SNAP Mock Test by SIBM Pune

Let the incomes of P, Q, and R be $$p,q$$ and $$r,$$ respectively.

We are given that the value of $$q$$ is $$25\%$$ less than $$r$$. So, the value of $$q$$ can be written as,

$$q\ =\left(1\ -\ \dfrac{25}{100}\right)r=\dfrac{3}{4}r$$.

We are also given that $$p$$ is $$20\%$$ higher than $$q$$. So, the value of $$p$$ can be written as,

$$p=\left(1\ +\ \dfrac{20}{100}\right)q=\dfrac{6}{5}q=\dfrac{6}{5}\times\ \dfrac{3}{4}r=\dfrac{9}{10}r$$

The percentage by which the salary of $$p$$ is less than $$r$$ can be calculated as,

Percentage $$=$$ $$\dfrac{r\ -\ p}{r}\times\ 100\ =\ \dfrac{\left(r\ -\ \dfrac{9}{10}r\right)}{r}\times\ 100\ =\ \dfrac{1}{10}\times\ 100\ =\ 10\%$$

Hence, the correct answer is option B.

For an amount to become three times itself, the interest gained on that amount must be twice its value.

Let the value of the amount be P, and the number of years it takes for the interest gained on that amount is 2P can be calculated as,

$$SI=\dfrac{P\times\ T\times\ R}{100}$$

$$2P=\dfrac{P\times\ T\times\ 25}{100}$$

$$T\ =\ 8$$ years.

Therefore, the time it takes for the amount to become three times its value is 8 years.

Hence, the correct answer is option B.

Let the three numbers be $$a,b$$ and $$c$$ in decreasing order. 

Half of the average of these 3 numbers is,

$$\dfrac{1}{2}\times\ \dfrac{a\ +\ b\ +\ c}{3}\ =\ 20$$

$$\ a\ +\ b\ +\ c\ =\ 120$$ --(1)

We can see that the first number mentioned is the highest in the question, and its value is given to be,

$$a=2b$$ --(2)

The second number value is given to be,

$$b=3c$$ --(3)

So, the value $$a$$ can also be written as, 

$$a=2b=2*3c=6c$$ --(4)

Substituting (3) and (4) in (1), we get,

$$6c+3c+c=120$$

$$10c=120$$

$$c=12$$

$$a=6c=72$$

Sum of smallest and largest numbers can be calculated as,

$$a+c=72+12=84$$.

Hence, the correct answer is option C.

Let the scheduled time be T and the distance that needs to be travelled be D.

While travelling at 50kmph, the train covers D in (T - 1) hours. The value D can also be written as,

D = speed * time = 50kmph * (T - 1)

We are also given that when the train runs at 8.33 m/s = $$\dfrac{25}{3}$$m/s = $$\dfrac{25}{3}\times\ \dfrac{18}{5}\ =\ 30$$kmph, train covers the distance is (T + $$\dfrac{300}{60}$$) hours = (T + 5) hours. The value of D can also be written as,

D = speed * time = 30kmph * (T + 5)

Equating both the values of D, we get,

50T - 50 = 30T + 150

20T = 200

T = 10 hours.

So, the correct time for the train to complete the journey is 10 hours.

Hence, the correct answer is option A.

The total area of square $$=\ S^2\ =\ \left(150cm\right)^2\ =\ 22500cm^2$$

The area of the parts cut from the edges $$=\ 4\ \times\ \dfrac{1}{4}\ \times\ \dfrac{22}{7}\ \times\ \left(14cm\right)^2\ =\ 22\ \times\ 14\ \times\ 2\ =\ 616\ cm^2\ $$

Area remaining $$=22500-616=21884\ cm^2$$

Hence, the correct answer is option C.

Let the total amount of work that needs to be done be $$W$$, and the efficiencies of $$A$$ and $$B$$ be $$a$$ and $$b$$, respectively. 

The amount of work done by $$A$$ and $$B$$ together in 12 days is,

$$12(a+b)=W$$ --(1)

The amount of work done by $$A$$ alone in 18 days is,

$$18(a)=W$$

$$a=\dfrac{W}{18}$$ --(2)

Substituting (2) in (1), we get,

$$12\left(\dfrac{W}{18}\ +\ b\right)\ =\ W$$

$$\dfrac{2W}{3}\ +\ 12b\ =\ W$$

$$12b\ =\ \dfrac{W}{3}$$

$$W\ =\ 36b$$

So, $$B$$ alone can complete the total work in 36 days.

Hence, the correct answer is option B.

Roots of a quadratic equation $$ax^2\ +\ bx\ +\ c\ =\ 0$$ can be calculated using the formula,

Roots $$=\ \dfrac{-b\ \pm\ \sqrt{\ b^2\ -\ 4ac}}{2a}$$

Roots of $$x^2−7x+12=0$$ can be calculated as,

Roots $$=\ \dfrac{-\left(-7\right)\ \pm\ \sqrt{\ \left(-7\right)^2\ -\ 4\left(1\right)\left(12\right)}}{2\left(1\right)}\ =\ \dfrac{7\ \pm\ \sqrt{\ 49\ -\ 48}}{2}\ =\ \dfrac{7\ \pm\ 1}{2}\ =\ \dfrac{7\ +\ 1}{2}\ or\ \dfrac{7\ -\ 1}{2}\ =\ 4\ or\ 3$$

So, the roots of the equation are (3, 4)

Hence, the correct answer is option A.

To find this number, we must first find the Least Common Multiple (LCM) of the divisors (12, 18, 27) and then add the required remainder 9.

Step 1: To find the LCM of 12, 18, and 27

We use prime factorisation to find the LCM:

• $$12=2^2\times3^1$$
• $$18 = 2^1 \times 3^2$$
• $$27 = 3^3$$

The LCM is the product of the highest power of all prime factors:

$$\text{LCM}(12, 18, 27) = 2^2 \times 3^3$$

$$\text{LCM} = 4 \times 27$$

$$\text{LCM} = 108$$

Step 2: Add the Remainder

The required number is (LCM) + (Remainder).

$$\text{Required Number} = 108 + 9$$

$$\text{Required Number} = 117$$

Hence, the correct answer is option A.

Let the initial length and breadth of the rectangle be $$L$$ and $$B$$.

We are told that the length is increased by 25% and the breadth is decreased by 20%. So, the new length and breadth become 

Length $$L\left(1\ +\dfrac{25}{100}\right)\ =\ \dfrac{5}{4}L$$

Breadth $$=\ B\left(1\ -\ \dfrac{20}{100}\right)\ =\ \dfrac{4}{5}B$$

Initial Area $$=\ L\times\ B$$

Final Area $$=\ \dfrac{5}{4}L\times\ \dfrac{4}{5}B\ =\ L\ \times\ B$$

We can see that there is no change in the Area. So the percentage change in area is 0%.

Hence, the correct answer is option A.

If the sum ₹12,000 is divided between A, B, and C in the ratio 2:3:5, then the share of B can be calculated as,

B's share $$=\ \dfrac{3}{2\ +\ 3\ +\ 5}\ \times\ 12000\ =\ \dfrac{3}{10}\times\ 12000\ =\ 3600$$

So, B's share is ₹3,600.

Hence, the correct answer is option B.