Mathematical Operations - 1 - CUET Commerce Free MCQ Test with solutions

By the Given data , We have the expression:

20 + 8 - 8 ÷ 4 × 2 = 20 + 8 - 2 × 2 = 20 + 8 - 4 = 24.

Let the runs scored by P, Q, R, S and T be  a, b, c, d and e

From given data,

a + b + c + d + e = 36 x 5 = 180
b + c = 107
Let a = x
e = x - 8
d = x - 3
now 3x - 11 + 107 = 180
3x = 84
x = 28
e's score = 28 - 8 = 20

Given 5x - 6 = 3x - 8

5x - 3x = -8 + 6

2x = -2

=> x = -1.

Using Correct Symbols , We have :

Given expression = 30 / 2 + 3 x 6 - 5 = 15 + 18 - 5 = 28

Let the least value of the prize = Rs. x

Then the next value of the prize is x + 30 , x + 60, x + 90, ....x + 240.

Given total amount of cash prizes = Rs.1890

--> x + (x + 30) + (x + 60) + (x + 90) + ....+ (x + 240) = 1890

--> 9x + (30 + 60 + 90 + 120 + 150 + 180 + 210 + 240) = 1890

--> 9x + 30(1 + 2 + 3 + 4....+ 8) = 1890

--> 9x + 30(36) = 1890

--> 9x = 810 --> x=90

Hence the least value of the prize x = 90

1. Focus on the last two digits of the base: 2151 → last two digits are 51.

2. Compute small powers of 51 and look at their last two digits:

   • 51¹ = 51 → ends in 51

   • 51² = 51 × 51 = 2601 → ends in 01

3. Notice the cycle:

   • Any odd power of 51 ends in 51

   • Any even power of 51 ends in 01

4. Since the exponent 415 is odd, the last two digits of 51⁴¹⁵ must be the same as those of 51¹, namely 51.

By the given data, we have  the expression :

(3 × 15 + 19) ÷ 8 - 6 = (45 + 19) ÷ 8 - 6 = 64 ÷ 8 - 6 = 8 - 6 = 2

After substituting the signs as given in the question, the equation becomes
81 ÷ 9 -10 x 6 + 5
Applying BODMAS rule
9 - 10 x 6 + 5
9 - 60 + 5
9 - 55
= -46

-4-(-10) = -4 + 10 = 6

-10-(-4) = -10 + 4 = -6

Therefore, 6-(-6) = 6 + 6 = 12