divelab/Sys2Bench huggingface dataset:
Question: If 1/3 of the passengers on a ship are from North America, 1/8 are Europeans, 1/5 are from Africa, 1/6 are from Asia and the remaining 42 people are citizens of other continents, then how many passengers are on board the ship?
Options:
A)240
B)250
C)280
D)300
E)320
Rationale: 1/3 + 1/8 + 1/5 + 1/6 = (40+15+24+20)/120 = 99/120
Let x be the number of passengers on the ship.
42 = (21/120)x = 7x/40
x = 240
The answer is A.
Answer: A
Question: Exactly 36% of the numbers in set S are even multiples of 3. If 40% of the even integers in set S are not multiples of 3, what percent of the numbers in set S are not even integers?
Options:
A)76%
B)60%
C)50%
D)40%
E)24%
Rationale: Let S be the total number of elements in set S.
We know that 0.36*S is the number of even multiples of three.
Let N be the number of even numbers.
We know 0.40*N are even numbers not multiple of three. This also means, 0.60*N are even numbers that are multiples of three.
Therefore:
0.06*N = 0.36*S
N = (0.36/0.60)*S = (36/60)*S = (6/10)*S = 0.60*S = 60% of S
So even numbers are 60% of S. That means, 40% of S are not even integers. Answer = D
Answer: D
Question: In a party there are 5 couples. Out of them 5 people are chosen at random. Find the probability that there are at the least two couples?
Options:
A)5/21
B)5/29
C)5/23
D)5/24
E)5/26
Rationale: Number of ways of (selecting at least two couples among five people selected) = (5C2 * 6⁶C1)
As remaining person can be any one among three couples left.
Required probability = (5C2 * 6C1)/10C5
= (10 * 6)/252 = 5/21
Answer:A
Answer: A
Question: Billy and Bonnie are opposing participants in an online auction. Each has a constant bidding strategy: Billy always bids up by raising $40 over any previous bid. Bonnie always bids up by doubling any previous bid. The auction begins with an initial bid by a third player, and progresses as follows: Billy bids up first, and Bonnie follows. Billy bids up again, and Bonnie follows and raises the bid to a final $600. What was the dollar value of the initial price?
Options:
A)464
B)46
C)90
D)466
E)4673
Rationale: Initial bid = p$
Billy = p+40
Bonnie = 2p
The sequence is R---> B ---> R -----> B ---> 600
Thus per the sequence we have: p+40, 2(p+40), 2p+80+40 , 2(2p+120)
Now given, 2(2p+120) = 600----> p =90$. Thus C is the correct answer.
Answer: C
Question: Last year Sandy saved 10% of her annual salary. This year, she made 10% more money than last year, and she saved 15% of her salary. The amount saved this year was what percent of the amount she saved last year?
Options:
A)142%
B)150%
C)165%
D)176%
E)188%
Rationale: Let last year's salary be x.
Last year, Sandy save 0.1x
This year, Sandy saved 0.15*1.1x = 0.165x
0.165x / 0.1x = 1.65 = 165%
The answer is C.
Answer: C
Question: Two, trains, one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is:
Options:
A)2:3
B)3:4
C)4:3
D)6:7
E)7:6
Rationale: Train from Howrah to Patna A and Patna to Howrah B.
Then,
(A's speed) : (B's speed) = b : a = 16 : 9 = 4 : 3.
ANSWER :C
Answer: C
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 28 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)01
Rationale: Explanation:
As there is meal for 70 adults and 28 have their meal, the meal left can be catered to 42 adults.
Now, 70 adults = 90 children
7 adults = 9 children
Therefore, 42 adults = 54 children
Hence, the meal can be catered to 54 children.
ANSWER: B
Answer: B
Question: Solve the equation
5(-3x - 2) - (x - 3) = -4(4x + 5) + 13
Options:
A)2
B)5
C)6
D)0
E)9
Rationale: Given the equation
5(-3x - 2) - (x - 3) = -4(4x + 5) + 13
Multiply factors.
-15x - 10 - x + 3 = -16x - 20 +13
Group like terms.
-16x - 7 = -16x - 7
Add 16x + 7 to both sides and write the equation as follows
0 = 0
correct answer D
Answer: D
Question: Can is filled with a mixture of water & milk in the ratio of 3 : 5. Find the quantity of mixture to be drawn off and replaced with water, in order to get the mixture ashalf milk and 1/2water.
Options:
A)1/5
B)2/3
C)3/7
D)1/9
E)2/5
Rationale: Correct option : (c)
A container contains milk and water in the ratio of 3 : 5. This means that the vessel contains 8 litres of mixture.
Assume that x litres of this mixture is replaced with water.
From the mixture containing water and milk of 3 : 5, x quantity of mixture is withdrawn and is replaced by water of the same quantity in the mixture. From the mixture (3x)/(8) part of water is removed and x quantity of water is added.
1) Quantity of water in the newly formed mixture = [3 – (	3x	) + x]
8
---- (3 is the quantity of water, x is the quantity of mixture replaced by water)
2) Similarly, quantity of milk in the newly formed mixture = [5 – (	5x	)]
8
---- (Here x is not added because only water was added in the mixture and not the syrup)
Therefore,
[ 3 – (	3x	) + x] = [5 – (	5x	)]
8	8
Solving 1 and 2, we get
5x + 24 = 40 -5x
x = 8 / 5
So the part of mixture replaced from 8 litres =	8/5x1/8	=	1/5
A
Answer: A
Question: A truck carrying cans of soup contains P stacks. Each stack contains Q cases, and each case contains 2P cans. How many cans of soup are contained in 5 trucks?
Options:
A)5/2(P^2)Q
B)2Q/5
C)2P^2/5Q
D)10(P^2)Q
E)10P^2/Q
Rationale: There are 2P cans in each case.
There are 2PQ cans in each stack.
There are 2(P^2)Q cans in each truck.
There are 10(P^2)Q cans in 5 trucks.
The answer is D.
Answer: D
Question: Rohith,who works in a parcel service, has a certain number of small plastic boxes to pack into parcels. If he packs 3,4,5 or 6 in a parcel, he is left with one over; if he packs 7 in a parcel, none is left over. What is the number of boxes, he may have to pack?
Options:
A)223
B)301
C)377
D)1368
E)222
Rationale: The required number of boxes is such that it leaves a remainder of 1 when divided by 3, 4, 5 or 6 and no remainder of 1 when divided by 3, 4, 5 or 6 and no remainder when divided by 7. Such a number is 301.
Answer:B
Answer: B
Question: The radius of a semi circle is 3.5 cm then its perimeter is?
Options:
A)32.8
B)32.4
C)32.1
D)18
E)32.9
Rationale: Diameter = 7 cm
1/2 * 22/7 * 7+7 = 18
Answer : D
Answer: D
Question: What is the 56743rd term in the series 1234567891011121314…….?
Options:
A)1
B)3
C)7
D)5
E)6
Rationale: 1 to 9=9 no.s 1 digit each no.
9 to 99=90no.s 2 digit each,total digit=90*2=180 terms
99 to 999=900 no.s 3 digit each,total digit=900*3=2700
999 to 9999=9000 no.s 4 digit each,total digit=9000*4=36,000
now till 999,we have=9+180+2700=2889 digits
till 9999 we hv=2889+36000=38889 digits
56743-38889=17854
after 9999 each no. has 5 digit
so 17584/5=3570 with remainder 4.
3570th no after 9999 =9999+3570=13569
next term=13570...4th digit=7 ..so ans=7
ANSWER:C
Answer: C
Question: A sock drawer contains six socks, each one a different solid color. The colors of the socks are red, orange, yellow, blue, indigo, and violet. If a person reaches into the drawer and pulls out two socks, how many different color combinations are possible in the selected pair of socks?
Options:
A)12
B)15
C)21
D)36
E)42
Rationale: Since here order of socks is not important then we'll have 6C2 i.e. 15 combinations.
Answer : B
Answer: B
Question: A cricketer has a certain average for 10 innings. In the eleventh inning, he scored 108 runs, there by increasing his average by 6 runs. His new average is:
Options:
A)17 runs
B)19 runs
C)10 runs
D)48 runs
E)12 runs
Rationale: Let average for 10 innings be x. Then,
(10x + 108)/11 = x + 6
= 11x + 66 = 10x + 108
= x = 42.
New average = (x + 6) = 48 runs.
Answer:A
Answer: A
Question: A cistern is filled by a tap in 3 1/2 hours. Due to leak in the bottom of the cistern, it takes half an hour longer to fill the cistern. If the cistern is full how long will it take the leak to empty it?
Options:
A)33
B)7
C)8
D)28
E)91
Rationale: 2/7 - 1/x = 1/4
x = 28
Answer: D
Answer: D
Question: The number 1254 is divisible by which number?
Options:
A)8
B)7
C)4
D)5
E)6
Rationale: Last 3 digit cannot divisible by 8 so,Reject option A
1254 is not divisible by 7
Last number is not 0 or 5, so neglect
4 is not divide 1254
Divisibility rule of 6 is a number divisible both by 3 and 2.
1254 attains both divisible.
So the number is divisible by 6
Option E
Answer: E
Question: If the product of two numbers is 62216 and their H.C.F. is 22, find their L.C.M.
Options:
A)2828
B)2850
C)1365
D)1574
E)None of these
Rationale: Explanation:
HCF * LCM = 62216, because we know
Product of two numbers = Product of HCF and LCM
LCM = 62216/22 = 2828
Option A
Answer: A
Question: Find the average of first 4 multiples of 12?
Options:
A)10
B)12.6
C)22.5
D)30
E)40.8
Rationale: average = (12+24+36+48)/4 = 30
Answer is D
Answer: D
Question: Foodmart customers regularly buy at least one of the following products: milk, chicken, or apples. 60% of shoppers buy milk, 50% buy chicken, and 20% buy apples. If 10% of the customers buy all 3 products, what percentage of Foodmart customers purchase exactly 2 of the products listed above?
Options:
A)5%
B)10%
C)15%
D)25%
E)30%
Rationale: The formula is
n(AUBUC) = n(A)+n(B)+n(C)-B+A-N
where B is both (it will sum of (AB), (BC), (CA)).
and A is All
and N is neither
so plugging in we get
100 = 60+50+20+10-B-0
B = 40.
exactly two = 40 - 3(10) = 10
subtracting 10% three times as this value is including in all the (AB), (BC), (CA).
Answer B
Answer: B
Question: If 125% of j is equal to 25% of k, 150% of k is equal to 50% of l, and 175% of l is equal to 75% of m, then 20% of m is equal to what percent of 200% of j ?
Options:
A)0.35
B)3.5
C)35
D)350
E)3500
Rationale: First of all, let us write the information in form of equations and numbers
125 J = 5K or 5J = K (i)
150 K = 50L or 3K = L (ii)
175L = 75M or 7L = 3M (iii)
We need to find a relation between J and M
From (i) and (ii), 15J = L
Multiplying this by 7, 105J = 7L = 3M
Hence, 35J = M
Now, we are asked: 20% of m is equal to what percent of 200% of j
Assume J = 1, hence M = 35
20% of M = 7 and 200% of J =2
20% of M = 3.5 (200% of J) = (350/100)* (200% of J)
Hence 20% of M = 350% of 200% of J.
Option D
Answer: D
Question: Initially, the men and women in a room were in the ratio of 9 : 5. After 5 minutes, 3 women entered the room and 2 men left the room. Now there are 47 men in the room. Then, the number of women doubled. How many women are currently in the room?
Options:
A)56
B)65
C)15
D)34
E)36
Rationale: Working backwards: 47 men - 2 men = 45 men in the room initially.
From ratio men : women = 9 : 5 = 1 : 5/9 we have:
women in the room initially = (5/9)(45) = 25.
From this we work forward: (2)(25 women + 3 women) = 56 women in the room.
Answer: A
Answer: A
Question: A boy has thirteen trousers and 15 shirts. In how many different ways can he select a trouser and a shirt?
Options:
A)186 ways
B)177 ways
C)188 ways
D)195 ways
E)189 ways
Rationale: The boy can select one trouser in thirteen ways.
The boy can select one shirt in 15 ways.
The number of ways in which he can select one trouser and one shirt is 13 * 15 = 195 ways.
Answer:D
Answer: D
Question: Find the fourth proportional to 2.5, 4.6 and 7.6?
Options:
A)13.98
B)14.98
C)15.18
D)15.34
E)16
Rationale: Formula = Fourth propotional = (b Ã— c)/a
A = 2.5 , B = 4.6 and C = 7.6
(4.6 Ã— 7.6)/2.5 = 13.98
A
Answer: A
Question: If x and y are positive integers, and x^3*y^4 = 648, which of the following is the value of xy?
Options:
A)6
B)10
C)14
D)15
E)21
Rationale: 648 = 2*324 = 2^2*162 = 2^3*81 = 2^3*3^4
xy = 2*3 = 6
The answer is A.
Answer: A
Question: 1/3 +1/2 - 5/6 + 1/5 + 1/4 - 9/20 -1=
Options:
A)1
B)2/15
C)2/5
D)9/20
E)5/6
Rationale: We need to determine the result of 1/3 + 1/2 - 5/6 + 1/5 + 1/4 - 9/20
Let’s add the given fractions in two groups. In the group of the first three fractions, notice that1/3 and 1/2 share a common denominator of 6 with 5/6.
1/2 + 1/3 = 3/6 + 2/6 = 5/6
Thus, 5/6 – 5/6 = 0
Looking at the 2nd group of the fractions (1/5, 1/4, and 9/20), notice that1/5 and 1/4 share a common denominator of 20 with 9/20.
1/5 + 1/4 = 4/20 + 5/20 = 9/20
Thus, 9/20 – 9/20 = 0.
Thus, the result of 1/3 + 1/2 – 5/6 + 1/5 + 1/4 – 9/20 is 1.
Answer: A
Answer: A
Question: How many of the following numbers are divisible by 24 ?
264, 396, 462, 792, 968, 2178, 5184, 6336
Options:
A)3
B)5
C)4.
D)6
E)7
Rationale: By using your calculator you can calculate that the following numbers are divisible by 24 : 264, 792, 5184, 6336
Required number of number = 4.
C
Answer: C
Question: A coin is weighted so that the probability of heads on any flip is 0.7, while the probability of tails is 0.3. If the coin is flipped 5 times independently, which of the following represents the probability that tails will appear no more than twice?
Options:
A)(0.7)^5 + 5(0.7)^4(0.3) + 10(0.7)^3(0.3)^2
B)(0.7)^5 + 4(0.7)^4(0.3) + 6(0.7)^3(0.3)^2
C)(0.7)^5 + 3(0.7)^4(0.3) + 2(0.7)^3(0.3)^2
D)(0.6)^5 + 2(0.6)^4(0.4) + (0.6)^3(0.4)^2
E)(0.6)^5 + (0.6)^4(0.4) + (0.6)^3(0.4)^2
Rationale: Probability of Head, P(H) = 0.7
Probability of Tail, P(T) = 0.3
Tail will appear NO more than twice
i.e. favourable cases
2 Tails and 3 Heads, Probability = 5C2*(0.7)^3*(0.3)^2
1 Tail and 4 Heads, Probability = 5C1*(0.7)^4*(0.3)^2
0 Tail and 5 Heads, Probability = (0.7)^5
Required Probability = Sum of all Favourable cases = (0.7)^5 + 5(0.7)^4(0.3) + 10(0.7)^3(0.3)^2
Answer: option A
Answer: A
Question: the banker's discount on a bill due 4 months hence at 15% is Rs.420. what is the true discount?
Options:
A)41
B)400
C)390
D)380
E)480
Rationale: TD= (BD * 100)/(100+TR)= (420 * 100)/100+((4/12)*15)= (420 * 100)/100+((1/3)*15)
=(420 * 100)/100+5= (84 * 100)/21 = 4*100=400
correct answer: (B)
Answer: B
Question: A geometric shape is obtained by combining a square of side 8 cm and a semicircle of diameter 8cm. What is the perimeter of the shape obtained?
Options:
A)28.32
B)36.57
C)44.57
D)48.93
E)54.24
Rationale: Circumference of the semicircle= Ï€xD/2 = (22/7)x4 = 12.57
Perimeter of the square plot = 3x8 = 24
Perimeter of the shape=36.57cm
Answer: B
Answer: B
Question: The average of a group of men is increased by 5 years when a person aged of 18 years is replaced by a new person of aged 38 years. How many men are there in the group?
Options:
A)3
B)4
C)5
D)6
E)7
Rationale: Solution: Let N be the no. of persons in the group.
Required number of person is given by;
Member in group* aged increased = difference of replacement
N*5 = 38-18
Or, 5N = 20
Or, N = 4.
Answer: Option B
Answer: B
Question: 1397 x 1397 = ?
Options:
A)1951798
B)1987209
C)1952909
D)1951609
E)1958735
Rationale: D
1951609
1397 x 1397 = (1397)2
= (1400 - 3)2
= (1400)2 + (3)2 - (2 x 1400 x 3)
= 1960000 + 9 - 8400
= 1960009 - 8400
= 1951609.
Answer: D
Question: If w and k are integers whose product is 400, which of the following statements must be true?
Options:
A)n + k > 0
B)n is not equal to k.
C)Either n or k is a multiple of 10.
D)If n is even, then k is odd.
E)If w is odd, then k is even.
Rationale: If w and k are integers whose product is 400, which of the following statements must be true?
A.n + k > 0n and k should have the same sign, but that means that they can also both be negative
B.n is not equal to k.20 x 20 = 400[i]
C.Either n or k is a multiple of 10. [i]25 x 16 = 400
D.If n is even, then k is odd.k can also be even since even x even will be even
E. If w is odd, then k is even.k must be even since odd x even = even
Answer: E
Question: A painting at an art gallery is framed such that the area of the square frame is 1/2 the area of the painting itself. If the diagonal line connecting corners of the frame has a length of 6, what is the area of the painting inside the frame?
Options:
A)12
B)20
C)30
D)40
E)50
Rationale: Suppose side of the painting(which is a square) is a and side of the outer square(painting+frame) is b
Area of painting = a^2 and we know that the area of the frame is (1/2) of that of the painting so
Area of frame = (1/2) * a^2
Area of Frame + Painting = a^2 + (1/2)*a^2 = (3/2)*a^2 which is equal to b^2
Line connecting the corners of the frame is the diagonal of the outer square which is equal to b*sqrt2
so, b * sqrt2 = 6
so, b = 3 * sqrt2
b^2 = 18
we know that b^2 = (3/2)*a^2
so, a^2 = (2/3)*b^2 = (2/3)*18 = 12
And area of painting = a^2 = 12
So, answer will be A.
Answer: A
Question: If Q is a positive integer greater than 1, what is the sum of the multiples of Q from Q to Q^2, inclusive?
Options:
A) Q(Q + 1)(Q-1)
B) Q^2(Q + 1)/2
C) Q^2(Q-1)
D) (x^3 + 2x)/2
E) x(x-1)^2
Rationale: Ans :- B.
I substituted 3 in all answer choices.
Option B and C were both satisfying the condition.
So i substituted 4 instead of 3 in all answer choices,
Only B option satisfied = Q^2(Q + 1)/2
Answer: B
Question: A train running at a speed of 36 kmph crosses an electric pole in 12 seconds. In how much time will it cross a 340 m long platform?
Options:
A)19
B)27
C)29
D)46
E)28
Rationale: Let the length of the train be x m.
When a train crosses an electric pole, the distance covered is its own length.
So, x = 12 * 36 * 5 /18 m = 120 m.
Time taken to cross the platform = (120 +340)/ 36 * 5/18
= 46 min.
Answer:D
Answer: D
Question: In a soccer bet,we play 4 teams.first team odd 1.29,second 5.23,third 3.25,fourth 2.05.we place the bet with 5.00 euros.
how much money we expect to win?
Options:
A)219.5
B)224.75
C)235.01
D)266.74
E)669.0
Rationale: in the case we won the bet,we have:
1.29* 5.23* 3.25* 2.0*5.00=219.50
we will win 224.75
so correct answer is B
Answer: B
Question: 15 men take 21 days of 8 hrs. each to do a piece of work. How many days of 6 hrs. each would it take for 21 women if 3 women do as much work as 2 men?
Options:
A)30
B)20
C)19
D)29
E)39
Rationale: Let 1Man does 1unit/hr of work
15M in 21 days of 8hrs will do (15*21*8) units
3W = 2M
1W=(2/3)units/hr
21W with 6hrs a day will take (15*21*8)/(21*6*(2/3)) days
=> 30 Days
ANSWER:A
Answer: A
Question: On dividinng 109 by a number, the quotient is 9 and the remainder is 1. Find the divisor.
Options:
A)12
B)14
C)15
D)16
E)17
Rationale: d = (D-R)/Q
= (109 - 1)/9
= 108/9 = 18
A
Answer: A
Question: A box contains 100 pairs of shoes (200 shoes in total). If two shoes are selected at random, what it is the probability that they are matching shoes?
Options:
A)1/190
B)1/20
C)1/199
D)1/10
E)1/9
Rationale: The problem with your solution is that we don't choose 1 shoe from 200, but rather choose the needed one AFTER we just took one and need the second to be the pair of it. So, the probability would simply be: 1/1*1/199(as after taking one at random there are 199 shoes left and only one is the pair of the first one)=1/199
Answer: C.
Answer: C
Question: The cost price of an article is 36% of the marked price. Calculate the gain percent after allowing a discount of 20%.
Options:
A)22.2%
B)36%
C)80%
D)122.2%
E)none
Rationale: Sol.
Let marked price = Rs. 100. Then, C.P. = Rs. 36. S.P = Rs. 80.
âˆ´ Gain % = [44/36 * 100]% = 122.2%.
Answer D
Answer: D
Question: A tradesman by means of his false balance defrauds to the extent of 20%? in buying goods as well as by selling the goods. What percent does he gain on his outlay?
Options:
A)49%
B)34%
C)44%
D)48%
E)42%
Rationale: g% = 20 + 20 + (20*20)/100
= 44%
Answer:C
Answer: C
Question: A school’s annual budget for the purchase of student computers increased by 40% this year over last year. If the price of student computers increased by 25% this year, then the number of computers it can purchase this year is what percent greater than the number of computers it purchased last year?
Options:
A)15%
B)14%
C)12%
D)13%
E)11%
Rationale: Pick numbers.
Lats year: budget = $100, price = $1 --> the number of computers purchased = 100/1 = 100;
This year: budget = $140, price = $1.25 --> the number of computers purchased = 140/1.25 = ~112.00
Increase in the number of computers purchased = 12%
Answer: C.
Answer: C
Question: If Albert’s monthly earnings rise by 30%, he would earn $598. If, instead, his earnings rise by only 35%, how much (in $) would he earn this month?
Options:
A)621
B)652
C)660
D)690
E)693
Rationale: =598/1.3∗1.35
=621=621
Answer is A
Answer: A
Question: Henry purchased a house for Rs.60000 and a site for Rs.20000 respectively, if he sold the house for Rs.70000 and the site for Rs.12000, then find the resultant percentage of gain?
Options:
A)2.1%
B)2.2%
C)2.3%
D)2.4%
E)2.5%
Rationale: Sol.
Total CP = 80000 & total SP = 82000 ⇒ gain = SP/CP = 82000/80000 = 1.025= 2.5%.
Answer E
Answer: E
Question: If the first 200 numbers are written down and those divisible by 2 are deleted and again those divisible by 5 are deleted, how many numbers are left out ?
Options:
A)80
B)150
C)200
D)160
E)None
Rationale: Total numbers divisible by 2 = 100
Total numbers divisible by 5 = 40
But there is double counting. So we have to subtract Total numbers which are divisible by both 2 and 5 i.e, 200 / 10 = 20
So Total numbers which are divisible by either 2 or 5 is 100 + 40 - 20 = 120
Number of numbers which are not divisible by any of those = 200 - 120 = 80
Correct Option: A
Answer: A
Question: On dividing 127 by a number, the quotient is 9 and the remainder is 1. Find the divisor.
Options:
A)A)12
B)B)14
C)C)16
D)D)18
E)E)22
Rationale: d = (D-R)/Q
= (127 - 1)/9
= 126/9 = 14
B)
Answer: B
Question: A feed store sells two varieties of birdseed: Brand A, which is 65% millet and 35% sunflower, and Brand B, which is 40% millet and 60% safflower. If a customer purchases a mix of the two types of birdseed that is 50% millet, what percent of the mix is Brand A?
Options:
A)40%
B)45%
C)50 %
D)60 %
E)55 %
Rationale: Yes there is a simple method :
Consider the following method
Brand A : 65% millet and 35% sunflower
Brand B : 40% millet and 60% safflower
Mix : 50% millet
Here the weighted average is 50%,
Now Brand A has 65% millet, which is 15% more than the weighted average of mix = + 0.15 A --------------- I
Similarly, Brand B has 40 % millet, which is 10 % less than the weighted average of mix = - 0.10B ------------ II
Now, both Brand A and Brand B are combined to give a 50% mix containing millet, so equate I and II
implies, 0.15 A = 0.10 B
Therefore A/B = 0.10/0.15 = 2/3
A : B : (A + B) = 2 : 3 : (3+2) = 2 : 3 : 5
We have to find, percent of the mix is Brand A i.e. A : (A + B) = 2 : 5 = (2 / 5) * 100 = 40 %
Here is a pictorial representation :
Brand A= 65%------------------------15% or 0.15 above average, A times-----------------Total below = + 0.15 A
----------------------------------------------------------------------------------------Average = 50% or 0.50
Brand B = 40%--------------------------10% or 0.10 below average, B times-----------------Total above = - 0.10 B
Since the amount below the average has to equal the average above the average; therefore,
0.15 A = 0.10 B
A/B = 2/3
A:B: Total = 2:3:5
Therefore
A/Total = 2:5 = 40 %
Answer: A
Question: How much is 35% of 40 is greater than 4/5 of 25?
Options:
A)29
B)776
C)66
D)14
E)99
Rationale: (80/100) * 40 – (4/5) * 25
35 - 20 = 14
Answer: D
Answer: D
Question: A car owner buys petrol at Rs 7.50, Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of petrol if he spends Rs. 4000 each year ?
Options:
A)7.99
B)7.98
C)7.91
D)7.92
E)7.9
Rationale: Explanation:
Total quantity of petrol consumed in 3 years = liters
= liters
= liters
Total amount spent = Rs. (3 x 4000) = Rs. 12000.
Average cost = Rs. = Rs. 7.98.
Answer: B) 7.98
Answer: B
Question: A publisher sells cipies of a book retail dealer at 5 Rs.per copy.but allow 25 copies to be countes as 24.if the retailer sells each of the 25 copies at Rs.6.his profit percentage
Options:
A)20%
B)25%
C)15%
D)35%
E)45%
Rationale: CP of 25 copies = 24 * 5 = 120 (25 copies to be counts as 24)
SP of 25 copies = 25*6 = 150
Profit = 150 - 120 = 30
Profit % = (30/120) * 100 = 25%
ANSWER:B
Answer: B