51. Alok and Bhanu play the following min-max game. Given the expression N = 32 + X* (Y - Z), where X, Y and Z are variables representing single digits (0 to 9), Alok would like to maximize N while Bhanu would like to minimize it. Towards this end, Alok chooses a single digit number and Bhanu substitutes this for a variable of her choice (X, Y or Z). Alok then chooses the next value and Bhanu, the variable to substitute the value. Finally Alok proposes the value for the remaining variable. Assuming both play to their optimal strategies, the value of N at the end of the game would be
Answer:
Explanation:For these type of ques, remember dis thumb rule.. X*Y-Z=18
X+Y-Z=11 X-Y-Z=2
So
N = 32 + X * Y - Z N= 32+18=50
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52. Determine the distance between the x-intercept and the z-intercept of the plane whose equation is 2x+9y- 3z=18
Answer:
Explanation:Putting y=z=0 x=9
Putting x=y=0 z=-6
Distance=(x^2+z^2) ^0.5=10.82 unit
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53. If a person makes a row of toys of 20 each, there would be 15 toys left. If they made to stand in rows of 25 each, there would be 20 toys left, if they made to stand in rows of 38 each, there would be 33 toys left and if they are made to stand in rows of 40 each, there would be 35 toys left. What is the minimum number of toys the person have?
Answer:
Explanation:Required number of toys =LCM (20, 25, 28, 38 and
40)-5
⇒ 3,800−5=3,800−5= 3,795.
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54. The IT giant Tirnop has recently crossed a head count of 150000 and earnings of $7 billion. As one of the forerunners in the technology front, Tirnop continues to lead the way in products and services in India. At Tirnop, all programmers are equal in every respect. They receive identical salaries and also write code at the same rate. Suppose 24 such programmers take 24 minutes to write 24 lines of code in total. How long will it take 72 programmers to write 72 lines of code in total?
Answer:
Explanation:Use the equation (men*time)/work (p1*t1)/l1= (p2*t2)/l2 (14*14)/14= (5*t2)/5 14=t2
Time taken = 14
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55. Consider a class of 40 students whose average weight is 40 kgs. m new students join this class whose average weight is n kgs. If it is known that m + n = 50, what is the maximum possible average weight of the class now?
Answer:
Explanation:If the overall average weight has to increase after the new people are added, the average weight of the new entrants has to be higher than 40.
So, n > 40
Consequently, m has to be < 10 (as n + m = 50)
Working with the "differences" approach, we know
that the total additional weight added by "m" students would be (n - 40) each, above the already existing average of 40. m (n - 40) is the total extra additional weight added, which is shared amongst 40 + m students. So, m
(n - 40) / (m + 40)
has to be maximum for the overall average to be maximum.
At this point, use the trial and error approach (or else, go with the answer options) to arrive at the answer. The maximum average occurs when m = 5, and n = 45 And the average is 40 + (45 - 40) * (5 / 45)
= 40 + (5 / 9)
= 40.56 kgs
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56. A greengrocer was selling apple at a penny each, chickoos at 2 for a penny and peanuts at 3 for a penny. A father spent 7 pennies and got the same amount of each type of fruit for each of his three children. What did each child get?
Answer:
Explanation:1 apple costs 1 penny ===> 3 apples for 3 pennies, 2 chickoos cost one penny ===> 3 kids * 2 chickoos each for 3 pennies
3 peanuts cost one penny ===> 3 peanuts from the balance penny
Spending 7 pennies and giving each child 1 apple, 2 chickoos and 1 peanut each.
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57. 1/3 of a number is 6 more than 1/6 of that number then what is the number?
Answer:
Explanation:Let the number be x
So, x*(1/3) = 6+[x*(1/6)] Solving this,
x=36
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58. John has chocolates of types A and B in the ratio 3 : 7, while Mike has chocolates of types B and C in the ratio 5 : 4, Ram has chocolates of types C and A in the ratio 3 : 5. If there are more chocolates of type C than of type B, and more of type B than of type A, what is the minimum possible number of chocolates overall?
Answer:
Explanation:Let john have 3x chocolates of type A and 7x of type B Let Mike have 5y chocolates of type B and 4y of type C Let john have 3z chocolates of type C and 5z of type A So in total A=3x+5z; B=7x+5y; C=4y+3z
Since C>B we get solving y<3z-7x —>(1) Since B>A we get solving 5y>5z-4x —> (2)
a. π : 36√3 b. π : 18√3
c. π : 27√3 d.π : 42√3
Answer: C
Explanation:
19. John has chocolates of types A and B in the ratio 3 : 7, while Mike has chocolates of types B and C in the ratio 5 : 4, Ram has chocolates of types C and A in the ratio 3 : 5. If there are more chocolates of type C than of type B, and more of type B than of type A, what is the minimum possible number of chocolates overall? a. 78 b. 40 c. 56 d. 72 Answer: A
What gets inferred from above 2 statements is z>=3. So when x=1, z=3, we get only y=1 as choice, for which second condition doesn't satisfy.
So, when x=1,z=4, we get y<5 from first condition and when y > 3.2 from second condition. So which gives choice the only y=4.
Hence x=1,y=4 and z=4 works and is the best possible answer.
For these values, we get A=23,B=27,C=28.
Minimum possible number of chocolates overall is 78.
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TCS Digital
59. Let n! = 1 x 2 x 3 x x n for integer n> 1. If p = 1!
+ (2 x 2!) + (3 x 3!) + …… (10 x 10!), then p+2 when divided by 11! Leaves remainder of
Answer:
Explanation:If P = 1! = 1
Then P + 2 = 3, when divided by 2! Remainder will be 1. If P = 1! + 2 × 2! = 5
Then, P + 2 = 7 when divided by 3! Remainder is still 1. Hence, P = 1! + (2 × 2!) + (3 × 3!)+ ……+ (10 × 10!)
Hence, when p + 2 is divided by 11! The remainder is 1
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60. Vodafone has come up with a new scheme "Pay Easy". They have decided to charge the first 100 calls of a Pay Easy customer @Rs.1/-call, the next 100 calls @Rs.1.25/-call and the next 100 calls @Rs.1.75/-call. Raj is a Pay Easy customer. He paid Rs.286.25/- as his mobile bill that month. How many calls did Raj make?
Answer:
Explanation:When you go through the option all the options are above 200
Total cost paid=Rs.286.25 For 1st 100 calls=Rs.100
2nd 100 calls=Rs.125 Still he has to pay Rs.61.25 For third 100 call=Rs.1.75 No of calls=61.25/1.75=35
Total number of calls =100+100+35=235 calls
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