**Maths Questions for TISSNET PDF**

Download important TISSNET Maths Questions PDF based on previously asked questions in TISSNET and other MBA exams. Practice Maths Questions and answers for TISSNET and other exams.

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**Question 1: **How many pairs of positive integers m, n satisfy 1/m + 4/n = 1/12 , where n is an odd integer less than 60?

a) 6

b) 4

c) 7

d) 5

e) 3

**Question 2: **Let T be the set of integers {3,11,19,27,…451,459,467} and S be a subset of T such that the sum of no two elements of S is 470. The maximum possible number of elements in S is

a) 32

b) 28

c) 29

d) 30

**Question 3: **Suppose n is an integer such that the sum of digits on n is 2, and $10^{10} < n < 10^{11}$. The number of different values of n is

a) 11

b) 10

c) 9

d) 8

**Question 4: **If a/b = 1/3, b/c = 2, c/d = 1/2 , d/e = 3 and e/f = 1/4, then what is the value of abc/def ?

a) 3/8

b) 27/8

c) 3/4

d) 27/4

e) 1/4

**Question 5: **What are the values of x and y that satisfy both the equations?

$2^{0.7x} * 3^{-1.25y} = 8\sqrt{6}/27$

$4^{0.3x} * 9^{0.2y} = 8*81^{1/5}$

a) x = 2, y = 5

b) x = 2.5, y = 6

c) x =3, y = 5

d) x = 3,y = 4

e) x = 5,y = 2

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**Question 6: **If R = $(30^{65}-29^{65})/(30^{64}+29^{64})$ ,then

a) $0<R\leq0.1$

b) $0.1<R\leq0.5$

c) $0.5<R\leq1.0$

d) $R>1.0$

**Question 7: **If x = $(16^3 + 17^3+ 18^3+ 19^3 )$, then x divided by 70 leaves a remainder of

a) 0

b) 1

c) 69

d) 35

**Question 8: **Let $n!=1*2*3* …*n$ for integer $n \geq 1$.

If $p = 1!+(2*2!)+(3*3!)+… +(10*10!)$, then $p+2$ when divided by 11! leaves a remainder of

a) 10

b) 0

c) 7

d) 1

**Question 9: **Let $x = \sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4- \ to \ infinity}}}}$. Then x equals

a) 3

b) $(\sqrt{13} – 1)/2$

c) $(\sqrt{13} + 1)/2$

d) $\sqrt{13}$

**Question 10: **Let N = 1421 * 1423 * 1425. What is the remainder when N is divided by 12?

a) 0

b) 9

c) 3

d) 6

**Answers & Solutions:**

**1) Answer (E)**

1/m + 4/n = 1/12

So, 1/m = 1/12 – 4/n

So, m = 12n/(n-48)

Since m is positive, n should be greater than 48

Also, since n is an odd number, it can take only 49, 51, 53, 55, 57 and 59

If n = 49, 51, 57 then m is an integer, else it is not an integer

So, there are 3 pairs of values for which the equation is satisfied

**2) Answer (D)**

No. of terms in series T , 3+(n-1)*8 = 467 i.e. n=59.

Now S will have atleast have of 59 terms i.e 29 .

Also the sum of 29th term and 30th term is less than 470.

Hence, maximum possible elements in S is 30.

**3) Answer (A)**

The sum of digits should be 2. The possibilities are 1000000001,1000000010,10000000100,..these 10 cases . Also additional 1 case where 20000000000. Hence option A .

**4) Answer (A)**

a/d = a/b * b/c * c/d = 1/3 * 2 * 1/2 = 1/3

Similarly, b/e and c/f are 3 and 3/8 respectively.

b/e = b/c*c/d*d/e = 3

c/f = c/d*d/e*e/f = 3/8

=> Value of abc/def = 1/3 * 3 * 3/8 = 3/8

**5) Answer (E)**

$2^{0.7x} * 3^{-1.25y} = 8\sqrt{6}/27$ => $2^{0.7x} * 3^{-1.25y}$ = $2^{3.5} * 3^{-2.5}$

=> 0.7x = 3.5 => x = 5

=> -1.25y = -2.5 => y = 2

$4^{0.3x} * 9^{0.2y} = 8*81^{1/5}$ => $2^{0.6x} * 3^{0.4y}$ = $2^3 * 3^{0.8}$

=> 0.6x = 3 => x = 5

=> 0.4y = 0.8 => y = 2

=> (5,2) is the solution.

**6) Answer (D)**

$\frac{(30^{65}-29^{65})}{(30^{64}+29^{64})} = ((30-29)*\frac{(30^{64}+30^{63}*29+….+29^{64})}{(30^{64}+29^{64})}$ , which is greater than 1 . Hence option D.

**7) Answer (A)**

We know that x = $16^3 + 17^3 + 18^3 + 19^3 = (16^3 + 19^3) + (17^3 + 18^3)$

= $(16 + 19)(16^2 – 16 * 19 + 19^2) + (17 + 18)(17^2 – 17 * 18 + 18^2)$ = 35 × odd + 35 × odd = 35 × even = 35 × (2k)

=> x = 70k

=> Remainder when divided by 70 is 0.

**8) Answer (D)**

According to given condiiton we have p = (1 × 1!) + (2 × 2!) + (3 × 3!) + (4 × 4!) + … + (10 × 10!) . So n × n! = [(n + 1) – 1] × n! = (n + 1)! – n!. So equation becomes p = 2! – 1! + 3! – 2! + 4! – 3! + 5! – 4! +… + 11! – 10!. So p = 11! – 1! = 11! – 1. p + 2 = 11! + 1 .So when it is divided by 11! gives a remainder of 1. Hence, option 4.

**9) Answer (C)**

$x = \sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4- \ to \ infinity}}}}$

=> $x = \sqrt{4+\sqrt{4-x}}$

=> $x^2 = 4 + \sqrt{4-x}$

=>$x^4 + 16 – 8x^2 = 4 – x$

=> $x^4 – 8x^2 + x +12 = 0$

On substituting options, we can see that option C satisfies the equation.

**10) Answer (C)**

The numbers 1421, 1423 and 1425 when divided by 12 give remainder 5, 7 and 9 respectively.

5*7*9 mod 12 = 11 * 9 mod 12 = 99 mod 12 = 3

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