divelab/Sys2Bench huggingface dataset:
Question: Tarun bought a T.V. with 20% discount on the labelled price. Had he bought it with 25% discount, he would have saved Rs. 500. At what price did he buy the T.V.?
Options:
A)Rs. 5000
B)Rs. 10000
C)Rs. 12000
D)Rs. 16000
E)None of these
Rationale: Let the labeled price be Rs. 100.
S.P. in 1st case=Rs. 80, S.P. in 2nd case=Rs. 75.
5 : 100 : : 500 : x.
So, x=[(100X500)/5]=10000.
i.e Required CP = 80 % of 10000 = 8000
ANSWER:E
Answer: E
Question: The smallest 3 digit prime number is?
Options:
A)99
B)101
C)105
D)107
E)109
Rationale: The smallest 3-digit number is 100, which is divisible by 2.
100 is not a prime number.
101 < 11 and 101 is not divisible by any of the prime numbers 2, 3, 5, 7, 11.
101 is a prime number.
Hence 101 is the smallest 3-digit prime number.
B)
Answer: B
Question: The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:
Options:
A)276
B)299
C)322
D)345
E)395
Rationale: Clearly, the numbers are (23 x 13) and (23 x 14).
Larger number = (23 x 14) = 322.
Answer: Option C
Answer: C
Question: A committee that includes 10 members is about to be divided into 2 subcommittees with 5 members each. On what percent of the possible subcommittees that Michael is a member of is Annie also a member?
Options:
A)48%
B)44%
C)40%
D)36%
E)32%
Rationale: Michael is a member of one of the subcommittees.
P(Annie is on the same subcommittee) = 4/9 which is about 44%.
The answer is B.
Answer: B
Question: Find the average marks of all the students in 2 separate classes, if the average marks of students in the first class of 55 students is 60 and that of another class of 48 students is 58
Options:
A)55.1
B)51.1
C)53.1
D)52.1
E)59.1
Rationale: Sum of the marks for the class of 55 students = 55 * 60 = 3300
Sum of the marks for the class of 48 students = 48 * 58 = 2784
Sum of the marks for the class of 103 students =
3300 + 2784 = 6084
Average marks of all the students = 6084/103
= 59.1
Answer:E
Answer: E
Question: A and B together can complete a work in 12 days. A alone complete it in 20 days. If B does the work only for half a day daily, then in how many days A and B together will complete the work?
Options:
A)110 days
B)11 days
C)15 days
D)20 days
E)17 days
Rationale: B's one day work= (1/12 - 1/20) = 1/30.
Now, (A+B)'s one day work = (1/20+1/60) = 1/15. [∵ B work for half day only].
So, A and B together will complete the work in 15 days.
CORRECT ANSWER: "C".
Answer: C
Question: Priya purchased n items from a catalog for $9 each. Postage and handling charges consisted of $6 for the first item and $1 for each additional item. Which of the following gives the total dollar amount for Kim’s purchase, including postage and handling, in terms of n ?
Options:
A)10n + 2
B)10n + 4
C)19n + 2
D)19n + 3
E)8n + 4
Rationale: its B
n items for $9 each
total price $9n
postage and handling of $5 for 1st item and $1 for the rest
total postage and handling = $ [5+ (n-1)]
= $ n+4
total cost
9n+n+4 = 10n+4
Answer: B
Question: Out of 5 women and 4 men, a committee of three members is to be formed in such a way that at least one member is woman. In how many different ways can this be done?
Options:
A)60
B)100
C)120
D)90
E)80
Rationale: We may have (1 woman and 2 men) or (2 women and 1 man) or (3 women).
So Required number of ways= (5C1x4C2) + (5C2x4C1) + (5C3) =(5*6) + (10*4) +(10) =80
ANSWER:E
Answer: E
Question: In a race of 1000 m, A can beat by 100 m, in a race of 800m, B can beat C by 100m. By how many meters will A beat C in a race of 600 m?
Options:
A)127.5
B)127.1
C)127.8
D)127.2
E)127.0
Rationale: When A runs 1000 m, B runs 900 m and when B runs 800 m, C runs 700 m.
When B runs 900 m, distance that C runs = (900 * 700)/800 = 6300/8 = 787.5 m.
In a race of 1000 m, A beats C by (1000 - 787.5) = 212.5 m to C.
In a race of 600 m, the number of meters by which A beats C = (600 * 212.5)/1000
= 127.5 m.
Answer: B
Answer: B
Question: If a(a - 6) = 27 and b(b - 6) = 27, where a ≠ b, then a + b =?
Options:
A)−48
B)−2
C)46
D)6
E)48
Rationale: i.e. if a =-3 then b = 9
or if a = 9 then b =-3
But in each case a+b = -3+9 = 6
Answer: option D
Answer: D
Question: A tank has 5 inlet pipes. Three pipes are narrow and two are wide. Each of the three narrow pipes works at 1/2 the rate of each of the wide pipes. All the pipes working together will take what fraction W of time taken by the two wide pipes working together to fill the tank?
Options:
A) 1/2
B) 2/3
C) 3/4
D) 3/7
E) 4/7
Rationale: I always plug in for these questions.. I find them easier
Let 2 large pipes each work fill 10lts/hr
small ones = 1/2*10 = 5
total 5 = 10+10+5+5+5 = 35 lts/hrs
now assume the total capacity = 140 lts ( LCM of 20,35)
Qn :All the pipes working together will take what fraction of time taken by the two wide pipes working together to fill the tank
all working together will take 140/35 = 4 hrs
two large pipes working will take 140/20 = 7 hrs
hence ratio W= 4/7 = E
Answer: E
Question: If the average (arithmetic mean) of a and b is 150, and the average of b and c is 90, what is the value of a − c?
Options:
A)−220
B)−100
C)120
D)135
E)It cannot be determined from the information given
Rationale: a+b = 300
b+c = 180
a-c = 120.
IMO option C.
Answer: C
Question: A starts business with a capital of Rs.1200 B and C join with some investments after 3 and 6 months respectively. If the end of a year, the profit is divided in the ratio 2:3:5 respectively. What is B's investment in the business?
Options:
A)2400
B)24087
C)2409
D)2405
E)2401
Rationale: 1200 * 12: x * 9 = 2:3
x = 2400
Answer:A
Answer: A
Question: How much interest can a person get on Rs. 45000 at 12.5% p.a. simple interest for a period of four years and six months?
Options:
A)45587.58
B)25587.52
C)35387.5
D)25312.5
E)37587.57
Rationale: I = (45000 * 12.5 * 4.5)/100
= (45000 * 25 * 9)/(100 * 2 * 2)
= Rs. 25312.50
Answer: D
Answer: D
Question: A certain function f satisfies the equation f(x)+2*f(6-x) = x for all real numbers x. The value of f(1) is
Options:
A)2
B)1
C)3
D)4
E)can't determine
Rationale: f(1)+2*f(6-1)=1......... (1)
f(5)+2*f(6-5)=5......... (2)
substituting we have (2) in (1) we have :-
-3f(1)=-9,
hence answer f(1)=3.
ANSWER:C
Answer: C
Question: Ajay can walk 4km in 1 hour. In how many hours he can walk 40km?
Options:
A)5hrs
B)10hrs
C)15hrs
D)20hrs
E)30hrs
Rationale: 1 hour he walk 4km
he walk 40km in = 40/4 * 1 = 10hours
Answer is B
Answer: B
Question: How many leading zeros (zeros after the decimal point but before the first nonzero digit of a number) are in 1/8!?
Options:
A)1
B)3
C)5
D)6
E)9
Rationale: 8! = 40320
1/100000 < 1/40320 < 1/10000
0.0001 < 1/40320 < 0.001
So 1/40320 has 5 leading zeros
Answer: C
Answer: C
Question: 108 Ã— 108 + 92 Ã— 92 = ?
Options:
A)20120
B)11500
C)20128
D)20660
E)20228
Rationale: Explanation :
(a + b)2 + (a âˆ’ b)2 = 2(a2 + b2)
(Reference : Basic Algebraic Formulas)
1082 + 922 = (100 + 8)2 + (100 âˆ’ 8)2 = 2(1002 + 82) = 2(10000 + 64) = 20128. Answer : Option C
Answer: C
Question: All of the students of the Music High School are in the band, the orchestra, or both. 80 percent of the students are in only one group. There are 105 students in the band. If 50 percent of the students are in the band only, how many students are in the orchestra only?
Options:
A)42
B)45
C)48
D)51
E)54
Rationale: 50% of students are in the band only, so 30% of students are in the orchestra only.
Since 20% of students are in both, then 70% of students are in the band which is 105 students.
The number of students in the orchestra is (30%/70%)*105 = 45.
The answer is B.
Answer: B
Question: The Shopkeeper increased the price of a product by 15% so that customer finds it difficult to purchase the required amount. But somehow the customer managed to purchase only 70% of the required amount. What is the net difference in the expenditure on that product?
Options:
A)A)12.5%
B)B)13%
C)C)11.5%
D)D)14%
E)E)15%
Rationale: Quantity X Rate = Price
1 x 1 = 1
0.7 x 1.15 = 0.805
Decrease in price = (0.115/1) × 100 = 11.5%
C)
Answer: C
Question: The median annual household income in a certain community of 21 households is $50,000. If the mean income of a household increases by 10% per year over the next 2 years, what will the median M income in the community be in 2 years?
Options:
A) $50,000
B) $60,000
C) $60,500
D) $65,000
E) Cannot be determined
Rationale: Answer is E, because there are different numbers in the set and we are not sure which side of the numbers in the set will be increased so the mean M is increase by 10%. It could be the case that small number of higher end incomes increased a little or many low end incomes increased - it cannot be identified.
Answer: E
Question: A certain class of students is being divided into teams. The class can either be divided into 16 teams with an equal number of players on each team or 24 teams with an equal number of players on each team. What is the lowest possible number of students in the class?
Options:
A)6
B)36
C)48
D)60
E)72
Rationale: Let total no of students in the class be N
So, we are told that N is divisible by both 16 24
So, Lets find the least common multiple of 16 24 , ie 48
So our answer is (C) 48
Answer: C
Question: In expressing a length 65.687 km as nearly as possible with three significant digits, find the percentage error?
Options:
A)0.019%
B)0.029%
C)0.035%
D)0.045%
E)0.023%
Rationale: Error = 65.7-65.687 = 0.013
Required percentage = (0.013/65.687)*100 = 0.019%
Answer is A
Answer: A
Question: The average weight of 20 oarsmen in a boat is increased by 2kg when one of the crew, who weight 40kg is replaced by a new man. Find the weight of the new man.
Options:
A)60kg
B)50kg
C)92kg
D)80kg
E)70kg
Rationale: Total weight increased = 2*20 = 40kg
Weight of the new man = 40+40 = 80 kg
Answer is D
Answer: D
Question: A student can select one of 11 different math book, one of 3 different chemistry book & one of 4 different science book.In how many different ways students can select book of math, chemistry & science.
Options:
A)22
B)72
C)199
D)237
E)264
Rationale: 11C1×3C1×4C1 = 11x3x4=264 ways
Answer:E
Answer: E
Question: A charity collected $2,199 from donors during the last month. If each donor gave at least $20, what is the greatest possible number of donors the charity could have?
Options:
A)104
B)100
C)107
D)109
E)101
Rationale: Min fund = 20
Total = 2199
The closest multiple of 20 to 2199 is : 108
Which implies 108 people gave 20 and 1 person gave 39
Ans. D
Answer: D
Question: A batsman scored 125 runs which included 5 boundaries and 5 sixes. What percent of his total score did he make by running between the wickets?
Options:
A)45%
B)500/11
C)600/11
D)55%
E)60%
Rationale: Explanation :
Number of runs made by running,
=>125−(5×4+5×6).
=>125-50
=>75
Hence, the required percentage is:-
=>75/125*100
=>60%
Answer : E
Answer: E
Question: Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 a.m. and travels towards B at 20 kmph. Another train starts from B at 8 a.m. and travels towards A at a speed of 25 kmph. At what time will they meet?
Options:
A)9 a.m.
B)10 a.m.
C)10.30 a.m.
D)11 a.m.
E)12 a.m.
Rationale: Given that distance between two stations is 110km.
First trains speed is 20kmph and starts at 7 am
2nd trains speeed is 25kmph and starts at 8 am.
So at 8 am, the distance covered by first train is 20 kms.
Now remaining distance to cover is (110-20 = 90 km).
The relative speed of both the trains is 20+45 = 45 kms.
So it will take 2 hours to cover 90 km.
Hence the trains will meet after 2 hours from 8am i.e. at 10 am.
ANSWER:B
Answer: B
Question: Three men sit at a table. One dollar is placed in front of each of them. When a buzzer sounds, the first man to touch the money in front of him keeps the money that is in front of him, while the other two receive nothing. If none of the three men touches the money in front of them, then the amount of money doubles for the next round. After four rounds of this activity, what is the maximum amount of money that any one man could receive?
Options:
A)$4
B)$5
C)$8
D)$15
E)$16
Rationale: The Optimized scenario
After 1st Round - $1. Let nobody touches the amount will double in next round.
After 2nd Round - $2. Let nobody touches the amount will double in next round.
After 3th Round - $4. Let Man A touches it and get $4
After 4th Round - $4. Let the same Man A touches it and get another$4.
So, after 4th round, the maximum amount a man receives is $8.
Answer (C)
Answer: C
Question: Mr. SIMPSON invested in Fund X and Fund Y. The total amount she invested, in both funds combined, was $200,000. In one year, Fund X paid 33% and Fund Y paid 18%. The interest earned in Fund Y was exactly $300 greater than the interest earned in Fund X. How much did Ms. Tom invest in Fund X?
Options:
A)50,000
B)60,000
C)70,000
D)100,000
E)80,000
Rationale: X+Y=200,000
0.18Y=0.33A+300
Take away decimals first: 18Y+33X+30,000
Isolate first equation to solve for X (your goal): Y=200,000-Y
Plug in for B:
18(200,000-X)=33X+30,000
3,600,000-18X=33X+30,000
3,570,000=51X
3,570,000/51=X
X=70,000=Answer Choice C
Answer: C
Question: The profit earned by selling an article for Rs. 832 is equal to the loss incurred when the same article is sold for Rs. 448. What should be the sale price for making 60% profit?
Options:
A)1024
B)960
C)277
D)266
E)121
Rationale: Let C.P. = Rs. x.
Then, 832 - x = x - 448
2x = 1280 => x = 640
Required S.P. = 160% of Rs. 640 = 160/100 * 640 = Rs. 1024.
Answer: A
Answer: A
Question: If 10 crates have 97 apples each and how many apples more is required in order to make 100 apples in each crate?
Options:
A)30
B)10
C)15
D)20
E)0
Rationale: Each crate requires 3 apples and totally there are 10 crates so required apples = 10 * 3 = 30
Answer: A
Answer: A
Question: A woman sold 100 oranges at $12.10, some at the rate of 3 for 35 cents and the rest at 4 for 85 cents. How many were sold at the first rate?
Options:
A)45
B)21
C)9
D)15
E)12
Rationale: This can be solved like a classical mixture problem but numbers are awkward to deal with.
It's easier to just look at the answer choices. You know that a multiple of 3 oranges has to be sold at the first rate, and a multiple of 4 at the second rate. You simple subtract the answer choices for the first rate from 100 and check whether the remainder (i.e. the number of oranges sold at the second rate) is a multiple of 4.
100 - 45 = 55 => not a multiple of 4 so exclude
100 - 21 = 79 => not a multiple of 4 so
100 -9 = 91 => nota multiple of 4 so exclude
100 - 15 = 85 => not a multiple of 4 so exclude
100 - 12 = 88 => a multiple of 4 so keep
Hence, answer choice 12 is correct.
E
Answer: E
Question: A mother is planning a party for her child and buys some snacks. She buys 3 bags of chips, 1 bag of pretzels, and 2 packages of cookies. She pays $3.00 per bag of chips, $2.50 per bag of pretzels, and $4.00 per package of cookies. What is the average price per snack the mother paid?
Options:
A)$2.25
B)$3.25
C)$2.00
D)$3.50
E)$3.00
Rationale: Explanation:
3 1 2
$3.00 $2.50 $4.00
$9.00 + $2.50 + $8.00 = $19.50/6 = $3.25
B
Answer: B
Question: Find the number of ways a batsman can score a double century only in terms of 4's & 6's?
Assume the batsman scored x 4's and y 6's.
Options:
A)22
B)16
C)77
D)91
E)13
Rationale: 4x + 6y = 200 ⇒2x+3y=100⇒2x+3y=100 ⇒x=100−3y2=50−32y⇒x=100−3y2=50−32y
As x is an integer, y should be a multiple of 2.
If
y = 0, x = 50
y = 2, x = 47
y = 4, x = 44
...
y = 32, x = 2
So total ways are (32-0)/2 + 1 = 17 ( if 0 6's are possible) otherwise 16
Answer:B
Answer: B
Question: A envelop weight 8.5 gm,if 820 of these envelop are sent with an advertisement mail. How much wieght?
Options:
A)6.6kg
B)6.8kg
C)6.7kg
D)6.97kg
E)7.8kg
Rationale: 820*8.5
6970.0gm
6.97kg
ANSWER:D
Answer: D
Question: Spheres A and B have their radil 40 cm and 10 cm respectively. The ratio of the surface area of A to the surface area of B is
Options:
A)1 : 4
B)1 : 16
C)4 : 1
D)16 : 1
E)None
Rationale: Solution
Let the radil of A and B be r and R respectively.
Required ratio	= 4Πr2/ 4ΠR2
‹=› r2 / R2
‹=› (r/R)2
‹=›(40/10)2
‹=› 16 : 1.
Answer D
Answer: D
Question: Fence X is twice as long as fence Y, and fence Y is 4 feet shorter than fence Z. If 3 feet were added to each fence, which of the following must be true?
I) X is twice as long as Y.
II) Y is 2 feet shorter than Z.
III) X is longer than Z.
Options:
A)I only
B)II only
C)III only
D)I and II
E)II and III
Rationale: If X= 6
Y= 3
Z= 7
Then, after adding 3 feet to each:
X= 9
Y=6
Z=10
And the answer would be B, which is wrong.
Answer: B
Question: There are ten 50 paise coins placed on a table. Six of these show tails and four show heads. A coin is chosen at random and flipped over (not tossed). This operation is performed seven times. One of the coins is then covered. Of the remaining nine coins, five show tails and four show heads. The covered coin shows
Options:
A)a head
B)a tail
C)more likely a head
D)more likely a tail
E)None
Rationale: Explanation :
H-head T-tail
10 coins – 6T and 4 H -initial
9 coins - 5T and 4H -final
7 coins can be flipped for 6T + 1 H resulting in 1T and 9H. Not possible
7 coins can be flipped for 3T + 4 H resulting in 7T and 3 H. Not possible
7 coins can be flipped for 4T + 3 H resulting in 5T and 5 H. Possible
7 coins can be flipped for 5T + 2 H resulting in 3T and 7 H. Not possilbe
Answer : A
Answer: A
Question: A solution of 80 litres contains oil and water in the ratio 10:x. If five litres of water is added to the solution, the ratio becomes 20:10, find the value of x?
Options:
A)4.1
B)4.9
C)4.7
D)5.7
E)8.5
Rationale: Exp. Total new quantity = original sol + water =80+5 =85
New ratio = 20:10, New quantity of oil =20/30*85 = 56.6 Lit,
New quantity of water = 10/30*85 =28.3 Lit
Water present initially = (28.3-5) = 23.3 Lit
Ratio = 56.6/23.3 =10/x There for x = 4.11
Answer: A
Answer: A
Question: A wheel that has 6 cogs is meshed with a larger wheel of 21 cogs. When the smaller wheel has made 21 revolutions, then the number of revolutions mad by the larger wheel is:
Options:
A)49
B)4
C)12
D)6
E)27
Rationale: Let the required number of revolutions made by larger wheel be x.
Then, More cogs, Less revolutions (Indirect Proportion)
21 : 6 :: 21 : x 21 * x = 6 x 21
x =	(6 x 21)/21
x = 6
Answer is D.
Answer: D
Question: howmany 3 digit numbers are there tens digit place is more than hundreds digit place and units place less than hundreds digit place?
Options:
A)80
B)100
C)110
D)120
E)140
Rationale: we have numbers {0,1,2,3,4,5,6,7,8,9}
now we would decide the cases on behalf of value of hundred place digit. and accordingly choose unit's digit and ten's digit.
value of hundred place digit must be choosen in such a manner that there should exist a greater digit for ten's place and a smaller digit for unit place.
thus we can have values of hundred's digit from 1,2,3,4,5,6,7,8 only
now starting from hundred's digit=8 :
hundred's digit possible values for ten's digit possible values for unit digit total ways
8 9 0,1,2,3,4,5,6,7 1*1*8=8
7 8,9 0,1,2,3,4,5,6 1*2*7=14
6 7,8,9 0,1,2,3,4,5 1*3*6=18
5 6,7,8,9 0,1,2,3,4 1*4*5=20
4 5,6,7,8,9 0,1,2,3 1*5*4=20
3 4,5,6,7,8,9 0.1.2 1*6*3=18
2 3,4,5,6,7,8,9 0,1 1*7*2=14
1 2,3,4,5,6,7,8,9 0 1*8*1=8
so, total no of cases=8+14+18+20+20+18+14+8 =120
hence total no of possible 3 digit numbers = 120
ANSWER:D
Answer: D
Question: If n is a positive integer, which of the following expressions must be even?
Options:
A)(n − 1)( n + 7)
B)(n − 2)( n + 7)
C)(n − 2)( n + 10)
D)(n − 3)( n + 7)
E)(n − 3)( n + 11)
Rationale: Whether n is even or odd, (n-2)(n+7) will have one odd factor and one even factor.
The product will be even.
The answer is B.
Answer: B
Question: What is the units digit of (3^13)(4^13)?
Options:
A)2
B)4
C)6
D)7
E)8
Rationale: -> The ones place of (~3)^n repeats after 4 times like 39713.
The ones place of (~4)^n repeats after 2 times like 464.
Then, 3^13=3^4*3+13^1.,
4^13=4^2*6+1=4^1=~4
which is (3^13)(4^13) (3^1)(~4)=(~3)(~4)=~2. Therefore, the answer is A.
Answer: A
Question: Two goods train each 500 m long, are running in opposite directions on parallel tracks. Their speeds are 45 km/hr and 30 km/hr respectively Find the time taken by the slower train to pass the driver of the faster one.
Options:
A)12 sec
B)24 sec
C)48 sec
D)60 sec
E)None
Rationale: Solution
Relative speed	= (45 + 30 )km/hr
= (75 x 5/18)m/sec
=(125 / 6) m/sec
Total Distance covered	= (500 + 500) m
= 1000 m
Required time	= (1000 x 6 /125)sec
= 48 sec
Answer C
Answer: C
Question: All the milk in container A which was filled to its brim was poured into two containers B and C. The quantity of milk in container B was 62.5% less than the capacity of container A. If 148 liters was now transferred from C to B, then both the containers would have equal quantities of milk. What was the initial quantity of milk in container A?
Options:
A)1184
B)1723
C)1129
D)2613
E)1372
Rationale: A
1184
B has 62.5% or (5/8) of the milk in A. Therefore, let the quantity of milk in container A(initially) be 8k.
Quantity of milk in B = 8k - 5k = 3k.
Quantity of milk in container C = 8k - 3k = 5k
Container: A B C
Quantity of milk: 8k 3k 5k
It is given that if 148 liters was transferred from container C to container B, then both the containers would have equal quantities of milk.
5k - 148 = 3k + 148 => 2k = 296 => k = 148
The initial quantity of milk in A = 8k = 8 * 148 = 1184 liters.
Answer: A
Question: How many seconds will a 770 m long train take to cross a man walking with a speed of 2.5 km/hr in the direction of the moving train if the speed of the train is 46.5 km/hr?
Options:
A)63
B)88
C)66
D)44
E)35
Rationale: Speed of train relative to man = 46.5 - 2.5 = 44 km/hr.
= 44 * 5/18 = 110/9 m/sec.
Time taken to pass the man = 770 * 9/110 = 63 sec.
Answer: A
Answer: A
Question: At a blind taste competition a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?
Options:
A)1/12
B)5/14
C)4/9
D)1/2
E)2/3
Rationale: "The probability that a contestant does not taste all of the samples" means that contestant tastes only 2 samples of tea (one sample is not possible as contestant tastes 4 cups>3 of each kind).
C23∗C46C49=514C32∗C64C94=514.
C23C32 - # of ways to choose which 2 samples will be tasted;
C46C64 - # of ways to choose 4 cups out of 6 cups of two samples (2 samples*3 cups each = 6 cups);
C49C94 - total # of ways to choose 4 cups out of 9.
Answer: B
Answer: B
Question: In a garment industry, 12 men working 8 hours per day complete a piece of work in 10 days. To complete the same work in 8 days, working 13.33 hours a day, the number of men required is:
Options:
A)4
B)5
C)6
D)8
E)9
Rationale: Explanation :
Let the required number of men be x.
Less days, More men (Indirect Proportion)
More working hrs per day, Less men (Indirect Proportion)
Days 8 : 10
Working Hrs 13.33 : 8 :: 12 : x
=> 8 x 13.33 x x = 10 x 8 x 12
=> x = 10 x 8 x 12 / (8 x 13.33)
=> x = 9
Answer : E
Answer: E
Question: If 9x-3y=12 and 3x-5y=7 then 6x-2y = ?
Options:
A)-5
B)4
C)2
D)8
E)6
Rationale: 9x-3y=12-----eq(1)
3x-5y=7------eq(2)
solving we get x=13/12 and y=-3/4
so, 6x-2y=6*(13/11)-2(-3/4)=8
ANSWER:D
Answer: D
Question: In Solution P, the proportion of water to alcohol is 4:1. In Solution Q, the proportion of water to alcohol is 2:3. If the two solutions are combined, what is the concentration of alcohol in the new solution if the original solutions were mixed in equal amounts?
Options:
A)40%
B)48%
C)50%
D)56%
E)60%
Rationale: Let V be the total volume of the new solution.
The amount of alcohol added from Solution P is (1/5)*V/2=(1/10)*V
The amount of alcohol added from Solution Q is (3/5)*V/2=(3/10)*V
The total amount of alcohol in the new solution is (1/10+3/10)*V=(2/5)*V
The alcohol concentration is 2/5=40%
The answer is A.
Answer: A