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Basic Maths: Test 53

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Question 1
Rahul's monthly salary is one and a half times Harsh's monthly salary, Nikhil's monthly salary is five-fourth of Rahul's monthly salary. If the total of the monthly salaries of all the three is Rs.1.83,750. what is Nikhil's monthly salary?
A
Rs.42,000
B
Rs.68,500
C
Rs.78,750
D
Rs.63,000
Question 1 Explanation:
$\begin{align} & Let\,\,\,Harsh's\,\,\,monthly\,\,\,income\,\,be\,\,Rs.\,x. \\ & Rahul's\,\,\,monthly\,\,\,income\,\,=Rs.\,\frac{3x}{2} \\ & Nikhil's\,\,monthly\,\,\,income\,\,=Rs.\,\frac{15x}{8} \\ & Therefore\,\,x+\frac{3x}{2}+\frac{15x}{8}=183750 \\ & \Rightarrow \frac{8x+12x+15x}{8}=183750 \\ & \Rightarrow 35x=183750\times 8 \\ & \Rightarrow x=\frac{183750\times 8}{35}=42000 \\ & Therefore\,\,Nikhil's\,\,\,monthly\,\,\,\,income\, \\ & =Rs.\,\left( \frac{15}{8}\times 42000 \right) \\ & =Rs.78750 \\ \end{align}$
Question 2
$\left( 41\times 71+123\times 23+82\times 25 \right)$ is equal to
A
779000
B
7090
C
7790
D
70900
Question 2 Explanation:
$\begin{align} & =\left( 41\times 71+123\times 23+82\times 25 \right) \\ & =41\,\left( 71+3\times 23+2\times 25 \right) \\ & =41\,\left( 71+69+50 \right) \\ & =41\times 190=7790 \\ \end{align}$
Question 3
$\left( \frac{3.77\times 3.77\times 3.77-1.77\times 1.77\times 1.77}{3.77\times 3.77+3.77\times 1.77+1.77\times 1.77} \right)$ is equal to
A
−2
B
0.2
C
−0.2
D
2
Question 3 Explanation:
Let 3.77 = a
And 1.77= b
Therefore expression
$\begin{align} & =\frac{{{a}^{3}}-{{b}^{3}}}{{{a}^{2}}+ab+{{b}^{2}}} \\ & =a-b=3.77-1.77=2.0 \\ \end{align}$
Question 4
$5\frac{2}{71}\times 9\frac{12}{17}\times 12\frac{10}{11}+4\frac{3}{4}=?$
A
603.75
B
635.75
C
634.75
D
605.75
Question 4 Explanation:
$\begin{align} & ?=\frac{357}{71}\times \frac{165}{17}\times \frac{142}{11}+4.75 \\ & =630+4.75=634.75 \\ \end{align}$
Question 5
Capture
A
386
B
312
C
314
D
320
Question 5 Explanation:
$\begin{align} & 7512\times \frac{5}{38}\times \frac{95}{100}=?+619 \\ & \Rightarrow 939=?+619 \\ & \Rightarrow ?=939-619=320 \\ \end{align}$
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