CAT рдкреНрд░рд╢реНрди рдФрд░ рдЙрддреНрддрд░ рдХрд╛ рдЕрднреНрдпрд╛рд╕ рдХрд░реЗрдВ

рдкреНрд░: A Product is supported each week by the same three Product supporters. Last month the first supporter took 440 calls, the second took 360 calls, and the third took 300 calls. This month the job will consists of 1500 calls. If the three supporters each increase their work proportionately, how many more calls will the second supporter take this month than last month ? 1509 0

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    131 calls
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    160 calls
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    491 calls
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    600 calls
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рдЙрддреНрддрд░ : 1. "131 calls"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :

Answer: A) 131 calls Explanation: 1st supporter recieve 440 calls 2nd supporter recieve 360 calls 3rd supporter recieve 300 calls So total calls = 1100 calls ; Calls this month= 1500 So remaining calls to be distributed is 400 So Now Ratio 1st:2nd:3rd ==> 440:360:300=> 22:18:15 Now No. of More Calls 2nd supporter will get => [18/(22+18+15)] x 400=> (18/55) x 400 => 131 CallsSo 131 more Calls than last month.

рдкреНрд░: Statements : In a recent survey report, it has been stated that those who undertake physical exercise for at least half an hour a day are less prone to have any heart ailments. Conclusions : a) Moderate level of physical exercise is necessary for leading a healthy life.b) All people who do desk-bound jobs definitely suffer from heart ailments. 2541 0

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    Only a follows
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    Both a & b follows
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    Only b follows
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    Neither a nor b follows
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рдЙрддреНрддрд░ : 1. "Only a follows"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :

Answer: A) Only a follows Explanation: The statement mentions that chances of heart ailments are greatly reduced by a regular half-hour exercise. So, a follows. However, it talks of only reducing the probability which does not mean that persons involved in sedentary jobs shall definitely suffer from heart ailments. So, b does not follow. Hence, concusion a only follows the given statement.

рдкреНрд░: If a number 72k23l is divisible by 88. Then find value of k and l ? 2313 0

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    k=8 & l=2
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    k=7 & l=2
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    k=8 & l=3
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    k=7 & l=1
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рдЙрддреНрддрд░ : 2. "k=7 & l=2"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :

Answer: B) k=7 & l=2 Explanation: If a number to be divisile by 88, it should be divisible by both "8" and "11" Check for '8' :For a number to be divisible by "8", the last 3-digit should be divisible by "8"Here 72x23y --> last 3-digit is '23y'So y=2 [ (i.e) 232 is absolutely divisible by "8"] Chech for '11' :For a number to be divisible by "11" , sum of odd digits - sum of even digits should be divisible by "11"(7 + x + 3) - (2 + 2 + y)(7 + x + 3) - (2 + 2 + 2)(10 + x) - 6 should be divisible by "11"for x = 7=> 17 - 6 = 11 [ which is absolutely divisible by "11"] So x = 7 , y= 2.

рдкреНрд░: A class consists of both boys and girls along with a teacher. After a class, the teacher drinks 9 litres of water, a boy drinks 7 litres of water and a girl drinks 4 litres of water. If after a class 42 litres of water was consumed, find the number of girls in the class ? 1139 0

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    8
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    6
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    5
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    3
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рдЙрддреНрддрд░ : 4. "3"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :

Answer: D) 3 Explanation: Given teacher drinks 9 ltrLet number of boys be 'A'.Let number of girls be 'B'.Boy drinks 7 ltr and girl drinks 4 ltrAfter class total water consumed = 42 ltrThen,9 + 7A + 4B = 42=> 7A + 4B = 33By trial and error method,The only integers which satisfy the equation is A = 3 and B = 3Therefore, number of girls in the class = 3.

рдкреНрд░: One girl can eat 112 chocolates in half a minute, and her boy friend can eat half as many in twice the length of time. How many chocolates can both boy and girl eat in 12 seconds ? 1726 0

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    44
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    32
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    56
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    49
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рдЙрддреНрддрд░ : 3. "56"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :

Answer: C) 56 Explanation: Girl eats 112 chocolates in 30 sec so she can eat in 12 sec is 12 x 112/30 = 44.8 chocolates.   Her boy friend can eat one-half of 112 in twice of 30 sec so he can eat 56 in 60 sec Then he can eat in 12 sec is 56 x 12/60 = 11.2 chocolates.   Hence, together they can eat = 44.8 + 11.2 =56 chocolates in 12 seconds.

рдкреНрд░: Out of sixty students, there are 14 who are taking Economics and 29 who are taking Calculus. What is the probability that a randomly chosen student from this group is taking only the Calculus class ? 1663 0

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    8/15
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    7/15
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    1/15
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    4/15
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рдЙрддреНрддрд░ : 2. "7/15"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :

Answer: B) 7/15 Explanation: Given total students in the class = 60Students who are taking Economics = 24 andStudents who are taking Calculus = 32Students who are taking both subjects = 60-(24 + 32) = 60 - 56 = 4Students who are taking calculus only = 32 - 4 = 28probability that a randomly chosen student from this group is taking only the Calculus class = 28/60 = 7/15.

рдкреНрд░: A batsman scores 26 runs and increases his average from 14 to 15. Find the runs to be made if he wants top increasing the average to 19 in the same match ? 2842 0

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    74
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    79
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    72
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    60
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рдЙрддреНрддрд░ : 1. "74"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :

Answer: A) 74 Explanation: Number of runs scored more to increse the ratio by 1 is 26 - 14 = 12To raise the average by one (from 14 to 15), he scored 12 more than the existing average.Therefore, to raise the average by five (from 14 to 19), he should score 12 x 5 = 60 more than the existing average. Thus he should score 14 + 60 = 74.

рдкреНрд░: When I was married 10 years ago my wife is the 6th member of the family. Today my father died and a baby born to me.The average age of my family during my marriage is same as today. What is the age of Father when he died ? 4462 0

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    50 yrs
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    60 yrs
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    70 yrs
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    65 yrs
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рдЙрддреНрддрд░ : 2. "60 yrs"
рд╡реНрдпрд╛рдЦреНрдпрд╛ :

Answer: B) 60 yrs Explanation: Let the Father be x years when he died Average Age 10 years ago be A Total Age 10 years ago = 6*A Total Age after 10 years(Just before father's Death) = 6A + 6*10 = 6A + 60 Father Died and Baby was born => the Total number of people in the family is Same (6) Baby born today so age of baby = 0 (6A +60 - x)/6 = 6A/6 => A + 10 -(x/6) = A => x/6 = 10 => x = 60 Therefore we can conclude that the father was 60 years old when he died.

  рддреНрд░реБрдЯрд┐ рдХреА рд░рд┐рдкреЛрд░реНрдЯ рдХрд░реЗрдВ

рдХреГрдкрдпрд╛ рд╕рдВрджреЗрд╢ рджрд░реНрдЬ рдХрд░реЗрдВ
рддреНрд░реБрдЯрд┐ рд░рд┐рдкреЛрд░реНрдЯ рд╕рдлрд▓рддрд╛рдкреВрд░реНрд╡рдХ рдЬрдорд╛ рд╣реБрдИ

  рддреНрд░реБрдЯрд┐ рдХреА рд░рд┐рдкреЛрд░реНрдЯ рдХрд░реЗрдВ

рдХреГрдкрдпрд╛ рд╕рдВрджреЗрд╢ рджрд░реНрдЬ рдХрд░реЗрдВ
рддреНрд░реБрдЯрд┐ рд░рд┐рдкреЛрд░реНрдЯ рд╕рдлрд▓рддрд╛рдкреВрд░реНрд╡рдХ рдЬрдорд╛ рд╣реБрдИ

  рддреНрд░реБрдЯрд┐ рдХреА рд░рд┐рдкреЛрд░реНрдЯ рдХрд░реЗрдВ

рдХреГрдкрдпрд╛ рд╕рдВрджреЗрд╢ рджрд░реНрдЬ рдХрд░реЗрдВ
рддреНрд░реБрдЯрд┐ рд░рд┐рдкреЛрд░реНрдЯ рд╕рдлрд▓рддрд╛рдкреВрд░реНрд╡рдХ рдЬрдорд╛ рд╣реБрдИ

  рддреНрд░реБрдЯрд┐ рдХреА рд░рд┐рдкреЛрд░реНрдЯ рдХрд░реЗрдВ

рдХреГрдкрдпрд╛ рд╕рдВрджреЗрд╢ рджрд░реНрдЬ рдХрд░реЗрдВ
рддреНрд░реБрдЯрд┐ рд░рд┐рдкреЛрд░реНрдЯ рд╕рдлрд▓рддрд╛рдкреВрд░реНрд╡рдХ рдЬрдорд╛ рд╣реБрдИ

  рддреНрд░реБрдЯрд┐ рдХреА рд░рд┐рдкреЛрд░реНрдЯ рдХрд░реЗрдВ

рдХреГрдкрдпрд╛ рд╕рдВрджреЗрд╢ рджрд░реНрдЬ рдХрд░реЗрдВ
рддреНрд░реБрдЯрд┐ рд░рд┐рдкреЛрд░реНрдЯ рд╕рдлрд▓рддрд╛рдкреВрд░реНрд╡рдХ рдЬрдорд╛ рд╣реБрдИ

  рддреНрд░реБрдЯрд┐ рдХреА рд░рд┐рдкреЛрд░реНрдЯ рдХрд░реЗрдВ

рдХреГрдкрдпрд╛ рд╕рдВрджреЗрд╢ рджрд░реНрдЬ рдХрд░реЗрдВ
рддреНрд░реБрдЯрд┐ рд░рд┐рдкреЛрд░реНрдЯ рд╕рдлрд▓рддрд╛рдкреВрд░реНрд╡рдХ рдЬрдорд╛ рд╣реБрдИ

  рддреНрд░реБрдЯрд┐ рдХреА рд░рд┐рдкреЛрд░реНрдЯ рдХрд░реЗрдВ

рдХреГрдкрдпрд╛ рд╕рдВрджреЗрд╢ рджрд░реНрдЬ рдХрд░реЗрдВ
рддреНрд░реБрдЯрд┐ рд░рд┐рдкреЛрд░реНрдЯ рд╕рдлрд▓рддрд╛рдкреВрд░реНрд╡рдХ рдЬрдорд╛ рд╣реБрдИ

  рддреНрд░реБрдЯрд┐ рдХреА рд░рд┐рдкреЛрд░реНрдЯ рдХрд░реЗрдВ

рдХреГрдкрдпрд╛ рд╕рдВрджреЗрд╢ рджрд░реНрдЬ рдХрд░реЗрдВ
рддреНрд░реБрдЯрд┐ рд░рд┐рдкреЛрд░реНрдЯ рд╕рдлрд▓рддрд╛рдкреВрд░реНрд╡рдХ рдЬрдорд╛ рд╣реБрдИ