Question
Solve: 0.6a +0.2a = 0.4 a + 8
✓
Answer
$\begin{aligned} & 0.6 a+0.2 a=0.4 a+8 \\ & \Rightarrow \frac{6}{10} a+\frac{2}{10} a=\frac{4}{10} a+\frac{8}{1} \\ & \Rightarrow \frac{6 a+2 a}{10}=4 a+80 \\ & \Rightarrow 6 a+2 a=4 a+80 \\ & \Rightarrow 6 a+2 a-4 a=80 \\ & \Rightarrow 4 a=80 \\ & \Rightarrow a=\frac{80}{4}=20 \\ & \therefore a=20\end{aligned}$
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A, B and C can do a piece of work in 12, 15 and 20 days respectively. How long will they take to do it working together?
In India, the Census is held every 10 years to count the whole population of the country. In a particular Census, the data of the number of children in 25 families of a village was provided as :
1, 2, 4, 2, 0, 2, 3, 3, 1, 0, 2, 3, 4, 3, 1, 1, 2, 1, 2, 3, 2, 4, 1, 2, 2.
Rahul works in the analytics department of the Census. He was given the task of analysing this data of the 25 families of this village.
Q.1. How many families have 3 children?
(a) 2 $\quad$$\quad$(b) 3$\quad$$\quad$ (c) 5$\quad$$\quad$ (d) 6
Q.2. The cumulative frequency of families having 3 children is :
(a) 13$\quad$$\quad$(b) 17$\quad$$\quad$ (c) 18$\quad$$\quad$ (d) 22
Q.3. The mean number of children in these 25 families is :
(a) 1.96 $\quad$$\quad$(b) 2.04$\quad$$\quad$ (c) 2.85$\quad$$\quad$ (d) 3.15
Q.4. The median of the number of children in these 25 families is :
(a) 1 $\quad$$\quad$(b) 2 $\quad$$\quad$(c) 3$\quad$$\quad$ (d) 4
→1, 2, 4, 2, 0, 2, 3, 3, 1, 0, 2, 3, 4, 3, 1, 1, 2, 1, 2, 3, 2, 4, 1, 2, 2.
Rahul works in the analytics department of the Census. He was given the task of analysing this data of the 25 families of this village.
Q.1. How many families have 3 children?
(a) 2 $\quad$$\quad$(b) 3$\quad$$\quad$ (c) 5$\quad$$\quad$ (d) 6
Q.2. The cumulative frequency of families having 3 children is :
(a) 13$\quad$$\quad$(b) 17$\quad$$\quad$ (c) 18$\quad$$\quad$ (d) 22
Q.3. The mean number of children in these 25 families is :
(a) 1.96 $\quad$$\quad$(b) 2.04$\quad$$\quad$ (c) 2.85$\quad$$\quad$ (d) 3.15
Q.4. The median of the number of children in these 25 families is :
(a) 1 $\quad$$\quad$(b) 2 $\quad$$\quad$(c) 3$\quad$$\quad$ (d) 4
Simplify, giving Solution with positive index
$\left(\frac{1}{4 a b^2 c}\right)^2 \div\left(\frac{3}{2 a^2 b^2}\right)^4$
→$\left(\frac{1}{4 a b^2 c}\right)^2 \div\left(\frac{3}{2 a^2 b^2}\right)^4$
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