Simple Equations Based Problems From Previous CAT Papers


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    let us solve some previous CAT questions which are based on basic concepts like forming an equation and solving it. We should be able to tackle them without using any high-funda concepts. I am not going to focus on short cuts and solving via options kind of stuffs here as the goal of this chapter is to get comfortable with forming equations and solving them.

    Three friends, returning from a movie, stopped to eat at a restaurant. After dinner, they paid their bill and noticed a bowl of mints at the front counter. Sita took 1/3 of the mints, but returned four because she had a momentary pang of guilt. Fatima then took 1/4 of what was left but returned three for similar reasons. Eswari then took half of the remainder but threw two back into the bowl. The bowl had only 17 mints left when the raid was over. How many mints were originally in the bowl?
    a. 38
    b. 31
    c. 41
    d. None of these (CAT 2001)

    Let's the initial count be X
    Sita took 1/3 and returned four = > Current count is X - X/3 + 4 = 2X/3 + 4
    Fatima took 1/4 and returned three = > Current count is 3/4 * (2X/3 + 4) + 3 = X/2 + 3 + 3 = X/2 + 6
    Eshwari took half of remaining and returned two = > 1/2 (X/2 + 6) + 2 = X/4 + 3 + 2 = X/4 + 5
    It is given that X/4 + 5 = 17
    X/4 = 12
    X = 48
    Answer : None of these.

    For all those people who feel we wasted lot of time here as Sita took 1/3rd and none of the option is a multiple of 3, Keep Calm and solve your equations.

    There were a hundred schools in a town. Of these, the number of schools having a play – ground was 30, and these schools had neither a library nor a laboratory. The number of schools having a laboratory alone was twice the number of those having a library only. The number of schools having a laboratory as well as a library was one fourth the number of those having a laboratory alone. The number of schools having either a laboratory or a library or both was 35. What was the ratio of schools having laboratory to those having library?
    (a) 1 : 2
    (b) 5 : 3
    (c) 2 : 1
    (d) 2 : 3 (CAT 1991)

    Total number of schools = 100
    Schools with Playgrounds = 30
    Number of schools with library alone = X
    Number of schools with laboratory alone = 2X
    Number of schools with library and laboratory = 1/4 * 2X = X/2
    Number of schools having either a library or laboratory or both = 35
    -- > X + 2X + X/2 = 35
    -- > X = 10
    Schools having Laboratory = 2 x 10 + 10/2 = 25
    Schools having Library = 10 + 5 = 15
    Required ration = 25/15 = 5:3

    My son adores chocolates. He likes biscuits. But he hates apples. I told him that he can buy as many chocolates he wishes. But then he must have biscuits twice the number of chocolates and should have apples more than biscuits and chocolates together. Each chocolate cost Re 1. The cost of apple is twice the chocolate and four biscuits are worth one apple. Then which of the following can be the amount that I spent on that evening on my son if number of chocolates, biscuits and apples brought were all integers?
    a. Rs. 34
    b. Rs. 33
    c. Rs. 8
    d. None of these (CAT 1998 )

    Let he bought x chocolates = > he bought 2x biscuits and 3x + y apples.
    Chocolate price = 1 re
    Apple price = 2 re
    Biscuit price = 1/4 of Apple price = 0.5 re
    -- > the amount is x + 0.5 * 2x + 6x + 2y = 8x + 2y
    Option c not possible as minimum amount is 10 (x = 1, y = 1)
    Option B not possible as the amount (8x + 2y) is always even
    Option A is possible as 8x + 2y = 34 is possible for x = 3, y = 5

    Using only 2, 5, 10, 25 and 50 paise coins, what will be the minimum number of coins required to pay exactly 78 paise, 69 paise, and Re. 1.01 to three different persons?
    a. 19
    b. 20
    c. 17
    d. 1 (CAT 2003)

    As we need the minimum number of coins go for the highest denomination first
    78 -- > 50 + 2 x 10 + 4 x 2 (7 coins)
    69 -- > 50 + 10 + 5 + 2 x 4 (5 coins)
    1.01 -- > 50 + 25 + 2 x 10 + 3 x 2 (7 Coins)
    Total = 7 + 5 + 7 = 19 coins.

    Once I had been to the post office to buy five-rupee, two-rupee and one-rupee stamps. I paid the clerk Rs. 20, and since he had no change, he gave me three more one-rupee stamps. If the number of stamps of each type that I had ordered initially was more than one, what was the total number of stamps that I bought?
    a. 10
    b. 9
    c. 12
    d. 8 (CAT 1996)

    Say, I bought x 5 re stamps, y 2 re stamps and z 1 re stamps
    As minimum 2 from each stamps were bought initially = > 5 x 2 + 2 x 2 + 2 x 1 = > minimum 16 rupees is spent.
    As the clerk gave three more 1 re stamp as balance we can easily see the initial amount is 17 which is only possible when x = 2, y = 2 and z = 3
    Total number of stamps (including the additional three 1 re stamps) = x + y + z + 3 = 2 + 2 + 3 + 3 = 10

    A transport company charges for its vehicles in the following manner If the driving is 5 hours or less, the company charges Rs. 60 per hour or Rs. 12 per km (whichever is larger) If driving is more than 5 hours, the company charges Rs. 50 per hour or Rs. 7.5 per km (whichever is larger). If Anand drove it for 30 km and paid a total of Rs. 300, then for how many hours does he drive?
    (1) 4
    (2) 5.5
    (3) 7
    (4) 6 (CAT 2002)

    Anand drove for 30 km. so If he paid per km he should have paid 30 x 12 = 360 rs (if drove for 5 hours of less) or 7.5 x 30 = 225 rs (if drove for more than 5 hours).
    So the payment is based on time and he drove for more than 5 hours as otherwise he should pay 360 (higher than 300). He paid 300 and 300 = 6 x 50 (6 hours)

    A shop stores x kg of rice. The first customer buys half this amount plus half a kg of rice. The second customer buys half the remaining amount plus half a kg of rice. Then the third customer also buys half the remaining amount plus half a kg of rice. Thereafter, no rice is left in the shop. Which of the following best describes the value of x?
    (1) 2 ≤ x ≤ 6
    (2) 5 ≤ x ≤ 8
    (3) 9 ≤ x ≤ 12
    (4) 11≤ x ≤ 14
    (5) 13 ≤ x ≤ 18 (CAT 2008 )

    First customer buys x/2 + 1/2 -- > Remaining x – (x/2 + 1/2) = x/2 – 1/2 = (x-1)/2
    Second customer buys 1/2(x-1)/2 + 1/2 -- > remaining (x-1)/2 – [(x-1)/4 + 1/2] = (x-1)/4 – 1/2 = (x – 3)/4
    Third customer buys 1/2(x-3)/4 + 1/2 -- > remaining (x-3)/4 – (x-3)/8 – 1/2 = (x-3)/8 – 1/2 = (x-7)/8
    Given (x-7)/8 = 0 = > x = 7

    A test has 50 questions. A student scores 1 mark for a correct answer, −1/3 for a wrong answer, and −1/6 for not attempting a question. If the net score of a student is 32, the number of questions answered wrongly by that student cannot be less than
    (1) 6
    (2) 12
    (3) 3
    (4) 9 (CAT 2003 Leaked)

    Let x be the correct answers, y be the wrong answers and z be the un attempted questions
    x + y + z = 50 --- (1)
    x –y/3 – z/6 = 32 = > 6x – 2y – z = 192 --- (2)
    (1) + (2) = 7x - y = 242 = > 7x = 242 + y
    Multiple of 7 > 242 is 245
    So minimum value of 242 + y is 245 = > y = 3

    Ten years ago, the ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born during the same year. After another three years, one more member died, again at 60, and a child was born during the same year. The current average age of this eight member joint family is nearest to:
    (1) 23 years
    (2) 22 years
    (3) 21 years
    (4) 25 years
    (5) 24 years (CAT 2007)

    Sum of all the ages 10 years ago = 231
    Three years later Sum of all the ages = 231 – 57 + 0 + 7 x 3 = 195 ( the person who died was 57 years old 3 years back and all other 7 members would have added 3 more years to the sum. New born child is of 0 age)
    Similarly, Another three years later, Sum of all the ages = 195 – 57 + 0 + 21 = 159
    Current age is again after 4 years (as we started 10 years ago) and every 8 members will add 4 more years to the sum. So total 4 x 8 = 32 years.
    So sum of all ages after 4 more years = 159 + 32 = 191
    Average = 191/8 (~ 24 years)

    A telecom service provider engages male and female operators for answering 1000 calls per day. A male operator can handle 40 calls per day whereas a female operator can handle 50 calls per day. The male and the female operators get a fixed wage of Rs. 250 and Rs. 300 per day respectively. In addition, a male operator gets Rs. 15 per call he answers and a female operator gets Rs. 10 per call she answers. To minimize the total cost, how many male operators should the service provider employ assuming he has to employ more than 7 of the 12 female operators available for the job?
    (1) 15
    (2) 14
    (3) 12
    (4) 10 (CAT 2005)

    Female operator – 50 calls per day, 300 fixed wage and 10 x 50 = 500 rs as bonus
    Total cost (female) – 800 rs/day
    Male operator – 40 calls per day, 250 fixed wage and 15 x 40 = 600 rs as bonus
    Total cost (male) = 250 + 600 = 850s
    So we need to maximize the female operators ( max is 12 ) and 12 female operators can handle 600 calls and remaining 400 calls can be handled by 10 male operators.


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