Tag: maths

Questions Related to maths

The construction of $\Delta LMN$ when $MN=6$ $cm$ and $m\angle M=45^\circ$ is not possible when difference between $LM$ and $LN$ is equal to:

  1. $6.9$ $cm$

  2. $5.2$ $cm$

  3. $5$ $cm$

  4. $4$ $cm$


Correct Option: A
Explanation:

In a triangle sum of length of $2$ sides is $>$ third side.

or 

Difference of length of two sides is less than third side.

$LM,MN,NL$ are the length of sides.

$|LM-NL|<MN$

$|LM-NL|<6$

So construction of triangle is not possible as $|LM-NL|=6.9cm$

To construct a triangle similar to a given triangle ABC with its sides 6/5th of the corresponding sides of $\Delta$ABC. Correct order of steps of construction -
(a) Draw a ray AX inclined at certain angle with AB on opposite side of C.
(b) Starting from A, cut off six equal line segments AX$ _1$, X$ _1$X$ _2$, X$ _2$X$ _3$, X$ _3$X$ _4$, X$ _4$X$ _5$ and X$ _5$X$ _6$ on AX.
(c) Draw a line B'C' parallel to BC to intersect AC produced at C'
(d) Join X$ _5$B and draw a line X6B' parallel to X5B, to intersect AB produced at B'.

  1. abcd

  2. acbd

  3. abdc

  4. adcb


Correct Option: C
Explanation:

To construct a triangle like the one given in the following question, the given steps are to be followed:

1. Draw a ray AX inclined at a certain angle with AB on opposite side of C.
2. Starting from A, cut off six equal line segments $AX _1, X _1X _2, X _2X _3, X _3X _4, X _4X _5 and X _5X _6$ on $AX$.
3. Join $X _5B$ and draw a line $X _6B'$ parallel to $X _5B$, to intersect $AB$ produced at $B'$.
4.Draw a line $B'C'$ parallel to $BC$ to intersect $AC$ produced at $C'$.

Construct a $\Delta ABC$, whose perimeter is $10.5  cm$ and base angles are $60^o$ and $45^o$. Find the third angle.

  1. $75^o$

  2. $45^o$

  3. $90^o$

  4. $60^o$


Correct Option: A
Explanation:
  1. Draw a line segment PQ such that AB + BC + AC = PQ.
    2. Construct $\angle XPQ =\angle B = { 60 }^{ 0 }\ and\ \angle YQP =\angle C =45^{ 0 }$
    3. Bisect $\angle XPQ and \angle YQP$ and their bisectors will meet at A.
    4. Draw the perpendicular bisector DE of AP which meets PQ  at B. 
    5. Draw the perpendicular bisector FG of AQ which meets PQ  at C.
    6. Join AB & AC.
    7. ABC is the required triangle.
    Sum of interior angle of triangle= $180^o$
    so third angle = $180^o-45^o-60^o\ = 75^o $

Express $5\dfrac {2}{3} hrs$ in minutes.

  1. $235\ mins$

  2. $320\ mins$

  3. $340\ mins$

  4. $523\ mins$


Correct Option: C
Explanation:

$5\dfrac {2}{3} hrs = \dfrac {17}{3}hrs$
$1\ hr = 60\ mins$
$\therefore \dfrac {17}{3} hrs = \dfrac {17}{3} \times 60\ mins = 340\ mins$.

Convert $2$ hours into minutes.

  1. $60\ min$

  2. $30\ min$

  3. $90\ min$

  4. $120\ min$


Correct Option: D
Explanation:

$1$ hour $=60$ minutes


$2$ hours $=2\times 60=120$ minutes

So option $D$ is correct.

What decimal of an hour is a second?

  1. $.0002\overline 9$

  2. $.00022\overline 8$

  3. $.0002\overline 7$

  4. $.0002\overline 6$


Correct Option: C
Explanation:

Given that one hour,

Now,

$1$ hour=$60\times 60$ second

$ 1s=\dfrac{1}{60\times 60}h $

$ =0.0002777777... $

$ =0.0002\overline{7} $

Hence, this is the answer.

$21$ months are equal to how many years?

  1. $1$

  2. $1\dfrac{1}{2}$

  3. $1\dfrac{3}{4}$

  4. $2$


Correct Option: C
Explanation:

$1$ year $=12$ months 

$1$ month $=\dfrac{1}{12}$ years
$21$ months $=\dfrac{1}{12}\times21=\dfrac{7}{4}=1\dfrac{3}{4}$
So option $C$ is correct.

State 'T' for true and 'F' for false.
(I) $12$ hours $:30$ hours $=8$km $:20$km
(II) The ratio of $1$ hour to one day is $1:1$
(III) The two terms of a ratio can be in two different units.

  1. (I)-T, (II)-T, (III)-T

  2. (I)-F, (II)-F, (III)-F

  3. (I)-T, (II)-F, (III)-F

  4. (I)-F, (II)-T, (III)-F


Correct Option: C
Explanation:

$i)$ $12$ hours $:$ $30$ hours$=8Km:20Km$

$\cfrac{12}{30}=\cfrac{2}{5}$ hours.
$\cfrac{8}{20}=\cfrac{2}{5}$ Km.
Hence, $True.$
$ii)$ The ratio of $1$ hour to $1$ day,
$\cfrac{1\quad hour}{1\quad day}=\cfrac{1}{24}\quad hours\neq 1$
Hence, $false.$
$iii)$ The two terms of a ratio can not be in two different units.
Hence, $false$.
$(I)-T, (II)-F,(III)-F$

A clock gains 5 minutes every hour.Then the angle traversed by the seconds hand in one minute will be 

  1. $390^0$

  2. $380^0$

  3. $360.5^0$

  4. $360^0$


Correct Option: A
Explanation:

Gain in 60 min=5min
in 1 min =$\dfrac{5}{60}min=\dfrac{5}{60}\times 60 sec=5 sec$
Angle traversed by second hand=$360^0+\dfrac{5}{60}\times 360^0=390^0$

$3$ hour $12$ minutes is equal to how many seconds?

  1. $10521\ seconds$

  2. $10510\ seconds$

  3. $11520\ seconds$

  4. $10600\ seconds$


Correct Option: C
Explanation:

$3$ hr $12$ min $=3\times 60\times 60+12\times 60$

$=3600\times3+720$ s
$=10800+720$ s $=11,520$ s