Quantitative Aptitude (QA) Test - 10

Let there be ‘x’ women assist one man and one kid to complete the given work in 3/5th day.
$$\frac{3}{5} * [\frac{1}{6} + \frac{x}{15} + \frac{1}{10}] = 1$$
$$\frac{3}{5} * [\frac{1}{6} + \frac{x}{15} + \frac{1}{10}] = 1$$
$$\frac{3}{5} * [\frac{8}{30} + \frac{x}{15} = 1$$
$$\frac{8}{50} + \frac{x}{25} = 1$$
$$x = \frac{42*25}{50} = 21$$
Hence, option D is the correct choice.

By pigeon hole principle, if n items are put into m containers, with n > m, then at least one container must contain more than one item.

In this case, we can compare suits as containers, and the number of cards can be compared as items. We know that in a deck, there are 4 suits.

To ensure that the number of cards (items) of the same suit (container) is at least 2, we need n>m => n>4 => the value of n must be at least 5.

Let the angles be of the form: $$a-d,\ a,\ a+d$$

Then, $$a-d+\ a+\ a+d\ =\ 180^{\circ\ }$$; $$\therefore\ a=\ 60^{\circ\ }$$

The angles take the form: $$60-d,\ 60,\ 60+d$$

It is given that the side corresponding to $$60^{\circ\ }$$ is 8 cm.

In any triangle, $$\frac{a}{\sin\ A}=\frac{b}{\sin\ B}=\frac{c}{\sin\ C}=2R$$ where R is the circumradius while a,b and c are the corresponding sides of angles A,B and C. 

Using this, we obtain: $$\frac{8}{\sin\ 60^{\circ\ }}=2R$$

$$\therefore\ R=\frac{8}{\sqrt{\ 3}}$$

Hence,  Option A is the correct choice.

Average speed = 100/(30/36+20/24+50/30) =30 Kmph

In the triangle
QP=21cm, QR=35cm
Therefore PR=$$^{\sqrt{\ 35^2-21^2}}$$
               PR=28cm

The given triangle is a right angled triangle .
Now if the triangle is rotated about the base PR 
We will obtain a right circular cone with 
radius =21cm and height =28cm 
Now volume of the cone obtained will be :
$$\frac{1}{3}\pi\ \times\ 21\times\ 21\times\ 28$$
=12,936 cm^3
If the triangle is rotated about PQ 
we will obtain a right circular cone with radius =28 and height=21
The volume of such a cone will be:
$$\frac{1}{3}\pi\ \times\ 28\times\ 28\times\ 21$$
17,248cm^3

Therefore maximum volume will be 17,248cm^3
so option B is the correct answer.

Raja has to use 3 litres of milk and 0.5 Kg of sugar. Apart from purchasing milk and sugar, he has to pay Rs.20 to his employee for making the sweet. 
Total cost incurred by Raja to make 1 Kg of milk sweet = 3*25 + 0.5*40 + 20 
                                                                                   = 75 + 20 + 20
                                                                                   = Rs. 115. 
Raja marks up the price by 20%. Therefore, the selling price of 1 Kg of milk sweet is 1.2*115 = Rs. 138. 
Profit obtained by selling 1 Kg of sweet = Rs. 138 - 115 =  Rs.23
Raja makes a profit of Rs. 460.
Therefore, Raja should have sold a total of 460/23 = 20 kg of milk sweet. 
To make 20 Kg of milk sweet, Raja has to buy 20*0.5 = 10 Kg of sugar. 
10 kg of sugar costs 10*40 = Rs.400
Therefore, option B is the right answer. 

Since the largest side is the diameter, the triangle should be right angled triangle.
=> $$a^{2}+7^{2}=b^{2}$$ => $$b^{2}-a^{2}=7^{2}$$ => $$(b+a)(b-a)=7^{2}$$
So, the ordered pairs for [(b+a),(b-a)] are (49,1), (7,7), (1,49)
For (49,1) we get (a,b) as (24,25).
For (7,7) we get (a,b) as (0,7) which is not possible since triangle’s side can’t be 0.
For (1,49) we get (a,b) as (-24,25) which is also not possible.
Thus, only one ordered pair (24,25) exists.

Since the triangle lies between two parallel lines, the area of the triangle will be half that of the parallelogram that lies between those parallel lines and has the same base as the triangle. In this case, the parallelogram is the rectangle. Hence the total area of the field allotted to the daughters is half the area of the total field.

The area of the field must be 26 units, as the only two-digit number that is preceded and followed by a perfect square (25) and a perfect cube number (27) respectively, is 26.

Thus the total area allotted to the daughters must be 13 units. And since the areas are integers, the sum of their simple ratio must be a factor of 13.

13 has only two factors: 1 and 13. The Sum of two positive integers cannot be 1 ( as areas have to be positive ), the sum of the antecedent and consequent of their simple ratio can only be 13.

Looking at the options, only C has the sum of the antecedent and the consequent as 13, and is our answer.

Let the human finish h units and a robot finish r units in an hour.
Hence, (5h + 4r) 24 = (20h + 5r) 17
120h + 96r = 340 h + 85r
11r = 220h
r = 20h
To complete the entire piece of work, we need to work for (20h + 5r)17 units, or 120 x 17h units.
Hence, the total number of hours to complete work = 2040.

Let $$p$$,$$q$$ and $$r$$ be the sizes of markets P,Q,R respectively. 

Given, $$q\ =\ \left(1-\dfrac{\left(33\ \dfrac{1}{3}\right)}{100}\right)p\ =\dfrac{2}{3}p\ =>\ p\ =\ \dfrac{3}{2}q\ $$ -----(1)

Also, $$r = (1+ \dfrac{12.5}{100})(q) = \dfrac{9}{8} q \Rightarrow r = \dfrac{9}{8}q$$ -----(2)

Let $$p'$$,$$q'$$ and $$r'$$ be the sizes of markets P,Q,R the next year respectively. 

Given, $$p' = (1+\dfrac{25}{100})(p) = \dfrac{5}{4} p = \dfrac{5}{4} \times \dfrac{3}{2} q = \dfrac{15}{8} q $$

$$q' = (1+\dfrac{50}{100})(q) = \dfrac{3}{2} q $$

$$r' = (1+ \dfrac{60}{100})(r) = \dfrac{8}{5} r = \dfrac{8}{5} \times \dfrac{9}{8} q = \dfrac{9}{5} q $$

Initial size of the company = $$(\dfrac{40}{100} \times \dfrac{3}{2} q)+(\dfrac{50}{100} \times q)+(\dfrac{80}{100} \times \dfrac{9}{8} q)$$

$$ = (\dfrac{3}{5} q) + (\dfrac{1}{2} q) + (\dfrac{9}{10} q) = \dfrac{20}{10} q $$ = $$2q$$

New size of the company = $$ (\dfrac{40}{100} \times \dfrac{15}{8} q) + (\dfrac{50}{100} \times \dfrac{3}{2} q) + (\dfrac{80}{100} \times \dfrac{9}{5} q) $$

$$ = (\dfrac{3}{4} q) + (\dfrac{3}{4} q) + (\dfrac{36}{25} q) $$

$$ = \dfrac{294}{100} q $$ = $$2.94q$$

So, overall growth percent of the company = $$\dfrac{\text{size of new company - size of old company}}{\text{size of old company}} \times 100 $$

=$$\dfrac{\left(2.94\ q-2\ q\right)}{2\ q}\times\ 100$$ = 47%

Let y be the third term and x be the fourth term. We have,

(x+y)(x-y) = 559 = 43*13

x - y = 13 and x + y = 43. So, x = 28 and y = 15. Hence the seventh term is 114

The product has to be positive. So, all three numbers can be positive or two numbers can be negative and one number positive. If all the three numbers are positive, then the product is maximum if the numbers are as close to each other as possible. Since the number can take negative values also, the product is maximum when the magnitudes of the negative numbers and the positive number are as high as possible. The maximum value of the positive number is 59.
So, the sum of the negative numbers is -19. These numbers can be -10 and -9.
So, the maximum value of the product is -9 * -10 * 59 = 5310

Here let the roots be 5x and 9x.
Difference of roots = $$\pm 4x$$ = 12
$$x = \pm3$$
But, the roots are positive as their sum is positive in the given equation.
$$x = 3$$
Hence, the roots are 15 and 27
The equation is $$(x-15)(x-27)=0$$
$$x^2-42x+405 = 0$$ or $$2x^2-84x+810 = 0$$
Hence a = 2 and c = 810
$$\frac{c}{a} =405$$

Let Enoch took a loan of $$\$100x$$ from the loan agency.

The rate of interest was $$10$$% per month, i.e., $$12\times 10 = 120$$% per annum.

The total amount he had to pay at the end of $$2$$ years = $$100x$$ + $$\frac{\left(100x\ \times\ 2\times\ 120\right)}{100}$$ = $$100x + 240x = 340x$$

Thus, the amount of loan he took from the Atlantic City bank = $$340x$$

He pays the bank back in $$2$$ years at a rate of $$10$$%pa compound interest.

Thus, total amount = $$340x \times (1 + \frac{10}{100})^2 = 340x \times 1.21 = 411.4x$$

Interest paid by Enoch to the loan agency = $$340x - 100x = 240x$$

Interest paid by Enoch to the Bank = $$411.4x - 340x = 71.4x$$

The difference in interest = $$33720$$

$$240x - 71.4x = 33720$$

$$168.6x = 33720$$

$$x = 200$$

Thus, the total amount paid to the loan agency = $$340x = 340\times 200 = \$68000$$

Hence, $$68000$$ is the required answer.

Let x = 1.1!+2.2!+3.3!+4.4!+.…+11.11!
Adding (1!+2!+3!+…+11!) on both sides we get,
x + 1!+2!+3!+…+11! = 1.1!+2.2!+3.3!+4.4!+.…+11.11! + 1!+2!+3!+…+11!
x + 1!+2!+3!+…+11! = 2.1!+3.2!+4.3!+….+12.11!
x + 1!+2!+3!+…+11! = 2!+3!+4!+…+12!
x = 12! - 1
Thus the remainder when x is divided by 12 is 11.

Let the weights of the individual pieces of the article after breaking be 2a and 3.1a
The original weight of the article is 5.1a
As the value if directly proportional to the square of the weight, the value of the article after the fall =  $$28611*\frac{2^2+3.1^2}{5.1^2} = 14971$$

Let SP = 100 => MP = 140.
New SP = 70.
Profit = 20% => CP = 70*100/120.
If the discount is 25%, the new SP is 105.
Therefore, profit % = $$\frac{105 - 70*100/120}{70*100/120} * 100$$% = 80%

Let the size of each container be 630 liters,
Therefore, in first container amount of milk and water is 360 and 270 liters respectively.
In Second container amount of milk and water is 350 and 280 liters respectively.
In Third container amount of milk and water is 210 and 420 liters respectively.
In Fourth container amount of milk and water is 180 and 450 liters respectively.
Size of the big vessel = 4.2*630 = 2646 liters.
Amount of milk in this vessel = 360+350+210+180 = 1100 liters and hence, amount of water in this vessel is = 2646-1100 = 1546 liters
Therefore, the ratio of milk to water in the big vessel = 1100:1546

Let the number of one rupee coins be a.
Let the number of 50 paisa coins be b.
Let the number of 25 paisa coins be c.

Therefore, a + b + c = 20 and a + b/2 + c/4 = 18.25
Or 2*b + 3*c = 7. The only non negative integral solution for this is b=2 and c=1

Among the given numbers the attribute of the chosen number must be a multiple of 3 and must be of the form 7k+2.

Since of the following attributes the numbers of form form 7k + 2 are lesser in number and hence must be counted.

The numbers less than 150 and are of the given form 7k+2 are :

2,9, 16, 23, 30, 37, 44, 51, 58, 65, 72, 79, 86, 93, 100, 107, 114, 121, 128, 135, 142, 149.

Among these numbers we need to select the numbers which are multiples of 3.

Hence we can only choose among 9, 30, 51, 72, 93, 114, 135.

Hence 7 of the 150 possibilities and the answer is 7/150

$$\log_{2} a + \log_{4} a + \log_{16} a + \log_{256} a + . . . = 1$$
So,$$\log_{2} a *(1 +1/2 +1/4 + 1/8 + . . .) = 1$$
So, $$\log_{2} a * 2 = 1$$.
Hence, $$a= \sqrt{2}$$.

Let 'x' km/hr be the speed of Mahesh. It is given that Mahesh started 20 minutes later. In these 20 minutes Ramesh and Suresh would have covered 20 km and 25 kms. Therefore,

Time taken by Mahesh to overtake Ramesh = $$\dfrac{20}{x- 60}$$ hours.

Time taken by Mahesh to overtake Suresh = $$\dfrac{25}{x- 75}$$ hours.

It is given that Mahesh overtook Suresh an hour after he overtook Ramesh. Therefore, 

$$\Rightarrow$$ $$\dfrac{25}{x-75}-\dfrac{20}{x-60}=1$$

$$\Rightarrow$$ $$25*(x-60)-20*(x-75)= (x-75)*(x-60)$$

$$\Rightarrow$$ $$x=50$$ or $$90$$ km/hr.

In order to overtake Suresh whose speed is 75 km/hr, Mahesh's speed should be greater than 75 km/hr. Hence, we can say that Mahesh's speed = 90 km/hr. 

Hence, we can that option B is the correct answer.