PE 1 - Numbers

This is a direct application of LCM Model II, as the complement of the remainder is the same in all the three cases.

Complement of remainder = Difference between divisor and remainder 

Case I       d₁ - r₁ = 25 – 15 = 10 = x (say)
Case II      d₂ - r₂ = 35 – 25 = 10
Case III     d₃ - r₃ = 45 – 35 = 10

Hence, smallest number = LCM(d₁, d₂, d₃) - x

In this case, LCM of 25, 35 and 45 = 1575

Required number = 1575 – 10 = 1565

Alternatively,
The answer should be such a number that when 10 is added to it, it is divisible by 25, 35 as well as 45, i.e. by 1575. 

Add 10 to each option and check which option is divisible by 1575.

Hence, option (a).

This is a direct application of LCM Model I, as the remainder is the same in all the three cases.

∴ Number of students = LCM(3, 4, 5) + 1 = 60 + 1 = 61

Hence, option (d).

288 = 2⁵ × 3²

The highest power of 3 in 35! is $R\left[\frac{35}{3}\right]$ + $R\left[\frac{35}{3^2}\right]$ + $R\left[\frac{35}{3^3}\right]$ = 11 + 3 + 1 = 15.

∴ The highest power of 3² will be $\frac{15}{2}$ = 7.

The highest power of 2 in 35! is $R\left[\frac{35}{2}\right]$ + $R\left[\frac{35}{2^2}\right]$ + $R\left[\frac{35}{2^3}\right]$ + $R\left[\frac{35}{2^4}\right]$ = 17 + 8 + 4 + 2 + 1 = 32.

∴ The highest power of 2⁵ will be $\frac{32}{5}$ = 6.

⇒ The highest power of 288 will be lower of 7 and 6 i.e., 6.

Hence, option (b).

$\frac{A}{2}$ = $\frac{B}{3}$ = $\frac{C}{4}$ = x

⇒ A = 2x; B = 3x; C = 4x

⇒  A : B : C = 2 : 3 : 4

Largest part = 369 × $\frac{4}{9}$ = 164

Hence, 164.

LCM of 3, 6, 9, 12, 15, 18 and 21 is 1260.

So, the bells will toll together after every 1260 s, i.e. 21 mins.

In 300 mins, they will toll together for (300/21) + 1 = 15 times.

Hence, option (d).

n = 5k + 3

⇒ n² = 25k² + 30k + 9

= $R\left[\frac{25k^2+30k+9}{5}\right]$ 

= $R\left[\frac{25k^2}{5}\right]$ + $R\left[\frac{30k}{5}\right]$ + $R\left[\frac{9}{5}\right]$ = 0 + 0 + 4 = 4.

Hence, option (d).

N = $\frac{1}{1\times 2}$ + $\frac{1}{2\times 3}$ + $\frac{1}{3\times 4}$ + ... + $\frac{1}{n\times (n+1)}$

N = $\frac{2-1}{1\times 2}$ + $\frac{3-2}{2\times 3}$ + $\frac{4-3}{3\times 4}$ + ... + $\frac{(n+1)-n}{n\times (n+1)}$

N = $\frac{2}{1\times 2}$ - $\frac{1}{1\times 2}$ + $\frac{3}{2\times 3}$ - $\frac{2}{2\times 3}$ + $\frac{4}{3\times 4}$ - $\frac{3}{3\times 4}$ + ... + $\frac{(n+1)}{n\times (n+1)}$ - $\frac{n}{n\times (n+1)}$

N = $\frac{1}{1}$ - $\frac{1}{2}$ + $\frac{1}{2}$ - $\frac{1}{3}$ + $\frac{1}{3}$ - $\frac{1}{4}$ + ... + $\frac{1}{n}$ - $\frac{1}{(n+1)}$

N = $\frac{1}{1}$ - $\frac{1}{n}$

Since 'n' is a natural numbers N will always be less than 1.

Hence, option (a).

Let tens digit be x. Then, units digit = 2x

Sum of the digits = x + 2x = 3x

Number = 10x + 2x = 12x

Now, $\frac{12x}{3x}$ = $\frac{4}{1}$

This is true for all values of x.

Hence, option (d).

The number of 5s in 30! is calculated as follows:

$R\left[\frac{40}{5}\right]$ + $R\left[\frac{40}{5^2}\right]$ = 8 + 1 = 9

Hence, option (e).

Suppose numbers are x and y, we have

x(x + y) = 2772            (i)
y(x + y) = 1584            (ii)

Adding (i) and (ii), we get
(x + y)² = 4356
∴ x + y = 66            (iii)

Subtracting (ii) from (i), we get
x² – y^{2​​​​​​​} = 1188
⇒ (x + y) × (x – y) = 1188

⇒ 66 × (x – y) = 1188         using (iii)
∴ x – y = 18

Hence, option (a).