7th Grade > Mathematics
INTEGERS MCQs
In a quiz, team A scored - 40, 10, 0 and team B scored 20, 0, - 40 in three successive rounds, team C scored 30,30,20 and team D scored 80, -20, 10.
(i) Which team scored the highest?
(ii) Which team scored the lowest?
(ii) What is the difference between the teams which scored the highest and the team which scored lowest?
(iii) Can we say that we can add integers in any order?
[4 MARKS]
:
Each question - 1 Mark
Team A - 40, 10, 0.
Total score = - 40 + 10 + 0 = - 30
Team B - 20, 0,- 40.
Total score = 20 + 0 +(- 40) = - 20
Team C - 30, 30, 20
Total score = 30 + 30 + 20 = 80
Team D - 80, -20, 10
Total score = 80 - 20 + 10 = 90
(i) So, team D scored the highest = 90
(ii) Team A scored the lowest =-30
(iii) The difference between the highest scorer and the lowest scorer
= 90 - (-30) = 90 + 30 = 120
(iv)Yes, we can add integers in any order. We had observed that the scores obtained by both teams in successive rounds were numerically equal but different in order. Yet, the total score of both teams was equal. It is the associative property of addition of integers.
A bird is flying at a height of 1200 m from the ground. A hunter shot the bird standing on a hill of height 600m. If the height from which he fired is 1 meter above the hill, find the distance travelled by the bullet to reach the bird? He missed in the first attempt and fired again but by then the bird was 1300 m from the ground. Find the distance travelled by the bullet in the second attempt.
[4 MARKS]
:
First attempt: 2 Marks
Second Attempt: 2 Marks
Distance of the bird from the ground surface in the first attempt = 1200m
Distance of the hunter from the ground surface = 600m
Distance of the bird from the hunter
= 1200m - 600m = 600m
But, when the hunter stands, the height of the gun from ground = 600 + 1 = 601m
So, distance traveled by bullet in the first attempt= 1200 - 601= 599m
Distance of the bird from the ground surface in the second attempt = 1300m
Distance traveled by the bullet in the second attempt= 1300 - 601 = 699m
:
Each option: 2 Marks
a) 625 × (−35) + (−625) × 65
= 625 × [(−35) + (−65)]
(∵ (a × b) + (a × c) = a × (b + c))
= 625 × (−100) = −62500
b) (−17) × (−29)
= (−17) × [−30 + 1]
= [(−17) × (−30)] + [(−17) × 1]
(∵ a × (b + c) = (a × b) + (a × c)
= (510) + (−17) = 493
:
Rule: 1 Mark
Value: 3 Mark
LHS = {10 × (5 + 2)}
= 10 × (7)
= 70
RHS = {(10 × 5 ) + (10 × 2 )}
= {50 + 20}
= 70
Hence , LHS = RHS
This rule is known as Distributivity of multiplication over addition.
We have to find out the value of
99×101 = 99×(100+1) = 9900 + 99 = 9999
In a class test (+3) marks are given for every correct answer and (−2) marks are given for every incorrect answer and no marks for not attempting a question. Rakesh scores 18 marks by attempting 16 questions. How many questions has he attempted correctly and how many has he attempted incorrectly? [4 MARKS]
:
Writing the equation: 1 Mark
Steps: 2 Marks
Result: 1 Mark
Marks obtained for 1 right answer = +3
Marks obtained for 1 wrong answer = −2
Total marks scored by Rakesh = 18
Number of questions attempted = 16
(No. of correct ans)(3) + (No. of incorrect ans)(-2)
= 18
⇒ (No. of correct ans)(3) + (16 - No. of correct ans)(-2)
= 18
⇒ (No. of correct ans)(3) + -32 + 2(No. of correct ans)
= 18
⇒ (No. of correct ans)(5) + -32 = 18
⇒ (No. of correct ans)(5) = 18 + 32 = 50
⇒ No. of correct ans = 10
∴ No. of incorrect ans = 16 - 10 = 6
∴ Total number of correct and incorrect answers scored by Rakesh is 10 and 6 respectively.