER:D
Answer: D
Question: There are some sweets in party, 1/3rd eaten by kids . Then 2/5th of the remaining by men , then 2/3rd of the remaining by women . At last 2 were remaining . How many sweets were in total ?
Options:
A)45
B)27
C)15
D)26
E)91
Rationale: Sol:
If x sweets were there in total , then
x × (1 – 1/3)× (1 – 2/5) ×(1 – 2/3) = 2
x×2/3 × 3/5 × 1/3 = 2
x = 15
Answer:C
Answer: C
Question: 11 friends went to a hotel and decided to pay the bill amount equally. But 10 of them could pay Rs. 60 each as a result 11th has to pay Rs. 50 extra than his share. Find the amount paid by him.
Options:
A)110
B)111
C)115
D)123
E)None
Rationale: Explanation:
Average amount paid by 10 persons = Rs. 60
Increase in average due to Rs. 50 paid extra by the 11th men = Rs. 50/10= Rs. 5
Therefore, Average expenditure of 11 friends = Rs. 60 + Rs. 5 = Rs. 65
Therefore, Amount paid by the 11th men = Rs. 65 + Rs. 50 = Rs. 115
Correct Option: C
Answer: C
Question: The average of 1st 3 of 4 numbers is 6 and of the last 3 are 5. If the sum of the first and the last number is 17. What is the last numbers?
Options:
A)4
B)5
C)6
D)7
E)8
Rationale: A + B + C = 18
B + C + D = 15
A + D = 17
A – D = 3
A + D = 17
2D = 14
D = 7
ANSWER:D
Answer: D
Question: complete the numerical series with the correct number
6+3+2=181227
7+2+4=142840
8+6+3=482466
9+4+5=364567
6+5+5= ?
Options:
A)468645
B)303055
C)454585
D)454386
E)444586
Rationale: 6+5+5=303055
Answer : B
Answer: B
Question: a television have side 6*5.its screen size is 5*4.calculate the percentage of unscreened part t.v.
Options:
A)30
B)25
C)33.33
D)44.44
E)28
Rationale: unscreened part=30-20=10.
%unscreened part=10/30*100=33.33%
answer C
Answer: C
Question: 28 machines can do a work in 10days. How many machines are needed to complete the work in 40days?
Options:
A)10
B)6
C)4
D)7
E)5
Rationale: Required number of machines = 28*10/40 = 7
Answer is D
Answer: D
Question: Solve the equation :
(2x - 1)3 + 2x(2x - 3).(3 - 2x) – (3x - 1)2 = 3x2 - 2
Options:
A)0
B)1
C)2
D)3
E)4
Rationale: Solution:
We open the brackets by using the formulas for multiplication:
8x3 - 3(2x)2.1 + 3.2x(1)2 – 13 - 2x(2x - 3)2 – (9x2 - 6x + 1) = 3x2 - 2 <=>
8x3 - 12x2 + 6x - 1 - 2x(4x2 - 12x + 9) - 9x2 + 6x - 1 = 3x2 - 2 <=>
8x3 - 21x2 + 12x - 8x3 + 24x2 - 9x = 3x2 <=>
3x2 + 3x = 3x2 <=>
3x = 0 <=> x = 0
Answer A
Answer: A
Question: A man can row downstream at 18 kmph and upstream at 10 kmph. Find the speed of the man in still water and the speed of stream respectively?
Options:
A)6
B)5
C)7
D)4
E)3
Rationale: Let the speed of the man in still water and speed of stream be x kmph and y kmph respectively.
Given x + y = 18 --- (1)
and x - y = 10 --- (2)
From (1) & (2) 2x = 28 => x = 14, y = 4.
Answer: D
Answer: D
Question: In a 2000 m race, A beats B by 100 meters or 20 seconds. Find the speed of B?
Options:
A)10m/s
B)2m/s
C)5m/s
D)7m/s
E)3m/s
Rationale: Since A beats B by 100 m or 20 seconds,
it implies that B covers 100 m in 20 seconds.
Hence speed of B = 100/20 = 5 m/s.
Answer: C
Answer: C
Question: A is twice as good as B. And together they finish a piece of work in 20days. In how many days will A alone finish the work
Options:
A)30
B)25
C)12
D)10
E)32
Rationale: (A's 1day work):(B's 1day work) = 2:1
A+B 1day work = 1/20
A's 1day work = (1/20)*(2/3) = 1/30
A alone can finish the work in 30 days
Answer is A
Answer: A
Question: Will decides to attend a basketball game with four friends.
If the party of five sits together in five consecutive seats, and Will must NOT sit in between two of his friends, how many ways
can the five friends be arranged?
Options:
A)24
B)36
C)48
D)72
E)120
Rationale: Will have to sit at either of the ends.
If will sits on the left most corner, then other 4 friends can be arranged in 4! = 24
Will can also sit on the other end, then another 4! ways we can arrange.
So total number of ways = 24+24 = 48
ANS:Option( C)
Answer: C
Question: In how many ways can the letters of the word SUBMITTED be arranged?
Options:
A)181215
B)181000
C)181450
D)181220
E)181440
Rationale: SUBMITTED has 9 words, but T is repeated two times. Thus ways = 9! / 2! = 181440 ways
E
Answer: E
Question: s(n) is a n-digit number formed by attaching the first n perfect squares, in order, into one integer. For example, s(1) = 1, s(2) = 14, s(3) = 149, s(4) = 14916, s(5) = 1491625, etc. How many digits are in s(99)?
Options:
A)350
B)353
C)354
D)356
E)357
Rationale: Focus on the points where the number of digits in squares change:
1, 2, 3 - Single digit squares. First 2 digit number is 10.
4 , 5,...9 - Two digit squares. To get 9, the last number with two digit square, think that first 3 digit number is 100 which is 10^2. so 9^2 must be the last 2 digit square.
10, 11, 12, ... 31 - Three digit squares. To get 31, think of 1000 - the first 4 digit number. It is not a perfect square but 900 is 30^2. 32^2 = 2^10 = 1024, the first 4 digit square.
32 - 99 - Four digit squares. To get 99, think of 10,000 - the first 5 digit number which is 100^2.
So number of digits in s(99) = 3*1 + 6*2 + 22*3 + 68*4 = 3 + 12 + 66 + 272 = 353;
ANSWER:B
Answer: B
Question: In the sequence 1, 2, 4, 8, 16, 32, …, each term after the first is twice the previous term. What is the sum of the 15th, 17th, and 18th terms in the sequence?
Options:
A)2^18
B)3(2^17)
C)7(2^16)
D)13(2^14)
E)7(2^15)
Rationale: the sequence comes out to be 2^0,2^1,2^2,2^3 and so on...
so 15th term will be 2^14
17th term 2^16
18th term 2^17
adding all three
we get
2^14+(2^15)*4+(2^15)*8
so answer is 2^15(1+4+8)=>13(2^15)
answer D
Answer: D
Question: The speed of a boat in still water in 42 km/hr and the rate of current is 3 km/hr. The distance travelled downstream in 44 minutes is:
Options:
A)86.6 km
B)46.6 km
C)33 km
D)35.6 km
E)26.6 km
Rationale: Speed downstream = (42 + 3) = 45 kmph
Time = 44 minutes =44/60 hour = 11/15 hour
Distance travelled = Time × speed =11/15×45 = 33 km
ANSWER:C
Answer: C
Question: If (-3)^(2x) = 3^(12 - x) and x is an integer, what is the value of x ?
Options:
A)5
B)4
C)3
D)2
E)1
Rationale: Since x is an integer, (-3)^(2x) is always positive.
So, 3^2x = 3^(12 - x)
2x = 12 - x
3x = 12
x = 4
Answer: B
Answer: B