Class 7 Exam  >  Class 7 Tests  >  Mathematics Olympiad Class 7  >   Olympiad Test: Integers -2 - Class 7 MCQ

Olympiad Test: Integers -2 - Class 7 MCQ


Test Description

10 Questions MCQ Test Mathematics Olympiad Class 7 - Olympiad Test: Integers -2

Olympiad Test: Integers -2 for Class 7 2024 is part of Mathematics Olympiad Class 7 preparation. The Olympiad Test: Integers -2 questions and answers have been prepared according to the Class 7 exam syllabus.The Olympiad Test: Integers -2 MCQs are made for Class 7 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Olympiad Test: Integers -2 below.
Solutions of Olympiad Test: Integers -2 questions in English are available as part of our Mathematics Olympiad Class 7 for Class 7 & Olympiad Test: Integers -2 solutions in Hindi for Mathematics Olympiad Class 7 course. Download more important topics, notes, lectures and mock test series for Class 7 Exam by signing up for free. Attempt Olympiad Test: Integers -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
Olympiad Test: Integers -2 - Question 1

72÷? = –4

Detailed Solution for Olympiad Test: Integers -2 - Question 1

72 ÷ x = – 4

Olympiad Test: Integers -2 - Question 2

What is the product of additive inverse of 7 and multiplicative inverse of 14?

Detailed Solution for Olympiad Test: Integers -2 - Question 2

Additive inverse of 7 = –7
Multiplicative inverse of

1 Crore+ students have signed up on EduRev. Have you? Download the App
Olympiad Test: Integers -2 - Question 3

[16–{5+(2–7)}]–[12–{7–(3–4)}]?

Detailed Solution for Olympiad Test: Integers -2 - Question 3

[16 – {5 + (2 – 7)}] – [12 – {7 – (3 – 4)}]
= [16 – {5 – 5}] – [12 – {7 + 1}]
= [16 – 0] – [12 – 8] = 16 – 4 = 12

Olympiad Test: Integers -2 - Question 4

[16÷{11–(2+5)}] =?

Detailed Solution for Olympiad Test: Integers -2 - Question 4

 [16 ÷ {11 – (2 + 5)}]
= [16 ÷ {11 – 7}] = [ 16 ÷ 4] = 4

Olympiad Test: Integers -2 - Question 5

25+16÷4×3–5+7 of (–3) = ?

Detailed Solution for Olympiad Test: Integers -2 - Question 5

25+16 ÷ 4×3 –5+7 of (–3)
= 25 + 16 ÷ 4×3 –5 + (–21)
= 25 + 4×3 –5 –21
= 25 + 12–26
= 37 – 26 = 11

Olympiad Test: Integers -2 - Question 6

Detailed Solution for Olympiad Test: Integers -2 - Question 6

Olympiad Test: Integers -2 - Question 7

A cement company earns a profit of  Rs 8 per bag of white cement sold and a loss of Rs 5  per bag of grey cement sold. The company sells 3000 bags of white cement and 5000 bags of grey cement in a month. What is its profit or loss?  

Detailed Solution for Olympiad Test: Integers -2 - Question 7

Profit on one white cement bag = Rs. 8

Loss on one grey cement bag = - Rs. 5 [Loss is considered to be negative]

Profit on 3,000 bags of white cement = 3,000 × Rs. 8 = Rs. 24,000

Loss on 5,000 bags of white cement = - 5,000 × Rs. 5 = - Rs. 25,000

Since the loss amount is greater than the profit amount,

Total loss = - Rs. 25,000 + Rs. 24,000 = - Rs. 1,000 

i.e. the total loss amount is Rs.1000.

Olympiad Test: Integers -2 - Question 8

In a class test containing 15 questions 4 marks are given for every correct answer and (–2) marks are given for every incorrect answer. Mohan attempts all questions but only 9 of his answers are correct. What is his total score?

Detailed Solution for Olympiad Test: Integers -2 - Question 8

Marks for 9 correct answers = 9 × 4 = 36
Marks for 6 incorrect answers = 6 × (–2) = –12
Mohan’s total score = 36 + (–12) = 36 –12 = 24

Olympiad Test: Integers -2 - Question 9

A shopkeeper earns a profit of Rs 1 by selling one copy and incure a loss of 40 paise per pen while selling pens of his old stock. In a particular month he incurs a loss of Rs 5. In this period he sold 45 copy. How many pens did he sell in this period?  

Detailed Solution for Olympiad Test: Integers -2 - Question 9

Profit earned by selling 45 copies = Rs 45
Total loss = Rs 5 Profit earned + loss incurred = Total loss.
⇒ 45 + Loss incurred = –5
⇒ Loss incurred = –5 – 45 = –50 = – 5000 Paise

Olympiad Test: Integers -2 - Question 10

The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2°C per hour until midnight, at what time would the temperature be 8°C below zero?

Detailed Solution for Olympiad Test: Integers -2 - Question 10

The temperature at 12 noon = 10°C (given)

The temperature decreases by 2°C in 1 hour (given)

Thus, the temperature decreases by 1°C in 1/2 hour

Temperature 10°C above zero - Temperature 8°C below zero = 10 - (- 8) = 10 + 8 = 18°C

The temperature decreases by 18°C in 1/2 × 18 = 9 hours

Thus, from 10°C above zero to 8°C below zero it takes 9 hours

Total time = 12 noon + 9 hours

= 21 hours = 9 pm

Thus, at 9 pm temperature would be 8°C below zero.

24 videos|51 docs|112 tests
Information about Olympiad Test: Integers -2 Page
In this test you can find the Exam questions for Olympiad Test: Integers -2 solved & explained in the simplest way possible. Besides giving Questions and answers for Olympiad Test: Integers -2, EduRev gives you an ample number of Online tests for practice

Up next

24 videos|51 docs|112 tests
Download as PDF

Up next