Number System Questions and Answers

Q: If x is a whole number, then \(x^{2}(x^{2}-1)\) is always divisible by:

Factorise the expression.\[ x^{2}(x^{2}-1) = (x-1)\,x\,x\,(x+1) \]\( (x-1)x(x+1) \) is a product of three consecutive integers, so it is divisible by both 2 and 3, hence by 6; the extra factor of \( x \) supplies a second factor of 2 whenever \( x \) is even, and when \( x \) is odd both neighbours are even.Either way the product is divisible by 12. It is not always divisible by 24: \( x = 2 \) gives 12.

Q: \((1000)^{9} \div 10^{24} = ?\)

Write the numerator as a power of ten.\[ (1000)^{9} = (10^{3})^{9} = 10^{27} \]\[ \frac{10^{27}}{10^{24}} = 10^{3} = 1000 \]

Q: Which one of the following can't be the square of natural number ?

A perfect square never ends in 2, 3, 7 or 8.\[ 143642 \text{ ends in } 2 \]so it cannot be a square. The last digit of a square is always 0, 1, 4, 5, 6 or 9 - a single glance settles this without any arithmetic.

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81.

The sum of the two digits of a number is 10. If the number is subtracted from the number obtained by reversing its digits, the result is 54. Find the number?

A.34
B.28
C.12
D.17

Answer: B

Any two digit number can be written as \((10P + Q)\), where \(P\) is the digit in the tens place and \(Q\) is the digit in the units place.
\(P + Q = 10\) ----- (1)
\((10Q + P) - (10P + Q) = 54\)
\(9(Q - P) = 54\)
\((Q - P) = 6\) ----- (2)
Solve (1) and (2) \(P = 2\) and \(Q = 8\)
The required number is \(= 28 \)

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82.

The sum of three consecutive even numbers is 42. Find the middle number of the three?

A.14
B.16
C.18
D.24

Answer: A

Three consecutive even numbers \((2P - 2), 2P, (2P + 2)\)
\((2P - 2) + 2P + (2P + 2) = 42\)
\(6P = 42 => P = 7\)
The middle number is: \(2P = 14\)

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83.

The numerator of a certain fraction is 8 less than the denominator. If 3 is added to the numerator and 3 is subtracted from the denominator, the fraction becomes \(\dfrac{3}{4}\). Find the original fraction?

A.\(\dfrac{2}{5}\)
B.\(\dfrac{7}{9}\)
C.\(\dfrac{3}{11}\)
D.\(\dfrac{8}{5}\)

Answer: C

The denominator be \(P\), the numerator will be \((P - 8)\)
The fraction will be \(\dfrac{(P - 8)}{P}\)
Adding 3 to the numerator and subtracting 3 from the denominator,
\(\dfrac{(P - 8 + 3)}{P - 3}= \dfrac{3}{4}\)
\(\dfrac{(P - 5)}{P - 3}= \dfrac{3}{4}\)
\(P = 20 - 9 \Rightarrow P = 11\)
The fraction is: \(\dfrac{3}{11}\)

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84.

The sum of the present ages of two persons A and B is 60. If the age of A is twice that of B, find the sum of their ages 5 years hence?

A.50
B.60
C.70
D.80

Answer: C

\(A + B = 60, A = 2B\)
\(2B + B = 60 => B = 20\) then \(A = 40\)
5 years, their ages will be 45 and 25
Sum of their ages \(= 45 + 25 = 70\)

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85.

If x is a whole number, then \(x^{2}(x^{2}-1)\) is always divisible by:

A.12
B.24
C.12-x
D.multiple of 12

Answer: A

Factorise the expression.

\[ x^{2}(x^{2}-1) = (x-1)\,x\,x\,(x+1) \]

\( (x-1)x(x+1) \) is a product of three consecutive integers, so it is divisible by both 2 and 3, hence by 6; the extra factor of \( x \) supplies a second factor of 2 whenever \( x \) is even, and when \( x \) is odd both neighbours are even.

Either way the product is divisible by 12. It is not always divisible by 24: \( x = 2 \) gives 12.

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86.

A number when divided by 6 leaves a remainder 3. When the square of the same number is divided by 6, the remainder is:

Answer: C

Let \(x=6q+3\). Then,
\(x^{2} = (6q+3)^{2} \\= 36q^{2}+36q+9 \\= 6(6q^{2}+6q+1)+3\)
So, when \(2n\) is divided by \(4\), remainder \(=3\)

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87.

\(\left \{(476 + 424)^{2} - 4 \times 476 \times 424 \right \} = ?\)

A.2906
B.3116
C.2704
D.2904
E.None of these

Answer: C

Given Expression = \([(a + b)^{2} - 4ab]\), where \(a = 476\) and \(b = 424\)
\(=[(476 + 424)^{2} - 4 \times 476 \times 424]\)
\(= [(900)^{2} - 807296]\)
\(= 810000 - 807296\)
\(= 2704\)

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88.

\((1000)^{9} \div 10^{24} = ?\)

A.10000
B.1000
C.100
D.10
E.None of these

Answer: B

Write the numerator as a power of ten.

\[ (1000)^{9} = (10^{3})^{9} = 10^{27} \]

\[ \frac{10^{27}}{10^{24}} = 10^{3} = 1000 \]

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89.

Which one of the following can't be the square of natural number ?

A.30976
B.75625
C.28561
D.143642
E.None of these

Answer: D

A perfect square never ends in 2, 3, 7 or 8.

\[ 143642 \text{ ends in } 2 \]

so it cannot be a square. The last digit of a square is always 0, 1, 4, 5, 6 or 9 - a single glance settles this without any arithmetic.

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90.

How many of the following numbers are divisible by 132 ?
264, 396, 462, 792, 968, 2178, 5184, 6336

Answer: A

\(132 = 4 \times 3 \times 11\)
So, if the number divisible by all the three number 4, 3 and 11, then the number is divisible by 132 also.
\(264 \Rightarrow 11,3,4\)
\(396 \Rightarrow 11,3,4\)
\(462 \Rightarrow 11,3\)
\(792 \Rightarrow 11,3,4\)
\(968 \Rightarrow 11,4\)
\(2178 \Rightarrow 11,3\)
\(5184 \Rightarrow 3,4\)
\(6336 \Rightarrow 11,3,4\)
Therefore the following numbers are divisible by 132 : 264, 396, 792 and 6336.
Required number of number = 4.

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