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 The IITian’s Prashikshan Kendra A Resonance of Brilliant Minds…! Answerkey - T es t – 1 (Jan. 2014) In-Program Test for PRODI! ("res#ers) Program $AT%&$ATI' Sol.1 (c) The answer is 5! = 5 X 4 X 3 X 2 X 1 = 120 Sol.2 (b) In option (a) 9 – 6 = 3  4 In option (b) 8 – 4 = 4 48 is also iisible b" 3 an 4# $en%e option (b) is %orre%t Sol.3 (b) c a c b b c b a a c a b  x  x  x  x  x  x  + + + + + + + +  1 1 1 1 1 1 c a c b b c b a a c a b  x  x  x  x  x  x  x  x  x  x  x  x + + + + + + + + = 1 1 1 1 1 1 a b c c c a b b c b a a  x  x  x  x  x  x  x  x  x  x  x  x + + + + + + + + = 1 = + + + + = c b a c b a  x  x  x  x  x  x Sol.4 (c) 0 1 2 2 1  2 = + = +  x  x  x  x  1 0 ) 1 (  2 = =  x  x  2 1 1 1 8 8 = + = +  x  x Sol.5 (c) &e'er the roi " *+ie 'or the sol+tion # Sol.6 (a) &easonin, -s the .an wal/s at 4 3 o' his +s+al rate the ti.e that he ta/es is 3 4 o' his +s+al ti.e#  3 4 o' his +s+al ti.e = +s+al ti.e 20 .in+tes i#e# 3 1 o' his +s+al ti.e =20 .in+tes his +s+al ti.e = 60 .in+tes = 1 ho+r# Sol.7 (d) &e'er the roi" *+ie 'or th e sol+tion Sol.8 (b) 5 2 2 2 2 32 2  = =   y  x  y  x  5 2  =  y  x  (1) -ain 4 2 2 16 2  = =  + +  y  x  y  x  4 = + y  x  (2) Page 1 of 5

Freshers Maths n Chem Soln

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STWT-5

The IITians Prashikshan Kendra

A Resonance of Brilliant Minds!

Answerkey - Test 1 (Jan. 2014)In-Program Test for PRODIGY (Freshers) Program

MATHEMATICSSol.1(c) The answer is 5! = 5 X 4 X 3 X 2 X 1 = 120

Sol.2(b) In option (a) 9 6 = 3 4

In option (b) 8 4 = 4

48 is also divisible by 3 and 4.

Hence option (b) is correctSol.3(b)

Sol.4(c)

Sol.5(c) Refer the Prodigy Guide for the solution.Sol.6(a) Reasoning: As the man walks at of his usual rate, the time that he takes is of his

usual time.

of his usual time = usual time + 20 minutes

i.e. of his usual time =20 minutes his usual time = 60 minutes = 1 hour.

Sol.7(d) Refer the Prodigy Guide for the solution

Sol.8(b)

(1)

Again

(2)

Adding (1) and (2), we get

Substituting this value for x in equation (2), we get

Now,

Sol.9(c) Let y be the additive inverse of. Then

Sol.10(a) Let,

SHAPE \* MERGEFORMAT

The two interior angles of the triangle are together equal to the exterior angle on the opposite side of the interior angles

EMBED Equation.3

But BO and CO are the bisectors of these angles

Hence, and

or each half of the angles is equal to or (as shown in the fig)

and

In

In

180

180 360

260 130260

13065Sol.11(c) In

For

In

so,

Sol.12 (b)

Divide by numerator and denominator both

But is given

Sol.13(b) 5 girls can sit in ways behind them 6 boys can stand in ways

Therefore, the total number of ways = = 86400.Sol.14(a)=

=

As remainder

EMBED Equation.3 is a factor of polynomial .Sol.15(b) Placing the, 1 as indicated in the problem, shifts the given digits to the left, so that t is

now the hundreds digit and u is now the tens digit.

CHEMISTRYSOLUTIONS :31(c)

32(c)

33(a)

34(b)

35(c)

36(d)

37(d)

38(b)

39(a)

40(a)

41(c)

42(b)

43(d)

44(c)

45(d)

X

2

B

x

q

A

90- EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

O

C

8

B

6

A

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

50

r

y

Y

C

O

p

Page 1 of 5The IITians Prashikshan Kendra A Resonance of Brilliant Minds!

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