Mathematics Class X

Question 1 of 36

1. Question
1 point(s)
Find the value of k so that the sum of roots of the quadratic equation
$\left( {k + 1} \right){x^2} + 2kx + 4 = 0$is equal to product of roots
- 2
- -2
- 1
- 0
Correct

Incorrect

Hint

Sum of roots $= \dfrac{{ - b}}{a}$
Product of roots $= \dfrac{c}{a}$
$\dfrac{{ - b}}{a} = \dfrac{c}{a} \Rightarrow c = - b \Rightarrow - 2k \Rightarrow k = - 2$
Question 2 of 36

2. Question
1 point(s)
Find one root of the$\left( {a – b} \right){x^2} + \left( {b – c} \right)x + \left( {c – a} \right) = 0$
- 0
- 1
- 1
- 2
Correct

Incorrect

Hint

x=1 satisfy the given equation.
Question 3 of 36

3. Question
1 point(s)
If$a{x^2} + bx + c = 0$has equal roots then c=
- $\dfrac{{ – b}}{{2a}}$
- $\dfrac{b}{{2a}}$
- $\dfrac{{ – {b^2}}}{{4a}}$
- $\dfrac{{{b^2}}}{{4a}}$
Correct

Incorrect

Hint

$\begin{array}{l}D = {b^2} - 4ac = 0\\{b^2} - 4ac = 0\\ \Rightarrow 4ac = {b^2}\\ \Rightarrow c = \dfrac{{{b^2}}}{{4a}}\end{array}$
Question 4 of 36

4. Question
1 point(s)
If$x – \dfrac{1}{{x – 2}} = 2 – \dfrac{1}{{x – 2}}$then x is equal to
- 1
- 2
- 3
- none of these
Correct

Incorrect

Hint

By solving we get x=a but x=2 equation is not define.
Question 5 of 36

5. Question
1 point(s)
The value of k which the roots of equation (x−1)(x−5)+k=0 differ by 2 is
- 3
- 6
- -3
- $\dfrac{1}{2}$
Correct

Incorrect

Hint

${x^2} - 6x + \left( {5 + k} \right) = 0$ and $a - b = 2$
$\begin{array}{l} \Rightarrow \alpha - \beta = \sqrt {{{\left( {\alpha + \beta } \right)}^2} - 4\alpha \beta } \\ \Rightarrow 4 = 36 - 4\left( {5 + k} \right)\\ \Rightarrow k = 3\end{array}$
Question 6 of 36

6. Question
1 point(s)
If one root of ${x^2} – 4x + k – 0$is 6 then the value of k is
- 12
- 2
- -2
- -12
Correct

Incorrect

Hint

$\begin{array}{l}{x^2} - 4x + k = 0\\ \Rightarrow {6^2} - 4 \times 6 + k = 0\\ \Rightarrow k = - 12\end{array}$
Question 7 of 36

7. Question
1 point(s)
The condition for the roots of equation${x^2} – lx + m = 0$to differ by one is
- ${l^2} = 4m + 1$
- ${l^2} + {m^2} = 1$
- ${m^2} = 4l + 1$
- $l = m + 1$
Correct

Incorrect

Hint

$\begin{array}{l} \Rightarrow a - b = 1\\ \Rightarrow \sqrt {{l^2} - 4m = 1} \\ \Rightarrow {l^2} = 1 + 4m\end{array}$
Question 8 of 36

8. Question
1 point(s)
a and b are two numbers such that a+b=6, a-b=8. Then, a and b are the roots of the quadratic equation.
- ${x^2} + 6x + 7 = 0$
- ${x^2} – 6x – 7 = 0$
- ${x^2} + 6x – 8 = 0$
- ${x^2} – 6x + 8 = 0$
Correct

Incorrect

Hint

$\begin{array}{l}a + b = 6,a - b = 8\\a = 7,b = - 1\\ \Rightarrow \left( {x - a} \right)\left( {x + b} \right) = 0\\ \Rightarrow \left( {x - 7} \right)\left( {x + 1} \right) = 0\\ \Rightarrow {x^2} - 6x - 7 = 0\end{array}$
Question 9 of 36

9. Question
1 point(s)
If one root of a quadratic equation is$\dfrac{1}{{\sqrt 4 – \sqrt 3 }}$then the quadratic equation can be
- ${x^2} – 2\sqrt 4 x + 1 = 0$
- ${x^2} – \sqrt 4 x + 1 = 0$
- ${x^2} + 2\sqrt 4 x + 1 = 0$
- ${x^2} – 2\sqrt 3 x + 1 = 0$
Correct

Incorrect

Hint

One root $= \dfrac{1}{{\sqrt 4 - \sqrt 3 }}$
Other root $= \sqrt 4 - \sqrt 3$
The quadratic equation
$\begin{array}{l} \Rightarrow \left[ {x - \left( {\sqrt 4 - \sqrt 3 } \right)} \right]\left[ {x - \left( {\sqrt 4 + \sqrt 3 } \right)} \right] = 0\\ \Rightarrow {x^2} - 2\sqrt 4 x + 1 = 0\end{array}$
Question 10 of 36

10. Question
1 point(s)
If sum of the roots of ${x^2} – kx + 6 = 0$is 2 then the value of k is
- 2
- ½
- 3
- none of these
Correct

Incorrect

Hint

Sum of roots $= \dfrac{{ - b}}{a} = k = 2$
Question 11 of 36

11. Question
1 point(s)
The value of k for which
${x^2} – 4x + k = 0$has coincident roots is
- 4
- 0
- -4
- -2
Correct

Incorrect

Hint

${4^2} = 4k \Rightarrow k = 4$
Question 12 of 36

12. Question
1 point(s)
The value of p so that${x^2} + 5px + 16 = 0$has no real root is
- Greater than 8/5
- Less than -8/5
- Lies between -8/5 and 8/5
- None of these
Correct

Incorrect

Hint

$\begin{array}{l}{\left( {5p} \right)^2} - 4\left( {16} \right) < 0\\{p^2} < \dfrac{{64}}{{25}} \Rightarrow \dfrac{{ - 8}}{5} < p < \dfrac{8}{5}\end{array}$
Question 13 of 36

13. Question
1 point(s)
For what value of ‘a’ will the sum of squares of roots of the equation${x^2} – \left( {a – 2} \right)x – a – 1 = 0$have the least value
- 0
- 1
- 2.5
- 6.4
Correct

Incorrect

Hint

$\begin{array}{l}a + b = a - 2\\a \times b = - \left( {a + 1} \right)\\{a^2} + {b^2} = {\left( {a - 2} \right)^2} - 2\left( {a - 1} \right)\\ = {a^2} - 2a + 6\\ = {\left( {a - 1} \right)^2} + 5\end{array}$
So, the minimum value of ${a^2} + {b^2},a = 1$
Question 14 of 36

14. Question
1 point(s)
The nature of roots of equation${x^2} + 2x\sqrt 3 + 3 = 0$is
- Real and equal
- Rational and equal
- Rational and unequal
- Irrational and unequal
Correct

Incorrect

Hint

$\begin{array}{l}{x^2} + 2x\sqrt 3 + 3 = 0\\x = - \sqrt 3 , - \sqrt 3 \end{array}$
Question 15 of 36

15. Question
1 point(s)
The length of a rectangular plot is 8m more than its width. If the length is reduced by 4m and width increased by 3m, the area remains the same. The dimension of the plot are
- 16,8
- 20,12
- 24,16
- 30,22
Correct

Incorrect

Hint

$\begin{array}{l}x\left( {x + 8} \right) = \left( {x + 8 - 4} \right)\left( {x + 3} \right)\\ \Rightarrow {x^2} + 8x = {x^2} + 7x + 12\\ \Rightarrow x = 12\end{array}$
Question 16 of 36

16. Question
1 point(s)
A dealer sells an article for Rs75 and gains as much percent and the cost price of the article. What is the cost price of the article?
- Rs50
- Rs30
- Rs40
- Rs45
Correct

Incorrect

Hint

$\begin{array}{l} \Rightarrow x\left( {1 + \dfrac{x}{{100}}} \right) = 75\\ \Rightarrow x = 50\end{array}$
Question 17 of 36

17. Question
1 point(s)
If ${\left( {x – \dfrac{1}{2}} \right)^2} – {\left( {x – \dfrac{3}{2}} \right)^2} = x + 2$then x=?
- 3
- 2
- 4
- none of these
Correct

Incorrect

Hint

$\begin{array}{l} \Rightarrow \left( {x - \dfrac{1}{2} - x + \dfrac{3}{2}} \right)\left( {x - \dfrac{1}{2} + x - \dfrac{3}{2}} \right) = x + 2\\ \Rightarrow 2x - 2 = x + 2\\ \Rightarrow x = 4\end{array}$
Question 18 of 36

18. Question
1 point(s)
A boat takes 90 min less to travel 36km downstream than to travel the same distance upstream. If the speed of the boat in still water is 10km/hr. the speed of the stream is
- 2Km/hr
- 2.5Km/hr
- 3Km/hr
- 4Km/hr
Correct

Incorrect

Hint

$\begin{array}{l} \Rightarrow \dfrac{{36}}{{10 - x}} - \dfrac{{36}}{{10 + x}} = \dfrac{{90}}{{60}}\\ \Rightarrow \dfrac{{36\left( {10 + x} \right) - 36\left( {10 - x} \right)}}{{\left( {10 - x} \right)\left( {10 + x} \right)}} = \dfrac{3}{2}\\ \Rightarrow x = 2km/hr\end{array}$
Question 19 of 36

19. Question
1 point(s)
If$\dfrac{{x + 3}}{{x – 2}} – \dfrac{{1 – x}}{x} = \dfrac{{17}}{4}$, then x=?
- 44441
- -2/9
- -2/9 or 4
- 5
Correct

Incorrect

Hint

$\begin{array}{l} \Rightarrow \dfrac{{x + 3}}{{x - 2}} - \dfrac{{1 - x}}{x} = \dfrac{{17}}{4}\\ \Rightarrow 8{x^2} + 8 = 17{x^2} - 34x\\ \Rightarrow 9{x^2} - 34x - 8 = 0\\ \Rightarrow \left( {x - 4} \right)\left( {9x + 2} \right) = 0\\ \Rightarrow x = 4,\dfrac{{ - 2}}{9}\end{array}$
Question 20 of 36

20. Question
1 point(s)
A cistern can be filled in 9hr but it takes 10 hr due to a leak. In how much time will the leak empty the full water?
- 60 hrs
- 70 hrs
- 80 hrs
- 90 hrs
Correct

Incorrect

Hint

$\begin{array}{l} \Rightarrow \dfrac{1}{x} = \dfrac{1}{9} - \dfrac{1}{{10}}\\ \Rightarrow \dfrac{{10 - 9}}{{90}} = \dfrac{1}{{90}}\\ \Rightarrow x = 90hrs\end{array}$
Question 21 of 36

21. Question
1 point(s)
The sum of the squares if two positive number is 3341 and the difference of their squares is 891. The two numbers are
- 46, 35
- 45, 46
- 36, 15
- 46, 36
Correct

Incorrect

Hint

$\begin{array}{l}{a^2} + {b^2} = 3341...(1)\\{a^2} - {b^2} = 891...(2)\end{array}$
On solving equation (1) and (2)
$\begin{array}{l}2{a^2} = 4232\\a = \sqrt {2116} = 46\\b = 35\end{array}$
Question 22 of 36

22. Question
1 point(s)
The solution of the equation${7^{1 + x}} + {7^{1 – x}} = 50$is
- 0
- 2
- ±1
- None of these
Correct

Incorrect

Hint

$\begin{array}{l}{7.7^x} + \dfrac{7}{{{7^x}}} = 50\\ \Rightarrow {7.7^{2x}} - 50 \times {7^x} + 7 = 0\\ \Rightarrow {7.7^{2x}} - 49 \times {7^x} - {7^x} + 7 = 0\\ \Rightarrow {7.7^x}\left( {{7^x} - 7} \right) - 1\left( {{7^x} - 7} \right) = 0\\ \Rightarrow x = 1, - 1\end{array}$
Question 23 of 36

23. Question
1 point(s)
If${x^2} + 2x + c = 0$and${x^2} + 4x – 5 = 0$have a common root which is greater than the other root of second equation then the value of c is
- 3
- -3
- 5
- -5
Correct

Incorrect

Hint

$\begin{array}{l}{x^2} - 4x - 5 = 0\\\left( {x + 5} \right)\left( {x - 1} \right) = 0\\x = 1, - 5\end{array}$
If x=1 is the common root then
$\begin{array}{l}{1^2} + 2\left( 1 \right) + c = 0\\c = - 3\end{array}$
Question 24 of 36

24. Question
1 point(s)
The value of x which satisfy the equation
$\dfrac{2}{{{x^2}}} – \dfrac{5}{x} + 2 = 0$are
- 2,1
- ½,3
- ½,2
- -1/2,-2
Correct

Incorrect

Hint

$\begin{array}{l}2 - 5x - 2{x^2} = 0\\ \Rightarrow x = - 2,\dfrac{{ - 1}}{2}\end{array}$
Question 25 of 36

25. Question
1 point(s)
If x=2 and x=-3/7 are roots of the equation$7{x^2} – kx + 2m = 0$, then (k, m)=
- (11,3)
- (11,-3)
- (-3,11)
- (-3,-11)
Correct

Incorrect

Hint

$\begin{array}{l}7 \times {\left( 2 \right)^2} - 2k + 2m = 0...(1)\\7 \times {\left( {\dfrac{{ - 3}}{7}} \right)^2} - k\left( {\dfrac{{ - 3}}{7}} \right) + 2m = 0...(2)\end{array}$
Solving (1) and (2) we get, $k = 11,m = - 3$
Question 26 of 36

26. Question
1 point(s)
The sum of squares of three consecutive positive integers is 50 then the integers are:-
- 3,4,5
- 4,5,6
- 2,3,4
- None of these
Correct

Incorrect

Hint

$\begin{array}{l}{x^2} + {\left( {x + 1} \right)^2} + {\left( {x + 2} \right)^2} = 50\\3{x^2} + 6x + 5 = 50\\{x^2} + 2x - 45 = 0\\\left( {x + 5} \right)\left( {x - 3} \right) = 0\\x = 3, - 5\end{array}$
Numbers are 3, 4, 5.
Question 27 of 36

27. Question
1 point(s)
The value of k for which${x^2} – 6x + k = 0$has coincident roots is
- 4
- -4
- 9
- -9
Correct

Incorrect

Hint

$\begin{array}{l}{6^2} = 4k\\k = 9\end{array}$
Question 28 of 36

28. Question
1 point(s)
If a, b are the roots of the equation${x^2} – 3x + 2 = 0$then the equation whose roots are (a+1) and (b+1) is
- ${x^2} + 5x + 6 = 0$
- ${x^2} – 5x – 6 = 0$
- ${x^2} + 5x – 6 = 0$
- ${x^2} – 5x + 6 = 0$
Correct

Incorrect

Hint

$\begin{array}{l}{x^2} - \left( {a + b + 1 + 1} \right)x + \left( {a + 1} \right)\left( {b + 1} \right) = 0\\ \Rightarrow {x^2} - \left( {3 + 2} \right)x + \left( {2 + 3 + 1} \right) = 0\\ \Rightarrow {x^2} - 5x + 6 = 0\end{array}$
Question 29 of 36

29. Question
1 point(s)
Id a, b are the roots of the equation${x^2} – 8x + p = 0$and${a^2} + {b^2} = 40$, then p is equal to
- 8
- 10
- 12
- 14
Correct

Incorrect

Hint

$\begin{array}{l}a + b = 8\\ab = p\\{a^2} + {b^2} = 40 = {\left( {a + b} \right)^2} - 2ab\\{8^2} - 2p = 40\\2p = 64 - 40 = 24\\p = 12\end{array}$
Question 30 of 36

30. Question
1 point(s)
A shopkeeper buys a number of books of Rs80. If he had bought 4 more books for the same amount it would have cost Rs1 less. How many books did he buy?
- 14
- 18
- 16
- none of these
Correct

Incorrect

Hint

$\dfrac{{80}}{x} - \dfrac{{80}}{{x + 4}} = 1$
On solving x=16
Question 31 of 36

31. Question
1 point(s)
One root of the equation$a{x^2} + bx + c = 0$is square of the other if
- ${a^2}c + {b^3} + a{c^2} = 3abc$
- ${a^3} + {b^3} + {c^3} = 3abc$
- ${b^2}c + {c^2}a + {a^2}b = abc$
- ${a^2} + {b^2} + {c^2} – ab – bc – ca = 0$
Correct

Incorrect

Hint

$\begin{array}{l}a + {a^2} = \dfrac{{ - b}}{a}\\ \Rightarrow a.{a^2} = \dfrac{c}{a}\\ \Rightarrow a = {\left( {\dfrac{c}{a}} \right)^{1/3}}\\ \Rightarrow a\left( {1 + a} \right) = \dfrac{{ - b}}{a}\\ \Rightarrow {a^3}\left[ {1 + {a^3} + 3a\left( {a + 1} \right)} \right] = \dfrac{{ - {b^3}}}{{{a^3}}}\\ \Rightarrow \dfrac{c}{a}\left[ {1 + \dfrac{c}{a} + 3\left( {\dfrac{{ - b}}{a}} \right)} \right] = \dfrac{{ - {b^3}}}{{{a^3}}}\\ \Rightarrow ac\left( {a + c - 3b} \right) + {b^3} = 0\\ \Rightarrow {a^2}c + {b^3} + a{c^2} = 3abc\end{array}$
Question 32 of 36

32. Question
1 point(s)
If one root of the quadratic equation is$\sqrt 3 + 1$then the equation whose coefficient are rational is
- ${x^2} – 2x – \sqrt 3 = 0$
- ${x^2} – 5x + 2 = 0$
- ${x^2} – 2x – 2 = 0$
- none of these
Correct

Incorrect

Hint

The other root will be $1 - \sqrt 3$
So, equation is ${x^2} - 2x - 2 = 0$
Question 33 of 36

33. Question
1 point(s)
If the ratio of the roots of the equation${x^2} – 2ax + b = 0$is equal to that of the roots
${x^2} – 2cx + d = 0$then
- ${a^2}b = {c^2}d$
- ${a^2}c = {b^2}d$
- ${a^2}d = {c^2}b$
- ${d^2}b = {c^2}a$
Correct

Incorrect

Hint

Let the roots be a, b, ka, kb then
$\begin{array}{l}a + b = 2a...(1),ab = b...(2)\\ \Rightarrow k\left( {a + b} \right) = 2c...(3)\\ \Rightarrow {k^2}ab = d...(4)\\ \Rightarrow k\left( {2a} \right) = 2c\\ \Rightarrow k = \dfrac{c}{a}...(5),{k^2}b = d...(6)\end{array}$
From (5) and (6),
$\begin{array}{l}{\left( {\dfrac{c}{a}} \right)^2} = \dfrac{d}{b}\\ \Rightarrow {a^2}d = b{c^2}\end{array}$
Question 34 of 36

34. Question
1 point(s)
The hypotenuse of a grassy land in the shape of a right triangle is 1 meter more than twice the shortest side. If the third side is 7meter more than the shortest side. Find the sides of the grassy land
- 8m, 17m, 15m
- 8m, 12m, 15m
- 9m, 18m, 25m
- none of these
Correct

Incorrect

Hint

$\begin{array}{l}{\left( {2x + 1} \right)^2} = {x^2} + {\left( {x + 7} \right)^2}\\ \Rightarrow 4{x^2} + 4x + 1 = 2{x^2} + 14x + 49\\ \Rightarrow 2{x^2} - 10x - 48 = 0\\ \Rightarrow {x^2} - 5x - 24 = 0\\ \Rightarrow \left( {x - 8} \right)\left( {x + 3} \right) = 0\\ \Rightarrow x = 8\end{array}$
Hypotenuse $= 2x + 1 = 2 \times 8 + 1 = 17m$
Third side $= x + 7 = 8 + 7 = 15m$
Question 35 of 36

35. Question
1 point(s)
If the sum of n successive odd natural numbers staring from 3 is 48. Find the number of terms.
- 6
- 7
- 8
- 3
Correct

Incorrect

Hint

$\begin{array}{l} \Rightarrow \dfrac{n}{2}\left[ {2 \times 3 + \left( {n - 1} \right) \times 2} \right] = 48\\ \Rightarrow n\left[ {3 + n - 1} \right] = 48\\ \Rightarrow {n^2} + 2n - 48 = 0\\ \Rightarrow \left( {n + 8} \right)\left( {n - 6} \right) = 0\\ \Rightarrow n = - 8,6 \Rightarrow n = 6\end{array}$
Question 36 of 36

36. Question
1 point(s)
If sin q and cos q are the roots of equation$a{x^2} + bx + c = 0$, then
- ${b^2} + {a^2} = 2ac$
- ${b^2} – {a^2} = ac$
- ${b^2} – {a^2} = 2ac$
- none of these
Correct

Incorrect

Hint

${a^2}{b^2} = 1$ , here
$\begin{array}{l}a = \sin q,b = \cos q\\ \Rightarrow {a^2} + {b^2} = {\left( {a + b} \right)^2} - 2ab\\ \Rightarrow 1 = {\left( {\dfrac{{ - b}}{a}} \right)^2} - \dfrac{{2c}}{a} \Rightarrow {b^2} - {a^2} = 2ac\end{array}$