Basic Maths: Test 50

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Basic Maths: Test 50

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Question 1

A Washing Machine consumes 8 units of electricity in 30 minutes and a tube light, consumes 18 units of electricity in 6 hours, How much total unit of electricity will both Washing Machine and Tube light consume in 8 days if they run 10 hours a day?

A	1280 units
B	1528 units
C	1248 units
D	1520 units

Question 1 Explanation:

Total electric consumption = (10 ×16×8+3×10×8) units
= (1280 +240) units = 1520 units

Question 2

On Independence Day sweets were to be equally distributed among 400 children. But on that particular day, 100 children remained absent. Thus, each child got 3 sweets extra. How many sweets did each child get?

A	6
B	12
C	9
D	Cannot be determined

Question 2 Explanation:

Number of sweets for 100 absent children = 300 × 3 =900
Therefore Number of sweets each students gets =

$\frac{900}{100}+3=12$

Question 3

There are some peacocks and some leopards in a forest. If the total number of animal heads in the forest are 858 and total number of animal legs are 1,746, what is the number of peacocks in the forest?

A	845
B	833
C	800
D	None of these

Question 3 Explanation:

If the number of peacocks in the forest be x, then number of leopards = 858 - x
Therefore

$\begin{align} & x\times 2+\left( 858-x \right)\times 4=1746 \\ & \Rightarrow 2x=3432-1746=1686 \\ & \Rightarrow x=\frac{1686}{2}=843 \\ \end{align}$

Question 4

$\frac{139\times 139+135\times 135+18765}{139\times 139\times 139-135\times 135\times 135}$ is equal to

A	4
B	270
C	$\frac{1}{4}$
D	$\frac{1}{270}$

Question 4 Explanation:

$\begin{align} & =\frac{139\times 139+135\times 135+18765}{139\times 139\times 139-135\times 135\times 135} \\ & \left[ 139\times 135=18765 \right] \\ & Let\,139=a\,\,and\,\,135=b \\ & Therefore\,\,\,\exp ression\,\,=\frac{{{a}^{2}}+{{b}^{2}}+ab}{{{a}^{3}}+{{b}^{3}}} \\ & =\frac{{{a}^{2}}+{{b}^{2}}+ab}{\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)} \\ & =\frac{1}{a-b} \\ & =\frac{1}{139-135} \\ & =\frac{1}{4} \\ \end{align}$

Question 5

$\left( \frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}+\frac{1}{15.17}+\frac{1}{17.19} \right)$ is equal to

A	6/133
B	2/133
C	12/133
D	1/133

Question 5 Explanation:

$\begin{align} & \left( \frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}+\frac{1}{15.17}+\frac{1}{17.19} \right) \\ & =\frac{1}{2}\left( \frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+\frac{2}{17.19} \right) \\ & =\frac{1}{2}\left( \frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19} \right) \\ & =\frac{1}{2}\left( \frac{1}{7}-\frac{1}{19} \right) \\ & =\frac{1}{2}\left( \frac{19-7}{133} \right) \\ & =\frac{1}{2}\times \frac{12}{133} \\ & =\frac{6}{133} \\ \end{align}$

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