Wipro Turbo Quiz on Flowcharts with Solutions

FALSE

Explanation.

even if Box 3 is updated instead of Box 2 in instruction C,

In instruction D, Number in Box 2 is compared with Box 7, and the value is anyway greater, so the result is YES there and thus exiting and finishing the loop.

Answer: Option D
Explanation:

None of the available answers is the correct:

step 1 : ins 1 : 2[Box 11] 3 = 5[Box 11]

step 2 : ins 2 : 8[Box 1] -> 8[Box 10]

step 3 : ins 3 : is 1 an odd number? YES

step 4 : ins 4 : go to instruction 6[Box 2[Box 6]]

step 5 : ins 6 : 4[Box 5] 5[Box 11] = 9[Box 11]

step 6 : ins 7 : change ins2: [Box 1 1] -> [Box 10 - 1]

step 7 : ins 8 : 9 < 8[Box 1] ? NO

step 8 : ins 9 : go to instruction 2

step 9 : ins 2 : 6[Box 2] -> 6[Box 9]

step 10 : ins 3 : is 2 an odd number? NO

step 11 : ins 6 : 4[Box 5] 9[Box 11] = 13[Box 11]

step 12 : ins 7 : change ins2: [Box 2 1]  ->  [Box 9 - 1]

step 13 : ins 8 : 8 < 8[Box 1] ? NO

step 14 : ins 9 : go to instruction 2

step 15 : ins 2 : 5[Box 3] -> 5[Box 8]

step 16 : ins 3 : is 3 an odd number? YES

step 17 : ins 4 : go to instruction 6[Box 2[Box 6]]

step 18 : ins 6 : 4[Box 5] 13[Box 11] = 17[Box 11]

step 19 : ins 7 : change ins2: [Box 3 1] -> [Box 8 - 1]

step 20 : ins 8 : 7 < 8[Box 1] ? YES

ANSWER:  5[Box 8] and 17[Box 11]

Instruction 1 : 0 [Box 3] + 1 [Box 11] = 1 [Box 11]
Instruction 2 : 1 [Box 1] + 1 [Box 11] = 2 [Box 5]
Instruction 3 : 3 [Box 4] * 3 [Box 4] = 9 [Box 4]
Instruction 4 : IS 2 [Box 5] = 6 [Box 10] ? NO
Instruction 5 : 9 [Box 4] - 6 = 3 [Box 4]
Instruction 6 : 2 [Box 5] + 2 [Box 2] = 4 [Box 5]
Instruction 3 : 3 [Box 4] * 3 [Box 4] = 9 [Box 4]
Instruction 4 : IS 4 [Box 5] = 6 [Box 10] ? NO
Instruction 5 : 9 [Box 4] - 6 = 3 [Box 4]
Instruction 6 : 4 [Box 5] + 2 [Box 2] = 6 [Box 5]
Instruction 3 : 3 [Box 4] * 3 [Box 4] = 9 {box 4]
Instruction 4 : IS 6 [Box 5] = 6 [Box 10] ? YES
Instruction 7 : 24 [Box 7] - 6 [Box 5] = 18 [Box 8]
Instruction 8 : 9 [Box 4] + 18 [Box 8] = 27 [Box 6]
ANSWER Box 6 = 27

Instruction 1 : 5 [Box 4] + 2 [Box 10] = 3 [Box 1]
Instruction 2 : 3 [Box 1] + 3 [Box 6] = 6 [Box 1]
Instruction 3 : IS 6 [Box 1] evenly divisible by 4? = NO
Instruction 4 : 6 [Box No.] + 3[Box 6] = 9 [Box No.]
Instruction 5 : IS 13 [Box 2] > 9[Box No.] ? = YES
Instruction 6 : 7[Box 9] - 1 = 6 [Box 9]
Instruction 2 : 6 [Box 1] + 6 [Box 9] = 12 [Box 1]
Instruction 3 : IS 12 [Box 1] evenly divisible by 4? = NO
Instruction 4 : 9[Box No.] + 3[Box 6] = 12[Box No.]
Instruction 5 : IS 13 [Box 2] > 12 [Box No.] ? = YES
Instruction 6 : 6 [Box 9] -1 = 5 [Box 9]
Instruction 2 : 12 [Box 1] + 2 {Box 12] =14[Box 1]
Instruction 3 : IS 14 [Box 1] evenly divisible by 4? = NO
Instruction 4 : 12 [Box No.] + 3 [Box 6] = 15 [Box No.]
Instruction 5 : IS 13 [Box 2] > 15 [Box No.] ? = NO
ANSWER Box 1 = 14

Assume Box 2 has number 2, Box 3 has number 3 and so on.
And Box 1 has number 10
Instruction 1 : 2 [Box 2] + 2[Box 2] = 4[Box 2]
Instruction 2 : 2 [Box No.] + 2 = 4 [Box No.]
Instruction 3 : IS 10 [Box 1] < 4 [Box No.] ? NO
Instruction 1 : 4 [Box 4] + 4{Box 4] = 8[Box 4]
Instruction 2 : 4 [Box No.] + 2 = 6 [Box No.]
Instruction 3 : IS 10 [Box 1] < 6 [Box No.] ? NO
Instruction 1 : 6 [Box 6] + 6 [Box 2] = 12[Box 6]
Instruction 2 : 6 [Box No.] + 2 = 8 [Box No.]
Instruction 3 : IS 10 [Box 1] < 8 [Box No.] ? NO
Instruction 1 : 8 [Box 8] + 8 [Box 8] = 16[Box 8]
Instruction 2 : 8 [Box No.] + 2 = 10 [Box No.]
Instruction 3 : IS 10 [Box 1] < 10 [Box No.] ? NO
Instruction 1 : 10 [Box 10] + 10 [Box 10] = 20[Box 10]
Instruction 2 : 10 [Box No.] + 2 = 12 [Box No.]
Instruction 3 : IS 10 [Box 1] < 12 [Box No.] ? YES
ANSWER is 10

Given Box 4 = A
Also, (1) means number in Box 1;
For A,
IS (3) > (1) ? NO --(1)
IS (2) > (1) ? NO --(2)
IS (3) > (2) ? NO --(3)
ANSWER
The largest number is in BOX 1,
Because number in BOX 1 is not smaller than number in Box 2 and 3(refer (1) and (2)),
And, The smallest number is in BOX 3,
Because number in BOX 3 is not greater than number in Box 1 and Box 2(refer (1) and (3)).

approved_amt = sal * 5 * (60 - age) = 15,000 * 5 * (60 - 35) = 18,75,000 Rs.

approved_amt = sal * 3 * (60 - age)
approved_amt = 60,000 * 3 * (60 - 28)

approved amt = 57,60,000 Rs. So, yes, he would get a loan equal to Rs. 57,60,000.

ShortCut By :: Laboni

Correct answer: Option A

approved_amt=60000*3*(60-28)=Rs. 5,760,000

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Currently there is no PrepInsta Explanation. Did you know?

You can also submit your own version of explanations under user explanation section and also view other submitted explanations by other users.