Quantitative Aptitude (QA) Test - 13

If there are x,y,z red, green and blue marbles then, the conditions give the following equality $$ ^xC_3 = y*^xC_2 = x*y*z$$. Hence z=(x-1)/2 and y=(x-2)/3. The smallest x to give integral values to y and z is 5.

Let the scores in first match of Prithvi and Rishabh be 23x and 26x respectively.
Let the decrease in number of runs for Prithvi and Rishabh in second match be y
So the score of Prithvi in second match becomes $$23x-y$$ and the score of Rishabh in second match becomes $$26x-y$$

$$\frac{\left(23x-y\right)}{\left(26x-y\right)}=\frac{13}{17}$$

$$391x-17y=338x-13y$$

$$4y=53x$$

$$y=\frac{53}{4}x$$

Rishabh's score in second match$$=26x-y=26x-\frac{53}{4}x=\frac{51}{4}x$$

Ratio$$=\frac{\frac{51}{4}x}{26x}=\frac{51}{104}$$

Number of odd numbers divisible by 3 = multiples of 3 - multiples of 6 = [1000/3] - [1000/6] where [x] is greater integer less than equal to x.

Number of odd numbers divisible by 3 = 333 - 166 = 167

Number of odd numbers divisible by 3 and 7 = multiples of 21 - multiples of 42 = [1000/21] - [1000/42] = 47 - 23 = 24

So, number of odd numbers divisible by only 3 and not 7 is 167-24 = 143

Percentage of milk in mixture A = 75 %
Percentage of milk in mixture B = 88 %
Let the contents of A and B be mixed in the ratio a:b
So we have
75a + 88b = 80(a + b)
=> 5a = 8b
=> a/b = 8 :5
Hence, the required ratio is 8:5. Thus, option A is the correct answer.

Let area of PON be x and area of POM be y 
Now in triangle QOR :NO:OR = 8/10 =4/5 
In triangle PNR ;
NO:OR = x :y+5 =4:5       (1)
Similarly In triangle QMR
MO:QO=1:2
So In triangle PMQ
MO:QO=y:x+8                    (2)
simplifying (1)& (2)
we get
x+8=2y and 5x=4y+20
solving we get x=12 and y= 8
so area of triangle PQR = 22+10+8+5=45cm$$^2$$

Let the salaries of Amit, Ramesh, and Suresh be $$4x$$, $$6x$$, and $$7x$$, respectively.

Let the three expenditures be $$y$$, $$2y$$ and $$3y$$, respectively.

We are given that, 

$$7x-y=4x+3y$$

$$3x=4y$$

$$y=\frac{3}{4}x$$

We are also given that,

$$(7x-3y)-(6x-2y)=1000$$

$$x-y=1000$$

$$\frac{x}{4}=1000$$

$$x=4000$$

Thus, $$y=3000$$

Thus, the expenditure of Ramesh = $$2y=2\times 3000=6000$$

Hence, the answer is option C.

Simple interest required for the elder son is 13.5 lakhs in 6 years and for the younger son is 13.5 lakhs in 9 years. Therefore, the interests required are 30% and 20%.

The second statement is true as $$57=20+19+18$$. So 57! will be bigger than 20!*19!*18! as for 57! we can verify as after 20! every term is greater than 20 while on the other side each term is less than 20.
The first statement is true because $$(23!)^2 = (23*1)*(22*2)*(21*3)...(1*23) > 23*23*...23 = 23^{23}$$
The third statement is false because $$(33!)^2 > 33^{33}$$.
So, $$(33!)^4 > 33^{66} > 33^{60}$$

$$log_a 7 = 1/x$$, $$\log_a 28 = 1/y$$
=> $$\log_a 4*7 = 1/y$$ => $$2\log_a 2 + \log_a 7 = 1/y$$
=> $$\log_a 2 = (1/y - 1/x)/2$$

$$\log_a 14$$ = $$\log_a 7 + \log_a 2$$
= 1/x + (1/y - 1/x)/2 = (x+y)/2xy

So, $$\log_a 14$$ = (x+y)/2xy

Let the initial price be 100k

The price at the end of the first month is 105k. At the end of second month, the price is 115.5k.

At the end of the third month, the price is 132.82k.

At the end of the fourth month, the price is 159.39k, and at the end of the 5th month, it is 199.23k.

So, percentage increase in price = (199.23k - 100k)/100k * 100% = 99.23%

Given a rectangle ABCD,

Let AB=CD=x and AD=BC=y

By the time P reaches D, Q reaches E. So the ratio of speeds of P and Q is 4:1.

Let the speeds of P and Q be 1 and 4 respectively after they interchange their speeds.

It is given that they reach point C simultaneously i.e time =  constant => $$\frac{dis\tan ce}{speed}=cons\tan t$$

So, $$\frac{x}{1}=\frac{\left(\frac{3y}{4}\right)}{4}$$ => $$\frac{y}{x}=\frac{16}{3}\ =>\ BC\ :\ AB\ =\ 16:3$$

The amount of work done by Ravi in 4 days is equal to the amount of work done by Sahil in 3 days. Hence if Ravi does 1 unit of work in 4 days, Sahil can do it in 3 days. In 20 days, Ravi would do 20/4 = 5 units of work.
The time needed by Vinod to complete 5 units of work = 5*3 = 15 days.

The ratio of speeds of A and B is 3 : 5. So, by the time bus A travels 375 km, bus B would have travelled 625 km. So, the first time that the two buses meet when travelling in opposite directions is at a distance of 125 km from city Y. Time taken for this meeting from the start of the journey = 6.25 hours.

For the two buses to meet again while moving in the opposite direction, they should first reach the two opposite ends and then again meet somewhere in the middle. Hence, together they should cover 2*500km = 1000 km.

Time taken to cover 1000 km while they are moving in opposite direction = 1000 / (100kmph + 60kmph) = 1000/160 hrs = 6.25 hrs.

Thus, from the first meeting, the two buses will meet each other while travelling in opposite directions every 6.25 hours. 

Therefore, if they travel infinite number of times, they meet at 6.25hr, 12.5hrs, 18.75hrs, 25 hrs etc.

Distance covered by B for each time period is 6.25*100 km, 12.5*100km, 18.75*100km, 25*100 km etc i.e. 625km, 1250 km, 1875 km, 2500 km etc.

But we know that B travels only 2 times between X to Y and Y to X. So max distance travelled by B is 4*500 km = 2000 km

Hence, A and B will meet at 3 points corresponding to the following distances travelled by B => 625km, 1250km, 1875km.

The first point of meeting is (625 mod 500) = 125 km from Y.

The second point of meeting is (1250 mod 500) = 250 km from Y i.e. mid point of X and Y.

The third point of meeting is (1875 mod 500) = 375 km from Y

This is the final meeting between the two buses. Bus B comes to rest before the next meeting between the buses can take place. So, the answer is 3.

We can solve the above using Cramer’s rule:

System of equations: 

$$2A+3B+2C\ =15$$

$$2A+2B+4C\ =20$$

$$5A+4B+2C\ =25$$

Coefficient matrix  = $$D$$ = $$\begin{vmatrix}2 & 3 & 2\\2 & 2 & 4\\5 & 4 & 2\end{vmatrix}$$ = 20

Determinant of x-matrix ($$D_x$$) = $$\begin{vmatrix}15 & 3 & 2\\20 & 2 & 4\\25 & 4 & 2\end{vmatrix}$$ = 60

Determinant of y-matrix ($$D_y$$) = $$\begin{vmatrix}2 & 15 & 2\\2 & 20 & 4\\5 & 25 & 2\end{vmatrix}$$ = 20

Determinant of z-matrix ($$D_z$$) = $$\begin{vmatrix}2 & 3 & 15\\2 & 2 & 20\\5 & 4 & 25\end{vmatrix}$$ = 20

According to Cramer's rule: $$x=\frac{D_x}{D};\ y=\frac{D_y}{D};\ z=\frac{D_z}{D}$$

Hence, we get $$x=3;\ y=1;\ z=3$$

Option (B) is the correct choice.

The area of the shaded region = $$\frac{\pi}{8}PQ^2-\frac{\pi}{8}PS^2-\frac{\pi}{8}SQ^2\ \ \ $$

= $$\frac{\pi}{8}\left\{\left(PS+SQ\right)^2-PS^2-SQ^2\right\}\ \ \ $$

= $$\frac{\pi}{8}\left\{\left(PS^2+SQ^2+2.PS.SQ\right)^{ }-PS^2-SQ^2\right\}\ \ \ $$

$$\therefore\ Required\ area\ =\frac{\pi}{8}\left\{2.PS.SQ\ \right\}\ =\frac{\pi}{4}\left(PS\times\ SQ\right)\ \ \ $$........(i)

On connecting P and Q to R, we obtain $$\ \triangle\ PQR$$, right-angled at R. 

$$\therefore\ PR^2+QR^2\ =PQ^2$$

Now, $$PR^2=RS^2+PS^2$$ and $$QR^2=RS^2+QS^2$$; also, $$PQ^2=PS^2+QS^2+2.\ PS.QS$$

Equating the above, we obtain $$RS^2=PS.QS$$..............(ii)

Substituting the above in (i), we get $$Shaded\ area\ =\ \frac{\pi}{4}RS^2=\frac{\pi}{4}\times\ 144\ =36\pi\ \ \ $$

Hence, K = 36

Let the total weight of rice that Rajesh purchased be x.

Since Rajesh got cheated by the wholesaler with a faulty weight which weighs 1000gm for 800gm,

CP of 1 kg rice for Rajesh = $$\frac{1000}{800}\times\ 100=125$$

Since he sold 750gm for the price of 1000 gm with a markup of 15%,

SP of Rajesh of 1kg = $$\frac{1000}{750}\times\ 1.15\times\ 100=153.333$$

Thus, total profit for 1 Kg = $$153.333-125\ =\ 28.333$$

Since total profit = 28333,

28.333x = 28333

x = 1000 kg.

Since the wholesaler sold 1000gm for 800gm,

the amount of rice Rajesh thought he purchased = $$1000\times\ \frac{1000}{800}\ =\ 1250$$

Thus, the correct answer is A.

Since it was given that the ice loses one-fourth of its quantity after every one hour.

The truck requires 2 hours of travel to reach the destined location.

In order to make sure the exact requirement is met :

If X is initially carried out at the time of supply he can give only :

$$X\left(\frac{3}{4}\right)\left(\frac{3}{4}\right)$$

Hence if he places X he can supply only 9X/16 and hence he must compensate for 7X/16 lost.

He can compensate by marking up on the cost price.

If Rs A is the cost price of X. He must sell 9X/16 which should originally cost Rs 9A/16 for  Rs A.

Hence the marked price is Rs A and the cost price is Rs 9A/16

$$\frac{\left(A\ -\ \frac{9A}{16}\right)}{\left(\frac{9A}{16}\right)}\cdot\ 100\ =\ \frac{\left(100\cdot7\right)}{9}=\ \frac{700}{9}=\ 77.77\%$$

$$1\cdot5^3+0\cdot5^2+1\cdot5^1+2\cdot5^{0\ }=2x^2+4x+6$$

$$2x^2+4x+6=132$$

=> x = -9 or 7

Since x cant be negative,

X=7

The average of a A.P. Series is half of the sum of the first and last numbers in the series.
X = [1+(2n-1)]/2 = n, Y = [2 + 2n]/2 = n+ 1.
Thus X - Y = -1.

18 identical pink balls can selected in 19 ways.
Similarly 14 identical violet balls can be selected in 15 ways and 8 identical orange balls can be selected in 9 ways.
A selection can be made between the 4 different grey balls in $$2^4=16$$ ways.
Hence, the required no. of ways is $$19*15*9*16 - 1$$

Buses A and B start $$1$$ hour apart from each other.

Thus, B overtakes A after = $$\frac{30\times 1}{45-30}=\frac{30}{15}=2$$hours, i.e., at $$12$$noon.

Buses A and D start $$2$$ hours apart from each other.

Thus, D overtakes A after = $$\frac{30\times 2}{75-30}=\frac{60}{45}=1$$ hours $$20$$ minutes, i.e., at $$12:20$$ PM.

Since we are given that bus C overtakes A at the time exactly between the times of D and B, it will overtake A at $$12:10$$ PM.

Distance travelled by bus A in $$3$$ hours $$10$$ minutes = $$30\times (3 + \frac{1}{6})=95$$ kms

Thus, the time taken by bus C to cover this distance = $$\frac{95}{60}=1$$ hour $$35$$ minutes

Hence, bus C will start from the station at $$10:35$$ AM.

The answer is option A.

(4,4,3) ; (5,3,3) ; (5,4,2) and (5,5,1) are the only four possible triangles with integer sides whose perimeter is 11.