Find two numbers whose mean proportional is 12 and third proportional is 96.
6,18
6,24
8,18
8,24
Explanation: Let the numbers be $a$ and $b$. Mean proportional: $\sqrt{ab}=12 \Rightarrow ab=144$. Third proportional: $\frac{b^{2}}{a}=96$. Substituting $a=\frac{144}{b}$ gives $\frac{b^{3}}{144}=96 \Rightarrow b^{3}=13824 \Rightarrow b=24$, and then $a=6$. The numbers are $6$ and $24$.
Question 2 of 6medium
If a:b = c:d = e:f = 2:3, then (a+c+e):(b+d+f) = ?
2:3
3:2
4:9
1:1
Explanation: Each ratio is 2/3, so sum of numerators/sum of denominators = same ratio = 2:3
Question 3 of 6medium
y varies directly as \(x^{2}\) and inversely as \(\sqrt{z}\). If y=8 when x=2 and z=9, find y when x=3 and z=16.
12
13.5
15
18
Explanation: Write $y=\frac{kx^{2}}{\sqrt{z}}$. Using $y=8,\ x=2,\ z=9$: $8=\frac{k \cdot 4}{3} \Rightarrow k=6$. Then at $x=3,\ z=16$: $y=\frac{6 \cdot 9}{\sqrt{16}}=\frac{54}{4}=13.5$.
Question 4 of 6medium
If 15 men working 8 hours a day can complete a work in 20 days, how many men working 10 hours a day can complete double the work in 16 days?
20
25
30
35
Explanation: Work = M×H×D. Case1: W=15×8×20=2400. Case2: 2W=2×2400=4800. 4800=M×10×16=160M→M=30
Question 5 of 6medium
Three numbers are in continued proportion. The sum of the first and third is 20 and the product of the first and third is 64. Find the second number.
8
10
12
16
Explanation: Let numbers a,b,c. \(b^{2}\)=ac, a+c=20, ac=64. \(b^{2}\)=64→b=8
Question 6 of 6medium
If x varies as y and y varies as z, then x varies as:
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