divelab/Sys2Bench huggingface dataset:
Question: John has 1,210 Dogs in his house. If exactly 40 percent of all dogs in home are tagged, what percent of the untagged dogs must be tagged so that half of all dogs in the home are tagged?
Options:
A)25.5
B)30.33
C)33
D)37
E)16.6
Rationale: All dogs =1210
Currently Tagged = 1210*40/100 = 484
So Untagged dogs = 1210 - 484 =726
Half of all dogs =605
The number of dogs to be tagged to make half of all dogs tagged = 605-484 = 121
So now the question remains - 121 is how much percentage of untagged dogs (726)
= 121 * 100 /726
= 16.6%
Answer (E)
Answer: E
Question: The unit's digit in the product 274 × 318 × 577 × 313 is
Options:
A)2
B)3
C)4
D)5
E)6
Rationale: Solution
Required digit	= Unit digit in(4×8×7×3).
= 2.
Answer A
Answer: A
Question: Pipes A and B can fill a tank in 7 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in
Options:
A)1 13/17 hours
B)2 8/11 hours
C)3 9/17 hours
D)4 8/19 hours
E)None
Rationale: Solution
Net part filled in 1 hour	= (1/7 + 1/6 -1/12)
= 19/84.
Therefore, the tank will be full in 84/19 hrs.	â€¹=â€º 4 8/19 hrs.
Answer D
Answer: D
Question: Two squares are chosen at random on a chessboard. What is the probability that they have a side in common?
Options:
A)1/18
B)64/4032
C)63/64
D)1/9
E)1/4
Rationale: Explanation :
No of ways of selecting two squares from 64 squares on the chess board is 64C2.
Three cases arise:
case 1: when first square is any of the four corner ones
In this case second square can be chosen in 2 ways
no of ways of selecting two squares: 4*2 = 8
case2: when first square is any of the 24 squares on the side of the chess board other than the corner ones
the second square can be chosen in 3 ways
no of ways of selecting two squares: 24*3 = 72
case 3: the first square is any of the 36 remaining squares
the second square can be chosen in 4 ways
no of ways of selecting two squares: 36*4 = 144
Total no of ways of selecting two adjacent squares = 8 + 72 + 144 = 224
So required probability = 224/64C2 = 224/4032 = 1/18
Answer : A
Answer: A
Question: The length of a rectangular plot is 10mtr more than its width. The cost of fencing the plot along its perimeter at the rate of Rs. 6.5mtr is Rs.2210. The perimeter of the plot is?
Options:
A)126
B)156
C)340
D)321
E)260
Rationale: Sol. Let width = x, Length = (10+x)
Perimeter = 2(x+(10+x))
= 2(2x=10)
& 2(2x+10)*6.5 = 2210
X = 80
Required perimeter = 2(80+90) = 340
C
Answer: C
Question: The radius of the two circular fields is in the ratio 4: 10 the area of the second field is what percent greater than the area of the first?
Options:
A)72%
B)96%
C)94%
D)84%
E)82%
Rationale: r = 4 Ï€r2 = 16
r = 10 Ï€r2 = 100
100 Ï€ â€“ 84 Ï€
100 ---- ? =>84%.Answer:D
Answer: D
Question: If y = 35p, and p is prime, what is the greatest common factor of y and 42p, in terms of p?
Options:
A)p
B)2p
C)5p
D)7p
E)p^2
Rationale: y = 35p = 5*7*p
42p = 2*3*7*p
The greatest common factor of 30p and 20p is the product of all the common prime factors, using the lower power of repeated factors. The greatest common factor is 7*p = 7p
The answer is D.
Answer: D
Question: A, B and C can do a piece of work in 24 days, 30 days and 40 days respectively. They began the work together but C left 4 days before the completion of the work. In how many days was the work completed?
Options:
A)11
B)28
C)27
D)29
E)271
Rationale: Explanation:
One day's work of A, B and C = (1/24 + 1/30 + 1/40) = 1/10
C leaves 4 days before completion of the work, which means only A and B work during the last 4 days.
Work done by A and B together in the last 4 days = 4 (1/24 + 1/30) = 3/10
Remaining Work = 7/10, which was done by A,B and C in the initial number of days.
Number of days required for this initial work = 7 days.
Thus, the total numbers of days required = 4 + 7 = 11 days.
Answer: A) 11 days
Answer: A
Question: The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:
Options:
A)1008
B)1015
C)1022
D)1032
E)1122
Rationale: Explanation :
Required number = (L.C.M. of 12,16, 18, 21, 28) + 7
= 1008 + 7
= 1015
ANSWER IS B
Answer: B
Question: If every 2 out of 3 readymade shirts need alterations in the sleeves, and every 4 out of 5 need it in the body, how many alterations will be required for 60 shirts ?
Options:
A)88
B)123
C)133
D)143
E)153
Rationale: no.of Alteration required in 1 shirt is (2/3)+(4/5)+(3/4)=133/60
for 60 shirts it will 60*(133/60)=133
ANSWER:C
Answer: C
Question: I bought two books; for Rs.480. I sold one at a loss of 15% and other at a gain of 19% and then I found each book was sold at the same price. Find the cost of the book sold at a loss?
Options:
A)280
B)670
C)230
D)67
E)120
Rationale: A
x*(85/100) = (480 - x)119/100
x = 280
Answer: A
Question: A boy has nine trousers and 12 shirts. In how many different ways can he select a trouser and a shirt?
Options:
A)198 ways
B)128 ways
C)100 ways
D)108 ways
E)208 ways
Rationale: The boy can select one trouser in nine ways.
The boy can select one shirt in 12 ways.
The number of ways in which he can select one trouser and one shirt is 9 * 12 =108 ways
Answer:D
Answer: D
Question: A and B invest in a business in the ratio 3: 2. Assume that 5% of the total profit goes to charity. If A's share is Rs. 855, what is the total profit?
Options:
A)1200
B)1100
C)1500
D)1700
E)1600
Rationale: Assume that the total profit is x
Since 5% goes for charity, 95% of x will be divided between A and B in the ratio 3:2
Therefore, A's profit =(95x/100)×(3/5)
Given that A's share is Rs. 855. Therefore,
(95x/100)×(3/5)=855
⇒(95x/100)=855×5/3=285×5=1425
⇒x=(1425×100)/95=(285×100)/19=1500
Hence the total profit = 1500
Answer is C.
Answer: C
Question: In a dairy farm, 50 cows eat 50 bags of husk in 50 days. In how many days one cow will eat one bag of husk ?
Options:
A)34 days
B)40 days
C)46 days
D)50 days
E)None of these
Rationale: Explanation:
Less Cows, More days (Indirect Proportion)
Less bags, Less days (Direct Proportion)
[Cows1 50
Bags 50 1]::50:x
=>x∗50∗1=50∗1∗50
=>x=50
Option D
Answer: D
Question: Given that 2x + 13 > 5 and 5x - 13 < 7, all values of x must be between which of the following pairs of integers?
Options:
A)-4 and -1
B)-4 and 4
C)-4 and 1
D)-2 and 5
E)2 and 5
Rationale: 2x + 13 > 5
i.e 2x >5-13
i.e. 2x > -8
i.e. x > -4
Also, 5x - 13 < 7
i.e. 5x <7+13
i.e 5x < 20
i.e x < 4
i.e. -4 < x < 4
Answer: Option B
Answer: B
Question: If (2/3)^n=(3/2)^3, what is the value of n?
Options:
A)-3
B)-2
C)-1
D)0
E)1
Rationale: (2/3)^n = (3/2)^3
= (2/3)^-3
n = -3 (Answer = A)
Answer: A
Question: A man took loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was?
Options:
A)Rs. 2000
B)Rs. 10,000
C)Rs. 15,000
D)Rs. 20,000
E)Rs. 25,000
Rationale: Principal =(100 x 5400)/(12 x 3) = Rs. 15000.
Answer: Option C
Answer: C
Question: Q. How many 8-digit even numbers less than 3,000,000 can be formed using the following digits: 1, 2, 2, 3, 5, 5, 6, 6?
Options:
A)180
B)240
C)210
D)270
E)300
Rationale: the number is less than 3000000 so we will have the digit on the left most side to be either 1 or 2
i.e your number can be 1 _ _ _ _ _ _ or 2 _ _ _ _ _ _
case 1 > 1 _ _ _ _ _ _ here the units digit can be either 2 or 6
when the units digit is 2 i.e 1 _ _ _ _ _ 2
number of ways this can be done would be 5! / 2! (as 5 is repeated twice) = 60
when the units digit is 6 i.e. 1 _ _ _ _ _ 6, number of ways would be 5!/(2! * 2!) {both 2 and 5 repeat twice} = 30
case 2 > 2 _ _ _ _ _ _ (here units digit may be 2 or 6)
number of ways this can be done would be 5! / (2!) for 2 =60
and 5! / 2! for 6 ... = 60
adding all of these ... gives the answer 270 ...
the explanation looks difficult but this technique helps if you are familiar with the concept of permutations when the numbers/alphabets repeat
D
Answer: D
Question: If a and b are odd integers, which of the following is an even integer?
Options:
A)3(2a+b)
B)3(2a+b)+2a
C)3(2a+b)+a+b
D)3(2a+b)+7a-b
E)3(2a+b)+ab
Rationale: 3(2a+b)+ab=odd(odd)+odd=odd+odd=even
The answer is E.
Answer: E
Question: If the probability of Sita getting selected to a school is 20% and she applied to 3 schools. What is the probability that she will get selected in at least one school?
Options:
A)24/125
B)21/125
C)61/125
D)12/125
E)18/125
Rationale: The probability that she will be selected by at least one school equals to the sum of the probabilities of the following three events:
1. She is selected by only one school: P(SRR)=3!/2!∗1/5∗4/5∗4/5=48/125 (S stands for selected and R stands for rejected). We are multiplying by 3!/2! since SRR scenario can occur in several ways: SRR, RSR, RRS, (so 3!/2! is # of permutations of 3 letters SRR out of which 2 R's are identical);
2. She is selected by only two school: P(SSR)=3!/2!∗1/5∗1/5∗4/5=12/125, the same reason of multiplying by 3!/2!;
3. She is selected by all three school: P(SSS)=(15)^3=1125, we are not multiplying this time since SSS can occur only in one way.
So P=48/125+12/125+1/125=61/125
Answer: C
Answer: C
Question: Find the largest number of four digits which is exactly divisible by 27,18,12,15
Options:
A)9700
B)9710
C)9720
D)9520
E)7500
Rationale: Explanation:
LCM of 27-18-12-15 is 540.
After dividing 9999 by 540 we get 279 remainder.
So answer will be 9999-279 = 9720
Option C
Answer: C
Question: Julie put half of her savings in a savings account that pays an annual simple interest and half in a savings account that pays an annual compound interest. After two years she earned $120 and $124 from the simple interest account and the compound interest account respectively. If the interest rates for both accounts were the same, what was the amount of Julie's initial savings?
Options:
A)600
B)720
C)1080
D)1200
E)1800
Rationale: $120 for 2 years = $60 per year.
Extra $4 yearned with the compound interest is the percent yearned on percent. So, $4 is yearned on $60, which means that the interest = 15%.
This on the other hand means that half of the savings = 60*15 = $900. Twice of that = $1,800.
Answer: E.
Answer: E
Question: A can do a piece of work in 12 days. He worked for 15 days and then B completed the remaining work in 10 days. Both of them together will finish it in.
Options:
A)12 1/2 days
B)5 days
C)8.5 days
D)2 days
E)11.5 days
Rationale: Answer: Option A
15/25 + 10/x = 1 => x = 25
1/25 + 1/25 = 2/25
25/2 = 12 1/2 days
Answer: A
Question: A can do a piece of work in 30 days. He works at it for 5 days and then B finishes it in 20 days. In what time can A and B together it?
Options:
A)13 1/0
B)13 1/3
C)13 1/2
D)13 1/1
E)13 1/7
Rationale: 5/30 + 20/x = 1
x = 24
1/30 + 1/24 = 3/40
40/3 = 13 1/3 days
Answer:B
Answer: B
Question: Students at a school were on average 180 cm tall. The average female height was 170 cm, and the average male height was 181 cms. What was the ratio of men to women?
Options:
A) 5:2
B) 5:1
C) 4:3
D) 4:1
E) 10:1
Rationale: We're given a few facts to work with:
1) The average height of the females is 170cm
2) The average height of the males is 181cm
3) The average of the GROUP is 180cm
We're asked for the RATIO of men to women.
W = number of Women
M = number of Men
(170W + 181M)/(W+M) = 180
170W + 181M = 180W + 180M
1M = 10W
M/W = 10/1
The ratio of men to women is 10 to 1.
E
Answer: E
Question: The greatest common factor of two numbers is 4!. Which of the following can be the second number, if one of the numbers is 8!?
Options:
A)3(5!)
B)4(5!)
C)5(5!)
D)6(5!)
E)4(4!)
Rationale: GCF is the product of common factors of the numbers involved.
GCF = 4!
a = 8! = 8*7*6*5*4!
b will certainly have 5! and cannot have any more common factors with a (as this will increase the GCF)
Looking at the answers only 4 (4!) and 8! will have GCF as 4!
Ans E
Answer: E
Question: Thomas bought a pack of eleven computers for Rs. 500000 and sold it for Rs.550000. What is the % of profit he is making per computer?
Options:
A)10
B)14
C)19
D)21
E)25
Rationale: The cost of buying a single computer= Rs(500000/11)=Rs. 45454.5
The cost of selling a selling computer=Rs(550000/11)=Rs. 50000.0
Profit per computer=Rs.(50000-45454.5)=Rs. 4545.5
The % of profit in selling a single computer=(4545.5/45454.5)*100=10%
Answer:A
Answer: A
Question: Find out the C.I on Rs.5000 at 4% p.a. compound half-yearly for 1 1/2 years.
Options:
A)306.08
B)306.03
C)306.04
D)306.01
E)306.09
Rationale: Explanation:
A = 5000(51/50)3
= 5306.04
5000
-----------
306.04
Answer:C
Answer: C
Question: Convert 600 miles into meters?
Options:
A)784596
B)845796
C)965604
D)784596
E)864520
Rationale: 1 mile = 1609.34 meters
600mile = 600*1609.34 = 965604 meters
Answer is C
Answer: C
Question: Insert the missing number.
3, 7, 6, 5, 9, 3, 12, 1, 15, (....)
Options:
A)3
B)7
C)-1
D)8
E)-45
Rationale: Explanation:
There are two series, beginning respectively with 3 and 7. In one 3 is added and in another 2 is subtracted.
The next number is 1 - 2 = -1.
Answer: C) -1
Answer: C
Question: If BOMBAY is written as MYMYMY, how will TAMIL NADU be written in that code?
Options:
A)TIATIATIA
B)IATIATIAT
C)MNUMNUMNU
D)ALDALDALD
E)None of these
Rationale: Letters at multiples of 3 is repeated in the code.
So,
In the code BOMBAY, 3rd and 6th letter repeated for the code.
Like that, in the code TAMIL NADU, 3rd 6th and 9th letter must be repeated.
So, TAMILNADU = MNUMNUMNU
ANSWER:C
Answer: C
Question: An bus covers a certain distance at aspeed of 240 kmph in 5 hours. To cover the same distance in 1hr, it must travel at a speed of?
Options:
A)700 km/hr
B)650 km/hr
C)690 km/hr
D)710 km/hr
E)720 km/hr
Rationale: Distance = (240 x 5) = 1200 km.
Speed = Distance/Time
Speed = 1200/(5/3) km/hr. [We can write 1 hours as 5/3 hours]
Required speed = 1200 x 3 km/hr = 720 km/hr.
5
E
Answer: E
Question: What is the present worth of Rs. 132 due in 2 years at 5% simple interest per annum?
Options:
A)128
B)277
C)120
D)888
E)171
Rationale: Let the present worth be Rs. x. Then,
S.I. = (132 - x)
(x * 5 * 2) / 100 = (132 - x)
10x = 13200 - 100x
110x = 13200 => x = 120.
Answer: C
Answer: C
Question: A man can row downstream at the rate of 26 Kmph and upstream at 12 Kmph. Find the man’s rate in still water and rate of current?
Options:
A)8.5
B)8.0
C)9.5
D)7.0
E)8.25
Rationale: Rate of still water = 1/2 (down stream + upstream)
= 1/2 (26 + 12) = 19 Kmph
rate of current = 1/2 (down stream - upstream)
= 1/2 (26 - 12)
= 1/2 (14)
= 7 kmph
Answer is D.
Answer: D
Question: A car is running at a speed of 120kmph. What distance will it cover in 12sec?
Options:
A)396m
B)267m
C)180m
D)200m
E)250m
Rationale: Speed = 120kmph = 120*5/18 = 33 m/s
Distance covered in 12sec = 33*12 = 396m
Answer is A
Answer: A
Question: The average of 11 results is 48, if the average of first six results is 49 and that of the last six is 52. Find the sixth result?
Options:
A)21
B)56
C)78
D)25
E)23
Rationale: 1 to 11 = 11 * 48 = 528
1 to 6 = 6 * 49 = 294
6 to 11 = 6 * 52 = 312
6th = 294 + 312 â€“ 528 = 78
Answer:C
Answer: C
Question: Tom opened a shop investing Rs.30000. Jose joined him 2 months later, investing Rs.45000. They earned a profit of Rs.54000 after completion of one year. What will be Jose's share of profit?
Options:
A)12000
B)23000
C)24000
D)25000
E)30000
Rationale: SoL = ~s- So Anju’s share = [5/9] x 54000 = 30000
E
Answer: E
Question: A trader mixes 25 kg of rice at Rs. 20 per kg with 25 kg of rice of other variety at Rs. 30 per kg and sells the mixture at Rs. 30 per kg. His profit percent is:
Options:
A)No profit, no loss
B)5%
C)8%
D)20%
E)None of these
Rationale: C.P. of 50 kg rice = Rs. (25 x 20 + 25 x 30) = Rs. (500 + 750) = Rs. 1250
S.P. of 50 kg rice = Rs. (50 x 30) = Rs. 1500
Gain =250/1250x100% = 20%.
Answer: Option D
Answer: D
Question: RAM and GOPAL invested in a business. They earned some profit which they divided in the ratio of 1:2. If RAM invested Rs.25, 000, the amount invested by GOPAL is:
Options:
A)Rs. 65,000
B)Rs. 50,000
C)Rs. 80,000
D)Rs. 90,000
E)Rs. 60,000
Rationale: Invested by Gopal = 25000/1*2 = 50000
ANSWER : B
Answer: B
Question: A company has 15 managers and 75 associates. The 15 managers have an average salary of $90,000. The 75 associates have an average salary of $30,000. What is the average salary for the company?
Options:
A)$35,000
B)$40,000
C)$55,000
D)$65,000
E)$75,000
Rationale: Another method is to get ratios
Say 30000=a and
We know the # of people are in 1:5 ratio
Average=(3a*1+a*5)/6= 8a/6=40000
Answer is B. $40,000
Answer: B
Question: Alice’s take-home pay last year was the same each month, and she saved the same fraction of her take-home pay each month. The total amount of money that she had saved at the end of the year was 2 times the amount of that portion of her monthly take-home pay that she did NOT save. If all the money that she saved last year was from her take-home pay, what fraction of her take-home pay did she save each month?
Options:
A)1/7
B)1/6
C)1/5
D)1/4
E)1/3
Rationale: Suppose the portion of her salary that she saved is X
and the remaining portion is Y
X+Y= total take home salary
given, 12X = 2Y
or 6X = Y
Total take-home salary= X+ 6X= 7X
hence, she saved 1/7-th portion of her take home salary
A
Answer: A
Question: In a fuel station the service costs $1.75 per car, every liter of fuel costs 0.65$. Assuming that a company owns 12 cars and that every fuel tank contains 30 liters and they are all empty, how much money total will it cost to fuel all cars?
Options:
A)320$
B)255$
C)420$
D)450$
E)480$
Rationale: Total Cost = ( 1.75*12 ) + ( 0.65 * 12 * 30 ) = 21 + 234 => 255
Hence answer will be (B)
Answer: B
Question: A cycle is bought for Rs.1120 and sold for Rs.1200, find the gain percent?
Options:
A)7%
B)2%
C)6%
D)1%
E)8%
Rationale: Explanation:
1120---- 80
100 ---- ? => 7%
Answer: A
Answer: A
Question: A man in train notice that he can count 51 telephone posts in 1 minute. If they are known to be 80 meters apart,then at what speed is the train traveling?
Options:
A)100km/hr
B)110km/hr
C)240km/hr
D)150km/hr
E)200km/hr
Rationale: Number of gaps between 51 telephone posts = 50
Distance traveled in 1 minute = 50*80 = 4000m = 4km
Speed = 60*4=240km/hr
Answer is C
Answer: C
Question: A certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 2 of 9 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)252
Rationale: Ans: 252
7C1 * 9C2
Answer E)
Answer: E
Question: A number is doubled and 9 is added. If the resultant is trebled, it becomes 75. What is that number?
Options:
A)3.5
B)6
C)8
D)9
E)None of these
Rationale: by option
8*2=16(number is doubled)
add 9(16+9=25)
if 25 trebled it becomes 75
ANSWER:C
Answer: C
Question: A car covers a distance of 720 km in10 hrs. What is its speed in mps.
Options:
A)40 m/s
B)25 m/s
C)20 m/s
D)15 m/s
E)16 m/s
Rationale: Explanation:
Distance = Speed x Time
Speed = (720/10) = 72 kmph
Spees in mps = 72 x 5/18 = 4 x 5 = 20 mps.
Answer: Option C
Answer: C
Question: The present ages of three persons in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 56. Find their present ages (in years).
Options:
A)8, 20, 28
B)16, 28, 36
C)20, 35, 45
D)20, 45, 55
E)None of these
Rationale: Explanation:
Let their present ages be 4x, 7x and 9x years respectively.
Then, (4x - 8) + (7x - 8) + (9x - 8) = 56
⇒ 20x = 80
⇒ x = 4.
⇒ Their present ages are 4x = 16 years, 7x = 28 years and 9x = 36 years respectively.
Answer: B
Answer: B
Question: If x, y, and z are all factors of 35, which of the following must be true?
Options:
A)The product xyz > 35
B)Either x = y, y = z, or x = z
C)x – y – z = even
D)The product xyz is odd
E)yz/x is an integer
Rationale: All the factors of 35 are odd numbers.
The product of odd numbers is odd.
The answer is D.
Answer: D
Question: In how many different ways can the letters of the word ‘PREPTY’ be arranged?
Options:
A)230
B)240
C)260
D)280
E)360
Rationale: Number of ways = 61/21 6X5X4X3X2X1 - 360
E
Answer: E
Question: The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, find out the value of x
Options:
A)13
B)14
C)15
D)16
E)17
Rationale: Explanation:
Let the Cost Price of one article = Rs. 1
CP of x articles = Rs. x
CP of 20 articles = 20
Selling price of x articles = 20
Profit = 25% [Given]
⇒(SP−CP /CP)=25/ 100=1/4
⇒(20−x)/x=1/4
⇒80−4x=x
⇒5x=80
Option D
⇒x=805=16
Answer: D
Question: What is the maximum number of pieces of birthday cake of size 3” by 3” that can be cut from a cake 21” by 21”?
Options:
A)5
B)7
C)49
D)20
E)25
Rationale: The prompt is essentially asking for the maximum number of 3x3 squares that can be cut from a larger 21 by 21 square. Since each 'row' and each 'column' of the larger square can be sub-divided into 7 'pieces' each, we have (7)(7) = 49 total smaller squares (at maximum).
C
Answer: C
Question: If an item that originally sold for r dollars was marked up by p percent and then discounted by q percent, which of the following expressions represents the final price of the item?
Options:
A)(10,000r + 100r(p – q) – pqr)/10,000
B)(10,000r + 100r(q – p) – pqr)/10,000
C)(100r(p – q) – pqr)/10000
D)(100r(q – p) – pqr)/10000
E)10000 /(p – q)
Rationale: Assume r=100, p=10, q=10. So final price is 99. Now put these value each option. Only option A will give right answer.
Answer: A
Question: The product of n consecutive positive integers is always divisible by?
Options:
A)n^2 -1
B)(n+1)!
C)2n +1
D)n!
E)n^2 + 1
Rationale: Plug in some vlues and check -
Product of first 2 number is 2*1
Product of first 3 number is 3*2*1
Product of first 4 number is 4*3*2*1
Product of first 5 number is 5*4*3*2*1
So, The product of first n natural numbers is always divisible by n! , answer will be (D)
Answer: D
Question: A bag marked at $240 is sold for $120. The rate of discount is?
Options:
A)10%
B)25%
C)20%
D)50%
E)45%
Rationale: Rate of discount = 120/240 * 100 = 50%
Answer is D
Answer: D
Question: What is the sum of odd integers from 20 to 80, inclusive?
Options:
A)1,550
B)1,500
C)1,240
D)1,120
E)1,100
Rationale: Number of odd integers = (80-20)/2 + 1
= 60/2 + 1
= 31
Sum of odd integers = (20+80)/2 * 31
= 50 * 31
= 1550
Answer A
Answer: A
Question: A box contains 2 white, 3 black and 5 red balls. In how many ways can three balls be drawn from the box if at least one black ball is to be included in the draw?
Options:
A)85
B)66
C)27
D)19
E)17
Rationale: Explanation:
Total balls in a box = 2W + 3B + 5R = 10 balls.
Therefore total number of ways of drawing 3 balls from 10 balls is 10C3 = 120 ways ........(1)
This includes all types of color combinations of white, black and red.
Now total balls in a box of white and red color = 2W + 5R = 7 balls.
Therefore total number of ways of drawing 3 balls from 7 balls is 7C3 = 35 ways ........(2)
This includes all types of color combinations of white and red only.
Therefore from (1) and (2), we get the total number of ways of drawing 3 balls, which includes at least one black ball = 120 – 35 = 85 ways.
ANSWER: A
Answer: A
Question: Adam, Bryan, Chris, and David are all basketball players. An unknown habit of Adam's is to score 7 more than half as many points as Bryan. Bryan is their star player and will always score at least 50 points plus Chris' score. Chris is a consistent player in that he will score twice as many points as David. Write a general equation that shows the total score of these four players in a game in terms of David, D
Options:
A)82 +6D
B)82 +4D
C)82 +8D
D)8D
E)50+4D
Rationale: A = B/2+7
B = 50+C
C = 2D
D = D
Now compile new equations
A+B+C+D = (B/2+7) +(50+C) +(2D) +(D)
= (B/2+7) +(50+C) +3D, look back at original equations
=(B/2+7) +(50+2D) +3D
=(B/2+7) +50+5D
= (50+2D)/2+7 +50 +5D
= 25 +D +7+50+5D
=82 +6D
Answer: A
Answer: A
Question: A mixture of 70 lts. Of wine and water contains 10% of water. How much water must be added to make 37% of the resulting mixture?
Options:
A)30
B)31
C)32
D)33
E)34
Rationale: Its given that currently water is 10% of the mixture.
=> (10/100)(70) = 7lts
lets say x litres of water is added.
=> 7+x = (37/100)(70+x)
=> x = 30 liters
ANSWER:A
Answer: A