Quant Boosters - Sagar Gupta - Set 1
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Number of Questions - 30
Topic - Quant Mixed Bag
Solved ? - Yes
Source - Compilation of posts from Sagar Gupta - 99.2 Percentile in CAT 2015 (Quant)
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Q1) What will be the reminder if (13^3 + 14^3 + 15^3 + 16^3 + 17^3+ .... 34^3) is divided by 35.
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[ 34 * 35 /2 ]^2 - [ 12 * 13/2 ]^2
(17 * 35)^2 - ( 6 * 13)^2 mod 35
(17 * 35)^2 mod 35 =0
-(6 * 13)^2 mod 35 = 169 mod 35 = - 29
-29 = 35-29 = 6
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Q2) Ram is very tall and so, he can climb 1, 2 or 3 steps of a certain staircase at a time. In how many different ways can he climb the staircase if it has 10 steps?
(Note that climbing 2 steps first and 1 step later is a different case from climbing 1 step first and 2 steps later)
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Case1:
3+3+3+1=10. So here, 4! / 3! =4Case2:
Using only two 3s here.
3 + 3 + 2 + 2=10
So,4!/2!.2!= 6
3 + 3 + 2 + 1 + 1=10
So,5!/2!.2!= 30
3 + 3 + 1 + 1 + 1 + 1=10
So,6!/2!.4!= 15Case3:
Using only 1 3 now
3,2,2,2,1
So,5!/3!= 20
3,2,2,1,1,1
So,6!/3!.2!= 60
3,2,1,1,1,1,1
So,7!/5!= 42
3,1,1,1,1,1,1,1
So,8!/7!= 8Similarly,make cases when we have no 3s.
So,that comes out to be total 6 subcases are--
1,1,1,1,1,1,1,1,1,1
2,2,2,2,1,1
2,2,2,1,1,1,1
2,2,1,1,1,1,1,1
2,1,1,1,1,1,1,1,1
2,2,2,2,2And their sum is 1+15+35+28+9+1=89
Total will be sum of all the cases=274
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Q3) If the product of four positive integers is 10! What is the smallest possible value their sum can have?
a. 175
b. 176
c. 180
d. 181
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45^4 = 2025^2 = 41xxxxxx > 10 !
now write 10 ! = 6* 4 * 5 * 6 * 7 * 8 * 9 * 10
make this expression as product of four numbers in 40's
40 * 42 * 45 * 48
so sum = 175
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Q4) A set of 28 books are kept side by side alphabetically. The width of each book according to its position is given by 2n+1 inches including the width of the front and the back page. The width of front end and back end of each cover is same and equals 2 inches. A worm starts eating from front end of the first book to the back end of the last book. How much does it eats away?
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2(2) + 1 +2
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2(27) + 1 + 1
1
=> 2 * (2 + 2 + .... + 27) + 26 * 3 + 1 = 834
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Q5) What will be the remainder when C(58, 29) is divided by 29?
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C(58, 29) can be written as
C(29, 0)^2 + C(29, 1)^2 + .... + C(29, 29)^2
Except first and last term rest all terms are divisible by 29
So, remainder will be 2.
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Q6) If abc not equal to 0 and (2a^2) + (17b^2) + (8c^2) - 6ab - 20bc = 0, then what is the value of (a+b-c)/(a+b+c) ?
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(2a^2-6ab+18b^2/4)+(50b^2/4-20bc+8c^2)=0
2(a-3b/2)^2+2(5b/2-2c)=0
so a=3b/2 and 5b/2 = 2c
a = 3b/2 = 6c/5
Solve and answer is 1/3
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Q7) A diamond expert cuts a huge cubical diamond into 960 identical diamond pieces in minimum no. of 'n' cuts. If he wants to maximize the no. of identical pieces making same no. of cuts to it. Then maximum no. of such pieces are ?
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960 = 8 * 10 * 12.
total cuts (8-1) + (10-1) + (12-1) = 27 = 9 + 9 + 9.
max pieces. 10 x 10 x 10 = 1000.
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Q8) What is the remainder when { 2^11 + 5^11 + 8^11 + 11^11 + .......... + 59^11} is divided by 610 ?
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Direct property: a^n + b^n + ... mod (a + b + c ..) = 0 if a, b, c... are in AP and n is odd
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Q9) If the expression ax^2 + bx + c is equals to 4 when x = 0 leaves a reminder 4 when divided by x+1 and a reminder 6 when divided by x+2, then the value of a,b and c are respectively
a) 1,1,4
b) 2,2,4
c) 3,3,4
d) 4,4,4
e) 2,3,4
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x = 0 => c = 4
x = -1 => a - b + 4 = 4 => a = b
x = -2 => 4a - 2b + 4 = 6 => a = b = 1
a,b,c = 1,1,4
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Q10) Find the number of negative integral solution in (x +1)/x + |x + 1| = (x + 1)^2/x
a) 0
b) 1
c) 2
d) 3
