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7th Grade > Mathematics

RATIONAL NUMBERS MCQs

Total Questions : 120 | Page 1 of 12 pages
Question 1. Compare the rational numbers and choose the correct option.
3926 ___64
  1.    >
  2.    =
  3.    None of the above
 Discuss Question
Answer: Option C. -> =
:
C
Reducing 3926 to its lowest form , we get,
39÷1326÷13 =32( 13 is the common factor)
Multiplying numerator and denominator by 2 we get
3×22×2 = 64
Hence,3926 is also equal to 64
So, both the fraction are equal
Hence, 3926=64
Question 2. (a) How many right angles are there in a cubical room? 
(b) What should be subtracted from (512+32) to get 2?
[3 MARKS]
 Discuss Question

:
(a) Answer: 1 Mark
(b) Steps: 1 Mark
Result: 1 Mark
(a) Each corner has three mutually perpendicular edges, hence forming three right angles in each corner. A room has 8 corners. Hence, there are 24 right angles in the room.
(b) Let the number to be subtracted be x.
512+32x=2
5+1812x=2
2312x=2
x=23122
x=232412
x=112
x=112
Question 3. The sum of the rational numbers 819 and 457 is _____.
  1.    −2857
  2.    −1257
  3.    −4247
  4.    −3247
 Discuss Question
Answer: Option A. -> −2857
:
A
819+457=(8×3)(19×3)+457
=2457+457
=2857
Question 4. Write a rational number equal to 47 with 63 as the denominator.
  1.    1263
  2.    2063
  3.    3663
  4.    6063
 Discuss Question
Answer: Option C. -> 3663
:
C
Consider given rational number47.
To make the denominator as 63, we have to multiply it with 9.
Multiplying both numerator and denominator by 9, we get
4×97×9=3663
Therefore, 3663 is anequivalent rational number of
47 with denominator equal to 63.
Question 5. Find the greatest rational  number among the given options.
  1.    −49
  2.    −512
  3.    −718
  4.    −23
 Discuss Question
Answer: Option C. -> −718
:
C
Note that LCM of 9, 12, 18, 3 is 36.
Using LCM to convert each of the given rational numbers to equivalent forms,
49 = 1636
512 = 1536
718 = 1436
23 = 2436
Since the denominators are equal, we now compare the rational numbers using its numerators.
Thus, 2436<1636<1536<1436
i.e., 23<49<512<718
718 is the greatest among the given options.
Question 6. Simplified form of 10812 is 8.
  1.    True
  2.    False
  3.    67
  4.    −67
 Discuss Question
Answer: Option B. -> False
:
B
12 × 9 = 108
10812=(12×9)12
= 9
Question 7. Find the result obained by the addition of
23  to the reciprocal of  23.
  1.    56
  2.    −56
  3.    67
  4.    −67
 Discuss Question
Answer: Option B. -> −56
:
B
Reciprocal of23is32.
2332=4696=(49)6=56
[LCM of 2 and 3 is 6]
Question 8. Add 213 to the reciprocal of 2612.
  1.    57
  2.    −56
  3.    67
  4.    −413
 Discuss Question
Answer: Option D. -> −413
:
D
Reciprocal of 2612 is 1226.
213+1226 [Taking LCM]
=4261226=41226=826=413
Question 9. (a) Give the additive inverse of: 
314,519
(b) Give the multiplicative inverse of 34
[2 MARKS]
 
 Discuss Question

:
Each part: 1 Mark
(a) Additive inverse of314=314
Additive inverse of 519=519
(b)Multiplicative inverse of 34=43
Question 10. Find the sum of: 
(a) 311,14 and 58
(b) 59,712 and  1118
[2 MARKS]
 Discuss Question

:
Each part: 1 Mark
(a) 311+14+58
3×8881×2288+5×1188
2422+5588
46+5588=988
(b) 59+712+1118
5×4367×336+11×236
2021+2236=1936

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