divelab/Sys2Bench huggingface dataset:
Question: An equilateral triangle T2 is formed by joining the mid points of the sides of another equilateral triangle T1. A third equilateral triangle T3 is formed by joining the mid-points of T2 and this process is continued indefinitely. If each side of T1 is 60 cm, find the sum of the perimeters of all the triangles.
Options:
A)180 cm
B)220 cm
C)240 cm
D)270 cm
E)360 cm
Rationale: We have 60 for first triangle, when we join mid-points of first triangle we get the second equilateral triangle then the length of second one is 30 and continues.
So we have 60,30,15,...
We have ratio = 1/2, and it is GP type.
Sum of infinite triangle is a/1-r = 60/1-(1/2) = 120
Equilateral triangle perimeter is 3a = 3*120 = 360.
So option E.
Answer: E
Question: A single discount equivalent to the discount series of 20%, 10% and 5% is?
Options:
A)31.7
B)31.1
C)31.6
D)31.4
E)31.8
Rationale: 100*(80/100)*(90/100)*(95/100)
= 68.4
100 - 68.4
= 31.6
Answer: C
Answer: C
Question: Let y = 2m + x^2 and m = 5x + 4. If m^2 =81, then which of the following can be a value of 2y + 3m?
Options:
A)65
B)130
C)141
D)142
E)160
Rationale: m^2 = 81 => m=9
9= 5x+4 => x=1
y= 2*9+1^2 = 19
2y+3m = 2*19+3*9 = 65
Answer : A
Answer: A
Question: A sells a bicycle to B and makes a profit of 60%. B sells the same bicycle to C at a profit of 25%. If the final S.P. of the bicycle was Rs. 225, find out the cost price of the bicycle for A.
Options:
A)237
B)126
C)971
D)611
E)112.5
Rationale: Explanation:
Let CP be 100
A sells at 60% profit so SP=160
B sells at 25% profit = 160 x (1+25/100) = 200
CP SP
100--- 200
x --- 225
CP = 225 x 100/200 = 112.5
ANSWER: E
Answer: E
Question: The ages of Patrick and Michael are in the ratio of 3 : 5 and that of Michael and Monica are in the ratio of 3 : 5. If the sum of their ages is 146, what is the difference between the ages of Patrick and Monica?
Options:
A)27
B)48
C)45
D)72
E)18
Rationale: Ages of P and Mi=3x+5x
Ages of Mi and Mo=3x:5x
Rationalizing their ages. ratio of their ages will be 9x:15x:25x
Sum=47x=146
x=3
Difference if ages of Pa and Mo=25x-9x=16x=16*3=48
Answer B
Answer: B
Question: At a certain conference, 150 students spoke one or more of the following languages: Spanish, English, and French. If 51 students spoke Spanish, 90 spoke English, 65 spoke French, and 32 spoke only 2 languages, how many students spoke three different languages?
Options:
A)7
B)9
C)12
D)15
E)18
Rationale: 150=51+90+65-32-2x
x=12
C. 12
Answer: C
Question: From the digits 2, 3, 4, 5, 6 and 7, how many 5-digit numbers can be formed that have distinct digits and are multiples of 12?
Options:
A)36
B)60
C)84
D)72
E)86
Rationale: DETAILED SOLUTION
Any multiple of 12 should be a multiple of 4 and 3. First, let us look at the constraint for a number being a multiple of 3. Sum of the digits should be a multiple of 3. Sum of all numbers from 2 to 7 is 27. So, if we have to drop a digit and still retain a multiple of 3, we should drop either 3 or 6.
So, the possible 5 digits are 2, 4, 5, 6, 7 or 2, 3, 4, 5, 7.
When the digits are 2, 4, 5, 6, 7. the last two digits possible for the number to be a multiple of 4 are 24, 64, 52, 72, 56, 76. For each of these combinations, there are 6 different numbers possible. So, with this set of 5 digits we can have 36 different numbers.
When the digits are 2, 3, 4, 5, 7. The last two digits possible for the number to be a multiple of 4 are 32, 52, 72, 24. For each of these combinations, there are 6 different numbers possible. So, with this set of 5 digits we can have 24 different numbers.
Overall, there are 60 different 5-digit numbers possible.
Correct Answer: B
Answer: B
Question: David invested certain amount in three different schemes. A, B and C with the rate of interest 10% p.a., 12% p.a. and 15% p.a. respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in scheme C was 150% of the amount invested in scheme A and 240% of the amount invested in scheme B, what was the amount invested in scheme B?
Options:
A)5000
B)5007
C)5006
D)5003
E)5009
Rationale: Let x, y and z be the amount invested in schemes A, B and C respectively. Then,
(x * 10 * 1)/100 + (y * 12 * 1)/100 + (z * 15 * 1)/100 = 3200
10x + 12y + 15z = 320000
Now, z = 240% of y = 12/5 y
And, z = 150% of x = 3/2 x
x = 2/3 z = ( 2/3 * 12/5) y = 8/5 y
16y + 12y + 36y = 320000
y = 5000
Sum invested in scheme B = Rs. 5000.
Answer:A
Answer: A
Question: A rope of which a calf is tied is increased from 12 m to 23 m, how much additional grassy ground shall it graze?
Options:
A)1218
B)1212
C)1210
D)277
E)291
Rationale: π (232 – 122) = 1210
Answer: C
Answer: C
Question: In ΔPQS above, if PQ =5 and PS = 6, then
Options:
A) 9/4
B) 12/5
C) 16/5
D) 30/7
E)50/3
Rationale: there are two ways to calculate area of PQS. area remains same, so both are equal.
5*6/2 = PR*7/2
PR = 30/7
D
Answer: D
Question: P and Q can do a work in 30 days. Q and R can do the same work in 24 days and R and P in 20 days. They started the work together, but Q and R left after 12 days. How many days more will P take to finish the work?
Options:
A)12
B)15
C)18
D)20
E)22
Rationale: Let work done by P in 1 day = p,
Work done by Q in 1 day = q,
Work done by R in 1 day = r
p + q = 1/30
q + r = 1/24
r + p = 1/20
Adding all the above, 2p + 2q + 2r = 1/30 + 1/24+ 1/20 = 15/120 = 1/8
=> p + q + r = 1/16
=> Work done by P,Q and R in 1 day = 1/16
Work done by P, Q and R in 12 days = 12 × (1/16) = 12/16 = 3/4
Remaining work = 1 = 3/4 = 1/4
Work done by P in 1 day = Work done by P,Q and R in 1 day - Work done by Q and R in 1 day
= 1/16 – 1/24 = 1/48
Number of days P needs to work to complete the remaining work = (1/4) / (1/48) = 12
Option A
Answer: A
Question: Some of the 50% solution of acid was removed and this was replaced with an equal amount of 10% solution of acid. As a result, a 20% solution of acid was obtained. What fraction of the original solution was replaced?
Options:
A)1/2
B)2/3
C)3/4
D)3/5
E)4/5
Rationale: Let x be the fraction of the original solution that was replaced.
0.5*(1-x)+0.1(x)=0.2
0.4x=0.3
x=3/4
The answer is C.
Answer: C
Question: Set C consists of the following unique integers: -2, 17, 3, n, 2, 15, -3, and -27; which of the following could be the median of set C?
Options:
A)5
B)6
C)7
D)1
E)766
Rationale: Whenever a question asks you to deal with the MEDIAN of a group of numbers, you MUST put the numbers in order from least to greatest (or at the very least,group offthe numbers so that you can restrict the potential value of the median).
Here, we have 8 values (one of the values is the variable N). We're told that the values are UNIQUE, so N CANNOT be any of the other 7 values listed. We're asked which of the following answers COULD be the median, so if we determine that one of the answer choices is a POSSIBLE median, then we can stop working....
Since the group includes 8 values, the median will be the AVERAGE of the 4th and 5th values...
Putting the 7 numbers in order, we have:
-27, -3, -2, 2, 3, 15, 17
This means that the2will either be the 4th term or the 5th term (depending on the value of N).
IF....
N is REALLY BIG, then the 4th and 5th terms will be 2 and 3, so the median will = 2.5
N is REALLY SMALL, then the 4th and 5th terms will be -2 and 2, so the median will = 0
Neither of those answers is among the choices though, so N must be something relativelycloseto 2.....
IF....
N = 0, then the 4th and 5th terms will be 0 and 2, so the median will = 1. That answer IS among the choices, so we're done.
D
Answer: D
Question: Printer P can print one lakh books in 8 hours. Printer Q can print the same number of books in 10 hours while Printer R can print the same in 12 hours. All the Printers started printing at 9 A.M. Printer P is stopped at 11 A.M. and the remaining 2printers complete work. Approximately at what time will the printing of 1 lac books be completed?
Options:
A)10 pm
B)1 pm
C)12 pm
D)2 pm
E)3 pm
Rationale: Work done by P in 1 hour = 1/8
Work done by Q in 1 hour = 1/10
Work done by R in 1 hour = 1/12
Work done by P,Q and R in 1 hour = 1/8 + 1/10 + 1/12 = 37/120
Work done by Q and R in 1 hour = 1/10 + 1/12 = 22/120 = 11/60
From 9 am to 11 am, all the Printer were operating.
Ie, they all operated for 2 hours and work completed = 2 × (37/120) = 37/60
Pending work = 1- 37/60 = 23/60
Hours taken by Q an R to complete the pending work = (23/60) / (11/60) = 23/11
which is approximately equal to 2
Hence the work will be completed approximately 2 hours after 11 am ; ie around 1 pm
B
Answer: B
Question: How many times the keys of a writter have to be pressed in order to write first 400 counting no.'s?
Options:
A)998
B)1000
C)1092 times
D)1100
E)1200
Rationale: 1 to 9 = 9 * 1 = 9
10 to 99 = 90 * 2 = 180
100 to 400 = 301 * 3 = 903
-----------
1092
C
Answer: C
Question: Three types of tea the a,b,c costs Rs. 95/kg,100/kg and70/kg respectively.How many kgs of each should be blended to produce 100 kg of mixture worth Rs.90/kg, given that the quantities of band c are equal
Options:
A)70,15,15
B)50,25,25
C)60,20,20
D)40,30,30
E)20,30,30
Rationale: given quantities of b and c are equal .
therefore instead of considering them as a different quantities take average of both and consider it as a single entity .
so the cost of the mixture "D" (equal quantities of "B" and "C") is 170/2=85rs/kg
now the tea contains only "A" of 95 rs/per and "D" of 85 rs/kg.
it is clearly seen that "A" is 5 more and "D" is 5 less than the required cost per kg .
therefore 50/50 will go well .
A=50kg and D=50kg. (D= B&C in equal ratio). therefore B&C=25kg each .
(A:B:C)=(50,25,25)
ANSWER:B
Answer: B
Question: There are four members in a family. Their average income is Rs.10000 per month. Three members get Rs.8000, Rs.15000, Rs.6000 respectively. Therefore what is the income of the fourth member of the family?
Options:
A)Rs. 10000
B)Rs. 11000
C)Rs. 12000
D)Rs. 9000
E)Rs. 8500
Rationale: EXPLANATION
Total income of 3 members = Rs. (8000 + 15000 + 6000) = Rs. 29000.
Required income = Rs. [ (10000 x 4) – 29000 ]
= Rs. (40000 – 29000)
= Rs. 11000.
Answer B
Answer: B
Question: The cost price of 15 articles is equal to the selling price of 11 articles. Find the profit percent?
Options:
A)18 2/91%
B)36 4/11%
C)18 9/11%
D)18 2/91%
E)12 2/11%
Rationale: 15 CP = 11 SP
11 --- 2 CP
100 --- ? =>36 4/11%
Answer:B
Answer: B
Question: How many different number-plates for cars can be made if each number-plate contains four of the digits 0 to 9 followed by a letter A to Z, assuming that
(a) no repetition of digits is allowed?
Options:
A)131040
B)116700
C)142369
D)147225
E)132547
Rationale: we want 4 digit of cars and repetition is not allowed
1st digit can be fill with 0 to 9 i.e = 10
2nd digit can be fill with 0 to 9 except earlier selected i.e = 9
3rd digit can be fill with 0 to 9 except earlier selected i.e = 8
4th digit can be fill with 0 to 9 except earlier selected i.e = 7
No of ways = 10 * 9 * 8 * 7 * (total aplphabets)
= 5040 * 26
= 1,31,040
Ans - A
Answer: A
Question: During a two-week period, the price of an ounce of silver increased by 30 percent by the end of the first week and then decreased by 20 percent of this new price by the end of the second week. If the price of silver was x dollars per ounce at the beginning of the two-week period, what was the price, in dollars per ounce, by the end of the period?
Options:
A)1.16x
B)1.13x
C)1.1x
D)1.07x
E)1.04x
Rationale: The price at the end is 0.8(1.3x)=1.04x
The answer is E.
Answer: E
Question: Two taps can separately fill a cistern 10 minutes and 15 minutes respectively and when the waste pipe is open, they can together fill it in 9 minutes. The waste pipe can empty the full cistern in?
Options:
A)18
B)17
C)16
D)19
E)15
Rationale: 1/10 + 1/15 - 1/x = 1/9
x = 18
Answer:A
Answer: A
Question: If john makes a contribution to a charity fund at school, the average contribution size will increase by 50% reaching $75 per person. If there were 3 other contributions made before john's, what is the size of his donation?
Options:
A)$100
B)$150
C)$200
D)$250
E)$450
Rationale: Cavg = average contribution before John
Cavg*1.5 = 75, therefore the average cont is $50 before John.
If he needs to increase the average contribution by $25, he must put in $25 for each of the 3 people.
so $75.
But, he also has to put in the average for himself (the fourth person), so add $75.
So $150 is your answer.
Answer B
Answer: B
Question: The average weight of 5 students decreases by 12 kg when one of them weighing 72 kg is replaced by a new student. The weight of the student is
Options:
A)12 kg
B)10 kg
C)20 kg
D)72 kg
E)None of these
Rationale: Explanation :
Let the weight of student be x kg.
Given, Difference in average weight = 12 kg
=> (72-x)/5 = 12
=> x = 12
Answer : A
Answer: A
Question: A container has a capacity of 20 gallons and is full of spirit. 4 gallons of spirit is drawn out and the container is again filled with water. This process is repeated 4 times. Find how much spirit is left in the resulting mixture finally?
Options:
A)8(257/525) gallons
B)8(346/625) gallons
C)8.192 gallons
D)8.25 gallons
E)None of these
Rationale: There is a direct formula to calculate such repetition:
Left QTY / Orig QTY = ( (Orig QTY - Rep QTY) / Orig QTY )^n
where n is number of times mixing done:
Left QTY / 20 = (20 -4)^4 / 20^4
Left QTY = 20 (16/20)^4
Left QTY = 20 ( 4/5 )^4
Left QTY = 8.192
Answer is C
Answer: C
Question: Find the cost of fencing around a circular field of diameter 28 m at the rate of Rs.1.50 a meter?
Options:
A)289
B)132
C)279
D)277
E)278
Rationale: 2 * 22/7 * 14 = 88
88 * 1 1/2
= Rs.132
Answer: B
Answer: B
Question: If the price of an article decreases by 111/9 % and the sale of the article increases by 12 ½ %, what is the net effect on revenue?
Options:
A)1% loss
B)1% gain
C)No loss or no gain
D)Cannot be determined
E)None of these
Rationale: M. F 1 = 100-100/9/100 =800/900 = 8/9
M. F 2 = 100+25/2/100 =225/200 =9/8
Total M. F =8/9*9/8 = 1
Overall % change = ( M. F – 1 ) * 100 = ( 1 – 1 ) * 100 = 0
ANSWER:C
Answer: C
Question: A train 280 m long, running with a speed of 63 km / hr will pass a tree in
Options:
A)12 sec
B)15 sec
C)16 sec
D)20 sec
E)None
Rationale: Sol.
Speed = (63 x 5/18) m/sec.
= 35 /2 m / sec.
Time taken = (280 x 2/35) sec = 16 sec
Answer C
Answer: C
Question: A salesman travels a distance of 50 km in 2 hours and 30 minutes. How much faster in kilometer per hour, on an average, must he travel to make such trip in 5/6 hour less time?
Options:
A)10km/hr
B)15km/hr
C)20km/hr
D)30km/hr
E)25km/hr
Rationale: Time required = 2 hrs 30 min - 50 min = 1 hr 40 min = 1 2/3 hrs
Required speed = 50*3/5 = 30km/hr
Original speed = 50*2/5 = 20 km/hr
Difference = 30-20 = 10 km/hr
Answer is A
Answer: A
Question: John tossed a fair coin 2 times. What is the probability that the coin landed heads up exactly twice?
Options:
A)0.5
B)0.55
C)0.6
D)0.65
E)0.7
Rationale: The probability is exactly same as the cion landed tail up once. That is 2*(1/2)^2 = 2/4 =0.5.
The answer is, therefore, (A).
Answer: A
Question: 0.002 x 0.5 = ?
Options:
A)0.0001
B)0.001
C)0.01
D)0.1
E)1.0
Rationale: 2 x 5 = 10.
Sum of decimal places = 4
0.002 x 0.5 = 0.001
Answer: Option B
Answer: B
Question: Andy completes a piece of work in 10 days, Raju completes the same work in 40 days. If both of them work together, then the number of days required to complete the work is
Options:
A)6 days
B)8 days
C)9 days
D)10 days
E)11 days
Rationale: If A 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 = 10 × 40/50 = 8 days.
B
Answer: B
Question: The price of commodity X increases by 40 paise every year, while the price of commodityY increases by 15 paise every year. If in 2001, the price of commodity X was Rs. 4.20 and that of Y was Rs. 6.30, in which year commodity X will cost 40 paise more than the commodity Y ?
Options:
A)2010
B)2011
C)2012
D)2013
E)None
Rationale: EXPLANATION
Suppose commodity X will cost 40 paise more than Y after z years.
Then, (4.20 + 0.40z) – (6.30 + 0.15z) = 0.40
0.25z = 0.40 + 2.10
x =2.50/ 0.25 =250/25 =10
X will cost 40 paise more than Y 10 years after 2001 i.e., 2011.
Answer B
Answer: B
Question: Let's find out how sharp your mind is. Let's see if you can find the missing numbers in the series:
192, 021, 222, 324, 252, 627, 2__, 9__?
Options:
A)281 and 930.
B)282 and 930.
C)283 and 930.
D)None
E)Cannot be determined
Rationale: Solution:
If you look closely, you will find a sequence in every alternating number.
192, 222 and 252 form one series where 30 is added each time.
021, 324 and 627 form one series where 303 is added each time.
Thus the next two numbers will be 282 and 930.
Answer B
Answer: B
Question: A watch was sold at a loss of 21%. If it was sold for Rs.140 more, there would have been a gain of 4%. What is the cost price?
Options:
A)560
B)882
C)799
D)778
E)901
Rationale: 79%
104%
--------
25% ---- 140
100% ---- ? => Rs.560
Answer: A
Answer: A
Question: Each factor of 200 is inscribed on its own plastic ball, and all of the balls are placed in a jar. If a ball is randomly selected from the jar, what is the probability that the ball is inscribed with a multiple of 42?
Options:
A)1/16
B)5/42
C)1/8
D)3/16
E)1/9
Rationale: 200=2*3*5*7, so the # of factors 210 has is (1+1)(1+1)(1+1)(1+1)=16 (see below);
42=2*3*7, so out of 16 factors only two are multiples of 42: 42 and 210, itself;
So, the probability is 2/16=1/9.
Answer: E.
Answer: E
Question: The radius of a cylinder is 8 m, height 13 m. The volume of the cylinder is:
Options:
A)2200
B)2614.9
C)3300
D)1100
E)4400
Rationale: Cylinder volume = Ï€r(power2)h
= 22/7 Ã— 8 Ã— 8 Ã— 13
= 2614.9 m(power3)
Answer is B.
Answer: B
Question: A started a business with an investment of Rs. 70000 and after 6 months B joined him investing Rs. 120000. If the profit at the end of a year is Rs. 26000, then the share of B is?
Options:
A)A)14500
B)B)14000
C)C)12000
D)D)13007
E)E)14098
Rationale: Ratio of investments of A and B is (70000 * 12) : (120000 * 6) = 7 : 6
Total profit = Rs. 26000
Share of B = 6/13 (26000)
= Rs. 12000
Answer:C
Answer: C
Question: Farm Tax is levied on the 60% of the cultivated Land. The tax department collected total $ 3840 through the farm Tax from the village of Mr. william. Mr. william paid only $ 480 as farm tax. The percentage of total land of Mr. willam over the total taxable land of the village is :
Options:
A)15%
B)25%
C)0.125%
D)0.2083%
E)none
Rationale: This will be equal to the percentage of total cultivated land he holds over the total cultivated land in the village. That leads to (480/3840)x100 =12.5%in percentage terms.
But the question asks ratio between his total land to total cultivated land. Hence the answer is 12.5%x (100/60) = 20.833% and the answer is not there in the options.
The correct answer is (E).
Answer: E
Question: If a+b+c=13, a2+b2+c2=69, then find ab+bc+ca
Options:
A)50
B)60
C)69
D)75
E)80
Rationale: (a+b+c)2= a2+b2+c2+2(ab+bc+ca)
2(ab+bc+ca)= (a+b+c)2- a2+b2+c2
=169-69
=100
ab+bc+ca=50
ANSWER A 50
Answer: A
Question: M = {-6, -5, -4, -3, -2}
T = {-2, -1, 0, 1, 2, 3, 4, 5}
If an integer is to be randomly selected from set M above and an integer is to be randomly selected from set T above, what is the probability that the product of the two integers will be negative?
Options:
A)5/8
B)1/3
C)2/5
D)1/2
E)3/5
Rationale: Answer A.
Total # of outcomes: 5*8= 40
# of outcomes where Product is -ve : (-6,1) ,(-6,2), (-6,3)... Hence, total: 25
Probability: 25/40 = 5/8
Answer: A
Question: Thabo owns exactly 280 books, and each book is either paperback fiction, paperback nonfiction, or hardcover nonfiction. If he owns 20 more paperback nonfiction books than hardcover nonfiction books, and twice as many paperback fiction books as paperback nonfiction books, how many hardcover books nonfiction books does Thabo own?
Options:
A)10
B)20
C)30
D)40
E)55
Rationale: I think we can use double-matrix method and solve using only one variable.
Our goal is to find the number of hardcover nonfiction books. Let that number be x. We are given that all 140 books are either paperback fiction, paperback nonfiction, or hardcover nonfiction. This implies that number of hardcover fiction books is 0.
Double-matrix: P = paperback; H = hardcover; F = fiction; NF = nonfiction
P H Total
F 2x+40 0
NF x+20 x
Total 3x+60 x 280
3x + 60 + x = 280
x = 55
Answer (E.)
Answer: E
Question: Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 108 as interest at the end of the loan period, what was the rate of interest?
Options:
A)3.6
B)3
C)18
D)24
E)None of these
Rationale: Let rate = R% and time = R years.
Then, (1200 x R x R)/100 = 108
12R2 = 108
R2 =9
R = 3.
Answer: Option B
Answer: B
Question: In a bag there are coins of 50 paisa, 25 paisa and one rupee in the proportion 5:6:2. If there are in all Rs.37, the number of 25 paisa coins is?
Options:
A)30
B)31
C)32
D)34
E)37
Rationale: 5x 6x 2x
50 25 100
250x + 150x + 200x = 3700
600x = 3700
x = 6.16 => 6x = 37
ANSWER:E
Answer: E
Question: A sum of money is to be distributed among A, B, C, D in the proportion of 5:2:4:3. If C gets Rs. 1000 more than D, what is B's share?
Options:
A)2778
B)2679
C)2000
D)2699
E)2661
Rationale: Let the shares of A, B, C and D be 5x, 2x, 4x and 3x Rs. respectively.
Then, 4x - 3x = 1000 => x = 1000.
B's share = Rs. 2x = 2 * 1000 = Rs.2000.
Answer:C
Answer: C
Question: When I acquired my Mercedes-Benz car in Germany, the first thing I had to do was to get a license plate. The plate I got had a peculiar number on it. It consisted of 5 different numbers and by mistake when I fixed it upside down the number could be still read, but the value had increased by 78633.
Options:
A)10966
B)10968
C)10967
D)10966
E)10965
Rationale: 5 differ numbers can be read upside down same are 0 , 1 , 6 , 8 , 9
try with these no.s
number is 10968.
10968
78633
-------
89601
--------
ANSWER:B
Answer: B
Question: A basket contains 10 apples, of which 1 is spoiled and the rest are good. If we select 2 apples from the basket simultaneously and at random, what is the probability that the 2 apples selected will include the spoiled apple?
Options:
A)2/9
B)3/8
C)2/7
D)1/6
E)1/5
Rationale: The total number of ways to choose 2 apples is 10C2 = 45
The number of ways that include the spoiled apple is 9C1 = 9
P(the spoiled apple is included) = 9/45 = 1/5
The answer is E.
Answer: E
Question: Ram has two dies which has a nos 2,3,4,6 when he throws the two dies together find the no of possibility so that the some of the two dies should be even...???
Options:
A)4/8
B)5/8
C)6/8
D)7/8
E)3/8
Rationale: total possible combinations=(2,2)(2,4)(2,6)(4,2)(4,4)(4,6)(3,3)(6,2)(6,4)(6,6)=10
so probability=10/16=5/8
ANSWER:B
Answer: B
Question: When positive integer N is divided by positive integer J, the remainder is 16. If N/J = 127.05, what is value of J?
Options:
A)300
B)320
C)370
D)400
E)450
Rationale: When a number is divided by another number, we can represent it as : Dividend = Quotient * Divisor + Remainder
So, Dividend/Divisor = Quotient + Remainder/Divisor
Given that N/J = 127.05
Here 127 is the quotient.
Given that Remainder = 16
So, 127.05 = 127 + 16/J
So, J = 320
Answer: B
Answer: B
Question: When Greenville State University decided to move its fine arts collection to a new library, it had to package the collection in 20-inch by 20-inch by 15-inch boxes. If the university pays $1.20 for every box, and if the university needs 3.06 million cubic inches to package the collection, what is the minimum amount the university must spend on boxes?
Options:
A)$255
B)$275
C)$510
D)$612
E)$2,550
Rationale: Total no. Of boxes = 3060000/(20×20×15) = 510
Total cost = 510×$1.20=$612
Answer D
Answer: D
Question: Which of the following describes all values of x for which 9–x^2 >= 0?
Options:
A) x >= 1
B) x <= –1
C) -3 <= x <= 3
D) x <= –1 or x >= 1
E) –1 <= x <= 1
Rationale: 9 - x^2 >=0
means x^2 - 9 <=0
=> (x-3)(x+3) <= 0
=> -3 <= x <= 3
Answer - C
Answer: C