# good quant questions

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ans

Q1. The number of two-digit numbers [Where neither digit is zero] whose product of the digits is a square are?
a. 16                   b. 17                 c. 18                  d. None of these

squre 1/4/9/16/25/36/49/64/81

so possible number 11,22,33,28,44,82,55,49,66,94,77,88,99

Q2. What is a percent of b divided by b percent of a?
(a) a        (b)b           (c)    1    (d)10           (e)    100

=(ab/100)/(ba/100)=1

Q3. There are four boys and five girls standing outside the new record shop. The sign in the window offers a prize to every couple (one boy and one girl) that enters the store. How many prizes can the nine people get?

1 boy goes with 5 girls in alternative so total 20 prizes nine people can get

Q4. While taking a class Sirji asked
“What is the highest number possible with 3 digits?”
My answer then was 9 to the 99th power (written mathematically of course) as this uses "3 digits" and is obviously a huge number. Is this correct?
a. yes
b. no
c. cant be determined
d. There is not a single highest number

ans is NO becasue 9^9^9 is highest number

Little Star

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Q1. The number of two-digit

Q1. The number of two-digit numbers [Where neither digit is zero] whose product of the digits is a square are?
a. 16                   b. 17                 c. 18                  d. None of these

Both the digits of the two digit number can be same, possible number of such 2 digit numbers = 9

And the 2 digit perfect square numbers are 81, 64, 49, 36, 25, 16, 9, 4, 1
Now
36 can be 9 x 4 (except 6 x 6)
16 = 8 x 2
9  = 9 x 1
4 = 4 x 1
i. e  8 different numbers (94, 49, 82, 28 etc)

So total numbers = 9 + 8 = 17

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Q5 Only I Kamal Lohia

Q5

Only I

Kamal Lohia

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Incognito
Sequential Numbers

Q5. Among 39 sequential natural numbers there always is a number with the sum of its digits divisible by
I. 11 II. 13 III. 17
a. only I
b. Both I & II
c. I, II & III
d. None of the above

39 consecutive numbers will always have addition of 3 to ten's and 0-9 in units place. Hence total of at least one among them will always be divisible by 11. Hence a.