Question

Maruki and Suzuki entered into a partnership 5 months ago. The ratio of profit claimed by Maruki and Suzuki is 6: 17. If Suzuki had just started business 12 months ago with Rs. 1275, what is the amount contributed by Maruti?
(a) Rs. 980
(b) Rs. 1080
(c) Rs. 1200
(d) Rs. 998

Answer 1

Hint: Here, we need to find the amount contributed by Maruti 5 months ago. We will assume the amount contributed by Maruti 5 months ago to be Rs. \[x\]. The ratio of profits is equal to the ratio of the products of investment and duration of investment of Maruti and Suzuki respectively. We will use this to form a linear equation in one variable in terms of \[x\]. We will solve this equation to obtain the value of \[x\], and hence, the amount contributed by Maruti 5 months ago.

Complete step-by-step answer:
The ratio of profits is equal to the ratio of the products of investment and duration of investment of Maruti and Suzuki respectively.
Let the amount contributed by Maruti 5 months ago be Rs. \[x\].
The duration of investment by Maruti is 5 months.
Therefore, we get
Investment amount \[ \times \] Duration of investment \[ = x \times 5\]
Multiplying the terms, we get
\[ \Rightarrow \] Investment amount \[ \times \] Duration of investment \[ = 5x\]
The amount contributed by Suzuki 12 months ago is Rs. 1275.
The duration of investment by Suzuki is 12 months.
Therefore, we get
\[ \Rightarrow \] Investment amount \[ \times \] Duration of investment \[ = 1275 \times 12\]
Multiplying the terms, we get
\[ \Rightarrow \] Investment amount \[ \times \] Duration of investment \[ = 15300\]
The ratio of profits is equal to the ratio of the products of investment and duration of investment of Maruti and Suzuki respectively.
It is given that the ratio of profits is 6: 17.
Therefore, we get the equation
\[\dfrac{{5x}}{{15300}} = \dfrac{6}{{17}}\]
This is a linear equation in one variable. We will solve this to find the value of \[x\].
Multiplying both sides of the equation by 17, we get
\[\begin{array}{l} \Rightarrow \dfrac{{5x}}{{15300}} \times 17 = \dfrac{6}{{17}} \times 17\\ \Rightarrow \dfrac{{5x}}{{900}} = 6\end{array}\]
Multiplying both sides of the equation by 900, we get
\[\begin{array}{l} \Rightarrow \dfrac{{5x}}{{900}} \times 900 = 6 \times 900\\ \Rightarrow 5x = 5400\end{array}\]
Dividing both sides of the equation by 5, we get
\[ \Rightarrow x = 1080\]
\[\therefore \] We get the amount contributed by Maruti 5 months ago as Rs. 1,080.
Thus, the correct option is option (b).

Note: We have formed a linear equation in one variable in terms of \[x\] in the solution. A linear equation in one variable is an equation of the form \[ax + b = 0\], where \[a\] and \[b\] are integers. A linear equation of the form \[ax + b = 0\] has only one solution. Linear equation has the highest degree of a variable as 1. There are different types of linear equations that are categorised based on the number of distinct variables in the expression, such as linear equation in one variable, linear equation in two variables and so on.