1Gopi promises to give Rs. 90000 plus one old scooter as salary to his servant for one year. The servant leaves after 9 months and receives Rs. 65000 and the scooter. Find the price of the scooter.
ARs. 25000
BRs. 10000
CRs. 20000
DRs. 15000
2There are two sections A and B for class X in a school. If 6 students are sent from A to B, the number of students in B is double that of A. Otherwise, if 12 students from B are sent to A the number of students would be equal in both the sections. Find the number of students in section B initially.
A84
B78
C57
D66
3Mahesh, Pawan, Charan are playing $Uno$ and each has some money with them. In each round the looser has to double the money with each of the remaining two players. In the first round, Mahesh lost the game and doubled the money of Pawan and Charan. In the second round Pawan lost and in the third round Charan lost the game. After the third round, they found that each has Rs.240 with them. Find the initial amount with Pawan.
A200
B230
C210
D180
4Bhavya, Divya, Navya, Sravya are best friends. Bhavya has bought $n$ chocolates on her birthday. She gives Divya half the number of chocolates plus half chocolate. Then she gives Navya half the remaining number of chocolates plus half chocolate. Finally, she gives half the remaining number of chocolates plus half chocolate. And she left with no chocolates. Then "n" lies in between?
A$2 \le n \le 6$
B$5 \le n \le 8$
C$9 \le n \le 12$
D$11 \le n \le 16$
5A man spent 1/6th of his life in child hood, 1/12th of his life as youngster and 1/7th of his life as a bachelor. After five years of his marriage, a son was born to him. The son died four years before the father died and at the time of his death, his age was half the total age of his father. What is the age of the father?
A60
B72
C84
D96
6Pawan has bought a new car. For the car registration, he has the following conditions to be met. The number should be a four digit number having non zero and distinct digits. The sum of digits at the units and tens position is equal to the sum of the remaining two digits. The sum of the digits in the middle positions is three times to the sum of the remaining digits. If the sum of the digits is not more than 20, then how many such four digit numbers are possible.
A1
B3
C6
D8
ARs. 25000
BRs. 10000
CRs. 20000
DRs. 15000
Answer: B
Explanation:
The servant is supposed to get 9 months salary and 9 months worth of scooter for his contribution.
Let the cost of scooter= $x$
$\therefore \dfrac{9}{{12}}\left( {90000 + x} \right) = 65000 + x$
$ \Rightarrow \dfrac{3}{4} \times 90000 + \dfrac{3}{4}x = 65000 + x$
$\Rightarrow \require{cancel}\dfrac{3}{\cancel{4}} \times \cancel{90000}^{22500}+ \dfrac{3}{4}x =$ $65000 + x$
$ \Rightarrow 3 \times 22500 - 65000 = x - \dfrac{3}{4}x$
$ \Rightarrow 67500 - 65000 = \dfrac{x}{4}$
$ \Rightarrow x = 4 \times 2500 = 10000$
Explanation:
The servant is supposed to get 9 months salary and 9 months worth of scooter for his contribution.
Let the cost of scooter= $x$
$\therefore \dfrac{9}{{12}}\left( {90000 + x} \right) = 65000 + x$
$ \Rightarrow \dfrac{3}{4} \times 90000 + \dfrac{3}{4}x = 65000 + x$
$\Rightarrow \require{cancel}\dfrac{3}{\cancel{4}} \times \cancel{90000}^{22500}+ \dfrac{3}{4}x =$ $65000 + x$
$ \Rightarrow 3 \times 22500 - 65000 = x - \dfrac{3}{4}x$
$ \Rightarrow 67500 - 65000 = \dfrac{x}{4}$
$ \Rightarrow x = 4 \times 2500 = 10000$
2There are two sections A and B for class X in a school. If 6 students are sent from A to B, the number of students in B is double that of A. Otherwise, if 12 students from B are sent to A the number of students would be equal in both the sections. Find the number of students in section B initially.
A84
B78
C57
D66
Answer: D
Explanation:
Let the number of students in class A are $x$ and class B are $y$.
$\begin{array}{*{20}{ccl}}
{\underline A }&{\underline B }\\[4px]
x&y&{}\\[4px]
{x - 6}&{y + 6}&{ \Rightarrow \dfrac{{y + 6}}{{x - 6}} = 2}\\[4px]
&&\Rightarrow y+6 = 2x - 12 \\[4px]
&&\Rightarrow 2x -y = 18 & (1) \\[4px]
\hdashline
{\underline A }&{\underline B }\\[4px]
x&y&{}\\[4px]
{x + 12}&{y - 12}&{ \Rightarrow x + 12 = y - 12}\\[4px]
&&\Rightarrow x -y = -24 & (2) \\[4px]
\end{array}$
Subtract equation (2) from (1),
$x = 42$
Substitute this value in equation (1),
$2 \times 42 - y = 18$
$ y = 84 - 18 = 66$
$\therefore$ Number of students in class B are 66.
Explanation:
Let the number of students in class A are $x$ and class B are $y$.
$\begin{array}{*{20}{ccl}}
{\underline A }&{\underline B }\\[4px]
x&y&{}\\[4px]
{x - 6}&{y + 6}&{ \Rightarrow \dfrac{{y + 6}}{{x - 6}} = 2}\\[4px]
&&\Rightarrow y+6 = 2x - 12 \\[4px]
&&\Rightarrow 2x -y = 18 & (1) \\[4px]
\hdashline
{\underline A }&{\underline B }\\[4px]
x&y&{}\\[4px]
{x + 12}&{y - 12}&{ \Rightarrow x + 12 = y - 12}\\[4px]
&&\Rightarrow x -y = -24 & (2) \\[4px]
\end{array}$
Subtract equation (2) from (1),
$x = 42$
Substitute this value in equation (1),
$2 \times 42 - y = 18$
$ y = 84 - 18 = 66$
$\therefore$ Number of students in class B are 66.
3Mahesh, Pawan, Charan are playing $Uno$ and each has some money with them. In each round the looser has to double the money with each of the remaining two players. In the first round, Mahesh lost the game and doubled the money of Pawan and Charan. In the second round Pawan lost and in the third round Charan lost the game. After the third round, they found that each has Rs.240 with them. Find the initial amount with Pawan.
A200
B230
C210
D180
Answer: C
Explanation:
This question can be solved easily if we calculate the shares from backwards.
$\begin{array}{*{20}{|c|c|c|}}\hline
&M &P &C \\[4px]\hdashline
After\\Round\,3&{240}&{240}&{240} \\[4px]\hline
Before \\
Round\,3& 120 &120& 240+(120+120) \\ &&&=480\\\hline
\end{array}$
$\begin{array}{*{20}{|c|c|c|}}\hline
&M &P &C \\[4px]\hdashline
After \\ Round\,2&{120}&{120}&{480} \\[4px]\hline
Before \\
Round\,2& 60 &120+(60+240) & {240}\\
&&=420 \\\hline
\end{array}$
$\begin{array}{*{20}{|c|c|c|}}\hline
&M &P &C \\[4px]\hdashline
After \\ Round\,1&{60}&{420}&{240} \\[4px]\hline
Before \\
Round\,1& 60+(210+120) &210 & {120}\\
&=390 \\\hline
\end{array}$
$\therefore$ Initial share of Pawan is Rs.210
Explanation:
This question can be solved easily if we calculate the shares from backwards.
$\begin{array}{*{20}{|c|c|c|}}\hline
&M &P &C \\[4px]\hdashline
After\\Round\,3&{240}&{240}&{240} \\[4px]\hline
Before \\
Round\,3& 120 &120& 240+(120+120) \\ &&&=480\\\hline
\end{array}$
$\begin{array}{*{20}{|c|c|c|}}\hline
&M &P &C \\[4px]\hdashline
After \\ Round\,2&{120}&{120}&{480} \\[4px]\hline
Before \\
Round\,2& 60 &120+(60+240) & {240}\\
&&=420 \\\hline
\end{array}$
$\begin{array}{*{20}{|c|c|c|}}\hline
&M &P &C \\[4px]\hdashline
After \\ Round\,1&{60}&{420}&{240} \\[4px]\hline
Before \\
Round\,1& 60+(210+120) &210 & {120}\\
&=390 \\\hline
\end{array}$
$\therefore$ Initial share of Pawan is Rs.210
4Bhavya, Divya, Navya, Sravya are best friends. Bhavya has bought $n$ chocolates on her birthday. She gives Divya half the number of chocolates plus half chocolate. Then she gives Navya half the remaining number of chocolates plus half chocolate. Finally, she gives half the remaining number of chocolates plus half chocolate. And she left with no chocolates. Then "n" lies in between?
A$2 \le n \le 6$
B$5 \le n \le 8$
C$9 \le n \le 12$
D$11 \le n \le 16$
Answer: B
Explanation:
After Bhavya gave chocolates to Sravya She left with none.
Let us assume she has $z$ chocolates before giving to Sravya.
$z - \left( {\dfrac{z}{2} + \dfrac{1}{2}} \right) = 0$
$ \Rightarrow \dfrac{z}{2} = \dfrac{1}{2}$
$ \Rightarrow z = 1$
$\therefore$ Bhavya has given only one chocolate to Sravya. One half of one chocolate and half cholocate is equal to 1.
Let us assume she has $y$ chocolates before giving Chocolates to Sravya.
$y - \left( {\dfrac{y}{2} + \dfrac{1}{2}} \right) = 1$
$ \Rightarrow \dfrac{y}{2} = \dfrac{3}{2}$
$ \Rightarrow y = 3$
$\therefore$ Bhavya has three chocolates and has given 2 chocolates to Sravya.
Let us assume she has $x$ chocolates before giving chocolates to Divya.
$x - \left( {\dfrac{x}{2} + \dfrac{1}{2}} \right) = 3$
$ \Rightarrow \dfrac{x}{2} = \dfrac{7}{2}$
$ \Rightarrow x = 7$
$\therefore$ Bhavya has 7 chocolates to start with.
Explanation:
After Bhavya gave chocolates to Sravya She left with none.
Let us assume she has $z$ chocolates before giving to Sravya.
$z - \left( {\dfrac{z}{2} + \dfrac{1}{2}} \right) = 0$
$ \Rightarrow \dfrac{z}{2} = \dfrac{1}{2}$
$ \Rightarrow z = 1$
$\therefore$ Bhavya has given only one chocolate to Sravya. One half of one chocolate and half cholocate is equal to 1.
Let us assume she has $y$ chocolates before giving Chocolates to Sravya.
$y - \left( {\dfrac{y}{2} + \dfrac{1}{2}} \right) = 1$
$ \Rightarrow \dfrac{y}{2} = \dfrac{3}{2}$
$ \Rightarrow y = 3$
$\therefore$ Bhavya has three chocolates and has given 2 chocolates to Sravya.
Let us assume she has $x$ chocolates before giving chocolates to Divya.
$x - \left( {\dfrac{x}{2} + \dfrac{1}{2}} \right) = 3$
$ \Rightarrow \dfrac{x}{2} = \dfrac{7}{2}$
$ \Rightarrow x = 7$
$\therefore$ Bhavya has 7 chocolates to start with.
5A man spent 1/6th of his life in child hood, 1/12th of his life as youngster and 1/7th of his life as a bachelor. After five years of his marriage, a son was born to him. The son died four years before the father died and at the time of his death, his age was half the total age of his father. What is the age of the father?
A60
B72
C84
D96
Answer: C
Explanation:
Let the father lived for $x$ years.
The age of the man at the time of marriage = $\dfrac{x}{6} + \dfrac{x}{{12}} + \dfrac{x}{7} = \dfrac{{33x}}{{84}}$
The age of the man at the time of the birth of son = $\dfrac{{33x}}{{84}} + 5$
The age of the son at the time of his death = $\left( {x - 4} \right) - \left( {\dfrac{{33x}}{{84}} + 5} \right)$ = $\dfrac{{51x}}{{84}} - 9$
Given that, $\dfrac{x}{2} = \dfrac{{51x}}{{84}} - 9$
$ \Rightarrow \dfrac{{51x}}{{84}} - \dfrac{x}{2} = 9$
$ \Rightarrow \dfrac{{51x - 42x}}{{84}} = 9$
$ \Rightarrow \dfrac{{9x}}{{84}} = 9$
$ \Rightarrow x = 84$
Explanation:
Let the father lived for $x$ years.
The age of the man at the time of marriage = $\dfrac{x}{6} + \dfrac{x}{{12}} + \dfrac{x}{7} = \dfrac{{33x}}{{84}}$
The age of the man at the time of the birth of son = $\dfrac{{33x}}{{84}} + 5$
The age of the son at the time of his death = $\left( {x - 4} \right) - \left( {\dfrac{{33x}}{{84}} + 5} \right)$ = $\dfrac{{51x}}{{84}} - 9$
Given that, $\dfrac{x}{2} = \dfrac{{51x}}{{84}} - 9$
$ \Rightarrow \dfrac{{51x}}{{84}} - \dfrac{x}{2} = 9$
$ \Rightarrow \dfrac{{51x - 42x}}{{84}} = 9$
$ \Rightarrow \dfrac{{9x}}{{84}} = 9$
$ \Rightarrow x = 84$
6Pawan has bought a new car. For the car registration, he has the following conditions to be met. The number should be a four digit number having non zero and distinct digits. The sum of digits at the units and tens position is equal to the sum of the remaining two digits. The sum of the digits in the middle positions is three times to the sum of the remaining digits. If the sum of the digits is not more than 20, then how many such four digit numbers are possible.
A1
B3
C6
D8
Answer: D
Explanation:
Let the number be $\color{blue}{a\,b\,c\,d}$
Given that, $a + b = c + d$ and $\dfrac{{b + c}}{{a + d}} = \dfrac{3}{1}$
Let $b+c=3x$ and $a+d=x$
Now $a + b + c + d = 4x$
From the above equation, we infer that the sum of the digits is a multiple of 4.
Now we check various possibilities for 4, 8, 12 ...till 20.
If sum is equal to 4, then all digits are same. So this is ruled out.
If sum is equal to 8, then $b + c=6$ and $a + d = 2$. Now $a, d$ are same.
For the remaining possibilities, refer to the following table.
If sum is equal to 12, then $b + c=9$ and $a + d = 3$
$(a, d) = (1, 2)$ or $(a,d) = (2, 1)$.
as $a + b = c + d$, we have to take $b$, $c$ in such a way that their difference is 1.
Let $b = x$ and $c = x + 1$
$b + c = 2x + 1 = 9$
$x = 4$
$\therefore$ $b = (4,5)$ or $(5, 4)$
Similarly we find the remaining possibilites.
$\begin{array}{*{20}{c|cc}}
&a &b &c &d\\[4px]\hdashline
12 & 1 & (4&5) & 2 \\[4px]
& 2 &(5 & 4) & 1 \\[4px]\hline
16 & 3 & (5 & 7) & 1 \\[4px]
& 1 &(7& 5) & 3 \\[4px]\hline
20 & 1 & (9 & 6) & 4 \\[4px]
& 4 &(6& 9) & 1 \\[4px]
& 2 & (7 & 8) & 3 \\[4px]
& 3 &(8 & 7) & 2 \\[4px]\hline
\end{array}$
Explanation:
Let the number be $\color{blue}{a\,b\,c\,d}$
Given that, $a + b = c + d$ and $\dfrac{{b + c}}{{a + d}} = \dfrac{3}{1}$
Let $b+c=3x$ and $a+d=x$
Now $a + b + c + d = 4x$
From the above equation, we infer that the sum of the digits is a multiple of 4.
Now we check various possibilities for 4, 8, 12 ...till 20.
If sum is equal to 4, then all digits are same. So this is ruled out.
If sum is equal to 8, then $b + c=6$ and $a + d = 2$. Now $a, d$ are same.
For the remaining possibilities, refer to the following table.
If sum is equal to 12, then $b + c=9$ and $a + d = 3$
$(a, d) = (1, 2)$ or $(a,d) = (2, 1)$.
as $a + b = c + d$, we have to take $b$, $c$ in such a way that their difference is 1.
Let $b = x$ and $c = x + 1$
$b + c = 2x + 1 = 9$
$x = 4$
$\therefore$ $b = (4,5)$ or $(5, 4)$
Similarly we find the remaining possibilites.
$\begin{array}{*{20}{c|cc}}
&a &b &c &d\\[4px]\hdashline
12 & 1 & (4&5) & 2 \\[4px]
& 2 &(5 & 4) & 1 \\[4px]\hline
16 & 3 & (5 & 7) & 1 \\[4px]
& 1 &(7& 5) & 3 \\[4px]\hline
20 & 1 & (9 & 6) & 4 \\[4px]
& 4 &(6& 9) & 1 \\[4px]
& 2 & (7 & 8) & 3 \\[4px]
& 3 &(8 & 7) & 2 \\[4px]\hline
\end{array}$
