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Simple interest

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Simple interest

by j_shreyans » Fri Sep 19, 2014 8:51 am
Samantha invests i1 dollars in bond X, which pays r1 percent simple interest annually, and she invests i2 dollars in bond Y, which pays r2 percent simple interest annually. After one year, will she have earned more interest, in dollars, from bond X than from bond Y?

(1) (r1)^2>(r2)^2

(2) The ratio of i1 to i2 is larger than the ratio of r1 to r2.



Guys ,

Can't we do this question by letting i1 and i2 with certain amount and r1 and r2 with certain interest.
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Source: — Data Sufficiency |
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by nipunranjan » Mon Nov 03, 2014 11:21 am
Interest from bond X : i1*r1/100
Interest from bond Y : i2*r2/100
we need to know if i1*r1 > i2*r2 ?

(1) r1^2 > r2^2 ... Not sufficient as it doesn't tell us anything about the actual amounts invested
(2) i1/i2 > r1/r2 which implies i1*r2 > i2*r1 ... doesnt help us solve our inequality
(1+2) r1^2 > r2^2
So r1 > r2 as interest rate on bond cannot be negative
i1*r1 > i1*r2
i2*r1 > i2*r2
Combining with condition known from Statement 2, we get i1*r1 > i2*r2

Answer is (C)


Putting number in such questions can be tricky because there is a big range of values. You can never be sure if you just test by putting one value. So, my suggestion would be that you solve such questions using algebra and not putting values.
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by Mathsbuddy » Wed Nov 19, 2014 8:46 am
(1) No information to compare i1 and i2: INSUFFICIENT

(2) i1/i2 > r1/r2, so 3 known variables required to know all 4
Also X gives i1 x (100 + r1) and Y gives i2 x (100 + r2), so 3 variables need to be known to compare X and Y
If we substitute, 1 variable will still be at large, so NOT SUFFICIENT

Combined, the one missing variable from (2) can be replaced using Statement (1) = SUFFICIENT
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by Mathsbuddy » Wed Nov 19, 2014 9:01 am
If you imagine plotting 2 simple straight line graphs, the y intercepts are the initial starting points i1 and i2. Then the gradients (slopes) are r1 and r2. So by imagining the lines, you can see that they might either diverge or converge, and might even intercept.

As we don't know in Statement 1 which line's y-intercept is higher, it is clearly NOT SUFFICIENT as the outcome is a product of the startpoint.

Statement 2 is trickier to visualise. Nonetheless, the ratio of i1/i2 alone bears no reflection on the outcome of the DIFFERENCES (INTERESTS earned).

Combined, we can sketch a simple picture: y1 > y2 AND Slope1 > Slope2, so VERY SUFFICIENT!
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by Mathsbuddy » Wed Nov 19, 2014 9:38 am
Another look at using straight-line graphs (as explained below) gives me a different answer:

Using y = mx + c
dy/dx = m
So dy = mdx
Hence yA-y1 = m1 x 1year
and yB-y2 = m2 x 1year

So we need to know if m1 or m2 is bigger

Statement 1: only gives a comparison of each C (y1,y2) value; INSUFFICIENT
Statement 2: y1/y2 > m1/m2, so m2 > (m1 x y2/y1); no data for y2 and y1 = INSUFFICIENT

Combined: y1>y2 AND y1/y2 > m1/m2, means that m1/m2 could be greater or less than 1: INSUFFICIENT!



Mathsbuddy wrote:If you imagine plotting 2 simple straight line graphs, the y intercepts are the initial starting points i1 and i2. Then the gradients (slopes) are r1 and r2. So by imagining the lines, you can see that they might either diverge or converge, and might even intercept.

As we don't know in Statement 1 which line's y-intercept is higher, it is clearly NOT SUFFICIENT as the outcome is a product of the startpoint.

Statement 2 is trickier to visualise. Nonetheless, the ratio of i1/i2 alone bears no reflection on the outcome of the DIFFERENCES (INTERESTS earned).

Combined, we can sketch a simple picture: y1 > y2 AND Slope1 > Slope2, so VERY SUFFICIENT!
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