Linear Equation


  1. The auto-rickshaw fare consists of a fixed charge together with the charge for the distance covered. For a journey of 10 km, the charge paid is 85 and for a journey of 15 km, the charge paid is 120. The fare for a journey of 25 km will be









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    Let use assume the fixed charge = a
    and charge for 1 km = b
    According to question,
    for 10 KM journey charge paid = 85
    a + 10 x b = 85
    a + 10b = 85 .........................(1)
    for 15 KM journey charge paid = 120
    a + 15b = 120.........................(2)
    Solve the equation (1) and (2) to get the answer.

    Correct Option: B

    Let use assume the fixed charge = a
    and charge for 1 km is = b
    According to question,
    for 10 KM journey charge paid = 85
    a + 10 x b = 85
    a + 10b = 85 .........................(1)
    for 15 KM journey charge paid = 120
    a + 15b = 120.........................(2)
    Subtract the equation (1) from equation (2). we will get,
    a + 15b - a - 10b = 120 - 85
    5b = 35
    b = 7
    Put the value of b in equation (1). we will get
    a + 10 x 7 = 85
    a = 85 - 70
    a = 15
    Charges for 25 km = a + 25 x b
    Put the value of a and b in above equation.
    Charges for 25 km =15 + 25 x 7 = 15 + 175 = 190
    Charges for 25 km =190


  1. The present ages of Vikas and Vishal are in the ratio 15:8. After ten years , their ages will be in the ratio 5:3. Find their present ages









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    Method 1
    Let us assume the ratio factor is x.
    Therefor the present ages of Vikas and Vishal be 15x years and 8x years.
    After 10 years
    vikas's age = 15x + 10 and Vishal' age = 8x + 10
    According to question,
    (15x+10)/(8x+10) = 5/3
    solve the equation and get the answer.

    Method 2
    Let us assume the present age of Vikas = x years and Vishal's present year = y years
    According to question,
    Present age of Vikas/ Present age of Vishal = 15/8
    ⇒ x/y = 15/8
    ⇒ 8x = 15y
    ⇒ 8x - 15y = 0
    ⇒ x = 15y/8 .................................(1)
    After 10 years Vikash age = x + 10 and vishal age = y + 10
    Ratio of age after 10 years = 5/3
    (x + 10)/(y + 10) = 5/3
    solve the equation and get the answer.



    Correct Option: A

    Method 1
    Let us assume the ratio factor is x.
    Therefor the present ages of Vikas and Vishal be 15x years and 8x years.
    After 10 years
    vikas's age = 15x + 10 and Vishal' age = 8x + 10
    According to question,
    (15x+10)/(8x+10) = 5/3
    ⇒ 3(15x + 10) = 5(8x + 10)
    ⇒ 45x + 30 = 40x + 50
    ⇒ 5x =20
    ⇒ x = 20/5
    ⇒ x = 4
    Therefore Present age of Vikas = 15x = 15 x 4 = 60 years
    and Present age of Vishal = 8x = 8 x 4 = 32 years

    Method 2
    Let us assume the present age of Vikas = x years and Vishal's present year = y years
    According to question,
    Present age of Vikas/ Present age of Vishal = 15/8
    ⇒ x/y = 15/8
    ⇒ 8x = 15y
    ⇒ 8x - 15y = 0
    ⇒ x = 15y/8 .................................(1)
    After 10 years Vikash age = x + 10 and vishal age = y + 10
    Ratio of age after 10 years = 5/3
    (x + 10)/(y + 10) = 5/3
    ⇒ 3(x + 10) = 5(y + 10)
    ⇒ 3x + 30 = 5y + 50
    ⇒ 3x - 5y = 50 - 30
    ⇒ 3x - 5y = 20 ................................(2)
    Put the value of x from equation (1) in above equation (2).
    ⇒ 45y/8 - 5y = 20
    ⇒ 45y - 40y = 20 x 8
    ⇒ 5y = 20 x 8
    ⇒ y = 4 x 8 = 32
    Put the value of Y in equation (1)
    x = 15 x 32/8 = 15 x 4 = 60

    Therefore Present age of Vikas = x = 60 years
    and Present age of Vishal = y = 32 years



  1. The sum of three consecutive multiples of 3 is 72. What is the largest number ?









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    Method 1
    Let us assume the number be 3p , 3(p+1) and 3(p+2)
    According to question,
    3p + 3(p+1) + 3(p+2) = 72
    Solve the equation and find the answer.

    Method 2
    Let us assume the numbers P , P + 3 ,P+ 6
    According to Question,
    sum of three numbers = 72
    P + P + 3 + P + 6 = 72
    Solve the equation and find the answer.

    Correct Option: C

    Method 1
    Let us assume the number be 3p , 3(p+1) and 3(p+2)
    According to question,
    3p + 3(p+1) + 3(p+2) = 72
    ⇒ 3p + 3p + 3 +3p + 6 = 72
    ⇒ 9p +9 = 72
    ⇒ 9p = 72 - 9
    ⇒ 9p = 63
    ⇒ p = 63/9 = 7
    Largest number = 3(p + 2)
    Put the value of p in above equation.
    ⇒ Largest number = 3 x ( 7 + 2 )
    ⇒ Largest number = 3 x 9
    ⇒ Largest number = 27

    Method 2
    Let us assume the numbers P , P + 3 ,P+ 6
    According to Question,
    sum of three numbers = 72
    P + P + 3 + P + 6 = 72
    ⇒ 3P + 9 = 72
    ⇒ 3P = 72 - 9
    ⇒ 3P = 72 - 9
    ⇒ P = 63/3
    ⇒ P = 21
    So largest Number = P + 6 = 21 + 6 = 27


  1. The sum of the ages of a father and his son is 4 times the age of the son. If the average age of the father and the son is 28 years, What is the son's age?









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    Let us assume father age is F and his son age is S.
    According to question,
    The sum of the ages of father and his son is 4 times the age of the son,
    F + S = 4S
    F = 3S.............. (1)
    Average age of father and son is 28.
    (F + S)/2 = 28
    Solve the above equation to find the age of son.

    Correct Option: A

    Let us assume father age is F and his son age is S.
    According to question,
    The sum of the ages of father and his son is 4 times the age of the son,
    F + S = 4S
    F = 3S.............. (1)
    Average age of father and son is 28.
    (F + S)/2 = 28
    F + S = 56
    Put the value of F from equation from (1),
    3S + S = 56
    4S = 56
    S = 14 years
    Age of Son = 14 years.



  1. The product of two numbers is 192 and the sum of these two numbers is 28. What is the smaller of these two numbers ?









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    Let the two numbers be x and y .
    According to question,
    xy = 192...................... (1)
    x + y = 28........................(2)
    As we know that,
    (x - y)2 = (x + y)2 - 4xy
    Solve the above equation and find the answer

    Correct Option: C

    Let the two numbers be x and y .
    According to question,
    xy = 192...................... (1)
    x + y = 28........................(2)
    As we know that,
    (x - y)2 = (x + y)2 - 4xy
    Put the value of xy and x + y from equation (1) and (2), we will get
    (x - y)2 = (28)2 - 4 x 192
    ⇒ (x - y)2 = 784 - 786
    ⇒ (x - y)2 = 16
    ⇒ x - y = 4......................(3)
    Add the equation (2) and (3), we will get
    x + y + x - y = 28 + 4
    2x = 32
    x = 16
    Put the value of x in equation (2), we will get
    16 + y = 26
    y = 28 - 16
    y = 12

    So numbers are x = 16 and y = 12.
    Smaller number is 12.