The correct answer is option 3 i.e. 242.

Given:

5^x = 3125

Calculations:

⇒ 5^x = 5⁵ 

Comparing both sides;

⇒ x = 5

Now the value of 3^x - 1 = 3⁵ - 1 

⇒ 243 - 1 = 242

The correct answer is Option 4 i.e. 25.

Let the cost price of the laptop = 100a

According to the question,

100a × 140/100 × (100 – x)/100 = 100a × 105/100

140 × (1 – x/100) = 105

140 – 105 = 140x/100

x = 35 × 5/7

x = 25

The correct answer is Option 1 i.e. Rs. 20.

x × 125/100 + (x – 50) × 120/100 = 2x + 30

5x/4 + (x – 50) × 6/5 = 2x + 30

5x/4 + (6x – 300)/5 = 2x + 30

25x + 24x – 1200 = 40x + 600

9x = 1800

x = 200

Cost price of each pen = 200/10 = Rs. 20

The correct answer is option 4 i.e. 8 

∵ 36 = 9 × 4

The last two numbers should be divisible by 4. So, the only possible value of x is '1' such that the last two digits become 16.

The number to get divisible by 9 the sum of the digits must be divisible by 9. Thus

1 + 6 + y + 9 + 6 + 1 + 6 = divisible by 9

29 + y = divisible by 9

Only possible value of y is '7'

Hence, (x + y)/(x)² = (7 + 1)/(1)² = 8

The correct answer is Option 1 i.e. 10080 and 99960.

To solve this,

L.C.M (15, 10, and 8) = 120

Smallest 5-digit number = 10000

Divide 10000 by 120 

Remainder = 40

Required number = 10000 + (120 - 40) = 10000 + 80 = 10080

Therefore, the smallest 5-digit number that is divisible by 15, 10, and 8 is 10080, and

Largest 5-Digit number = 99999

Divide 99999 by 120

Remainder = 39

Required number = 99999 - 39 = 99960

The largest 5-digit number that is divisible by 15, 10, and 8 is 99960

Hence, the smallest and largest 5-digit numbers are 10080 and 99960 respectively.

The correct answer is Option 2 i.e. 25.

Let the number be ‘x’.

⇒ x – √x = 20

⇒ x – 20 = √x

Squaring both sides, we have;

⇒ (x – 20)² = (√x)²

⇒ x² + 400 – 40x = x

⇒ x² – 41x + 400 = 0

⇒ x² – 25x – 16x + 400 = 0

⇒ x(x – 25) – 16(x – 25) = 0

⇒ (x – 25)(x – 16) = 0

⇒ x = 25 or x = 16

But, since 16 – √16 = 16 – 4 = 12 (here difference must be 20)

Only 25 is a valid solution

The correct answer is Option 1 i.e. 16 (1/4) days.

Let total work = 720 units (LCM of 36, 48 and 60)

Efficiency of Faruk = 720/36 = 20 units

Efficiency of Bhim = 720/48 = 15 units

Efficiency of Chand = 720/60 = 12 units

Work completed by Faruk, Bhim and Chand together in 15 days = 15 × (20 + 15 + 12) = 705 units

Amount of time taken Chand to complete the remaining work = (720 – 705)/12 = 5/4 days

Required time = 15 + 5/4 = 65/4 = 16 (1/4) days

The correct answer is Option 3 i.e. Rs. 1500.

A, B and C’s efficiency ratio = (1/12) : (1/16) : (1/20)

= (20 : 15 : 12)/240

= 20 : 15 : 12

47’s = 3525

1’s = 75

The share of A = 75 × 20 = Rs. 1500