DS: Ratio & Proportion – Level 3
1.
A club has men and women. If 5 women leave, what is the new ratio of men to women?
(1) Originally, the ratio of men to women was 2:3.
(2) Originally, there were 15 men in the club.
(1) Originally, the ratio of men to women was 2:3.
(2) Originally, there were 15 men in the club.
Solution (C):
To find the new ratio \(\frac{M}{W-5}\), we need specific numbers.
Analyze Statement (1): \(M:W = 2:3\). This allows for (20M, 30W) -> New 20:25, or (200M, 300W) -> New 200:295. Ratios differ. Not Sufficient.
Analyze Statement (2): \(M = 15\). No info on W. Not Sufficient.
Combine: \(M=15\). From (1), \(15/W = 2/3 \implies 2W = 45 \implies W=22.5\). (Assuming the problem implies integer humans, this scenario is impossible, but mathematically we found a unique value for W). New ratio = \(15 : (22.5 – 5) = 15 : 17.5\). Sufficient.
Analyze Statement (1): \(M:W = 2:3\). This allows for (20M, 30W) -> New 20:25, or (200M, 300W) -> New 200:295. Ratios differ. Not Sufficient.
Analyze Statement (2): \(M = 15\). No info on W. Not Sufficient.
Combine: \(M=15\). From (1), \(15/W = 2/3 \implies 2W = 45 \implies W=22.5\). (Assuming the problem implies integer humans, this scenario is impossible, but mathematically we found a unique value for W). New ratio = \(15 : (22.5 – 5) = 15 : 17.5\). Sufficient.
2.
Is \(x > y\)?
(1) \(\frac{x+5}{y+5} > \frac{x}{y}\) and \(y > 0\).
(2) \(x\) and \(y\) are positive integers.
(1) \(\frac{x+5}{y+5} > \frac{x}{y}\) and \(y > 0\).
(2) \(x\) and \(y\) are positive integers.
Solution (A):
Analyze Statement (1): \(\frac{x+5}{y+5} > \frac{x}{y}\). Cross multiply (valid since \(y>0\) and if we assume \(y+5>0\)): \(y(x+5) > x(y+5) \implies xy + 5y > xy + 5x \implies 5y > 5x \implies y > x\). The question asks “Is \(x > y\)?”. We found \(y > x\), so the answer is definitively “No”. Sufficient.
Analyze Statement (2): Just says they are positive. No comparison. Not Sufficient.
Analyze Statement (1): \(\frac{x+5}{y+5} > \frac{x}{y}\). Cross multiply (valid since \(y>0\) and if we assume \(y+5>0\)): \(y(x+5) > x(y+5) \implies xy + 5y > xy + 5x \implies 5y > 5x \implies y > x\). The question asks “Is \(x > y\)?”. We found \(y > x\), so the answer is definitively “No”. Sufficient.
Analyze Statement (2): Just says they are positive. No comparison. Not Sufficient.
3.
What is the ratio of the area of Circle A to Circle B?
(1) The radius of Circle A is 5 more than the radius of Circle B.
(2) The circumference of Circle A is 20\(\pi\).
(1) The radius of Circle A is 5 more than the radius of Circle B.
(2) The circumference of Circle A is 20\(\pi\).
Solution (C):
Area ratio = \(\left(\frac{r_A}{r_B}\right)^2\). We need the ratio of radii.
Analyze Statement (1): \(r_A = r_B + 5\). Ratio = \(\frac{r_B+5}{r_B}\). This value changes depending on \(r_B\). Not Sufficient.
Analyze Statement (2): \(2\pi r_A = 20\pi \implies r_A = 10\). No info on B. Not Sufficient.
Combine: \(r_A = 10\). From (1), \(10 = r_B + 5 \implies r_B = 5\). Ratio of radii is \(10:5 = 2:1\). Area ratio is \(4:1\). Sufficient.
Analyze Statement (1): \(r_A = r_B + 5\). Ratio = \(\frac{r_B+5}{r_B}\). This value changes depending on \(r_B\). Not Sufficient.
Analyze Statement (2): \(2\pi r_A = 20\pi \implies r_A = 10\). No info on B. Not Sufficient.
Combine: \(r_A = 10\). From (1), \(10 = r_B + 5 \implies r_B = 5\). Ratio of radii is \(10:5 = 2:1\). Area ratio is \(4:1\). Sufficient.
4.
Is \( \frac{a}{b} > \frac{c}{d} \)? (assume \(b, d > 0\))
(1) \(a > c\).
(2) \(ad > bc\).
(1) \(a > c\).
(2) \(ad > bc\).
Solution (B):
We want to know if \( \frac{a}{b} – \frac{c}{d} > 0 \), which simplifies to \(\frac{ad – bc}{bd} > 0\). Since \(b,d > 0\), the denominator is positive. We just need to know if the numerator \(ad – bc > 0\), which means \(ad > bc\).
Analyze Statement (1): \(a > c\). Does not tell us about \(b\) or \(d\). Not Sufficient.
Analyze Statement (2): \(ad > bc\). This is exactly the condition derived above. Sufficient.
Analyze Statement (1): \(a > c\). Does not tell us about \(b\) or \(d\). Not Sufficient.
Analyze Statement (2): \(ad > bc\). This is exactly the condition derived above. Sufficient.
5.
A mixture contains chemical A and chemical B in ratio 2:3. If \(x\) liters of A are added, is the new ratio 1:1?
(1) The original volume was 50 liters.
(2) \(x = 10\).
(1) The original volume was 50 liters.
(2) \(x = 10\).
Solution (C):
Original: \(A = 2k, B = 3k\).
New A: \(2k + x\). New B: \(3k\).
New Ratio 1:1 means \(2k + x = 3k \implies x = k\).
The question is: Does \(x\) equal the original “unit part” \(k\)?
Analyze Statement (1): Total = 50. \(5k = 50 \implies k=10\). We know \(k=10\), but we don’t know \(x\). Not Sufficient.
Analyze Statement (2): \(x=10\). We don’t know \(k\). Not Sufficient.
Combine: \(k=10\) and \(x=10\). Since \(x=k\), the ratio becomes 1:1 (Yes). Sufficient.
Analyze Statement (1): Total = 50. \(5k = 50 \implies k=10\). We know \(k=10\), but we don’t know \(x\). Not Sufficient.
Analyze Statement (2): \(x=10\). We don’t know \(k\). Not Sufficient.
Combine: \(k=10\) and \(x=10\). Since \(x=k\), the ratio becomes 1:1 (Yes). Sufficient.
Score: 0 / 0