XAT Averages, Ratio and Proportion Questions

Let the total number of jewels be X. 
The jewels in the first box would be X/3
The jewels in the second box would be kX/5
And the number of jewels in the third box is 66

The jewels of the second and third boxes should add up to 2X/3 jewels, giving us the relation. 
$$\frac{kX}{5}+66=\frac{2X}{3}$$
$$66=\frac{X\left(10-3k\right)}{15}$$
$$X\left(10-3k\right)=2\times\ 3^2\times\ 5\times\ 11$$

We are given that k is a positive integer. 
Taking k= 1, we get (10 - 3k) to be 7, which is not present on the right-hand side. 
Taking k = 2, we get (10 - 3k) to be 4, which is again not present on the right-hand side. 
Taking k = 3, we get (10 - 3k) to be 1, which is possible on the right-hand side and would give the value fo X to be 990. 

Taking any further value of k would give a negative value of (10 - 3k), which would not be possible. 

Therefore, the king must have had 990 jewels. 

Hence, Option A is the correct answer.  

Option A) False, as it is mentioned no-one scored exactly 40 marks.

Option B) False, as only 1 student  or only 2 students can score above 40 and group average can be 40.

                 Case i:- 44, 39, 39, 39 & 39.

                 Case ii:- 42, 41, 39, 39 & 39.

Option C) False,  student's scores could be 44, 39, 39, 39 & 39.

Option D) False, student's scores could be 44, 43, 42, 32 & 39.

Option E) True, if scores of all students are more than 40. The average will be more than 40.

Let the number of eggs bought = 9x+2

number of eggs left after throwing away 2 = 9x

number of eggs kept in fridge = 5x

number of eggs brought to his mothers' house = 4x

number of eggs left after cooking 2 which are kept in fridge = 4x-2

Given, 4x-2 <=5

=> x$$\le\frac{7}{4}$$

Hence the max value of x is 1

Max number of eggs bought = 11

Let us assume there are three buses, each carrying 30 passengers.

Now it is given that one-third of the buses from City A to City B stop at City C, while the rest go non-stop to City B. This means that one bus stops at C while two buses go directly to B. This means that $$2\times 30=60$$ passengers directly reach to B.

For the buses that stops at C, One-third of the passengers continue to City B, that is 10 passengers in each bus continue to city B, while the rest alight at City C. Since there is only one bus which stops at city C, this means that $$1\times 20=20$$ passengers alight at C, while $$1\times 10=10$$ passengers travel from city C to city B.

We are asked what proportion of the passengers going to City B from City A travel by a bus stopping at City C. So, in total, we can see the total number of passengers who are travelling to City B is 70, while the number of passengers travelling by the buses that stop at City C is 10.

So, the proportion of the passengers going to City B from City A who travel by a bus stopping at City C = $$\dfrac{10}{70}=\dfrac{1}{7}$$

Let four friends Anita, Bina, Chaitra and Divya be represented as A,B,C,D respectively. 

From the given information, Messages sent by A are 150 out of which 90 are Jokes

Messages sent by B are 180 out of which 72 are Jokes

Messages sent by C are 150 out of which 120 are Jokes

Messages sent by D are 120 out of which 60 are Jokes

.'. Percentage of jokes that were neither sent by D or B is 210*100/342=61.4

Given, z/4 > x/6 > y/5 ...(i)

and (x+y+z)/15= 2*y/5 => x+z=5y

The value of y can only be 1.

=> x+z=5.

If z = 1, then z/4 is less than x/6.

If z = 2, then z/4 is equal to x/6.

If z = 4 or 5 then y/5 is greater than x/6.

The only possible value to satisfy (i) condition is z=3. and x=2.

Let the initial salary of A,B,C,D,E be $$a,b,c,d,e$$ respectively and let the final salary of everyone be $$x$$ 

Now, $$a*(1+ \frac{p}{100})*(1+ \frac{p+1}{100}) = x$$

$$\Rightarrow a = \frac{x}{(1+ \frac{p}{100})*(1+ \frac{p+1}{100})}$$

$$\Rightarrow a = \frac{x*100*100}{(100+p)*(100+p+1)}$$

$$\Rightarrow a = \frac{x*100*100}{(p+100)*(p+101)}$$

Similarly, $$b= \frac{x}{(1+ \frac{p+2}{100})*(1+ \frac{p-1}{100})}$$

$$\Rightarrow b = \frac{x*100*100}{(p+102)*(p+99)}$$

Similarly, $$c= \frac{x}{(1+ \frac{p+3}{100})*(1+ \frac{p-2}{100})}$$

$$\Rightarrow c = \frac{x*100*100}{(p+103)*(p+98)}$$

Similarly, $$d= \frac{x}{(1+ \frac{p+4}{100})*(1+ \frac{p-3}{100})}$$

$$\Rightarrow d = \frac{x*100*100}{(p+104)*(p+97)}$$

Similarly, $$e= \frac{x}{(1+ \frac{p+5}{100})*(1+ \frac{p-4}{100})}$$

$$\Rightarrow e = \frac{x*100*100}{(p+105)*(p+96)}$$

The numerators of the fractions are same, therefore the one with the smallest value of denominator will have the greatest value. If we compare the denominators, we can find out the fraction with the highest value. The person with the highest initial salary got the least raise, as we know that the final salary of all the candidates is same. 

Thus, denominator of a, $$a_{den}= (p+100)*(p+101)= p^{2}+201p+100*101$$

Similarly, $$b_{den}= (p+102)*(p+99)= p^{2}+201p+102*99$$

$$c_{den}= (p+103)*(p+98)= p^{2}+201p+103*98$$

$$d_{den}= (p+104)*(p+97)= p^{2}+201p+104*97$$

$$e_{den}= (p+105)*(p+96)= p^{2}+201p+105*96$$

We see that we need to compare only the last terms of the denominators as the other terms are same. 

Thus, last term of a, $$a_{lt}= 100*101= 100.5^{2}-0.5^{2}$$

 last term of b, $$b_{lt}= 102*99= 100.5^{2}-1.5^{2}$$

 last term of c, $$c_{lt}= 103*98= 100.5^{2}-2.5^{2}$$

 last term of d, $$d_{lt}= 104*97= 100.5^{2}-3.5^{2}$$

 last term of e, $$d_{lt}= 105*96= 100.5^{2}-4.5^{2}$$

Thus, we can see that since denominator of $$e$$ is the smallest, therefore E has the highest initial salary.

Inverse proportionality is expressed as follows : 

$$a\propto \frac{1}{b}$$

$$\Rightarrow$$ a*b = constant 

When value of a changes, the value of b changes accordingly such that their product remains same.

Thus, new value of a $$a^{'} = a + a = 2a $$

Thus, the new value of b $$b^{'}$$ can be solved by : 

$$a^{'}*b^{'}=a*b$$

$$\Rightarrow 2a*b^{'}= a*b $$

$$\Rightarrow b^{'}= \frac{b}{2} $$

Thus b decreases by 50%. 

Let the number of girls in the school be G. 
=> Number of boys = G+30.
Some girls joined the class and the number of boys and girls became 3:5. 
Let the number of girls who joined the class be 'X'.
It has been given that (G+30)/(G+X) = 3/5
5G + 150 = 3G + 3X
2G + 150 = 3X
=> X = (2G/3) + 50.
2G has to be divisible by 3. 
Therefore, the least value that G can take is 3. 
When G = 3, X = 2 + 50
X = 52. 
The least number of girls who could have joined is 52. 
Therefore, option E is the right answer.

Let the age of the father be 'f', mother be 'm', Hari be 'h', Chari be 'c'. It has been given that Chari is 4 years older than Gouri. Therefore, the age of Gouri is c-4. 

When Chari was born, the sum of the ages of Hari, his father and mother was 70 years. 
If Chari's age is 'c' now, then Chari's father's age when Chari was born would have been 'f-c' (i.e, Current age - the number of years that has passed after Chari's birth). The same holds true for all the family members. 

=> f - c + m - c + h - c = 70
f+m+h-3c = 70 -------------(1)

The sum of the ages of the 4 family members when Gouri was born was twice the sum of the ages of the father and mother at the time of Hari's birth. 

=> f-(c-4) + m -(c-4) + h -(c-4) + c - (c-4) = 2(f-h+m-h)
=> f + m + h + c - 4(c-4) = 2f + 2m - 4h
f+m+h-3c + 16 = 2f+2m-4h
Substituting (1), we get,
70 +16 = 2f+2m-4h
43 + 2h= f+m ---------------(2)
Substituting (2) in (1), we get, 
43 + 2h + h - 3c = 70
3h - 3c = 27
=> h-c = 9
Therefore, the difference between the age of Hari and Chari is 9 years. Therefore, option E is the right answer.

XYZ running cost = Rs 400/student

Each student pays Rs 2000

Profit = 2000 -  400 = 1600

ABC receives 60% of profit, i.e. 1600*0.6 = Rs 960

XYZ begins to make profit from first student itself.

For ABC to recover its investment, number of students should be 100000/960 = 104.16

Therefore, partnership needs at least 105 students to reach break even point. 

The answer is option C.

Expression : $$S=\frac{\alpha\times\omega}{\tau+\rho\times\omega}$$

=> $$\frac{1}{S} = \frac{\tau+\rho\times\omega}{\alpha\times\omega}$$

=> $$\frac{1}{S} = \frac{\tau}{\alpha \omega} + \frac{\rho}{\omega}$$

Since, $$\tau, \rho$$ and $$\alpha$$ are constant,

=> $$\frac{1}{S} = \frac{k_1}{\omega} + k_2$$

Thus, $$S \propto \omega$$

$$\therefore$$ When $$\omega$$ increases, S increases.

Grade A $$\geq$$ 90 and Grade B = 87 to 89

If Ramesh scores 70 instead of 97, => Change of marks = 97 - 70 = 27

It creates a change from grade A to B, this means an overall change in average by

= Minimum marks for grade A - Minimum marks for Grade B = 90 - 87 = 3

$$\therefore$$ Number of subjects = $$\frac{27}{3} = 9$$

Let $$x, y, z$$ be the number of students getting 6, 8 and 20 marks respectively.

=> $$6x + 8y + 20z = 504$$ -------------Eqn(I)

Since, the mode is 8, => Most populated category = $$y$$

=> $$y = x + 2z$$

=> $$x = y - 2z$$ -------------Eqn(II)

Substituting it in eqn(I), we get : 

=> $$(6y - 12z) + 8y + 20z = 504$$

=> $$14y + 8z = 504$$

By hit and trial, since $$x,y,z$$ are integers, we get : $$y = 32$$ and $$z = 7$$

Putting it in Eqn(II), => $$x = 32 - 2 \times 7 = 18$$

$$\therefore$$ Total students = $$x + y + z = 18 + 32 + 7 = 57$$

We know that Ramesh bought 6 fruits in total. 
If the number of apples is 1, then the number of oranges required to get an equal amount of juice will be 0. Therefore, we can eliminate this case. 

If the number of apples is 2, then the number of oranges required to get an equal amount of juice will be 1. We know that Ramesh had 1 more orange than needed. The total number of fruits in this case is 2+1+1 = 4. Therefore, we can eliminate this case too.

If the number of apples is 3, then the number of oranges required to get an equal amount of juice will be 2. We know that Ramesh had 1 more orange than needed. The total number of fruits in this case is 3+2+1 = 6. This satisfies the condition. 

The quantity of juice from 3 apples is equal to the quantity of juice from 2 oranges. 
Therefore, the proportion of apple juice in the initial mixture is 2/(2+3) = 2/5 = 40%. (2+3 is used since we are finding the quantity of juice. 2 denotes the quantity of juice obtained from 3 apples)

Therefore, option E is the right answer. 

Median = 5 , Mode = 8

Mean = 5, => Sum of 5 salaries = $$25$$

As mode is 8, it will occur 2 times but not 3 ($$\because$$ sum is 25)

Also, median is 5, the third salary is 5 and the first two are less than 5

=> Sum of third , fourth and fifth salary = 5 + 8 + 8 = 21

Sum of first two = 25 - 21 = 4

First and second salaries cannot be same ($$\because$$ mode is 8)

=> First and second salary = 1 and 3

$$\therefore$$ Sum (in lakh) of the highest and the lowest salaries = 1 + 8 = 9

Let the software fails $$a, b$$ and $$c$$ times in a single stage, in two stage and in all stages respectively.

Given : $$c = 4$$

To find : $$a = ?$$

Solution : $$a + 2b + 3c = 15 + 12 + 8$$

=> $$a + 2b + 3c = 35$$ -------------Eqn(I)

Also, $$b + 3c = 6 + 7 + 4$$

=> $$b + (3 \times 4) = 17$$

=> $$b = 17 - 12 = 5$$

Using eqn(I),

=> $$a + (2 \times 5) + (3 \times 4) = 35$$

=> $$a + 10 + 12 = 35$$

=> $$a = 35 - 22 = 13$$

Number of teams in the bottom group $$= 10$$
Let the total number of teams in top group = $$n$$
Total number of teams $$ = 10 + n$$

=> Number of matches played amongst the bottom group teams = $$^{10}C_2$$
= $$\frac{10 \times 9}{1 \times 2} = 45$$

Number of points bottom group teams get playing amongst themselves $$= 45\cdot1=45$$

Let the total number of teams in top group = $$n$$

wkt, "Each of the bottom ten teams earned half of their total points against the other nine teams in the bottom ten"
i.e. they get half the total points by playing amongst themselves and the other half of total points by playing with the top group teams.

Since they got 45 points playing amongst themselves, the bottom teams get 45 points from their matches against top group teams, => 45 out of $$10 n$$ points
Total points by matches between top teams and bottom teams $$= 10\cdot{n}\cdot1 =10n$$ points

Number of points that top group teams get from matches playing amongst themselves = $$^nC_2$$

Number of points that top group gets against the bottom group = $$10n - 45$$

wkt, "Exactly half of the points earned by each team were earned in games against the ten teams which finished at the bottom of the table.". Therefore the top n teams obtained half points by playing amongst themselves and half the points by playing against bottom 10 teams.

=> $$^nC_2 = 10n - 45$$

=> $$n (n - 1) = 20n - 90$$

=> $$n^2 - 21n + 90 = 0$$

=> $$(n - 6) (n - 15) = 0$$

If, $$n = 6$$, top group would get = $$C^n_2 + 10n - 45$$

= $$C^6_2 + 60 - 45 = 30$$

Average points per game = $$\frac{30}{6} = 5$$

Bottom teams will get on an average = $$\frac{45 + 45}{10} = 9$$

This is not possible.

=> $$n = 15$$

$$\therefore$$ Total number of teams = $$15 + 10 = 25$$

Marks = 71, 76, 80, 82 and 91

For average to be an integer each time, the sum of numbers entered after 2nd entry should be divisible by 2, after third entry should be divisible by 3 and so on.

The first two numbers have to be both odd or both even, so that their sum is even and can be divisible by 2.

Also, the sum of first four numbers should also be even.

=> Numbers entered are either $$OOEEE$$ or $$EEOOE$$   (where O -> odd and E -> even)

Case 1 : First two numbers are 71 and 91 in any order.

Average = $$\frac{71 + 91}{2} = 81$$

Now, sum of 71 and 91 is multiple of 3, so the third number has to be a multiple of 3, which is not possible.

Case 2 : First two numbers can be = $$(76,80) , (76,82) , (80,82)$$

Now, following above criteria, only 2nd option is possible 

So, third number has to be 91, average = $$\frac{76 + 82 + 91}{3} = 83$$

$$\therefore$$ The fourth and fifth marks that Prof. Bee entered = 71 and 80

Let total employees in Sun Metals = $$100x$$

Number of employees who are general graduates = $$\frac{40}{100} \times 100x = 40x$$

=> Number of employees who are engineers = $$100x - 40x = 60x$$

Now, number of engineers who earn more than Rs. 5 lakh/year = $$\frac{75}{100} \times 60x = 45x$$

Number of employees (both general graduates and engineers) who earn more than Rs. 5 lakhs/year = $$\frac{50}{100} \times 100x = 50x$$

=> Number of general graduates who earn more than Rs. 5 lakhs/year = $$50x - 45x = 5x$$

Thus, number of general graduates who earn less than Rs. 5 lakhs/year = $$40x - 5x = 35x$$

$$\therefore$$ Proportion of the general graduates employed by the organisation earn Rs. 5 lakh or less

= $$\frac{35 x}{40 x} = \frac{7}{8}$$