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Reasonin Rakesh Yadav

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SOLUTION<br />

1. (d) From the two different views<br />

of the dice it is clear that '6'<br />

lies opposite to '5'.<br />

2. (b)<br />

nks vyx&2 fLFkfr;ksa ls ;g vFkZ fudyrk<br />

gS fd 5 ds foijhr 6 gksxkA<br />

First Layer<br />

Second Layer<br />

Third Layer<br />

The number of cubes which<br />

are painted on 2 sides:<br />

12 (n – 2).<br />

n = length of bigger cube ÷<br />

length of smaller cube<br />

So, 12 (3 – 2)<br />

12 × 1 = 12<br />

,sls ?kuksa dh la[;k ftldh nks lrg jaxh<br />

gqbZ gSa%<br />

12 (n – 2).<br />

n = cM+s ?ku dh Hkqtk dh yEckbZ ÷<br />

NksVs ?ku dh Hkqtk dh yEckbZ<br />

vr% 12 (3 – 2)<br />

12 × 1 = 12<br />

3. (c) From the two views of cube it<br />

is clear the '3' lies opposite to<br />

'1'<br />

,d ?ku dh nks fLFkfr;ksa ls ;g lkiQ gksrk<br />

gS fd 3, 1 ds foijhr gSA<br />

1<br />

4. (b) If there is one triangle at<br />

the bottom. there would be<br />

three triangle on the top.<br />

5. (b)<br />

6<br />

;fn ?ku dh fupyh lrg ij ,d f=kHkqt gS<br />

rks mldh Åijh lrg ij 3 f=kHkqt gksus<br />

pkfg,A<br />

2<br />

6. (b) Number of cubes having two<br />

colours<br />

7. (a)<br />

= 4×3 = 12<br />

nks jxksa ds ?kuksa dh la[;k = 4×3 = 12<br />

White<br />

Red<br />

Yellow<br />

The cubes of middle row will have<br />

no red colour 9 cubes<br />

The Central cube will have no<br />

colour<br />

Now, out of 8 cubes, 4 cubes have<br />

either yellow or white colour.<br />

,d ?ku ds chp dh ykbu esa fcuk yky jax ds ?kuksa<br />

dh la[;k 9 ?ku<br />

dsUnz ds ?ku es a dksbZ jax ugha gksxkA<br />

8 ?ku esa ls pkj ?ku ;k rks ihyh ;k liQsn jax ds<br />

gSA<br />

8. (b) The numbers 2, 4, 5 and 6<br />

cannot be on the face<br />

opposite to 3.<br />

The numbers 1, 3, 4 and 6<br />

cannot be on the face<br />

opposite to 5.<br />

Therefore, 2 lies opposite 5.<br />

Clearly, 4 lies opposite 6.<br />

fn;s x;s ?kuksa es a 2, 4, 5 vkSj 6, 3 ds<br />

foijhr ugha gks ldrs gS<br />

rFkk 1, 3, 4 vkSj 6, 5 ds foijhr ugha<br />

gks ldrk gSA<br />

<strong>Rakesh</strong> <strong>Yadav</strong> Sir<br />

blfy,, 5 ds foijhr 2 gksxk rFkk 6 ds<br />

foijhr 4 gksxkA<br />

9. (b) There are four cubes in<br />

Layer-I and four cubes in<br />

Layer IV which have only<br />

one face painted red and all<br />

other faces not painted at<br />

all. Thus there are eight<br />

such cubes.<br />

,d ?ku ds ijr& I es a 4 ?ku gS rFkk<br />

ijr& IV esa Hkh pkj ?ku gS ftlls dsoy<br />

lrg dks yky jax ls jaxk tkrk gS rks lHkh<br />

nwljh lrg dks ugh jaxk tkrk gS blfy,<br />

10. (d)<br />

yky jax ls dsoy 8 ?ku gh jaxs gksxsaA<br />

Layer-1<br />

Layer-2<br />

Layer-3<br />

Layer-4<br />

Layer-1<br />

Layer-2<br />

Layer-3<br />

Layer-4<br />

Green<br />

Red<br />

Black<br />

11. (c) The numbers 1, 2, 4 and 5 lie<br />

on the faces adjacent to 3.<br />

Therefore, the number 6<br />

lies opposite 3.<br />

The numbers 3, 4, 5 and 6 lie<br />

on the faces adjacent to 1.<br />

Therefore, the number 2<br />

lies opposite 1.<br />

Now, the number 4 lies<br />

opposite 5.<br />

lrg la[;k 3, la[;k,sa 1, 2, 4 vkSj 5<br />

ij lyaXu djrh gSA<br />

;|fi la[;k 6, 3 ds foijhr fLFkr gSA<br />

la[;k 1, la[;k 3, 4, 5 vkSj 6 ij<br />

lyaXu djrh gSA<br />

;|fi la[;k 2, 1 ds foijhr fLFkr gSA<br />

vc la[;k 4, 5 ds foijhr gSA<br />

12. (a) From the two views of dice.<br />

it is clear that number '1'<br />

lies opposite to number '4'.<br />

,d ?ku dh pkj fLFkfr;ksa ls ;g irk<br />

pyrk gS fd 4 ds foijhr 1 gksxkA<br />

13. (c) The numbers 1, 2, 3 and 6<br />

are on adjacent faces of the<br />

number 5. Therefore, the<br />

number 3 lies opposite to 5.<br />

bles a 5 dh lyaXu la[;k,sa 1, 2, 3 vkSj<br />

6 gS blfy, 5 ds foijhr 3 gksxkA<br />

226 <strong>Rakesh</strong> <strong>Yadav</strong> Readers Publication Pvt. Ltd

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