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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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