\(\left ( 64 - 12 \right )^{2} + 4 × 64 × 12 = ?\)
Answer: D
Given statement is like \(\left ( a - b \right )^{2} + 4ab\) where \(a=64\) and \(b=12\)
\(\left ( a - b \right )^{2} + 4ab\)
\(=\left (a^{2} - 2ab+b^{2}\right )+4ab\)
\(=a^{2}+2ab+b^{2}\)
\(=\left ( a + b \right )^{2}\)
Hence,
\(\left ( 64 - 12 \right )^{2} + 4 × 64 × 12\)
\(=\left ( 64 + 12 \right )^{2}\)
\(=76^{2}\)
\(=5776\)
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\((1000)^{9} \div 10^{24} = ?\)
Answer: B
\(=\dfrac{(1000)^{9}}{(10)^{24}}\\ =\dfrac{(10^{3})^{9}}{(10)^{24}}\\ =\dfrac{(10)^{27}}{(10)^{24}}\\ =10^{(27-24)}\\ =10^{3}\\ = 1000\)
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Minimum number of square marbles required to tile a floor of length 5 metres 78 cm and width 3 metres 74 cm is ?
Answer: B
The marbles used to tile the floor are square marbles.
Therefore, the length of the marble=width of the marble.the length of the marble=width of the marble.
As we have to use whole number of marbles, the side of the square should a factor of both 5 m 78 cm and 3m 74. And it should be the highest factor of 5 m 78 cm and 3m 74.
5 m 78 cm = 578 cm and 3 m 74 cm = 374 cm.
The HCF of 578 and 374 = 34.
Hence, the side of the square is 34.
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How many five digit multiples of 11 are there, if the five digits are 3, 4, 5, 6 and 7?
Repetition of digits is now allowed?
Answer: C
A number is divisible by 11, if the difference between the sum of digits at even places and odd places is either 0 or divisible by 11.
Numbers 5,3,6,4,7 is a multiple of 11
∵(5+6+7)−(3+4)=11, which is a multiple of 11
The number at odd places 5,6,7 can be arranged in 3! ways and 3,4 can be arranged in 2! ways.
Therefore, the total number of ways of such numbers =3!×2!=12
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If n=1+x, where x is a product of four consecutive positive integers, then which of the following is true?
A. n is odd
B. n is prime
C. n is a perfect square
Answer: A
Since xx is the product of four consecutive integers, it is always divisible by 4, i,.e., it is always even. So, 1+x is always odd.
n=1+x
x=(y−1)(y)(y+1)(y+2)=y(y2−1)(y+2)=(y3−y)(y+2)=y4+2y3−y2−2y
⇒1+x=y4+2y3−y2−2y+1=y4+y2+1+2y3−2y2−2y=(y2+y−1)2
So, 1+x is a perfect square as we can see. Hence, option A is the correct choice.
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Which one of the following can't be the square of natural number ?
Answer: D
The square of a natural number nerver ends in 2.
Thus, 143642 is not the square of natural number.
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If we write down all the natural numbers from 259 to 492 side by side get a very large natural number 259260261....491492259260261....491492.
How many 8's will be used to write this large natural number?
Answer: D
58, there are 200 natural numbers so there will be 2×20=40 8's
From 459 to 492 we have 13 more 8's and so answer is 40+13= 53
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A boy writes all the numbers from 100 to 999. The number of zeroes that he uses is 'aa', the number of 5's that he uses is 'bb' and the number of 8's he uses is 'cc'.
What is the value of b+c−ab+c−a?
Answer: B
We can see by symmetry b=cb=c and hence all we need to calculate bb and aa
b=280 and a=180b=280 and a=180
⇒ 2b−a=2b−a= 380
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If aa and bb are positive integers, and x=2×3×7×ax=2×3×7×a , and y=2×2×8×by=2×2×8×b , and the values of both xx and yy lie between 120 and 130 (not including the two), then a–b=a–b=
Answer: B
We are given that x=2×3×7×a=42ax=2×3×7×a=42a and y=2×2×8×b=32by=2×2×8×b=32b
We are given that the values of both xx and yy lie between 120 and 130 (not including the two).
The only multiple of 42 in this range is 42×3=12642×3=126.
Hence, x=126x=126 and a=3.a=3.
The only multiple of 32 in this range is 32×4=12832×4=128.
Hence, y=128y=128 and b=4b=4.
Hence, a−b=3−4=a−b=3−4= -1
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What is the remainder left after dividing 1!+2!+3!.....+100! by 7?
Answer: B
7!+8!+9!.......+100! is completely divisible by 7.
Now, 1!+2!+3!....+6!=873
When 873 is divided by 7 it leaves a remainder =5.
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