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1

JEE Main 2021 (Online) 27th August Evening Shift

Numerical
3 $$\times$$ 722 + 2 $$\times$$ 1022 $$-$$ 44 when divided by 18 leaves the remainder __________.
Your Input ________

Answer

Correct Answer is 15

Explanation

3(1 + 6)22 + 2 . (1 + 9)22 $$-$$ 44 = (3 + 2 $$-$$ 44) = 18 . I

= $$-$$ 39 + 18 . I

= (54 $$-$$ 39) + 18(I $$-$$ 3)

= 15 + 18I1

$$\Rightarrow$$ Remainder = 15
2

JEE Main 2021 (Online) 26th August Evening Shift

Numerical
Let $$\left( {\matrix{ n \cr k \cr } } \right)$$ denotes $${}^n{C_k}$$ and $$\left[ {\matrix{ n \cr k \cr } } \right] = \left\{ {\matrix{ {\left( {\matrix{ n \cr k \cr } } \right),} & {if\,0 \le k \le n} \cr {0,} & {otherwise} \cr } } \right.$$

If $${A_k} = \sum\limits_{i = 0}^9 {\left( {\matrix{ 9 \cr i \cr } } \right)\left[ {\matrix{ {12} \cr {12 - k + i} \cr } } \right] + } \sum\limits_{i = 0}^8 {\left( {\matrix{ 8 \cr i \cr } } \right)\left[ {\matrix{ {13} \cr {13 - k + i} \cr } } \right]} $$ and A4 $$-$$ A3 = 190 p, then p is equal to :
Your Input ________

Answer

Correct Answer is 49

Explanation

$${A_k} = \sum\limits_{i = 0}^9 {{}^9{C_i}} {}^{12}{C_{k - i}} + \sum\limits_{i = 0}^8 {{}^8{C_i}} {}^{13}{C_{k - i}}$$

$${A_k} = {}^{21}{C_k} + {}^{21}{C_k} = 2.{}^{21}{C_k}$$

$${A_4} - {A_3} = 2\left( {{}^{21}{C_4} - {}^{21}{C_3}} \right) = 2(5985 - 1330)$$

$$190p = 2(5985 - 1330) \Rightarrow p = 49$$
3

JEE Main 2021 (Online) 25th July Evening Shift

Numerical
If the co-efficient of x7 and x8 in the expansion of $${\left( {2 + {x \over 3}} \right)^n}$$ are equal, then the value of n is equal to _____________.
Your Input ________

Answer

Correct Answer is 55

Explanation

$${}^n{C_7}{2^{n - 7}}{1 \over {{3^7}}} = {}^n{C_8}{2^{n - 8}}{1 \over {{3^8}}}$$

$$\Rightarrow$$ n $$-$$ 7 = 48 $$\Rightarrow$$ n = 55
4

JEE Main 2021 (Online) 25th July Evening Shift

Numerical
Let n$$\in$$N and [x] denote the greatest integer less than or equal to x. If the sum of (n + 1) terms $${}^n{C_0},3.{}^n{C_1},5.{}^n{C_2},7.{}^n{C_3},.....$$ is equal to 2100 . 101, then $$2\left[ {{{n - 1} \over 2}} \right]$$ is equal to _______________.
Your Input ________

Answer

Correct Answer is 98

Explanation

1. $${}^n{C_0} + 3.{}^n{C_1} + 5.{}^n{C_2} + ... + (2n + 1).{}^n{C_n}$$

$${T_r} = (2r + 1){}^n{C_r}$$

$$S = \sum {{T_r}} $$

$$S = \sum {(2r + 1){}^n{C_r}} = \sum {2r{}^n{C_r} + \sum {{}^n{C_r}} } $$

$$S = 2(n{.2^{n - 1}}) + {2^n} = {2^n}(n + 1)$$

$${2^n}(n + 1) = {2^{100}}.101 \Rightarrow n = 100$$

$$2\left[ {{{n - 1} \over 2}} \right] = 2\left[ {{{99} \over 2}} \right] = 98$$

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