divelab/Sys2Bench huggingface dataset:
Question: A recipe requires 2 1/2 (mixed number) cups of flour 2 3/4 (mixed number) cups of sugar and 1 1/3 (mixed number) cups of milk to make one cake. Victor has 15 cups if flour, 16 cups of sugar and 8 cups of milk. What is the greatest number of cakes tom can make using this recipe?
Options:
A)5
B)6
C)7
D)8
E)9
Rationale: Less work up front: Go through each item and see what the greatest number of cakes you can make with each. The lowest of these will be the right answer.
Flour: 15 cups, we need 2.5 cups each. Just keep going up the line to see how many cakes we can make: That means I can make 2 cakes with 5 cups, so 6 cakes overall with 15 cups. I've already got the answer narrowed to either A or B.
Sugar: 16 cups, we need 2.75 cups each. Same principle. I can make 2 cups with 5.5 cups, so to make 6 cakes I'd need 16.5 cups. I don't have that much sugar, so we're limited to 5 cakes. No need to even do milk because we're already at 5. Sugar will be the limiting factor.
answer is A
Answer: A
Question: If the average (arithmetic mean) of a and b is 45 and the average of b and c is 85, what is the value of c âˆ’ a?
Options:
A)25
B)80
C)90
D)140
E)It cannot be determined from the information given
Rationale: the arithmetic mean of a and b =(a+b)/2=45 -- a+b=90 --1
similarly for b+c=170 --2
subtracting 1 from 2 we have
c-a=80;
ANSWER:B
Answer: B
Question: Eighty percent of the lights at Hotel California are on at 8 p.m. a certain evening. However, forty percent of the lights that are supposed to be off are actually on and ten percent of the lights that are supposed to be on are actually off. What percent L of the lights that are on are supposed to be off?
Options:
A)22(2/9)%
B)16(2/3)%
C)11(1/9)%
D)10%
E)5%
Rationale: Is the Answer D.
let me try..
Let the light which are supposed to be OFF = SO
Let the light which are supposed to be ON = SN
Let the light which are ACTUALLY OFF = AO
Let the light which are ACTUALLY ON = AN
Let the total no. of lights be 100,
So ACTUALLY ON lights = 80
And ACTUALLY OFF lights = 20
Also given >> forty percent of the lights that are supposed to off are actually on >>> (40/100)*SO are ACTUALLY ON
it means >>> (60/100)*SO are ACTUALLY OFF
Also given >> ten percent of the lights that are supposed to be on are actually off >>> (10/100)*SN are ACTUALLY OFF
it means >>> (90/100)*SN are ACTUALLY ON
So, Total ACTUALLY ON lights = (40/100)*SO + (90/100)*SN = 80
and Total ACTUALLY OFF lights = (60/100)*SO + (10/100)*SN = 80
From here we get SO = 20
we need to find: What percent of the lights that are on are supposed to be off >>> So light ACTUALLY ON are 80 and light which are ACTUALLY ON, which are supposed to be OFF = (40/100)*SO = 8.
So (8/80)*100 L= 10%.D
Answer: D
Question: P is three times as fast as Q and working together, they can complete a work in 24 days. In how many days can Q alone complete the work?
Options:
A)16
B)77
C)32
D)55
E)778
Rationale: P = 3Q
P + Q = 3Q + Q = 4Q
These 4Q people can do the work in 24 days, which means Q can do the work in 96 days.
Hence, P can do the work in 32 days.
Answer:C
Answer: C
Question: In a room with 9 people, 6 people have exactly 1 friend in the room and 5 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if Jane is Paul's friend, Paul is Jane's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?
Options:
A)5/21
B)3/7
C)4/7
D)5/7
E)1
Rationale: (6/9)(7/8) + (5/9)(6/8) if you choose one of the 6 with one other friend, then you have a 7/8 chance of not picking their friend 2nd. If you choose one of the 5 with 2 friends, you have a 6/8 chance of not picking one of their friends second. Add them up.
42/72 + 30/72
72/72 = 1
E. 1
Answer: E
Question: Divide Rs. 1500 among A, B and C so that A receives 1/3 as much as B and C together and B receives 2/3 as A and C together. A's share is?
Options:
A)326
B)372
C)375
D)381
E)312
Rationale: A+B+C = 1500
A = 1/3(B+C); B = 2/3(A+C)
A/(B+C) = 1/3
A = 1/4 * 1500 => 375
Answer: C
Answer: C
Question: If 40% of a number is equal to two-third of another number, what is the ratio of first number to the second number?
Options:
A)2:5
B)1:4
C)3:7
D)5:3
E)2:3
Rationale: Let 40% of A = 2/3 B
then 40A/100 = 2B/3
2A/5 = 2B/3
A/B = 5/3
A:B = 5:3
Answer is D
Answer: D
Question: Jim has arrived late for his golf tee time with his friends. When he arrives at the course, he discovers that his friends have already walked 180 yards ahead on the course and that they are continuing to walk forward at a rate of 1.5 yards per second. A bystander with a golf cart takes pity on Jim and drives him to meet his friends at the rate of 4 yards per second. Once Jim meets his friends, he exits the cart and walks with them at their rate for the next 120 yards. What is Jim’s approximate average speed for his total trip?
Options:
A)1.8 yards per second
B)2.0 yards per second
C)2.4 yards per second
D)2.7 yards per second
E)3.0 yards per second
Rationale: Given Jim's friends has a lead of 180 yards and moving at 1.5 yards/sec speed
Jim on golf cart with speed 4 yards/sec has to meet this friends, assuming he meets them after "t" secs
So the total distance traveled by Jim and his friends should be same so as to meet them=> 180+1.5t=4t
=> 180=2.5t
=>72 secs, so Jim meets his friends after traveling 72 secs at 4 yards/sec speed---First part of his journey (288 yards in 72 secs)
Now, Jim travels 120 yards with speed 1.5 yards/sec in 120/1.5=80 secs---second part of his journey (120 yards in 80 secs)
So avg speed of Jim= total distance traveled/ total time
=> (288+120)/(72+80)=408/152
=>2.7 yards/sec approx
Answer: D
Answer: D
Question: What is the minimum value of z for which z^2 + z - 3/8> 0 is not true?
Options:
A) -5/2
B) -3/2
C) -1/2
D) -3/4
E) 1/2
Rationale: I think D -3/4 is the best ans...
Answer: D
Question: A set of 16 different integers has median of 25 and a range of 25. What is greatest possible integer that could be in this set?
Options:
A)32
B)37
C)40
D)44
E)50
Rationale: The correct answer is D.
A set of 16 different integers has median of 25 and a range of 25. What is greatest possible integer that could be in this set?
Given 15 different integers, lets say
n1, n2, n3, n4, n5, n6, n7, n8, n9, n10, n11, n12, n13, n14, n15
Also given median is 25 i.e. n8 = 26
n1, n2, n3, n4, n5, n6, n7, 25, n9, n10, n11, n12, n13, n14, n15
As each integer is different we need to find the maximum values for all those numbers before the median.
the maximum value n7 can have is one less then the median i.e. 24, similarly n6 will be one less than 24 i.e. 23 ... using this process the values for all number before the median would be..
18, 19, 20, 21, 22, 23, 24, 25, n9, n10, n11, n12, n13, n14, n15
Also given the range is 25 i.e. n15 - n1 (18) = 26
The maximum value n15 can have is 26 + n1 (18) = 44
D
Answer: D
Question: The forth proportional to 2,4,6 is?
Options:
A)8
B)12
C)15
D)18
E)22
Rationale: Let the fourth proportional to 2,4,6 be x
then 2:4::6:x
2x = 4*6
x = 12
Answer is B
Answer: B
Question: 2, 3, 5, 16, 231, ?
Options:
A)53789
B)53210
C)53051
D)52109
E)53105
Rationale: 2
3
32 - 22 = 5
52 - 32 = 16
162 - 52 = 231
2312 - 162 = 53105
Answer is E
Answer: E
Question: Ramesh can finish a piece of work in 10 days. Rohan is twice as efficient as Ramesh.If they work together how many days will they need to finish the same amount of work?
Options:
A)8/3 days
B)9/3 days
C)10/3 days
D)11/3 days
E)7/3 days
Rationale: Ramesh can finish a piece of work in 10 days. Rohan is twice as efficient as Ramesh.
So Rohan can finish a piece of work in 5 days.
days needed to finish the same amount of work = 1/(1/5+1/10)=10/3 days
ANSWER:C
Answer: C
Question: A boat crossed a lake from North to South at the speed of 4 km/h, entered a river and covered twice as much distance going upstream at 3 km/h. It then turned around and stopped at the south shore of the lake. If it averaged 3.2 km/h that day, what was its approximate downstream speed?
Options:
A)3
B)5
C)6
D)7
E)8
Rationale: Another way will be:
Total time for the trip = distance covered in Lake/speed in still water +distance covered upstream/upstream speed + distance covered downstream/downstream speed
If 5D is the total distance, then distance covered in lake = D, distance covered upstream = 2D and distance covered downstream = 2D
=>5D/3.2 = D/4 + 2D/3 + 2D/x (where x is the downstream speed)
=> 5/3.2 = 1/4 + 2/3 +2/x (div. both sides by D)
=> 1.56 = .25+.66 +.65/x => x=3.09 ~ 3
option A
Answer: A
Question: The speed of a boat in still water is 60kmph and the speed of the current is 20kmph. Find the speed downstream and upstream?
Options:
A)80,70 kmph
B)80,40 kmph
C)10,40 kmph
D)80,80 kmph
E)84,40 kmph
Rationale: Speed downstream = 60 + 20
= 80 kmph
Speed upstream = 60 - 20
= 40 kmph
Answer: B
Answer: B
Question: There are 6 gentlemen and 4 ladies to dine at a round table. In how many ways can they seat themselves so that no two ladies are seated together.
Options:
A)14400
B)36400
C)86400
D)65600
E)88600
Rationale: First arrange the Gents. This is a circular arrangment. 6 Gents can be arranged in 5! ways.
Now there are5 spaces (space in between two gents) and we have to seat 4 ladies there.
First select 4 places out of 5 available places. can be done in 5C4 ways. Then arrange 4 Ladiess there in 4! ways.
Total: 5!*5C4*4! = 14400
A is the answer.
Answer: A
Question: How many times are the hands of a clock at right angles in a day?
Options:
A)42
B)44
C)49
D)41
E)47
Rationale: In 12 hours, they are at right angles 22 times.
==> In 24 hours, they are at right angles 44 times.
Answer is B.
Answer: B
Question: The function f(Y) represents the number of ways that prime numbers can be uniquely summed to form a certain number Y such that Y = a + b + c + d… where those summed variables are each prime and a ≤ b ≤ c ≤ d ... For instance f(8) = 3 and the unique ways are 2 + 2 + 2 + 2 and 2 + 3 + 3 and 3 + 5. What is f(12)?
Options:
A)4
B)5
C)6
D)7
E)8
Rationale: It is so better to start with 2 and check whether sum of two primes is primes is even.
1) 2(6 times)
2) 2(3 times) + 3(2 times)
3) 2(2 times) + 3 + 5
4) 2 + 3 + 7
5) 2 + 5 + 5
6) 3(4 times)
7) 5 + 7
Answer: D
Answer: D
Question: In what time will Rs.4400 lent at 3% per annum on simple interest earn as much interest as Rs.5000 will earn in 5 years at 4% per annum on simple interest?
Options:
A)7 4/3
B)7 4/6
C)7 4/7
D)7 1/2
E)7 1/5
Rationale: (4400*3*R)/100 = (5000*5*4)/100
R = 7 4/7
Answer: C
Answer: C
Question: 1,257
1,275
1,527
........
........
+7,521
The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,5 and 7 exactly once in each integer. What is the sum T of the 24 integers ?
Options:
A) 26,996
B) 44,406
C) 60,444
D) 66,660
E) 99,990
Rationale: This is the way i solved it :
Total no. of different combos = 24, total digits 4 therefore each digit must repeat 24/4 = 6 times in each row .. So the right most row would add up to 1x6 + 2x6 + 5x6 + 7x6 = 6+12+30+42 = 90 .. Each row would add up to 90, so 90 in the first means we have 9 that carries over and we get 0 , the second time its 90+9 and 9 stays and one 9 goes to the row to the left, so the last two digits of the SUM T should be 90 (E) .. We could go on and solve the exact number but since only one answer choice has the last digits as 90 we needn't go any further..
Answer: E
Question: In Josephs convent school,80% are boys at the same time 75% are girls students.If 88% are boys students then, what is percentage of the girls students in the particular school?
Options:
A)54
B)41
C)50
D)39
E)63
Rationale: Percentage of boys students = 80%; Percentage of girls students = 70% (ie) 0.75; Total = 0.7*100; Percentage of boys students = 88-80 = 8; Percentage of girls students =100-70 =30; So total Percentage of girls students =0.70*30=21;
Total Percentage of girls students = 30+21=41 Answer: B
Answer: B
Question: A can have a piece of work done in 10 days, B can work two times faster than the A, C can work three times faster than A. How many days will they take to do the work together ?
Options:
A)3/5 days
B)5/2 days
C)4/3 days
D)can't say
E)none
Rationale: A's 1 day work is 1/15
B's 1 day work is 2/15
C's 1 day work is 3/15
(A+B+C)'s 1 day work(1/15+2/15+3/15) is 2/5
so, they can finish the work in 5/2 days
ANSWER:B
Answer: B
Question: A card player is dealt a poker hand of 5 cards. Considering that there are 12 face cards and 40 other cards in a deck of cards, how many hands are there in which the player will be dealt 2 face cards and 3 other cards?
Options:
A)410,860
B)580,420
C)652,080
D)712,600
E)880,420
Rationale: (12c2)*(40c3)
i.e.
[12!/(2!*10!)]*[40!/(3!*37!)]
66*9,880
652,080
C
Answer: C
Question: The ratio of flour to water to sugar in a recipe is 10:6:3. The ratio in a new recipe calls for a doubling of the ratio of flour to water from the original recipe and a halving of the ratio of flour to sugar. If the new recipe calls for 2 cups of water, how much sugar is required?
Options:
A)3
B)4
C)5
D)6
E)7
Rationale: The ratio of flour to water is 10:3.
The ratio of flour to sugar is 5:3=10:6.
The new ratio of flour to water to sugar is 10:3:6
If we need 2 cups of water, then we need 4 cups of sugar.
The answer is B.
Answer: B
Question: A father said to his son, "I was as old as you are at the present at the time of your birth". If the father's age is 20 years now, the son's age five years back was:
Options:
A)9 years.
B)8 years.
C)7 years.
D)5 years.
E)6 years.
Rationale: Let the son's present age be x years. Then, (20- x) = x
2x = 20
x = 10.
Son's age 5 years back (10 - 5) = 5 years.
Answer: Option D
Answer: D
Question: A and B walk around a circular track. They start at 9 a.m. from the same point in the opposite directions. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. How many times shall they cross each other before 10:00 a.m.?
Options:
A)8
B)7
C)6
D)5
E)10
Rationale: Sol.
Relative speed = (2 + 3) = 5 rounds per hour.
So, they cross each other 5 times in an hour and 5 times in NEXT hour.
Hence, they cross each other 10 times before 10:00 a.m.
Answer E
Answer: E
Question: A rectangular lawn of dimensions 100 m * 60 m has two roads each 10 m wide running in the middle of the lawn, one parallel to the length and the other parallel to the breadth. What is the cost of traveling the two roads at Rs.3 per sq m?
Options:
A)A)3820
B)B)930
C)C)9309
D)D)4500
E)E)8302
Rationale: Explanation:
Area = (l + b – d) d
(100 + 60 – 10)10 => 1500 m2
1500 * 3 = Rs.4500
Answer: Option D
Answer: D
Question: For positive integers p and q, which of the following can be written as q^2?
Options:
A)p^2+12
B)p^2+11
C)p^2+8
D)p^2+7
E)p^2-7
Rationale: Plug values: if p=4, then 4^2-7=9
The question asks which of the following CAN be written as q^2. If p=4, then p^2-7 CAN be written as 3^2.
Answer: E
Answer: E
Question: Find the compound ratio of (4:3), (1:3) and (2:3) is
Options:
A)16:27
B)12:13
C)13:14
D)14:15
E)31:27
Rationale: Required ratio = 4/3 * 1/3 * 2/3 = 16/27 = 16:27
Answer is A
Answer: A
Question: A train 400 m long can cross an electric pole in 20 sec and then find the speed of the train?
Options:
A)77
B)88
C)72
D)66
E)12
Rationale: Length = Speed * time
Speed = L/T
S = 400/20
S = 20 M/Sec
Speed= 20*18/5 (To convert M/Sec in to Kmph multiply by 18/5)
Speed = 72 Kmph.Answer:C
Answer: C
Question: In a game of billiards, A can give B 40 points in 80 and he can give C 70 points in 80. How many points can B give C in a game of 100?
Options:
A)18
B)27
C)25
D)75
E)17
Rationale: A scores 80 while B score 40 and C scores 10.
The number of points that C scores when B scores 100 = (100 * 10)/40 = 25
In a game of 100 points, B gives (100 - 25) = 75 points to C.
Answer: D
Answer: D
Question: There are 4 children and 3 chairs. In how many ways children can occupy the chairs?
Options:
A)76
B)84
C)56
D)24
E)80
Rationale: total children = 4
total chairs = 3
1st chair can be occupy in 4ways
2nd chair can be occupy in 3ways
3rd chair can be occupy in 2ways
total number of ways = 4*3*2 = 24 ways
correct option is D
Answer: D
Question: Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats?
Options:
A)4:5:6
B)3:4:5
C)1:2:3
D)2:3:4
E)3:2:1
Rationale: Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.
Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).
= (140/100 x 5x), (150/100 x 7x) and (175/100 x 8x)
= 7x, 21x/2 and 14x.
Therefore, The required ratio = 7x : 21x/2 : 14x
= 14x : 21x : 28x
= 2 : 3 : 4.
Answer is D.
Answer: D
Question: What is the least value of x. So that 24x57 is divisible by 3.
Options:
A)2
B)0
C)1
D)3
E)4
Rationale: Explanation:
The sum of the digits of the number is divisible by 3, then the number is divisible by3.
2 + 4 + x + 5 + 7 = 18 + x
Least value of x may be 0
Therefore 18 + 0 = 18 is divisible by 3.
Answer: Option B
Answer: B
Question: At what rate percent on simple interest will Rs.850 amount to Rs.950 in 5 years?
Options:
A)2.35%
B)5.95%
C)4.35%
D)5.33%
E)6.33%
Rationale: 100 = (850*5*R)/100
R = 2.35%
ANSWER:A
Answer: A
Question: The work done by a man, a woman and a child is in the ratio of 3:2:1. There are 20 men, 30 women & 48 children in a factory. Their weekly wages amount to Rs.840, which is divided in the ratio of work done by the men, women and children. What will be the wages of 15 men, 21 women & 30 children for 2 weeks?
Options:
A)Rs.1000
B)Rs.1100
C)Rs.1170
D)Rs.1190
E)Rs.1200
Rationale: sax Ratio of wages of 20 men, 30 women
and 45 chﬂdren perweek
: (3 x 2o):(2 x 30) (1 x 43) 5
Total wages or 20 men per week
= R5 300
Therefore, wages of a man per week : Rs 15,
sirmlarly, wages of woman per week
: Rs 10 and wages ofchwld perweek Rs 5
Total wages of (15 men, 21 women and 30
chﬂdrenj perweek
= 15x15 + 21x10 + 30x5 = 585
Total wages for 2 weeks = Rs.1170
C
Answer: C
Question: When x divided by 288, the remainder is 47. Find the remainder when the same x is divided by 24?
Options:
A)20
B)21
C)23
D)25
E)26
Rationale: 23
Option C
Answer: C
Question: Find the surface area of a 10 cm x 8 cm x 3 cm brick.
Options:
A)84 cu cm.
B)124 cu cm.
C)164 cu cm.
D)180 cu cm.
E)268 cu cm.
Rationale: Sol.
Surface area = [2(10 x 8 + 8 x 3 + 10 x 3)]
= (2 x 134)
= 268 cu cm
Answer E
Answer: E
Question: The average weight of 8 person's increases by 2.5 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)76 kg
B)76.5 kg
C)85 kg
D)None of these
E)Data inadequate
Rationale: Total weight increased = (8 x 2.5) kg = 20 kg.
Weight of new person = (65 + 20) kg = 85 kg.
Answer: Option C
Answer: C
Question: 34.95 + 240.016 + 23.98 = ?
Options:
A)298.946
B)289.946
C)298.964
D)298.496
E)None of them
Rationale: 34.95
240.016
+ 23.98
--------
298.946
Answer is A.
Answer: A
Question: What is the difference between the CI on Rs. 5000 for 1 years at 4% per annum CI yearly and half yearly?
Options:
A)Rs. 2.04
B)Rs. 2.12
C)Rs. 2.23
D)Rs. 3
E)Rs. 3.04
Rationale: C.I. when interest
compounded yearly
= Rs. 5000 x 1 +	4 x 1 + x 4
100	100
= Rs. 5000 x	26	x	51
25	50
= Rs. 5304.
C.I. when interest is
compounded half-yearly
= Rs. 5000 x 1 +	2 3
100
= Rs. 5000 x	51	x	51	x	51
50	50	50
= Rs. 5306.04
Difference = Rs. (5306.04 - 5304) = Rs. 2.04
A
Answer: A
Question: Which of the following describes all values of x for which 1–x^2 < 0?
Options:
A) x > 1
B) x < –1
C) 0 <= x <= 1
D) x < –1 or x > 1
E) –1 <= x <= 1
Rationale: Which of the following describes all values of x for which 1–x^2 >= 0?
(A) x > 1
Plugged in 2. 1–(2)^2 >= 0 -3>=0? yes. But are these all the values? No. Wrong
(B) x < –1
Plugged in -2. 1–(-2)^2 >= 0 -3>=0? yes. But are these all the values? No. Wrong.
(C) 0 <= x <= 1
Start with 0. 1-0 = 1. Wrong. All the other values wrong as well.
(D) x <= –1 or x >= 1
This is the Correct Answer as it captures both A and B. Answer is D.
(E) –1 <= x <= 1
Plug in 0. Value is greater than 0. Wrong.
Which of the following describes all values of x for which 1–x^2 >= 0?
(A) x >= 1
Plugged in 2. 1–(2)^2 >= 0 -3>=0? yes. But are these all the values? No. Wrong
(B) x <= –1
Plugged in -2. 1–(-2)^2 >= 0 -3>=0? yes. But are these all the values? No. Wrong.
(C) 0 <= x <= 1
Start with 0. 1-0 = 1. Wrong. All the other values wrong as well.
(D) x <= –1 or x >= 1
This is the Correct Answer as it captures both A and B. Answer is C.
(E) –1 <= x <= 1
Plug in 0. Value is greater than 0. Wrong.
Which of the following describes all values of x for which 1–x^2 >= 0?
(A) x >= 1
Plugged in 2. 1–(2)^2 >= 0 -3>=0? yes. But are these all the values? No. Wrong
(B) x <= –1
Plugged in -2. 1–(-2)^2 >= 0 -3>=0? yes. But are these all the values? No. Wrong.
(C) 0 <= x <= 1
Start with 0. 1-0 = 1. Wrong. All the other values wrong as well.
(D) x <= –1 or x >= 1
This is the Correct Answer as it captures both A and B. Answer is C.
(E) –1 <= x <= 1
Plug in 0. Value is greater than 0. Wrong.
Correct Answer is D.
Answer: D
Question: AV + CD = AAA, where AV and CD are two-digit numbers and AAA is a three digit number; A, V, C, and D are distinct positive integers. In the addition problem above, what is the value of C?
Options:
A) 1
B) 3
C) 7
D) 9
E) Cannot be determined
Rationale: Since AV and CD are two-digit integers, their sum can give us only one three digit integer of a kind of AAA: 111.
So, A=1 and we have 1V+CD=111
Now, C can not be less than 9, because no two-digit integer with first digit 1 (1V<20) can be added to two-digit integer less than 90, so that to have the sum 111 (if CD<90, so if C<9, CD+1V<111).
Hence C=9.
Answer: D.
Answer: D
Question: The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 45. What is the difference between the two digits of that number?
Options:
A)4
B)5
C)6
D)8
E)9
Rationale: Sol.
Let the ten’s digit be x and unit’s digit be y,
Then, (10x + y) - (10y + x) = 45 ⇔ 9 (x - y) = 45
⇔ x - y = 5
Answer B
Answer: B
Question: How many 4's are there in the following series which are preceded by 7, but are not preceded by 8?
3 4 5 7 4 3 7 4 8 5 4 3 7 4 9 8 4 7 2 7 4 1 3 6
Options:
A)1
B)2
C)3
D)4
E)More than four
Rationale: count 1 - - - - - - 7 4 3
count 2------------748
count 3------------749
count 4------------741
ANSWER:D
Answer: D
Question: The average of seven numbers is 24. The average of first three numbers is 14 and the average of last three numbers is 19. What is the middle number?
Options:
A)27
B)29
C)31
D)34
E)69
Rationale: The total of seven numbers = 7X24 = 168
The total of first 3 and last 3 numbers is = 3 X 14+3 X 19 = 99
So, the middle number is (168 - 99 ) = 69
E
Answer: E
Question: Suresh started a business, investing Rs.18000. After 3 months and 4 months respectively, Rohan and Sudhir joined him with capitals of 12000 and 9000. At the end of the year the total profit was Rs.3850. What is the difference between Rohan’s and Sudhir’s share in the profit?
Options:
A)s.345
B)s.350
C)s.352
D)s.362
E)s.368
Rationale: Suresh : Rohan : Sudhir
Ratio of their investments = 18000 × 12 : 12000 × 9 : 9000 × 8
= 6 : 3: 2
The difference between Rohan’s and Sudhir’s share = 1 share
:.i.e. = Rs. 3850 × 1/11 = Rs.350.
B
Answer: B
Question: Find the area of the quadrilateral of one of its diagonals is 20 cm and its off sets 9 cm and 6 cm?
Options:
A)128
B)150
C)882
D)272
E)121
Rationale: 1/2 * 20(9 + 6) = 150 cm2
Answer: B
Answer: B
Question: A team of 8 persons joins in a shooting competition. The best marksman scored 85 points. If he had scored 92 points, the average score for the team would have been 84. The number of points, the team scored was :
Options:
A)665
B)376
C)998
D)1277
E)1991
Rationale: Explanation:
Let the total score be x.
(x + 92 - 85) / 8 = 84.
So, x + 7 = 672 => x = 665.
Answer: A) 665
Answer: A
Question: If e≠f and e·f≠0, which of the following may be true?
Options:
A)e=f
B)e.f=0
C)(e+f)(e-f)=0
D)e^2 = f^2
E)e^2+f^2 =0
Rationale: except c all conditions are not as per question
C
Answer: C