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Maths M.C.Q (1 Marks)

MODEL PAPER 1 (BASIC) · CBSE STD 10 (CBSE - English Medium). 18 questions.

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M.C.Q (1 Marks)

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18 questions · timed · auto-graded

MCQ 11 Mark
If the mean of a frequency distribution is 8.1 and $\sum f _{ i } x _{ i }=132+5 k , \sum f _{ i }=20$ then k =
  • A
    3
  • B
    4
  • C
    5
  • ✓
    6
Answer
Correct option: D.
6
(D) 6
Explanation: Mean = 8.1
$\Sigma f _{ i } x _{ i }=132+5 k$
$\sum f_i=20$
$\therefore$ Mean $=\frac{\Sigma f_i x_i}{\Sigma f_i} \Rightarrow 8.1=\frac{132+5 k}{20}$
$\Rightarrow 132+5 k=8.1 \times 20=162$
$\Rightarrow 5 k =162-132=30$
$\Rightarrow k =\frac{30}{5}=6$
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MCQ 21 Mark
A cylindrical tub of radius 5 cm and height 9.8 cm is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed into the tub. If the radius of the hemisphere is 3.5 cm and the height of the cone outside the hemisphere is 5 cm, then find the volume of water left in the tub.(Take $\pi=\frac{22}{7}$ )
  • A
    $716 cm^3$
  • B
    $616 cm^3$
  • C
    $600 cm^3$
  • D
    $535 cm^3$
Answer
(B) $616 cm^3$
Image
Volume of water in the cylindrical tub = Volume of the tub
$=\pi r^2 h=\left(\frac{22}{7} \times 5 \times 5 \times 9.8\right) cm ^3=770 cm^3$
Volume of the solid immersed in the tub = Volume of the hemisphere + Volume of the cone
$=\left[\left(\frac{2}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2}\right)+\left(\frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 5\right)\right] cm ^3$
$=\left(\frac{539}{6}+\frac{385}{6}\right) cm ^3=\left(\frac{924}{6}\right) cm ^3=154 cm^3$
Volume of water left in the tub = Volume of the tub - Volume of solid immersed
$=(770-154) cm ^3=616 cm^3$
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MCQ 31 Mark
If 35 is removed from the data: 30, 34, 35, 36, 37, 38, 39, 40, then the median increases by:
  • ✓
    0.5
  • B
    1.5
  • C
    2
  • D
    1
Answer
Correct option: A.
0.5
(A) 0.5
Explanation:Given data = 30, 34, 35, 36, 37, 38, 39, 40
Here n = 8 which is even
$\therefore$ Median $=\frac{1}{2}\left[\frac{n}{2}\right.$ th $+\left(\frac{n}{2}+1\right)$ th $]$ term $=\frac{1}{2}(4$ th +5 th term $)$
$=\frac{1}{2}(36+37)=\frac{73}{2}=36.5$
After removing 35, then n = 7
$\therefore$ New median $=\frac{7+1}{2}$ th term $=4$ th term $=37$
$\therefore$ Increase in median $=37-36.5=0.5$
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MCQ 41 Mark
Raju bought a fish from a shop for his aquarium. The shop keeper takes out one fish from a tank containing 15 male fish and 18 female fish. The probability that the fish taken out is a male fish is
  • ✓
    $\frac{5}{11}$
  • B
    $\frac{6}{11}$
  • C
    $\frac{5}{12}$
  • D
    $\frac{7}{11}$
Answer
Correct option: A.
$\frac{5}{11}$
(A) $\frac{5}{11}$
Explanation: Total number of fish = 15 + 18 = 33
Male fish = 15
Number of possible outcomes = 15
Number of total outcomes = 15 + 18 = 33
Required Probability = $\frac{15}{33}=\frac{5}{11}$
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MCQ 51 Mark
A chord of a circle subtends an angle of $60^{\circ}$ at the centre. If the length of the chord is 100 cm, find the area of the major segment.
  • A
    $30391.7 cm^2$
  • B
    $30720.5 cm^2$
  • C
    $30520.61 cm^2$
  • D
    $31021.42 cm^2$
Answer
(C) $30520.61 cm^2$
Explanation: In $\triangle OAB$, by angle sum property
Image
$60^{\circ}+\angle O A B+\angle O B A=180^{\circ}$
$\Rightarrow 2 \angle O A B=120^{\circ} \Rightarrow \angle O A B=60^{\circ}$
$\Rightarrow \triangle O A B$ is an equilateral triangle.
$\Rightarrow r =100 cm$
Area of major segment
= Area of major sector + Area of $\triangle OAB$
$=\frac{\left(360^{\circ}-60^{\circ}\right)}{360^{\circ}} \times \pi r^2+\frac{\sqrt{3}}{4} r^2$
$=\frac{300}{360} \times \frac{22}{7} \times(100)^2+\frac{\sqrt{3}}{4} \times 100^2$
$=\frac{5}{6} \times \frac{22}{7} \times(100)^2+\sqrt{3} \times 2500$
= 26190.48 + 4330.13 = 30520.61$cm ^2$
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MCQ 71 Mark
$\frac{\sin \theta}{1+\cos \theta}$ is equal to
  • A
    $\frac{1-\sin \theta}{\cos \theta}$
  • B
    $\frac{1-\cos \theta}{\cos \theta}$
  • ✓
    $\frac{1-\cos \theta}{\sin \theta}$
  • D
    $\frac{1+\cos \theta}{\sin \theta}$
Answer
Correct option: C.
$\frac{1-\cos \theta}{\sin \theta}$
(C) $\frac{1-\cos \theta}{\sin \theta}$
Explanation: We have, $\frac{\sin \theta}{1+\cos \theta}=\frac{\sin \theta(1-\cos \theta)}{(1+\cos \theta)(1-\cos \theta)}$
$=\frac{\sin \theta(1-\cos \theta)}{1-\cos ^2 \theta}=\frac{\sin \theta(1-\cos \theta)}{\sin ^2 \theta}$
$=\frac{1-\cos \theta}{\sin \theta}$
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MCQ 81 Mark
An electric pole is $10 \sqrt{3}$ m high and its shadow is 10 m in length, then the angle of elevation of the sun is
  • A
    $45^{\circ}$
  • B
    $15^{\circ}$
  • C
    $30^{\circ}$
  • D
    $60^{\circ}$
Answer
(D) $60^{\circ}$
Explanation:
Image
Let AB be the electric pole of height $10 \sqrt{3} m$ and its shadow be BC of length 10 m. And the angle of elevation of the sun be $\theta$.
$\therefore \tan \theta=\frac{ AB }{ BC }$
$\Rightarrow \tan \theta=\frac{10 \sqrt{3}}{10}$
$\Rightarrow \tan \theta=\sqrt{3}$
$\Rightarrow \tan \theta=\tan 60^{\circ}$
$\Rightarrow \theta=60^{\circ}$
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MCQ 91 Mark
$(\operatorname{cosec} \theta-\cot \theta)^2=?$
  • A
    $\frac{1+\sin \theta}{1-\sin \theta}$
  • ✓
    $\frac{1-\cos \theta}{1+\cos \theta}$
  • C
    $\frac{1-\sin \theta}{1+\sin \theta}$
  • D
    $\frac{1+\cos \theta}{1-\cos \theta}$
Answer
Correct option: B.
$\frac{1-\cos \theta}{1+\cos \theta}$
(b) $\frac{1-\cos \theta}{1+\cos \theta}$
Explanation: $(\operatorname{cosec} \theta-\cot \theta)^2=\left(\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}\right)^2=\frac{(1-\cos \theta)^2}{\sin ^2 \theta}=\frac{(1-\cos \theta)^2}{\left(1-\cos ^2 \theta\right)}=\frac{(1-\cos \theta)}{(1+\cos \theta)}$
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MCQ 101 Mark
The distance between two parallel tangents of a circle of radius 3 cm is
Image
  • ✓
    6 cm
  • B
    3 cm
  • C
    4.5 cm
  • D
    12 cm
Answer
Correct option: A.
6 cm
(A) 6 cm
Explanation: Since the distance between two parallel tangents of a circle is equal to the diameter of the circle.
Given: Radius (OP) = 3 cm
$\therefore$ Diameter $=2 \times$ Radius $=2 \times 3=6 cm$
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MCQ 111 Mark
In the adjoining figure P and Q are points on the sides AB and AC respectively of $\triangle A B C$ such that AP = 3.5 cm, PB = 7cm, AQ = 3 cm, QC = 6 cm and PQ = 4.5 cm. The measure of BC is equal to
Image
  • ✓
    13.5 cm.
  • B
    12.5 cm.
  • C
    9 cm.
  • D
    15 cm
Answer
Correct option: A.
13.5 cm.
(A) 13.5 cm
Explanation: In $\triangle ABC$,
$\Rightarrow \frac{ AQ }{ QC }=\frac{ AP }{ PB } \Rightarrow \frac{3}{6}=\frac{3.5}{7} \Rightarrow \frac{1}{2}$
Since $\frac{ AQ }{ QC }=\frac{ AP }{ PB }$,
therefore, $QP \| BC$
$\therefore \frac{ AQ }{ AC }=\frac{ QP }{ BC }$
$\Rightarrow \frac{1}{3}=\frac{4.5}{ BC }$
$\Rightarrow BC =13.5 cm$
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MCQ 121 Mark
In triangles ABC and DEF, $\angle A =\angle E =40^{\circ}, AB : ED = AC : EF$ and $\angle F =65^{\circ}$, then $\angle B =$
  • A
    $75^{\circ}$
  • B
    $85^{\circ}$
  • C
    $35^{\circ}$
  • D
    $65^{\circ}$
Answer
(A) $75^{\circ}$
Explanation: In $\triangle ABC$ and $\triangle DEF$
$\angle A=\angle E=40^{\circ}$
$A B: E D=A C: E F, \angle F=65^{\circ}$
Image
$\Rightarrow \frac{ AB }{ ED }=\frac{ AC }{ EF }$
$\because$ In $\triangle ABC$ and $\triangle EDF$
$\angle A =\angle E \quad\left(\right.$ each $\left.=40^{\circ}\right)$
$\frac{A B}{E D}=\frac{A C}{E F}$
$\therefore \triangle ABC \sim \triangle EDF$ (SAS criterion)
$\therefore \angle C=\angle F=65^{\circ}$
and $\angle B =\angle D$
But $\angle A +\angle B +\angle C =180^{\circ}$ (Sum of angles of a triangle)
$\Rightarrow 40^{\circ}+65^{\circ}+\angle C=180^{\circ}$
$\Rightarrow 105^{\circ}+\angle C=180^{\circ}$
$\therefore \angle C=180^{\circ}-105^{\circ}=75^{\circ}$
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MCQ 131 Mark
Two vertices of $\triangle ABC$ are A (-1, 4) and B(5, 2) and its centroid is G(0, -3). Then, the coordinates of C are
  • A
    (4, 3)
  • B
    (4, 15)
  • ✓
    (-4, -15)
  • D
    (-15, -4)
Answer
Correct option: C.
(-4, -15)
(C) (-4, -15)
Explanation:Let the vertex C be C (x,y). Then
$\frac{-1+5+x}{1}$= 0 and $\frac{4+2+y}{3}=-3 \Rightarrow x+4=0$ and $6+y=-9$
$\therefore \quad x=-4$ and $y=-15$
so, the coordinates of C are (-4, -15).
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MCQ 141 Mark
$x^2-6 x+6=0$ have
  • A
    Real and Equal roots
  • B
    Real roots
  • C
    No Real roots
  • D
    Real and Distinct roots
Answer
(D) Real and Distinct roots
Explanation: Comparing the given equation to the below equation
$a x^2+b x+c=0$
$a=1, b=-6, c=6$
$D=b^2-4 a c$
$D=(-6) 2-4 \times 1 \times 6$
$D=36-24$
D = 12
$D>0$.
If $b^2-4 a c>0$, then the equation has real and distinct roots Hence Real and Distinct roots.
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MCQ 151 Mark
The angles of a triangle are $x^0, y^0$ and $40^{\circ}$. The difference between the two angles x and y is $30^{\circ}$, then
  • A
    $x^0=75^{\circ}$ and $y^0=45^{\circ}$
  • ✓
    $x^0=85^{\circ}$ and $y^0=55^{\circ}$
  • C
    $x^0=95^{\circ}$ and $y^0=35^{\circ}$
  • D
    $x^0=65^{\circ}$ and $y^0=95^{\circ}$
Answer
Correct option: B.
$x^0=85^{\circ}$ and $y^0=55^{\circ}$
(B) $x^{\circ}=85^{\circ}$ and $y^{\circ}=55^{\circ}$
Explanation: According to the question,
$x^{\circ}+y^{\circ}+40^{\circ}=180^{\circ}$
$x^0+y^0=140^{\circ}$... (i)
and $x^{\circ}+y^0=30^{\circ}$... (ii)
and $y^{\circ}=55^{\circ}$
On solving eq. (i) and eq. (ii),
$x+y+x-y=140+30$
$2 x=170$
$x =85^{\circ}$
Putting the value of x in equation (i), we get
$85^{\circ}+y=140^{\circ}$
$y=140^{\circ}-85^{\circ}$
$y=55^{\circ}$
we get $x^{\circ}=85^{\circ}$ and $y^{\circ}=55^{\circ}$

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MCQ 161 Mark
If x = 3 is a solution of the equation $3 x^2+(k-1) x+9=0$ then k = ?
  • A
    13
  • ✓
    -11
  • C
    11
  • D
    -13
Answer
Correct option: B.
-11
(B) -11
Explanation: $3 x^2+(k-1) x+9=0$
x = 3 is a solution of the equation means it satisfies the equation
Put x = 3, we get
$3(3)^2+(k-1) 3+9=0$
$27+3 k-3+9=0$
$27+3 k+6=0$
$3 k=-33$
k = - 11
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MCQ 171 Mark
Given that H.C.F. (306, 954, 1314) = 18, find L.C.M. (306, 954, 1314).
  • A
    1183234
  • B
    1123328
  • ✓
    1183914
  • D
    1123238
Answer
Correct option: C.
1183914
(C) 1183914
Explanation: L.C.M. (306, 954, 1314)
$=\frac{306 \times 954 \times 1314 \times \text { H.C.F. }(306,954,1314)}{\text { H.C.F. }(306,954) \times \text { H.C.F. }(954,1314) \times \text { H.C.F. }(306,1314)}$
$=\frac{306 \times 954 \times 1314 \times 18}{18 \times 18 \times 18}=1183914$
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MCQ 181 Mark
The $\text{HCF}$ of $95$ and $152,$ is
  • A
    $57$
  • ✓
    $19$
  • C
    $38$
  • D
    $1$
Answer
Correct option: B.
$19$
Using the factor tree for $95,$ we have:
Image
Therefore,
$95=5 \times 19$
$152=2^3 \times 19$
$\operatorname{HCF}(95,152)=19$
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