Q.1) If 5^x-3^y=13438 and 5^(x-1)+3^(y+1)=9686 then find x+y.
a)13
b)14
[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]a)13[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]
Solution:
Here we can go by hit n trial
13438 is 5 digit number so 5^x must be a 5 digit number
5^6=15625
So putting x=6 in the first equation we get 3^y=2187
So y=7
Putting this in eq 2
5^5 +3^8=3125+6561=9686
So x+y=6+7=13
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Q.2) 
a)0
b)10
c)1
d)-1
[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]c)1[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]
Solution:
a1=1
a1-a2=2
a2=-1
a1-a2+a3=3
a3=1
a1-a2+a3-a4=4
a4=-1
1,-1,1,-1….so on
so sum of every two consecutive terms will be 0
so only a1023 will remain which is odd numbered term having value 1
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Q.3) Amal invests Rs 12000 at 8% interest, compounded annually, and Rs 10000 at 6% interest, compounded semi-annually, both investments being for one year. Bimal invests his money at 7.5% simple interest for one year. If Amal and Bimal get the same amount of interest, then the amount, in Rupees, invested by Bimal is
a)20920
b)10920
[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]a)20920[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]
Solution:
Interests for both are same so equate their interests
Semi-annually for Amal so rate becomes half (i.e.3% )and time will become double i.e. 2 years
Interest for Amal= (12000*8*1)/100 +10000[1.03]^2 -10000
Interest for Bimal= (P*7.5*1)/100
960+609=(7.5P)/100
156900=7.5P
P=20920
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Q.4) Anil alone can do a job in 20 days while Sunil alone can do it in 40 days. Anil starts the job, and after 3 days, Sunil joins him. Again, after a few more days, Bimal joins them and they together finish the job. If Bimal has done 10% of the job, then in how many days was the job done?
a)14
b)15
c)13
d)12
[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]c)13[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]
Solution:
Work=lcm(20,40,3)=120
A=6 units per day
S=3 units per day
Anil does =6*3=18 units in 3 days
Bimal does 10% of the job=12 units
So anil and sunil will do 90% of the job=108 units
Anil has already done 18 units
So anil and sunil will do 90 units in, Say X days
So
X(A+S)=90
Putting values of A and S
X=10 days
So B will also work for 10 days
So total=3+10=13 days were required to complete the work
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Q.5)
a)1
b)2
c)3
d)4
[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]d)4[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]
Solution:
Using options, only option 4 satisfies because (4x-x^2)>0 for those values of x only
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Q.6)
a)4
b)5
[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]a)4[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]
Solution:
This can be rewritten as n^2-m^2=105
(n+m)(n-m)=105
We have an odd product in the RHS
So in LHS we must have only odd numbers that give a product of 105
Which are (1,105), (3,35), (5,21), (7,15)
So only 4 pairs will satisfy
iQuanta Shortcut: Number of factors/2= 8/2=4
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Q.7) In 2010, a library contained a total of 11500 books in two categories – fiction and non-fiction. In 2015, the library contained a total of 12760 books in these two categories. During this period, there was 10% increase in the fiction category while there was 12% increase in the non-fiction category. How many fiction books were in the library in 2015?
a)6160
b)6600
c)6000
d)5500
[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]b)6600[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]
Solution:
Fiction….F
Non fiction….N
In 2010…..
F be the num of fiction books
11500-F be the number of non-fiction books
In 2015….
1.1F+1.12(11500-F)=12760
Calculating F in the equation above we get
F=6600
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Q.8) The salaries of Ramesh, Ganesh and Rajesh were in the ratio 6:5:7 in 2010, and in the ratio 3:4:3 in 2015. If Ramesh’s salary increased by 25% during 2010-2015, then the percentage increase in Rajesh’s salary during this period is closest to
a)9
b)8
c)7
d)10
[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]c)7[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]
Solution:
Salaries of Ramesh, Ganesh and Rajesh were 6x:5x:7x in 2010
Ratio of their salaries is 3y:4y:3y in 2015
Salary of Ramesh after 25% increase, salary=6x*1.25=7.5x
So 3y=7.5x
y=2.5x
so Rajesh’s salary in 2015=3y=7.5x
so % increase in Rajesh’s salary=(7.5x-7x)/7x *100=50/7=7.1% so 7% is closest.
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Q.9)
a)4851
b)5851
[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]a)4851[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]
Solution:
From 1 to 47 we have 24 terms
48n+12(48)=5280
4n+48=440
4n=392
n=98
so 1+2+3+4+…..+98
98/2(99)=49*99=4851
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Q.10) Two circles, each of radius 4 cm, touch externally. Each of these two circles is touched externally by a third circle. If these three circles have a common tangent, then the radius of the third circle, in cm, is
a)1
b)2
c)4
d)5
[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]a)1[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]
Solution:
r is the radius of the smaller circle.
In right triangle ABC
AB^2+BC^2=AC^2
(4-r)^2+4^2=(4+r)^2
r=1
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Q.11) Let A and B be two regular polygons having a and b sides, respectively. If b = 2a and each the interior angle of B is 3/2 times each interior angle of A, then each interior angle, in degrees, of a regular polygon with a + b sides is
a)140
b)150
[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]b)150[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]
Solution:
the angle of B=1.5 * angle of A
so the angle of B is greater
(b-2)*180/b=1.5*(a-2)*180/a
(b-2)/b=1.5(a-2)/a
ab-2a=1.5ab-3b
0.5ab=3b-2a
½=3/a-2/b
b=2a
so a=4….b=8
a+b=12
so each interior angle of polygon having 12 sides is (12-2)*180/12=150
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Q.12) John jogs on track A at 6 kmph and Mary jogs on track B at 7.5 kmph. The total length of tracks A and B is 325 metres. While John makes 9 rounds of track A, Mary makes 5 rounds of track B. In how many seconds will Mary make one round of track A?
a)34
b)48
c)45
d)50
[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]b)48[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]
Solution:
A+B=325…eq1
9A/6=5B/7.5
A=4B/9
So putting in eq1
4B/9+B=325
B=225m
A=100 m
Speed of Mary=7.5 kmph=37.5/18 m/sec
So Mary completes one round of A i.e 100 m in =100/(37.5/18)=18000/375=48 sec
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Q.13) In a triangle ABC, medians AD and BE are perpendicular to each other, and have lengths 12 cm and 9 cm, respectively. Then, the area of triangle ABC, in sq cm, is
a)50
b)60
c)72
d)80
[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]c)72[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]
Solution:
Area(ABC)=2*area(ADB)
Since median intersect at centroid and centroid divides median into 2:1
So
AG=8….GD=4
BG=6…GE=3
Area of AGB=1/2*AD*BG=1/2*12*6=36
So Area of ABC=2*36=72
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Q.14)
a)k>5/7
b)k<=5/7
c)k=0
d)None of these
[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]c)k=30[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]
Solution:
a,b=+-3, +-4
x,y=+-12, +-5
But no combination of these values gives us 65
But a=5, b=0…x=13….y=0
This satisfies ax+by=65
So k=0
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Q.15)The base of a regular pyramid is a square and each of the other four sides is an equilateral triangle, length of each side being 20 cm. The vertical height of the pyramid, in cm, is
a)10√2
b)10
c)20
d)20√2
[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]a)10√2[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]
Solution:
Since each side is an equilateral triangle of side 20 cm ,
Hence the side of base square will also be 20 each.
So in right triangle ABC
BC^2=AB^2+AC^2
AC= ½ * diagonal of square
Diagonal of square= 20√2
So AC=10√2
So AB^2=20^2-(10√2)^2=200
AB=10√2
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Q.16) The quadratic equation x^2+ bx + c = 0 has two roots 4a and 3a, where a is an integer. Which of the following is a possible value of b^2+ c?
a)427
b)549
c)366
d)305
[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]b)549[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]
Solution:
So roots will be 4a,3a
So equation will be x^2-7ax+12a^2=0
b^2+c=(7a)^2+12a=49a^2+12a^2=61a^2
so answer must be a multiple of 61 multiplied by a perfect square which is only among the 549
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Q.17) Two ants A and B start from a point P on a circle at the same time, with A moving clock-wise and B moving anti-clockwise. They meet for the first time at 10:00 am when A has covered 60% of the track. If A returns to P at 10:12 am, then B returns to P at
a)10:14
b)10:27
c)10:15
d)10:29
[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]b)10:27[/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]
Solution: