Question

Find the value of 48: 122::168:?
(a) 284
(b) 286
(c) 288
(d) 290

Answer 1

Hint: To solve this question we will try to make a pattern for all the set of given numbers and then follow the same pattern to find the number of question marks. To do, we will observe that the first number ‘48’ can be obtained by adding or subtracting the square or cube of some numbers. By this method, we will make the pattern to solve.

Complete step by step answer:
We will consider 48 first. We observe that,
\[48={{7}^{2}}-1\]
As, \[{{7}^{2}}=49\] and \[{{7}^{2}}-1=49-1=48.\]
Therefore,
\[48={{7}^{2}}-1\]
Now we will try to form a similar pattern on the second number of 48: 122:: 168: ? that is to 122. We observe that,
\[122={{\left( 11 \right)}^{2}}+1\]
\[\Rightarrow 122=121+1\]
\[\Rightarrow 122=122\]
Now, because we need a pattern, so we will relate \[{{7}^{2}}-1\] and \[{{11}^{2}}+1.\] \[{{11}^{2}}+1\] can be rewritten as \[{{\left( 7+4 \right)}^{2}}+1.\]
Therefore, we have,
\[48={{7}^{2}}-1.......\left( i \right)\]
\[122={{\left( 7+4 \right)}^{2}}+1.......\left( ii \right)\]
So we can observe that we are getting a certain pattern. Now consider that the third number is 168. We can write 168 as
\[168={{13}^{2}}-1\]
The pattern followed in (i) and (ii) will be followed in the next two numbers. Therefore, the last number will be
\[\Rightarrow {{\left( 13+4 \right)}^{2}}+1\]
\[\Rightarrow {{\left( 17 \right)}^{2}}+1\]
\[\Rightarrow 290\]
Therefore the last number is 290.
Hence, the right option is (d).

Note:
The number is of the form a:b::c:d is said to be categorized into two groups a, b and c, d. The property which is followed in a and b is similarly followed in c and d. Therefore for solving the above property or pattern in which is followed in 48:122 needs to be followed in 168: ?.