TCS latest Questions with Answers - 27

1. I bought a certain number of marbles at a rate of 27 marbles for rupees 2 times M,where M is an integer.  I divided these marbles into 2 parts of equal numbers,one part of which I sold at the rate of 13 marbles for Rs.M and the other at the rate of 14 marbles for Rs.M.I spent and received an integral no of rupees,but bought the least number of marbles.How many did I buy?
a. 870
b. 102660
c. 1770
d. 9828
Explanation:
Let he bought 2x marbles.
27 marbles costs = Rs.2M so 1 marble costs = Rs.2M/27
Therefore, x marbles costs = Rs. (2M × x) / 27
Now we calculate the selling prices.
He sold x marbles at the rate of 13 for Rs.M so 1 marble selling price = M/13
x marbles selling price = x × M/13
He sold another x marbles at the rate of 14 for Rs.M so 1 marble selling price = M/14
x marbles selling price = x × M/14
Now $\frac{2Mx}{27},\frac{xM}{13},\frac{xM}{14}$ are integers.
So x marbles must be divisible by 27, 13, 14.  LCM of 4914
So 2x = 9828

2. How many different integers can be expressed as the sum of three distinct numbers from the set {3, 10, 17, 24, 31, 38, 45, 52}?
option
a) 8
b) 56
c) 16
d) 15
Explanation:
Interesting question.  If you think that ${}^{8}{C}_{3}$ = 56 is correct then it is wrong answer.  We are not asked how many ways we can select 3 numbers out of 8. But how many different numbers can be expressed as a sum of three numbers from the given set.  For example, 3 + 10 + 31 = 3 + 17 + 24 = 47.  So 47 can be expressed as a sum of 3 numbers in two different ways but 47 should be considered as only one number.
Now the minimum number that can be expressed as a sum of 3 numbers = 30. The next number is 37.  Similarly the largest number is 38 + 45 + 52 = 135.
So there exists many numbers in between, with common difference of 7.
Total numbers = $\frac{l-a}{d}+1=\frac{135-30}{7}+1$ = 16.

3. How many different integers can be expressed as the sum of three distinct numbers from the set {3, 8, 13, 18, 23, 28, 33, 38, 43, 48}?
option
a) 8
b) 56
c) 120
d) 22
Explanation:
From the above discussion, minimum number = 24 and maximum number = 129.  So there exist many numbers in between these two numbers, with common difference of 5.  All these numbers can be expressed as a sum of 3 different integers from the given set.
Total numbers =  $\frac{l-a}{d}+1=\frac{129-24}{5}+1$ = 22.

4. A owes B Rs.50. He agrees to pay B over a number of consecutive days starting on a Monday, paying single note of Rs.10 or Rs.20 on each day. In how many different ways can A repay B.
Explanation:
He can pay by all 10 rupee notes in 5 days = 1 way
3 Ten rupee + 1 twenty rupee = $\frac{4!}{3!×1!}$ = 4 ways
1 Ten rupee + 2 twenty rupee notes = $\frac{3!}{2!×1!}$ = 3 ways
Total ways = 1 + 4 + 3 = 8

5. HCF of 2472,1284 and a third number 'n'is 12.If their LCM is 8*9*5*103*107.then the number 'n'is..
a. 2^2*3^2*5^1
b. 2^2*3^2*7^1
c. 2^2*3^2*8103
d. None of the above.
Explanation:
2472 = ${2}^{3}×3×103$
1284 = ${2}^{2}×3×107$
HCF = ${2}^{2}×3$
LCM = ${2}^{3}×{3}^{2}×5×103×107$
HCF of the numbers is the highest number which divides all the numbers.  So N should be a multiple of ${2}^{2}×3$
LCM is the largest number that is divided by the given numbers.  As LCM contains ${3}^{2}×5$ these two are from N.
So N = ${2}^{2}×{3}^{2}×{5}^{1}$

6. What is the value of 77!*(77!-2*54!)^3/(77!+54!)^3+54!*(2*77!-54!)^3/(77!+54!)^3
a. 77! - 54!
b. 77! + 54!
c. 77!^2 - 54!^2
d. 77!
Explanation:
The above question can be written as $\frac{77!{\left(77!-2\ast 54!\right)}^{3}}{{\left(77!+54!\right)}^{3}}$$\frac{54!{\left(2\ast 77!-54!\right)}^{3}}{{\left(77!+54!\right)}^{3}}$
Let A = 77!, B = 54!
Then equation in the form
$\frac{a{\left(a-2b\right)}^{3}}{{\left(a+b\right)}^{3}}+\frac{b{\left(2a-b\right)}^{3}}{{\left(a+b\right)}^{3}}$
$a{\left(a-2b\right)}^{3}$ = $a\left({a}^{3}-6{a}^{2}b+12a{b}^{2}-8{b}^{3}\right)$ = ${a}^{4}-6{a}^{3}b+12{a}^{2}{b}^{2}-8a{b}^{3}$
$b{\left(2a-b\right)}^{3}$ = $b\left(8{a}^{3}-12{a}^{2}b+6a{b}^{2}-{b}^{3}\right)$ = $8{a}^{3}b-12{a}^{2}{b}^{2}+6a{b}^{3}-{b}^{4}$
Grouping similar terms, ${a}^{4}-6{a}^{3}b+12{a}^{2}{b}^{2}-8a{b}^{3}$ + $8{a}^{3}b-12{a}^{2}{b}^{2}+6a{b}^{3}-{b}^{4}$
${a}^{4}-{b}^{4}+\left(-6{a}^{3}b+8{a}^{3}b\right)$ + $\left(12{a}^{2}{b}^{2}-12{a}^{2}{b}^{2}\right)$ + $\left(-8a{b}^{3}+6a{b}^{3}\right)$
${a}^{4}-{b}^{4}+2{a}^{3}b-2a{b}^{3}$
$\left({a}^{2}-{b}^{2}\right)\left({a}^{2}+{b}^{2}\right)$+$2ab\left({a}^{2}-{b}^{2}\right)$
$\left({a}^{2}-{b}^{2}\right)\left[\left({a}^{2}+{b}^{2}\right)+2ab\right]$
$\left({a}^{2}-{b}^{2}\right)\left(a+b{\right)}^{2}$
$\left(a-b\right)\left(a+b\right)\left(a+b{\right)}^{2}$
$\left(a-b\right)\left(a+b{\right)}^{3}$
Therefore, $\frac{a{\left(a-2b\right)}^{3}}{{\left(a+b\right)}^{3}}+\frac{b{\left(2a-b\right)}^{3}}{{\left(a+b\right)}^{3}}$ = $\frac{\left(a-b\right){\left(a+b\right)}^{3}}{{\left(a+b\right)}^{3}}=a-b$

Shortcut:
If you try to solve this questions using above method, its almost impossible.  The best way is take a = 4, and b = 2. and substitute in the given equation. 0+ $\frac{2{\left(8-2\right)}^{3}}{{6}^{3}}=2$.  Now substitute a, b values in the given options and check where it is equal to 2.  Option a satisfies. If you like this shortcut, +1 this!!

7. The marked price of coat was 40% less than the suggested retail price. Eesha purchased the coat for half of the marked price at the 15th anniversary sale. What percent less than the suggested retail price did Eesha pay?
a) 60%
b) 20%
c) 70%
d) 30%
Explanation:
Let the retail price = 100
So the market price will be = (100 - 40)% (100) = 60
Easha purchased price = 60/2 = 30
So she bought it for 70% less than retail price.

8. In a city there are few engineering, MBA and CA candidates. Sum of four times the engineering, three times the MBA and 5 times CA candidates is 3650. Also three times CA is equal to two times MBA and three times engineering is equal to two times CA. In total how many MBA candidates are there in the city?
a. 200
b. 300
c. 450
d. 400
Explanation:
Let e = Number of engineering students, m = Number of MBA students and c = Number of CA students.
Given that,
4e + 3m + 5c = 3650- - - - (1)
3c = 2m , therefore $c=\frac{2m}{3}$
3e = 2c ⇒ $e=\frac{2c}{3}=\frac{2}{3}×\frac{2m}{3}=\frac{4m}{9}$
Substituting values of c and e in the given equation,
$4×\frac{4m}{9}+3m+5×\frac{2m}{3}=3650$
$⇒\frac{16m+27m+30m}{9}=3650$
$⇒\frac{76m}{9}=3650$
$⇒m=450$

9. A rectangle is divided into four rectangles with area 70, 36, 20, and x.  The value of x is
a. 350/9
b. 350/7
c. 350/11
d. 350/13
Explanation:
Areas are in proportion.
$\frac{70}{x}=\frac{36}{20}⇒x=\frac{350}{9}$

10. If a ladder is 10 m long and distance between bottom of ladder and wall is 6 m.  What is the maximum size of cube that can be placed between the ladder and wall.
a. 34.28
b. 24.28
c. 21.42
d. 28.56
Here a = 6, and c = 10. b = 8 ($\because$ using Pythagorean theorem)
The maximum side of the square which can be inscribed in a right angle triangle = $\frac{abc}{{a}^{2}+{b}^{2}+ab}$ ($\because$ see 7th questionfor formula)
So side = $\frac{10×6×8}{{6}^{2}+{8}^{2}+6×8}=3.243$
Volume of the cube = ${3.243}^{3}=34.075$