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| From: rajagopal_1024 | Reply 61 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 43 |
4:1
Posted at: Mon Jan 25 14:59:46 2010 (GMT)
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| From: dj_glitzzz | Reply 62 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 5 |
p=5
Posted at: Fri Jan 29 10:32:24 2010 (GMT)
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| From: dj_glitzzz | Reply 63 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 7 |
19.35 sq.cm
Posted at: Fri Jan 29 10:56:42 2010 (GMT)
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| From: uma11 | Reply 64 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 7 |
19.35sq.cm
area of rectangle-(ar of 1circle+area of 2circle+ar of semicircle)
=(15*6)-28.26+28.26+14.13)
=90-70.65
=19.35sq.cm
Posted at: Fri Feb 5 14:36:37 2010 (GMT)
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| From: ashsang123 | Reply 65 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 1 |
4250
SI = PXRXT/100
340=PX4X2/100
P=340X100/8
P=4250
Posted at: Wed Feb 10 11:19:16 2010 (GMT)
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| From: kmouli | Reply 66 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 4 |
p=5
Posted at: Thu Feb 11 12:48:49 2010 (GMT)
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| From: kmouli | Reply 67 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 17 |
55
let the angle CAD be x
then 25+x=80
therefore x=80-25=55
Posted at: Thu Feb 11 12:52:17 2010 (GMT)
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| From: kmouli | Reply 68 of 193 | Reply | |
| Subject: ur ans is wrong |
yhe correct ans = 55
better luck next time Posted at: Thu Feb 11 12:53:18 2010 (GMT)
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| From: kmouli | Reply 69 of 193 | Reply | View replies (1)
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| Subject: Contributed Answer/Explanation to Q. 30 |
Rs.660
4000*110*15/100=
4*11*15
=660
Posted at: Thu Feb 11 12:56:32 2010 (GMT)
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| From: kmouli | Reply 70 of 193 | Reply | |
| Subject: hi |
you are mind blowing
the ans is right Posted at: Thu Feb 11 12:57:16 2010 (GMT)
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| From: preetie | Reply 71 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 1 |
RS.4250
S.I = RS.340 , TIME = 2YRS , RATE = 4 %
LET SUM BE S
INT = S * 2 * .04 = .08S
.08S = 340
OR S = 340 / .08 = RS.4250
Posted at: Fri Feb 12 06:45:40 2010 (GMT)
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| From: preetie | Reply 72 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 4 |
5
X-2 = 0
X= 2
F(x) = 2*2*2*2-2*2-P2-2
0= 16-4-P2-2
0= 10-P2
P2= 10
P=10/2=5
Posted at: Fri Feb 12 06:49:10 2010 (GMT)
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| From: preetie | Reply 73 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 7 |
19.35 CM SQ
LENGTH OF RECTANGLE = 15CM
BREADTH OF RECTANGLE = 6CM ]
AREA OF RECTANGLE = 15*6 = 90CMS SQARE
AREA OF 1 CIRCLE OF RADIUS 3 CM = 3.14*3*3 =28.26CM SQUARE
AREA OF 2 CIRCLE = 28.26 *2 =56.52CM SQ
AREA OF SEMICIRCLE = (3.14*3*3)/2 = 14.13CM SQ
THEREFORE TOT ARE OF 2 CIRCLES + 1 SEMICIRCLE = 70.65CM SQ
THEREFORE AREA OF REMAINING UNSHADED PORTION = 90 - 70.65 =19.35 CM SQ.
Posted at: Fri Feb 12 07:00:20 2010 (GMT)
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| From: preetie | Reply 74 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 7 |
19.35 CM SQ
LENGTH OF RECTANGLE = 15CM
BREADTH OF RECTANGLE = 6CM ]
AREA OF RECTANGLE = 15*6 = 90CMS SQARE
AREA OF 1 CIRCLE OF RADIUS 3 CM = 3.14*3*3 =28.26CM SQUARE
AREA OF 2 CIRCLE = 28.26 *2 =56.52CM SQ
AREA OF SEMICIRCLE = (3.14*3*3)/2 = 14.13CM SQ
THEREFORE TOT ARE OF 2 CIRCLES + 1 SEMICIRCLE = 70.65CM SQ
THEREFORE AREA OF REMAINING UNSHADED PORTION = 90 - 70.65 =19.35 CM SQ.
Posted at: Fri Feb 12 07:00:22 2010 (GMT)
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| From: preetie | Reply 75 of 193 | Reply | View replies (1)
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| Subject: Contributed Answer/Explanation to Q. 8 |
RS. 1356.80
PRINTED PRICE OF THE CAMERA = RS.1600
DISCOUNT = .2 *1600 = RS.320
NEW DISCOUNTED PRICE IS = RS (1600-320) = RS.1280
SALES TAX IS 6%
THEREFORE
NEW PRICE AFTER ALLOWING THE DISCOUNT = .06* 1280 + 1280. = 1356.8
Posted at: Fri Feb 12 07:05:56 2010 (GMT)
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| From: preetie | Reply 76 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 16 |
1
SIN 25 / SEC 65 + COS 25 / COSEC 65
= SIN (90-65) / (1/COS 65) + COS (90-65) / (1/ SIN 65)
= COS 65.COS 65 +SIN 65.SIN 65
= COS^2 65 + SIN ^2 65
= 1
Posted at: Fri Feb 12 07:12:31 2010 (GMT)
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| From: preetie | Reply 77 of 193 | Reply | View replies (1)
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| Subject: Contributed Answer/Explanation to Q. 17 |
75
ANG BCD = 180-80= 100
ANG ACD = 25 [ALTERNATE ANGLES ARE EQUAL]
ANG BCA = 100 - 25 =75
THEREFORE
ANG ANG CAD = 75 [ALTERNATE ANGLES ARE EQUAL]
Posted at: Fri Feb 12 07:15:27 2010 (GMT)
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| From: preetie | Reply 78 of 193 | Reply | View replies (1)
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| Subject: Contributed Answer/Explanation to Q. 44 |
7M
VOL OF A CONICAL TENT = 1232M CUBE
AREA OF THE BARE FLOOR = 154 M SQ
A.T.P.
22/7 * R*R = 154
R*R = 154 *7/22 = 49 M SQ
R = 7 M
Posted at: Fri Feb 12 07:29:49 2010 (GMT)
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| From: preetie | Reply 79 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 44 |
7M
VOL OF A CONICAL TENT = 1232M CUBE
AREA OF THE BARE FLOOR = 154 M SQ
A.T.P.
22/7 * R*R = 154
R*R = (154 *7)/22 = 49 M SQ
R = 7 M
Posted at: Fri Feb 12 07:30:18 2010 (GMT)
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| From: preetie | Reply 80 of 193 | Reply | |
| Subject: Contributed Answer/Explanation to Q. 45 |
24m
VOL OF THE CONOCAL TENT = 1232M CUBE
ATP
1/3 PIE R*R*H =1232
OR H = (1232 *3)/ 22*7
OR H = 24M
Posted at: Fri Feb 12 07:33:10 2010 (GMT)
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