# Olympiad Test: Algebraic Expressions - 2

## 10 Questions MCQ Test Mathematics Olympiad Class 7 | Olympiad Test: Algebraic Expressions - 2

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Attempt Olympiad Test: Algebraic Expressions - 2 | 10 questions in 20 minutes | Mock test for Class 7 preparation | Free important questions MCQ to study Mathematics Olympiad Class 7 for Class 7 Exam | Download free PDF with solutions
QUESTION: 1

### Find the difference of  x2 + xy + y2 + yz and 2x2 – 3xy – y2 – yz.

Solution:

2x2 – 3xy – y2– yz – (x2– xy + y2 + yz)
= 2x2 – 3xy – y2– yz – x2 + xy – y2– yz
= x2 – 2xy – 2y2– 2yz

QUESTION: 2

### What is the result when the sum of (–5x2 + 7xy + 2y2) and (x2 – 3xy – y2) is subtracted from –7?

Solution:

Sum = –5x2 + 7xy + 2y2 + x2 – 3xy – y2
= – 4x2 + 4xy + y2
= –7 –(– 4x2 + 4xy + y2)
= –7 + 4x2– 4xy – y2)

QUESTION: 3

### What is the product of (0.8m – 0.7n) and (1.5n – 1.7m)?

Solution:

(0.8m – 0.7n) (1.5n – 1.7m)
= (0.8m)(1.5n) + (0.7n) (1.7m) – (0.7n)(1.5n) – (0.8m) (1.7m)
= 1.2mn + 1.19mn –1.05n2 –1.36m2
= 2.39mn – 1.05n2 – 1.36m2

QUESTION: 4

Find the sum of 0.3x2– 3xy + 0.8y2 and 4x2 – 2y2 + 0.7xy?

Solution:

0.3x2 – 3xy + 0.8y2 + 4x2 – 2y2 + 0.7xy
= (4 + 0.3) x2 + (0.7 – 3) xy + (0.8 – 2)y2
= 4.3x2 – 2.3xy –1.2y2

QUESTION: 5

Find the product of (7x2 – x  + 11) and (x2 – 3).

Solution:

(7x4– x + 11) (x2– 3)
= 7x4 – 21x2– x3 + 3x + 11x2 – 33
= 7x4– x3 – 10x2 + 3x – 33

QUESTION: 6

What is the product of 1.5 a(10a2b – 100ab2)

Solution:

1.5a(10a2b –100ab2)
= 15a3b – 150a2b2

QUESTION: 7

Find the simplified expression of 7x2 – [x2– 3x –{x + y}] – (5x – 3y + 3).

Solution:

7x2 – [x2– 3x –{x + y}] – (5x – 3y + 3)
= 7x2 – x2 – 3x – x + y – 5x + 3y – 3
= 6x2 – x + 4y – 3

QUESTION: 8

Find the result of x(x + 4) + 3x (2x2 – 3) + 4x2 + 5?

Solution:

x(x + 4) + 3x (2x2– 3) +4x2 + 5
= x 2 + 4x + 6x3– 9x + 4x2 + 5
= 6x3 + 5x2– 5x + 5

QUESTION: 9

4mn (m – n) – 6m2(n – n2)  – 3n2(2m2 – m) = ?

Solution:

4mn(m – n) – 6m2(n – n2) –3n2(2m2 – m)
= 4 m2n – 4m – n2 – 6m2n + 6m2 n2 – 6n2m2 + 3n2m
= –2m2n – n2m = –mn(2m + n)

QUESTION: 10

a2b(a – b2) – ab2 (3ab – a2) – a3b (1– 2b) = ?

Solution:

a2b(a – b2) –ab2(3ab – a2) – a3b (1 – 2b)
= a3b – a2b3 – 3a2b3 + a3b2 – a3b + 2a3b2
= 3a3b2– 4a2b3

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