# TCS latest Questions with Answers – 27

a. 870

b. 102660

c. 1770

d. 9828

Answer: d

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 2Mx27,xM13,xM142Mx27,xM13,xM14 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

Answer: c

Explanation:

Interesting question. If you think that 8C38C3 = 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 = l−ad+1=135−307+1l−ad+1=135−307+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

Answer: d

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 = l−ad+1=129−245+1l−ad+1=129−245+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 = 4!3!×1!4!3!×1! = 4 ways

1 Ten rupee + 2 twenty rupee notes = 3!2!×1!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.

Answer:

Explanation:

2472 = 23×3×10323×3×103

1284 = 22×3×10722×3×107

HCF = 22×322×3

LCM = 23×32×5×103×10723×32×5×103×107

HCF of the numbers is the highest number which divides all the numbers. So N should be a multiple of 22×322×3

LCM is the largest number that is divided by the given numbers. As LCM contains 32×532×5 these two are from N.

So N = 22×32×5122×32×51

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!

Answer: a

Explanation:

The above question can be written as 77!(77!−2∗54!)3(77!+54!)377!(77!−2∗54!)3(77!+54!)3+ 54!(2∗77!−54!)3(77!+54!)354!(2∗77!−54!)3(77!+54!)3

Let A = 77!, B = 54!

Then equation in the form

a(a−2b)3(a+b)3+b(2a−b)3(a+b)3a(a−2b)3(a+b)3+b(2a−b)3(a+b)3

a(a−2b)3a(a−2b)3 = a(a3−6a2b+12ab2−8b3)a(a3−6a2b+12ab2−8b3) = a4−6a3b+12a2b2−8ab3a4−6a3b+12a2b2−8ab3

b(2a−b)3b(2a−b)3 = b(8a3−12a2b+6ab2−b3)b(8a3−12a2b+6ab2−b3) = 8a3b−12a2b2+6ab3−b48a3b−12a2b2+6ab3−b4

Grouping similar terms, a4−6a3b+12a2b2−8ab3a4−6a3b+12a2b2−8ab3 + 8a3b−12a2b2+6ab3−b48a3b−12a2b2+6ab3−b4

= a4−b4+(−6a3b+8a3b)a4−b4+(−6a3b+8a3b) + (12a2b2−12a2b2)(12a2b2−12a2b2) + (−8ab3+6ab3)(−8ab3+6ab3)

= a4−b4+2a3b−2ab3a4−b4+2a3b−2ab3

= (a2−b2)(a2+b2)(a2−b2)(a2+b2)+2ab(a2−b2)2ab(a2−b2)

= (a2−b2)[(a2+b2)+2ab](a2−b2)[(a2+b2)+2ab]

= (a2−b2)(a+b)2(a2−b2)(a+b)2

= (a−b)(a+b)(a+b)2(a−b)(a+b)(a+b)2

= (a−b)(a+b)3(a−b)(a+b)3

Therefore, a(a−2b)3(a+b)3+b(2a−b)3(a+b)3a(a−2b)3(a+b)3+b(2a−b)3(a+b)3 = (a−b)(a+b)3(a+b)3=a−b(a−b)(a+b)3(a+b)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+ 2(8−2)363=22(8−2)363=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%

Answer:

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

Answer: c

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=2m3c=2m3

3e = 2c ⇒ e=2c3=23×2m3=4m9e=2c3=23×2m3=4m9

Substituting values of c and e in the given equation,

4×4m9+3m+5×2m3=36504×4m9+3m+5×2m3=3650

⇒16m+27m+30m9=3650⇒16m+27m+30m9=3650

⇒76m9=3650⇒76m9=3650

⇒m=450⇒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

Answer: a

Explanation:

Areas are in proportion.

70x=3620⇒x=350970x=3620⇒x=3509

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

Answer:

Explanation:

Here a = 6, and c = 10. b = 8 (∵∵ using Pythagorean theorem)

The maximum side of the square which can be inscribed in a right angle triangle = abca2+b2+ababca2+b2+ab (∵∵ see 7th questionhere for formula)

So side = 10×6×862+82+6×8=3.24310×6×862+82+6×8=3.243

Volume of the cube = 3.2433=34.0753.2433=34.075**Note:**

The maximum side of a square is obtained when two sides of the square matches with a and b. In this case side = ab/(a+b) = 3.428 which is higher than 3.243.