divelab/Sys2Bench huggingface dataset:
Question: In a bag, there are coins of 25 p, 10 p and 5 p in the ratio of 1 : 2 : 3. If there is Rs. 30 in all, how many 5 p coins are there?
Options:
A)50
B)100
C)150
D)200
E)250
Rationale: Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively.
25p of x,10p of 2x,5p of 3x;
25p=25/100=1/4;
10p=10/100=1/10;
5p=5/100=1/20;
1x/4 + 2x/10 + 3x/20=30;
5x+4x+3x=600;
12x=600 (3*50)
x=150 answer : C
Answer: C
Question: The amount of water (in ml) that should be added to reduce 9 ml. Lotion, containing 50% alcohol, to a lotion containing 30% alcohol, is?
Options:
A)5
B)2
C)3
D)4
E)6
Rationale: 4.5 4.5
30% 70%
30% ----- 4.5
70% ------? => 10.5 - 4.5 = 6 ml
ANSWER E
Answer: E
Question: What is the perimeter of a regular polygon with sides of length 10 cm and internal angles measuring 140° each?
Options:
A)90 cm
B)108 cm
C)120 cm
D)132 cm
E)144 cm
Rationale: Sum of all interior angles is given by:
180(n-2) = n is number of sides
180(n-2) = 140*n
from here
n = 9
So perimeter becomes 10*9= 90
Answer: A
Answer: A
Question: If n = 2×3×5×7×11×13×18, then which of the following statements must be true?
I. n^2 is divisible by 600
II. n + 19 is divisible by 19
III. (n + 4)/2 is even
Options:
A)I only
B)II only
C)III only
D)I and III
E)None of the above
Rationale: Statement I: this is true.
Statement 2 this is NOT true.
Statement 3: (n+4)/2 = n/2 + 2, So this is true.
Hence option (D).
Answer: D
Question: If $x is invested at a constant annually compound interest rate of k percent, what is the ratio of the total amount including interest after 6n years to that after 3n years?
Options:
A)(1+n/100)^k
B)(1+k)^n
C)(1+kn/100)
D)(1+k/100)^3n
E)(1+kn)
Rationale: Amount after 6n years will be = x ( 1 + k/100)^6n
Amount after 3n years will be = x ( 1 + k/100)^3n
So, required ratio will be -
{ x ( 1 + k/100)^6n } / {x ( 1 + k/100)^3n } = (1+k/100)^3n
Hence answer will be a. (1+k/100)^3n
D
Answer: D
Question: What profit percent is made by selling an article at a certain price, if by selling at 2/3rd of that price, there would be a loss of 30%?
Options:
A)20%
B)26%
C)42%
D)5%
E)22%
Rationale: SP2 = 2/3 SP1
CP = 100
SP2 = 70
2/3 SP1 = 70
SP1 = 105
100 --- 5 => 5%
Answer: D
Answer: D
Question: A can complete the job in 3 hours and B can complete the same job in 3 hours. A works for 1 hour and then B joins and both complete the job. What fraction of the job did B complete
Options:
A)1/6
B)3/10
C)1/2
D)5/6
E)8/9
Rationale: 1/6=A
Answer: A
Question: Twelve machines, each working at the same constant rate, together can complete a certain job in 12 days. How many additional machines, each working at the same constant rate, will be needed to complete the Job in 8 days?
Options:
A)2
B)3
C)4
D)6
E)8
Rationale: Another solution which is faster is Since each machine works at a constant rate. The time needs to bought down from 12 to 8. So the new time is 2/3 of the original time. Thus to achieve this we need the rate to be 3/2 of original.
So 3/2*12 =18
So we need 18-12 = 6 more machines.
Answer : D
Answer: D
Question: The current of a stream at 1 kmph. A motor boat goes 35 km upstream and back to the starting point in 12 hours. The speed of the motor boat in still water is?
Options:
A)6 kmph
B)7 kmph
C)9 kmph
D)4 kmph
E)2 kmph
Rationale: S = 1
M = x
DS = x + 1
US = x - 1
35/(x + 1) + 35/(x - 1) = 12
x = 6
Answer: A
Answer: A
Question: The average mark of the students of a class in a particular exam is 80. If 5 students whose average mark in that exam is 40 are excluded, the average mark of the remaining will be 90. Find the number of students who wrote the exam.
Options:
A)87
B)27
C)25
D)26
E)24
Rationale: Let the number of students who wrote the exam be x.
Total marks of students = 80 x.
Total marks of (x - 5) students = 90(x - 5)
80x - (5 * 40) = 90(x - 5)
250 = 10x => x = 25
Answer: C
Answer: C
Question: A and B finish the job in 15 days.while A,B and C can finish it in 5 days . C alone will finish the job in
Options:
A)4.50 days
B)3.50 days
C)6.0 days
D)7.5 days
E)5.50 days
Rationale: Explanation:
5 = (15 * x)/(15 + x)
75+5x=15x
10x=75
X=7.5
Answer: Option D
Answer: D
Question: A car averages 55 mph for the first 4 hours of a trip and averages 70 mph for each additional hour. The average speed for the entire trip was 60 mph. How many hours long is the trip?
Options:
A)6
B)8
C)11
D)12
E)14
Rationale: In 4hr journey = 55*4 = 220mph
Let in the next additional hour journey = 70*x =70x mph
So, (220+70x)/(4+x)= 60
=> 220+70x = 240+60x
=> 10x = 20
=> x = 2h
So the trip is (4+x)=> 6 hours long
ANSWER:A
Answer: A
Question: GOOD is coded as 164 then BAD coded as 21.if ugly coded as 260 then DUMP?
Options:
A)200
B)210
C)216
D)225
E)240
Rationale: GOOD=(7+15+15+4)*4=164
BAD= (2+1+4)*3=21
UGLY= (21+7+12+25)*4=260
coding follows
(sum of position of alphabets)*(no. of letters in the given word)
so,
DUMP=(4+21+13+16)*4= 216
ANSWER:C
Answer: C
Question: We bought 85 hats at the store. Blue hats cost $6 and green hats cost $7. The total price was $600. How many green hats did we buy?
Options:
A)A)36
B)B)90
C)C)40
D)D)42
E)E)44
Rationale: Let B be the number of blue hats and let G be the number of green hats.
B+G=85.
B=85-G.
6B+7G=600.
6(85-G)+7G=600.
510-6G+7G=600.
G=600-510=90.
The answer is B.
Answer: B
Question: A part of certain sum of money is invested at 8% per annum and the rest at 12% per annum, if the interest earned in each case for the same period is equal, then ratio of the sums invested is?
Options:
A)4:6
B)3:2
C)4:3
D)4:1
E)4:2
Rationale: 12:8
= 3:2
Answer: B
Answer: B
Question: 12. What is the dividend. divisor 17, the quotient is 9 and the remainder is 6.
Options:
A)150
B)152
C)154
D)159
E)144
Rationale: D = d * Q + R
D = 17 * 9 + 6
D = 153 + 6
D = 159
ANSWER D
Answer: D
Question: If LCM of two number is 693, HCF of two numbers is 11 and one number is 99, then find other
Options:
A)34
B)77
C)12
D)45
E)None of these
Rationale: Explanation:
For any this type of question, remember
Product of two numbers = Product of their HCF and LCM
So Other number =
693×1199
= 77
Answer: Option B
Answer: B
Question: 1000 men have provisions for 15 days. If 200 more men join them, for how many days will the provisions last now?
Options:
A)10.5
B)12.5
C)13.5
D)11.5
E)11
Rationale: 1000*15 = 1200*x
x = 12.5
ANSWER B
Answer: B
Question: Viswak was born 3 years after his father's marriage. His mother is 8 years younger than his father but 20 years older than Viswak who is 10 years old. At what age did the father get married ?
Options:
A)23 years
B)25 years
C)26 years
D)24 years
E)22 years
Rationale: Explanation:
Viswak's present age = 10 years.
His mother's present age = (10 + 20) years = 30 years.
His father's present age = (30 + 8) years = 38 years.
His father's age at the time of Viswak's birth = (38 - 10) years = 28 years.
Therefore Viswak's father's age at the time of marriage = (28 - 3) years = 25 years.
Answer: B
Answer: B
Question: If a person walks at 15 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance traveled by him is?
Options:
A)50 km
B)76 km
C)18 km
D)60 km
E)97 km
Rationale: Let the actual distance traveled be x km. Then,
x/10 = (x + 20)/15
5x - 300 =>x = 60 km.
Answer: D
Answer: D
Question: If x and y are both odd prime numbers and x < y, how many distinct positive integer Q factors does 2xy have?
Options:
A)3
B)4
C)6
D)8
E)12
Rationale: Since 2xy prime Q factors are x^1*y^1*2^1, its total number or factors must be (1+1)(1+1)(1+1)=2^3=8. Thus, I think D would be the correct answer.
Answer: D
Question: A grocer purchased 80 kg of sugar at 13.50 per kg and mixed it with 120 kg sugar at 16 per kg. At what rate should he sell the mixture to gain 17%?
Options:
A)17 per kg
B)17.55 per kg
C)16.5 per kg
D)16 per kg
E)None of these
Rationale: C.P. of 200 kg of mixture = (80 × 13.50 + 120 × 16)
= 3000.
S.P. = 117% of 3000 = (117⁄100 × 3000) = 3510
∴ Rate of S.P. of the mixture = (3510⁄200) per kg
= 17.55 per kg.
Answer B
Answer: B
Question: A train of 24 carriages, each of 60 meters length, when an engine also of 60 meters length is running at a speed of 60 kmph. In what time will the train cross a bridge 1.5 km long?
Options:
A)1 mins
B)3 mins
C)8 mins
D)6 mins
E)4 mins
Rationale: D = 25 * 60 + 1500 = 3000 m
T = 3000/60 * 18/5 = 180 sec
= 3 mins
Answer: B
Answer: B
Question: Find out the wrong number in the given sequence of numbers.
1, 2, 6, 15, 31, 57
Options:
A)2
B)57
C)15
D)31
E)91
Rationale: 1(result) + (1*1) = 2.
2(result) + (2*2) = 6.
6(result) + (3*3) = 15.
15(result) + (4*4) = 31.
31(result) + (5*5) = 56.
Now we are getting 56 not 57.. So 57 is the wrong number of the given.
answer : B
Answer: B
Question: If C(n, 7) = C(n, 5), find n
Options:
A)15
B)12
C)18
D)-1
E)-2
Rationale: We know that C(n,r) = n! / r!(n - r)!
Now C(n,7) = C(n,5)
n! / 7! (n - 7)! = n! / 5! (n - 5)!
=> 5! (n - 5)! = 7! (n - 7)!
=> [5!] x [(n - 5)(n - 6)[(n - 7)!] = 7 x 6 x 5! x (n - 7)!
=> n2 - 11n + 30 = 42
=> n2 - 11n - 12 = 0
=> (n - 12)(n + 1) = 0
=> n = 12 or n = - 1
But n = - 1 is rejected as n is a non negative integer. Therefore, n = 12
ANSWER:B
Answer: B
Question: A sum of money at simple interest amounts to Rs. 2240 in 2 years and to Rs. 2600 in 5 years. What is the principal amount
Options:
A)1000
B)1500
C)2000
D)2500
E)3500
Rationale: Explanation:
SI for 3 year = 2600-2240 = 360
SI for 2 year 360/3 * 2 = 240
principal = 2240 - 240 = 2000
Answer: Option C
Answer: C
Question: Amanda goes to the toy store to buy 1 ball and 3 different board games. If the toy store is stocked with 3 types of balls and 6 types of board games, how many different selections Q of the 4 items can Amanda make?
Options:
A)9
B)12
C)14
D)15
E)60
Rationale: Amanda goes to the toy store to buy 1 ball and 3 different board games. If the toy store is stocked with 3 types of balls and 6 types of board games, how many different selections of the 4 items can Amanda make?
3! / 1!2! * 6! / 3!3!
Q=3*20=60
E. 60
Answer: E
Question: A wheel rotates 10 times every minute and moves 30 cm during each rotation. How many metres does
the wheel move in two hour?
Options:
A)6 metre
B)12 metre
C)120 metre
D)876 metre
E)360 metre
Rationale: Expl : Number of times wheel moves in 1 hour = 10 * 120 = 1200
:. Distance moves = (1200 *30) cms = 36000 cms
In metres = 360 metre
Answer: E
Answer: E
Question: Look at this series: 31, 29, 24, 22, 17, ... What number should come next?
Options:
A)11
B)15
C)16
D)17
E)18
Rationale: B
15
This is a simple alternating subtraction series, which subtracts 2, then 5.
Answer: B
Question: A Man travelled a distance of 80 km in 10 hours. He travelled partly on foot at 5 km/hr and partly on bicycle at 9 km/hr. What is the distance travelled on foot?
Options:
A)16 km
B)14 km
C)12 km
D)10 km
E)None of these
Rationale: Explanation:
Let the time in which he travelled on foot = x hour
Time for travelling on bicycle = (9 - x) hr
Distance = Speed * Time, and Total distance = 80 km
So,
5x + 10(9-x) = 80
=> 5x = 10
=> x = 2
So distance traveled on foot = 5(2) = 10 km
Answer: Option D
Answer: D
Question: John makes $40 a week from his job. He earns a raise and now makes $70 a week. What is the % increase?
Options:
A)16%
B)16.66%
C)17.9%
D)18.12%
E)75%
Rationale: Increase = (30/40)*100 = (3/4)*100 = 75%.
E
Answer: E
Question: At a conference, the only events scheduled for the day are seminars and breaks. All seminars are the same length and are 10 times as long as each break. A break must be between two consecutive seminars. If the first seminar begins at 9:00 a.m. and the third one ends at 11:20 a.m., how long will the conference last if there are a total of 6 seminars that day?
Options:
A)hours and 15 minutes
B)hours and 20 minutes
C)hours and 25 minutes
D)hours and 30 minutes
E)hours and 35 minutes
Rationale: Well its like this S B S B S B S B S B S
the interval between 9 am to 11:40 am is 160 min having 3 seminars and 2 breaks .If 1 break is of 5 min than time for 3 seminars = 160 - 2*5=150
Time of 1 seminar =150 /3=50 min .
So calculating for 6 seminars and 5 breaks ,total time comes out to be 5 hrs and 20 min .
Answer is B.
Answer: B
Question: When n is divided by 19, the remainder is 7. When n is divided by 11, the quotient is equal to Z, and the remainder is equal to 7. Z must be divisible by which of the following numbers?
Options:
A)16
B)18
C)20
D)19
E)22
Rationale: N = 19 q + 7
N = 11 z + 76
Equating
19 Q + 7 = 11z+7
11z=19q
11z/19 = q
since q is an integer Z must be divisible by 19.
Hence D
Answer: D
Question: In the formula A = (2r)^3, if r is halved, then A is multiplied by ?
Options:
A)64
B)1/8
C)1
D)8
E)1/64
Rationale: Say r=2 => A1 = 64
when r=1; V2 = 8
V2 = 1/8*V1.
Answer : B
Answer: B
Question: In a 8 x 8 chess board what is the total number of squares.
Options:
A)238
B)204
C)678
D)169
E)161
Rationale: Ans: The total number of squares in a n x n chess board is equal to "the sum of first n natural number squares"
i.e., n(n+1)(2n+1)6n(n+1)(2n+1)6
So Substituting 8 in the above formula we get 204
Answer:B
Answer: B
Question: Two circular signs are to be painted. If the radius of the larger sign is three times that of the smaller sign, how many times more paint is needed to paint the larger sign? (We can assume that a given amount of paint covers the same area on both signs.)
Options:
A)3
B)9
C)12
D)15
E)18
Rationale: Let R be the radius of the smaller sign.
The area A of the smaller sign is A = piR^2.
The area of the larger sign is pi(3R)^2=9piR^2=9A.
Since the area is 9 times larger, we need 9 times more paint for the larger sign.
The answer is B.
Answer: B
Question: If 20 machine can finish a job in 36 days, then how many more machines would be needed to finish the job in one -third less time?
Options:
A)a. 4
B)b. 8
C)c. 10
D)d. 12
E)e. 16
Rationale: You might think of this in a management context - we can use the principle of 'person-hours' to solve any problem where we have identical workers. So, using simpler numbers, suppose you know that 6 identical employees, working simultaneously, would finish a job in 5 hours. Then that job requires 6*5 = 30 total hours of person-work. If instead you wanted the job done in 3 hours, you'd assign 30/3 = 10 employees to do the job, because you want to get a total of 30 hours of work from the employees.
We can solve this problem identically. If 20 machines (identical ones, I assume) work simultaneously for 36 days, they will do a total of 20*36 machine-days of work. So the job requires 20*36 days of machine work in total. We instead want the job done in 1/3 less time, so in 24 days. So we'll need 20*36/24 = 30 machines, or 10 additional machines.
C
Answer: C
Question: In terms of percentage profit, which among following the best transaction.
Options:
A)C.P. 36, Profit 17
B)C.P. 50, Profit 24
C)C.P. 40, Profit 19
D)C.P. 60, Profit 29
E)C.P. 70, Profit 39
Rationale: Explanation:
Hint: Calculate profit percent as
Profit% = (profit/cost) * 100
ANSWER IS D
Answer: D
Question: A and B can do a piece of work in 3 days, B and C in 4 days, C and A in 6 days. How long will C take to do it?
Options:
A)18 days
B)20 days
C)24 days
D)30 days
E)32 days
Rationale: 2c = ¼ + 1/6 – 1/3 = 1/12
c = 1/24 => 24 days
ANSWER:C
Answer: C
Question: A, B, C and D enter into partnership. A subscribes 1/3 of the capital B 1/4, C 1/5 and D the rest. How much share did A get in a profit of Rs.2460?
Options:
A)14028
B)14029
C)14027
D)14000
E)14298
Rationale: 25*12: 30*12: 35*8
15:18:14
14/47 * 47000 = 14000
Answer: D
Answer: D
Question: Two cars A and B are running towards each other from different places 88 km apart. If the ratio of the speeds of the cars A and B is 5 : 6 and the speed of the car B is 90 km per hour then after how long will the two meet each other?
Options:
A)2263 minutes
B)24 minutes
C)32 minutes
D)36 minutes
E)None of these
Rationale: Speed of the car A = 5⁄6 × 90 = 75 km/hr
\ Reqd time = 88/90+75×60=32minutes
Answer C
Answer: C
Question: If A can do a work in 8 days and B can do the same work in days,in how many days A and B together can finish the work
Options:
A)3 3/7
B)4 3/7
C)5 3/7
D)6 3/7
E)None of these
Rationale: no of days taken to finish the work=8*6/(8+6)
Instead of solving this directly take the LCM of 8 and 6
8*6/(8+6)=LCM(8,6)/(LCM(8,6)/8 + LCM(8,6)/6)=24/(3+4)=24/7=3 3/7
Answer A
Answer: A
Question: On Saturday, the original price of a shirt was discounted by 20%. On Monday, the price of the shirt was marked up p percent so that the new price on Monday was 20% more than the original price before the discount. What is the value of p?
Options:
A)20%
B)30%
C)33 1⁄3%
D)40%
E)50%
Rationale: Let Original price be O
Saturday price = 0.8 x O
Monday price = 1.P x 0.8x O
Monday price is 20 percent higher than original.price
So 1.P x 0.8 X O = 1.2 x O
1.P = 1.5
Hence P = 50 %
ANSWER:E
Answer: E
Question: The ratio of length of two refill is 9:8.2nd refill got damaged when only half length of 2nd refill get used.The ratio of total no. of pages written by both refill.
Options:
A)8:3
B)6:5
C)9:8
D)9:4
E)5:4
Rationale: the no. of page written is directly proportional to length of refill.
so, initial no. of page to be write =9x,8x.
now,2nd refill only can write 4x page
so new ratio of page written =9x/4x=9:4.
answer D
Answer: D
Question: if {x+y}/{x-y}= 4/3 and x≠0, then what percentage (to the nearest integer) of x+3y is x–3y ?
Options:
A)40%
B)30%
C)55%
D)36%
E)None of these
Rationale: Explanation :
=>(x+y)/(x-y)=4/3.
=>(x/y)+1/(x/y)−1=4/3.
Cross-multiplying this equation yields
=>3x/y+3=4x/y−4.
Solving for x/y yields x/y=7.
Now, the percentage of x+3y the expression x−3y makes is
=>x−3y /x+3y∗100
Dividing both the numerator and the denominator of the expression by y, gives:-
=>((x/y)−3/(x/y)+3))×100.
=>((7−3)/(7+3))×100.
=>40%.
Answer : A
Answer: A
Question: How long does a train 200 m long running at the speed of 72 km/hr takes to cross a bridge 132 m length?
Options:
A)82.1 sec
B)12.1 sec
C)16.1 sec
D)13.1 sec
E)16.6 sec
Rationale: Speed = 72 * 5/18 = 20 m/sec
Total distance covered = 200 + 132 = 332 m.
Required time = 332/20 = 16.6 sec.
Answer:E
Answer: E
Question: Two trains leave the same train station at 6:00 am and 6:30 am and they travel at 100 kph and 120 kph respectively. How many kilometers from the train station will the two trains be together?
Options:
A)300
B)320
C)340
D)360
E)380
Rationale: At 6:30, the first train is 50 km ahead.
The second train gains 20 km each hour.
The time it takes the second train to catch the first train is 50 /20 = 2.5 hours.
In this time, the second train travels 2.5*120=300 km.
The answer is A.
Answer: A
Question: One hour after Matthew started waking from e to y, a distance of 45km, Johnny started walking along the same road from y to e. Matthew's walking rate was 3 km per hour and Johnny's was 4km per hour, how many km had Johnny walked when they met?
Options:
A)24
B)23
C)22
D)21
E)19.5
Rationale: Just an alternative method...
After the first hour the distance e is 42 km (45-3). Now the problem can be treated as if bothof them started at the same time. Since the speeds are in the ratio 3 : 4, the distances will also be in the same ratio. Splitting 42 in that ratio we get 18 : 24. So answer is 24.
Hence A.
Answer: A
Question: what is the Max number of rectangular boxes, each measuring 4 inches by 6 inches by 10 inches, that can be packed into a rectangular packing box measuring 16 inches by 18 inches by 30 inches, if all boxes are aligned in the same direction?
Options:
A)36
B)14
C)12
D)15
E)11
Rationale: The 4 inch side should be aligned to the 16 inch side ( 4 layer)
6 inch side should be aligned to the 18 inch side. (3 layer)
and 10 inch side should be aligned to the 30 inch side.(3 layer)
Maximum number of rectangles =4*3*3=36
Answer is A
Answer: A
Question: In an arithmetic progression the difference between the any two consecutive terms is a constant. What is the arithmetic mean of all of the terms from the first to the 29th in an arithmetic progression if the sum of the 13th and 17th terms of the sequence is 102?
Options:
A)43
B)47
C)51
D)55
E)59
Rationale: Let x be the difference between any two consecutive terms.
The mean of a sequence like this is the middle term, thus the 15th term in the sequence.
Then the mean of the 14th and 16th term is also equal to the overall mean, because the 14th term is (15th term - x) and the 16th term is (15th term + x).
Similarly the mean of the 13th and 17th term is also equal to the mean.
etc...
Thus the mean is 102/2 = 51
The answer is C.
Answer: C
Question: Hariprasad and Madhusudan started a business, investing sums in the ratio of 2 : 3. If Hariprasad had invested an additional amount of 10,000 the ratio of Hariprasad's investment to Madhusudan's investment would have been 3 : 2. What was the amount invested by Hariprasad?
Options:
A)8000
B)12000
C)9000
D)Data inadequate
E)None of these
Rationale: Let the initial investments of Hariprasad and Madhusudan be 2x and 3x,respectively.
From the question,
2x+10000/3x=32
or, 4x + 20000 = 9x
\ x = 4000
\ Amount invested by Hariprasad = 2x = 8000
Answer A
Answer: A
Question: In 1950, Richard was 4 times as old as Robert. In 1955, Richard was 3 times as old as Robert. In which year was Richard 1.5 times as old as Robert?
Options:
A)1960
B)1965
C)1970
D)1975
E)2000
Rationale: In 1950: Ri = 4 Ro -------------- eq1
In 1955: Ri +5 = 3(Ro+5)---------eq2
Thus in 1950, solving eq1 and eq2
Ro= 10 , Ri =40
Now for each year we can calculate:
2000: Ri= 60, Ro = 90
Thus Ans:E
Answer: E
Question: In a hot dog eating contest the average number of hot dogs that were eaten by boys was 7 each, and the average number eaten by girls was 5 each. If the total number of hot dogs eaten during the contest was 50, what was the total number of boys and girls in the contest?
Options:
A)6
B)7
C)8
D)9
E)10
Rationale: We can determine quickly that total number should range between 50/7<= N <= 50/5 , so ans should be between 7 and 10.
Now solving the expression 7A+5B =50
decreasing 50 by multiples of 7 and checking divisibility of that number by 5, we get
5 boys and 3 girls will total 50 hot dogs, 5*7 + 3*5 = 50
Hence total number is 5 + 3 = 8
Ans C
Answer: C
Question: A certain restaurant offers 8 different salads, 6 different main courses, 6 different desserts. If customers choose one salad, one main course and two different desserts for their meal, how many different meals are possible?
Options:
A)120
B)240
C)480
D)600
E)1200
Rationale: MAGOOSHOFFICIAL SOLUTION:
To count the number of meals, we have to count the possibilities for the three components, and then, according to theFundamental Counting Principle, we multiply. For more on the FCP, seethis blog.
For salads, there are 8 possibilities. Done.
For main courses, there are 5 different possibilities. Done.
For desserts, there are 6 choices, but we pick two different choices. This would be 6C2. To calculate this, we can usethe shortcut, nC2 = (n*(n-1))/2.
6C2 = (6*5)/2 = 15
So, there are 15 combinations of two different desserts.
For the number of meals, we multiply 8*6*15=720
Answer = A
Answer: A
Question: A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 650 sq. feet, how many feet of fencing will be required?
Options:
A)34
B)40
C)85
D)88
E)None
Rationale: EXPLANATION
We have: l = 20 ft and lb = 650 sq. ft.
So, b = 32.5 ft.
Length of fencing = (l + 2b) = (20 + 65) ft = 85 ft.
Answer C
Answer: C
Question: If an article is sold at 19% profit instead of 12% profit, then the profit would be Rs. 105 more. What is the cost price?
Options:
A)1533
B)1543
C)1500
D)1278
E)1202
Rationale: Explanation:
Let the cost price of an article be Rs. x.
(19% of x) - (12% of x) = 105
19x/100 - 12x/100 = 105 => 7x = 105 * 100
=> x = 1500
Cost price = Rs. 1500
Answer:C
Answer: C