Quantitative Aptitude (QA) Test - 16

Given that the average weight of boys is 40 and girls is 60 . Let us assume the number of boys be x and the number of girls be y .

40(x) +60(y)/x+y = 45

5x = 15y 

x = 3y 

Now if we add the 8 new joinees  :

(40(3y) +80(8) + 60(y))/(4y+8) = 52 .

y = 8 , x = 24 .

$$\log_{8}{\dfrac{8}{x^2}} = 3 (\log_{8}{x})^2$$

=>$$\log_{8}{8} - \log_{8}{x^2} = 3 (\log_{8}{x})^2$$

Let $$\log_{8}{x}$$ be $$y$$

Then, $$1 - 2y = 3y^2$$

or, $$3y^2 + 2y - 1 = 0$$

on solving, we get 

$$y = -1$$ or $$y = \dfrac{1}{3}$$

=>$$\log_{8}{x} = -1$$ or $$\log_{8}{x} = \dfrac{1}{3}$$

=>$$x = \dfrac{1}{8}$$ or $$x = 2$$

Thus, two values of $$x$$ are possible.

Let S = $$ 1 + \frac{2}{5} + \frac{4}{5^2} + \frac{7}{5^3} $$... -(1)
Dividing by 5 we get $$\frac{S}{5} = \frac{1}{5} + \frac{2}{5^2} + \frac{4}{5^3} + \frac{7}{5^4} + \frac{4}{5^5} $$..... (2)
Subtracting (1) from (2) we get: $$\frac{4S}{5} = 1 + \frac{1}{5} + \frac{2}{5^2} + \frac{3}{5^3} + \frac{4}{5^4} + \frac{5}{5^5} $$..... (3)
Dividing (3) by 5 we get: $$\frac{4S}{25} = \frac{1}{5} + \frac{1}{5^2} + \frac{2}{5^3} + \frac{3}{5^4} + \frac{4}{5^5} + \frac{5}{5^6} $$..... (4)
Subtracting (4) from (3) we get: $$\frac{16S}{25} = 1 + \frac{1}{5^2} + \frac{1}{5^3} + \frac{1}{5^4} + \frac{1}{5^5} + \frac{1}{5^6} $$.....
The above equation is in G.P with a = $$\frac{1}{5^2}$$ and r = $$\frac{1}{5}$$
Thus, $$\frac{16S}{25} = 1 + \frac{1/5^2}{1 - 1/5} = 1 + \frac{1}{20}$$ = $$\frac{21}{20}$$
S = $$\frac{105}{64}$$

Rs 5000 at the end of three years amounts to
$$5000*(1+4/100)^6$$ = Rs 6326.6

Rs 1000 is paid at the end of the first year.
Interest incurred on this payment at the end of the 3 years = $$1000*(1+4/100)^4$$ = Rs 1169.9
Another installment of Rs 1000 is paid at the end of the second year
Interest incurred on this payment at the end of the 3 years = $$1000*(1+4/100)^2$$ = Rs 1081.6
So, amount that has to be paid at the end of the third year = 6326.6 - (1169.9 + 1081.6) = Rs 4075.1

So, amount that has to be paid at the end of the third year so as to clear all dues = Rs 4075.1

The given expression equals $$1+\frac{8}{y^2+12y+24}$$ This expression equals a prime number only if $${y^2+12y+24}$$ equals either 2 or 4 or 8. Solving the equation, we get y=-2 or y=-10 as the only integral solutions

Area of a cyclic quadrilateral is given by the ancient Indian mathematician Bramhagupta. 

When a cyclic quadrilateral has sides a,b,c,d, its area is given by $$\sqrt{\ \left(s-a\right)\left(s-b\right)\left(s-c\right)\left(s-d\right)}$$

where s=$$\frac{\left(a+b+c+d\right)}{2}$$

.'. s= 36/2 =18.

Area = $$\sqrt{\ \left(18-10\right)\left(18-6\right)\left(18-8\right)\left(18-12\right)}$$ =$$24\sqrt{\ 10}$$

For each jump, kangaroo rolls back 1m. Net distance travelled for each jump = 4m

But at the last jump, it will travel a total of 5m.

=14*4+5=61m

Option A

Picks formula states that for any lattice polygon: A=I+$$\frac{1}{2}P$$-1

Where A is the area of the polygon, I is the number of interior lattice points of the polygon and P is the number of boundary points of the polygon. For answering the question, we need to find the interior lattice points I as well as the points on x-axis and y-axis between the two lines.

The area between the lines is shown by the purple area in the following figure

$$x \geq 0$$ and $$y \geq 0$$ ==> We need to consider only the first quadrant

4x+5y=100 intersects the axes at (25,0) and (0,20)

2x+3y=30 intersects the axes at (15,0) and (0,10)

Hence Area A = Difference in area of two triangles = 1/2 * 20 * 25 -1/2* 10 *15 = 250-75 = 175

Number of boundary pts P = Pts on 4x+5y=100 + Pts on 2x+3y=30 + Pts on x axis + Pts on y axis

4x+5y=100 => y = 20 - 4x/5

Hence, we will have integral coordinates when x=0, 5, 10, 15, 20 and 25 => 6 points

2x+3y=30 => y = 10 - 2x/3

Hence, we will have integral coordinates when x=0, 3, 6, 9, 12 and 15 => 6 points

Pts on x axis = (16,0), (17,0) . . (24,0) = 9 points. We dont consider (15,0) and (25,0) as they already have been added earlier with the points on the lines

Similarly, Pts on y axis = (0,11) . .. (0, 19) = 9 points

P = 6 + 6+ 9 + 9 = 30

By Pick's formula, A = I + P/2 -1

=> 175 = I + 30/2 - 1

=> I =161

Hence, number of points with integral co-ordinates = I + pts. on x axis + pts on y axis = 161 + 9 + 9 = 179

Total number of rectangles on a square board having distinct length and breadth = 9C2*9C2 - ($$1^2+2^2+....+8^2$$) = (36*36) - (8*9*17/6) = 1296 - 204 = 1092.
Now we want the are 18 and dimensions 6 and 3.
Let the breadth be 3 units. If we consider first 3 rows as a breadth, then the total number of rectangles formed is 3.
The breadth of 3 units can be taken in 6 ways.
Total number of rectangles formed = 6*3 = 18

Similarly if we consider length 3 units, Total number of rectangles formed = 18

The number of rectangles with area 18 sq units with one side as 6 units = 18+18 = 36

Probability = 36/1092 = 3/91.

Let Raju complete 10 units of work in a day, then, Ramya will do 40 and Ranvijay will do 48. Since, it takes seven days for all of the working together, the total units of work are=98*7=686 units.
Starting with Raju, Ranvijay and Raju will finish 638 units in 22 days. There are 48 units of work remaining. Raju will do 10 units of work on the 23rd day, leaving 38 units for Ranvijay. Ranvijay will take $$\dfrac{38}{48}$$ of the day to complete this. Hence, the answer is Option C.

The numbers that can be made using 0, 1 and 5 less than 1 million are $$3 ^ 6 =729$$.
Of these the numbers that are less than 1000 are $$3^3=27$$. Hence the numbers between 1000 and 1 mn are 729 - 27 -1 (removing 1000) = 701.

It is given that $$a = -|x|*y$$
Multiplying both sides by $$-y$$ we get $$-ay = |x|*y^{2}$$
Adding $$x$$ to both sides, we get $$x - ay = x + |x|*y^{2}$$
Note that $$x + |x|*y^2 \geq 0$$ as $$y^2>1$$
Therefore, $$x-ay \geq 0$$

Now it is given that :

Now let each side be a
As it is a regular octagon 
Each angle will be 135 degrees
In triangle BMC:
BM=CM=asin45=$$\frac{a}{\sqrt{\ 2}}$$
In rectangle CMPD
MP=CD=a
Therefore In triangle BPG
BP=a+$$\frac{a}{\sqrt{\ 2}}$$
similarly PG will be a+$$\frac{a}{\sqrt{\ 2}}$$
Now area of figure BCDG will be
Area of triangle BMC +Area of rectangle CMPD+ Area of triangle BPG
⇒$$\frac{1}{2}\times\ \frac{a}{\sqrt{\ 2}}\times\ \frac{a}{\sqrt{\ 2}}$$+$$a\times\ \frac{a}{\sqrt{\ 2}}$$+$$\frac{1}{2}\times\ \left(a+\frac{a}{\sqrt{\ 2}}\right)^{^2}$$
⇒we get $$\frac{a^2}{4}$$+$$\frac{a^2}{\sqrt{\ 2}}$$+$$\frac{1}{2}+\frac{1}{4}+\frac{1}{\sqrt{\ 2}}$$
⇒we get ($$\sqrt{\ 2}+1$$) $$a^2$$
Now area of an octagon :
Let center of octagon be O each side will subtend an angle of 45 at the center
Now let us consider 1 triangle AOB

OP is perpendicular to AB
Now AP=a/2 
let OP be x
Now Angle AOP=45/2=22.5
Now tan22.5=a/2x
x=a/2tan22.5
so x=$$\left(\frac{a}{2\left(\left(\sqrt{\ 2}-1\right)\right)}\right)$$
so x= $$\left(\frac{a\left(\sqrt{\ 2}+1\right)}{2}\right)$$
Now area of octagon =8 times area of AOB
So area of octagon = $$8\times\ \frac{1}{2}\times\ a\times\ \frac{a\left(\sqrt{\ 2}+1\right)}{2}$$
= 2($$\sqrt{\ 2}+1\ $$) a$$^2$$
Therefore ratio of areas = 1:2

Assume the coordinates of point Q as (0,0), R (a,0), P (0,a).
This means that coordinates of A are (a,a). This means PQRA is a square of side a and ABCD is a square of side a/2.
Now, $$AQ=PR=a\sqrt{2}$$ => $$QC=PC=\frac{a\sqrt{2}}{2}=\frac{a}{\sqrt{2}}$$
So, perimeter of $$PQC= a+a\sqrt{2}$$ and perimeter of the square is (a/2) X 4= 2a
The ratio is $$\frac{1+\sqrt{2}}{2}$$

Let us assume the cost price of one orange to be Rs. 100 for ease of calculation.
Total cost incurred by the shopkeeper = 80*100 = Rs.8000

20% of the oranges get spoiled. Therefore, the shopkeeper would have 0.8*80 = 64 apples left.
He sells 12.5% of these oranges at 0.8*100 = Rs. 80
Therefore, the shopkeeper would have sold 8 oranges at Rs.80 netting Rs.640.
He must sell the remaining 56 oranges at Rs. 8000 - 640 = Rs. 7360 to break even.

Therefore, the selling price of 1 orange must be 7360/56 = 131.42

As we can see, the markup percentage must approximately be equal to 31.42%.
Hence, option D is the right answer.

Let A and B meet of the nth page. Time taken by A = n/10
Number of pages read by B = 1000 - n
Time taken by B = (1000-n)/5
So, n/10 = (1000-n)/5
=> n = 2000 - 2n
=> 3n = 2000
=> n = 666.67

So, A and B meet on the 667th page of the document.

Let the marbles with Amit and Suresh be 7$$x$$ and 16$$x$$, respectively. Thus, the total number of marbles = 7$$x$$ + 16$$x$$ = 23$$x$$

Let the marbles with Amit and Suresh after the sharing be 9y and 4y, respectively. Thus, the total number of marbles = 4y + 9y = 13y.

23$$x$$ = 13y

y = $$\frac{23x}{13}$$

Marbles with Suresh after giving = 4y = 4 x $$\frac{23x}{13}$$ = $$\frac{92x}{13}$$

A = $$\frac{16x-\frac{92x}{13}}{16x}$$ = $$\frac{208x-92x}{208x}$$ = $$\frac{116}{208}$$ = $$\frac{29}{52}$$

B = $$\frac{9y}{16x-\frac{92x}{13}}$$ = $$\frac{9\times\frac{23x}{13}}{\frac{208x-92x}{13}}$$ = $$\frac{9\times23}{116}$$ = $$\frac{207}{116}$$

A : B =  $$\frac{29}{52}$$ : $$\frac{207}{116}$$ = 29 x 116 : 207 x 52 = 841:2691

Hence, option B is the answer.

The numerator can be factorized as follows:

$$4^{82}$$ - $$3^{82}$$ = (4-3)*($$4^{81}$$ +$$4^{80}*3$$ + . . + $$3^{81}$$)

Therefore, S = ($$4^{81}$$ + $$4^{80}*3$$ + . . + $$3^{81}$$)/( $$4^{81}$$ - $$3^{81}$$ ) > 1

In this question, it is given that during the initial journey from A to B => It takes 1 hour travelling at 30 kmph => distance between A and B = d = 30 kms.

Let us assume the speed of the train during the return journey is 's' kmph.

We know that the average speed is the ratio of total distance through out the journey to the total time taken.

Waiting time of the time during the return journey = 4 * 3 mins = 12 mins = $$\dfrac{12}{60}=\dfrac{1}{5}$$ hr

Therefore, Average Speed = Total distance/ Total speed

=> $$40=\dfrac{\left(30+30\right)}{1+\dfrac{1}{5}+\dfrac{30}{s}}$$

=> $$\dfrac{6}{5}+\dfrac{30}{s}=\dfrac{3}{2}$$

=> $$\dfrac{30}{s}=\dfrac{3}{2}-\dfrac{6}{5}=\dfrac{3}{10}$$ => s = 100 kmph.

Let the price of the TV be Rs. 100. Then, the cost of the fridge and the AC becomes Rs.40 and Rs.60 respectively. 

Initially, the discounted prices were supposed to be as follows:

Fridge (10% discount) = Rs.36 ; AC (20% discount) = Rs.48; TV (25% discount) = Rs.75; Total = (36+48+75) = Rs. 159

After misinterpretation, the prices became:

Fridge (25% discount) = Rs.30 ; AC (20% discount) = Rs.48; TV (10% discount) = Rs.90; Total = (30+48+90) = Rs. 168

% increase = $$\frac{\left(168-159\right)}{159}\times\ 100\approx\ 6\%$$

Now 
Let us draw the fig

Now PO1Q is a right triangle 
so we get O1P =40cm
Now let radius of smaller circle be r
We can say
triangles O1QP and O2NP are similar(AA)
so we get
$$\ \frac{24}{r}=\frac{40}{16-r}$$
solving we get r = 6cm
Now we know that
PD.PE=PN^2
Now PN =$$\ \sqrt{\ PO2^2-r^2}$$
⇒PN =$$\ \sqrt{\ 10^2-6^2}$$
=8cm
Now PD.PE=64
⇒PE =$$\ \frac{64}{5}$$
      = 12.8cm
So DE =12.8-5 =7.8cm
⇒x=7.8
Therefore 5x= 39cm

Initially in Jar A the water and Juice are in the ratio 2 : 3.

The total volume of Jar A is 1 litre and hence the volume of the water and Juice are 2/5 litre and 3/5 litre.

Similarly for Jar B the water and Juice are in the ratio 3 : 4 and a total volume of 2 litres and hence :

6/7 litre water and 8/7 litre juice.

Now if they are mixed the new solution contains :

Water = 2/5+ 6/7 litres of water = 44/35 litres of water.

Juice = 3/5 + 8/7 = 61/35 litres of Juice.

Of which 1 litre is taken out = Hence this contains :

44/105 litres of water and 61/105 litres of juice.

This is mixed in jar of volume 1 litre which contains water and Juice in the ratio 1 : 4.

Hence 1/5 litre water and 4/5 litre juice this can be written as 21/105 litre water and 84/105 litre Juice .

Hence the final mixture has :

(44+21)/105 litre water and (61+84)/105 litre Juice.

65/105 : 145/105

13: 29