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Choose the correct option.

The letters in word ADOPTS are permuted in all possible ways and arranged in the alphabetical order. Find the word at position 42 in the permuted alphabetical order.


AAOTDSP

BAOSTPD

CAOTDPS

DAOTPDS

Answer: Option B

Explanation:

No. of words starting with AD=4!=24, AO=4!
Thus the word has to start with AO because the last word with AO will be the 48th word.
Now no. of words formed with AOD=3!=6, AOP=3!=6, AOS=3!=6 Thus the last word formed with AOS is=4!+3!+3!+3!=24+6+6+6=42
Thus the last word formed with AOS is AOSTPD

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SHSTTON
9
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8 year old Eesha visited her grandpa. He gave her this riddle. I started working at 13. I spent 1/6 of my working life in a factory. I spent 1/4 of my working life in an office, and I spent 1/4 of my working life as a school caretaker. For the last 32 years of my working life I've been doing social service. How old am I?


A109

B102

C105

D113

Answer: Option A

Explanation:

Let x be the number of years he worked.
=> x/6 + x/4 + x/4 + 32 = x
=> x = 96
So, his age = 96 + 13 = 109

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SHSTTON
6
Solv. Corr.
11
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There is a set of 36 distinct points on a plane with the following characteristics:
* There is a subset A consisting of fourteen collinear points.
* Any subset of three or more collinear points from the 36 are a subset of A.
How many distinct triangles with positive area can be formed with each of its vertices being one of the 36 points? (Two triangles are said to be distinct if at least one of the vertices is different)


A7140

B4774

C1540

D6776

Answer: Option D

Explanation:

The given data indicates that 14 points are collinear and remaining 22 points are non collinear.
A triangle can be formed by taking 1 points from 14 and 2 points from 22
OR
2 points from 14 and 1 points from 22
OR
3 points from 22

=> 14C1*22C2+14C2*22C1+22C3 = 6776

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SHSTTON
6
Solv. Corr.
6
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What is the number of ways of expressing 3600 as a product of three ordered positive integers (abc, bca etc. are counted as distinct).
For example, the number 12 can be expressed as a product of three ordered positive integers in 18 different ways.


A441

B540

C84

D2100

Answer: Option B

Explanation:

3600 = 2^4 * 3^2 * 5^2
Let abc = 2^4 * 3^2 * 5^2
We have to distribute four 2's to three numbers a, b, c in (4+3-1)C(3-1)=6C2 = 15 ways.
Now two 3's has to be distributed to three numbers in (2+3-1)C(3-1)=4C2 = 6 ways
Now two 5's has to be distributed to three numbers in (2+3-1)C(3-1)=4C2 = 6 ways
Total ways = 15 * 6 * 6 = 540

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SHSTTON
7
Solv. Corr.
5
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How many 6 digit even numbers can be formed from digits 1, 2, 3, 4, 5, 6, and 7 so that the digit should not repeat and the second last digit is even?


A6480

B320

C2160

D720

Answer: Option D

Explanation:

If the we have to form even numbers, units digit must be 2, 4, 6. i.e., 3 ways.


Also 5th digit should be even.


So it can be filled in 2 ways. Now remaining 5 digits can be filled in 5! ways. So total 5! * 3 * 2 = 720 ways.

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NA
SHSTTON
31
Solv. Corr.
49
Solv. In. Corr.
80
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In the town of Unevenville, it is a tradition to have the size of the front wheels of every cart different from that of the rear wheels. They also have special units to measure cart wheels which is called uneve. The circumference of the front wheel of a cart is 133 uneves and that of the back wheel is 190 uneves. What is the distance traveled by the cart in uneves, when the front wheel has done nine more revolutions than the rear wheel?


A570

B1330

C3990

D389

Answer: Option C

Explanation:

LCM of 133 and 190 is 1330.


So, to cover this distance, front wheel takes 10 rounds, and back wheel takes 7 rounds.


So, for 3 rounds extra, 1330 uneves distance has to be travelled.


To take 9 rounds extra, 1330 * 3 = 3990 uneves has to be traveled.

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SHSTTON
21
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22
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How many 2's are there between the terms 112 to 375?


A313

B159

C156

D315

Answer: Option C

Explanation:

Let us calculate total 2's in the units place. (122, 132, 142 ... 192), (201, 212, 222, ... 292), (302, 312, ... 372) = 8 + 10 + 8 = 26
Total 2's in tenth's place, (120, 121, 122, ..., 129) + (220, 221, ..., 229) + (320, 321, ..., 329) = 30
Total 2's in hundred's place = (200, 201, ... 299) = 100.
Total 2's between 112 and 375 = 26 + 30 + 100 = 156

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SHSTTON
2
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Consider the sequence of numbers 0, 2, 2, 4,... Where for n > 2 the nth term of the sequence is the unit digit of the sum of the previous two terms. Let sn denote the sum of the first n terms of this sequence. What is the smallest value of n for which sn >2771?


A692

B693

C694

D700

Answer: Option B

Explanation:

[0, 2, 2, 4, 6, 0, 6, 6, 2, 8, 0, 8, 8, 6, 4, 0, 4, 4, 8, 2], 0, 2, 2...this series repeats after every 20 terms.


Sum of these 20 terms = 80


So 2771 =34*80 + 51


Sum of 13 terms = 52


So we have to use 34 times 20 terms = 34*20 = 680


680+13 = 693

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SHSTTON
6
Solv. Corr.
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Three generous friends, each with some money, redistribute the money as follows: Sandra gives enough money to David and Mary to double the amount of money each has. David then gives enough to Sandra and Mary to double their amounts. Finally, Mary gives enough to Sandra and David to double their amounts. If Mary had 11 rupees at the beginning and 17 rupees at the end, what is the total amount that all three friends have?


A105

B60

C88

D71

Answer: Option D

Explanation:

Let Sandra, David and Mary each has s, d and 11 respectively.
After the first distribution,
David has d + d = 2d, Mary has 11 + 11 = 22 and Sandra has s - d - 11.
After the second distribution,
Sandra has 2*(s - d - 11) , mary has 2*22 = 44 and david has 2d - (s - d - 11) - 22=3d - s -11.
After the third distribution,
Sandra has 2*2(s - d - 11), david has 2*(3d - s - 11) and mary has 44 - 2(s - d - 11) - (3d - s - 11) = 77 - s - d
It is given that finally Mary has Rs.17. So, 77 - s - d=17
=> s + d = 60
=> s + d + 11 = 60 + 11 = 71.

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SHSTTON
14
Solv. Corr.
23
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Babla alone can do a piece of work in 10 days. Ashu alone can do it in 15 days. The total wages for the work is Rs.5000. How much should be Babla be paid if they work together for an entire duration of work.


A2000

B4000

C5000

D3000

Answer: Option D

Explanation:

Bablu work = 1/10
Ashu work = 1/15
Total work = 1/10+1/15 = 3+2/30 = 5/30
Total work = 6 days
Bablu paid = 6/10*5000 = 3000

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