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Question Detail
\begin{aligned}
(3\frac{1}{4}\div \{1\frac{1}{4} - \frac{1}{2}(2\frac{1}{2} - \overline {\frac{1}{4} - \frac{1}{6}} ) \} )
\end{aligned}
- 78
- 88
- 98
- 108
Answer: Option A
Explanation:
Tip:
As you can see, there is bar over
\begin{aligned}
\overline{\frac{1}{4}-\frac{1}{6}}
\end{aligned}
So their sign will be changed from - to + as
\begin{aligned}
\frac{1}{4}+\frac{1}{6}
\end{aligned}
1. \begin{aligned} 3034 -(1002 \div 20.04) = ? \end{aligned}
- 1964
- 1984
- 2964
- 2984
Answer: Option D
Explanation:
\begin{aligned}
= 3034 -( \frac{1002}{2004} \times 100)
\end{aligned}
\begin{aligned}
= 3034 - 50 = 2984
\end{aligned}
2. \begin{aligned}
\frac{4+4 \times 18 -6 - 8}{123 \times 6 - 146 \times 5 }
\end{aligned}
- 7.50
- 7.75
- 8
- 8.05
Answer: Option B
Explanation:
\begin{aligned}
\frac{4+72-6-8}{738-730}
\end{aligned}
\begin{aligned}
= \frac{76-14}{8}
\end{aligned}
\begin{aligned}
= \frac{62}{8} = 7.75
\end{aligned}
3. If a - b = 3 and \begin{aligned} a^2 + b^2 = 29 \end{aligned}, then find the value of ab
- 7
- 8.5
- 10
- 12
Answer: Option C
Explanation:
Remember the formula, if not then cram it :)
\begin{aligned} 2ab = (a^2 + b^2)- (a-b)^2 \end{aligned}
=> 2ab = 29 - 9 = 20
=> ab = 10
4. 5.8 * 2.5 + 0.6 * 6.75 + 139.25= ?
- 157.60
- 147.80
- 147.60
- 157.80
Answer: Option D
5. The number of girls in a class are 7 times the number of boys, which value can never be the of total students
- 40
- 48
- 24
- 30
Answer: Option D
Explanation:
Let the boys are X, then girls are 7X, total = X+7X = 8X
So it should be multiple of 8, 30 is not a multiple of 8.
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