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171 / 483
A rectangular block has the dimensions 5*6*7 cm it is dropped into a cylindrical vessel of radius 6cm and height 10 cm.if the level of the fluid in the cylinder rises by 4 cm, what portion of the block is immersed in the fluid.
A22/7*15/18
B22/7*24/35
C22/7*22/34
D22/7*12/15
Answer: Option B
Explanation:Since level of water increased in cylinder by height 4.
This is because of the rectangular block .
Therefore , area of rectangular block immersed in water is = 22/7 * (6)^2 * 4
Thats why portion of block immersed in water is= (22/7 * 36* 4) / total vol of rectangle
= (22/7 *36*4)/ (7*5*6)
= 22/7 *24/35.
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172 / 483
Two boys and a girl can do a work in 5 days, while a boy and 2 girls can do it in 6 days. if the boy is paid at the rate of $28 a week. What should be the wages of the girls?
A$12
B$16
C$23
D$17
ENone of these
Answer: Option B
Explanation:2b+1g=1/5=>10b+5g=1---------(1)
1b+2g=1/6=>6b+12g=1---------(2)
from (1)&(2)
10b+5g=6b+12g
4b=7g
b:g=7:4
but above ratio is reverse b:g=4:7
wages of the girls=(4*28)/7
=4*4=16
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173 / 483
What would be the value of (log a)2 -(log b)2
Alog(a+b)log(ab)
B3
Clog(ab)log(a/b)
DNone of these
Answer: Option C
Explanation:Question is in the form of a2 - b2 so the formulae is
(a+b)(a-b)=a2 - b2 where a=loga,b=logb
substitute in the formulae we get
(loga+logb)(loga-logb)
(log(ab))(log(a/b))
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174 / 483
A salesman makes a profit of 25% on all sales.How many fax machines will he have to sell for $375 each to make a total commission of at least $500?
A4
B5
C6
D15
E20
Answer: Option C
Explanation:sp=375
cp=100/125*375=300,
profit=375-300=75
no. of fax mac. req=500/75=6.666
So, approximate ans is 6
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175 / 483
In triangle PQR, PQ=6 cm,PR=8 cm and QR=12 cm. Calculate the area of the triangle PQR.
A23.33 cm^2
B17.5 cm^2
C21.33 cm^2
D28.67 cm^2
Answer: Option C
Explanation:s=(a+b+c+)/2 [Formula]
Here a=6cm, b=8cm , c=12cm
Therefore s=(6+8+12)/2
=26/2
=13
Area of tringle=sqrt [ s (s-a) (s-b) (s-c) [Formula]
=sqrt [ 13* (13-6) *(13-8)* (13-12)
=sqrt [ 13*7*5*1]
=sqrt [455]
=21.33 cm^2
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176 / 483
Consider the following two curves in the X-Y plane.
y = x^3 + x^2 + 5
y = x^2 + x + 5
Which of the following statements is true for -2 < x < 2?
AThe two curves do not intersect.
BThe two curves intersect thrice.
CThe two curves intersect twice.
DThe two curves intersect once.
ENone of the above
Answer: Option B
Explanation:Here x^3 + x^2 + 5 = x^2 + x + 5
=> x^3 - x = 0
Factor: x(x^2 - 1) = 0
Factor: x(x+1)(x-1) = 0
x= -1, 0, 1
Three points of intersection.
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177 / 483
A man engaged a servant on a condition that he will pay Rs 90 and also give him a bag at the end of the year . He worked for 9 months and was paid Rs 55 and a turban. So the price of the turban is?
A10
B29
C11
DNone of these
Answer: Option A
Explanation:12*x=90+y
9*x=45+y
By, Solving the above equation
y= 10
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178 / 483
Find the remainder when ( X^3 + 4 X^2 + 6X - 2 ) is divided (X+5)
A-57
B57
C34
DNone of these
Answer: Option A
Explanation:X+5) X^3 + 4 X^2 + 6 X -2 ( - X^2 + X -11
-X^3 - 5 X^2
..................................
- X^2 + 6 X
+ X^2 + 5 X
........................................
+ 11 X -2
- 11 X - 55
...........................................
- 57
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179 / 483
What will be the minimum value of X in \(X^{3}-7X^{2}+11X-5\). So we get a positive number.
A4
B5
C8
DNone of these
Answer: Option C
Explanation:If we substitute 4 in the given equation then it becomes
(4)^3-(7*(4)^2)+(11*4)-5=-9
then substitute 5 in the equation
(5)^3-(7*(5)^2)+(11*5)-5=0
then substitute 8 in the equation
(8)^3-(7*(8)^2)+(11*8)-5=147 which is a positive integer.
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180 / 483
In how many ways can 5 members be selected out of 10 members,so that any two particular members must always be excluded?
A52
B56
C60
D64
Answer: Option B
Explanation:Two particular members must be excluded so member=10-2=8
so the answer will be = 8! / 5! (8-5)!=8*7*6*5!/5!*3!=8*7*6/3*2*1=8*7=56
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