0

Calculations and mental strategies 1 - class-VIII

Attempted 0/116 Correct 0 Score 0

Four years ago, the father's age was three times the age of his son. The total of the age of the father and the son after four years will be $64$ years. What is the father's age at present?

  1. $32$ years

  2. $36$ years

  3. $44$ years

  4. $40$ years


Correct Option: D
Explanation:

fathers present age $= x$ years.

sons present age $= y$ years.  

four years ago, fathers age $=3 \times $ sons age.
$(x-4)=(y-4)\times 3$
$x=3y-8$.....(i) 

after 4 years, fathers age$ +$ sons age $= 64 $
$x+4+y+4=64 $
$x+y=56$.....(ii) 

solve for $x$
$x=3\times 56-3x-8$ 
$4x=160$ 
$x=40$ years  

D is correct.   

The area of a field in the shape of a trapezium measures $1440{m}^{2}$. The perpendicular distance between its parallel sides is $24m$. If the ratio of the parallel sides is $5:3$, the length of the longer parallel side is:

  1. $45m$

  2. $60m$

  3. $75m$

  4. $120m$


Correct Option: C
Explanation:
Parallel sides $= 5x,\, 3x$
area $=\dfrac{24}{2}(5x+3x)=1440 $
$12(8x)=1440$
$x=\dfrac{120}{8}=15$ 
$5x=15\times 5$ 
     $=75m$

If $\left(\dfrac { 3 } { 4 }\right)^{th}$ of $x$ of $\left(\dfrac { 1 } { 4 }\right)^{th}$ of $35600 = 1668.75 ,$ find $x$

  1. $\dfrac { 2 } { 3 }$

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

  3. $\dfrac { 2 } { 5 }$

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


Correct Option: D
Explanation:

$\dfrac { 3 }{ 4 } \times x\times \dfrac { 1 }{ 4 } \times 35600=1668.75$

$\Rightarrow x=\dfrac { 1668.75\times 16 }{ 3\times 35600 } =\dfrac { 1 }{ 4 } $      [D]

If a function $f$ is linear with $f(0)=5$ and $f(2)=9$ then find $f(9)$

  1. 23

  2. 25

  3. 26

  4. none of these


Correct Option: A
Explanation:

$f$ is a linear function 

So Let $f=ax+b$
$f(0)=a(0)+b=5\\implies b=5\cdots(1)\f(2)=9\a(2)+5=9\\implies a=2\f(x)=2x+5\f(9)=2(9)+5\18+5=23$

If $(30 + x) : (23 + x) = 5 : 4$ then find the value of x.

  1. 5

  2. 6

  3. 4

  4. 7


Correct Option: A
Explanation:

$ => 30 + x : 23 + x = 5:4 $


$ => \frac {30 + x}{23 + x} = \frac {5}{4} $

Cross multiplying, 

$ 120 + 4x = 115 + 5x $


$ 120 - 115 = 5x - 4x $


$ x = 5 $

Translate the following sentence into an algebraic equation: "13 diminished from twice a number is equal to 17"

  1. $2x-13=17$

  2. $13-2x=17$

  3. $2x-17=13$

  4. $2x+13=17$


Correct Option: A
Explanation:

Let  the  number  be  $x.$


Twice  of  the  number  is  $2x$  and  $13$  is  diminished  from  $2x$  means  $(2x-13)$  and  it  is  equal   to   $17. $
 So  answer  is  option  A

'8 is less than twenty times a' is written as 

  1. 8 - 20a

  2. 20ac

  3. 8 < 20 $\times$ a - 8

  4. 20a - 8


Correct Option: D
Explanation:

Eq : 20a - 8

Algebraic expression for the statement 8 times a taken away from 60 

  1. 8a - 60

  2. 60 - 8a

  3. 0

  4. 60a - 8


Correct Option: B
Explanation:

Eq : 20a - 8

The equation for the statement : 'half of a number added to 20 is 25' 

  1. $\displaystyle{\frac{x}{4}}$ + 20 = 25

  2. $\displaystyle{\frac{x}{5}}$ + 25 = 20

  3. $\displaystyle{\frac{x}{2}}$ + 20 = 25

  4. $\displaystyle{\frac{x}{3}}$ + 20 = 25


Correct Option: C
Explanation:

$\displaystyle{\frac{x}{2}}$ + 20 = 25 

Half of the number added to 20 will give the answer 25
$\dfrac{10}{2} +20=25$

Equation of the statement : 'Thrice the length of a room is $240$ metres.' 

  1. $3l = 240$

  2. $3 + l = 40$

  3. $3l = 420$

  4. None of these


Correct Option: A
Explanation:

Let $l$ be the length of a room 

therefore according to statement, required equation is   
$3l = 240 $

Equation for the statement: 'thrice the length (L) of a room is 340 metres'

  1. $3L = 43$

  2. $3L = 340$

  3. $3 + L= 34$

  4. None of these


Correct Option: B
Explanation:

$\Rightarrow$  The length of room is $L.$

$\Rightarrow$  So, thrice the length means $3L.$

The equation for thrice the length of the room is $340 \ meters$

$\Rightarrow$  $3L=340\ meters.$

The equation for the statement:"half a number added to 10 is 15".

  1. $\dfrac{X}{2}\,+ 10\,+ 5$

  2. $\dfrac{X}{2}\,+ 10\,= 5$

  3. $\dfrac{X}{2}\,+ 15\,= 10$

  4. $\dfrac{X}{2}\,= -10\,+ 15$


Correct Option: D
Explanation:

Let the number be $X.$

Half of the number is $\dfrac{X}{2}$
Half a number added to $10$ is $\dfrac{X}{2}+10$
So, according to the question,
$\Rightarrow$  $\dfrac{X}{2}+10=15$

$\Rightarrow$  $\dfrac{X}{2}=-10+15$

In three coloured boxes - Red, Green and Blue, $108$ balls are placed. There are twice as many balls in the green and red boxes combined as there are in the blue box and twice as many in the blue box as there are in the red box. How many balls are there in the green box?

  1. $18$

  2. $36$

  3. $45$

  4. None of these


Correct Option: D

The following expressions in statement is given as:
$x+3$

3 is added to x

  1. True

  2. False


Correct Option: A
Explanation:

Yes the  statement is correct. This defines the expression in words.

The following expressions in statement is given as:
$y-7$ 


7 is subtracted from y

  1. True

  2. False


Correct Option: A
Explanation:

The satement is true.  7 is being subtracted from y.

The $ \dfrac{x}{5}$ expression in the statement is given as $x$ is divided by $5$.

Check whether the statement is true or false.

  1. True

  2. False


Correct Option: A
Explanation:

The statemnt is true.  x is being divided by 5.

Twelve less than $3$ times a number is $27$. The number is ___________.

  1. $29$

  2. $39$

  3. $65$

  4. $13$


Correct Option: D
Explanation:

Let the number be $x$.
According to question,
$3x - 12 = 27$

$\Rightarrow 3x = 39$
$\Rightarrow x = \dfrac {39}{3} = 13$

The equation for the statement; 'Half of a number added to $10$ is $15$' is __________.

  1. $\dfrac {x}{2} = 10 + 5$

  2. $\dfrac {x}{2} + 10 = 15$

  3. $\dfrac {x}{2} = 15 + 10$

  4. $\dfrac {x}{2} = 10 + 15$


Correct Option: B
Explanation:

Let, the number be $x$
Half of a number $=$ $\dfrac{x}{2}$
Therefore, the required equation is $\dfrac{x}{2}$ $+10 = 15$.

Kapil bought some packets of pens. He bought one packet containing $5$ pens and $n$ packets containing $2$ pens each. Which of the following expressions could be used to find the total number of pens that Kapil bought ? 

  1. $7n$

  2. $10n$

  3. $5 + 2n$

  4. $2 + 5n$


Correct Option: C
Explanation:

Number of packets bought $= 1 + n$

Given that, 
Single packet has $5$ pens and $n$ packets have $2$ pens each
$\therefore$ Total number of pens in $n$ packets $= 2n$

$\therefore$ Total number of pens bought by Kapil $= 2n + 5$

If five times of a number is $80$, then the number is ________.

  1. $-16$

  2. $1$

  3. $16$

  4. $40$


Correct Option: C
Explanation:

Let, the number be $x$
$5 $ times of a number $=$ $5\times x$
Therefore, $5x=80$
$\Rightarrow x =\dfrac{80}{5}$ $=16$

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)&amp;(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 $5kg$

For a journey the cost of a child ticket is $\cfrac { 1 }{ 3 } $ of the cost of an adult ticket. If the cost of tickets for $4$ adults and $5$ children is Rs. $85$, the cost of a child ticket is

  1. Rs. $15$

  2. Rs. $6$

  3. Rs. $10$

  4. Rs. $5$


Correct Option: A
Explanation:

Let the cost of an adult ticket be $y$ $\Rightarrow$ Child's ticket costs $\cfrac{1}{3}y$
$\Rightarrow 4y+\cfrac { 5 }{ 3 } y=85\Rightarrow 17y=225\Rightarrow y=Rs. 15\quad $
$\therefore$ cost of child's ticket $=Rs.15$

If  $2x  + 2(4 + 3x) < 2 + 3x > 2x + \dfrac{x}{2}$ then $x$ can take which of the following values. 

  1. $-3$

  2. $1$

  3. $0$

  4. $-1$


Correct Option: A
Explanation:

Given,
$2x+2(4+3x) < 2+3x > 2x+\dfrac{x}{2}$

Solving $1^{st}$ inequality

$2x+8+6x < 2+3x$

$5x < -6$

$x < \dfrac{-6}{5}$

Solving $2^{nd}$ inequality

$2+3x > 2x+\dfrac{x}{2}$

$\dfrac{x}{2} > -2$

$x > -4$

$x\in \left (-4, \dfrac{-6}5\right )$

$x$ can take values $-3$

$A$ is correct.

Divide $Rs\ 1,545$ between three people $A,B$ and $C$ such that $A$ gets three-fifths of what $B$ gets and the ratio of the share of $B$ to $C$ is $6:11$.
what amount will each person get ?

  1. $A=240, \ B= 440, \ C=825$

  2. $A=230, \ B= 440, \ C=825$

  3. $A=270, \ B= 450, \ C=825$

  4. $A=245, \ B= 440, \ C=825$


Correct Option: C
Explanation:
$3\times \dfrac {6x}{5}+6x+11x=1545$
$\dfrac {18x+30x+55x}{5}=1545$
$\dfrac {103x}{5}=1545$
$x=\dfrac {5\times 1545}{103}$
$x=75$
$A=\dfrac {18}{5}\times 75 =270$
$B=6\times 75 =450$
$C=11\times 75=825$


If $(a+b):(b+c):(c+a)=6:7:8$ and $(a+b+c)=14$ , then the value of $c$ is

  1. $6$

  2. $7$

  3. $8$

  4. $14 $


Correct Option: A
Explanation:
$a+b:b+c:c+a=6:7:8$

$a+b+c=14$

$\therefore \dfrac{a+b}{b+c}=\dfrac{6}{7}$ ……..$(1)$

$\therefore \dfrac{b+c}{c+a}=\dfrac{7}{8}$ ……..$(2)$

$\therefore \dfrac{a+b}{c+a}=\dfrac{6}{8}$ ...........$(3)$

From $(1)$ & $(2)$

$\left.\begin{matrix}If & a+b=6x\\ then & b+c=7x\end{matrix}\right\}\rightarrow x\in R$

then $c+a=8x$

$\therefore 2(a+b+c)=6x+7x+8x$

$\therefore a+b+c=\dfrac{21x}{2}$

$\therefore a+b+c=10.5x$

$\therefore c=10.5x-6x$

$\therefore c=4.5x$

Also, $10.5x=14$

$\therefore x=\dfrac{14}{10.5}$

$\therefore c=\dfrac{4.5\times 14}{10.5}$

$\therefore c=6$.

In a two digit no. the digit at units place is $\left| x-2 \right| $ and digit at tens place is $\left| x+1 \right| $, then that number is

  1. 2x+3

  2. 11x+12

  3. 10x+1

  4. 11x+21


Correct Option: A

if $76$ is divided into four parts proportional to $7,5,3.4$ then the smallest part is:

  1. $12$

  2. $15$

  3. $16$

  4. $19$


Correct Option: A
Explanation:
Four parts proportional to 

$7,5,3$ and $4$

$\therefore $  Four part are equal to$7x, 5x, 3x,$ and $4x.$

$\therefore7x+5x+3x+4x=76$

$\therefore 19x=76$

$\therefore x=\dfrac {76}{19}$

$\therefore x=4$

$\therefore 7x=4\times7=28$

$\therefore 5x=5\times4=20$

$\therefore 3x=3\times 4=12$

$\therefore 4x=4\times4=16$

$\therefore $ For parts are $28, 20, 12$ and $16$.

The ratio of the volumes of water and glycerine in 240cc of mixture is 1:3 .The quantity of water (in cc) that should be added to the mixture so the volumes of water and glycerine 2:3 is: 

  1. $55$

  2. $60$

  3. $62.5$

  4. $64$


Correct Option: B
Explanation:
The ratio of the volume of water and Glycerine in $240$cc of mixture is $1:3$. 

The quantity of water(in cc) that should be added to the mixture so that the new ratio of the volume of water and glycerine becomes $2:3$

Let the ratio of the volume of water and glycerine be $x$ and $3x$.

$x+3x=240$

$\Rightarrow 4x=240$

$\Rightarrow x=60$

So the volume of water is $60cc$

Volume of glycerine is $60cc$

Volume of glycerine is $(3\times 60)cc=180$cc

Let x litre of water is added

$\dfrac{60+x}{180}=\dfrac{2}{3}$

$\Rightarrow 3(60+x)=360$

$\Rightarrow 180+3x=360$

$\Rightarrow 3x=360-180$

$\Rightarrow 3x=180$

$\Rightarrow x=60$.

$8$% of the voters in an election did not cast their votes. In this election, there were only two candidates. The winner by obtaining $48$% of the total votes, defeated his contestant by $1100$ votes. The total number of voters in the election was?

  1. $21000$

  2. $23500$

  3. $22000$

  4. $27500$


Correct Option: D

Which of the equation satisfy the given statement ?
"Perimeter of an equilateral triangle is three times its side l" 

  1. $l \times l \times l$

  2. $3 \times 3 \times 3$

  3. $3 \times l$

  4. $3 + l$


Correct Option: C
Explanation:
An equilateral triangle is a triangle in which all three sides are equal.Let $l$ be the length of the side of a triangle.
Thus, perimeter is equal to sum of all its sides$3\times side=3\times l$

Equation of the statement 'Thrice the length (l) of room is $340$ metres' is ____ 

  1. $3l=430$

  2. $3l=340$

  3. $3+l=340$

  4. $3l+340=0$


Correct Option: B
Explanation:

Let $l$ be the length of the room

Thrice the length of the room$=3l$
Given:Thrice the length of the room$=340\ m$
Equation is $3l=340\ m$

An algebraic expression, $11-y$ can be written in the statement form as  ____ 

  1. $11$ less than $y$

  2. $y$ less than $11$

  3. $y$ more than $11$

  4. $y$ divided by $11$


Correct Option: B
Explanation:
$y$  less than $11$

If fives of a number is $80$, then the number is ___ 

  1. $-16$

  2. $1$

  3. $16$

  4. $40$


Correct Option: C
Explanation:
Let $x$ be a number.
$\Rightarrow\,5x=80$
$\Rightarrow\,x=\dfrac{80}{5}=16$
$\therefore\,x=16$
The number is $16$

If a $\in { 1,2,3,4 } ,$ then number of equations of the form $x ^ { 2 } + a x + 1 = 0$ having real roots is

  1. $1$

  2. $2$

  3. $3$

  4. $4$


Correct Option: C

The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digit of that number ?

  1. 3

  2. 4

  3. 9

  4. Cannot be determined

  5. None of these


Correct Option: B
Explanation:
Let the ten's digit be $x$ and unit's digit be $y$.

Then, $(10x + y) - (10y + x) = 36$

$9(x - y) = 36$

$x - y = 4$

Therefore, the difference between the two digit of that number is $4$

The sum of the digit of a two-digit number is $\frac { 1 }{ 5 } $ of the difference between the number and the number obtained by interchanging the positions of the digit. What is definitely the difference between the digit of that number ?

  1. 5

  2. 7

  3. 9

  4. Data inadequate

  5. None of these


Correct Option: A

The sum of the digits of a two-digit number is 9 less then the number. Which of the following digits is at unit's place of the number ?

  1. 1

  2. 2

  3. 4

  4. Data inadequate


Correct Option: A
Explanation:
Let $x,y$ are the digits of a number. Then the number $= 10x+y$

as per condition : $10x+y-9 = x+y$

Then as per the given problem,

$x+y=10x+y-9$

or, $9x=9$

or, $x=1$

So, we can say that for any value of y from 1 to 9, the condition satisfies.

for eg: if the number is 18

18–9 = 1+8

9=9 , satisfied.


So the tenth's place digit is $1$.

Get the algebraic expression in the following cases using variables, constant and arithmetic operations.
Subtraction of $z$ from $y$.

  1. $y+z$

  2. $y-z$

  3. $z-y$

  4. $None\ of\ these$


Correct Option: B
Explanation:

Subtraction of $z$ from $y$ implies that a number of value $z$ is to be subtracted from another number of $y$ value $y$ which means $y-z$.

Two years ago the mean age of $40$ people was $11$ years. Now a person left the group and the mean age is changed to $12$ years, then the age of the person who left the group is?

  1. $52$ years

  2. $48$ years

  3. $54$ years

  4. None of these


Correct Option: A

Five years ago Anjali's age was twice Sarita's age. Five years hence, their ages will be $4 : 3$. Anjali's present age is 

  1. $15$ years

  2. $10$ years

  3. $12$ years

  4. $16$ years


Correct Option: A

Kiran is 24 years older than Rakesh. 10 years back Evans age was five times the age of Rakesh. Find the present age of Rakesh.

  1. $40 years $

  2. $16 years$

  3. $20 years$

  4. $36 years


Correct Option: A

A game is played by three players. The loser has to triple the money of each of the other players has. Three games are played and each one losses a game. At the end all have the same amount namely $Rs.\ 54$. The amount the first loser has at the begining is

  1. $Rs.\ 120$

  2. $Rs.\ 112$

  3. $Rs.\ 110$

  4. $Rs.\ 90$


Correct Option: A

When a two digit number is reversed, the new number is reduced by $63$ and the sum of digits of these individual numbers is $9$. Now, if these numbers are converted into base $'x'$ the larger number becomes $5$ times that of smaller one, then the value of $x$ is :

  1. $36$

  2. $13$

  3. $15$

  4. $8$


Correct Option: A

A tree, in each year, grows $5\ cm$ less than it grew in the previous year. If it grew half a meter per year, then the height of the tree (in meters), when it ceases to grow, is

  1. $2.50$

  2. $2.00$

  3. $3.00$

  4. $2.75$


Correct Option: A

The sum of present ages of a father and his son is $10$ years more than the present age o mother. If the mother is $15$ years older than the son, then find the age of father after $5$ years.

  1. $20$ years

  2. $25$ years

  3. $40$ years

  4. $30$ years


Correct Option: A

If a number, when divided by $4$, is reduced by $21$, the number is:

  1. $18$

  2. $20$

  3. $28$

  4. $38$


Correct Option: C
Explanation:
Let $x$ be the number.
If a number is divided by the $4$ the expression becomes $\dfrac{x}{4}$
If a number, when divided by $4$, is reduced by $21$, then the equation is
$\dfrac{x}{4}=x-21$
$\Rightarrow\,\dfrac{x}{4}-x=-21$
$\Rightarrow\,\dfrac{x-4x}{4}=-21$
$\Rightarrow\,-3x=-21\times 4$
$\Rightarrow\,x=\dfrac{21}{3}\times 4=7\times 4=28$
$\therefore\,$,the number is $28$

The product of two numbers is 1296. If one number is 16 times the other, then the two numbers are 

  1. 144, 25

  2. 81, 49

  3. 144, 9

  4. none of these


Correct Option: C
Explanation:

Let the first number be $x$. Then, other number is $16x$.


Now, $x \times 16x =1296$

$16x^2=1296$
$x^2=81$
$x=9$
$16x=16\times 9=144$

Thus, the two numbers are $9$ and $144.$

The average age of the mother and her six children is 12 years which is reduced by 5 years if the age of mother is excluded. How old is the mother?

  1. 40 years

  2. 42 years

  3. 48 years

  4. 50 years


Correct Option: A

Two years ago, a father was five times as old as his son. Two years later, father age will be 8 more than three times the age of the son. Find the sum of present ages of father and son.

  1. 62 years

  2. 52 years

  3. 42 years

  4. 34 years


Correct Option: A

If the sum of five consecutive even numbers is $240$ then find the middle number.

  1. $45$

  2. $50$

  3. $44$

  4. $48$


Correct Option: A

If $50$ is subtracted from two-third of a number, the result is equal to sum of $40$ and one-fourth of that number. What is the number? 

  1. $174$

  2. $216$

  3. $246$

  4. $336$


Correct Option: A

The average of 9  consecutive numbers with difference of 9 is 79 , then with smallest number  among them ?

  1. 43

  2. 34

  3. 52

  4. 44


Correct Option: A

The two parts of $100$ for which the sum of double of first and square of part is minimum. value of the smaller part is:

  1. $48$

  2. $1$

  3. $32$

  4. $16$


Correct Option: A

find the sum of two numbers ,whose ratio is 7:19 and their difference is 12

  1. 22

  2. 26

  3. 24

  4. 23


Correct Option: A

A number is $\dfrac{2}{5}$ of the second number seum of the two number is $70$; then the number are 

  1. $20,30$

  2. $20,40$

  3. $50,20$

  4. $20,15$


Correct Option: A

In an examination Karan got  $10$  marks more than Bhavna. Isha got  $5$  marks less than Bhavna. If Trio get a total of  $170 ,$ then what is the marks obtained by Isha?

  1. $65$

  2. $55$

  3. $50$

  4. $45$


Correct Option: A

Half the number is added to $18$, then sum is $46$, find the number?

  1. $26$

  2. $46$

  3. $36$

  4. $56$


Correct Option: D
Explanation:
Let $x$ be the number

$\dfrac{1}{2}x+18=46$(given)

$\Rightarrow \dfrac{x+36}{2}=46$

$\Rightarrow x+36=92$

$\Rightarrow x=92-36=56$

$\therefore $ the number is $56$

After $12$ years I shall be $3$ times as old as I was $4$ years ago. Find my present age.

  1. $12$

  2. $13$

  3. $14$

  4. $15$


Correct Option: A
Explanation:
Let my present age$=x$ years

After $12$ years my age$=x+12$ years

$4$ years ago my age $=x-4$ years

According to the question,

$x+12=3\left(x-4\right)$

$\Rightarrow x+12=3x-12$

$\Rightarrow 3x-x=12+12$

$\Rightarrow 2x=24$ or $x=\dfrac{24}{2}=12$ years.

$\therefore$ my present age is $12$ years.

If six times of a number is $48$, then the number is

  1. $2$

  2. $4$

  3. $6$

  4. $8$


Correct Option: D
Explanation:
Let $x$ be the number

$6\times x=48$

$\Rightarrow x=\dfrac{48}{6}=8$

$\therefore$ the number is $8$

The number of girls in a class is $5$ times the number of boys.Which of the following can be the total number of children in a class?

  1. $x+5y$

  2. $y-5x$

  3. $x-5y$

  4. $x+y=5$


Correct Option: A
Explanation:
Let the number of girls in a class be $x$

Number of boys be $y$

Given: Number of girls in a class is $5\times$ number of boys

$\therefore x=5y$ 

$\therefore$ total number of children$=x+5y$

Rams age is $1/3$ of his fathers age. Rams fathers age will be $12$ years more than twice of Shyams age after $10$ years. If Shyams $8th$ bright day was celebrated $3$ years before, then what is Rams age at present?

  1. $14\ years$

  2. $16\ years$

  3. $11\ years$

  4. $13\ years$


Correct Option: A

If $x$ toffees are distributed equally among $5$ children, then each child gets $5x$ toffees.

  1. True

  2. False


Correct Option: A

In a two digits number, the digit at tens place is $7$ times the digits at unit place, then the number is

  1. $70$

  2. $71$

  3. $17$

  4. $78$


Correct Option: A

$15$ years hence a men will be just $4$ times as old as he was $15$ years ago. His present age is _________ years

  1. $25$

  2. $20$

  3. $15$

  4. $30$


Correct Option: A

If the sum and the product of the digits of two digit number is the same, then the number is_____________

  1. $11$

  2. $23$

  3. $10$

  4. $22$


Correct Option: A

The number of different necklaces formed by using $2n$ identical diamonds and $3$ different jewels when exactly two jewels are always together is ____________.

  1. $ 6n - 3$

  2. $ 6n $

  3. $none of these$

  4. $ 6n - 6$


Correct Option: A

20 years ago when my parents got married their average age was 23 years, now the average age of my family consisting of my self and my parents is 34 years, then my present age is?

  1. 14 years

  2. 16 years

  3. 18 years

  4. 36 years


Correct Option: A

If  $12$  years hence of Ravi will he  $9$  times his age  $12$  years ago, find the present age of Ravi

  1. $12 years$

  2. $15 years$

  3. $18 years$

  4. $20 years$


Correct Option: A

$A$  is  $3$ times as old as  $B$  and  $C$  together. If after  $10$  years  $A's$  age would be equal sum of the ages of  $B$  and  $C .$  find the present age of  $A$

  1. $15$ years

  2. $20$ years

  3. $21$ years

  4. $25$ years


Correct Option: A

The number of students scoring less than 60 marks is ___________.

  1. 40

  2. 45

  3. 50

  4. 55


Correct Option: A

The number of students scoring less than 80 marks but not less than 40 marks is _______.

  1. 45

  2. 40

  3. 35

  4. 30


Correct Option: A

Three times a number is equal to two times of the other. Find the ratio of 3 times the sum and 5 times the difference of the two numbers.

  1. 2

  2. 4

  3. 1

  4. 3


Correct Option: A

In a two-digit number, the tens digit is one more than twice the units digit. The sum of the digits is $36$ less than the number formed by reversing the digits. Find the product of the digits.

  1. $56$

  2. $12$

  3. $24$

  4. $36$


Correct Option: A

There are m men and two women participating in a chess tournament. Each participate plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by $84$, then the value of m is?

  1. $9$

  2. $7$

  3. $11$

  4. $12$


Correct Option: A

x years ago, the difference between the ages of Hari and Sadu was y years. Then after z years what will be the difference between their ages?

  1. x years

  2. y years

  3. z years

  4. $(x+y+z)$ years


Correct Option: A

The present age of Sohan is $20$ years more than the present age of Sachin. $3$ years hence, it will be thrice that of Sachin. Find the present age of Sohan.

  1. $23$ years

  2. $27$ years

  3. $30$ years

  4. $37$ years


Correct Option: A

There are $2$ boxes A and B. If we take out $10$ apples from A box & put these apples in B box then the number of apples in B box will be $4$ times of A box. If we take out $5$ apples from B box & put these apples into A box then the number of apples in both A & B boxes will be same in numbers. Find out the total apples in both the boxes :

  1. $20$

  2. $30$

  3. $50$

  4. $60$


Correct Option: C

In an examination, the ratio of passes to failures was 4 : 1. Had 30 less appeared and 20 less passed, the ratio of passes to failures would have been 5 : 1. Find the number of students who appeared for the examination.

  1. $260$

  2. $112$

  3. $2166$

  4. $150$


Correct Option: D
Explanation:

$
Let\quad the\quad pass:\quad fail\quad =\quad 4:1\ If\quad fail\quad students\quad =x,\quad pass\quad students\quad =4x\ Total\quad students\quad =\quad 5x\ If\quad 30\quad less\quad appeared\quad and\quad 20\quad less\quad passed..\ Students\quad appearing\quad =\quad 5x\quad -30\ Students\quad passing\quad =\quad 4x\quad -\quad 20\ Students\quad failing\quad =\quad 5x\quad -\quad 30\quad -(4x\quad -20)\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad x\quad -10\ 4x\quad -20\quad :\quad x-10\quad ::\quad 5:1\ 4x\quad -20\quad =\quad 5(x-10)\ 4x\quad -\quad 20\quad =\quad 5x\quad -\quad 50\ x\quad =\quad 30\ Students\quad failing\quad =\quad 30\ Students\quad passing\quad \quad =\quad 4x\quad =120\ Totla\quad students\quad =\quad 5x\quad =\quad 150\ \ \ \ 
$

$A$ is older than $B$. Taking present ages of $A$ and $B$ as $x$ years and $y$ years respectively, find in terms of $x$ and $y$. The difference between the ages of A and B.

  1. $(2x - y) years$

  2. $(x - 3y) years$

  3. $(x - y) years$

  4. None of these


Correct Option: C
Explanation:
Given, 
$A$ is older than $B$. Taking present ages of $A$ and $B$ as $x$ years and $y$ years respectively.
The difference between the ages of A and B.

$ = (x - y) $

A boy was asked to multiply a certain number by $25$. He multiplied it by $52$ and got his answer more than the correct one by $324$. The number to be multiplied was:

  1. $12$

  2. $15$

  3. $25$

  4. $52$


Correct Option: A
Explanation:

Suppose  the number to be multiplied $= x$
As per Question
$\Rightarrow 52\times x = 25\times x + 324$
$ \Rightarrow 52x = 25x + 324$
$\Rightarrow  52x - 25x = 324$
$\Rightarrow 27x = 324$
$ x = 12 $
Hence, option 'A' is correct.

An old rhyme and problem:

If to my age, there added be,
One half of it and three times three,
Four score and seven my age will be,
How old am I, pray tell me?

  1. $52$ years

  2. $50$ years

  3. $48$ years

  4. $54$ years


Correct Option: A
Explanation:

$1$ score $= 20$ years

Let the age be $x$.
Then, $x + \dfrac{1}{2}x + 3\times 3 = 4 \times 20 + 7$
$\Rightarrow  \dfrac{3}{2}x + 9 = 80 + 7$
$\Rightarrow  \dfrac{3}{2}x = 78$
$\Rightarrow  x = 78 \times \dfrac{2}{3}$
$\Rightarrow  x = 52$
Therefore, the age is $52$ years.

A can do a piece of work in $24$ days. If B is $60\%$ more efficient then the number of days required by B to do the twice as large as the earlier work is-

  1. $24$

  2. $36$

  3. $15$

  4. $30$


Correct Option: D
Explanation:
According to the question-
A can do a piece of work in $24$ days.
$\therefore$ work done by A in $1$ day  $=\cfrac{1}{24}$
Given that, B is $60 \%$ more efficient, i.e., B can do $60 \%$ more work than work done by A in $1$ day-
Work done by B in $1$ day  $=\cfrac{1}{24} + \cfrac{60}{100} \times {1}{24} = \cfrac{1}{15}$
$\therefore$ number of days required by B to do the same work $= 15 $days
$\therefore$ number of days required by B to do the twice as large as the earlier work  $=2 \times 15 = 30$ days

A and B can do a job in $12$ days, B and C can do it in $16$ days. After A has worked for $5$ days and B has worked for $7$ days, C can finish the rest in $13$ days. In how many days can C do the work alone?

  1. $16$ days

  2. $24$ days

  3. $36$ days

  4. $48$ days


Correct Option: B
Explanation:

Solution:-

Let the amount of work done by $A, B,$ and $C$ per day be $x, y$ and $z$ respectively.
Case I:-

$A$ and $B$ can do the job in $12$ days.
$\therefore$ work done by $A$ and $B$ in one day  $=\cfrac{1}{12}$
$\Rightarrow \; x + y = \cfrac{1}{12}$
Case II:-
B and C can do the job in $16$ days.
$\therefore$ work done by B and C in one day  $=\cfrac{1}{16}$
$\Rightarrow \; y + z = \cfrac{1}{16}$
As given, A has worked for $5$ days and B has worked for $7$ days, C can finish the rest in $13$ days.
$5x + 7y + 13z = 1$
$\Rightarrow$ $5x + 5y + 2y + 2z + 11z = 1$
$\Rightarrow$ $5(x + y) + 2(y + z) + 11z = 1$
$\Rightarrow \; 5 \times \cfrac{1}{12} + 2 \times \cfrac{1}{16} + 11z = 1$
$\Rightarrow \; 11z = 1 - \cfrac{5}{12} - \cfrac{1}{8}$
$\Rightarrow \; 11z = \cfrac{22}{48}$
$\Rightarrow \; z = \cfrac{1}{24}$
Therefore, work done by $C$  $1$ day  $=\cfrac{1}{24}$
Hence, C alone can finish the work in $24$ days.

A thief escaped from police custody. Since he was sprinter he could clock $40\ km/hr$. The police realized it after $3$ hr and started chasing him in the same direction at $50\ km/hr$. The police had a dog which could run at $60\ km/hr$. The dog could run to the thief and then return to the police and then would turn back towards the thief. If kept on doing so till the police caught the thief. Find the total distance travelled by the dog in the direction of the thief.

  1. $720 km$

  2. $600 km$

  3. $660 km$

  4. $360 km$


Correct Option: C
Explanation:
Speed of thief $=40 \  km \ per\ hr=v _{t}$
Speed of police $=50 \ km\ per\ hr=v _{p}$
Police started after 3hrs of start of thief
$50t=40t+40(3)$
$\therefore   t=12 hr$
Time taken by police to catch thief $=12 hr$
Distance travelled by dog $=12\times 60$
$=720 km$
$\Rightarrow$   Distance travelled by police + To and from distance by Dog $=720$
Distance travelled by police $=50\times 12$
$=600 km$
$\Rightarrow$   $600+2(\text{from distance by dog})=720$
"Fro" distance (distance from thief to police)="to"distance(distance from police to thief after once it returns)
$600+2(\text{'fro' distance by dog})=720$
$\therefore$ 'fro' distance $=60$
$\therefore$ Total distance towards thief $=600+'to"\ distance$
$=720-$ 'fro' distance
$=660 km$

Grass in lawn grown equally thick and in a uniform rate. It takes $24$ days for $70$ cows and $60$ days for $30$ cows to eat the whole of the grass. How many cows are needed to eat the grass in $96$ days?

  1. $20$ cows

  2. $24$ cows

  3. $28$ cows

  4. $32$ cows


Correct Option: A
Explanation:
Let $n, g, r,$ and $c$ be the no. of cows, grass initially, the rate at which grass grow/day, and grass eaten by cow/day respectively.
Now according to the question-
It takes $24$ days for $70$ cows to eat the whole of the grass.
$g + 24r = 70 \times 24c = 1680c \longrightarrow \left( i \right)$
$g + 60r = 60 \times 30c = 1800c$
Again, given that it takes 60 days for 30 cows to eat the whole of the grass.
$\therefore \; g = 1800c - 60r \longrightarrow \left( ii \right)$
From ${eq}^{n} \left( i \right) \& \left( ii \right)$, we have
$c = \cfrac{3}{10} r \longrightarrow \left( iii \right)$
Now, to eat the grass in 96 days-
$g + 96r = 96 \times nc$
$\Rightarrow \; 96 \times nc = 1800c - 60r + 96r = 1800c + 36r = 1800c + 120c = 1920c$
$\Rightarrow \; n = 20$
Hence, $20$ cows are needed to eat the grass in $96$ days.

A large tanker can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the empty tanker if only B is used in the first-half of the time and A and B are both used in the second-half of the time?

  1. $15$

  2. $20$

  3. $27.5$

  4. $30$


Correct Option: D
Explanation:

Let x minute will be taken. In one minute A can fill the $\dfrac {1}{60}$ part of tanker and in one minute B can fill the $\dfrac {1}{40}$ part.
Both can fill in t
$\dfrac {t}{60}+\dfrac {t}{40}=1$
$t=\dfrac {60\times 40}{100}$
$t=24$
both can fill in one minute $\dfrac {1}{24}$ part of tanker.
$1=\left (\dfrac {x}{2}\right )\dfrac {1}{40}+\dfrac {x}{2}\left (\dfrac {1}{24}\right )$
$1=\dfrac {x}{80}+\dfrac {x}{48}$
$x=\dfrac {80\times 48}{128}=30$

Two birds are flying in opposite directions over the edge of a circle-shaped forest of radius 4 km. Both start off from the same point simultaneously and both have to go to the same nest. Who reaches the nest first?
I. Speed of bird A is 60 km/hr and speed of bird B is 50 km/hr.
II. The nest is diametrically opposite to the starting point of the flight of the two birds, on the circumference of the forest.

  1. If the question can be answered by anyone of the statements alone, but cannot be answered by using the other statement alone.

  2. If the question can be answered by using either statement alone.

  3. If the question can be answered by using both the statements together, but cannot be answered by using either statement alone.

  4. If the question. cannot be answered even by using both the statements together.


Correct Option: D
Explanation:

In the problem statement, the distance is given. 
In the first statement, the speeds are given.
In the second statement, nothing substantial can be concluded.
Since information on the paths of flight is not given, we can not determine which bird reaches first. D,

To find the present age of Rishi, which statements can be dispensed.
I. In ten years,Richard will be twice as old as Rishi was 10 years ago.
II. Richard is now 9 years older than Rishi.
III. Five years ago, Rishi was 9 years younger than Richard.

  1. Only II

  2. Only III

  3. Either I or II

  4. None of the three statements can be dispensed with


Correct Option: C
Explanation:
There are 2 unkown variables, Rishi's age x and Richard's age y.
Each of the following statements give relation between x and y.

Statement I  says: $\quad y+10=2x.$
Statement II says: $\quad y=x+9$

Statement III says:  $y-5=(x-5)+9$= Statement II.
To solve for 2 unknown we require 2 equations.

Since, Statement II and Statement III are equivalent , either can be dispensed with.

Helen is determining how much money she and her friends will need to go for a movie.
Each person going will buy a ticket, a bag of popcorn, and a drink. Helen writes the given formula that will represent the situation.
$m = q (t + p + d)$
Which description represents the meaning of the variable $q$?

  1. The price of a ticket

  2. The cost of concessions

  3. The number of people going

  4. The amount of money required


Correct Option: C
Explanation:

From the formula $ m = q \left(t+p+d \right)$
t represent the money required for  each ticket
p represent the money required for popcorn for each person
d represent the money required for drink for each person
If we multiply this with the total no of persons going to watch movie we can get total money require.

A clock is set right at $8:00\ a.m$. The clock gains $10$ minutes in $24$ hours. What will be the right time when this clock indicates $1\ p.m$ on the following day?

  1. $11.40\ p.m$

  2. $12:00\ p.m$

  3. $10:00\ p.m$

  4. $12:48\ p.m$

  5. $None\ of\ these$


Correct Option: D
Explanation:

The clock gains $10$ mins in $24$ hours,
It will gain 1 min in every $2.4$ hours.
Difference in time the clock indicates=$1$ p.m.-$8$ a.m. (the next day)
=$29$ hours,
Time gained=$\dfrac {29}{2.4}=12$ mins.
Actual time=$12:48$ p.m.

Three years ago, the average age of the family of 5 members was 17 years. A baby having been born, the average age of the family is the same today. What is the baby today?

  1. 4 years

  2. 3 years

  3. 2 years

  4. 1 year


Correct Option: C
Explanation:

Present total ages of six members $=17\times6$ i.e., 102 years.
Present ages of 5 members $=5\times(17+3)$ i.e. 100 years.
$\implies$ Baby is 2 years old to-day.

Henry just set up direct deposit from his employer to his checking account. The equation $\displaystyle y=360x-126.13$ represents the balance in Henry's account if he deposit his weekly paycheck for x weeks. Based on this equation, which of the following statements is true

  1. Henry earns $126.13 per week.

  2. Henry made an initial deposit of $126.13.

  3. Before setting up the direct deposit, Henry had overdrawn his account.

  4. When Henry set up the direct deposit, he already has $360 in his account.


Correct Option: C
Explanation:

The balance in Henry's account follows the relation $y = 360x - 126.13$, where $y$ is the balance in Henry's checking account and $x$ is the number of weeks passed. 

The difference in the balance between any two successive weeks is $ $360 $. This shows that Henry earns $ $360$ per week.
The other options can be checked simultaneously. After putting $x = 0$, we get $y = -126.13$. This shows that only option C is correct.

Ajit is $21$ years younger than his father. What is their total age in $7$ years time?

  1. $(x + 28)$ years

  2. $(x + 35)$ years

  3. $(2x + 28)$ years

  4. $(2x + 35)$ years


Correct Option: D
Explanation:
Let the age of Ajit be $x$ years.
$\therefore$ Age of Ajit's father $= x + 21$
After $7$ years-
Age of Ajit $= (x + 7)$ years
Age of Ajit's father $= x + 21 + 7 = x + 28$
Total age of Ajit and Ajit' father $= x + 7 + x + 28 = (2x + 35)$ years

Reema bought $x$ pens at Rs.$2.60$ each and $y$ greeting cards at $80$ paise each. If the pens cost Rs.$12$ more than the cards, the equation involving $x$ and $y$ is

  1. $13x-4y=6$

  2. $13x-4y=60$

  3. $260x-8y=100$

  4. $260x-8y=12$


Correct Option: B
Explanation:

Given that Reema bought $x$ pens at Rs. $2.60$ each and $y$ greeting cards at $80$ paise each

Then cost of x pens $ =2.60 x$ RS 
And cost of y greeting cards $=0.80 y$ Rs
But given that total cost of pens is Rs $12$ more than the cards

$\therefore 2.60 x=0.80 y+12$
Mltiply by 10 both sides we get
$2.60x\times 10=0.80y\times 10+12\times 10$
$\Rightarrow 26x=8y+120$
$\Rightarrow 26x-8y=120$
Divided by 2 both sides we get
$13x-4y=60$

Equation for the statement 'Thrice the length ($l$) of a room is $340$ metres' is _____ .

  1. $3l=430$

  2. $3l=340$

  3. $3+l=340$

  4. $3l+340=0$


Correct Option: B
Explanation:

$\Rightarrow$   The length of room is $l$.

$\Rightarrow$   So, thrice the length means $3l$.
$\therefore$    According to the statement given in question,
$\Rightarrow$  $3l=340$

Ashima bought $23$ things from the market. She bought five more jeans than shirts and two fewer watches than jeans. If $x$ represents the number of shirts in total, then which sentence can be used to find how many of each thing are bought?

  1. $x + (x + 5) + (x + 3) = 23$

  2. $x + (x - 5) + (x - 3) = 23$

  3. $(x + 5) + (x + 3) = 23$

  4. $x + (x + 3) = 23$


Correct Option: A
Explanation:

Given: Number of shirts = $x$ 

Number of Jeans = $x+5$
Number of watches= $(x+5)-2=x+3$
$\therefore$  According to question,
$x+(x+5)+(x+3)=23$

You are decorating a gift pack with 15 flowers. You want an equal number of flowers in each of the 3 rows on the gift pack. Which equation would you use to find the number of flowers, r, in each row?

  1. r + 3 = 15

  2. 15 + r = 3

  3. 3r = 15

  4. $\dfrac {3}{r}$ = 15


Correct Option: C
Explanation:

We have to decorate a gift pack with 15 flowers. 

$\Rightarrow$  There are 3 rows on gift pack and we want equal number of flowers in each row.
$\Rightarrow$  Let $r$ be number of flowers in each row.
$\therefore$   Number of rows $\times$ Number of flowers in each row = Number of flowers.
$\Rightarrow$  $3\times r=15$
$\therefore$     $3r=15$

A shopkeeper sells bananas in two types of boxes, one small and one large. A large box contains as many as 6 small boxes plus 2 loose bananas. Form an equation which gives the number of bananas in each small box, if the number of bananas in 1 large box is 50.

  1. 3x + 1 = 50

  2. x + 1 = 20

  3. 6x + 2 = 50

  4. 2x + 1 = 20


Correct Option: C
Explanation:
 Let the number of bananas in each small box be $x.$
$\Rightarrow$  Number of small boxes $=6$
$\therefore$   According to question, $6x + 2 = 50$

The teacher tells the class that the highest marks obtained by a student in her class is four times the lowest marks plus 6. The highest score is 65. Form the equation which will calculate the lowest marks.

  1. 6m + 4 = 65

  2. 4m + 65 = 6

  3. 4m + 6 = 65

  4. 6m + 65 = 4


Correct Option: C
Explanation:
 Let the lowest marks obtained by a student be $m.$
$\Rightarrow$  The highest score is $65.$
$\Rightarrow$  According to question, $4m+6=65$
$\therefore$   The equation to calculate lowest marks is $4m+6=65$.

Ram's father's age is 3 years more than two times Ram's age. Ram's father is 45 years old. Form an equation to find Ram's age.

  1. 2x + 3 = 45

  2. 3x + 2 = 45

  3. 6x + 3 = 45

  4. 5x + 1 = 45


Correct Option: A
Explanation:
 Let Ram's age be $x$ years.
$\Rightarrow$  Ram's father's age = $(2x + 3)$ years
$\therefore$    According to question, $2x + 3 = 45$

A group of students decided to collect as many paise from each member of the group as is the number of members in the group. If the total collection amounts to Rs.$22.09$, the number of members in the group is.

  1. $37$

  2. $47$

  3. $39$

  4. $49$


Correct Option: B
Explanation:

Let the total number of members = $x$


So, money collected from each member = $x\ paise=Rs.\ 0.01x$

Total money collected = $x\times0.01x=0.01\ x^2$

$\Rightarrow0.01\ x^2=22.09$

$\Rightarrow x^2=\dfrac{22.09}{0.01}=2209$

$\Rightarrow x=\sqrt{2209}=47$

So, there are  $47$  members in the group.   $[B]$

Each child from a certain school can make $5$ items of handicraft in a day. If $1125$ handicraft items are to be displayed in an exhibition, then in how many days can $25$ children make these items?

  1. $6$ days

  2. $7$ days

  3. $8$ days

  4. $9$ days


Correct Option: D
Explanation:

Number of handicrafts made by one student in a day = $5$


Total number of students = $25$

So, total number of handicrafts made in a day = $25\times5=125$

Total number of handicrafts required = $1125$

$\therefore$  Number of days required = $\dfrac{Total\ number\ of\ handicrafts\ required}{Number\ of\ handicrafts\ made\ in\ a\ day}=\dfrac{1125}{125}=9\ days$.

So, we can make the required number of handicrafts in $9\ days$.   $[D]$

Kiran spent $Rs. 6x$ on a book, $Rs. 6$ on food and had $Rs.18$ left. what was the sum of money she had at first? Express your answer in terms of $x$.

  1. $Rs. (6x+18)$

  2. $Rs. (6x+24)$

  3. $Rs. 64x$

  4. $Rs. 24x$


Correct Option: B
Explanation:

Amount spent on book $=Rs. 6x$

Amount spent on food $=Rs. 6$
Total amount spent $=Rs. (6x+6)$
Amount left with Kiran $=Rs. 18$
$\therefore$ Amount she had at first $=Rs. (6x+6+18)$
                                            $=Rs. (6x+24)$

A person walks from his house at a speed of $4$ km/hr and reaches his schools $5$ minutes late. If his speed has been $5$ km/hr, he would have reached $10$ minutes earlier. The distance of the school from his house is

  1. $5$ km

  2. $6$ km

  3. $7$ km

  4. $8$ km


Correct Option: A
Explanation:

Let the correct time to reach the school be ‘t’ hrs.

Then,the time taken by the man when he wakes at a speed of $4km$ will be ‘t’ hrs+5 mins,

which is nothing but $(t+\dfrac{5}{60})$hrs.

And,the time taken by the man when he walks at 5km/hr will similarly be $(t-\dfrac{10}{60})$hours.

As the distance in both the cases is same,

$4(t+\dfrac{5}{60}$)hrs=$5(t-\dfrac{10}{60})$hrs

$4t+\dfrac{20}{60}$=$5t-\dfrac{50}{60}$

 Therefore$ t=\dfrac{70}{60}$=$\dfrac{7}{6}$hrs

Actual distance=$4(t+\dfrac{5}{60})$

=  $4(\dfrac{70}{60}+\dfrac{5}{60})$

=$4\times \dfrac{75}{60}$=$5kms.$

If $12$ men complete a work in $20$ days. If only $8$ men are  employed, then the time required  to complete  the same work is

  1. $24$ days

  2. $25$ days

  3. $30$ days

  4. $35$ days


Correct Option: C
Explanation:

As $12$ men complete work in $20$ days, $1$ man will complete the same work in

$20\times12=240$ days.

Time required by $8$ men to complete the work=$\dfrac{240}{8}$=$30$ days..

There are $50$ plates in a crate and there are $11$ such crates. When they are opened $\dfrac{2}{5}$ of the plates are found to be broken. How many plates are left intact?

  1. $220$

  2. $330$

  3. $530$

  4. $320$


Correct Option: B
Explanation:
$\Rightarrow$Total number of crates$=11$

$\Rightarrow$one crates contain$= 50$plates

$\Rightarrow$so total number plates=$11\times50=550$

$\Rightarrow$the broken part of plates is $\dfrac{2}{5}$

$\Rightarrow$so the number of plates$=550\times\dfrac{2}{5}=220$

$\Rightarrow$the number plats left intact$=550-220=330$

Three friends, Mala, Leela and Sheela, divided a box of apples weighing $15\dfrac{9}{10}$ kg equally between the three of them. How many kgs of apples did each get?

  1. $5\dfrac{3}{10} kg$

  2. $3\dfrac{3}{10}  kg $

  3. $3\dfrac{9}{10}  kg $

  4. $5\dfrac{9}{10}  kg $


Correct Option: A
Explanation:
Total weight of the apple $15\dfrac{9}{10}kg=\dfrac{(15\times10)+9}{10}=\dfrac{159}{10}$

There are three, they divide the apple equally between them

so one person will get the weight of apple $=\dfrac{159}{3\times10}=\dfrac{159}{30}=5\dfrac{3}{10}kg$
- Hide questions