Tag: maths

Questions Related to maths

A television set is sold for Rs. $10000$ cash on Rs. $2000$ cash down followed by six equal instalments of Rs. $1600$ each. What is the rate of interest?

investement and financial planning banking compound interest comparing quantity maths

  1. $50\%$

  2. $60\%$

  3. $70\%$

  4. $80\%$

Show answer
Correct Option: B
Explanation:

Given, $n = 6, I =$ Rs. $1600$, 

$E = 2000 + 6 \times  1600 - 10000 = $ Rs. $1600$
We know $R = \dfrac{2400E}{n(n+1)I-2E}$
$\Rightarrow R = \dfrac{2400\times 1600}{6(6+1)1600-2\times 1600}$
$\Rightarrow R = 60\%$
Thus, the rate of interest is $60\%$.

Raghav buys a shop for $Rs. 1,20,000$. He pays half of the amount in cash and agrees to pay the balance in $12$ annual installments of $Rs. 5000$ each. If the rate of interest is $12\%$ and he pays with the installment the interest due on the unpaid amount find the total cost of the shop.

investement and financial planning banking compound interest comparing quantity maths

  1. $Rs. 1,60,800$

  2. $Rs. 1,66,800$

  3. $Rs. 1,68,800$

  4. $Rs. 1,60,000$

Show answer
Correct Option: B
Explanation:
Given that: 
Raghav buys a shop for $Rs.1,20,000.$
He pays half of the amount in cash $= \dfrac{120000}{2}\Rightarrow Rs.60,000$

Balance amount to be paid $= 120000 - 60000 \Rightarrow Rs. 60000.$

Given that amount of each installment $=Rs. 5000.$

He agrees to pay the balance in $12$ annual installments with interest of $12\%.$

 Amount of the $1^{st}$ installment 
$\Rightarrow 5000 + \dfrac{12}{100}\times   60000$

$\Rightarrow 5000 + 600 \times 12$

$\Rightarrow 5000 + 7200$

$\Rightarrow Rs. 12,200.$


 Amount of the $2^{nd}$ installment
$ \Rightarrow 5000 + \dfrac{12}{100} \times (60000 - 5000)$
$\Rightarrow 5000 + \dfrac{12}{100}\times  55000$
$\Rightarrow 5000 + 550 \times 12$
$\Rightarrow 5000 + 6600$
$\Rightarrow Rs. 11,600.$

As the amount paid for installment is $12200,11600,....... $ so It forms an $AP.$

The first term $a = 12,200$
Common Difference $d =  11600 - 12200\Rightarrow-600$
Total number of terms $n = 12.$

We know that sum of $n$ terms in $AP$
$\Rightarrow \dfrac{n}{2}[2a + (n-1) d]$

 Therefore the total cost of the shop
 $\Rightarrow 60000 +\dfrac{ 12}{2}[2(12200) + (12-1) \times (-600)]$

$\Rightarrow 60000 + 6(24400 - 6600)$
$\Rightarrow 60000 + 6 \times 17800$
$\Rightarrow 60000 + 106800$
$=Rs. 1,66,800.$

Hence, the total cost of the shop $= Rs.1,66,800.$

What sun will become Rs 9826 in 18 months if the rate of interest is $\displaystyle 1\frac{1}{2}$% per annum and the interest is compounded half-yearly?

investement and financial planning banking compound interest comparing quantity maths

  1. Rs 9466.54

  2. Rs 9646.54

  3. Rs 9566.54

  4. Rs 9456.54

Show answer
Correct Option: A
Explanation:

r $ = 2\dfrac {1}{2} $ % $ = \dfrac {5}{2} $ % n $ = 18 $ months $ = \dfrac {3}{2} $ years

When the interest is compounded half yearly,

$ A=P\left( 1+\dfrac { r }{ 2\times 100 }  \right) ^{ n\times 2 } $

$ => 9826 = P\left( 1+\dfrac { \dfrac {5}{2} }{ 2\times 100 }  \right) ^{ \dfrac {3}{2}\times 2 } $
$ => 9826 = P( \dfrac {81}{80}) ^{ 3 } $
$ => P =Rs 9466.54 $

The NAV of a unit in mutual fund scheme is Rs $10.65$, then find the amount required to buy $500$ such units.

investement and financial planning banking compound interest comparing quantity maths

  1. Rs.$5325$

  2. Rs.$5235$

  3. Rs.$53250$

  4. Rs.$5350$

Show answer
Correct Option: A
Explanation:

Amount required to buy $500$ such units

$=10.62\times500$
 $=Rs.5325$

Two line segments, each $9\ cm$ long, bisect each other at right angles. Their end points are joined together. The shape formed is a:

maths construction of polygons construction of parallelograms and rectangles construction of special quadrilaterals constructions related to a quadrilateral

  1. Square

  2. Kite

  3. Trapezium

  4. rhombus

Show answer
Correct Option: A
Explanation:

Image result for Two line segments, each 9 cm long, bisect each other at right angles. Their end points are joined together. The shape formed is a:

Let $PS$ and $QR$ are the two line segments, each of $9$cm, and bisect each other at right angles.
By joining the end points of these line, we get a shape given in the figure.
In $\triangle POQ$, $\angle POQ=90^{o}$, $OP=OQ=4.5$
By using Pythagoras theorem,
$PQ^{2}=OP^{2}+OQ^{2}$
         $=(4.5)^2+(4.5)^2=40.5$
$\therefore\ PQ=6.36$
Similarly, $QS=6.36=RS=PS$
Thus, length of all sides is same and all angles are right angle.
Hence, the shape formed is square.

A square with side given can be constructed by using the property of its diagonals.

maths construction of polygons construction of parallelograms and rectangles construction of special quadrilaterals constructions related to a quadrilateral

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

This statement is true 

We can use property that diagonals are at 45 degree with side and diagonals bisect each other at 90 degree.for construction of square.

Can we construct a rhombus $ABCD$ with $AB=4\ cm$? Its diagonal intersect at the point $O$ and $\angle OAB = 60^0$.

maths construction of polygons construction of parallelograms and rectangles construction of special quadrilaterals constructions related to a quadrilateral

  1. Yes

  2. No

  3. Sometimes yes

  4. Can't say

Show answer
Correct Option: A
Explanation:

Given : $AB=4$cm

Diagonal intersect at $O$ and $\angle OAB=60^{o}$ ....... $(1)$
Draw side $AB$ of $4$cm.
In a rhombus, all sides are equal and diagonals bisect the opposite angles
From $(1)$ we get, $\angle A=120^{o}$
$\implies \angle B=60^{o}$ ........... (Adjacent angles are supplementary)
Draw a side $AD$ from A of $4$cm such that $\angle BAD=120^{o}$
Now, from $D$, draw side $DC = 4$cm such that $\angle ADC=60^{o}$
And then join $B-C$ such that $BC=4$cm and $\angle DCB=120^{o}$.
At last we get a rhombus $ABCD$ with length of each side is $4$ cm and diagonals $AC$ and $BD$.
Hence, we can construct a rhombus with $AB=4\ cm$.

We cannot construct a square if:

maths construction of polygons construction of parallelograms and rectangles construction of special quadrilaterals constructions related to a quadrilateral

  1. a side is given

  2. a diagonal is given

  3. one angle is $90^0$

  4. None of these

Show answer
Correct Option: C
Explanation:

If a side is given then we can draw a square with the same side as given.

If diagonals are given, by joining the endpoints we can draw the square.
In square, all angles are of $90^{o}$.
If one angle is $90^{o}$ is given, we can't directly conclude that all the angles are $90^{o}$.

Hence, if one angle is $90^{o}$ then we cannot construct a square.

When given a square, the construction of an angle bisector at any vertex will create the diagonal of the square. 

maths construction of polygons construction of parallelograms and rectangles construction of special quadrilaterals constructions related to a quadrilateral

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

This is statement is true

We know that diagonals of square bisects the angle. So angle bisector will be diagonal.

You are given the length of a diagonal of a rhombus and one of the angles of the rhombus. Which property of the rhombus will be used in the construction of this rhombus?

maths construction of polygons construction of parallelograms and rectangles construction of special quadrilaterals constructions related to a quadrilateral

  1. The lengths of the sides of a rhombus are equal.

  2. The angles of a rhombus are $90^\circ$

  3. Diagonal of a rhombus bisects the opposite angles.

  4. Diagonals of a rhombus are perpendicular bisectors of each other.

Show answer
Correct Option: C
Explanation:

$\Rightarrow$   We have given the length of diagonal of rhombus and one of angles of rhombus.

$\Rightarrow$  To construct an rhombus we will use the property that the diagonal of a rhombus bisect the opposite angle.
Because we know opposite angles of rhombus are equal, so it will be easier to construct rhombus.