SNAP OMETs - Quant Based Reasoning Questions

X = 3, Y = 2 and N = 100

Step 1. X=6 Y=3

Step 2. X=18 Y=4

Step 3. X=72 Y=5

Step 4. X=360 Y=6

Hence Option C is correct.

X = 3, Y = 2 and N = 100

Step 1. X=6 Y=3

Step 2. X=18 Y=4

Step 3. X=72 Y=5

Step 4. X=360 Y=6

Option D

X = 3, Y = 2 and N = 100

Step 1. X=6 Y=3

Step 2. X=18 Y=4

Step 3. X=72 Y=5

Step 4. X=360 Y=6

Hence Option C is correct.

X = 3, Y = 2 and N = 100

Step 1. X=6 Y=3

Step 2. X=18 Y=4

Step 3. X=72 Y=5

Step 4. X=360 Y=6

Step 5. X=2160 Y=7

Hence Option D is correct.

1. X=2, Y=3

X = XY and Y = Y + 1

2. X=6, Y=4

3.X=24, Y=5

4. X=120, Y=6

5. X= 720, Y=7

Hence X can take any value above 120

Hence among the given values Option A is the correct choice.

Let selling price per lemon of Jamuna be 'j' and that of Seema's be 's'.

Jamuna sells 2 lemons for Rs. 0.5 which makes j=(1/4)

Seema sells 3 lemons for Rs. 0.5 which makes s=(1/6)

After Seema leaves, Jamuna sells 5 lemons for a rupee and let it be denoted by t=(1/5)

Let the no. of lemons with each one of them be N.

If Seema would've sold her N lemons, she would've earned Rs. (N/6) and similarly, Jamuna would have earned Rs. (N/4).

When their individual stock is sold together, they earn Rs. (2N/5).

When counting, they observe that there's shortage of Rs. 3.50

So the eqn we can form is:

(2N/5)-(N/4)-(N/6) = 3.50

=> (N/60) = 3.5

=> N = 210 lemons

So, if Jamuna didn't sell both stocks together, she'd have earned Rs. (210/4) = Rs. 52.5

But on selling the stocks together and dividing the money equally, she gets Rs. ((2x210)/5) = Rs. 42 

$$\therefore\ $$ Jamuna suffers a loss of Rs. 10.5

Let the orange juice be in cup A initially and lemonade in B.

Let their vol. be 40L.

Let the vol. of the teaspoon be 10L.

On taking out 10L of orange juice from A and putting it in B, vol. of A becomes 30L and that of B becomes 50L.

In B, the ratio of vol. of orange juice to lemonade is 1:4.

Further when we take a teaspoon of the mixture from B, the vol. of liquid in the teaspoon will also have orange juice to lemonade in the ratio 1:4 i.e. 2L of orange juice and 8L of lemonade which is put into A.

So finally A contains 8L of lemonade and B contains 8L of orange juice because the teaspoon initially put in 10L and then took 2L out.

Let $$x$$ and $$y$$ be the 2 unequal number in which he divides the gold such that x>y.

Total coins with him = $$x+y$$

As per the merchant

$$48\left(x-y\right)\ =\ x^2-y^2$$

$$48\left(x-y\right)\ =\left(x-y\right)\left(x+y\right)$$

Either $$x+y = 48$$ or $$ x=y$$(rejected as it is given that they are unequal

hence merchant has 48 coins.

For Day 1 his income was = Rs 1

For Day 2 his income was  = Rs 1*2 = Rs. 2

For Day 3 his income was  = Rs 2*2 = Rs 4 = Rs 1 * $$2^2$$

Similarly for nth day his income = Rs 1* $$2^{n-1}$$

Hence for 10th  day his overall income is Rs 1*$$2^{10-1}$$ = $$2^9$$

We'll approach this question by checking the options.

When the 1st boy woke up, he ate 1 and hid a third of the remaining, which leaves two-thirds.

So, if the initial no. of apples was N, after the actions of the 1st boy who woke up, now remaining apples are:

The 2nd boy repeats the process which makes the remaining no. of apples :

Now the 3rd guy repeats the process as well. No. of apples remaining after the 3rd operation:

Now this number has to be an integer and one less than this will be divisible by 3.

On putting the value of N as 67, 79 and 85 we can observe that only 79 satisfies the conditions.