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A pack of coffee powder contains a mixture of x grams of coffee & y grams of choco. The amount of coffee is greater than that of choco and each coffee powder pack is atmost 10 grams. Which of the following inequations describe the conditions.
Amount of coffee= x gms
Amount of choco= y gms
If \(a,b,c \in R, \text {s.t.} a > b, c > 0\) then
If \(a,b,c \in R, \text {s.t.} a > b, c < 0, \) then
Let c=p
If p>0, q<0, then pq is
If p>0, q<0, then p+q is
If p>0, q<0, then pq is
If \( p>0, q<0, \) then \( \dfrac pq \) is
A company manufactures cassettes. Its cost and revenue functions are C(x)=26000+30x & R(x)=43x, respectively, where x is the number of cassettes produced and sold in week. How many cassettes must be sold by the company to realise some profit?
The water acidity in a pool is considerad normall when the average pH reading of three daily measurements is between 8.2 and 8.5. If the first two pH reading are 8.48 and 8.35, then find the range of pH value for the third reading that will result in the acidity level being normal.
A solution is be kept between \(40^\circ C \text{&} 45^\circ C\). What is the temperature range in degree fahrenheit, if the conversion forula is \(F = \dfrac{9}{5}C + 32\)
The longest side of a triangle is twice the shortest side & the third side is 2 cm longer than the shortest side. If the perimeter of the triangle is more than 166 cm, then find the minimum length of the shortest side.
Let the length of shortest side is x cm
Longest side= 2x cm
Third side= (2+x) cm
In drilling world’s deepest hole, it was found that the temperature T in degree celcius, x km below the earth’s surface was given by \(T = 30 + 25(x – 3), 3 \le x \le 15\). At what depth will the temperature be between \({155^ \circ }C \) and \( {205^\circ }C\)?
\(\dfrac{4}{{x + 1}} \le 3 \le \dfrac{6}{{x + 1}}(x > 0)\), Solve for x
From (i) & (ii), we get
\(\dfrac{{\left {x – 2} \right – 1}}{{\left {x – 2} \right – 2}} \le 0\), Solve for x
and y2=0
y = 1 and y = 2
and
\(\) \Rightarrow x \in \left( {0,1} \right] \cup \left[ {3,4} \right)/latex]
Solve:\(\dfrac{1}{{\left x \right – 3}} \le \dfrac{1}{2}\), for x is
\(\)\therefore x \in \left( { – \infty , – 5} \right] \cup \left( { – 3,3} \right) \cup \left[ {5,\infty } \right)/latex]
solve: \(\left {x – 1} \right \le 5,\left x \right \ge 2\), for x is
and
Solve: \( – 5 \le \dfrac{{2 – 3x}}{4} \le 9\) for x is
and
\(4x + 3 \ge 2x + 17, 3x – 5 \le – 2\), solve for x
Also,
From (i) & (ii) we have
If \(x < – 5\), then
If \(\left {x – 7} \right > 23\), then x is belongs to
If x and b are real numbers, where b>0 and \(\left x \right > b\) then x is
& b>0 or x>b
If xy>0, then
xy>0, when x>0, y>0
or x<0, y<0
Solve : \(\dfrac{{\left {x – 5} \right – 2}}{{\left {x – 5} \right – 5}} \le 0\), for x is
Let
\(\) \Rightarrow {\text{x}} \in \left[ {0,3} \right] \cup \left[ {7,10} \right)/latex]
Solve: \( – 4x + 1 \ge 0,3 – 4x < 0\)
Hence solution is
which means no solution exist.
Identify solution set for \(\left {7 – x} \right + 3 < 11\) is