Simplify 1 + 2 + 3 1 - 2 + 3 by rationalizing the denominator.

We have 3 + 1 + 2 3 + 1 - 2 = 3 + 1 + 2 3 + 1 - 2 × 3 + 1 + 2 3 + 1 + 2 = 3 + 1 + 2 2 3 + 1 - 2 3 + 1 + 2 Now, using a - b a + b = a 2 - b 2 and a + b + c 2 = a 2 + b 2 + c 2 + 2 a b + 2 b c + 2 c a , we get = 3 + 1 + 2 + 2 3 + 2 2 + 2 6 3 + 1 2 - 2 2 Again, using a + b 2 = a 2 + 2 a b + b 2 , we get = 6 + 2 3 + 2 2 + 2 6 3 + 1 + 2 3 - 2 = 2 3 + 3 + 2 + 6 2 1 + 3 = 3 3 + 1 + 2 1 + 3 1 + 3 = 3 + 1 3 + 2 3 + 1 = 3 + 2 .

Simplify the expression x 2 - 4 x + 3 + x + 1 x 2 - 9 x 2 + 4 x + 3 + x - 1 x 2 - 9 , where x > 3 .

We have x 2 - 4 x + 3 + x + 1 x 2 - 9 x 2 + 4 x + 3 + x - 1 x 2 - 9 , where x > 3 ⇒ x - 3 > 0 . Factorising the quadratic by splitting its middle term, we get = x 2 - x - 3 x + 3 + x + 1 x + 3 x - 3 x 2 + x + 3 x + 3 + x - 1 x + 3 x - 3 = x x - 1 - 3 x - 1 + x + 1 x + 3 · x - 3 x x + 1 + 3 x + 1 + x - 1 x + 3 · x - 3 = x - 3 x - 1 + x + 1 x + 3 · x - 3 x + 3 x + 1 + x - 1 x + 3 · x - 3 = x - 3 · x - 1 x - 3 + x + 1 x + 3 x + 3 · x + 1 x + 3 + x - 1 x - 3 = x - 3 x + 3 = x - 3 x + 3 × x + 3 x + 3 = x - 3 x + 3 x + 3 = x 2 - 9 x + 3 [ Using, a + b a - b = a 2 - b 2 ] .

Evaluate 5 + 6 + 7 5 + 6 - 7 5 - 6 + 7 - 5 + 6 + 7

We have to evaluate, 5 + 6 + 7 5 + 6 - 7 5 - 6 + 7 - 5 + 6 + 7 5 + 6 + 7 5 + 6 - 7 × 7 + 5 - 6 7 - 5 - 6 Using, a + b a - b = a 2 - b 2 , we get = 5 + 6 2 - 7 2 × 7 2 - 5 - 6 2 Now, using a ± b 2 = a 2 + b 2 ± 2 a b , we get = 5 + 6 + 2 30 - 7 × 7 - 5 - 6 + 2 30 = 2 30 + 4 × 2 30 - 4 Again using, a + b a - b = a 2 - b 2 , we get = 2 30 2 - 4 2 = 120 - 16 = 104 .

Suppose that a = 6 - 2 and b = 2 2 - 6 . Then A a > b , B a = b , C a < b , D b = 2 a , E a = 2 b .

Here, a = 6 - 2 and b = 2 2 - 6 On rationalising both the numbers, we get a = 6 - 2 6 + 2 6 + 2 and b = 2 2 - 6 2 2 + 6 2 2 + 6 Now, using x + y x - y = x 2 - y 2 , we get ⇒ a = 6 2 - 2 2 6 + 2 and b = 2 2 2 - 6 2 2 2 + 6 ⇒ a = 6 - 4 6 + 2 and b = 8 - 6 2 2 + 6 ⇒ a = 2 6 + 2 and b = 2 8 + 6 Now, we know that, 8 > 2 ⇒ 8 + 6 > 6 + 2 Also, we know that if x and y are positive real numbers and if x > y , then 1 x < 1 y . Hence, we have 2 6 + 2 > 2 8 + 6 . ∴ a > b .

Arrange the three values a = 27 - 26 , b = 28 - 27 , c = 29 - 28 in ascending order.

We have a = 27 - 26 , b = 28 - 27 and c = 29 - 28 . On rationalising all the three numbers, we get a = 27 - 26 27 + 26 27 + 26 , b = 28 - 27 28 + 27 28 + 27 and c = 29 - 28 29 + 28 29 + 28 Now, using x + y x - y = x 2 - y 2 , we get ⇒ a = 27 2 - 26 2 27 + 26 , b = 28 2 - 27 2 28 + 27 and c = 29 2 - 28 2 29 + 28 ⇒ a = 27 - 26 27 + 26 , b = 28 - 27 28 + 27 and c = 29 - 28 29 + 28 ⇒ a = 1 27 + 26 , b = 1 28 + 27 and c = 1 29 + 28 As, we know that 29 > 28 > 27 > 26 ⇒ 29 > 27 and 28 > 26 ⇒ 29 + 28 > 27 + 28 and 28 + 27 > 26 + 27 ⇒ 29 + 28 > 28 + 27 > 27 + 26 ⇒ 1 29 + 28 < 1 28 + 27 < 1 27 + 26 We know that, if x , y , z are positive real numbers such that x > y > z , then 1 x < 1 y < 1 z . ∴ c < b < a .

E M B I B E

Non-Routine Mathematics Resource Book-1 for PRMO>Chapter 2 - Number System>EXERCISE - 2.3>Q 103

EASY

IOQM - PRMO and RMO

IMPORTANT

Earn 100

Two integers have a sum of $26 .$ When two more integers are added to the first two integers the sum is $41 .$ Finally, when two more integers are added to the sum of the previous four integers the sum is $57 .$ What is the minimum number of even integers among the $6$ integers?

50% studentsanswered this correctly

Important Questions on Number System

EASY

IOQM - PRMO and RMO

IMPORTANT

Non-Routine Mathematics Resource Book-1 for PRMO>Chapter 2 - Number System>EXERCISE - 2.3>Q 104

How many ordered pairs of positive integers

(M, N)

satisfy the equation

\frac{M}{6} = \frac{6}{N}

?

EASY

IOQM - PRMO and RMO

IMPORTANT

Non-Routine Mathematics Resource Book-1 for PRMO>Chapter 2 - Number System>EXERCISE - 2.3>Q 105

Geeta and Babeeta play the following game. An integer between

0

and

999

, inclusive, is selected and given to Geeta. Whenever Geeta receives a number, she doubles it and passes the result to Babeeta. Whenever Babeeta receives a number, she adds

50

to it and passes the result to Geeta. The winner is the last person who produces a number less than

1000 .

Let

N

be the smallest initial number that results in a win for Geeta. What is the sum of the digits of

N

?

MEDIUM

IOQM - PRMO and RMO

IMPORTANT

Non-Routine Mathematics Resource Book-1 for PRMO>Chapter 2 - Number System>EXERCISE - 2.3>Q 106

A majority of the

30

students in Mrs. Bhatnagar's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than