Mixture and Comparison Problems - SSAT Middle Level Quantitative

1. State the equation for pure solute after adding $w$ L of water to $V$ L of $c\%$ solution.

   $\frac{c}{100}V=\frac{c_{\text{new}}}{100}(V+w)$. Conserves the solute amount while increasing the volume by added water to relate initial and new concentrations.

2. Find the ratio of $10\%$ to $40\%$ needed to obtain a $25\%$ mixture.

   $1:1$. Applies alligation with differences $40-25=15$ and $25-10=15$ to find the equal ratio for the target mixture.

3. Find $x$: How many liters of $30\%$ solution with $2$ L of $60\%$ make $40\%$?

   $x=4$. Uses the equation $0.3x + 1.2 = 0.4(x + 2)$ to balance solutes for the specified final concentration.

4. Find $x$: How many liters of $5\%$ solution with $2$ L of $35\%$ make $20\%$?

   $x=2$. Solves $0.05x + 0.7 = 0.2(x + 2)$ to find the volume that yields the target concentration in the mixture.

5. Find $x$: How many liters of $10\%$ solution with $5$ L of $40\%$ make $25\%$?

   $x=5$. Solves the mixture balance $0.1x + 2 = 0.25(x + 5)$ to find the volume that achieves the desired average concentration.

6. What is the ratio form (alligation) for amounts when mixing $p\%$ and $q\%$ to get $r\%$?

   $\text{amount at }p: \text{amount at }q=(q-r):(r-p)$. Applies the alligation method, using concentration differences to determine the balancing ratio for the target percentage.

7. Find $x$: How many lb of $2 candy with $5$ lb of $8 candy make a $5 mix?

   $x=5$. Uses $2x + 40 = 5(x + 5)$ to find the amount that achieves the desired average cost in the mix.

8. What is the weighted-average equation for a mixture percent $r$ from parts $a$ and $b$?

   $r=\frac{pa+qb}{a+b}$. Derives the final concentration as the weighted average of the individual concentrations, weighted by their respective volumes.

9. State the equation for total pure solute when mixing $a$ L of $p\%$ with $b$ L of $q\%$.

   $\text{solute}=\frac{p}{100}a+\frac{q}{100}b$. Sums the solute contributions from each solution based on their concentrations and volumes to find the total pure solute in the mixture.

10. State the equation for a two-item mixture cost: $a$ lb at $p$ and $b$ lb at $q$ gives average $r$.

    $pa+qb=r(a+b)$. Balances total cost from each component to equal the average cost times total weight in the mixture.

11. Find $A$ and $B$: $A$ is $12$ more than $B$, and $A$ is $3$ times $B$.

    $A=18,\;B=6$. Solves system $A = B + 12$ and $A = 3B$ by substitution to find values satisfying both conditions.

12. Find the ratio of $20\%$ to $50\%$ needed to obtain a $30\%$ mixture.

    $2:1$. Uses alligation: differences $50-30=20$ and $30-20=10$ give the ratio of $20\%$ to $50\%$ solution amounts.

13. Find the new concentration: $4$ L of $30\%$ solution is diluted with $2$ L of water.

    $20\%$. Maintains initial solute $1.2$ while increasing volume to $6$ L, yielding the diluted concentration.

14. Find $x$: How many liters of $15\%$ solution with $3$ L of $45\%$ make $30\%$?

    $x=3$. Applies the solute balance $0.15x + 1.35 = 0.3(x + 3)$ to compute the required volume for the target mixture.

15. Find $A$ and $B$: Two numbers are in ratio $3:5$ and have sum $64$.

    $A=24,\;B=40$. Divides sum $64$ by $8$ parts ($3+5$) and scales by ratio components to find each value.

16. What is the equation form for "$A$ is $k$ more than $B$" and "$A$ is $m$ times $B$"?

    $A=B+k$ and $A=mB$. Expresses the relationships directly as equations to model the comparison and multiple conditions.

17. State the equation that relates final concentration $r\%$ to total volume when two solutions are mixed.

    $\frac{p}{100}a+\frac{q}{100}b=\frac{r}{100}(a+b)$. Equates the total solute from the initial solutions to the solute in the final mixture, accounting for the combined volume.

18. Find $A$ and $B$: Two amounts are in ratio $2:7$ and differ by $45$.

    $A=18,\;B=63$. Sets difference of $45$ to $5$ parts ($7-2$), finds part value $9$, and scales for each amount.

19. Find the new concentration: $6$ L of $25\%$ solution is diluted with $2$ L of water.

    $18.75\%$. Preserves solute $1.5$ across new volume $8$ L to calculate the reduced concentration after dilution.

20. Find $x$: How many lb of $4 nuts with $3$ lb of $10 nuts make a $7 mix?

    $x=3$. Solves $4x + 30 = 7(x + 3)$ to balance costs for the target average price per pound.

21. Find $x$: How many liters of $20\%$ solution with $3$ L of $50\%$ make $30\%$?

    $x=6$. Balances the solute equation $0.2x + 1.5 = 0.3(x + 3)$ to determine the volume needed for the target concentration.

22. State the equation for solute after removing $r$ L from $V$ L of $c\%$ solution (same concentration removed).

    $\text{solute new}=\frac{c}{100}(V-r)$. Subtracts the removed solute, proportional to the concentration, from the initial solute to find the remaining amount.

23. State the equation to compare two quantities in ratio $m:n$ given total $T$.

    $\text{parts}=\frac{T}{m+n},\;A=m\cdot\text{parts},\;B=n\cdot\text{parts}$. Divides the total into parts proportional to the ratio and multiplies to find each quantity.

Study these flashcards