# Important Aptitude Test Questions and Answers for freshers

1.

When a number is subtracted from the number 8,12 and 20, the remainders are in continued proportion, Find the number ?

A.) 4
B.) 5
C.) 6
D.) 7

2.

Aditya and Bhushan invested 10000 each in scheme A and scheme B respectively for 3 years. Scheme A offers Simple interest @ 12% per annum and scheme B offers compound interest @ 10%. After 3 years, who will have larger amount and by how much?

C.) Bhushan, 280
D.) Bhushan, 290

Lets first calculate the total rate % that Aditya will have after 3 years:

As per the question Aditya invested at rate of 12% pa simple interst

So, for 3 years tenure he will get = 12 * 3 = 36%

And the amount that Bhushan invested at rate of 10% pa compound interest

By net% effect formula, we can calculate the total perecntage for 3 years tenure = 33.1% (sub details)

So, the difference between SI and CI = 36% – 33.1% = 2.9% (SI is more)

Here Aditya will get, 2.9% of 10000 = 290

So Aditya will have Rs. 290 more than Bhushan.

---------------------------------------------------------------------------------
Sub-details:-

Net% effect = x + y = xy/100 %

For the first 2 years: Here, x = y = 10%

= 10 + 10 = 10* 10 = 21 /100 %

And for the next year: Here x = 21% and y = 10%
= 21 + 10 = 21 * 10 = 33.1/100 %

Hence, option C is correct. 3.

Shantanu borrowed Rs. 2.5 lakh from a bank to purchase one car. If the rate of interest be 6% per annum compounded annually, what payment he will have to make after 2 years 6 months?

A.) 1,89,315
B.) 4,89,425
C.) 2,89,325
D.) 3,89,335

2,89,325

----> CI for 2 years 6 months at the rate of 6, applying the net% effect for first 2 years
----> = 6 + 6 + (6 * 6)/100 = 12.36%
----> Rate of interest for 6 months = (6/12) * 6 = 3%

----> For next 6 months = 12.36 + 3 + (12.36 * 3) /100 = 15.36 + 0.37% = 15.73%
----> Here, we can see that in 2 years 6 months the given compound rate of interest is approximate 15.73%.
----> Now, 115.73% of 250000 = (115.73 * 250000)/100 = 289,325.
----> Hence, option D is correct. 4.

In a mixture, the ratio of the alchohol and water is 6 : 5. When 22 litre mixture are replaced by water, the ratio becomes 9 : 13. Find the quantity of water after replacement.

A.) 56 litre
B.) 52 litre
C.) 58 litre
D.) 54 litre

5.

A person closes his account in an investment scheme by withdrawing Rs 10000. One year ago, he had withdrawn Rs 6000. Two years ago he had withdrawn Rs 5000. Three years ago he had not withdrawn any money. How much money had he deposited approximately at the time of opening the account 4 years ago, if the annual rate of compound interest is 10%.

A.) Rs. 15470
B.) Rs. 14250
C.) Rs. 15200
D.) Rs. 16680

Rs. 15470

Suppose the person has deposited Rs. X at the time of opening account.
---> After one year, he had
----> Rs. (x + x * 10 * 1/100) = Rs. (11x/10)
----> After two years, he had
-----> Rs.((11x/10)+(11x/10) * (10 * 1/100)) = (121x/100)
----> After withdrawing Rs 5000 from Rs. (121x/100) ,the balance
----> = Rs. (121x - 500000/100)
---> After 3 years, he had
----> ((121x - 500000/100) + (121x - 500000/100) * (10 * 1/100))
----> = 11(121x - 500000/1000)
----> After withdrawing 6000 from above, the balance
----> = Rs.(1331x/1000) â€“ 11500
----> After 4 years, he had
----> Rs.(11/10) ((1331x/1000) â€“ 11500) - 10000 = 0
----> Rs 15470 6.

In a mixture of milk and water of volume 30 litres, the ratio of milk and water is 7 : 3. How much quantity of water is to be added to the mixture to make the ratio of milk and water 1 : 2?

A.) 30 liters
B.) 32 liters
C.) 35 liters
D.) 33 liters

33 liters

In 30 liter of mixture quantity of milk = (7*30)/10 = 21 liter
Hence quantity of water = 30 - 21 = 9 liter
Now let x liter of water is added in mixture to make the ratio of milk and water 1 : 2.
Hence now quantity of water = (9 + x) L
Milk = 21 L
Therefore (21 : (9 + x)) = 1 : 2
21/(9 + x) = 1/2
42 = 9 + x
x = 42 - 9 = 33
Hence 33 liters of water is added in the mixture 7.

A container contains pure milk. From these 4 litres of milk taken out and replaced by water. This process is repeated for one more time and the remaining milk in the container is 12.8 litres. What is the initial quantity of milk in the container?

A.) 15 litres
B.) 20 litres
C.) 24 litres
D.) 25 litres

20 litres

Let us take initial quantity of a container be x
Remaining milk = Initial (1 - Replaced/Initial)n
12.8 = x (1 - 4/x)2
12.8 = x (1 + 16/x2 - 8x)
12.8 = x [(x2 + 16 - 8x) / x2]
12.8x = x2 + 16 - 8x
5x2 - 104x + 80 = 0
Simplify the above equation, we get x= 20 and 0.8 (Eliminate) 8.

Pankaj borrowed a total amount of Rs.32500 from his three friends Raj, Akash and Suresh. All of his friends apply different rates of interest in such a way that Raj applies 12%, Akash applies 16% and Suresh applies 18% interest rate respectively and total he gives Rs.5090 as interest. If the amount that Pankaj had taken from Raj is (18/25) of the amount taken from Suresh, then find that what amount Pankaj has taken from Akash?

A.) Rs. 13000
B.) Rs. 11000
C.)

Rs. 10000

D.) Rs. 12000

9.

Ratio of numerical value of rate of interest and time period is 4 : 1. Man invested Rs. 2400 and gets Rs. 864 as simple interest. Find the value of X, if man invested Rs. (2400 + X) at same rate of interest on C.I. for two years and get Rs. 814.08 as interest ?

A.) Rs. 400
B.) Rs. 600
C.) Rs. 800
D.) Rs. 200

Rs. 800

Let man invested at the rate of 4x % per annum and for the period of time is x yr
ATQ
(2400 * 4x * x)/100 = 864
4x2 = 864/24
4x2 = 36
x2 = 9
x = 3
So, Rate of interest = 4 * 3 = 12% per annum
Time of period = 3 years.
Equivalent C.I. of two year at the rate of 12% per annum
CI = P[(1 + r/100)n - 1]
814.08 = (2400 + x) [(1 + 12/100)2 - 1]
814.08 = (2400 + x) [(112/100)(112/100) - 1]
814.08 = (2400 + x) [(28/25)(28/25) - 1]
814.08 = (2400 + x) [784/625 - 1]
814.08 = (2400 + x) [(784 - 625)/625]
814.08 = (2400 + x) (159/625)
814.08*(625/159) = 2400 + x
3200 = 2400 + x
x = 3200 - 2400 = 800 10.

Rs. 39,030 is divide between A and B in such a way that amount given to A on C.I. in 7 years is equal to amount given to B on C.I. in 9 years. Find the part of A. If the rate of interest is 4%.

A.) Rs. 20,200
B.) Rs. 20,280
C.) Rs. 20,100
D.) Rs. 20,900

Rs. 20,280

Given:
Let the principal of A and B be 'a' and 'b' respectively.
Amount = Rs. 39,030
r = 4%
Amount given to A for 7 yrs = Amount given to B for 9 yrs
WKT, Amount = P[1+ (r/100)]n
a[1 + (4/100)]7 = b[1 + (4/100)]9
a/b = [1 + (4/100)]9/[1 + (4/100)]7
a/b = [1 + (4/100)]2/1
a/b = [26/25]2/1
a : b = 676/625
a : b = 676 : 625
Part of A = 39030*(676/1301) = Rs. 20,280. 11.

Smallest side of a right angled triangle is 13 cm less than the side of a square of perimeter 72 cm. Second largest side of the right angled triangle is 2 cm less than the length of the rectangle of area 112 cmÂ² and breadth 8 cm. What is the largest side of the right angled triangle?

A.) 13 cm
B.) 17 cm
C.) 19 cm
D.) 15 cm

12.

18 cylindrical water bottles with height same as that of the radius are emptied into a spherical earthen pot. The bottles fill half of the initially empty earthen pot. Find the ratio of the radius of the cylinder and that of the pot?

A.) (1:8)
B.) (1:4)
C.) (1:3)
D.) (2:3)

18 cylindrical water bottles with height same as that of the radius are emptied into a spherical earthen pot. The bottles fill half of the initially empty earthen pot. Find the ratio of the radius of the cylinder and that of the pot:

Reference:

---> Let the height and radius of bottles be r
---> Volume of all bottles = 18(πr2*r) = 18πr3
---> Let the radius of pot be P.
---> Half of the volume of the pot = (2πP3/3)
---> (2πP3/3) = 18πr3
---> πP3 = 27πr3
---> P = 3r
---> Ratio = (1:3)

Hence the answer is : (1:3) 13.

A mixture contains 200 litres Milk and 40 litres water, ______ litres of mixture are removed and ________ litres of pure water were added to it. If the final quantity of milk is 124 litres more than the final quantity of water. The values given in which of the following options will fill the blanks in the same order in which it is given to make the above statement true:

A) 30, 20
B) 18, 24
C) 24, 20
D) 36, 16

A.) A and B only
B.) C and B only
C.) B, C and D
D.) B, D and A only

14.

A number when divided by the sum of 555 and 445 gives two times their difference as quotient and 30 as the remainder. The number is _________.

A.) 22030
B.) 1220
C.) 1250
D.) 220030

220030

Let the required number be 'N' which is the 'dividend'.
Given, Divisor = sum of 555 and 445 = 555 + 445 = 1000
Quotient = 2 * (555 - 445) = 2 * 110 = 220
Remainder = 30

W.K.T: Dividend = Divisor * Quotient + Remainder
---> Dividend ie., Required Number = 1000 * 220 + 30
---> Required Number = 220000â€¬ + 30
---> Required Number = 220030 15.

An intelligence agency forms a code of two distinct digits selected from 0, 1, 2, …., 9 such that the first digit of the code is nonzero. The code, handwritten on a slip, can however potentially create confusion, when read upside down-for example, the code 91 may appear as 16. How many codes are there for which no such confusion can arise?

A.) 80
B.) 78
C.) 71
D.) 69

71

The available digits are 0,1,2, …9.
The first digit can be chosen in 9 ways (0 not acceptable), the second digit can be accepted in 9 ways (digits repetition not allowed).
Thus, the code can be made in 9 × 9 = 81 ways.
Now there are only 4 digits 1, 6, 8, 9 which can create confusion.
Hence, the total number of codes which create confusion are = 4 × 3 = 12.
Out of these 12 codes 69 and 96 will not create confusion.
Hence, in total 12 – 2 = 10 codes will create confusion.
Hence, the total codes without confusion are 81 – 10 = 71. 16.

There is meeting of 20 delegates that is to be held in a hotel. In how many ways these delegates can be seated along a round table, if three particular delegates always seat together?

A.) 17! 3!
B.) 17! 4!
C.) 18! 3!
D.) 18! 4!

17! 3!

Give
Total 20 persons, 3 always seat together, 17 + 1 =18 delegates can be seated in (18 -1)! Ways = 17!
And now that three can be arranged in 3! Ways.
So, 17! 3! is the correct answer. 17.

In a party there were totally 20 people, each person shook his hands with the other person. How many hand shakes would have taken place ?

A.) 400
B.) 190
C.) 40
D.) 39

190

There are 20 people. Every person has to shake hands with the other person. Which means we have to find the number of ways of choosing 2 people from the 20. The number of ways it ca happen = 20C2 = ( 20 x 19 ) / (1 x 2) = 190 18.

The speed of a motorboat in upstream is 12 km per hour while river is flowing with a speed of 2 km per hour. If the motorboat takes 3 hours more to travel x km in still water than to travel (x – 24) km in downstream. Find the value of x?

A.) 154 km
B.) 175 km
C.) 168 km
D.) 182 km

168 km

Let the speed of the motorboat in still water be 'u' km/hr.
Given:
Speed of the river = 2 km/hr
Speed of the motorboat in upstream = 12 km/hr
Upstream speed = u - v
u - v = 12
u - 2 = 12
u = 14 km/hr
Therefore, speed of the motorboat in still water is 14 km/hr.

Downstream speed = u + v
= 14 + 2 = 16 km/hr

WKT, Time = Distance/Speed
As per the question, motorboat takes 3 hours more to travel x km in still water than to travel (x – 24) km in downstream
(x/14) - ((x-24)/16) = 3
(8x - 7(x - 24))/112 = 3
8x - 7x + 168 = 336
x = 168 km. 19.

The taxi charges in a city consist of fixed charges and additional charges per kilometer. The fixed charges are for a distance of up to 5 km and additional charges are applicable per kilometer thereafter. The charge for a distance of 10 km is Rs. 350 and for 25 km is Rs. 800. The charge for a distance of 30 km is

A.) Rs. 800
B.) Rs.900
C.) Rs. 750
D.) Rs.950

Rs.950

Let's take the fixed charges = Rs. x (for the first 5 km)
And the additional charges = Rs. y /km
As per the question,
Charge for distance 10 km = Rs. 350
So for charge for first 5 km is Rs.x and additional charge for next 5 km is Rs. 5y which can be written as follows,
x + 5y = 350 ....(i)

Charge for distance 25 km = Rs. 800
So for charge for first 5 km is Rs.x and additional charge for next 20 km is Rs. 20y which can be written as follows,
x + 20y = 800 ...(ii)

On solving eqn. (i) and (ii), we get
x = 200, y = 30
Therefore, charge for a distance of 30 km = Charge for first 5 km is Rs.x and additional charges for next 25 km is Rs. 25y
= x + 25y
= 200 + 25(30) = Rs. 950 20.

Salman was travelling on one side of the Yamuna expressway with a constant speed of 120 kmph in his car. Govinda was travelling with a constant speed of 80 kmph in the opposite direction. When they crossed each other, Salman decided to take a U-turn and meet him. But before taking a U turn, Salman had to travel for another 3 minutes. How long will it take for Salman to meet Govinda? [Assume time taken by Salman to take U turn is negligible]

A.) 29 minutes
B.) 28 minutes
C.) 18 minutes
D.) 30 minutes

18 minutes

Distance Travelled by Salman in 3 minutes = speed * time
= (3/60)*120 = 6 km
Distance travelled by Govinda in 3 minutes = (3/60)*80 = 4 km
Hence, Salman will have to travel totally 10 km distance before catching up with Govinda.
Relative speed of Salman to that of Govinda after taking U turn = 120 - 80 = 40 kmph
Therefore, time taken by Salman after taking U turn = Distance/Relative speed
= 10/40 = 15 minutes
Total time taken by Salman to meet Govinda = time taken for U turn + time taken after U turn
= 3 + 15 = 18 minutes 21.

A group of students decided to collect as many paise from each member of group as is the number of members. If the total collection amounts to Rs. 59.29, the number of the member is the group is:

A.) 57
B.) 67
C.) 77
D.) 87

Money collected
= (59.29 x 100) paise
= 5929 paise.
Number of members
= sqrt(5929)
= 77. 22.

How many natural number lie between the square of the following numbers 12 and 13.

A.) 24
B.) 25
C.) 28
D.) 26

24
We know that between n2 and (n+1)2 there are 2n non-perfect square numbers. So, between 122 and 132, there are 2(12) , that is 24 natural numbers. 23.

Two urns contain 5 white and 7 black balls and 3 white and 9 black balls respectively. One ball is transferred to the second urn and then one ball is drawn from the second urn. Find the probability that the first ball transferred is black, given that the ball drawn is black?

A.) 13/23
B.) 14/23
C.) 7/23
D.) 11/23

24.

A box contains either blue or red flags. The total number of flags in the box is an even number. A group of children are asked to pick up two flags each. If all the flags are used up in the process such that 60% of the children have blue flags, and 55% have red flags, what percentage of children have flags of both the colors?

A.) 5%
B.) 15%
C.) 20%
D.) 10%

In this problem, let A be the event that a child picks up a blue flag, B be the event that a child picks up a red flag. Then, A∪B is the event that the child picks up either a blue or a red flag, and A∩B is the event that the child picks up both a blue and a red flag.
From the problem statement we know:
I. P(AUB) = 100% (Since there are even number of flags and all the flags in the box are taken by the children)
II. P(A) = 60%
III.P(B) = 55%
IV. P(A∩B)=?
We are trying to find the percentage of children who have flags of both the colors. Substituting the values from (3.) into the equation given in (1.) we get:
P(A∪B) = P(A) + P(B) - P(A∩B)
100% = 60% + 55% - P(A∩B)
P(A∩B) = (60% +55%)-100%
P(A∩B) =115%-100%
P(A∩B) =15%
Therefore the percentage of children who got both a red flag and a blue flag is 15%. Thus, C is the correct answer choice. 25.

A box contains 4 black balls, 3 red balls and 4 green balls. Two balls are drawn at random. What is the probability that both the balls are of same color?

A.) 17/55
B.) 13/55
C.) 19/66
D.) 19/55

13/55

Total balls = 4+3+4=11 2 balls drawn from 11. Numbebr of possible ways = 11C2 = 11x10/2 = 55 ways Both are same color = 4C2+3C2 + 4C2 =6 + 3 + 4 = 13 Probability = 13/55 26.

There are 7 red balls and 8 yellow balls in a bag. Two balls are simultaneously drawn at random. What is the probability that both the balls are of same colour ?

A.) 3/18
B.) 7/15
C.) 13/30
D.) 7/18

Total balls in the bag = 7 + 8 = 15
Total possible outcomes = Selection of 2 balls out of 15 balls
= 15C2 = (15*14) / (1*2) = 105
Total favourable outcomes = Selection of 2 balls out of 8 yellow balls + Selection of 2 balls out of 7 red balls
= 8C2 + 7C2 = [(8*7)/(1*2)] + [(7*6)/(1*2)]
= 28 + 21 = 49
Required probability = 49/105 = 7/15 27.

Mr. Phanse invests an amount of Rs. 24,200 at the rate of 4% per annum for 6 yr to obtain a simple interest. Later he invests the principal amount as well as the amount obtained as simple interest for another 4 yr at the same rate of interest. What amount of simple interest will he obtain at the end of the last 4 yr?

A.) Rs. 4800
B.) Rs. 4801.28
C.) Rs. 4700
D.) Rs. 4850.32

Rs. 4801.28

Given: Principal - Rs. 24,200; R - 4%; N - 6yr
WKT, SI = PNR/100
SI = (24200 x 6 x 4)/100
= Rs. 5808

He invests the principal amount as well as the amount obtained as simple interest for another 4 yr at the same interest.
Therefore, new pricipal = 24200 + 5808 = Rs. 30008
SI = (30008 x 4 x 4)/100
SI = Rs. 4801.28 28.

A sum of Rs.725 is lent in the beginning of the year at a certain rate of interest. After 8 months, a sum of Rs.362.50 more is lent but at a rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest?

A.) 3.46%
B.) 2.50%
C.) 6.25%
D.) 6%

3.46%

Assume,
Original rate is for 1 year and the new rate is for only 4 months i.e. 1/3 year(s).
Let the original rate be R%
Then, new rate = (2R)%
Formula to calculate Simple Interest:
SI = PRT/100
SI= Simple Interest; P=Principal
r=Interest rate; t=time
Now,
â¸«[(725 x R x 1)/(100)]+[(362.50 x 2R x 1)/(100 x 3)]= 33.50
Taking LCM, we get 300
(2175 + 725) R = 33.50 x 300
(2175 + 725) R = 10050
(2900)R = 10050
R = 10050/2900 = 3.46
Original rate = 3.46% 29.

A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both as interest. The rate of interest per annum is

A.) 57/8%
B.) 10%
C.) 5%
D.) 7%

10%

WKT, SI = PNR/100
A lent Rs. 5000 to B for 2 years,
= [(5000 x r x 2)/100]

A lent Rs. 3000 to C for 4 years,
= [(3000 x r x 4)/100]

As per the question,
[(5000 x r x 2)/100] + [(3000 x r x 4)/100] = 2200
[100 x r] + [120 x r] = 2200
r x 220 = 2200
r = 10% 30.

Chiku, Tipu and Pinku have some candies with each. Five times the number of candies with Pinku equals seven times the number of candies with Chiku while five times the number of candies with Chiku equals seven times the number of candies with Tipu. What is the minimum number of candies that can be there with all three of them put together?

A.) 209
B.) 109
C.) 119
D.) 97

109

--> Let the candies with Chiku, Pinku and Tipu be a,b and c respectively.

---> Given, 5b = 7a, and 5a = 7c
---> 25b = 35a & 35a = 49c

---> 25b = 35a = 49c => (b/49) = (a/35) = (c/25)

---> The least possible integral values for a,b & c will be a = 35 , b = 49 & c = 25

---> Total = 35+49+25 = 109 31.

One-third of the contents of a container evaporated on the 1st day, three-fourths of the remaining evaporated on the second day. What part of the contents of the container is left at the end of the second day?

A.) One-fourth
B.) One-eighteenths
C.) One-half
D.) One-sixth

One-sixth

After first day, 2/3 rd of the contents remain (i.e., 1 - (1/3) = 2/3)
After second day 2/3 â€“ (3/4) x (2/3) = 1/6 of the content remains 32.

A person spends one-third of the money with him on clothes, one-fifth of the remaining on food and one-fourth of the remaining on travel. Now, he is left with Rs.100. How much did he have with him in the beginning?

A.) 200
B.) 250
C.) 450
D.) 300

250

Initial amount be x.
Money spent on cloths = x/3.

Now,Balance = x - (x/3) = 2x/3
Money on food, (1/5) x(2x/3) = 2x/15

Now,Balance = (2x/3) - (2x/15) = 8x/15
Money spent on travel = (1/4) x (8x/15) = 2x/15

Now,Balance = (8x/15) - (2x/15) = 6x/15 = 2x/5

Now, he is left with Rs.100
=> 2x/5 = 100
=> x = 250 =>Initial amount. 33.

Eight people are planning to share equally the cost of a rental car. If one person withdraws from the arrangement and the others share equally the entire cost of the car, then the share of each of the remaining persons increased by

A.) One-ninth
B.) One-seventh
C.) Seven-eighths
D.) One-eighth

One-seventh

When there are 8 people, the share of each person is 1/8
When there are 7 people, the share of each person is 1/7
Increase in the share of each person is 1/7 - 1/8 = 1 / 56 , Which is 1/7 of 1/8 of the original share of each person. 34.

When a bus started from the first stop, the number of male passengers to the number of female passengers was 3: 1. At the first stop, 16 passengers got down and 6 more female passengers got in. The ratio of the male to female passengers now became 2:1. What was the total number of passengers in the bus when it started from the first stop?

A.) 64
B.) 72
C.) 48
D.) 54

64

Males = 3x, females = x.
First stop, m males left, f females left the bus and 6 females got in.
No. of females = x – f + 6 => (3x-m)/(x-f+6) = 2/1 => 3x -2 = 2x – 2f + 12 Also m+f=16 => X = 28 – 3f, => f=4 and x= 16
Number of passengers in the beginning = 4x = 64 35.

When a bus started from the first stop, the number of male passengers to the number of female passengers was 3: 1. At the first stop, 16 passengers got down and 6 more female passengers got in. The ratio of the male to female passengers now became 2:1. What was the total number of passengers in the bus when it started from the first stop?

A.) 64
B.) 54
C.) 72
D.) 48

64

Solution is
Given , the ratio of number of male passengers to female passengers = 3 : 1
Let initially , number of males = 3x
number of females = x
Given ,at first stop, passengers got down = 16
female passengers got in = 6
At first stop , supposemmales andffemales got down from the bus =>m + f = 16
So at first stop , number of males = 3x - m
number of females = ( x - f ) + 6
Given , the ratio of male to female passengers (now)= 2 :1
(No. of male passengers) / (No. of female passengers) = 2 / 1
=> ( 3x - m ) / ( x - f ) + 6 = 2 / 1
=> 3x - m = 2 [ (x - f) + 6 ]
=> 3x - m = 2x - 2f + 12
=> 3x - 2x = m - 2f + 12 ( i.e m + f = 16 => m= 16 - f )
=> x = 16 - f - 2f + 12
=16 - 3f + 12
x = 28 - 3f
For f = 4,
x = 28 - 3 (4)
= 28 - 12
x = 16
Therefore total number of passengers in the beginning = No. of ( males + females )
= 3x + x
= 4x = 4 * 16
64
[ For no other value of f, any of the other alternatives holds good ] 36.

When a bus started from the first stop, the number of male passengers to the number of female passengers was 3: 1. At the first stop, 16 passengers got down and 6 more female passengers got in. The ratio of the male to female passengers now became 2:1. What was the total number of passengers in the bus when it started from the first stop?

A.) 64
B.) 54
C.) 72
D.) 48

64

Males = 3x, Females = x.

First stop, m males left, f females left the bus and 6 females got.

No. of females = x - f + 6

(3x-m) / (x-f+6) = 2/1

=> 3x -2 = 2x - 2f + 12

Also m + f=16

x = 28 - 3f,

f=4 and x= 16 Number of passengers in the beginning = 4x = 64 37.

The ratio of males and females in a city is 7 : 8 and the percentage of children among males and females is 25% and 20% respectively. If the number of adult females in the city is 156800 what is the total population?

A.) 245000
B.) 367500
C.) 171500
D.) 196000

367500

Males = 7x
Females = 8x
Adult females = 8x * 80/100
= 32x/5
32x/5 = 156800
x = (156800*5)/32
= 24500
Total population = 15*24500
= 367500 38.

Jaya's attendance for first two semesters out of four was 60% and 70%, respectively. What is the minimum attendance required in third semester so that her average attendance will be 80% throughout four semesters? (Assume equal number of days among the four semesters)

A.) 85%
B.) 95%
C.) 90%
D.) 70%

90%

Let, there are 100 days in each semester then,
Jaya's total attendance for four semesters = 4*(80% of 100)
= 4*80 = 320 days
To know the minimum attendance in 3rd semester, we must assume 100% attendance in 4th semester.
Thus, minimum attendance required in 3rd semester
= 320 - (60 + 70 + 100) days = 90 days
= i.e. 90%
So, the minimum attendance required in third semester = 90% 39.

Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are

A.) 42, 30
B.) 42, 33
C.) 41,30
D.) 41, 32

42, 33

Let their marks be x + 9 and x.
Then, x+9 = (56/100) (x+9+ x)
=> 25(x + 9) = 14(2x + 9)
=> 25x + 225 = 28x + 126
=> 28x - 25x = 225 - 126
=>3x = 99
x = 33
Another student mark x + 9 = 33 + 9 = 42
So, their marks are 42 and 33 40.

50 % of 1250 + 18 % of2200 = ?

A.) 1031
B.) 1041
C.) 1021
D.) 1061 