Tag: equations from statements

Questions Related to equations from statements

Anu had 6 pairs of shoes. She let Mohit borrow $x$ pairs of shoes, and then she had 4 pairs left. Which equation model shows this solution ? 

  1. $6 - x = 4$

  2. $6 + x = 4$

  3. $6x = 4$

  4. $4x = 6 $


Correct Option: A
Explanation:

Number of pair of shoes Anu had$=6$

Number of pair of shoes Anu has now, after giving $x$ number of pairs to Mohit$=6-x$
Now, in the question, it is given that Anu has $4$ pairs of shoes.
Thus, $6-x=4$ will satisfy the solution for the given problem.

Jaya's score in Mathematics is $30$ more than two third of her score in English. If her score in English is $x$, find her score in Mathematics.

  1. $\dfrac {2}{3}(x + 30)$

  2. $\dfrac {2x}{3} + 30$

  3. $\dfrac {2x}{3} - 30$

  4. $30 - \dfrac {2x}{3}$


Correct Option: B
Explanation:

Given: Jaya's score in English $=x$

Jaya's score inMathematics $=30+ \dfrac{2}{3}x$
                                               $=\dfrac{2x}{3}+30$

Preeti travelled $3x$ km distance by walk, $9y$ km by cycle and $5$ km by bus. The total distance covered by Preeti is ____________.

  1. $(3x - 9y + 5)$ km

  2. $(3x + 9y + 5) $ km

  3. $(3x - 9y - 5) $ km

  4. $(9x + 3y - 5) $ km


Correct Option: B
Explanation:

Distance covered by walk $=$ $3x$ km

Distance covered by cycle $=$ $9y$ km
Distance covered by bus $=$ $5$ km
Therefore, total distance covered by Preeti $=$ $(3x+9y+5)$ km.

Meera bought packs of trading cards that contain $10$ cards each. She gave $7$ cards to her friend.
$x =$ Number of packs of trading cards
Which expression shows the number of cards left with Meera?

  1. $10x - 7$

  2. $7x - 8$

  3. $5 - 10x$

  4. $8 - 5x$


Correct Option: A
Explanation:

Number of cards in 1 pack $= 10$

Number of cards in $x$ packs $=10x$
Number of cards given to friend $=7$
Number of cards left with Meera $=10x-7$

The algebraic expression for the statement 'thrice of $x$ is added to $y$ multiplied by itself' is __________.

  1. $3x + 2y$

  2. $3x + y^{2}$

  3. $3(x + y^{2})$

  4. $3x + y$


Correct Option: B
Explanation:

Thrice of $x$ $=$ $3x$


$y$ multiplied by itself $=$ $y\times y=y²$
 
According to question, we have

The algebraic expression is $3x+ y²$.

Aanya got $5$ marks in Science test. Her friend Sneha got $'x'$ marks more than Aanya. How many marks did they get altogether?

  1. $5\times x$

  2. $5 + x$

  3. $10 + x$

  4. $2x + 5$


Correct Option: C
Explanation:

Marks obtained by Aanya $= 5$
Marks obtained by Sneha $= x + 5$
$\therefore$ Mark obtained by both of them
$= 5 + (x + 5) = x + 10$.

Match the following.

Column - I Column II
(i) The total mass of $3$ boxes is $5\ kg$. The mass of two of the boxes is $x\ kg$ each. The mass of third box is (a) $x - 11$
(ii) Sid had $x$ eggs. He used $5$ eggs to bake a cake and gave $6$ eggs to his neighbour. The number of eggs left with him is (b) $\dfrac {x}{3}$
(iii) Mohit had $Rs. x$. He gave the money to his $3$ sisters equally. Each girl get Rs. (c) $5 - 2x$
  1. $(i)\rightarrow (c), (ii) \rightarrow (a), (iii) \rightarrow (b)$

  2. $(i)\rightarrow (b), (ii) \rightarrow (c), (iii) \rightarrow (a)$

  3. $(i)]\rightarrow (c), (ii) \rightarrow (b), (iii) \rightarrow (a)$

  4. $(i)\rightarrow (a), (ii) \rightarrow (b), (iii) \rightarrow (c)$


Correct Option: A
Explanation:

(i) Given: Total mass of $3$ boxes is $5$ kg. 

Mass of each boxes out of two $=x$ kg each
$\therefore$ total mass of two boxes $=2x $ kg
Mass of $2$ boxes $+$ Mass of 3rd box $= 5$ kg
$\therefore$ Mass of 3rd Box $=5- 2x$ kg (c)

(ii) No of eggs Sid had $=x$ 
No of eggs used to bake a cake $= 5$
No of eggs given to neighbour $= 6$
$\therefore$ No of eggs left $= x-5-6= x-11$ (a)

(iii) Amount of money Mohit had $=Rs. x$
Amount of money divided equally among sisters = $Rs \dfrac{x}{3}$ (b)

Find the cost of $3$ notebooks and $1$ magazine in terms of y, if the cost of $1$ notebook is Rs.$2y$ and that of $1$ magazine is $Rs. (y+3)$.

  1. $Rs.(3y+3)$

  2. $Rs.(6y+3)$

  3. $Rs.(7y+3)$

  4. $Rs.(8y+3)$


Correct Option: C
Explanation:
 The no. of notebooks is =3 then
total cost of notebooks is = $3\times 2y=6y$
cost of one magazine is =(y+3)
hence total cost =$6y+y+3=(7y+3)$
then optiuon $C$is correct. 

In the last three months Mr.Sharma lost $5\frac{1}{2}$kg, gained $2\frac{1}{4}$ kg and then lost $3\frac{3}{4}$ kg weight. If he now weighs $95$kg, then how much did Mr.Sharma weighs in beginning?

  1. $100$kg

  2. $102$kg

  3. $106.5$kg

  4. $104$kg


Correct Option: B
Explanation:

Let $x$ be the weight of Mr. Sharma in beginning.


According to given details,
$x-5.5+2.25-3.75=95$
$\Rightarrow x=95+5.5+3.75-2.25=95+7=102\ kg$.

So, the answer is $102\ kg$.  $[B]$

In an alloy of zinc, tin and lead, quantity of zinc is $\left(\cfrac{3}{4}\right)^{th}$ that of tin and quantity of tin is $\left(\cfrac{4}{5}\right)^{th}$ that of lead. How much of each metal will be there in $12$ kg of the alloy?

  1. $5,4,3$

  2. $3,4,5$

  3. $4,5,3$

  4. $5,3,4$


Correct Option: D
Explanation:

  • Let the quantity of the tin be $'x'$ , zinc be $'y'$ and lead be $'z'$.
Given quantity of zinc is $\cfrac{3}{4}th$ of tin,
$\implies y=\cfrac{3}{4}(x)--------(1)$
Quantity of tin is $\cfrac{4}{5}th $ of lead,
$\implies x=\cfrac{4}{5}(z)------(2)$
But given total weight $=12kg$
$\implies x+y+z=12$
From $(1)&(2) $equations
$x+\cfrac{3}{4}x+\cfrac{5x}{4}=12$
$\cfrac{12x}{4}=12$
$\implies x=4$
if $x=4, y=3,z=5$
So the quantity of tin is $4kg$ , zinc is $3kg$ , lead is $5kg$