Mixture and Comparison Problems - SSAT Middle Level: Quantitative
Card 1 of 23
State the equation for total pure solute when mixing $a$ L of $p%$ with $b$ L of $q%$.
State the equation for total pure solute when mixing $a$ L of $p%$ with $b$ L of $q%$.
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$\text{solute}=\frac{p}{100}a+\frac{q}{100}b$. Sums the solute contributions from each solution based on their concentrations and volumes to find the total pure solute in the mixture.
$\text{solute}=\frac{p}{100}a+\frac{q}{100}b$. Sums the solute contributions from each solution based on their concentrations and volumes to find the total pure solute in the mixture.
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Find $x$: How many lb of $\$2$ candy with $5$ lb of $$8$ candy make a $\$5$ mix?
Find $x$: How many lb of $\$2$ candy with $5$ lb of $$8$ candy make a $\$5$ mix?
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$x=5$. Uses $2x + 40 = 5(x + 5)$ to find the amount that achieves the desired average cost in the mix.
$x=5$. Uses $2x + 40 = 5(x + 5)$ to find the amount that achieves the desired average cost in the mix.
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State the equation to compare two quantities in ratio $m:n$ given total $T$.
State the equation to compare two quantities in ratio $m:n$ given total $T$.
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$\text{parts}=\frac{T}{m+n},;A=m\cdot\text{parts},;B=n\cdot\text{parts}$. Divides the total into parts proportional to the ratio and multiplies to find each quantity.
$\text{parts}=\frac{T}{m+n},;A=m\cdot\text{parts},;B=n\cdot\text{parts}$. Divides the total into parts proportional to the ratio and multiplies to find each quantity.
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Find $A$ and $B$: Two numbers are in ratio $3:5$ and have sum $64$.
Find $A$ and $B$: Two numbers are in ratio $3:5$ and have sum $64$.
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$A=24,;B=40$. Divides sum $64$ by $8$ parts ($3+5$) and scales by ratio components to find each value.
$A=24,;B=40$. Divides sum $64$ by $8$ parts ($3+5$) and scales by ratio components to find each value.
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Find $A$ and $B$: Two amounts are in ratio $2:7$ and differ by $45$.
Find $A$ and $B$: Two amounts are in ratio $2:7$ and differ by $45$.
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$A=18,;B=63$. Sets difference of $45$ to $5$ parts ($7-2$), finds part value $9$, and scales for each amount.
$A=18,;B=63$. Sets difference of $45$ to $5$ parts ($7-2$), finds part value $9$, and scales for each amount.
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What is the equation form for “$A$ is $k$ more than $B$” and “$A$ is $m$ times $B$”?
What is the equation form for “$A$ is $k$ more than $B$” and “$A$ is $m$ times $B$”?
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$A=B+k$ and $A=mB$. Expresses the relationships directly as equations to model the comparison and multiple conditions.
$A=B+k$ and $A=mB$. Expresses the relationships directly as equations to model the comparison and multiple conditions.
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Find $A$ and $B$: $A$ is $12$ more than $B$, and $A$ is $3$ times $B$.
Find $A$ and $B$: $A$ is $12$ more than $B$, and $A$ is $3$ times $B$.
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$A=18,;B=6$. Solves system $A = B + 12$ and $A = 3B$ by substitution to find values satisfying both conditions.
$A=18,;B=6$. Solves system $A = B + 12$ and $A = 3B$ by substitution to find values satisfying both conditions.
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Find $x$: How many lb of $\$4$ nuts with $3$ lb of $$10$ nuts make a $\$7$ mix?
Find $x$: How many lb of $\$4$ nuts with $3$ lb of $$10$ nuts make a $\$7$ mix?
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$x=3$. Solves $4x + 30 = 7(x + 3)$ to balance costs for the target average price per pound.
$x=3$. Solves $4x + 30 = 7(x + 3)$ to balance costs for the target average price per pound.
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State the equation that relates final concentration $r%$ to total volume when two solutions are mixed.
State the equation that relates final concentration $r%$ to total volume when two solutions are mixed.
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$\frac{p}{100}a+\frac{q}{100}b=\frac{r}{100}(a+b)$. Equates the total solute from the initial solutions to the solute in the final mixture, accounting for the combined volume.
$\frac{p}{100}a+\frac{q}{100}b=\frac{r}{100}(a+b)$. Equates the total solute from the initial solutions to the solute in the final mixture, accounting for the combined volume.
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What is the weighted-average equation for a mixture percent $r$ from parts $a$ and $b$?
What is the weighted-average equation for a mixture percent $r$ from parts $a$ and $b$?
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$r=\frac{pa+qb}{a+b}$. Derives the final concentration as the weighted average of the individual concentrations, weighted by their respective volumes.
$r=\frac{pa+qb}{a+b}$. Derives the final concentration as the weighted average of the individual concentrations, weighted by their respective volumes.
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Find $x$: How many liters of $20%$ solution with $3$ L of $50%$ make $30%$?
Find $x$: How many liters of $20%$ solution with $3$ L of $50%$ make $30%$?
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$x=6$. Balances the solute equation $0.2x + 1.5 = 0.3(x + 3)$ to determine the volume needed for the target concentration.
$x=6$. Balances the solute equation $0.2x + 1.5 = 0.3(x + 3)$ to determine the volume needed for the target concentration.
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Find $x$: How many liters of $10%$ solution with $5$ L of $40%$ make $25%$?
Find $x$: How many liters of $10%$ solution with $5$ L of $40%$ make $25%$?
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$x=5$. Solves the mixture balance $0.1x + 2 = 0.25(x + 5)$ to find the volume that achieves the desired average concentration.
$x=5$. Solves the mixture balance $0.1x + 2 = 0.25(x + 5)$ to find the volume that achieves the desired average concentration.
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Find $x$: How many liters of $30%$ solution with $2$ L of $60%$ make $40%$?
Find $x$: How many liters of $30%$ solution with $2$ L of $60%$ make $40%$?
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$x=4$. Uses the equation $0.3x + 1.2 = 0.4(x + 2)$ to balance solutes for the specified final concentration.
$x=4$. Uses the equation $0.3x + 1.2 = 0.4(x + 2)$ to balance solutes for the specified final concentration.
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Find $x$: How many liters of $15%$ solution with $3$ L of $45%$ make $30%$?
Find $x$: How many liters of $15%$ solution with $3$ L of $45%$ make $30%$?
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$x=3$. Applies the solute balance $0.15x + 1.35 = 0.3(x + 3)$ to compute the required volume for the target mixture.
$x=3$. Applies the solute balance $0.15x + 1.35 = 0.3(x + 3)$ to compute the required volume for the target mixture.
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Find $x$: How many liters of $5%$ solution with $2$ L of $35%$ make $20%$?
Find $x$: How many liters of $5%$ solution with $2$ L of $35%$ make $20%$?
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$x=2$. Solves $0.05x + 0.7 = 0.2(x + 2)$ to find the volume that yields the target concentration in the mixture.
$x=2$. Solves $0.05x + 0.7 = 0.2(x + 2)$ to find the volume that yields the target concentration in the mixture.
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What is the ratio form (alligation) for amounts when mixing $p%$ and $q%$ to get $r%$?
What is the ratio form (alligation) for amounts when mixing $p%$ and $q%$ to get $r%$?
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$\text{amount at }p: \text{amount at }q=(q-r):(r-p)$. Applies the alligation method, using concentration differences to determine the balancing ratio for the target percentage.
$\text{amount at }p: \text{amount at }q=(q-r):(r-p)$. Applies the alligation method, using concentration differences to determine the balancing ratio for the target percentage.
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Find the ratio of $20%$ to $50%$ needed to obtain a $30%$ mixture.
Find the ratio of $20%$ to $50%$ needed to obtain a $30%$ mixture.
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$2:1$. Uses alligation: differences $50-30=20$ and $30-20=10$ give the ratio of $20%$ to $50%$ solution amounts.
$2:1$. Uses alligation: differences $50-30=20$ and $30-20=10$ give the ratio of $20%$ to $50%$ solution amounts.
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Find the ratio of $10%$ to $40%$ needed to obtain a $25%$ mixture.
Find the ratio of $10%$ to $40%$ needed to obtain a $25%$ mixture.
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$1:1$. Applies alligation with differences $40-25=15$ and $25-10=15$ to find the equal ratio for the target mixture.
$1:1$. Applies alligation with differences $40-25=15$ and $25-10=15$ to find the equal ratio for the target mixture.
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State the equation for pure solute after adding $w$ L of water to $V$ L of $c%$ solution.
State the equation for pure solute after adding $w$ L of water to $V$ L of $c%$ solution.
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$\frac{c}{100}V=\frac{c_{\text{new}}}{100}(V+w)$. Conserves the solute amount while increasing the volume by added water to relate initial and new concentrations.
$\frac{c}{100}V=\frac{c_{\text{new}}}{100}(V+w)$. Conserves the solute amount while increasing the volume by added water to relate initial and new concentrations.
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Find the new concentration: $4$ L of $30%$ solution is diluted with $2$ L of water.
Find the new concentration: $4$ L of $30%$ solution is diluted with $2$ L of water.
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$20%$. Maintains initial solute $1.2$ while increasing volume to $6$ L, yielding the diluted concentration.
$20%$. Maintains initial solute $1.2$ while increasing volume to $6$ L, yielding the diluted concentration.
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Find the new concentration: $6$ L of $25%$ solution is diluted with $2$ L of water.
Find the new concentration: $6$ L of $25%$ solution is diluted with $2$ L of water.
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$18.75%$. Preserves solute $1.5$ across new volume $8$ L to calculate the reduced concentration after dilution.
$18.75%$. Preserves solute $1.5$ across new volume $8$ L to calculate the reduced concentration after dilution.
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State the equation for solute after removing $r$ L from $V$ L of $c%$ solution (same concentration removed).
State the equation for solute after removing $r$ L from $V$ L of $c%$ solution (same concentration removed).
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$\text{solute new}=\frac{c}{100}(V-r)$. Subtracts the removed solute, proportional to the concentration, from the initial solute to find the remaining amount.
$\text{solute new}=\frac{c}{100}(V-r)$. Subtracts the removed solute, proportional to the concentration, from the initial solute to find the remaining amount.
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State the equation for a two-item mixture cost: $a$ lb at $p$ and $b$ lb at $q$ gives average $r$.
State the equation for a two-item mixture cost: $a$ lb at $p$ and $b$ lb at $q$ gives average $r$.
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$pa+qb=r(a+b)$. Balances total cost from each component to equal the average cost times total weight in the mixture.
$pa+qb=r(a+b)$. Balances total cost from each component to equal the average cost times total weight in the mixture.
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