CoCubes Flowcharts-Visual Reasoning-DI Question and Answers Quiz-1

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.

None of the available answers is the correct:

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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 ins 2: [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

Option : A

Given Box 4 = EITHER C or F
Also, (1) means number in Box 1.
For C,
IS (3) > (1) ? NO --(1)
IS (2) > (1) ? YES --(2)
For F,
IS (3) > (1) ? YES --(3)
IS (2) > (1) ? YES --(4)
IS (3) > (2) ? YES --(5)
ANSWER
Number in BOX 1 cannot be largest(refer (1), (2), (3) and (4)),
And, Number in BOX 3 cannot be smallest(refer (1), (3) and (5)).

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 b

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

For value of N equal to 0, the flowchart gets struck in an infinite loop. Hence answer to this question is option D (none of these)

Option : B