Mathematics Questions and Answers – Finding the Square of a Number

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Finding the Square of a Number”.

1. What can be general formula to find square of any number?
a) (a±b)2
b) a2
c) a
d) b
View Answer

Answer: a
Explanation: The general formula of finding square of any number is (a±b)2. Option a2 cannot be correct because if we need to find square of number such as 23 the method using the general formula (a±b)2 would be more easier. Hence the correct option is (a±b)2.
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2. How can one represent the square of 103?
a) (100+3)2
b) (100-3)2
c) (100+3)
d) (100-3)
View Answer

Answer: a
Explanation: The number 103 can be represented in form 100+3, this will help us to calculate the square of the number. This method would help us eliminate the traditional method of multiplying the number with itself.

3. 892= ____
a) (90-1)2
b) 902
c) 12
d) 92
View Answer

Answer: a
Explanation: If we have to find the square of the number which is near 0 we use the second variation i.e. (90-1)2. This would help us and make the calculations easier.
Therefore (90-1)2= 902-2×90×1+12
Therefore (90-1)2= 8100-180+1
Therefore (90-1)2=7921.
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4. If a student uses (a-b)2 to calculate the square of a number, then the number is _______
a) 121
b) 112
c) 199
d) 201
View Answer

Answer: c
Explanation: A student uses this variant of the formula for the numbers which are near 0. Here there is only one number which is close to 0 and hence 199 is the correct answer. The other options also can be solved using this method but that would be complicated.

5. Calculate the square of 201.
a) 40401
b) 39393
c) 40426
d) 41042
View Answer

Answer: a
Explanation: In the multiple choice question student can eliminate three options by mere observation. The student should notice that option 41042 contains 2 in its unit place so it cannot be a square of number ending with 1. Similarly, the other two options 39393 & 40426 ending with 3 and 6 whereas square of numbers ending with 1 have only 1 in its unit place.
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6. 342=_____
a) 1156
b) 1256
c) 1356
d)1456
View Answer

Answer: a
Explanation: In order to calculate the square of 34 we use the formula,
(a+b)2, where we consider a=30 and b=4
Therefore (30+4)2 = 302+2×30×4+42
Therefore (30+4)2 = 900+240+16
Therefore (30+4)2 = 1156.

7. How many numbers lie between the squares of 12 and 13?
a) 25
b) 26
c) 24
d) 28
View Answer

Answer: a
Explanation: If we need to find the number of non-square numbers, we can use the formula (2n+1).
Here when we apply this formula we get, (2×12+1)=25. Hence the answer would be 25 and the other options would be incorrect.
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8. Express the square of 11 in terms of sum of odd numbers.
a) 1+3+5+7+9+11+13+15+17+19+21
b) 1+2+3+5+7+9+11+13+15+17+19
c) 1+3+5+7+9+11+13+15+17+21+23
d) 1+2+3+4+5+6+7+8+9+10+11+12
View Answer

Answer: a
Explanation: The correct option is the one with the first 11 odd numbers as the sum of first n odd numbers gives n2. Hence the correct answer is the one with the first 11 odd number.

9. Find the square of 25.
a) 625
b) 525
c) 635
d) 615
View Answer

Answer: a
Explanation: There is a beautiful pattern followed by all the numbers ending with 5. We can get the square of any number ending with 5 without actually calculating it. All the numbers ending with 5 shows 25 at the end of their squares, and the other places can be filled by multiplying the next consecutive number. For example: 252 = (2×3)25 i.e.625 (the step shown is just for understanding).
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10. Find the square of 225.
a) 505625
b) 49625
c) 50625
d) 51625
View Answer

Answer: c
Explanation: There is a beautiful pattern followed by all the numbers ending with 5. We can get the square of any number ending with 5 without actually calculating it. All the numbers ending with 5 shows 25 at the end of their squares, and the other places can be filled by multiplying the next consecutive number. For example: 2252 = (22×23)25 i.e.50625 (the step shown is just for understanding).

Sanfoundry Global Education & Learning Series – Mathematics – Class 8.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter