RS Aggarwal 2021-2022 for Class 6 Maths Chapter 4 - Integer
RS Aggarwal Class 6 Math Solution Chapter 4- Integer Test Paper is available here. RS Aggarwal Class 6 Math Solutions are solved by expert teachers in step by step, which help the students to understand easily. RS Aggarwal textbooks are responsible for a strong foundation in Maths. These textbook solutions help students in exams as well as their daily homework routine. The solutions included are easy to understand, and each step in the solution is described to match the students’ understanding.
Rs Aggarwal Class 6 Math Solution Chapter 4- Integer
Test Paper
1. What are integers? Write all integers from – 5 to 5.
Solution:
The numbers.-4,-3,-2,-1, 0, 1, 2, 3, 4. are integers.
The group of positive and negative numbers including 0 is called integers
All integers from – 5 to 5 are – 5, – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4, 5
2. In each of the pairs given below, find the larger integer
(i) 0 – 3
(ii) – 4, – 6
(iii) – 99, 9
(iv) – 385, – 615
Solution:
(i) 0 is larger, because 0 is greater than any negative integer.
(ii) – 4 is larger, because smaller negative number is larger than bigger negative number.
(iii) 9 is larger, because positive integer is greater than any negative integer.
(iv) – 385 is larger, because smaller negative number is larger than bigger negative number.
3. Write the following integers in increasing order:
– 18, 16, 0, – 5, 8, – 36, – 1, 1
Solution:
In increasing order:- – 36, – 18, – 5, – 1, 0, 1, 8, 16
4. Find the value of
(i) 9 – |– 6 |
(ii) 6 + | – 4 |
(iii) – 8 – | – 3 |
Solution:
(i) 9 – |–6|
= 9 – (6)
= 9 – 6
= 3
(ii) 6 + |–4|
= 6 + (4)
= 6 + 4
= 10
(iii) –8 – |–3|
= –8 – 3
= –11
5. Write four integers less than – 6 and four integers greater than – 6.
Solution:
Four integers less than –6 are –7, –8, –9 and –10.
Four integers greater than –6 are –5, –4, –3 and –2.
6: Evaluate:
(i) 8 + ( –16)
(ii) ( – 5) + ( – 6)
(iii) (– 6) x ( – 8)
(iv) ( – 36) ÷ 6
(v) 30 – (– 50)
(vi) (– 40) ÷ (– 10)
(vii) 8 × (–5)
(viii) (– 30) – 15
Solution:
(i) 8 + (–16)
= 8 – 16
= –8
(ii) (–5) + (–6)
= –5 – 6
= –11
(iii) (–6) × (–8)
= (6 × 8)
= 48
7. The sum of two integers is – 12. If one of them is 34, find the other.
Solution:
Let the other integer be x.
∴ 34 + x = –12
or, x = –12 – 34
or, x = –46
Therefore, the other integer is –46.
8. Simplify:
(i) (–24) × (68) + (–24) × 32
(ii) (–9) × 18 – (–9) × 8
(iii) (–147) ÷ (–21)
(iv) 16 ÷ (–1)
Solution:
(i) (–24) × (68) + (–24) × 32
= –(24) × (68+32)
= –24 × 100
= –2400
(ii) (–9) × 18 – (–9) × 8
= (–9 ) × [18 – 8]
= –9 × 10
= –90
(iii) (–147) ÷ (–21)
= `frac{–147}{–21}`
= 7
(iv) 16 ÷ (–1)
= `frac{16}{–1}`
= –16
9. The successor of – 89 is
(a) – 90
(b) – 88
(c) 90
(d) 88
Solution: Correct potion is (b) −88
The successor of −89 is (−89 + 1) = −88
10. The predecessor of – 99 is
(a) – 98
(b) – 100
(c) 98
(d) 100
Solution: Correct potion is (d) −100
The successor of −99 is (−99 − 1) = − 100
11. Additive inverse of – 23 is
(a) `frac\{– 1}{23}`
(b) `frac\{1}{23}`
(c) 23
(d) – 23
Solution: Correct potion is (c) 23
Additive inverse of a number added to the number gives 0.
−23 + 23 = 0
Hence, 23 is the additive inverse of −23.
12. If ( – 13 + 6) ▭ – 25 – ( – 9), then the correct symbol in the place holder is
(a) <
(b) >
(c) =
(d) none of these
Solution: Correct potion is (b) >
L.H.S. = (−13 + 6)
= −7
R.H.S. = −25 − (−9)
= −25 + 9
= −16
−7 > −16
L.H.S. > R.H.S.
13. ? + ( – 8) = 12
(a) – 4
(b) ( – 20)
(c) 20
(d) 4
Solution: Correct potion is (c) 20
x + (−8) = 12
=> x − 8 = 12
=> x = 12 + 8
=> x = 20
14. The integer which is 5 more that ( – 7) is
(a) – 12
(b) 12
(c) – 2
(d) 2
Solution: Correct potion is (c) – 2
5 more than (−7) means 5 added to (−7).
5 + (−7)
= 5 − 7
= −2
15. What should be added to 16 to get ( – 31) ?
(a) 15
(b) – 15
(c) 47
(d) – 47
Solution: Correct potion is (d) −47
Let the number to be added to 16 be x.
x + 16 = (−31)
=> x = (−31)−16
=> x = −47
16. When 34 is subtracted from – 36, we get
(a) 2
(b) – 2
(c) 70
(d) – 70
Solution: Correct potion is (d) −70
−36 − 34
= −70
17. Fill in the blanks
(i) – 23 – (?) = 15
(ii) The largest negative integer is …….
(iii) the smallest positive integer is ……(iv) (– 8) + (– 6) – (– 3) = ……
(v) The predecessor of – 200 is ……..
Solution:
(i) Let the required number be x.
−23 − x = 15
=> −23 = 15 + x
=> 15 + x = −23
=> x = −15 −23
=> x = −38
(ii) The largest negative integer is − 1.
(iii) The smallest positive integer is 1.
(iv) (−8) + (−6) − (−3)
= (−8) + (−6) +3
= −8 − 6 + 3
= −11
(v) The predecessor of −200:
(−200 − 1)
= −201
18. Write "T' for true and 'F' for false in each of the following:
(i) 0 is neither positive nor negative.
(ii) – ( – 36) – 1 = – 37.
(iii) On the number line – 10 lies to the right of – 6
(iv) 0 is an integer.
(v) – | – 15 | – 15
(vi) | – 40 | + 40 = 0.
Solution:
(i) T
(ii) F
−(−36) − 1
= 36 − 1
= 35
(iii) F, This is because −10 is less than −6.
(iv) T
(v) T
−|−15|
= −(15)
= −15
(vi) F
|−40| + 40
= 40 + 40
= 80
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