120+ SNAP 2025 Quant Questions PDF with Video Solutions

Let the price of 1kg of rice be x.
The shopkeeper got 1.1 kg for x. So, CP per kg for shopkeeper is x/1.1
Since he makes 10% profit, the SP/MP of the rice will be x/1.1 X 1.1 =x
So, the price remains the same. So, he marked up the price by 0%.

Let CP of the watch be x.
When he earns 20% profit, the SP is 2400.
=> 1.2x=2400 or x=2000
This means that he earns Rs 400 profit by selling each watch.
Now the CP of the watch increases by 10% i.e. the new CP is Rs 2200.
He want to earn the same profit as before i.e. Rs 400.
=> The new SP is Rs 2600.

The interest is calculated quarterly.
Hence, total period = 4*2 = 8.
Effective rate for each period = $$\frac{16}{4}$$ = 4%
Therefore, the interest paid by Ravi = $$6300 \times (1+\frac{4}{100})^8 - 6300$$ = $$8621.98 - 6300$$ = $$Rs. 2321.95 \approx Rs. 2320$$.
Therefore, option A is the right answer.

In order to calculate this type of questions we have a simple formula or shortcut which is

$$n_2 = n_1 ^ { \left(\dfrac{t2}{t1}\right) }$$

Where n = no of times

t = years

Here $$n_1 = 2(Twice), n_2=8, t_1 = 4,  t_2 = ?$$

On substituting the values we get

$$8 = 2 ^ { \left(\dfrac{?}{4}\right) }$$

$$2^3 = 2 ^ { \left(\dfrac{?}{4}\right) }$$

=> Since the bases are equal, the powers are also equal

So, 3 = $$ \dfrac{?}{4} $$

=> ? = 3 x 4 = 12 years

Total time = 12 years

Required time = 12-4 = 8 years

$$\text{Average speed} = \dfrac{\text{Total distance}}{\text{Total time}}$$

Speed of car = $$\dfrac{1040}{13} = 80 \ \text{kmph} \ $$

So, speed of truck = $$30 \ \text{kmph}$$

Speed of train = $$120 \ \text{kmph}$$

So, average of the speed = $$\frac{\left(30+120\right)}{2}=75$$ km/hr

The number of goals scored by team C = 150 - 60 - 50 = 40
Number of matches played by C = 20+20 = 40 (20 matches with A and 20 matches with B)

So, average number of goals scored by C per match = 40/40 = 1

The ratio of the speeds of A to B = 3:12 = 1:4
Therefore the number of distinct points at which they meet on the circular track = 1+4 = 5.
So the correct option to choose is D -5.

He moves 3 steps forward and 1 backward. This means that in 8 seconds, he moves only 2 steps. So for 16 steps he take 64 seconds. After this when he moves forward 2 steps he will fall down and won’t be able to come back. Thus the time after which he falls down is 64 + 2 X 2=68 seconds.

Distance travelled in first half-n-hour = 4km.
Distance travelled in second half-n-hour = 3.5km and so on….
==> 4+3.5+3+2.5..n terms = 16
For n=5, the sum of the terms = 15.
And in the next half-n-hour amith’s speed = 3km/hr
Time taken by Amith to cover a distance of 1km = 1/3 hr.
==> Total time taken by Amith to cover a distance of 16 km = $$2\dfrac{1}{2}+\dfrac{1}{3}$$hr = $$2\dfrac{5}{6}$$ hr.

Relative speed of the trains => ( 90 - 60 ) = 30 km/h
=> Speed in m/sec => ( 25/3 ) m/sec
Total Distance covered in 24 sec => ( ( 25/3 ) m/sec ) x 24 sec = 200 m
Since, it would include length of both the trains, length of one train => 200 $$ \div $$ 2 = 100 m
=> Option A is correct

In the time that Raman runs 200 m, Sumit runs 160 m. Hence, the ratio of their speeds is 5:4. Now, in the time that Raman runs 200 m, Sumit runs 170 m. Hence, the ratio of their speeds is 20:17
Thus, we can say that the ratio of the speeds of all three people will be 25:20:17
Hence, the ratio of the speeds of Raman and Sumit is 25:17.
Hence, in the time that Raman travels 500 m, Sumit will travel 340 m. Thus, Raman will beat Sumit by 160 m.

Let the efficiency of a boy be ‘x’ and the efficiency of a girl be ‘y’
7x = 11y => x = (11/7)y
Total work to be done = 7x * 15 = 11y * 15 = 165y
Efficiency of one girl and one boy = x + y = (18/7)y
So, time required = 165/(18/7) = 165*7/18 = 55*7/6 = 385/6 = $$64 \frac{1}{6}$$ days

20 men can reap a field in 20 days.
⇒ 1 man can reap that field in (20 × 20) days = 400 days
Let 5 men leave the field after x days, so that the remaining 15 men can complete the work field in 24 days.
20x+ 15×24 = 400 ⇒ x = 2 days
∴ 5 men must leave the work after 2 days

Let him do 1 unit of work on day 1. Subsequently, he will do $$\dfrac{3}{4}, \dfrac{9}{16}, \dfrac{27}{64}$$ units of work on day 2, day 3, and day 4.

Total work = $$\left(1 - \left[\dfrac{3}{4}\right]^4\right) * 4=2.73$$ units

If he were working at the original efficiency of 1 unit/day, then he would have taken 2.73 days to complete the work.

Increase in time $$=\frac{4-2.73}{2.73}\times100=46.52\%\approx46\%$$

Considering the current age of Ram  = P, the current age of Lakshman = Q.

It has been given that : $$\dfrac{\left(P-4\right)}{\left(Q-4\right)}=\ \dfrac{2}{1}$$

P-4 =2Q - 8.

2Q-P = 4. ( 1)

Using the other information.

$$\dfrac{\left(P+5\right)}{\left(Q+5\right)}=\ \dfrac{3}{2}$$

2P+10 = 3Q+15.

2P-3Q = 5.  (2)

Solving 1 and 2 we get :

Q = 13, P = 22.

X+14 = 22, X =8.

Y+5 = 13, Y = 8.

Hence X = Y.

We are given the equation,
$$x^3-13x^2+50x-56$$

Since the constant term is a multiple of 2, we see if this particular equation is divisible by 2 or not. By substituting it we get,
8-52+100-56=0

Dividing the above equation by long division we get the quotient as $$x^2-11x+28$$
Therefore, the roots being 2, 4, 7

Applying the inequality for quadrilaterals, we have
$$a+b+c>d$$
Keeping the fourth side as x, we get the following inequalities,
$$x<2+4+7$$
$$x<13$$

And we also get, $$7<4+2+x$$
$$x>1$$

The final inequality is,

$$1<x<13$$

Integral values are, 2,3,4,5,6,7,8,9,10,11,12

There are 11 values

$$\dfrac{1}{3\left|x\right|-2}-\dfrac{4}{15}<0$$
$$\dfrac{\left(15-12\left|x\right|+8\right)}{15\left(3\left|x\right|-2\right)}<0$$
$$\dfrac{\left(23-12\left|x\right|\right)}{15\left(3\left|x\right|-2\right)}<0$$
Now for this equation to satisfy, either the numerator should be positive and the denominator negative or the numerator negative and denominator positive
Case 1: $$23-12\left|x\right|>0$$ and $$3\left|x\right|-2<0$$
Case 2: $$23-12\left|x\right|<0$$ and $$3\left|x\right|-2>0$$
We get the relation, $$\dfrac{23}{12}<\left|x\right|$$ and $$\left|x\right|<\dfrac{2}{3}$$
$$\left(-\infty\ ,\ -\dfrac{23}{12}\right)U\left(-\dfrac{2}{3},\dfrac{2}{3}\right)U\left(\dfrac{23}{12},\ \infty\ \right)$$
Only integers not satisfying this range is -1, 1
Hence sum is zero

Let the cost of a shirt be S and the cost of a pant be P.
Hence, 15S + 8P = 593
And, 8S + 15P = 649
So, 23S + 23P = 1242
Therefore, S+P = 54
Based on this information, we can calculate that S = 23 and P = 31
Hence, the cost of 2 shirts and one pant is 23*2 + 31 = 77

As we must fit all the petrol in the minimum number of containers, the volume of each container should be equal to
$$HCF(253, 345, 391) = 23 L$$
Thus number of containers required will be
= $$\dfrac{(253 + 345 + 391)}{23}$$
= $$43$$

It is given that N leaves remainder 3, 4 and 5 when it is divided by 6, 7 and 8 respectively. We can also say that N leaves -3 as the remainder when it is divided by 6, 7 and 8.
Therefore, the smallest number which is of this kind = (LCM of 6, 7, 8) - 3 = 168 - 3 = 165
Next such number = 165 + 168 = 333
We can say that
$$\Rightarrow$$ N $$\leq$$ 9999
$$\Rightarrow$$ 165+(n - 1)168 $$\leq$$ 9999
$$\Rightarrow$$ n < 59.53
Therefore, we can say that $$n_{max}$$ = 59.
Therefore, N = 165+58*168 = 9909
Hence, the sum of the digits of N = 9 + 9 + 0 + 9 = 27. (Option : E)

It has been given that the number leaves the same remainder when dividing 237 and 269. Therefore, the difference between the 2 numbers (269 - 237 = 32) must be divisible by the number. Therefore, the number must be a factor of 32.

The factors of 32 are 1, 2, 4, 8, 16 and 32. The number can take 6 values. Therefore, option C is the right answer.

Number of days in March = 31
Number of days in April = 30
Number of days till 15th May = 15
Total number of days till his birthday = 31+30+15 = 76
He starts with Rs. 3 and keeps on increasing the amount he saves by 3 i.e. the series is an A.P. with a = 3, d = 3 and n = 76
Sum of the money he has on his birthday =
$$\frac{n}{2}$$*(2a+(n-1)*d) = $$\frac{76}{2}$$*(2*3+(76-1)*3)
= 8778
He needs 10000 Rupees. Therefore, he will need 10000-8778 = 1222 Rupees.
Therefore, our answer is option ‘c’.

As the equation is given terms of prime factors x and y. Expressing 41472 in terms of prime factors we have
41472=512*81
=$$2^{9}\times 3^{4}$$
Therefore x=9,y=4
Required is $$(x+y)^{2}$$
=$$(9+4)^{2}$$
=169

The largest scale size must be the HCF of (42, 84, 98) i.e. 14 m. Thus, E is the correct answer.

The total number of ways in which Raman can pick two chocolates from the bag is $$^{15}C_2 = 15*14/2 = 105$$
The number of ways in which Raman can pick two chocolates of type B from the bag is $$^5C_2 = 5*4/2 = 10$$

Hence, the required probability is $$\dfrac{10}{105} = \dfrac{2}{21}$$

For B to win,
Case-1: A must get a head in his first chance and B must get a tail in his first chance
Case-2: A must get a head in his first chance and B must get a head in his first chance and A must get a head in his second chance and B must get a tail in his second chance
.
.
And so on.
Probability of B winning
= $$(\dfrac{1}{2})*(\dfrac{1}{2})+(\dfrac{1}{2})*(\dfrac{1}{2})*(\dfrac{1}{2})*(\dfrac{1}{2})+…$$

= $$\dfrac{1}{3}$$

The members can be selected in $$^9C_3 \times ^6C_3$$ ways, but since all the teams are of 3 members each, this number should be divided by 3!.
Thus, the number of ways are- $$\dfrac{^9C_3 \times ^6C_3}{3!} =280$$

Total number of ways of arrangement of AMBIGUITY is 9!/2! (divided by 2! because there are 2 I). Total number of words where 2 I are together are 8! (considering 2 I to be one letter and calculating all the possible permutation). So total number of words where 2 I are not together are- 9!/2!-8!= $$\dfrac{7 \times 8!}{2}$$

The different cases arising when we draw two balls randomly are- { (R,R), (R,G), (R,O) , (G,G), (G,O), (O,O) }

Thus, the probability that both the balls drawn are not red = 5/6.

The four boys can be seated in 4! ways. There are 5 spaces adjacent to the boys where the girls can be seated, so they don't sit next to each other. So, the girls can be seated in 5! ways. Total ways = 4!*$$^5C_3$$*3!=1440

The number of ways 5 similar chocolates can be distributed among 3 children = $$^{5 + 3 - 1}C_{3 - 1} = ^7C_2$$

The number of terms of the expansion $$(a + b + c)^5$$ is $$^{5 + 3 - 1}C_{3 - 1} = ^7C_2$$

Hence, option E is the correct answer.

There should be at least 2 6s. The other 9 can be obtained in 6+3, 5+4 = 2 ways
So, the combinations are 6+6+6+3, which can be arranged in 4!/3! = 4 ways
6+6+5+4, which can be arranged in 4!/2! = 12 ways.
So, the total number of ways = 4 + 12 = 16 ways

The possible A.P types,

For d=1: (1, 2, 3), (2, 3, 4), (3, 4, 5), (4, 5, 6), (5, 6, 7)

For d=2: (1, 3, 5), (2, 4, 6), (3, 5, 7)

For d =3: (1, 4, 7)

The number of required A.P is 9.

The possibilities for getting a sum of 14 are:
6, 6, 2 -> 3 arrangements possible
6, 5, 3 -> 6 arrangements possible
6, 4, 4 -> 3 arrangements possible
5, 5, 4 -> 3 arrangements possible

Starting with 5, for the set 6, 5, 3, there are 2 arrangements possible and for the set 5, 5, 4, there are 2 arrangements possible.
So, required probability = (2+2)/15 = 4/15

Let the breadth and length of the rectangle be 3x and 4x, respectively.
Given,
4x = 3x + 5
x = 5

Hence the breadth is 15cm and the length is 20 cm.
Diagonal = $$\sqrt{15^2 + 20^2}$$ = 25 cm

Side of the square = 25
Perimeter = 100 cm

Let the width of the footpath be w.
So, area of the footpath = (30+2w)*(20+2w) - 20*30 = 216
From the options, w = 2 satisfies the equation. So, option a) is the answer.

Since the distance between any 2 poles is the same, they form the vertices of an equilateral triangle. The field is the circumcircle of the equilateral triangle.

Radius of circumcircle of an equilateral triangle with side ‘a’ = $$ \dfrac{a}{\sqrt{3}}$$

We know that $$ \dfrac{a}{\sqrt{3}}$$  = $$173.2$$

=> $$a = {173.2}*{1.732}$$ = $$300 m.$$

The shortest length of rope required will be the perimeter of the equilateral triangle = 3*300 = 900 m.

2 new rectangular areas are formed due to the cut.
Therefore, the change in area will be $$2*2r*h = 4rh = 4*7*5 = 140 cm^2$$
Area of the original cylinder = $$2 \pi rh + \pi*r^2$$
= $$2*\frac{22}{7} * 7 *5 + 2\frac{22}{7}*7^2$$
=$$220 + 308$$
=$$528 cm^2$$
%age change = $$\frac{140}{528}$$ = $$27$$% (approx). Therefore, option A is the right answer.

17+13+7+X = 50

X = 13

So the answer is option A.

Let P denote the set of children enrolled in painting, M in music, and D in dance.

We use the formula: $$n(P\cup M\cup D)=n(P)+n(M)+n(D)-n(P\cap M)-n(M\cap D)-n(P\cap D)+n(P\cap M\cap D)$$

Substituting the given values: $$n(P\cup M\cup D)=35+28+20-10-8-5+0=60$$

So, 60 children are enrolled in at least one field.

Those in exactly two fields = 10+8+5−3×0=23

Therefore, children in only one field = 60 - 23 = 37.

We are said that 20% of students like Pokemon and 20% of the students like none. So, the remaining 60% students are those who like only Naruto and who like both. We can also say that the remaining students like Naruto. We are given the value of the students who like Naruto i.e. 150

60% of students = 150.

From this, the number of students in the class is $$\frac{150}{0.6}$$ = 250.

46 people like all the three drinks. We make following venn diagram -

Since 62 people like both Tea and Hot chocolate
62 - 46 = 16 people like only Tea and Hot chocolate

Since 53 people like both Hot chocolate and Coffee
53 - 46 = 7 people like only Hot chocolate and Coffee

Since 89 people like both Tea and Coffee
89 - 46 = 43 people like only Tea and Coffee.

Thus, number of people who like only Tea = 120 - 46 - 16 - 43 = 15

Thus, number of people who like only Coffee= 160 - 43 - 46 - 7 = 64

Number of people who like only Hot chocolate = 350 - 160 - 16 - 15 = 159

We get following venn diagram -

Hence, option B is the correct choice.

46 people like all the three drinks. We make following venn diagram -

Since 62 people like both Tea and Hot chocolate

62 - 46 = 16 people like only Tea and Hot chocolate

Since 53 people like both Hot chocolate and Coffee

53 - 46 = 7 people like only Hot chocolate and Coffee

Since 89 people like both Tea and Coffee

89 - 46 = 43 people like only Tea and Coffee.

Thus, number of people who like only Tea = 120 - 46 - 16 - 43 = 15

Thus, number of people who like only Coffee= 160 - 43 - 46 - 7 = 64

Number of people who like only Hot chocolate = 350 - 160 - 16 - 15 = 159

We get following venn diagram -

Hence, option D is the correct option.

46 people like all the three drinks. We make following venn diagram -

Since 62 people like both Tea and Hot chocolate

62 - 46 = 16 people like only Tea and Hot chocolate

Since 53 people like both Hot chocolate and Coffee

53 - 46 = 7 people like only Hot chocolate and Coffee

Since 89 people like both Tea and Coffee

89 - 46 = 43 people like only Tea and Coffee.

Thus, number of people who like only Tea = 120 - 46 - 16 - 43 = 15

Thus, number of people who like only Coffee= 160 - 43 - 46 - 7 = 64

Number of people who like only Hot chocolate = 350 - 160 - 16 - 15 = 159

We get following venn diagram -

Hence, option C is the correct choice.

46 people like all the three drinks. We make following venn diagram -

Since 62 people like both Tea and Hot chocolate

62 - 46 = 16 people like only Tea and Hot chocolate

Since 53 people like both Hot chocolate and Coffee

53 - 46 = 7 people like only Hot chocolate and Coffee

Since 89 people like both Tea and Coffee

89 - 46 = 43 people like only Tea and Coffee.

Thus, number of people who like only Tea = 120 - 46 - 16 - 43 = 15

Thus, number of people who like only Coffee= 160 - 43 - 46 - 7 = 64

Number of people who like only Hot chocolate = 350 - 160 - 16 - 15 = 159

We get following venn diagram -

Hence, option A is the correct choice.

46 people like all the three drinks. We make following venn diagram -

Since 62 people like both Tea and Hot chocolate

62 - 46 = 16 people like only Tea and Hot chocolate

Since 53 people like both Hot chocolate and Coffee

53 - 46 = 7 people like only Hot chocolate and Coffee

Since 89 people like both Tea and Coffee

89 - 46 = 43 people like only Tea and Coffee.

Thus, number of people who like only Tea = 120 - 46 - 16 - 43 = 15

Thus, number of people who like only Coffee= 160 - 43 - 46 - 7 = 64

Number of people who like only Hot chocolate = 350 - 160 - 16 - 15 = 159

We get following venn diagram -

Hence, option E is the correct choice.

From Monday 6:00 AM to Thursday 6:00 AM is exactly 3 days, or 3*24 = 72 hours.

From Thursday 6:00 AM to Thursday 8:00 AM is an additional 2 hours.

So, in total, 12 hours have elapsed.

The clock gains 5 minutes for every 12 seconds.

So, the rate of gain is 5/12 minutes per hour.

The total time gained over 74 hours is:

74*(5/12) = 185/6 minutes

185/6 minutes = 30 minutes + 5/6 minute

5/6 minute = 50 seconds.

Hence, total time gained: 30 minutes and 50 seconds.

So, Clock time: Thursday, 8:30:50 AM

After 12 hours, it was showing 10 PM. This means that the clock is losing 2  hours every 12 hours.

How many hours will it be after 3 and a half days?

$$3\times\ 24\ +\ 12=84\ hrs$$

So when it loses 2 hours every 12 hours, it will lose x hours in 84 hours. We can use cross multiplication.

$$\dfrac{2}{12}=\dfrac{x}{84}\longrightarrow\ 14\ hrs$$

So the clock should be 14 hrs before the actual time.

12 AM - 14 hours  = 10 AM

Hence, the time it will be showing is 10:00 AM.

In 12 hours, the hand turns $$360^\circ$$.
Here, the difference between time = 2 hours
Then, Required angle $$= \dfrac{360}{12}\times2 = 60^\circ$$

The clock makes one revolution, covering nine hours. This means that it covers $$360^{\circ\ }$$ degrees or one revolution in nine hours.

Angle covered per hour = $$\frac{360^{\circ\ }}{9}=40^{\circ\ }$$

It means the hour hand will cover $$40^{\circ}$$ every hour.

Similarly, there are 30 minutes in an hour. It means the minute hand covers $$360^{\circ\ }$$ in 1 hour or 30 minutes.

Angle covered every minute = $$\frac{360}{30}=12^{\circ\ }$$

Now, we need to find the angle between 5:15.

To reach 5 hours, the hour hand would have covered an angle of $$5\times\ 40^{\circ\ }\ =\ 200^{\circ\ }$$

But here, the hour hand is also covering the 15 minutes or half an hour; it will move forward with the angle of half-hour more, i.e. $$\frac{40}{2}=20^{\circ\ }$$

Hence, the angle made by the hour  hand at 5:15 = $$200+20=220^{\circ\ }$$

Similarly, for the minute hand to reach 15 minutes, it would have covered $$15\times\ 12^{\circ\ }=180^{\circ\ }$$.

Hence, the minor angle between them $$40^{\circ\ }$$.

1 hour has 60 minutes. An hour hand covers 12 hours to complete one revolution of $$360^{\circ}$$.

Converting in minutes, an hour hand covers 12*60 = 720 minutes to cover $$360^{\circ}$$.

It means the hour hand covers $$\frac{360}{720}=0.5^{\circ\ }$$ per minute.

The hour hand-covered $$92.5^{\circ}$$ during the movie.

It means the movie's duration was $$\frac{92.5}{0.5}=185$$ minutes or 3 hours 5 minutes.

This is the movie duration; however, the movie had an interval of 15 minutes.

Hence, excluding the interval duration, the movie will be of 2 hour and 50 minutes.

From (3) => B is wearing Yellow.
From (2) => C and E are wearing either Blue or Black
From (4) and (1) => G cannot wear Green and G is not wear Red or White, therefore, G is wearing Pink
So, A has to wear Green.
Therefore, D or F are wearing Red or White

Since, B can wear Yellow or the color of T-Shirt worn by D or E => Blue/Black/Red/White
=> B can wear five colors
But among these, B cannot wear Blue or Black as it contradicts the first condition. Hence, B can wear only 3 colours.

=> Option A is correct

From (3) => B is wearing Yellow.
From (2) => C and E are wearing either Blue or Black
From (4) and (1) => G cannot wear Green and G is not wear Red or White, therefore, G is wearing Pink
So, A has to wear Green.
Therefore, D or F are wearing Red or White

A is wearing Green Colored T-Shirt
=> Option E is correct

From (3) => B is wearing Yellow.
From (2) => C and E are wearing either Blue or Black
From (4) and (1) => G cannot wear Green and G is not wear Red or White, therefore, G is wearing Pink
So, A has to wear Green.
Therefore, D or F are wearing Red or White

There are four different arrangements possible
=> Option D is correct

From (3) => B is wearing Yellow.
From (2) => C and E are wearing either Blue or Black
From (4) and (1) => G cannot wear Green and G is not wear Red or White, therefore, G is wearing Pink
So, A has to wear Green.
Therefore, D or F are wearing Red or White

T-Shirt color for A, B and G can be determined
=> Option A is correct

From (3) => B is wearing Yellow.
From (2) => C and E are wearing either Blue or Black
From (4) and (1) => G cannot wear Green and G is not wear Red or White, therefore, G is wearing Pink
So, A has to wear Green.
Therefore, D or F are wearing Red or White

If B and G interchange their T-Shirt , B will be wearing Pink colored T-shirt
=> Option C is correct

From 7, Q lived in Chennai and P lived in Kolkata. From 4, R had a degree in Mathematics. From 3, V had a degree in History and Q had a degree in Philosophy.
From 5, S lived in Hyderabad and had a degree in Physics.

From 2, the person who has a degree in English, lives in Delhi. Also, from 8, T did not live in Delhi. Hence, U lived in Delhi.

From 8, R did not live in Pune.
From 1, V did not live in Mumbai. From 6, T cannot live in Bangalore.
Hence the following 4 cases are possible:

Hence, option E is the correct choice.

From 7, Q lived in Chennai and P lived in Kolkata. From 4, R had a degree in Mathematics. From 3, V had a degree in History and Q had a degree in Philosophy.
From 5, S lived in Hyderabad and had a degree in Physics.

From 2, the person who has a degree in English, lives in Delhi. Also, from 8, T did not live in Delhi. Hence, U lived in Delhi.

From 8, R did not live in Pune.
From 1, V did not live in Mumbai. From 6, T cannot live in Bangalore.
Hence the following 4 cases are possible:

Hence, option D is the correct choice.

From 7, Q lived in Chennai and P lived in Kolkata. From 4, R had a degree in Mathematics. From 3, V had a degree in History and Q had a degree in Philosophy.
From 5, S lived in Hyderabad and had a degree in Physics.

From 2, the person who has a degree in English, lives in Delhi. Also, from 8, T did not live in Delhi. Hence, U lived in Delhi.

From 8, R did not live in Pune.
From 1, V did not live in Mumbai. From 6, T cannot live in Bangalore.
Hence the following 4 cases are possible:

Hence, option C is the correct choice.

From 7, Q lived in Chennai and P lived in Kolkata. From 4, R had a degree in Mathematics. From 3, V had a degree in History and Q had a degree in Philosophy.
From 5, S lived in Hyderabad and had a degree in Physics.

From 2, the person who has a degree in English, lives in Delhi. Also, from 8, T did not live in Delhi. Hence, U lived in Delhi.

From 8, R did not live in Pune.
From 1, V did not live in Mumbai. From 6, T cannot live in Bangalore.
Hence the following 4 cases are possible:

Hence, option E is the correct choice.

From 7, Q lived in Chennai and P lived in Kolkata. From 4, R had a degree in Mathematics. From 3, V had a degree in History and Q had a degree in Philosophy.
From 5, S lived in Hyderabad and had a degree in Physics.

From 2, the person who has a degree in English, lives in Delhi. Also, from 8, T did not live in Delhi. Hence, U lived in Delhi.

From 8, R did not live in Pune.
From 1, V did not live in Mumbai. From 6, T cannot live in Bangalore.
Hence the following 4 cases are possible:

Hence, option A is the correct choice.

(1) It is known that the number of cars which are facing north direction is 3 more than the number of cars which are facing south direction.
(2) Baleno is parked in the middle and it is equidistant from both Vento and Duster.
(3) The cars which are parked on extreme ends are facing opposite direction.
(4) Swift is parked second to left of Fortuner and both the cars are not facing the same direction.
(5) Duster is the only car parked between Wagon R and Polo. Both Duster and Polo are facing the same direction.
(6) Both the cars which are parked adjacent to Swift are facing the same direction as that of Swift.
(7) Fortuner is parked at one of the extremes.

We have a total of 7 cars and it is known that the number of cars which are facing north direction is 3 more than the number of cars which are facing south direction. Hence, we can say that 5 cars are facing in north direction whereas 2 cars are facing south direction.

Both the cars which are parked adjacent to Swift are facing the same direction as that of Swift. This is only possible when Swift and both the cars which are parked adjacent to swift are facing north direction as only 2 cars are facing south direction.

Swift is parked second to left of Fortuner and both the cars are not facing the same direction. Hence, we can say that Fortuner is facing south direction.

In the arrangement, we can see that Fortuner can’t occupy the 7th spot as Swift is parked in the left direction to Fortuner. Therefore, we can say that Fortuner is parked at the 1st spot. Consequently, we can say that Swift is parked on the 3rd spot.

From the statement (2) we can see that Baleno is parked at 4th spot. Also we can say that Vento and Duster occupy 2nd and 6th spot in any order.
But from statement (5), we can say that Vento occupies 2nd spot and Duster occupies 6th spot. Also Polo and Wagon R occupies 5th and 7th spot in any order.

From the statement (3), we can say that the car parked at 7th spot and faces north direction.

In statement (3) it is given that both Duster and Polo are facing the same direction. Hence, we can say that Wagon R occupies 5th spot and Polo occupies 7th spot. Also, both Duster and Polo are facing north direction.

From the arrangement, we can see that Wagon R faces south direction. Therefore, option C is the correct answer.

(1) It is known that the number of cars which are facing north direction is 3 more than the number of cars which are facing south direction.
(2) Baleno is parked in the middle and it is equidistant from both Vento and Duster.
(3) The cars which are parked on extreme ends are facing opposite direction.
(4) Swift is parked second to left of Fortuner and both the cars are not facing the same direction.
(5) Duster is the only car parked between Wagon R and Polo. Both Duster and Polo are facing the same direction.
(6) Both the cars which are parked adjacent to Swift are facing the same direction as that of Swift.
(7) Fortuner is parked at one of the extremes.

We have a total of 7 cars and it is known that the number of cars which are facing north direction is 3 more than the number of cars which are facing south direction. Hence, we can say that 5 cars are facing in north direction whereas 2 cars are facing south direction. Both the cars which are parked adjacent to Swift are facing the same direction as that of Swift. This is only possible when Swift and both the cars which are parked adjacent to swift are facing north direction as only 2 cars are facing south direction.

Swift is parked second to left of Fortuner and both the cars are not facing the same direction. Hence, we can say that Fortuner is facing south direction.

In the arrangement, we can see that Fortuner can’t occupy the 7th spot as Swift is parked in the left direction to Fortuner. Therefore, we can say that Fortuner is parked at the 1st spot. Consequently, we can say that Swift is parked on the 3rd spot.

From the statement (2) we can see that Baleno is parked at 4th spot. Also we can say that Vento and Duster occupy 2nd and 6th spot in any order.  But from statement (5), we can say that Vento occupies 2nd spot and Duster occupies 6th spot. Also Polo and Wagon R occupies 5th and 7th spot in any order.

From the statement (3), we can say that the car parked at 7th spot and faces north direction.

In statement (3) it is given that both Duster and Polo are facing the same direction. Hence, we can say that Wagon R occupies 5th spot and Polo occupies 7th spot. Also, both Duster and Polo are facing north direction.

From the arrangement, we can see that Polo is parked third to the right of Baleno. Therefore, option D is the correct answer.

(1) It is known that the number of cars which are facing north direction is 3 more than the number of cars which are facing south direction.
(2) Baleno is parked in the middle and it is equidistant from both Vento and Duster.
(3) The cars which are parked on extreme ends are facing opposite direction.
(4) Swift is parked second to left of Fortuner and both the cars are not facing the same direction.
(5) Duster is the only car parked between Wagon R and Polo. Both Duster and Polo are facing the same direction.
(6) Both the cars which are parked adjacent to Swift are facing the same direction as that of Swift.
(7) Fortuner is parked at one of the extremes.

We have a total of 7 cars and it is known that the number of cars which are facing north direction is 3 more than the number of cars which are facing south direction. Hence, we can say that 5 cars are facing in north direction whereas 2 cars are facing south direction. Both the cars which are parked adjacent to Swift are facing the same direction as that of Swift. This is only possible when Swift and both the cars which are parked adjacent to swift are facing north direction as only 2 cars are facing south direction.

Swift is parked second to left of Fortuner and both the cars are not facing the same direction. Hence, we can say that Fortuner is facing south direction.

In the arrangement, we can see that Fortuner can’t occupy the 7th spot as Swift is parked in the left direction to Fortuner. Therefore, we can say that Fortuner is parked at the 1st spot. Consequently, we can say that Swift is parked on the 3rd spot.

From the statement (2) we can see that Baleno is parked at 4th spot. Also we can say that Vento and Duster occupy 2nd and 6th spot in any order.  But from statement (5), we can say that Vento occupies 2nd spot and Duster occupies 6th spot. Also Polo and Wagon R occupies 5th and 7th spot in any order.

From the statement (3), we can say that the car parked at 7th spot and faces north direction.

In statement (3) it is given that both Duster and Polo are facing the same direction. Hence, we can say that Wagon R occupies 5th spot and Polo occupies 7th spot. Also, both Duster and Polo are facing north direction.

From the arrangement, we can see that only 1 car is parked between Baleno and Vento. Therefore, option A is the correct answer.

(1) It is known that the number of cars which are facing north direction is 3 more than the number of cars which are facing south direction.
(2) Baleno is parked in the middle and it is equidistant from both Vento and Duster.
(3) The cars which are parked on extreme ends are facing opposite direction.
(4) Swift is parked second to left of Fortuner and both the cars are not facing the same direction.
(5) Duster is the only car parked between Wagon R and Polo. Both Duster and Polo are facing the same direction.
(6) Both the cars which are parked adjacent to Swift are facing the same direction as that of Swift.
(7) Fortuner is parked at one of the extremes.

We have a total of 7 cars and it is known that the number of cars which are facing north direction is 3 more than the number of cars which are facing south direction. Hence, we can say that 5 cars are facing in north direction whereas 2 cars are facing south direction. Both the cars which are parked adjacent to Swift are facing the same direction as that of Swift. This is only possible when Swift and both the cars which are parked adjacent to swift are facing north direction as only 2 cars are facing south direction.

Swift is parked second to left of Fortuner and both the cars are not facing the same direction. Hence, we can say that Fortuner is facing south direction.

In the arrangement, we can see that Fortuner can’t occupy the 7th spot as Swift is parked in the left direction to Fortuner. Therefore, we can say that Fortuner is parked at the 1st spot. Consequently, we can say that Swift is parked on the 3rd spot.

From the statement (2) we can see that Baleno is parked at 4th spot. Also we can say that Vento and Duster occupy 2nd and 6th spot in any order.  But from statement (5), we can say that Vento occupies 2nd spot and Duster occupies 6th spot. Also Polo and Wagon R occupies 5th and 7th spot in any order.

From the statement (3), we can say that the car parked at 7th spot and faces north direction.

In statement (3) it is given that both Duster and Polo are facing the same direction. Hence, we can say that Wagon R occupies 5th spot and Polo occupies 7th spot. Also, both Duster and Polo are facing north direction.

From the arrangement, we can see that Wagon R is parked to the immediate right of Baleno. Therefore, option E is the correct answer.

(1) It is known that the number of cars which are facing north direction is 3 more than the number of cars which are facing south direction.
(2) Baleno is parked in the middle and it is equidistant from both Vento and Duster.
(3) The cars which are parked on extreme ends are facing opposite direction.
(4) Swift is parked second to left of Fortuner and both the cars are not facing the same direction.
(5) Duster is the only car parked between Wagon R and Polo. Both Duster and Polo are facing the same direction.
(6) Both the cars which are parked adjacent to Swift are facing the same direction as that of Swift.
(7) Fortuner is parked at one of the extremes.

We have a total of 7 cars and it is known that the number of cars which are facing north direction is 3 more than the number of cars which are facing south direction. Hence, we can say that 5 cars are facing in north direction whereas 2 cars are facing south direction. Both the cars which are parked adjacent to Swift are facing the same direction as that of Swift. This is only possible when Swift and both the cars which are parked adjacent to swift are facing north direction as only 2 cars are facing south direction.

Swift is parked second to left of Fortuner and both the cars are not facing the same direction. Hence, we can say that Fortuner is facing south direction.

In the arrangement, we can see that Fortuner can’t occupy the 7th spot as Swift is parked in the left direction to Fortuner. Therefore, we can say that Fortuner is parked at the 1st spot. Consequently, we can say that Swift is parked on the 3rd spot.

From the statement (2) we can see that Baleno is parked at 4th spot. Also we can say that Vento and Duster occupy 2nd and 6th spot in any order.  But from statement (5), we can say that Vento occupies 2nd spot and Duster occupies 6th spot. Also Polo and Wagon R occupies 5th and 7th spot in any order.

From the statement (3), we can say that the car parked at 7th spot and faces north direction.

In statement (3) it is given that both Duster and Polo are facing the same direction. Hence, we can say that Wagon R occupies 5th spot and Polo occupies 7th spot. Also, both Duster and Polo are facing north direction.

From the arrangement, we can see that only 3 cars are parked between Swift and Polo. Therefore, option C is the correct answer.

From condition 3, the single-digit prime number that can be expressed as the sum of 3 distinct numbers is 7. This can be written as 7=1+3+4. Therefore, the numbers picked in round 1 are 1, 2, and 4.

From condition 4, the only 3 prime numbers less than 10 that are in arithmetic progression are 3, 5, and 7.

Given that A picked the smallest and C picked the largest, A picked 3 in round 2, and C picked 7. Consequently, B picked 5 in round 2.

From condition 2, A must pick a perfect square in rounds 1 and 3 since he didn't pick a perfect square in round 2. Therefore, A can either pick 1 or 4 in round 1, and in round 3, A can only pick 9.

Here we are given the two possible scenarios given the clues, the numbers remaining in either situation is 6,8. Now given that B didn't win any rounds, the maximum possible sum of numbers that B can get is in the first scenario, 2+5+8=15

From condition 3, the single-digit prime number that can be expressed as the sum of 3 distinct numbers is 7. This can be written as 7=1+3+4. Therefore, the numbers picked in round 1 are 1, 2, and 4.

From condition 4, the only 3 prime numbers less than 10 that are in arithmetic progression are 3, 5, and 7.

Given that A picked the smallest and C picked the largest, A picked 3 in round 2, and C picked 7. Consequently, B picked 5 in round 2.

From condition 2, A must pick a perfect square in rounds 1 and 3 since he didn't pick a perfect square in round 2. Therefore, A can either pick 1 or 4 in round 1, and in round 3, A can only pick 9.

The only situation where A could have the highest possible score is in scenario 2, and that also only when C picks number 6, bringing C's total points to 15 and A's total to 16.

No actor is a producer. Some directors are producers.
The following diagram represents a possibility.

Directors, who are producers, cannot be actors. Therefore, some directors are not actors. Hence, conclusion I is definitely true.
Directors, who are not producers, can be actors. Therefore, we cannot say conclusion II to be definitely true. Therefore, only conclusion I follows. Hence, option A is the right answer.

Consider the following possibility.

Both the conclusions are just possibilities. None of the conclusions can be definitely said to be true.
Hence, option D is the correct option.

Consider the following case:

From this we can observe that all cups being spoons is a possibility.

Consider the following case:

From this we can observe that some spoons are saucers need not be true.
So, the correct answer is option a).

The arrangement as mentioned in the statements will be as follows.

Hence, we can't say for sure that either Conclusion 1 or Conclusion 2 will follow.

1) All cities are states and no state is a street. Hence, no city will be street. The Conclusion I does not follow
2) All districts are countries does not follow.
3) Some streets being countries is a possibility. Conclusion III follows.
4) All cities are states and all states are countries. Hence, some countries will be cities is true in all possibilities.

Hence, option C is the correct answer.

The numbers are following a particular series, where
2x2+1=5
5x3+1=16
16x4+1=65
65X5+1=326, so the next number in the series will be
326x6+1=1957
Letters are following a series where the number of letters skipped is changing,
There is one letter between A and C, There is again one letter between C and E, There are two letters between E and H, and there are three letters between H and L, this is clearly following a fibonacci series of skipping alphabets, so the next letter will be R.
Answer will be 1957R

We see that 987 and 119 are related as

(9+8)*7 = 17*7 = 119

Similarly, 676

(6+7)*6 = 13*6 = 78

Option C is the correct answer.

625 : 49

We see the sqauare root of 625 is 25.
Now, 25 is related to 49 in sense i.e. $$\left(2+5\right)^2$$ = 49

Similarly, 729 is the squae of 27.
Therefore, the denominator should be $$\left(2+7\right)^2$$ = 81

The series follows the following pattern:
432 : 432 - ($$4^2 + 3^2 + 2^2$$)

Hence, the answer will be 718 - ( $$7^2 + 1^2 + 8^2$$)

$$18 + 1^2 + 8^2 = 83$$
$$46 + 4^2 + 6^2 = 98$$
Hence, option B is the correct option.

49 + (4*9) = 85
82 + (8*2) = 98
Hence, option B is the correct option.

The given two equations can be considered as straight lines of the form ax + by + c = 0.

Two equations have a solution if they meet at a point, no solution if they are parallel but not coinciding, and infinite solutions if both equations are the same.

The slope of a straight line = -a/b

=> For the two given equations, comparing slopes, we get

5x + 4y - 13 = 0 => Slope = -5/4

8x + ky - 14 = 0 => Slope = -8/k

Equating slope, we get => -5/4 = -8/k => k = (8 * 4)/5 = 32/5 = 6.4

Also, we can see that 5x + 4y - 13 = 0 and 8x + 6.4y - 14 = 0 are not the same equation, as the ratio of intercepts 13/14 is not in the same ratio of coefficients of x and y => They will have no solution for k = 6.4

2a + 5b = 108
108 - 5b = 2a
We have to get an even number by subtracting a multiple of 5 from 108. So b must be an even number.
There are 21 multiple of 5 below 108. Out of those 11 are odd, and 10 are even.
For only 10 even multiples will get subtraction as an even number.
So b = 2, 4, 6, …, 20
Hence, option C is the right answer.

Let the price of one pant be ‘p’, one shirt be ‘s’ and one towel be ‘t’.
2p + 2s + t = 300 ------(1)
It is given that for Rs. 60 more, he could have gotten thrice the number of towels I.e., 3 towels.
=> 2 towels cost Rs. 60 (1 towel out of the 3 towels is already included in the equation).
=> t = 30.
4p + 2s + 2t = 500 -------(2)
Subtracting (1) from (2), we get,
2p + t = 200
We know that t = 30
=> 2p = 170
p = 85
Substituting p and t, we get price of shirt, s as Rs. 50.

Based on the given information the following equations could be derived
2(pens) + 3(erasers) + 5(pencils) =32 ---- equation (1)
5(pens) + 7(erasers) + 4(pencils) = 46 ----- equation (2)
4* equation (2) - 3* equation (1) ==> 14(pens) + 19(erasers) +1*(pencil)=88
==>14 pens, 19 erasers, one pencil cost 88 rupees.
So the correct option to choose is B - Rs 88.

For a quadratic equation to have real roots, its discriminant must be greater than or equal to 0.
Let’s consider both the equations.
Equation 1) Discriminant = $$9^2-4*3*5=81-60=21$$
Thus equation 1 has real roots.
Equation 2) Discriminant = $$6^2-4*1*9=36-36=0$$
Thus equation 2 also has real roots.

$$x^2-\left(a-2\right)x-a-1=0$$
Sum of the squares of the roots can be written as $$\left(sum\ of\ roots\right)^2-2\left(product\ of\ roots\right)$$
$$\left(a-2\right)^2+2\left(a+1\right)$$
$$a^2+4-4a\ +2a+2$$
Minimum value of this quadratic equation occurs at -b/2a
$$-\frac{\left(-2\right)}{2}=1$$
Hence, for a=1, the sum of the squares of the roots will be minimum

Let the quadratic equation whose roots are squares of the roots of the quadratic equation $$2x^2+\left(k-3\right)x+6k=0$$ be $$px^2+qx+r=0$$.

We know for a quadratic equation $$ax^2+bx+c=0$$ the sum of roots of the quadratic equation is $$-\dfrac{b}{a}$$ and the product of roots of the quadratic equation is $$\dfrac{c}{a}$$.

For $$2x^2+\left(k-3\right)x+6=0$$, let the roots be $$\alpha,\ \beta\ $$

So, $$\alpha+\beta=-\ \dfrac{\left(k-3\right)}{2}$$

and $$\alpha\beta=\dfrac{6k}{2}$$

We want the roots of $$px^2+qx+r=0$$ to be $$\alpha^2$$ and $$\beta^2$$

So the product of roots = $$\alpha^2\beta^2=\dfrac{r}{p}$$

putting the values of $$\alpha\beta$$ in the equation we get

$$\dfrac{r}{p}=\left(\alpha\beta\right)^2=\left(\dfrac{6k}{2}\right)^2=\dfrac{36k^2}{4}$$

Similarly sum of roots = $$\alpha^2+\beta^2=\left(\alpha+\beta\right)^2-2\alpha\beta=-\frac{q}{p}$$

$$-\dfrac{q}{p}=\dfrac{\left(k-3\right)^2}{2^2}-2\times\dfrac{6k}{2}$$

$$-\dfrac{q}{p}=\dfrac{k^2-6k+9-24k}{4}$$

$$\dfrac{q}{p}=\dfrac{-\left(k^2-30k+9\right)}{4}$$

We get

$$p=4$$, $$q=-\left(k^2-30k+9\right)$$ and $$r=36k^2$$

So, the quadratic equation is $$4x^2-\left(k^2-30k+9\right)x+36k^2=0$$

$$x^2 - x - 20$$ = 0
=>$$(x + 4)(x - 5)$$ = 0
=> $$x$$ = -4 or $$x$$ = 5
Similarly, $$x^2 + 7x + 12$$ = 0
=> $$(x + 4)(x + 3)$$ = 0
=> $$x$$ = -4 or $$x$$ = -3
Since, $$x$$ = -4 is the value which satisfies both equations, option B is correct

Let the roots of the quadratic equation be a and b.
Then we know that a + b = 5
a*b = -14 => a*(5 - a) = -14
=> $$5a - a^2 = -14$$
=> $$a^2 - 5a - 14 = 0$$
=> (a - 7)(a + 2) = 0
=> a = 7 or a = -2
Hence the two roots are 7 and -2. Hence the difference between the two roots would be 7 - (-2) = 9.

$$S = \frac{n}{2}(a+l)$$ = $$\frac{7}{2}(11+43) = \frac{7}{2}(54)$$ = $$189$$

So the answer is option A.

$$T_{2} = 6 \Rightarrow a+d = 6$$-------(1)

$$T_{4} = 12 \Rightarrow a+3d = 12$$-------(2)

On solving (1) and (2)

a = 3, d = 3

$$S_{n} = \frac{n}{2} \times [2a+(n-1)d]$$

$$S_{9} = \frac{9}{2} \times [2(3)+(8)(3)]$$

$$S_{9} = \frac{9}{2} \times [30]$$

$$S_{9} =135$$

So the answer is option C.

Let the first term of the GP be $$a$$ and the common ratio be $$r$$

So the GP is of form $$a,\ ar,\ ar^2,\ ar^3,\ ....$$

Given, $$n^{th}$$ term of the Geometric Progression equals one-third of the sum of all terms after it, i.e.

$$a=\dfrac{1}{3}\left(ar+ar^2+ar^3+....\right)$$ ....(1)

The sum of infinite GP with first term $$ar$$ and common ratio $$r$$ is $$\dfrac{ar}{1-r}$$

putting in equation (1), we get

$$a=\dfrac{1}{3}\times\dfrac{ar}{1-r}$$

$$3-3r=r$$

$$r=\dfrac{3}{4}$$

Given the sum of the first four terms is 350, i.e.

$$a+ar+ar^2+ar^3=350$$

$$a\left(1+r+r^2+r^3\right)=350$$

putting value of $$r$$ we get

$$a\left(1+\dfrac{3}{4}+\left(\dfrac{3}{4}\right)^2+\left(\dfrac{3}{4}\right)^3\right)=350$$

$$a\left(1+\dfrac{3}{4}+\dfrac{9}{16}+\dfrac{27}{64}\right)=350$$

$$a\left(\dfrac{175}{64}\right)=350$$

$$a=128$$

So the sum of all terms of the GP is

$$\dfrac{a}{1-r}=\dfrac{128}{1-\dfrac{3}{4}}=512$$

Hence the answer is 512.

The series, 2,4,7... is a series with increasing difference. Such series can be represented in the form of a quadratic equation, $$ax^2+bx+c$$
$$2=a+b+c$$
$$4=4a+2b+c$$
$$7=9a+3b+c$$
Solving the above equations we get the expression for the nth term as $$T_n=\frac{n^2}{2}+\frac{n}{2}+1$$
We need to find the sum of the first 10 terms,
$$S_n=\Sigma\ \left(\frac{n^2}{2}+\frac{n}{2}+1\right)$$
$$\frac{\left(n\right)\left(n+1\right)\left(2n+1\right)}{12}+\frac{\left(n\right)\left(n+1\right)}{4}+n$$
n=10,
$$\frac{7\left(55\right)}{2}+\frac{55}{2}+10$$
$$230$$

We need to put $$g(x)$$ in $$f(x)$$.

$$f(g(x))$$ = $$\dfrac{\left(\dfrac{x}{x+1}+1\right)}{\dfrac{x}{x+1}}\ =\ \dfrac{\left(\dfrac{x+x+1}{x+1}\right)}{\dfrac{x}{x+1}}=\dfrac{\dfrac{2x+1}{x+1}}{\dfrac{x}{x+1}}$$

We can cancel the denominator. We get $$f(g(x))$$ = $$\dfrac{\left(2x+1\right)}{x}$$

We are given that $$f(g(x))=x$$

Therefore, $$\dfrac{\left(2x+1\right)}{x}=x$$

$$2x+1=x^2$$

$$x^2-2x-1=0$$

Upon solving, we get no integral value of x that satisfies the equation. Hence, the answer is zero.

The expression can be written as:

$$\dfrac{1}{1+a^{y-x}+a^{z-x}} + \dfrac{1}{1+a^{x-y}+a^{z-y}} + \dfrac{1}{1+a^{x-z}+a^{y-z}}$$

= $$\dfrac{a^x}{a^x + a^y + a^z} + \dfrac{a^y}{a^x + a^y + a^z} + \dfrac{a^z}{a^x + a^y + a^z}$$

= $$\dfrac{a^x + a^y + a^z}{a^x + a^y + a^z}$$ = 1

Let $$a^{3}$$ = $$b^{4}$$ = $$c^{5}$$ = $$d^{6}$$ = k.

=> $$a = k^\frac{1}{3}, b=k^\frac{1}{4}, c= k^\frac{1}{5}, d=k^\frac{1}{6}$$

Then, $$\log{abc}$$ = $$(\frac{1}{3} + \frac{1}{4} +\frac{1}{5} )\log{k}$$

= $$\frac{47}{60}\log{k}$$

$$\log{d}$$ = $$\frac {\log{k}}{6}$$

The ratio $$\frac{\log{abc}}{\log {d}}$$ = 47/10 = 4.7

$$\log_{42}147$$ can be written as $$\dfrac{\log_2147}{\log_242}$$

$$\dfrac{\log_2147}{\log_242}$$=$$\dfrac{\log_23\times\ 49}{\log_22\times\ 3\times\ 7}$$

=$$\dfrac{\log_23\ +\log_249\ }{1+\log_23+\log_27}$$

=$$\dfrac{\log_23\ +2\log_27\ }{1+\log_23+\log_27}$$

=$$\dfrac{a\ +2b\ }{1+a+b}$$

The expression can be written as follows:

$$2 \log_{10} t - \log_t 0.01$$ = $$2 \log_{10} t - \log_t 10^{-2}$$
= $$2 \log_{10} t + 2 * \log_t 10$$
= $$2 * (\log_{10} t + 1 / \log_{10} t)$$

Minimum value of $$(\log_{10} t + 1 / \log_{10} t)$$ = 2
So, the minimum value of the expression is 2*2 = 4

Let’s assume the number of bears in the zoo two years ago is ‘y’.
y of (100+35)% of (100-16.67)% = 405
$$y\times1.35\times\frac{5}{6} = 405$$
6.75y = 2430
y = 360
20% of the number of bears in the zoo two years ago = 20% of 360
= 72
Hence, option e is the correct answer.

In 2014, the population was 1,00,000.
It has been given that the population increases by 10% every year. Therefore, by the beginning of 2015, the population will be 1.1*1,00,000 = 1,10,000
Population by the beginning of 2016 = 1.1* 1,10,000 = 1,21,000.
The epidemic wiped off 20% of the population.
Therefore, population by the end of 2016 = 0.8*1,21,000 = 96,800.
By the beginning of 2017, the population would have again increased by 10%.
Therefore, population by the end of 2017 = 1.1*96,800 = 1,06,480.
Option D is the right answer.

560 of (100+14.28)% of (100+y)% of (100-(y-5))% = 624
560 of 114.28% of (100+y)% of (100-y+5)% = 624
560 of 114.28% of (100+y)% of (105-y)% = 624
$$560\times\frac{8}{7}\times\frac{(100+y)}{100}\times\frac{(105-y)}{100} = 624$$
$$640\times\frac{(100+y)}{100}\times\frac{(105-y)}{100} = 624$$
$$\frac{(100+y)}{100}\times\frac{(105-y)}{100} = 0.975$$
(100+y) (105-y) = 9750
$$y^{2}-5y-750=0$$
$$y^{2}-(30-25)y-750=0$$
$$y^{2}-30y+25y-750=0$$
y(y-30)+25(y-30) = 0
(y-30) (y+25) = 0
y = 30, -25
value of ‘y’ = 30 (Negative value of y is not possible.)
Hence, option d is the correct answer.

Let the amount obtained by P = 72p
The amount obtained by Q is 11.11% less than the amount obtained by P.
Amount obtained by Q = $$\frac{8}{9}$$ of the amount obtained by P
= $$\frac{8}{9}\times$$72p
= 64p
The amount obtained by P is 14.28% more than the amount obtained by R.
Amount obtained by P = $$\frac{8}{7}$$ of the amount obtained by R
72p = $$\frac{8}{7}\times$$Amount obtained by R
Amount obtained by R = 63p
Total amount divided among them is ₹1194
72p + 64p + 63p 1194
199p = 1194
p = 6
Amount obtained by R = 63p
= 63(6)
= ₹378
Hence, the correct answer is Option D

total minutes in a day = (24 × 60) minutes

So 45 minutes will make = $$\frac{45}{24 \times 60}$$

= $$\frac{1}{32}$$ part of a day

Share of $$A = \dfrac{4}{11} \times 73689 = 26796$$

Thrice share of $$A = 80388$$

Share of $$B = \dfrac{7}{11} \times 73689 = 46893$$

Twice the share of $$B = 93786$$

Difference = $$13398$$

The actual area is 8000 $$km^2$$
1 km = 1 lakh cms, so 1 $$km^2$$ = 1 $$lakh^2 cm^2$$
Since we are using the areas and the scale of a map is expressed in linear quantities, the required scale = $$\sqrt{20:8,000 * 1 lakh^2}$$
Also, we are scaling down the actual area to a small map. So, the ratio will have a smaller number on the left and a larger number on the right side.
So, the required scale is 1: $$\sqrt{400 * lakh^2}$$ = 1: 20 lakh

Based on the given information, the following equations could be derived.
A:B = 4:7
1.5A:0.75B = 48:42 = 8:7

So, the correct answer is option (B).

By taking the amount of water in each sample we have $$\frac{2}{5},\frac{5}{8},\frac{4}{9}$$
and alcohol as $$\frac{3}{5},\frac{3}{8},\frac{5}{9}$$
Total water content in new sample is $$\left(\left(\frac{2}{5}\right)+\left(\frac{5}{8}\right)+\left(\frac{4}{9}\right)\right)$$ = $$\frac{551}{360}$$
Total alcohol content in new sample is $$\left(\left(\frac{3}{5}\right)+\left(\frac{3}{8}\right)+\left(\frac{5}{9}\right)\right)=\frac{529}{360}$$
Ratio = 529 : 551

Let the quantity of milk and water in the container be 2x litres and 3x litres respectively.

Let the quantity taken out from the container be ‘5a’ litres.
Out of ‘5a’ litres, 2a litres will be milk and 3a litres will be water.

$$\dfrac{2x-2a}{3x-3a+5a} = \dfrac{2x-2a}{3x+2a}$$

Total quantity remaining = 2x-2a+3x+2a = 5x litres.

$$5x = 5a\times6$$

$$x = 6a$$

Quantity of water removed = 3a litres.

Quantity of milk initially = 2x = 12a litres.

Required percentage = $$\dfrac{3a}{12a}\times100 = 25$$%.

In a mixture, the ratio of milk and water is 5:3 respectively.
Let’s assume the initial quantity of milk and water is 40y and 24y respectively.
If 12.5% of mixture is taken out and 14 litres of milk is poured into the remaining mixture. Then the new ratio between milk and water is 9:5 respectively.

$$\dfrac{35y+14}{21y} = \dfrac{9}{5}$$

175y+70 = 189y

189y-175y = 70

14y = 70

y = 5
The initial quantity of mixture = (40y+24y) = 64y = $$64\times5$$ = 320 litres
Hence, option a is the correct answer.

Let’s assume the initial quantity of oil in the tank is 100y litres.
In an oil tank, 14% of oil is taken out and replaced the same quantity with water. Again 14% of mixture is taken out and replaced the same quantity with water.
Remaining quantity of mixture = $$100y \times \left(1-\dfrac{14}{100}\right)^{2}$$

= $$100y \times \left(\dfrac{86}{100}\right)^{2}$$

= $$100y \times 0.7396$$

= 73.96y

If the total 651 litres of oil is taken out.
100y-73.96y = 651
26.04y = 651
y = 25
Initial quantity of oil in the tank = 100y = $$100\times25$$ = 2500 litres
Hence, option b is the correct answer.

if the difference between the initial quantity of milk and water is 420 litres.
In a mixture, the ratio between the quantity of milk and water is 11:7 respectively.

Let's assume the initial quantity of milk and water is 11z and 7z respectively.

11z-7z = 420

4z = 420

z = 105

If ‘y’ and (y+20) litres of milk and water is poured into the mixture, then the quantity of milk will be 50% more than the quantity of water.

$$\dfrac{105\times11 + y}{105\times7 + (y+20)} = \dfrac{3}{2}$$

$$\dfrac{1155 + y}{735 + (y+20)} = \dfrac{3}{2}$$

$$2310+2y = 2205+3y+60$$

$$3y-2y = 2310-2265$$

$$= 45$$

Hence, option e is the correct answer.

In a class, the number of girls is 14 less than the number of boys.
Let’s assume the number of girls in the class is y.
Number of boys = (14+y)
The average weight of the class is 68 kg.
The total weight of the whole class = $$68\times (14+y+y)$$
= 68(14+2y) Eq.(i)
If the average weight of girls is 6.8 less than the average weight of the whole class.
The average weight of girls in the class = 68-6.8 = 61.2
The total weight of girls in the class = 61.2y Eq.(ii)
The average weight of boys is 10.8 more than the average weight of girls in the class.
The average weight of boys = 61.2+10.8 = 72
The total weight of boys in the class = 72(14+y) Eq.(iii)
Eq.(i) = Eq.(ii) + Eq.(iii)
68(14+2y) = 61.2y + 72(14+y)
952+136y = 61.2y + 1008 + 72y
952+136y = 133.2y + 1008
136y-133.2y = 1008-952 = 56
2.8y = 56
y = 20
Total number of students in the class = (14+y)+y = (14+20)+20 = 54
Hence, option e is the correct answer.

Let the initial number of men in the group be ‘n’.
Let the initial total weight of the group be ‘A’ kg.
Given, $$\dfrac{A}{n} = x$$
$$A = nx$$ → (1)

Let the weight of each person who left the group be ‘m’ kg.
$$\dfrac{A-4m}{n-4} = x-1.5$$

$$A-4m = (n-4)(x-1.5)$$
(1) ⇒ $$nx-4m = nx-1.5n-4x+6$$
$$-1.5n-4x+4m+6=0$$
$$-1.5n+4(m-x)+6=0$$
The initial average weight of the group of men is 9 kg less than the weight of each person who left the group.
⇒ m - x = 9
$$-1.5n+4\times9+6=0$$
$$-1.5n+36+6 = 0$$
$$1.5n = 42$$
$$n = 28$$
Hence, The initial number of men in the group = 28.

Let the six numbers are a, b, c, d, e and f respectively.

The average of the first four numbers is equal to 35.
$$\Rightarrow$$ $$\frac{a+b+c+d}{4}$$ = 35
$$\Rightarrow$$ a + b + c + d = 140…………(1)

The average of the last four numbers is 40.
$$\Rightarrow$$ $$\frac{c+d+e+f}{4}$$ = 40
$$\Rightarrow$$ c + d + e + f = 160…………(2)

The average of the first and sixth number is 45.
$$\Rightarrow$$ $$\frac{a+f}{2}$$ = 45
$$\Rightarrow$$ a + f = 90…………(3)

The average of the second and fifth number is 55.
$$\Rightarrow$$ $$\frac{b+e}{2}$$ = 55
$$\Rightarrow$$ b + e = 110…………(4)

Adding (1) and (2),
a + b + c + d + c + d + e + f = 140 + 160
(a + f) + (b + e) + 2(c + d) = 300
90 + 110 + 2(c + d) = 300
2(c + d) = 100
c + d = 50…………..(5)

Average of the first, third, fourth and sixth numbers = $$\frac{a+c+d+f}{4}$$
= $$\frac{a+f+c+d}{4}$$
= $$\frac{90+50}{4}$$
= $$\frac{140}{4}$$
= 35
Hence, the correct answer is Option A

Let the total number of students = T

Without error

Average of 96 students = 84

Sum of 96 students = 84 x 96

Let average of remaining T-96 students = p

Sum of the T-96 students = p(T-96)

Total marks = (84 x 96) + p(T-96)

Let the average of total students = q

$$\Rightarrow$$  (84 x 96) + p(T-96) = Tq.........(1)

With error

Average of 96 students = 72

Sum of 96 students = 72 x 96

Average of T-96 students = (p+3.5)

Sum of T-96 students = (p+3.5)(T-96)

Total marks = (72 x 96)+(p+3.5)(T-96)

Average of total students = (q-4.5)

$$\Rightarrow$$  (72 x 96)+(p+3.5)(T-96) = T(q-4.5)

$$\Rightarrow$$  (72 x 96)+(p+3.5)(T-96) = Tq - 4.5T

$$\Rightarrow$$  (72 x 96)+(p+3.5)(T-96) = (84 x 96) + p(T-96) - 4.5T [From (1)]

$$\Rightarrow$$  (p+3.5)(T-96) - p(T-96) + 4.5T = (84 x 96) - (72 x 96)

$$\Rightarrow$$  (T-96)(p+3.5-p) + 4.5T = 96(84 - 72)

$$\Rightarrow$$  (T-96)3.5 + 4.5T = 96 x 12

$$\Rightarrow$$  3.5T - 96 x 3.5 + 4.5T = 96 x 12

$$\Rightarrow$$  8T = 96 x 15.5

$$\Rightarrow$$  T = 186

Total number of students = 186

Hence, the correct answer is Option C