Test: Mathematical Operations (March 6) - CAT MCQ

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

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.

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

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

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

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

By the given data, we have  the expression :

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

Using Correct Symbols , We have :

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

Unit digit of this expression is always 1 as the base ends with 1. 

For the tenth place digit we need to multiply the digit in the tenth place of the base and unit digit of the power and take its unit digit.

i.e, tenth place digit in 2151 is 5 and

tenth place digit in power 415 is 1

And the units digit in the product of 5 x 1 = 5

Therefore, last two digits of 2151⁴¹⁵ is 51.

By the Given data , We have the expression:

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