Mathematics Class X

Question 1 of 25

1. Question
1 point(s)
Hitasha is four years older than her friend Sulekha. Hitasha’s mother is twice as old as Hitasha and Sulekha is twice as old as her brother. The age of Hitasha’s mother is 41 years more than the age of Sulekha’s brother. Find the age of Hitasha and Sulekha.
- 24 and 20 years
- 22 and 18 years
- 28 and 24 years
- 26 and 22 years
Correct

Incorrect

Hint

Let the ages of Hitasha and Sulekha be x and y years respectively.
y+x=4
Hitasha’s mother(z)=2x
Sulekha’s brother $W = \dfrac{y}{2}$
$\begin{array}{l} \Rightarrow z + 41 = W\\ \Rightarrow 2x + 41 = \dfrac{y}{2}\\ \Rightarrow 2x + 41 = \dfrac{{x + 4}}{2}\\ \Rightarrow 4x + 82 = x + 4\\ \Rightarrow 3x = 78\\ \Rightarrow x = 26\\ \Rightarrow y = 26 - 4 = 22\end{array}$
Question 2 of 25

2. Question
1 point(s)
Solve by the method of cross multiplication
$\begin{array}{l}5x – 8y – 58 = 0\\3x + 9y – 21 = 0\end{array}$
- $x=10,y=-1$
- $x=-22,y=-13$
- $x=-15,y=-11$
- $x=-13,y=-3$
Correct

Incorrect

Hint

$\begin{array}{l} \Rightarrow \dfrac{x}{{ - 8 \times \left( { - 21} \right) - 9 \times \left( { - 58} \right)}} = \dfrac{y}{{ - 58 \times 3 - \left( { - 21} \right) \times 5}} = \dfrac{1}{{5 \times 9 - 3 \times \left( { - 8} \right)}}\\ \Rightarrow \dfrac{x}{{168 + 522}} = \dfrac{1}{{45 + 24}},\dfrac{y}{{ - 174 + 105}} = \dfrac{1}{{69}}\\ \Rightarrow \dfrac{x}{{690}} = \dfrac{1}{{69}},\dfrac{y}{{ - 69}} = \dfrac{1}{{69}}\\ \Rightarrow \dfrac{x}{{690}} = \dfrac{1}{{69}},\dfrac{y}{{ - 69}} = \dfrac{1}{{69}}\\ \Rightarrow x = 10,y = - 1\end{array}$
Question 3 of 25

3. Question
1 point(s)
On selling a holder box at 40% gain and bowl at 50% gain a seller gains Rs16. If he sells the holder box at 40% gain and bowl box at 70% gain he gains Rs22. Find the actual price of holder box and bowl box.
- Rs7.5, Rs35
- Rs17.5, Rs45
- Rs12.5, Rs40
- Rs2.5, Rs30
Correct

Incorrect

Hint

$\begin{array}{l} \Rightarrow \left( {\dfrac{{140x}}{{100}} + \dfrac{{150y}}{{100}}} \right) - \left( {x + y} \right) = 16\\ \Rightarrow \dfrac{{40x}}{{100}} + \dfrac{{50y}}{{100}} = 16...(1)\end{array}$
Similarly,
$\dfrac{{40x}}{{100}} + \dfrac{{70y}}{{100}} = 22...(2)$
Solving 2 equations we get x=2.5, y=30
Question 4 of 25

4. Question
1 point(s)
Solve the following system of equations by the method of substitution.
$\begin{array}{l}\left( {m – n} \right)x + \left( {m + n} \right)y = {m^2} + {n^2}\\\left( {m + n} \right)x + \left( {m – n} \right)y = {m^2} + {n^2}\end{array}$
- $x = \dfrac{{{m^2} + {n^2}}}{{2m}},y = \dfrac{{{m^2} + {n^2}}}{{2m}}$
- $x = \dfrac{{{m^2} – {n^2}}}{{2m}},y = \dfrac{{{m^2} + {n^2}}}{{2m}}$
- $x = \dfrac{{{m^2} + {n^2} + 1}}{{2m}},y = \dfrac{{{m^2} + {n^2}}}{{2m}}$
- None of these
Correct

Incorrect

Hint

$x = \dfrac{{{m^2} + {n^2} - \left( {m + n} \right)y}}{{m - n}}$
Put this value of x in equation (2),
$\begin{array}{l} \Rightarrow \dfrac{{\left( {m + n} \right)\left( {{m^2} + {n^2} - my - ny} \right)}}{{m - n}} + my - ny = {m^2} + {n^2}\\ \Rightarrow {m^3} + m{n^2} - {m^2}y - mny + n{m^2} + {n^3} - nmy - {n^2}y + {m^2}y - nmy - nmy + {n^2}y = \left( {{m^2} + {n^2}} \right)\left( {m + n} \right)\end{array}$ On solving we get
$x = \dfrac{{{m^2} + {n^2}}}{{2m}},y = \dfrac{{{m^2} + {n^2}}}{{2m}}$
Question 5 of 25

5. Question
1 point(s)
Points P and Q are 44km apart on a highway. A car starts from P and another car starts from Q at the same time. If they travel in the same direction they meet in 4 hours. But if they travel towards each other they meet in one hour. What are their speeds?
- 27.5 Km/hr and 16.5 Km/ hr
- 28.5 Km/ hr and 17.5 Km/ hr
- 29.5 Km/hr and 18.5 Km/hr
- 26.5 Km/hr and 15.5 Km/hr
Correct

Incorrect

Hint

$\begin{array}{l}\dfrac{{44}}{x} + \dfrac{{44}}{y} = 4...(1)\\\dfrac{{44}}{x} - \dfrac{{44}}{y} = 1...(2)\end{array}$
Adding equations (1) and (2),
$\dfrac{{88}}{x} = 5$
x=16.5Km/hr, y=27.5Km/hr
Question 6 of 25

6. Question
1 point(s)
If the length of a rectangle is increased by 22 units and its breadth is decreased by 11 units then the area of the rectangle is increased by 55 square units. However if we decrease its length by 11 units and increase breadth by 5 units. Its area is decreased by 275 square units. Find the length and breadth of the rectangle.
- 143 units and 85 units
- 19 units and 85 units
- 143 units and 89 units
- 142 units and 87 units
Correct

Incorrect

Hint

Let the length be x and breadth be y. Area =xy
According to question,
$\begin{array}{l}\left( {x + 22} \right)\left( {y - 11} \right) = xy + 55...(1)\\\left( {x - 11} \right)\left( {y + 5} \right) = xy - 275...(2)\end{array}$
Solving 2 equations we get, x=143, y=85
Question 7 of 25

7. Question
1 point(s)
If a two digit number, its unit’s length is 3 more than ten’s digit. If sum of this number and the number obtained by reversing the order of its digits is 143. Find the number.
- 85
- 58
- 69
- 47
Correct

Incorrect

Hint

Let the ten’s digit be x
Unit’s digit be x+3
Number=10x+x+3=11x+3
Number obtained by reversing the digit
$\begin{array}{l} = 10x + 30 + x\\ = 11x + 30\end{array}$
According to question,
$\begin{array}{l} \Rightarrow 11x + 3 + 11x + 30 = 143\\ \Rightarrow 22x = 110\\ \Rightarrow x = 5\end{array}$
Number is 58.
Question 8 of 25

8. Question
1 point(s)
It is given the sum of the digits of a two digit number is 10. If 54 is added to the number the digits interchange their place. Find the number.
- 28
- 82
- 37
- 19
Correct

Incorrect

Hint

x+y=10…(1)
$\begin{array}{l} \Rightarrow 10x + y + 54 = 10y + x\\ \Rightarrow 9x - 9y = - 54\\ \Rightarrow x - y = - 6...(2)\end{array}$
On solving 2nd equations,
$2x = 4 \Rightarrow x = 2,y = 8$
Number is 28.
Question 9 of 25

9. Question
1 point(s)
Solve the following system of linear equations by equating the coefficient.
$\begin{array}{l}\dfrac{7}{x} + \dfrac{8}{y} = 402\\\dfrac{5}{x} + \dfrac{6}{y} = 298\end{array}$
- $x=8 \ \text{and} \ y=14$
- $x=\dfrac{1}{14} \ \text{and}, y=\dfrac{1}{38}$
- $x=14 \ \text{and} \ y=38$
- $x=13 \text{and} \ y=37$
Correct

Incorrect

Hint

$\begin{array}{l} \Rightarrow \dfrac{7}{x} + \dfrac{8}{y} = 402...(1) \times 5,\dfrac{5}{x} + \dfrac{6}{y} = 298...(2) \times 7\\ \Rightarrow \dfrac{{35}}{x} + \dfrac{{40}}{y} = 2010...(3),\dfrac{{35}}{x} + \dfrac{{42}}{y} = 2086...(4)\end{array}$
Solving equation (3) and equation (4),
$x = \dfrac{1}{{14}},y = \dfrac{1}{{38}}$
Question 10 of 25

10. Question
1 point(s)
Solve the following pair by the method of cross multiplication.
$\begin{array}{l}\dfrac{x}{g} + \dfrac{y}{h} = g + h\\\dfrac{x}{{{g^2}}} + \dfrac{y}{{{h^2}}} = 0\end{array}$
- $x = \dfrac{{ – 9{g^2}h + {g^3}}}{{g – h}},y = \dfrac{{9g{h^2} – {h^3}}}{{g – h}}$
- $x = \dfrac{{ – 9{g^2}h + {g^4}}}{{g + h}},y = \dfrac{{9g{h^2} + {h^3}}}{{g – h}}$
- $x = \dfrac{{ – 9{h^3} + {g^3}}}{{{g^2} – {h^2}}},y = \dfrac{{9g{h^2} – {h^3}}}{{g – h}}$
- $x = \dfrac{{ – 9{g^3}h + {g^4}}}{{{g^2} – h}},y = \dfrac{{9{g^2}{h^2} – {h^3}}}{{g – h}}$
Correct

Incorrect

Hint

$\begin{array}{l}\dfrac{x}{g} + \dfrac{y}{h} - \left( {g + h} \right) = 0...(1),\dfrac{x}{{{g^2}}} + \dfrac{y}{{{h^2}}} - 10 = 0...(2)\\ \Rightarrow \dfrac{x}{{\dfrac{1}{h} \times \left( { - 10} \right) - \left( {\dfrac{1}{{{h^2}}} \times \left( { - g - h} \right)} \right)}} = \dfrac{y}{{ - \left( {g + h} \right) \times \dfrac{1}{{{h^2}}} - \left( { - 10 \times \dfrac{1}{y}} \right)}}\\ \Rightarrow \dfrac{1}{{\dfrac{1}{g} \times \dfrac{1}{{{h^2}}} - \dfrac{1}{{{g^2}}} \times \dfrac{1}{h}}}\\ \Rightarrow \dfrac{x}{{\dfrac{{ - 10}}{h} - \dfrac{{\left( { - g - h} \right)}}{{{h^2}}}}} = \dfrac{1}{{\dfrac{1}{{g{h^2}}} - \dfrac{1}{{{g^2}h}}}}\\ \Rightarrow \dfrac{y}{{\dfrac{{ - \left( {g + h} \right)}}{{{h^2}}} + \dfrac{{10}}{y}}} = \dfrac{1}{{g{h^2} - \dfrac{1}{{{g^2}h}}}}\\ \Rightarrow x = \dfrac{{ - 10h + g + h}}{{{h^2}}} \times \dfrac{{{g^2}{h^2}}}{{g - h}}\\ \Rightarrow x = \dfrac{{ - 9{g^2}h + {g^3}}}{{g - h}},y = \dfrac{{9g{h^2} - {h^3}}}{{g - h}}\end{array}$
Question 11 of 25

11. Question
1 point(s)
Rohit went to withdraw Rs.8500 from the bank. He asked the cahier to give Rs.500 and Rs.1000 notes only. Rohit got 12 notes in all. Find the number of Rs.500 and Rs.1000 notes he received.
- 7 and 5
- 10 and 7
- 8 and 8
- 9 and 6
Correct

Incorrect

Hint

Let the notes of Rs500 be x and Rs1000 be (12-x)
$\begin{array}{l} \Rightarrow 500x + 1000(12 - x) = 8500\\ \Rightarrow 500x + 12000 - 1000x = 8500\\ \Rightarrow - 500x = - 3500\\ \Rightarrow x = 7\\ \Rightarrow y = 12 - 7 = 5\end{array}$
Question 12 of 25

12. Question
1 point(s)
It is given that the sum of digits of a two digit number is 8. If 36 is added to the number, the digits interchange their place. Find the number.
- 35
- 62
- 17
- 26
Correct

Incorrect

Hint

$\begin{array}{l} \Rightarrow x + y = 8...(1)\\ \Rightarrow 10x + y + 36 = 10y + x\\ \Rightarrow 9x - 9y = - 36\\ \Rightarrow x - y = - 4...(2)\end{array}$
Solving 2 equations we get
$\begin{array}{l}2x = 4 \Rightarrow x = 2\\y = 6\end{array}$
The number is 26.
Question 13 of 25

13. Question
1 point(s)
If thrice the daughter’s in years is added to mother’s age the sum is 66. If thrice the mother’s age is added to the daughter’s age the sum is 110. Find the age of mother and daughter.
- 34 and 11 years
- 32 and 10 years
- 33 and 10 years
- 33 and 11 years.
Correct

Incorrect

Hint

Let the daughter’s age be x and mother’s age be y.
According to question,
$\begin{array}{l}3y + y = 66...(1)\\3y + x = 110...(2)\end{array}$
On solving x=11, y=33
Question 14 of 25

14. Question
1 point(s)
The value of k for which the equation $x – 7y = 4$ and$2x = ky = 8$ has a unique solution is
- k≠14
- k=17
- k=14
- k≠17
Correct

Incorrect

Hint

$\begin{array}{l} \Rightarrow \dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}\\ \Rightarrow \dfrac{1}{2} \ne \dfrac{{ - 7}}{{ - k}} \Rightarrow k \ne 14\end{array}$
Question 15 of 25

15. Question
1 point(s)
Find the value of k for which two equations $ – 3p + kq + 1 = 0$ and $ – 3p + q – 2 = 0$ will have no solution.
- -1
- 2
- 1
- 3
Correct

Incorrect

Hint

$\begin{array}{l} \Rightarrow \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}\\ \Rightarrow \dfrac{{ - 3}}{{ - 3}} = \dfrac{k}{1}\\ \Rightarrow k = 1\end{array}$
Question 16 of 25

16. Question
1 point(s)
Find the value of k such that equations $ – 6p – 3q – 6 = 0$ and $ – 2p + kq – 2 = 0$ represents coincident lines.
- -2
- -3
- -1
- 0
Correct

Incorrect

Hint

$\begin{array}{l} \Rightarrow \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}\\ \Rightarrow \dfrac{{ - 6}}{{ - 2}} = \dfrac{{ - 3}}{k} = \dfrac{{ - 6}}{{ - 2}}\\ \Rightarrow k = - 1\end{array}$
Question 17 of 25

17. Question
1 point(s)
Two years ago Archana was four times older than her daughter. After two years Archana will be four years more than two times the age of her daughter. Find the present age of Archana and her daughter.
- 17 and 5 years
- 18 and 6 years
- 19 and 7 years
- 20 and 8 years
Correct

Incorrect

Hint

Let the present age of Archana be x and daughter be y years respectively.
According to question,
$\begin{array}{l}x - 2 = \left( {y - 2} \right) \times 4...(1)\\x + 2 = 2 \times \left( {y + 2} \right) + 4...(2)\end{array}$
Solving equation (1) and (2) we get, x=18, y=6
Question 18 of 25

18. Question
1 point(s)
If 6 is added to the numerator and denominator of a dfraction, the dfraction becomes $\dfrac{{27}}{{31}}$and if 2 is subtracted from its numerator and denominator it becomes $\dfrac{{19}}{{23}}$. Find the dfraction.
- $\dfrac{{22}}{{26}}$
- $\dfrac{{21}}{{25}}$
- $\dfrac{{25}}{{21}}$
- $\dfrac{{26}}{{22}}$
Correct

Incorrect

Hint

$\begin{array}{l}\dfrac{{x + 6}}{{y + 6}} = \dfrac{{27}}{{31}}\\31x + 186 = 27y + 162...(1)\\\dfrac{{x - 2}}{{y - 2}} = \dfrac{{19}}{{23}}\\23x - 46 = 19y - 38...(2)\end{array}$
Solving equation (1) and (2) we get,
x=21, y=25
Question 19 of 25

19. Question
1 point(s)
The ratio between the two digits number and the sum of its digits of that number is 4:1. If the digit in the unit place is 3 more than the digit in the tenth place. What is the number?
- 40
- 24
- 63
- 36
Correct

Incorrect

Hint

$\begin{array}{l}\dfrac{{10x + y}}{{x + y}} = \dfrac{4}{1} \Rightarrow 10x + y = 4x + y\\ \Rightarrow 6x - 3y = 0...(1)\end{array}$
Let the ten’s place be x,
$\begin{array}{l}y = x + 3\\y - x = 3...(2)\end{array}$
Solving 2 equations,
$\begin{array}{l} \Rightarrow 6x - 3y = 0\\ \Rightarrow \dfrac{{ - 6x + 6y = 18}}{{3y = 18}}\\ \Rightarrow y = 6\\ \Rightarrow 6 - x = 3\\ \Rightarrow x = 3\end{array}$
Number is 36.
Question 20 of 25

20. Question
1 point(s)
The value of k for which the system of equations $x+2y−3=0$ and $5x+ky+7=0$ has no solution is
- 3
- 1
- 10
- 6
Correct

Incorrect

Hint

$\begin{array}{l} \Rightarrow \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}\\ \Rightarrow \dfrac{1}{5} = \dfrac{2}{k}\\ \Rightarrow k = 10\end{array}$
Question 21 of 25

21. Question
1 point(s)
Value of x in pair of linear equations. $170x−396y=-169$ and $-396x+170y=−735$ is
- 3.2
- 2.5
- 1.3
- 2.4
Correct

Incorrect

Hint

$\begin{array}{l}170x - 396y = - 169 \Rightarrow x = \dfrac{{ - 169 + 396y}}{{170}}...(1)\\ - 396x + 170y = - 735...(2)\end{array}$
Solving 2 equations we get,
$\begin{array}{l} \Rightarrow - 396\left( {\dfrac{{ - 169 + 396y}}{{170}}} \right) + 170y = - 735\\ \Rightarrow 66924 - 156816y + 28900y = - 124950\\ \Rightarrow - 127916y = - 191874\\ \Rightarrow y = 1.5,x = 2.5\end{array}$
Question 22 of 25

22. Question
1 point(s)
The value of k for which the system of equations $5x+2y=0$ and $kx+20y=0$ has a non-zero solution
- 50
- 60
- 20
- 80
Correct

Incorrect

Hint

$\begin{array}{l}\dfrac{5}{k} = \dfrac{2}{{20}}\\k = \dfrac{{5 \times 20}}{2} = 50\end{array}$
Question 23 of 25

23. Question
1 point(s)
The value of k for which the system of equations $x+ky+30=0$ and $5x+30y−150=0$ represent parallel lines is
- 13
- 9
- 16
- 6
Correct

Incorrect

Hint

$\begin{array}{l} \Rightarrow \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}\\ \Rightarrow \dfrac{1}{5} = \dfrac{k}{{30}}\\ \Rightarrow 5k = 30\\ \Rightarrow k = 6\end{array}$
Question 24 of 25

24. Question
1 point(s)
Solve for $x \text{and} \ y$
$\begin{array}{l}\dfrac{{50}}{{x – y}} + \dfrac{{66}}{{x + y}} = 30\\\dfrac{{60}}{{x – y}} + \dfrac{{77}}{{x + y}} = 33\end{array}$
- $x=5, \ y=2$
- x=$\dfrac{{ – 1}}{5}$, y=$\dfrac{{ – 14}}{{15}}$
- $x=5, \ y=6$
- $x=2, \ y=\dfrac{4}{5}$
Correct

Incorrect

Hint

Let
$\begin{array}{l}\dfrac{1}{{x - y}} = U,\dfrac{1}{{x + y}} = V\\50U + 66V = 30...(1) \times 6\\60U + 77V = 33...(2) \times 5\end{array}$
On solving we get,
$\begin{array}{l}300U + 396V = 180...(3)\\300U + 385V = 165...(4)\end{array}$
Subtracting equation (3) from (4),
$\begin{array}{l} \Rightarrow 11V = 15\\ \Rightarrow V = \dfrac{{15}}{{11}},U = \dfrac{{ - 6}}{5}\\ \Rightarrow x - y = \dfrac{{11}}{{15}}\\ \Rightarrow x + y = \dfrac{{ - 5}}{6}\\ \Rightarrow 2x = \dfrac{{11}}{5} + \left( {\dfrac{{ - 5}}{6}} \right) = \dfrac{{ - 3}}{{30}} = \dfrac{{ - 1}}{{10}}\\ \Rightarrow x = \dfrac{{ - 1}}{5}\\ \Rightarrow \dfrac{{ - 1}}{5} - y = \dfrac{{11}}{{15}}\\ \Rightarrow - y = \dfrac{{11}}{{15}} + \dfrac{1}{5} = \dfrac{{11 + 3}}{{15}} = \dfrac{{14}}{{15}}\\ \Rightarrow y = \dfrac{{ - 14}}{{15}}\end{array}$
Question 25 of 25

25. Question
1 point(s)
The denominator of a rational number is greater than its numerator by 4. If 4 is subtracted from the numerator and 6 is added to the denominator, the new number becomes $\dfrac{1}{4}$. The original number was
- $\dfrac{{13}}{{15}}$
- $\dfrac{{14}}{{17}}$
- $\dfrac{{13}}{{19}}$
- $\dfrac{{17}}{{12}}$
Correct

Incorrect

Hint

Let the numerator be x
Denominator by x+4
$\begin{array}{l}\dfrac{{x - 4}}{{x + 4 + 6}} = \dfrac{1}{4}\\ \Rightarrow 4\left( {x - 4} \right) = x + 10\\ \Rightarrow 4x - 16 = x + 10\\ \Rightarrow 3x = 26\\ \Rightarrow x = \dfrac{{26}}{3}\\ \Rightarrow y = \dfrac{{26}}{3} + 4 = \dfrac{{26 + 12}}{3} = \dfrac{{38}}{3}\\ \Rightarrow \dfrac{x}{y} = \dfrac{{26}}{{38}} = \dfrac{{13}}{{19}}\end{array}$