divelab/Sys2Bench huggingface dataset:
Question: The value of x is one quarter of z. The sum of x, y, and z is equal to 26. If the value of y is twice the value of z, what is the largest factor of the sum of y and z?
Options:
A)2
B)3
C)8
D)12
E)24 IS ANSWER
Rationale: While most Test Takers would treat this as a 'system' algebra question and solve it that way, there are some interesting Number Property patterns in this prompt that will actually allow you to use the answer choices (and some pattern-matching) to get to the solution.
To start, I'm going to deal with the last piece of information first:
Y is twice Z
The prompt asks for the BIGGEST factor of the SUM of Y and Z. Since Y = 2Z, the sum of Y and Z is (2Z) + Z = 3Z. This means that the BIGGEST factor is a MULTIPLE OF 3 (and likely one of the larger multiples of 3 in the answer choices).
The second piece of information tells us...
X+Y+Z = 26
By extension...
X+(Y+Z) = 26
X+(larger multiple of 3) = 26
From the answer choices, there are really only 2 options that make any sense: Answers D and E....
With Answer D:
IF....Y+Z = 12, then X= 14, but....
The first piece of information tells us that X = 1/4 of Z, so X has to be relatively small....In this example, it's LARGER than Z, so D is NOT the answer. At this point, the answer would have to be E, but here's the proof...
IF...Y+Z = 24 and Y= 2Z....
Y = 16
Z = 8
X is 1/4 of Z, so X = (1/4)(8) = 2
X+Y+Z=26
2+16+8=26
All of the numbers 'fit' what we were told, so this MUST be the answer.
E
Answer: E
Question: A merchant mixes three varieties of rice costing Rs.20/kg, Rs.24/kg and Rs.30/kg and sells the mixture at a profit of 20% at Rs.30 / kg. How many kgs of the second variety will be in the mixture if 2 kgs of the third variety is there in the mixture?
Options:
A)1 kgs
B)3 kgs
C)5 kgs
D)6 kgs
E)None of these
Rationale: Explanation :
If the selling price of mixture is Rs.30/kg and the merchant makes a profit of 20%, then the cost price of the mixture = 30/1.2 = Rs.25/kg.
We need to find out the ratio in which the three varieties are mixed to obtain a mixture costing Rs.25 /kg.
Let variety A cost Rs.20/kg, variety B cost Rs.24 / kg and variety C cost Rs.30/kg. The mean desired price falls between B and C.
Step 1:
Find out the ratio Qa:Qc using alligation rule.
Qa/Qc=(30−25)/(25−20)=1/1.
Step 2:
Find out the ratio Qb:Qc using alligation rule.
Qb/Qc=30−25/(25−24)=5/1.
Step 3:
Qc is found by adding the value of Qc in step 1 and step 2 =1+1=2.
Therefore, the required ratio =1:5:2.
If there are 2 kgs of the third variety in the mixture, then there will be 5 kgs of the second variety in the mixture.
Answer : C
Answer: C
Question: Find the area of a parallelogram with base 24 cm and height 12 cm?
Options:
A)198 cm2
B)288 cm2
C)279 cm2
D)128 cm2
E)297 cm2
Rationale: Area of a parallelogram
= base * height
= 24 * 12 = 288 cm2
Answer:B
Answer: B
Question: Arjun started a business with Rs. 20,000 and is joined afterwards by Anoop with Rs.40, 000. After how many months did Anoop join if the profits at the end of the year are divided equally?
Options:
A)3
B)4
C)5
D)6
E)7
Rationale: Suppose Anoop joined after 3 months.
Then, 20000 * 12 = 40000 * (12 – x)
=>
=> x = 6.
ANSWER:D
Answer: D
Question: If a is a positive integer, which of the following could NOT be the square of a?
Options:
A)2432
B)7426
C)5966
D)3415
E)2155
Rationale: A number ending in 2 cannot be a perfect square.
2432 ends in a number 2, making that the correct answer
(A) 2432
Answer: A
Question: The difference between the simple interest received from two different sources on Rs. 1400 for 3 years is Rs. 13.50. The difference between their rates of interest is?
Options:
A)A)2.4%
B)B)2.6%
C)C)0.2%
D)D)2.0%
E)E)1.3%
Rationale: (1400 * R1 * 3)/100 - (1400 * R2 * 3)/100 = 13.50 4200(R1 - R2) = 1350
R1 - R2 = 0.2%
Answer:C
Answer: C
Question: A 1200 m long train crosses a tree in 120 sec, how much time will I take to pass a platform 600 m long?
Options:
A)176 Sec
B)180 Sec
C)178 Sec
D)267 Sec
E)276 Sec
Rationale: L = S*T
S= 1200/120
S= 10 m/Sec.
Total length (D)= 1800 m
T = D/S
T = 1800/10
T = 180 Sec
Answer:B
Answer: B
Question: A pack of baseball cards consists of 12 outfielder cards and 8 infielder cards. What is the lowest number of outfielder cards that would have to be removed from the pack so that no more than 25 percent of the pack would be outfielder cards?
Options:
A)10
B)5
C)6
D)7
E)8
Rationale: let the number of O card to be removed=X
then remaning cards=12-X
now this 12-X cards should count less than 25% of toatal Ocards
12-X<.25*12
X>12-3
X>9
X=10
Ans A
Answer: A
Question: The average of the marks of 10 students in a class is 80. If the marks of each student are doubled, find the new average?
Options:
A)80
B)120
C)160
D)270
E)110
Rationale: Sum of the marks for the 10 students
= 10 * 80= 800. The marks of each student are doubled, the sum also will be doubled.
The new sum = 800 * 2 = 1600. So, the new average
= 1600/10 = 160.
Answer:C
Answer: C
Question: What least number must be subtracted from 427398 so that remaining no.is divisible by 11
Options:
A)3
B)4
C)6
D)7
E)8
Rationale: Explanation:
On dividing 427398 by 11 we get the remainder 4, so 4 should be subtracted
Answer: Option B
Answer: B
Question: Water tax is increased by 20% but its consumption is decreased by 20%. Then, the increase or decrease in the expenditure of the money is :
Options:
A)No change
B)5% decrease
C)4% increase
D)4% decrease
E)None
Rationale: Explanation:
Let tax=Rs.100 and
Consumption=100 units
Original expenditure = Rs.(100×100)=Rs.10000(100×100)=Rs.10000
New expenditure = Rs.(120×80)=Rs.9600(120×80)=Rs.9600
Decrease in expenditure
= (400/10000×100)%=4%
Correct Option: D
Answer: D
Question: A shopkeeper buys mangoes at the rate of 5 a rupee and sells them at 3 a rupee. Find his net profit or loss percent?
Options:
A)66 2/3 %
B)33 7/3 %
C)33 1/3 %
D)33 5/8 %
E)34 1/3 %
Rationale: The total number of mangoes bought by the shopkeeper be 15.
If he buys 5 a rupee, his CP = 3
He selling at 3 a rupee, his SP = 5
Profit = SP - CP = 5 - 3 = 2
Profit Percent = 2/3 * 100 = 66 2/3 %
Answer: A
Answer: A
Question: For real number x, int(x) denotes integer part of x.int(x) is the largest integer less than or equal to x.int(1,2)=1,int(-2,4)=-3. Find the value of int(1/2)+int(1/2+ 100)+int(1/2+2/100)+....+int(1/2+99/100)
Options:
A)150
B)98
C)278
D)277
E)1811
Rationale: int (1/2) = 0
int (1/2 + 100 ) = 100
into (1/2 + 2/100) = 0
......
int ( 1/2 + 50/100 ) = 1
int (1/2 + 51 /100) = 1
.......
int (1/2 + 99/100) = 1
So 100 + 1 + 1 + .....50 times = 150
Answer:A
Answer: A
Question: when fair coin is tossed 9 times find the probability of getting head at least once.
Options:
A)1/512
B)511/512
C)9/512
D)503/51
E)503/510
Rationale: 1-1/2^9= 1-1/512 = 511/512
ANSWER:B
Answer: B
Question: A certain sum amount to $1400 in 2 yrs and $2000 in five years on certain fixed simple interest. In how many yrs would the amount be $2040 had it been put at the rate of interest compounded annually?
Options:
A)5
B)4
C)3
D)2
E)1
Rationale: We can form equations as following:
Initial amount = P, Total amount after interest = T, Interest per year = I
T = P+I
Given T1 = P+2I = 1400T2 = P+ 5I = 2000
From this we get, P = 1000I = 200 $
Now we know that simple interest is<Compound interest ( CI = SI only after 1st year if CI is compounded annually)
consider simple interest - after 1st yer it is 200$ , after 2nd years it is 400$ , after 3rd year it is 600$.and after 4rd year it is 800$.
To get a 640 $ compound interest the duration have to be > 2 year and < 4 years.
Only option fitting : Option C.
Answer: C
Question: If x and y are positive real numbers such that x + y = 1, which of the following could be the value of 100x + 600y?
I. 90
II. 250
III. 599
Options:
A)II only
B)III only
C)I and II
D)I and III
E)II and III
Rationale: We know that x and y are both between 0 and 1.
The minimum limit of 100x + 600y is 100, when x is close to 1 and y is close to 0.
The maximum limit of 100x + 600y is 600, when x is close to 0 and y is close to 1.
The expression can take on any value between 100 and 600.
The answer is E.
Answer: E
Question: 0.35 represents what percent of 0.0007?
Options:
A)0.05%
B)0.5%
C)5%
D)500%
E)50000%
Rationale: One more method
0.35 represents what percent of 0.0007?
Adjusting the decimal
3500 represents what percent of 7?
Divide by 7
500 represents what percent of 1?
Answer = 500*100 = 50000% = E
Answer: E
Question: Every year an amount increases by 1/8th of itself. How much will it be after two years if its present value is Rs.57600?
Options:
A)81000
B)81028
C)27772
D)72900
E)22312
Rationale: 57600* 9/8 * 9/8 = 72900.Answer:D
Answer: D
Question: In what time will a train 100 meters long cross an electric pole, if its speed is 360 km/hr
Options:
A)1 second
B)4.5 seconds
C)3 seconds
D)2.5 seconds
E)None of these
Rationale: Explanation:
First convert speed into m/sec
Speed = 360*(5/18) = 100 m/sec
Time = Distance/speed
= 100/100 = 1 second
Answer: A
Answer: A
Question: The tax on a commodity is diminished by 25% and its consumption increased by 13%. The effect on revenue is?
Options:
A)18
B)16
C)10
D)15
E)14
Rationale: 100 * 100 = 10000
75 * 113 = 8475
-----------
10000-----------1525
100-----------? => 15% decrease
Answer:D
Answer: D
Question: Two employees A and B are paid a total of Rs. 550 per week by their employer. If A is paid 120 percent of the sum paid to B, how much is B paid per week?
Options:
A)Rs.220
B)Rs.250
C)Rs.260
D)Rs.270
E)Rs.280
Rationale: Let the amount paid to A per week = x
and the amount paid to B per week = y
Then x + y = 550
But x = 120% of y = 120y/100 = 12y/10
∴12y/10 + y = 550
⇒ y[12/10 + 1] = 550
⇒ 22y/10 = 550
⇒ 22y = 5500
⇒ y = 5500/22 = 500/2 = Rs.250
B
Answer: B
Question: Machine A takes 100 hours to complete a certain job and starts that job at 7AM. After two hour of working alone, machine A is joined by machine B and together they complete the job at 12PM. How long would it have taken machine B to complete the jobif it had worked alone for the entire job?
Options:
A)100.11
B)10.98
C)18.07
D)45.01
E)35.09
Rationale: Let us assume total job = 100 units
A finishes 100 units in 100 hrs(given)
Hence A (working rate)= 1 units /hr
Now Given that A works for 2 hr( so 2 units done) then A and B finish total work in 15 hours. Hence A and B finish 98 units in 15 hours. Of these 1x15=15 units were done by A. Hence B did 83units in 15 hours.
Hence B(working rate)= 83/15 units /hr
Hence B takes 100x15/83 = 18.07 hours to complete the job.
Answer C.
Answer: C
Question: A certain university will select 2 of 6 candidates eligible to fill a position in the mathematics department and 3 of 5 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)150
D)165
E)315
Rationale: 6C2*5C3 = 15*10 = 150
the answer is (C)
Answer: C
Question: If 0 < L < Y, L is an odd number and Y is a prime number, which of the following can be the value of L + Y ?
Options:
A)11.
B)13.
C)17.
D)10.
E)7.
Rationale: If L is odd ,implies l(odd)+y(odd/even prime)=(Even/Odd)
If Y is even, the 2 is the even prime , therefore y=2l=1 since 0<L<Y therefore L+Y =3. But 3 is not an option
which implies y is odd,
Now L(odd)+Y(odd)=Even
10 is the only option that is even=D
Answer: D
Question: A sells a bicycle to B at a profit of 60% and B sells it to C at a loss of 60%. Find the resultant profit or loss.
Options:
A)-4%
B)0%
C)-5%
D)6%
E)-36%
Rationale: The resultant profit or loss = 60 - 60 - (60*60)/100 = -36%
loss = 36%
Answer is E
Answer: E
Question: An amount of Rs.2,000 is to be distributed amongst P, Q, R and S such that “P” gets twice as that of “Q” and “S” gets four times as that of “R”. If “Q” and “R” are to receive equal amount, what is the difference between the amounts received by S and P?
Options:
A)2378
B)268
C)269
D)500
E)6971
Rationale: Explanation:
We have, P = 2Q & S = 4R
Further Q = R & P + Q + R + S = 2,000
Thus we get, 2Q + Q + Q + 4Q = 2,000
8Q = 2,000 or Q = Rs. 250
Thus, R = Rs. 250, P = 500 & S = Rs. 1000
Hence, the required difference = (S – P) = (1000 – 500) = Rs. 500
ANSWER:D
Answer: D
Question: Pascal has 96 miles remaining to complete his cycling trip. If he reduced his current speed by 4 miles per hour, the remainder of the trip would take him 16 hours longer than it would if he increased his speed by 50%. What is his current speed Z?
Options:
A) 6
B) 8
C) 10
D) 12
E) 16
Rationale: Let the current speed be x miles per hour.
Time taken if speed is 50% faster (i.e. 3x/2 = 1.5x) = 96/1.5x
Time taken if speed is reduced by 4 miles/hr (i.e. (x-4)) = 96/(x-4)
As per question, 96/(x-4) - 96/1.5x = 16
Solving this Z we get x = 8.
B.
Answer: B
Question: A 180 meter long train crosses a man standing on the platform in 6 sec. What is the speed of the train?
Options:
A)48 kmph
B)52 kmph
C)108 kmph
D)100 kmph
E)98 kmph
Rationale: S = 180/6 * 18/5 = 108 kmph
answer :C
Answer: C
Question: For some integer y, y^2 - 5 is divisible by all of the following EXCEPT
Options:
A)29
B)30
C)31
D)38
E)41
Rationale: Hint: y^2-5 (y is an integer) is never multiple of 3 (try to prove this), hence 30 is out.
Answer: B
Answer: B
Question: A searchlight on top of the watchtower makes 2 revolutions per minute. What is the probability that a man appearing near the tower will stay in the dark for at least 5 seconds?
Options:
A)5/6
B)5/12
C)7/12
D)11/12
E)11/15
Rationale: The searchlight completes one revolution every 30 seconds.
The probability that the man's area will be lit up is 5/30=1/6.
The probability that he will stay in the dark is 1-1/6=5/6
The answer is A.
Answer: A
Question: If t and y are different prime numbers less than ten, what must be true about the sum of t and y?
Options:
A)the sum is even
B)the sum is odd
C)the sum is less than ten
D)the sum is greater than ten
E)the sum is less than 13
Rationale: This has to be solved by checking for each option:
Prime numbers less that 10 = 2, 3, 5, 7
A. the sum is even
May or may not be true. t = 3, y, = 5 -> True.
t = 2, y = 3 - True
B. the sum is odd
May or may not be true. t = 3, y, = 5 -> False.
t = 2, y = 3 - True
C. the sum is less than ten
May or may not be true t= 5, y = 7, Sum = 12
t = 2, y = 3. Sum = 5
D. the sum is greater than ten
May or may not be true t= 5, y = 7, Sum = 12
t = 2, y = 3. Sum = 5
E. the sum is less than 13
This will always be true.
Maximum sum = 7 + 5 = 12.
This is always less than 13.
Correct Option: E
Answer: E
Question: The height of the wall is 4 times its width and length of the wall is 3 times its height .if the volume of the wall be 10368 cu.m.its width is
Options:
A)4m
B)5m
C)6m
D)7m
E)8m
Rationale: Explanation:
Let width = x
Then, height=4x and length=12x
12x Ã— 4x Ã— x = 10368
x = 6
Answer: C
Answer: C
Question: A military garrison in a wild frontier has 500 men who have provisions to last 26 weeks. At the end of 6 weeks they are joined by another 300 men. How long will the provisions last?
Options:
A)13 1/3
B)12 1/3
C)11 3/3
D)12 1/2
E)12 3/4
Rationale: Total provisions=500*26*7=91000
Provisions used for in 6 weeks=6*7*500=21000
remaining provisions=70000
remaining provisions need to be split between 800(500+300) people.
Number of provisions per person= 70000/800
we need to find the answer in terms of weeks=(70000/800)/7=25/2
Answer:D
Answer: D
Question: A watch was sold at a loss of 10%. If it was sold for Rs.140 more, there would have been a gain of 4%. What is the cost price?
Options:
A)1000
B)3783
C)1837
D)3799
E)3786
Rationale: 90%
104%
--------
14% ---- 140
100% ---- ? =>
Rs.1000
Answer: A
Answer: A
Question: A family has two children. find the probability that both the children are girls given that at least
one of them is a girl?
Options:
A)1/2
B)1by3
C)3/5
D)3/7
E)3/11
Rationale: Let b stand for boy and g for girl. The sample space of the experiment is
S = {(g, g), (g, b), (b, g), (b, b)}
Let E and F denote the following events :
E : ‘both the children are girls’
F : ‘at least one of the child is a girl’
Then E = {(g,g)} and F = {(g,g), (g,b), (b,g)}
Now E n F = {(g,g)}
Thus P(F) = 3/4
and P (E n F )= 1/4
Therefore P(E|F) = P(E ∩ F)/P(F) = (1/4)/(3/4) = 1/3
B)
Answer: B
Question: There are 11 women and 9 men in a certain club. If the club is to select a committee of 2 women and 2 men, how many different such committees are possible?
Options:
A)120
B)720
C)1,060
D)1,520
E)1980 committees
Rationale: these are two independent events. meaning picking men from a group of men has no effect on picking women from group of women. so you have to multiply here.
11c2*9c2 =1980
Answer is E.
Answer: E
Question: Two trains, one from P to Q and the other from Q to P, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 25 hours respectively. The ratio of their speeds is
Options:
A)4:1
B)4:2
C)3:5
D)5:3
E)4:6
Rationale: Ratio of their speeds = Speed of first train : Speed of second train
= √25−−√9
= 5:3
Answer is D.
Answer: D
Question: What is the greatest possible (straight line) distance, between any two points on a hemisphere of radius 6?
Options:
A)0.12
B)1
C)π/2
D)2
E)12
Rationale: Maximum distance straight line is diameter
d = 2r = 12..
ANS option E.
Answer: E
Question: A sum was put at simple interest at a certain rate for 6 years Had it been put at 4% higher rate, it would have fetched 144 more. Find the sum.
Options:
A)500
B)600
C)700
D)800
E)900
Rationale: Difference in S.I. = P×T/100(R1−R2)
⇒ 144 = P×6x4/100(∵ R1 - R2 = 2)
⇒ P = 144×100/6×4=600
Answer B
Answer: B
Question: A bus travel 5/7 of its speed covers 42 km in 1 hr 40 min48 sec. What is theactual speed of the bus ?
Options:
A)25 km/hr
B)35 km/hr
C)42 km/hr
D)46 km/hr
E)55 km/hr
Rationale: Time = 1 hr 40 min 48 sec = 1hr +4060hr+483600hr=1+23+175=12675hrdistance = 42 kmspeed=distancetime=42(12675)=42×75126⇒57 of the actual speed = 42×75126⇒actual speed = 42×75126×75=42×1518=7×153=7×5=35 km/hr
B
Answer: B
Question: Avg cost of 5 pens and 4 pencils isRs. 36.The avg cost of 5 pens and 9 pencils is Rs.56.Find the total cost of 25 pens and 25 pencils?
Options:
A)80
B)90
C)100
D)110
E)120
Rationale: Avg cost of 5 pens and 4 pencils is Rs. 36
(5 pen + 4 pencils) 9=36
therefore 1 pen cost is 36/9=4, so 5 pen cost is 5x4=20 and 4 pencils cost is 4x4=16
Again, avg cost of 5 pens and 9 pencils is Rs.56
(5 pen + 9 pencils) 14=56
therefore 1 pen cost is 56/14=4, so 5 pen cost is 5x4=20 and 9 pencils cost is 9x4=36
So 25 pen cost is (25/a)=100 and 25 pencils cost is (25x4)=100
ANSWER:C
Answer: C
Question: The average weight of a group of boys is 30 kg. After a boy of weight 35 kg joins the group, the average weight of the group goes up by 1 kg. Find the number of boys in the group originally ?
Options:
A)4
B)5
C)6
D)7
E)9
Rationale: Let the number off boys in the group originally be x.
Total weight of the boys = 30x
After the boy weighing 35 kg joins the group, total weight of boys = 30x + 35
So 30x + 35 + 31(x + 1) = > x = 4.
Answer: A
Answer: A
Question: Three coins are tossed. Find the probability of at least 1 head and 1 tail ?
Options:
A)1/2
B)1/4
C)3/4
D)7/8
E)1/8
Rationale: n(s) = 2^3 = 8 ....(3 coins tossed)
let E is the event of getting at least 1 head and 1 tail
n(E) = TTH , HTT, THT, HHT, HTH, THH = 6
P(E) = n(E) / n(s)
= 6/8
= 3/4
Ans- C
Answer: C
Question: Calculate the value of 5.11 x 10^8 ?
Options:
A)511000000
B)5110000
C)511000
D)51100000
E)51100
Rationale: 5.11 x 10^6 = 5.11 x 100000000 = 511000000.
Answer = A
Answer: A
Question: The ratio of two quantities is 5 to 6. If each of the quantities is increased by 15, what is the ratio of these 2 new quantities?
Options:
A)5:6
B)25:27
C)15:16
D)20:21
E)It cannot be determined from the information given
Rationale: The ratio cannot be straight way added any quantity...
5:6 means 5x:6x...
So when you add a quantity, it becomes 5x + 15 : 6x +15..
So value of x is must..
Ans E
Answer: E
Question: In how many ways a committee, Consisting of 3 men and 5 women can be formed from 8 men and 10 women?
Options:
A)266
B)86400
C)11760
D)14112
E)14720
Rationale: To choose Required number of ways to form a committee:
Select 3 men out of 8 men and 5 women out of 10 women.
That is 8C3 * 10C6.
8C3=8!/3!*5!
=56
10C5=10!/5!*5!
=252
The Required number of ways=252*56=14112
D
Answer: D
Question: A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron
weighs 8g/ cm^3 , then the weight of the pipe is
Options:
A)4.996 kg
B)5.696 kg
C)3.796 kg
D)3.696 kg
E)3.996 kg
Rationale: External radius = 4 cm,
Internal radius = 3 cm.
Volume of iron = [ 22/7 x (4 ^2 x 3 ^2) 21] cm^3 = 462 cm^3
Weight of iron = (462 x 8)gm = 3696 gm = 3.696 kg
ANSWER:D
Answer: D
Question: Mario and Nina each have a bag of marbles, each of which contains 10 blue marbles, 10 red marbles, an 10 white marbles. If Mario and Nina each select one marble from their respective bags, what is the probability that either Mario or Nina select a red marble?
Options:
A)3/4
B)2/9
C)1/4
D)1/8
E)1/16
Rationale: Mario and Nina each select one marble from their respective bags.
Probability that either Mario or Nina select a red marble = Probability that Mario selects a red marble + Probability that Nina selects a red marble
Probability that either Mario or Nina select a red marble = (10/30)*(10/30) + (10/30)*(10/30) = 2*(1/9)
Probability that either Mario or Nina select a red marble = 2/9
Answer would be B.
Answer: B
Question: John tossed a fair coin 5 times. What is the probability that the coin landed heads up exactly twice?
Options:
A)0.156
B)0.256
C)0.35
D)0.456
E)0.555
Rationale: The probability is exactly same as the cion landed tail up once. That is 5*(1/2)^5 = 5/32 =0.156.
The answer is, therefore, (A).
Answer: A
Question: The bankers discount and the true discount of a sum at 20% per annum simple interest for the same time are Rs.200 and Rs.150 respectively. What is the sum and the time?
Options:
A)Sum = Rs.600 and Time = 1.6 years
B)Sum = Rs.500 and Time = 1.6 years
C)Sum = Rs.400 and Time = 1.6 years
D)Sum = Rs.600 and Time = 3 years
E)None of these
Rationale: Explanation :
BD = Rs.200
TD = Rs.150
R = 20%
F = BD ×TD/(BD – TD)=200×150/(200–150)=200×150/50=Rs. 600
BD = Simple Interest on the face value of the bill for unexpired time =FTR/100
⇒200 = (600×T×20)/100
⇒200 = 6 × T × 20 ⇒20 = 12 × T
⇒ T = 20/12 = 1.6 years Answer : Option A
Answer: A
Question: P and Q had the same number of candies. P gave Q 16 candies so that now Q has five times as many candies as P has. How many candies does Q have now?
Options:
A)25
B)30
C)35
D)40
E)45
Rationale: Let P and Q be the number of candies they had at the start.
P = Q
5(P-16) = Q+16
5(Q-16) = Q+16
4Q = 96
Q = 24
Now person Q has 24+16 candies which is 40 candies.
The answer is D.
Answer: D
Question: The average weight of 8 persons increases by 3.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person ?
Options:
A)75 kg
B)65 kg
C)55 kg
D)85 kg
E)93 kg
Rationale: Total weight increased = (8 x 3.5) kg = 28 kg.
Weight of new person = (65 + 28) kg = 93 kg.
Answer: E
Answer: E
Question: D and E are two points respectively on sides AB and AC of triangle ABC such that DE is parallel to BC. If the ratio of area of triangle ADE to that of the trapezium DECB is 144:25 and DE=13 cm, then find the length of BC.
Options:
A)5
B)13
C)14
D)11
E)15
Rationale: ABC and ADE are similar triangles.
So (side of ABC/side of ADE)^2=25/169
side of ABC/side of ADE = 5/13
So the length of BC = 5
Answer - A
Answer: A
Question: 0.99999999/1.0001−0.99999991/1.0003
Options:
A)10^(-8
B)3*10^(-8)
C)3*10^(-4)
D)2*10^(-4)
E)10^(-4)
Rationale: look at the options closely
A. 0.00000001
B. 0.00000003
C. 0.0003
D. 0.0002
E. 0.0001
Only one option is an even number, the rest odd (an esoteric sort of even/odd number)
both fractions in the question are odds
odd minus odd is always even.
So answer must be even.
Only D is even.
Answer: D