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## Algebra: Linear Equation Test-1

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Question 1 |

Children had fallen-in for a drill. If each row contained 4 children less, 10 more rows would have been made. But if 5 more children were accommodated in each row the number of rows would have reduced by 5. The number of children in the school is

A | 200 |

B | 150 |

C | 300 |

D | 100 |

Question 1 Explanation:

Suppose there are p rows & each row contains y no of children.

Then total no of children would be py.

Now according to question: (p+10) (y-4) =py

=10y-4p=40……………………………………a

And (p-5) (y+5) = py.

5p -5y = 25

p – y =5………………………….b

After solving these two equations a and b, we get p=15 & y=10

Then total no of children would be 15 x 10=150

Then total no of children would be py.

Now according to question: (p+10) (y-4) =py

=10y-4p=40……………………………………a

And (p-5) (y+5) = py.

5p -5y = 25

p – y =5………………………….b

After solving these two equations a and b, we get p=15 & y=10

Then total no of children would be 15 x 10=150

Question 2 |

There are two examination rooms A and B. If 10 candidates are sent from A to B, the number of students in each room is the same. If 20 candidates are sent from B to A, the number of students in A is double the number of students in B. How many students are there in rooms A and B respectively?

A | 100 and 80 |

B | 120 and 60 |

C | 80 and 100 |

D | 140 and 60 |

Question 2 Explanation:

Suppose there are x candidates in A and y candidates in B.

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

x = y +20…………………………..f

(x+20)=2(y-20)

x = 2y – 60………………………..g

After solving these two equation f and g we get x=100 & y=80

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

x = y +20…………………………..f

(x+20)=2(y-20)

x = 2y – 60………………………..g

After solving these two equation f and g we get x=100 & y=80

Question 3 |

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Sanchit paid Rs.45 for a book kept for 7 days, while Karan paid Rs.25 for the book he kept for 5 days. The fixed charge and the charge for each extra day is

A | Rs.5 and Rs.50 |

B | Rs.10 and Rs.5 |

C | Rs.15 and Rs.5 |

D | Rs.5 and Rs.15 |

Question 3 Explanation:

Suppose fixed charge is x & additional charge for one day is y.

Then according to question (x+4y)=45

And (x+2y)=25. These 2 equations will give x=5 & y=10.

But in option b order has been changed.

Then according to question (x+4y)=45

And (x+2y)=25. These 2 equations will give x=5 & y=10.

But in option b order has been changed.

Question 4 |

The basic one-way railway fare for a child aged between 3 and 10 yr costs half the regular fare for an adult plus a reservation charge that is the same on the child's ticket as on the adult’s ticket. One reserved ticket for an adult costs Rs.216 and the cost of a reserved ticket for an adult and a child (aged between 3 and 10) costs Rs.327. What is the basic fare for the journey for an adult?

A | Rs.210 |

B | Rs.52.50 |

C | Rs.111 |

D | Rs.58.50 |

Question 4 Explanation:

Suppose basic fare for a child is Rs. x and basic fare for an adult is Rs. y.

Reservation charge is fixed for child, adult which is Rs. c .

Then according to equation

y +c=216

x+y+2c=327.

After solving these equations we get x=Rs. 105 and y=Rs. 210

Therefore the correct answer is option a

Reservation charge is fixed for child, adult which is Rs. c .

Then according to equation

y +c=216

x+y+2c=327.

After solving these equations we get x=Rs. 105 and y=Rs. 210

Therefore the correct answer is option a

Question 5 |

One-fourth of Nikhil's money is equal to one-sixth of Yogesh's money. If both together have Rs.600, what is the difference between their amounts?

A | Rs.160 |

B | Rs.240 |

C | Rs.200 |

D | Rs.120 |

Question 5 Explanation:

Suppose Nikhil has Rs. n and Yogesh has Rs. y

Then (n/4)=(y/6)

it is 3n=2y , both have 600 together ie. n +y=600

Then n=240 & y=360.

Difference would be Rs. 120

Then (n/4)=(y/6)

it is 3n=2y , both have 600 together ie. n +y=600

Then n=240 & y=360.

Difference would be Rs. 120

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