Quantitative Aptitude (QA) Test - 11

Let the instalment amount be x.

Since the amount have to be paid in the equal 3 instalments at the rate of 10%,

$$x+x\left(1.1\right)+x\left(1.1\right)^2=33100\left(1.1\right)^3$$

$$x=13310$$

Outstanding amount after 1 year = $$33100\times1.1\ -13310=23100$$

Outstanding amount after 2 years = $$23100\times1.1\ -13310=12100$$

Now, since 12100 is to be paid in one installment after two years, and a penalty of 15% is levied on it,

Outstanding amount paid after 4 years = $$12100\times\ \left(1.1\right)^2\times\ 1.15\ =\ 16837.15$$

Thus, the amount paid in the 4th year will be 16837.

The correct option is A.

There are a total of 36 possible outcomes. The sum of the dice is greater than 7 in 15 outcomes, less than 7 in 15 outcomes and equal to 7 in 6 outcomes. Hence, the expected payoff = (5*15+0*6+-5*15)/36= 0

SI = p * 2 * r/100

CI = $$p*(1+R/100)^2-p$$= $$p(2R/100 + (R/100)^2)$$

Since they are equal, $$p * 2 * r/100 = p(2R/100 + (R/100)^2)$$

$$R^2 + 200R - 200r = 0$$

Solving the equation, we get $$R = -100 \pm \sqrt{10000 + 200r}$$

Since, R cannot be negative the only possible value of $$R = -100 + \sqrt{10000 + 200r}$$

=> $$ R + 100 = \sqrt{10000 + 200r}$$

Required ratio = 111% * (1.1) : 121% * (3) : 68% * (3.3) : 99% * (4)

= 37 : 110 : 68 : 120

The ratio of the amount is $$\dfrac{1}{9} :\dfrac{1}{17} : \dfrac{1}{11} : \dfrac{1}{13}$$

LCM of 9, 17, 11 and 13 is 21879

We can write the ratio as $$\dfrac{21879}{9} :\dfrac{21879}{17} : \dfrac{21879}{11} : \dfrac{21879}{13}$$ 

which is 2431 : 1287 : 1989 : 1683

Let the amounts distributed be 2431k, 1287k, 1989k, 1683k

It is given that 2431k - 1287k = 1144k = 5720

So, k = 5

Therefore, the total amount distributed = (2431 + 1287 + 1989 + 1683)*5 = Rs. 36950

Hence, 36950 is the correct answer.

Let the actual price of 100 gm goods be Rs. 100 

The dealer got 20% discount and also he cheats the wholesaler by getting 20% extra.

So, we can conclude that he purchases 120 gm of goods and pays Rs.80 for it.

CP of 1 gm of good = $$\frac{80}{120}=\frac{2}{3}$$

MP of 100 gm of goods= 1.8$$\times$$100 = Rs. 180

Discount = 25% = 0.25$$\times$$180 = 45

SP of 100 gm of goods = 180 - 45 = 135

But since the dealer cheats the customer by giving him 10% less than the actual quantity so basically the dealer sells 90 gm of goods for Rs. 135.

SP for 1 gm of goods = $$\frac{135}{90}=\frac{3}{2}$$

Profit percentage = $$\frac{\left(\frac{3}{2}-\frac{2}{3}\right)}{\frac{2}{3}}\times100\%=\frac{5}{4}\times100\%=125\%$$

The stone will travel 5t^2
 $$5t^2$$ m in $t$ seconds. 
By the end of the first second, the stone would have travelled 5 m.
By the end of the second second, the stone would have travelled 20 m.
=> Distance travelled in the second second = 20 - 5 = 15 m.

Distance travelled in 7 seconds = 5*49 = 245 m.
Distance travelled in 5 seconds = 5*25 = 125 m.
=> Distance travelled in last 2 seconds = 245-125 = 120 m.

Initially, Solution =1000ml

A:B=2:3

Thus, A=400ml and B= 600ml

The volume of A after the replacements with liquid B are over = 400 * (50/100)*(60/100)*(70/100) = 210000/1000000 * 400 = 84 ml

Volume of B = 916 ml

A = 84 ml B= 916 ml 

We need extra 316 ml of A, thus B should reduce by 316 ml.

We need to take out 316ml of B from the solution.

So out of 916 ml of B present if we take out 316 ml of B. 

Then from 1000ml of the solution, we should take out x ml of solution.

x/1000= 316/916

x=344.97= 345ml

So, the required percentage= (345/1000)*100= 34.5%

(Thus, if we take out 345 ml of the solution then it 316ml is B and 29ml is A. Then we replace it with 345ml of A.
Finally A = 84-29+345=400ml)

Alternate explanation:

The volume of A after the replacements with liquid B are over = 400 * (50/100)*(60/100)*(70/100) = 210000/1000000 * 400 = 84 ml

Volume of B = 916 ml

Let x % of this solution is replaced with liquid A.

After replacement, Volume of A = $$84(1-\frac{x}{100}) +  1000\times \frac{x}{100}$$                  Volume of B = $$916(1-\frac{x}{100})$$

Ratio = $$\frac{84(1-\frac{x}{100}) +  1000\times \frac{x}{100}}{916(1-\frac{x}{100})}$$ = $$\frac{2}{3}$$

$$\frac{84(1-\frac{x}{100}) + 10x}{916(1-\frac{x}{100})}$$ = $$\frac{2}{3}$$

Adding 1 to both sides, we get $$\frac{1000}{916(1-\frac{x}{100})}$$ = $$\frac{5}{3}$$ 

or, $$916(1-\frac{x}{100})$$ = 600

or, x= 34.49%

Data Table Explanation:

Now in the final solution B is 916ml.
In the initial solution B was 600ml.
Since the total qty is the same in both cases, we need to take out 316ml (916-600)ml of B.
This means we have to take out $$\left(\frac{316\ }{916}\times\ 100\right)\ \%\ $$ = 34.4% of the total liquid and replace it with A.
Rounding off we get 34%.
Hence the correct answer is Option B.

Initially, TAXe = MAXe + 300
After 100 mL deodorant escapes both bottles, (TAXe - 100) = (MAXe - 100) *2
Solving both these equations, TAXe = 700 and MAXe = 400 mL

After the leak, the amount of deodorant in TAXe = 600 mL

Let the side of the given square be 14 units
Thus, AE = 10 units, EB = 4 units, BF = 12 units, FC = 2 units, DG = 7 units, GC = 7 units, HD = 8 units and HA = 6 units.
Area of square ABCD = $$14^2 = 196$$ $$units^2$$
Area of triangle AHE = $$\dfrac{1*10*6}{2} = 30$$ $$units^2$$
Area of triangle EBF = $$\dfrac{1*4*12}{2} = 24$$ $$units^2$$
Area of triangle FCG = $$\dfrac{1*2*7}{2} = 7$$ $$units^2$$
Area of triangle GHD = $$\dfrac{1*7*8}{2} = 28$$ $$units^2$$
Thus, the area of polygon EFGH = Area of square ABCD - (Area of triangle AHE+Area of triangle EBF+Area of triangle FCG+Area of triangle GHD) = $$196-(30+24+7+28) = 107$$ $$units^2$$
Hence, the required ratio = $$\dfrac{107}{196}$$
Hence, option D is the correct answer.

Let inradius be r and perimeter be 2s.
We know that the area of a triangle = r*s
So we are asked to find the value of 2*area
s = 24
area = $$\sqrt{24*10*8*6}$$ = $$48\sqrt{5}$$
Required product = $$96\sqrt{5}$$

When Madhan starts, Karan is 60m ahead of him and charan is 35m ahead of him.
Given, charan finishes the line 11 sec after Madhan. When Madhan started charan is 35m ahead of him.

So, 11+ (d+60)/10.5=(d+25)/7

0.5d=138 => d=276

Now, Karan takes 276/3 = 92 sec to reach the end point
CHaran takes (276+25)/7 = 43 sec more to reach end point.
So, Karan reaches endpoint after (92-43) = 49 sec after Karan reaches endpoint.

Consider the expression $$x^2 - 6x + 14$$
Discriminant = $$(-6)^2 - 4*1*14$$ = 36 - 56 = -20 < 0
Since the coefficient of $$x^2$$ is positive, the value of the expression is always positive.
Minimum value of $$x^2 - 6x + 14$$ = $$\frac{4*1*14 - (-6)^2}{4*1}$$ = 5
So, maximum value of $$\frac{1}{x^2 - 6x + 14}$$ = 1/5

It is given that the depth at one bank is 50 m, which is gradually increasing to 100 m at the other bank, which implies the average depth of the lake is $$\ \frac{\ 50+100}{2}=\frac{150}{2}=75$$ m (Since it increases gradually).

The volume of the water = height * length * breadth

$$=>\ 75\times\ 300\times\ 200\ \text{m}^3$$

$$=>\ 4500000\ \text{m}^3$$

The correct option is D.

There is a circularity for the power of $$2$$. For example, the remainder when $$2^3$$ is divided by 6 is 2.
Hence, the remainder when $$2^{144} = 2^3 \times 2^3 \times 2^3 ...$$ (48 times)
So, the remainder when $$2^{144}$$ is divided by 6 is the same as the remainder when $$2^{48}$$ is divided by 6.
Similarly, $$2^{48} = 2^3 \times 2^3 \times 2^3 ...$$ (16 times)
So, the remainder when $$2^{48}$$ is divided by 6 is the same as the remainder when $$2^{16}$$ is divided by 6.
Similarly, $$2^{16} = 2^3 \times 2^3 \times 2^3 ...$$ (5 times) $$\times 2$$
So, the remainder when $$2^{16}$$ is divided by 6 is the same as the remainder when $$2^{5}\times 2 = 64$$ is divided by 6.
So, the answer is 4

Each of the chocolates can be distributed in 3 ways => $$3^5$$ Remove the cases where one person gets 0 chocolates => $$3*2^5$$ Add the cases where 2 persons get 0 chocolates as they are removed twice in the previous step => $$3^5-(3*2^5)+(3*1^5)$$ = 243-96+3 = 150.

Alternate explanation:

The chocolate (not identical) can be distributed in the number (3,1,1) or (1,2,2) among the friends. {since each of the friends got at least 1 chocolate.}

The number of ways such that the chocolate can be distributed in the number $$(3,1,1)$$ is $$3\left(^5C_1\cdot^4C_1\cdot^3C_3\right)=60$$

The number of ways such that the chocolate can be distributed in the number $$(2,2,1)$$ is $$3\left(^5C_1\cdot^4C_2\cdot^2C_2\right)=90$$

The total number of ways is $$(60+90)=150$$

If the side of the equilateral triangle is a, then its inradius is $$\ \frac{\ a}{2\sqrt{\ 3}}$$ and its circum radius is $$\ \frac{\ a}{\sqrt{\ 3}}$$.

=> Clearly, circumradius is two times the inradius, hence the inradius = 3$$\sqrt{\ 3}$$ 
Area if the incircle = $$\pi\ \times\ \left(3\sqrt{3}\right)^2$$ = 27$$\pi\ $$

First we have to plot the region enclosed by the lines of the given equation.

The equations are y = x -5 for x > 5, y = -(x - 5) for x < 5 and y = 5.

The plot of the above lines is shown below:

The region enclosed by the lines is a triangle in the first quadrant formed by the points (0,5), (5,0) and (10,5). The number of coordinates in the region with x coordinate are as follows
x=0, 1 point,
x=1, 2pts,
x=2, 3pts,
x=3, 4pts,
x=4, 5pts,
x=5, 6pts and so on.
Total points=36

Let $$f_1$$ and $$f_2$$ be the first terms of their salary of A and B, and $$d_1$$ and $$d_2$$ be their common difference.
Ratio of sum of salaries in $$x$$ years = $$\frac{x}{2} (2f_1+(x-1)d_1)$$ : $$\frac{x}{2}(2f_2+(x-1)d_2)$$ = $$\frac{2x+1}{3x+2}$$
$$\frac{(2f_1+(x-1)d_1)}{(2f_2+(x-1)d_2)}$$ = $$\frac{2x+1}{3x+2}$$

Ratio of the the amount they earned in 9th year = $$\frac{f_1+8d_1}{f_2+8d_2}$$

= $$\frac{2f_1+(17-1)d_1}{2f_2+(17-1)d_2}$$ = $$\frac{2*17+1}{3*17+2}$$ = $$\frac{35}{53}$$ = 35:53

With the given condition given the set of a = ( 1, 2, 3, 4, 5, 6, 7, 8)

For the set of b = ( 6, 7, 8, 9, 10, 11, 12)

For the roots of the equation to be real and distinct :the condition is that for a quadratic equation of the form

$$ax^2+bx+c\ =\ 0\ ,\ b^2-4ac>0$$.

For b = 6 the equation is of the form 36 - 12a >0. a will take the values of (1,2)

For b = 7 the equation is of the form 49 - 12a >0. a will take the values of (1,2, 3, 4)

For b = 8 the equation is of the form 64 - 12a >0. a will take the values of (1,2, 3, 4, 5)

For b = 9 the equation is of the form 81 - 12a >0. a will take the values of (1,2, 3, 4, 5, 6,)

For b = 10 the equation is of the form 100 - 12a >0. a will take the values of (1,2, 3, 4, 5, 6, 7, 8)

For b = 11 the equation is of the form 100 - 12a >0. a will take the values of (1,2, 3, 4, 5, 6, 7, 8)

The minima will be present at x = -b/2a which should be greater than -1

for b = 6 for this to happen the the value of 2a should be greater than 6 and hence a >3 and no cases are possible.

for b = 7 for this to happen the the value of 2a should be greater than 6 and hence a >3.5 and hence a = 4,  (1 case)

for b = 8 for this to happen the the value of 2a should be greater than 6 and hence a >4 and hence a = 5,      (1 case)

for b = 9 for this to happen the the value of 2a should be greater than 6 and hence a >4.5 and hence a = 5, 6   ( 2 cases)

for b = 10 for this to happen the the value of 2a should be greater than 6 and hence a > 5 and hence a = 6, 7, 8 ( 3 cases)

for b = 11 for this to happen the the value of 2a should be greater than 6 and hence a > 5.5 and hence a = 6, 7, 8   (3 cases)

Total of = (1+1+2+3+3) = 10 cases

Let the number of workers on each day be x, x+6, x+12, x+18, x+24
Let the amount of work done per worker each day be y, y-5, y-10, y-15, y-20
Total amount = 500
xy+(x+6)(y-5)+(x+12)(y-10)+(x+18)(y-15)+(x+24)(y-20) = 500
5xy+60y-50x-900 = 500
xy+12y-10x = 280
xy+12y-10x-120 = 280-120 = 160
(x+12)(y-10) = 160

Let the rate at which a man can work be x, and rate at which a woman can work be y.
Hence, 3x + 4y = $$\frac{1}{5}$$
and 4x + 3y = $$\frac{1}{6}$$
Adding both the equations we get,
7x + 7y = $$\frac{1}{5} + \frac{1}{6}$$
=> 4x + 4y = $$\frac{4}{7}(\frac{1}{5} + \frac{1}{6})$$ = $$\frac{44}{210}$$
hence, the number of days required by 4 men and 4 women = 210/44 days = 105/22days