divelab/Sys2Bench huggingface dataset:
Question: A “palindromic integer” is an integer that remains the same when its digits are reversed. So, for example, 43334 and 516615 are both examples of palindromic integers. How many 6-digit palindromic integers are both even and greater than 600,000?
Options:
A)150
B)200
C)240
D)300
E)480
Rationale: The first digit and last digit are the same so the 2 possibilities are 6 or 8.
The second and third digits can be any number from 0 to 9.
The total number of palindromic integers is 2*10*10 = 200
The answer is B.
Answer: B
Question: The number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 6, respectively. If the sixth car through the production line has either 3, 7, or 19 defects, for which of theses values does the mean number of defects per car for the first six cars equal the median?
I. 3
II. 7
III. 12
Options:
A)I only
B)II only
C)III only
D)I and III only
E)I, II, and III
Rationale: B
for 7 we will have mean as 7.16(approx) and median as 7 but for 3 we will have mean and median = 6.5 and for 12 we will have median and mean =8
Answer: B
Question: A company bought for its 7 offices 3 computers of brand N and 3 computers of brand M. In how many ways could computers be distributed among the offices if each office can have only 1 computer.
Options:
A)196
B)140
C)256
D)292
E)312
Rationale: This problem has to do withcombinations. Here's the general idea: if you have a set of n elements, and you are going to choose r of them (r < n), then the number of combinations of size r one could choose from this total set of n is:
# of combinations = nCr = (n!)/[(r!)((n-r)!)]
where n! is the factorial symbol, which means the product of every integer from n down to 1. BTW, nCr is readn choose r.
In this problem, let's consider first the three computers of brand M. How many ways can three computer be distributed to seven offices?
# of combinations = 7C3 = (7!)/[(3!)(4!)] = (7*6*5*4*3*2*1)/[3*2*1)(4*3*2*1)]
= (7*6*5)/(3*2*1) = (7*6*5)/(6) = 7*5 = 35
There are 35 different ways to distribute three computers to 7 offices. (The massive amount of cancelling that occurred there is very much typical of what happens in the nCr formula.)
One we have distributed those three M computers, we have to distribute 3 N computers to the remaining four offices. How many ways can two computer be distributed to four offices?
# of combinations = 4C3 = 4
For each of the 35 configurations of distributing the M computers, we have 4 ways of distributing the N computers to the remaining offices. Thus, the total number of configurations is 35*4 = 140. Answer choice =B
Answer: B
Question: A bookstore has a shelf that contains biographies which normally sell for $20 each and mysteries that normally sell for $12 each. During a sale, the biographies and mysteries are discounted at different rates so that a customer saves a total of $19 from the normal price by buying 5 discounted biographies and 3 discounted mysteries. If the sum of the discount rates for the two types of books is 27 percent, what is the discount rate on mysteries?
Options:
A)10%
B)11%
C)12.5%
D)13%
E)14%
Rationale: Let B be the discount on biographies and M be the discount on mysteries
So., B + M = 0.27 -----(1)
And
(20*5 + 12*3) - (20*5*(1-B) + 12*3*(1-M)) = 19
--> 100(1-(1-B)) + 36(1-(1-M) = 19
100B + 36M = 19 ------(2)
Solving 12., we get M = 0.125 = 12.5%
C
Answer: C
Question: If the circumference of a circle increases from 4pi to 8 pi, what change occurs in the area ?
Options:
A)Area is quadrupled
B)Area is tripled
C)Area is doubles
D)Area become half
E)None of these
Rationale: Explanation:
2πR1=4π
=>R1=22πR2=8π
>R2=4Original Area =4π∗22=16πNew Area =4π∗42=64π
So the area quadruples.
Option A
Answer: A
Question: Which of the following is equivalent to the pair of inequalities x + 5 > 10 and x - 1 <= 5 ?
Options:
A) 2 < x < 16
B) 2 <= x < 4
C) 2 < x <= 8
D) 4 < x <= 8
E) 5 < x <= 6
Rationale: Solution:
Let’s isolate x in both inequalities, starting with x + 5 > 10.
x + 5 > 10
x > 5
Next we isolate x in the inequality x – 1  5.
x – 1 <= 5
x <= 6
Bringing these two inequalities together we know:
5 < x <= 6
The answer is E.
Answer: E
Question: A man rows his boat 75 km downstream and 45 km upstream, taking 2 1/2 hours each time. Find the speed of the stream?
Options:
A)5 kmph
B)6 kmph
C)9 kmph
D)8 kmph
E)1 kmph
Rationale: Speed downstream = d/t = 75/(2 1/2) = 30 kmph
Speed upstream = d/t = 45/(2 1/2) = 18 kmph
The speed of the stream = (30 - 18)/2 = 6 kmph
Answer:B
Answer: B
Question: The length of minute hand of a clock is 5.4 cm. What is the area covered by this in 10 minutes
Options:
A)15.27
B)16.27
C)17.27
D)18.27
E)19.27
Rationale: area of circle is pi*r^2
but in 10 minutes area covered is (10/60)*360=60 degree
so formula is pi*r^2*(angle/360)=3.14*(5.4^2)*(60/360)=15.27 cm^2
ANSWER:A
Answer: A
Question: If the remainder is 12 when the integer n is divided by 22, what is the remainder when 2n is divided by 11?
Options:
A)0
B)2
C)3
D)6
E)7
Rationale: n = 22k+12
2n = 2(22k+12) = 4*11k + 24 = 4*11k + 2*11 + 2 = 11j+2.
The answer is B.
Answer: B
Question: A particular library has 75 books in a special collection, all of which were in the library at the beginning of the month. These book are occasionally loaned out through an inter-library program. If, by the end of the month, 70 percent of books that were loaned out are returned and there are 66 books in the special collection at that time, how many books of the special collection were loaned out during that month?
Options:
A)20
B)30
C)35
D)40
E)55
Rationale: Total = 75 books.
65% of books that were loaned out are returned --> 100%-70%=30% of books that were loaned out are NOT returned.
Now, there are 66 books, thus 76-68=7 books are NOT returned.
{loaned out}*0.30=7 --> {loaned out}=30.
Answer: B.
Answer: B
Question: A is twice as good a workman as B and they took 9 days together to do the work B alone can do it in?
Options:
A)25 days
B)88 days
C)21 days
D)27 days
E)13 days
Rationale: WC = 2:1
2x + x = 1/9
x = 1/27 => 27 days
Answer: D
Answer: D
Question: When 2/3 of the garments in the shipment were inspected, 18 of the garments passed inspection and the remaining 2 garments failed. How many of the uninspected garments must pass inspection in order that 70 percent of the garments in the shipment pass?
Options:
A)10
B)3
C)8
D)7
E)5
Rationale: B = 9
2/3x= 20
x=30
for 90% approval we need 21 garments approved.
already approved = 18
we need 3 more.
Answer : B
Answer: B
Question: David has d books, which is 1/2 times as many as Jeff and 3 as many as Paula. How many books do the three of them have altogether, in terms of d?
Options:
A) 5/6*d
B) 10/3*d
C) 7/3*d
D) 7/2*d
E) 9/2*d
Rationale: Although we could plug in a real value for d, the problem can be just as easily solved by setting up equations. However, let’s start by defining some variables. Since we are given that David has d books, we can use variable d to represent how many books David has.
number of books David has = d
number of books Jeff has = j
number of books Paula has = p
We are given that David has 3 times as many books as Jeff. We can now express this in an equation.
d = 3p
d/3 = p
We are also given that David has ½ as many books as Paula. We can also express this in an equation.
d = (1/2)j
2d = j
Notice that we immediately solved forj in terms of d and p in terms of d. Getting j and p in terms of d is useful when setting up our final expression. We need to determine, in terms of d, the sum of the number of books for David, Jeff, and Paula. Thus, we have:
d + d/3 + 2d
Getting a common denominator of 3, we have:
3d/3 + d/3 + 6d/3 = 10d/3 = 10/3*d
The answer is B
Answer: B
Question: 2 : 3 : : 23 : ?
Options:
A)25
B)28
C)22
D)29
E)30
Rationale: D
29
3 is the next prime number after 2. Similarly, 29 is the next prime number after 23.
Answer: D
Question: If two-third of a bucket is filled in 100 seconds then the time taken to fill the bucket completely will be .
Options:
A)90 seconds
B)150 seconds
C)60 seconds
D)100 seconds
E)120 seconds
Rationale: 2/3 filled in 100 seconds
1/3 filled in 50 secs
then 2/3+1/3=100 + 50 seconds = 150 seconds
ANSWER: B
Answer: B
Question: Find the product of the place value and face value of 3 in 5769354
Options:
A)900
B)9000
C)90
D)9
E)0.9
Rationale: Explanation:
Place value = Local value
Face value = Absolute value
The place value of 3 in 5769354 is 3 x 100 = 300
The face value of 3 in 5769354 is nothing but 3.
=> 300 x 3 = 900
Answer: Option A
Answer: A
Question: Find the length of the wire required to go 15 times round a square field containing 69696 m2.
Options:
A)15840
B)3388
C)2667
D)8766
E)66711
Rationale: a2 = 69696 => a = 264
4a = 1056
1056 * 15 = 15840
Answer: A
Answer: A
Question: If 25 percent of p is equal to 10 percent of q, and pq#0, then p is what percent of q ?
Options:
A)2.5%
B)15%
C)20%
D)35%
E)40%
Rationale: p/4 = q/10
p = 0.4q (40%)
Answer (E).
Answer: E
Question: If x and y are positive odd integers, then which of the following must also be an odd integer? I. x^(y+2) II. x(y+4) III. (y+1)^(x-1) + 5
Options:
A)I only
B)II only
C)III only
D)I and III
E)All options are applicable
Rationale: let x= 5, y= 3
I. 5^5 = > Odd
II. 5*7 = > Odd
III. 4^4 +1 = > Odd
Answer : E
Answer: E
Question: Anthony and Michael sit on the six-member board of directors for company T. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?
Options:
A)20%
B)30%
C)40%
D)50%
E)60%
Rationale: Soln: Let the commitees be I and II
Assuming that Anthony and Michael T go into commitee I, there is just one more place left to be filled in that commitee and it can be taken by any of the 4 remaining people. Thus 4 ways.
Since the Anthony and Michael can also go into commitee II, we get 4 ways for that commitee also.
So in total = 8 ways
Now total number of ways of choosing 3 from 6 people is
= 6C3
= 6 * 5 * 4/3!
= 20 ways
Therefore Probability is = (8 /20) * 100 = 40%
Answer: C
Question: The cash difference between the selling prices of an article at a profit of 10% and 6% is Rs. 3. The ratio of the two selling prices is?
Options:
A)52:56
B)55:53
C)52:50
D)22:56
E)52:51
Rationale: Let C.P. of the article be Rs. x.
Then, required ratio = 110% of x / 106% of x
= 110/106 = 55/53 = 55:53
Answer: B
Answer: B
Question: If n = 2gh, where g and h are distinct prime numbers greater than 2, how many different positive even divisors does n have, including n ?
Options:
A) Two
B) Three
C) Four
D) Six
E) Eight
Rationale: Answer - 'C' - Four.
For a number 2gh with primes gh,there will be four even divisors - 2,2*g,2*h,2*g*h
Answer: C
Question: A sum of Rs.63000 is divided into three parts such that the simple interests accrued on them for three, six and nine years respectively may be equal. Find the amount deposited for 3 years.
Options:
A)6000
B)2000
C)6500
D)4500
E)3000
Rationale: Let the amounts be x, y, z in ascending order of value. As the interest rate and interest accrued are same for 3 years 6 years and 9 years i.e. 3x = 6y = 9z = k.
L.C.M. of 3,6,9 = 18 So x:y:z: = 6000 : 3000 :2000
The amount deposited for 3 years = 6000
ANSWER:A
Answer: A
Question: A train travelled from station P to Q in 8 hours and came back from station Q to P is 6 hours. What would be the ratio of the speed of the train while traveling from station P to Q to that from station Q to P?
Options:
A)3:6
B)3:1
C)3:8
D)3:4
E)3:2
Rationale: Since S # 1/t
S1 : S2 = 1/t1 : 1/t2
= 1/8 : 1/6
= 3 : 4
Answer: D
Answer: D
Question: Evaluate
30!/28!
Options:
A)970
B)870
C)770
D)670
E)None of these
Rationale: Explanation:
=30!/28!
=30∗29∗28!/28!=30∗29=870
Option B
Answer: B
Question: The cost of 10 kg of mangos is equal to the cost of 24 kg of rice. The cost of 6 kg of flour equals the cost of 2 kg of rice. The cost of each kg of flour is $20.50. Find the total cost of 4 kg of mangos, 3 kg of rice and 5 kg of flour?
Options:
A)347.4
B)987.4
C)877.4
D)637.4
E)667.4
Rationale: C
$877.40
Let the costs of each kg of mangos and each kg of rice be $a and $r respectively.
10a = 24r and 6 * 20.50 = 2r
a = 12/5 r and r = 61.5
a = 147.6
Required total cost = 4 * 147.6 + 3 * 61.5 + 5 * 20.5
= 590.4 + 184.5 + 102.5 = $877.40
Answer: C
Question: Krishan and Nandan jointly started a business. Krishan invested four times as Nandan did and invested his money for double time as compared to Nandan. Nandan earned Rs. 4000. If the gain is proportional to the money invested and the time for which the money is invested then the total gain was?
Options:
A)Rs.20000
B)Rs.24000
C)Rs.28000
D)Rs.36000
E)Rs.35000
Rationale: 4:1
2:1
------
8:1
1 ----- 4000
9 ----- ? => Rs.36,000
Answer: D
Answer: D
Question: This week, Florry bought the same number of music albums she bought last week, but for a third of the total price. This week, the average (arithmetic mean) price per album
Options:
A)is twice last week's average
B)is three times last week's average
C)is a half of last week's average
D)is a third of last week's average
E)is the same as last week's average
Rationale: let the number of music albums brought by Florry last week be '3'
let the total price of the music albums brought by Florry last week be $30 (i.e., $10 per album)
The average price per album purchased last week= $10
Florry brought same number of albums this week but for a total price of $10 ( i.e., one third the total price last week)
Therefore this weeks average price per album= $10/3 one-third of last week's average
Hence D
Answer: D
Question: On Sunday morning, Pugsley and Wednesday are trading pet spiders. If Pugsley were to give Wednesday two of his spiders, Wednesday would then have eight times as many spiders as Pugsley does. But, if Wednesday were to give Pugsley ten of her spiders, Pugsley would now have fourteen fewer spiders than Wednesday had before they traded. How many pet spiders does Pugsley have before the trading game commences?
Options:
A)4
B)6
C)8
D)10
E)12
Rationale: If Pugsley were to give Wednesday four of his spiders, Wednesday would then have five times as many spiders as Pugsley does:
(w + 2) = 8(p - 2)
If Wednesday were to give Pugsley four of her spiders, Pugsley would now have four fewer spiders than Wednesday had before they traded:
p + 14 = w - 10
Solving gives p = 6 and w = 30.
Answer: B.
Answer: B
Question: A rectangular garden is to be twice as long as it is wide. If 300 yards of fencing, including the gate, will completely enclose the garden, what will be the length of the garden, in yards?
Options:
A)40
B)50
C)60
D)70
E)100
Rationale: ALTERNATE APPROACH
Backsolving ( Using answer options to reach the correct answer ) can work wonders here if one is fast in calculations.
Given Perimeter is 300
So, 2 ( L + B ) = 300
Or, L + B = 150
Now use the answer options ( Given Length ; Breath will be half the length)
(A) 40
L = 40 ; B = 20
L + b = 60
(B) 50
L = 50 ; B = 25
L + b = 75
(C) 60
L = 60; B = 30
L + b = 90
(D) 70
L = 70; B = 35
L + b = 105
(E) 100
L = 100; B = 50
L + b = 150
Thus you see no, need of any calculations, U can reach the correct option only by checking options ; correct answer will be (E)
Answer: E
Question: The average weight of A, B and C is 30 kg. If the average weight of A and B be 25 kg and that of B and C be 28 kg, then the weight of B is:
Options:
A)14 kg
B)15 kg
C)18 kg
D)19 kg
E)16 kg
Rationale: EXPLANATION
Let A, B, C represent their respective weights. Then, we have:
A + B + C = (30 x 3) = 90 â€¦. (i)
A + B = (25 x 2) = 50 â€¦. (ii)
B + C = (28 x 2) = 56 â€¦.(iii)
Adding (ii) and (iii), we get: A + 2B + C = 106 â€¦. (iv)
Subtracting (i) from (iv), we get : B = 16.
Bâ€™s weight = 16 kg.
Answer E
Answer: E
Question: A group of 55 adults and 70 children go for trekking. If there is meal for either 70 adults or 90 children and if 14 adults have their meal, find the total number of children that can be catered with the remaining food.
Options:
A)33
B)54
C)18
D)17
E)72
Rationale: Explanation:
As there is meal for 70 adults and 14 have their meal, the meal left can be catered to 56 adults.
Now, 70 adults = 90 children
7 adults = 9 children
Therefore, 56 adults = 72 children
Hence, the meal can be catered to 72 children.
ANSWER: E
Answer: E
Question: The average age of students of a class is 15.8 years. The average age of boys in the class is 16.4 years and that of the girls is 15.4 years, The ratio of the number of boys to the number of girls in the class is :
Options:
A)2:9
B)2:3
C)2:1
D)2:2
E)2:6
Rationale: Explanation:
Let the ratio be k : 1. Then,
k * 16.4 + 1 * 15.4 = (k + 1) * 15.8
<=> (16.4 - 15.8) k = (15.8 - 15.4) <=> k = 0.4/0.6 = 2/3.
Required ratio = 2/3 : 1 = 2 : 3.
Answer: B
Answer: B
Question: When the positive integer r is divided by 9, the remainder is 5. What is the remainder when 3r is divided by 9?
Options:
A)0
B)1
C)3
D)4
E)6
Rationale: Lets take the number as r
when r is divided by 9 the remainder is 5 hence r can be written as
r=9k +5
Multiplying by 3 will give
3r = 27k + 15
we can also write
3r = 27k + 9 + 6
Now 27k and 9 are divisible by 9 leaving the remainder as 6 hence E is the answer.
Answer: E
Question: David gets on the elevator at the 11th floor of a building and rides up at the rate of 57 floors per minute. At the same time, Albert gets on an elevator at the 51st floor of the same building and rides down at the rate of 63 floors per minute. If they continue travelling at these rates, then at which floor will their paths cross ?
Options:
A)30
B)20
C)10
D)50
E)40
Rationale: Suppose their paths cross after x minutes.
Then, 11 + 57x = 51 - 63x = 120x = 40
x = 1/3
Number of floors covered by David in (1/3) min. = (1/3 x 57)= 19.
So, their paths cross at (11 +19) i.e., 30th floor.
Answer is A.
Answer: A
Question: the numbers 272738 and 232342,when divided by n, a 2 digit number leave a remainder 13 and 17 respectively.Find the sum of digits of n?
Options:
A)4
B)5
C)6
D)7
E)8
Rationale: We have to find the h.c.f of both
1.272738-13
2.232325-17
H.c.f is 25
Ans is 7
ANSWER:D
Answer: D
Question: Calculate the sum of first 60 natural numbers.
Options:
A)1839
B)2830
C)1830
D)1831
E)1835
Rationale: Solution
We know that(1+2+3+.....+60)	= n(n+1)/2
Therefore (1+2+3+....+60)	=(60×61 / 2)
= 1830.
Answer C
Answer: C
Question: 1.14 expressed as a per cent of 1.9 is:
Options:
A)6%
B)10%
C)60%
D)90%
E)None
Rationale: Solution: Required percentage = (1.14*100)/1.9 = 60%.
Answer: Option C
Answer: C
Question: Jane biked 21 2/3 miles in 3 hours and 20 minutes. What was her average rate of speed in miles per hour?
Options:
A)6 1/2
B)7
C)7 1/2
D)7 3/4
E)8
Rationale: D = 21(2/3) = 65/3
T = 3(1/3) = 10/3
S = D/T = 6 1/2
Answer = A
Answer: A
Question: At 15:00 there were 20 students in the computer lab. At 15:03 and every three minutes after that, 3 students entered the lab. If at 15:10 and every ten minutes after that 7 students left the lab, how many students were in the computer lab at 15:44 ?
Options:
A)7
B)14
C)25
D)27
E)30
Rationale: Initial no of students + 3 * (1 + No of possible 3 minute intervals between 15:03 and 15:44) -
8 *(1 + No of possible 10 minute intervals between 15:10 and 15:44)
20 + 3*14 -8 * 4 = 25
C
Answer: C
Question: Out of 10 persons working on a project, 4 are graduates. If 3 are selected, what is the probability that there is at least one graduate among them?
Options:
A)5/8
B)5/2
C)5/1
D)5/6
E)5/5
Rationale: P(at least one graduate) = 1 - P(no graduates)
= 1 - 6C3/10C3 = 1 - (6 * 5 * 4)/(10 * 9 * 8)
= 5/6
Answer: D
Answer: D
Question: A 250 m long train is running at a speed of 55 Km/hr. It crossed a platform of length 300 m in ?
Options:
A)41.1 sec
B)20.2 sec
C)36 sec
D)50.4 sec
E)none of the above
Rationale: Speed = 55 Km/hr
(to convert km/hr in to M/s)
= 55 x 5/18 M/s
Distance = 250 m + 300 m ( If questions is about train crossing a post you need to consider only the length of Train, )
= 550 m
Time = Distance / Speed
= 550 x 18 / (5 x 55)
= 36 sec
Ans is :C
Answer: C
Question: The difference between a two-digit number and the number obtained by interchanging the digits is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ?
Options:
A)4
B)8
C)16
D)20
E)None of these
Rationale: Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit.
Let ten's and unit's digits be 2x and x respectively.
Then, (10 x 2x + x) - (10x + 2x) = 36
9x = 36
x = 4.
Required difference = (2x + x) - (2x - x) = 2x = 8.
Answer:B
Answer: B
Question: If a sum of money borrowed at 5 percent/annum SI amounts to Rs.1020 in 4 yrs, then the sum of money borrowed is?
Options:
A)Rs. 850
B)Rs. 900
C)Rs. 910
D)Rs. 1000
E)Rs. 1050
Rationale: Sol We have, A : Rs.1U2D
Let the pr xnpal be R5 x
Then mterest u — 12 : wzo — x
Therefore, by u xglurmula.
_umxx
’ Rx’: (
, _ma>< 1nzn—xv
\4Ve|\a\e,Jc— 5“
:> ~(:51DD—S*(or6x:51UU
or, J: : w 350
. The sum of money burrowed : Rs. 850
A)
Answer: A
Question: What was the day of the week on 15th August, 1947 ?
Options:
A)Wednesday
B)Tuesday
C)Friday
D)Thursday
E)None of these
Rationale: 15th August, 1947 = (1946 years + Period from 1st Jan.,1947 to 15th Aug., 1947)
Counting of odd days :
1600 years have 0 odd day. 300 years have 1 odd day.
47 years = (11 leap years + 36 ordinary years)
= [(11 × 2) + (36 × 1)]odd days
= 58 odd days
⇒ 2 odd days.
= 227 days = (32 weeks + 3 days) = 3 odd days.
Total number of odd days = (0 + 1 + 2 + 3) odd days = 6 odd days.
Hence, the required day was ‘Friday’.
Answer C
Answer: C
Question: A multiple choice test consists of 4 questions, and each question has 5 answer choices. In how many ways can the test be completed if every question is unanswered?
Options:
A)24
B)120
C)625
D)720
E)1024
Rationale: 5 choices for each of the 4 questions, thus total of 5*5*5*5 = 5^4 = 625 ways.
Answer: C.
Answer: C
Question: A shopping cart contains only apples, oranges, and pears. If there are twice as many oranges as apples, and twice as many pears as oranges, then the apples are equivalent to what fraction of the pears?
Options:
A)1/4
B)1/3
C)1/2
D)2/3
E)3/4
Rationale: O = 2A
P = 2O = 4A
A = P/4
The answer is A.
Answer: A
Question: Kelly and Jody packed boxes with books. If Jody packed 10% of the total number of boxes, what was the ratio of the number of boxes Kelly packed to the number that Jody packed?
Options:
A)1 to 6
B)1 to 4
C)2 to 5
D)9 to 1
E)2 to 3
Rationale: Explanation:
If Jody packed 10% of the boxes then Kelly packed 90%. The ratio of the number of boxes Kelly packed to the number Jody packed is thus 90%/10%=9/1
Answer: Option D
Answer: D
Question: A Certain car costs Rita $1500 afterwards she spent $50 over it. Now at what price must Rita sell this car in order to make a profit of 7% ?
Options:
A)$1712
B)$1715
C)$1720
D)$1725
E)$1730
Rationale: Cost Price = $1500 + $50 = $1600
Profit % = 7
Selling Price = ?
Selling Price = (100 + Profit %) * Cost Price /100
= (100 + 7) * 1600 / 100
= 1712
Answer is A
Answer: A
Question: A group of 7 students bought movie tickets in one row next to each other. If Bob and Lisa are in this group, what is the probability that Bob and Lisa will each sit next to only one of the six other students from the group?
Options:
A)5.76%
B)4.76%
C)15.75%
D)20.67%
E)25.15%
Rationale: The question basically asks about the probability that Bob and Lisa sit at the ends.
The total # of sitting arrangements is 7!.
Desired arrangement is either BVWXYZL or LVWXYZB. Now, VWXYZ can be arranged in 5! ways, therefore total # of favorable arrangements is 2*5!.
P=(favorable)/(total)=(2*5!)/7!=1/21.
Answer: B.
Answer: B
Question: Pipe A can fill a tank in 5 hours, pipe B in 10 hours and pipe C in 30 hours. If all the pipes are open, in how many hours will the tank be filled?
Options:
A)2
B)2.5
C)3
D)3.5
E)None
Rationale: Sol.
Part filled by (A + B + C) in 1 hour = (1/5 + 1/10 + 1/30) = 1/3.
∴ All the three pipes together will fill the tank in 3 hours.
Answer C
Answer: C