[2 Mark Question Answer]

Question 12 Marks

The sum S of first n even natural numbers is given by the relation $S = n(n + 1)$. Find $n$, if the sum is $420.$

Answer

$S = n(n + 1)$
Given, $S = 420$
$n(n + 1) = 420$
$n^2 + n – 420 = 0$
$n^2 + 21n – 20n – 420 = 0$
$n(n + 21) – 20(n + 21) = 0$
$(n + 21) (n – 20) = 0$
$n = -21, 20$
Since, $n$ cannot be negative.
Hence, $n = 20.$

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Question 22 Marks

The ages of two sisters are $11$ years and $14$ years. In how many years’ time will the product of their ages be $304?$

Answer

The ages of two sisters are $11$ years and $14$ years.
Let in x number of years the product of their ages be $304.$
$\therefore (11 + x)(14 + x) = 304$
$154+11 x+14 x+x^2=304$
$X^2+25 x-150=0$
$(x+30)(x-5)=0$
$X=-30,5$
But, the number of years cannot be negative. So, $x = 5.$
Hence, the required number of years is $5$ years.

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Question 32 Marks

The sum S of n successive odd numbers starting from 3 is given by the relation n(n + 2). Determine n, if the sum is 168.

Answer

From the given information, we have:
n(n + 2) = 168
n² + 2n – 168 = 0
n² + 14n – 12n – 168 = 0
n(n + 14) – 12(n + 14) = 0
(n + 14) (n – 12) = 0
n = -14, 12
But, n cannot be negative.
Therefore, n = 12.

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Question 42 Marks

Find two consecutive positive odd numbers, the sum of whose squares is $74.$

Answer

Let the consecutive positive odd numbers be x and $x + 2.$
From the given information,
$x^2 + (x + 2)^2 = 74$
$2x^2 + 4x + 4 = 74$
$2x^2 + 4x – 70 = 0$
$x^2 + 2x – 35 = 0$
$(x + 7)(x – 5) = 0$
$x = -7, 5$
Since, the numbers are positive, so,$ x = 5.$
Thus, the numbers are $5$ and $7.$

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Question 52 Marks

The sum of the squares of two consecutive positive even numbers is $52$. Find the numbers.

Answer

Let the consecutive positive even numbers be x and $x + 2.$
From the given information,
$x^2 + (x + 2)^2 = 52$
$2x^2 + 4x + 4 = 52$
$2x^2 + 4x – 48 = 0$
$x^2 + 2x – 24 = 0$
$(x + 6) (x – 4) = 0$
$x = -6, 4$
Since, the numbers are positive, so $x = 4.$
Thus, the numbers are $4$ and $6.$

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Question 62 Marks

The sum of the squares of two consecutive natural numbers is $41$. Find the numbers.

Answer

Let the numbers be $x$ and $x + 1.$
From the given information,
$x^2 + (x + 1)^2 = 41$
$2x^2 + 2x + 1 – 41 = 0$
$x^2 + x – 20 = 0$
$(x + 5) (x – 4) = 0$
$x = -5, 4$
But, $-5$ is not a natural number. So, $x = 4.$
Thus, the numbers are $4$ and $5.$

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Question 72 Marks

The product of two consecutive integers is $56$. Find the integers.

Answer

Let the two consecutive integers be $x$ and $x + 1.$
From the given information,
$x(x + 1) = 56$
$x^2 + x – 56 = 0$
$(x + 8) (x – 7) = 0$
$x = -8$ or $7$
Thus, the required integers are $– 8$ and $-7; 7$ and $8.$

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Question 82 Marks

Out of three consecutive positive integers, the middle number is $p.$ If three times the square of the largest is greater than the sum of the squares of the other two numbers by $67$; calculate the value of $p.$

Answer

Let the numbers be $p – 1, p$ and $p + 1.$
From the given information,
$3(p + 1)^2 = (p – 1)^2 + p^2 + 67$
$3p^2 + 6p + 3 = p^2 + 1 – 2p + p^2 + 67$
$p^2 + 8p – 65 = 0$
$(p + 13)(p – 5) = 0$
$p = -13, 5$
Since, the numbers are positive so p cannot be equal to $-13.$
Thus, $p = 5.$

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Question 92 Marks

Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by $60.$
Assume the middle number to be x and form a quadratic equation satisfying the above statement. Hence; find the three numbers.

Answer

Let the numbers be $x – 1, x$ and $x + 1.$
From the given information,
$x^2 = (x + 1)^2 – (x – 1)^2 + 60$
$x^2 = x^2 + 1 + 2x – x^2 – 1 + 2x + 60$
$x^2 = 4x + 60$
$x^2 – 4x – 60 = 0$
$(x – 10) (x + 6) = 0$
$x = 10, -6$
Since, x is a natural number, so $x = 10.$
Thus, the three numbers are $9, 10$ and $11.$

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