SNAP 2024 Question Paper

$$N^3=a^3\cdot b^3\ $$ and $$\frac{N^3}{b^2}=a^3\cdot b^{ }\ $$

The number of factors of $$ a^3\cdot b^{ }\ $$ = 4*2=8.

Through options check,

If x = 3, LCM of 36 & 54 = 108

If x = 4, LCM of 46 & 54 = 1242

If x = 1, LCM of 26 & 54 = 702

If x = 0, LCM of 6 & 54 = 54

So the answer is option A.

If the root is p then the sum of the roots = 3p/2 = -b/a => p=-2b/3a, and product of roots is $$p^2/2=c/a$$. $$(-2b/3a)^2=2c/a $$=>$$ 4b^2/9a^2 = 2c/a => 2b^2=9ac$$

Let the number be x y. Hence $$|x-y| = 4$$ and 10(10x+y)=14(10y+x+x+y)= 14(11y+2x) = 154y+28x => 100x+10y=154y+28x = 72x = 144y => x=2y.
Hence y=4 => x=8. Hence the number is 84

Volume of a cube = $$a^3$$
Surface area of a cube = $$6a^2$$.
Side of the large cube = $$64^{1/3}$$ = $$4 m$$.
Surface area of the larger cube = $$6*4^2$$ = $$96 m^2$$.

The large cube is broken down to small cubes of volume $$8 m^3$$.
Number of small cubes = $$\frac{64}{8}$$ = $$8$$.
Side of the small cube = $$8^{1/3}$$ = $$2 m$$.
Surface area of the small cube = $$6*2^2$$ = $$24 m^2$$
Total surface area of the smaller cubes = $$8*24 = 192 m^2$$
%age increase in surface area = $$\frac{192-96}{96}$$
= $$\frac{96}{96}$$
= $$100$$%.

Hence, option B is the right answer.

Given that A and B are supposed to complete $$\dfrac{17}{28}th$$ of work together.
Let the total work be 28 units.
17 units will be completed by A and B together.
Work done by C = 28-17 = 11 units
28 units → Rs.700
1 unit → Rs.25

Then, for 11 units, C will be paid 25*11 = Rs.275

Let us assume the capacity of the tank to be 100 litres.
C can fill the tank in 10 hours. Therefore, C can fill 10 litres every hour.
B can empty 5 litres every hour and A can fill 2.5 litres every hour.
Together, the 3 pipes can fill 2.5 - 5 + 10 = 7.5 litres evey hour.
They can fill 75 litres (three quarters) in 10 hours.
Therefore, option B is the right answer.

Sam runs 100 m by the time Ram runs 100-20 = 80 m.
Their speeds are in the ratio 5:4

Ram was given by a head start by Sam.
Now, both of them meet at 75 m mark.
Let the head start provided be x m.

(75-x)/4 = 75/5
=> 75-x = 60
x = 15 m.

Ram has to run 85 m. But, by the time he runs 80m, Sam would have completed the race. Therefore, Ram will lost the race by 5m.

Therefore, option A is the right answer.

cosec(30°) = 2 while cot(60°) = $$\frac{1}{\sqrt{3}}$$
cosec(30°) + cot(60°) = 2 + $$\frac{1}{\sqrt{3}}$$ = $$ \frac{2\sqrt{3} + 1}{\sqrt{3}} $$
Hence, option A is the correct option.

We have, $$a^2=b^3=c^4$$, $$b^2=d^5$$
=> d=$$b^{\frac{2}{5}}$$
Also, c= $$b^{\frac{3}{4}}$$ and a = $$b^{\frac{3}{2}}$$
Now, $$\log_ab$$*$$\log_cd$$ = $$\frac{\log b}{\log a}\times\ \frac{\log d}{\log c}$$
= $$\dfrac{\log\ b}{\log b^{\frac{3}{2}}}\times\ \dfrac{\log b^{\frac{2}{5}}}{\log b^{\frac{3}{4}}}$$

= $$\frac{\log\ b}{\frac{3}{2}\log b}\times\ \frac{\frac{2}{5}\log b}{\frac{3}{4}\log b}$$ = $$\frac{2}{3}\times\ \frac{2\times\ 4}{3\times\ 5}=\ \frac{\ 16}{45}$$