JIPMAT 2023 (QA) - <p>Given below are two statements:</p> <p>Statement (I): The difference, the sum, and the product of two numbers are in the ratio 1: 5: 12. The product of the two numbers is 18.</p><p>Statement (II): The last digit in the decimal representation of 2<sup>91</sup> is 6.</p> <p>In light of the above statements, choose the most appropriate answer from the options given below.</p> | PYQs + Solutions | AfterBoards
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JIPMAT 2023 (QA) PYQs

JIPMAT 2023

Number System
>
Miscellaneous

Easy

Given below are two statements:

Statement (I): The difference, the sum, and the product of two numbers are in the ratio 1: 5: 12. The product of the two numbers is 18.

Statement (II): The last digit in the decimal representation of 291 is 6.

In light of the above statements, choose the most appropriate answer from the options given below.

Correct Option: 2
Let me solve this step by step:
Statement I: \newline Let two numbers be x and y \newline Given their difference:sum:product =1:5:12= 1:5:12 \newline And product =18= 18
So, xy:(x+y):xy=1:5:12|x-y| : (x+y) : xy = 1:5:12 \newline Since xy=18xy = 18, and using ratios: \newline xy=1812=32|x-y| = \frac{18}{12} = \frac{3}{2}
x+y=18×512=152x+y = \frac{18×5}{12} = \frac{15}{2}
From these equations: \newline When x>yx > y: \newline xy=32x - y = \frac{3}{2} and x+y=152x + y = \frac{15}{2} \newline Solving gives x=4.5,y=3x = 4.5, y = 3
Numbers are not integers, Statement I is false.
Statement II: \newline To find last digit of 2912^{91}: \newline Pattern of last digit of powers of 2:2,4,8,6,2,4,8,6...2: 2,4,8,6,2,4,8,6... \newline 9191 mod 44 = 33 \newline So last digit is 88, not 66
Therefore Statement II is also false.

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