SNAP 2022 Question Paper

Let x be the actual cost of the table.
Selling price of the table when sold at 20% profit = 1.2x - (i)
Selling price of the table when sold at 20% loss = .8x -(ii)
Given, the difference between the two is Rs.500. Thus, 1.2x - 8x = 500
=>x = 500/.4x = Rs.1250

5 men working together can finish the work in 10 days. Let each man does x units of work in 1 day. So total work to be done = 50x
Now 10 women finish the work in 7.5 days. Hence, if y is the amount of work done by 1 women in 1 day then total work = 10*7.5 = 75y
50x = 75y
=> x/y = 3/2
So, let y = 2x/3
When 3 men and 3 women work, then the amount of work done by them in 1 day = 3x + 2x = 5x
Hence it will take 10 days to complete 50 units of work.

Let the distance between Ram’s office and home = 150d
Then his original speed = $$\frac{150d}{30}$$ = 5d

Distance that Ram covered at (1/3)rd of original distance = $$\frac{150d}{3}$$ = 50d
(4/5)th of original speed = $$\frac{4}{5} * 5d$$ = 4d
Time taken by Ram to cover this distance = $$\frac{50d}{4d} = 12.5$$minutes

Remaining distance = 150d - 50d = 100d
(2/5)th of original speed = $$\frac{2}{5}*5d$$ = 2d
Time taken by Ram to cover the remaining distance = $$\frac{100d}{2d}$$ = 50 minutes

Hence total time taken by Ram = 12.5 + 50 = 62.5 minutes

Let the two digit number be 10x and y.
ATQ,
(10y+x) = (100+75)% of (10x+y)
10y+x = 7/4 (10x+y)
10y - 7y/4 = 35x/2 - x
33y/4 = 33x/2
y = 2x ---- (a)
And from the question it says that the difference between the digits of the number is 3 so
y-x = 3 ---(b)
So from eq (a) and (b)
We get, y= 6 and x = 3
So the original number is 36.

We know that the number leaves the same remainder when divided by 2,3,4 and 5.
Therefore, the number should be of the form LCM(2,3,4,5) + k.
LCM of 2,3,4,5 is 60.
The largest multiple of 60 below 10,000 is 9960.
2 can leave a remainder of 0 or 1.
Therefore, the largest possible value of k is 1.
The largest 4 digit number that leaves the same remainder when divided by 2, 3, 4 and 5 is 9960 + 1 = 9961.

Therefore, option D is the right answer.

When IMP are always together, consider them as one group.
Number of ways of arranging IMP and other letters = 7!/2! = 7!/2!
IMP can be arranged in 3! ways.
Total number = 3!*7!/2! = 3*7!

Probability of drawing a green marble from the box is $$\frac{1}{5}$$
We know that the number of yellow marbles are twice as many as green marbles. Hence the probability of yellow marbles would be $$\frac{2}{5}$$
Since there are only 3 types of marbles in the box, so the probability of red marbles would be $$ 1 - \frac{1}{5} - \frac{2}{5} = \frac{2}{5}$$
Let the total number of marbles be k, then we have
$$\frac{10}{k} = \frac{2}{5}$$
=> k = 25

Natural numbers less than 201 and divisible by 5 are 5, 10, . . .200. Here, 5+200 = 10+195  . . = 100+105. Hence, there are 40/2 = 20 pairs totalling 205.

Sum of all natural numbers less than 201 divisible by 5 is 20*205 = 4100

Similarly, natural numbers less than 201 and divisible by 6 are 6, 12, . . .  198. 6+198= 12+192= . . . = 96+108  and 102. Hence, there are $$\lfloor201/6 /2 \rfloor$$ = 16 such pairs and 102.

Sum of all natural numbers less than 201, divisible by 6 is 16*204 + 102 = 3366

Similarly, Sum of all natural numbers divisible by both 5 and 6 is 6*210/2 = 630. These numbers have been summed in both the above totals. Hence, we need to deduct 2 times 630.

So, required total is 4100+3366-2*630 = 6206

Let the number of students who took exactly one exam be ‘x’.
The number of students who took exactly two exams be ‘y’ and the number of students who took exactly three exams be ‘z’.
So, x + y + z = 200
x + 2y + 3z = 112 + 160 + 128 = 400
From these two equations, we get, y + 2z = 200
To minimize ‘z’, we have to maximize ‘y’. If z = 0, y = 200. In this case, x = 0
So, the minimum number of students who took all the three tests is 0.

Let r and h be the radius and height of the cone.
Initial volume = $$\frac{1}{3}*\pi*r^2*h$$
Final volume = $$\frac{1}{3}*\pi*(1.1r)^2*(1.2h)$$
Therefore, final volume = $$1.452*\frac{1}{3}*\pi*r^2*h$$
Therefore, percentage increase = $$\frac{1.452*\frac{1}{3}*\pi*r^2*h - \frac{1}{3}*\pi*r^2*h}{\frac{1}{3}*\pi*r^2*h}*100 = 45.20$$%
Hence option D is the correct answer.