{"id":4350,"date":"2020-05-12T08:19:08","date_gmt":"2020-05-12T02:49:08","guid":{"rendered":"https:\/\/iquanta.in\/blog\/?p=4350"},"modified":"2021-12-27T12:48:14","modified_gmt":"2021-12-27T07:18:14","slug":"cat-2019-quant-slot-2","status":"publish","type":"post","link":"https:\/\/www.iquanta.in\/blog\/cat-2019-quant-slot-2\/","title":{"rendered":"CAT 2019 Quant – Slot 2"},"content":{"rendered":"

Q.1) If 5^x-3^y=13438 and 5^(x-1)+3^(y+1)=9686 then find x+y.<\/strong><\/p>\n

a)13<\/span><\/p>\n

b)14<\/span><\/p>\n

[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” ]<\/p>\n

Solution:<\/span><\/p>\n

Here we can go by hit n trial<\/span><\/p>\n

13438 is 5 digit number so 5^x must be a 5 digit number<\/span><\/p>\n

5^6=15625<\/span><\/p>\n

So putting x=6 in the first equation we get 3^y=2187<\/span><\/p>\n

So y=7<\/span><\/p>\n

Putting this in eq 2<\/span><\/p>\n

5^5 +3^8=3125+6561=9686<\/span><\/p>\n

So x+y=6+7=13<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

Q.2) <\/strong><\/span><\/p>\n

a)0<\/span><\/p>\n

b)10<\/span><\/p>\n

c)1<\/span><\/p>\n

d)-1<\/span><\/p>\n

[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” ]<\/p>\n

Solution:<\/span><\/p>\n

a1=1<\/span><\/p>\n

a1-a2=2<\/span><\/p>\n

a2=-1<\/span><\/p>\n

a1-a2+a3=3<\/span><\/p>\n

a3=1<\/span><\/p>\n

a1-a2+a3-a4=4<\/span><\/p>\n

a4=-1<\/span><\/p>\n

1,-1,1,-1\u2026.so on<\/span><\/p>\n

so sum of every two consecutive terms will be 0<\/span><\/p>\n

so only a1023 will remain which is odd numbered term  having value 1<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

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<\/strong><\/p>\n

a)20920<\/span><\/p>\n

b)10920<\/span><\/p>\n

[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” ]<\/p>\n

Solution:<\/span><\/p>\n

Interests for both are same so equate their interests<\/span><\/p>\n

Semi-annually for Amal so rate becomes half  (i.e.3% )and time will become double i.e. 2 years<\/span><\/p>\n

Interest for Amal= (12000*8*1)\/100 +10000[1.03]^2 -10000<\/span><\/p>\n

Interest for Bimal= (P*7.5*1)\/100<\/span><\/p>\n

960+609=(7.5P)\/100<\/span><\/p>\n

156900=7.5P<\/span><\/p>\n

P=20920<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

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?<\/strong><\/p>\n

a)14<\/span><\/p>\n

b)15<\/span><\/p>\n

c)13<\/span><\/p>\n

d)12<\/span><\/p>\n

[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” ]<\/p>\n

Solution:<\/span><\/p>\n

Work=lcm(20,40,3)=120<\/span><\/p>\n

A=6 units per day<\/span><\/p>\n

S=3 units per day<\/span><\/p>\n

Anil does =6*3=18 units in 3 days<\/span><\/p>\n

Bimal does 10% of the job=12 units <\/span><\/p>\n

So anil and sunil will do 90% of the job=108 units<\/span><\/p>\n

Anil  has already done 18 units <\/span><\/p>\n

So anil and sunil will do 90 units in, Say X days<\/span><\/p>\n

So<\/span><\/p>\n

X(A+S)=90<\/span><\/p>\n

Putting values of A and S<\/span><\/p>\n

X=10 days<\/span><\/p>\n

So B will also work for 10 days<\/span><\/p>\n

So total=3+10=13 days were required to complete the work<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

Q.5)<\/strong><\/span><\/p>\n

<\/p>\n

a)1<\/p>\n

b)2<\/p>\n

c)3<\/p>\n

d)4<\/p>\n

[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” ]<\/p>\n

Solution:<\/p>\n

Using options, only option 4 satisfies because (4x-x^2)>0 for those values of x only<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

Q.6)<\/span><\/p>\n

a)4<\/span><\/p>\n

b)5<\/span><\/p>\n

[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” ]<\/p>\n

Solution:<\/span><\/p>\n

This can be rewritten as n^2-m^2=105<\/span><\/p>\n

(n+m)(n-m)=105<\/span><\/p>\n

We have an odd product in the RHS<\/span><\/p>\n

So in LHS we must have only odd numbers that give a product of 105<\/span><\/p>\n

Which are (1,105), (3,35), (5,21), (7,15)<\/span><\/p>\n

So only 4 pairs will satisfy<\/span><\/p>\n

iQuanta Shortcut: Number of factors\/2= 8\/2=4<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

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?<\/strong><\/p>\n

a)6160<\/span><\/p>\n

b)6600<\/span><\/p>\n

c)6000<\/span><\/p>\n

d)5500<\/span><\/p>\n

[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” ]<\/p>\n

Solution:<\/span><\/p>\n

Fiction\u2026.F<\/span><\/p>\n

Non fiction\u2026.N<\/span><\/p>\n

In 2010\u2026..<\/span><\/p>\n

F be the num of fiction books<\/span><\/p>\n

11500-F be the number of non-fiction books<\/span><\/p>\n

In 2015\u2026.<\/span><\/p>\n

1.1F+1.12(11500-F)=12760<\/span><\/p>\n

Calculating F in the equation above we get<\/span><\/p>\n

F=6600<\/span><\/p>\n

[\/bg_colllapse]<\/p>\n

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\u2019s salary increased by 25% during 2010-2015, then the percentage increase in Rajesh\u2019s salary during this period is closest to<\/strong><\/p>\n

a)9<\/span><\/p>\n

b)8<\/span><\/p>\n

c)7<\/span><\/p>\n

d)10<\/span><\/p>\n

[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” ]<\/p>\n

Solution:<\/span><\/p>\n

Salaries of Ramesh, Ganesh and Rajesh were  6x:5x:7x in 2010<\/span><\/p>\n

Ratio of their salaries is 3y:4y:3y in 2015<\/span><\/p>\n

Salary of Ramesh after 25% increase, salary=6x*1.25=7.5x<\/span><\/p>\n

So 3y=7.5x<\/span><\/p>\n

y=2.5x<\/span><\/p>\n

so Rajesh\u2019s salary in 2015=3y=7.5x<\/span><\/p>\n

so % increase in Rajesh\u2019s salary=(7.5x-7x)\/7x *100=50\/7=7.1% so 7% is closest.<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

Q.9)<\/strong><\/span><\/p>\n

a)4851<\/span><\/p>\n

b)5851<\/span><\/p>\n

[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” ]<\/p>\n

Solution:<\/span><\/p>\n

From 1 to 47 we have 24 terms<\/span><\/p>\n

48n+12(48)=5280<\/span><\/p>\n

4n+48=440<\/span><\/p>\n

4n=392<\/span><\/p>\n

n=98<\/span><\/p>\n

so 1+2+3+4+\u2026..+98<\/span><\/p>\n

98\/2(99)=49*99=4851<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

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<\/strong><\/p>\n

a)1<\/p>\n

b)2<\/p>\n

c)4<\/p>\n

d)5<\/p>\n

[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” ]<\/p>\n

Solution:<\/span><\/p>\n

r<\/b> is the radius of the smaller circle.<\/span><\/p>\n

In right triangle ABC<\/span><\/p>\n

AB^2+BC^2=AC^2<\/span><\/p>\n

(4-r)^2+4^2=(4+r)^2<\/span><\/p>\n

r=1<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

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<\/strong><\/p>\n

a)140<\/span><\/p>\n

b)150<\/span><\/p>\n

[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” ]<\/p>\n

Solution:<\/span><\/p>\n

the angle of B=1.5 * angle of A<\/span><\/p>\n

so the angle of B is greater<\/span><\/p>\n

(b-2)*180\/b=1.5*(a-2)*180\/a<\/span><\/p>\n

(b-2)\/b=1.5(a-2)\/a<\/span><\/p>\n

ab-2a=1.5ab-3b<\/span><\/p>\n

0.5ab=3b-2a<\/span><\/p>\n

\u00bd=3\/a-2\/b<\/span><\/p>\n

b=2a<\/span><\/p>\n

so a=4\u2026.b=8<\/span><\/p>\n

a+b=12<\/span><\/p>\n

so each interior angle of polygon having 12 sides is (12-2)*180\/12=150<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

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?<\/strong><\/p>\n

a)34<\/span><\/p>\n

b)<\/span>48<\/span><\/p>\n

c)45<\/span><\/p>\n

d)50<\/span><\/p>\n

[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” ]<\/p>\n

Solution:<\/span><\/p>\n

A+B=325\u2026eq1<\/span><\/p>\n

9A\/6=5B\/7.5<\/span><\/p>\n

A=4B\/9<\/span><\/p>\n

So putting in eq1<\/span><\/p>\n

4B\/9+B=325<\/span><\/p>\n

B=225m<\/span><\/p>\n

A=100 m<\/span><\/p>\n

Speed of Mary=7.5 kmph=37.5\/18  m\/sec<\/span><\/p>\n

So Mary completes one round of A i.e 100 m in =100\/(37.5\/18)=18000\/375=48 sec<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

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<\/strong><\/p>\n

a)50<\/span><\/p>\n

b)60<\/span><\/p>\n

c)72<\/span><\/p>\n

d)80<\/span><\/p>\n

[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” ]<\/p>\n

Solution:<\/span><\/p>\n

<\/p>\n

Area(ABC)=2*area(ADB)<\/span><\/p>\n

Since median intersect at centroid and centroid divides median into 2:1<\/span><\/p>\n

So <\/span><\/p>\n

AG=8\u2026.GD=4<\/span><\/p>\n

BG=6\u2026GE=3<\/span><\/p>\n

Area of AGB=1\/2*AD*BG=1\/2*12*6=36<\/span><\/p>\n

So Area of ABC=2*36=72<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

Q.14)<\/strong><\/span><\/p>\n

a)k>5\/7<\/span><\/p>\n

b)k<=5\/7<\/span><\/p>\n

c)k=0<\/span><\/p>\n

d)None of these<\/span><\/p>\n

[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” ]<\/p>\n

Solution:<\/span><\/p>\n

a,b=+-3, +-4<\/span><\/p>\n

x,y=+-12, +-5<\/span><\/p>\n

But no combination of these values gives us 65<\/span><\/p>\n

But a=5, b=0\u2026x=13\u2026.y=0<\/span><\/p>\n

This satisfies ax+by=65<\/span><\/p>\n

So k=0<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

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<\/strong><\/p>\n

a)10\u221a2<\/span><\/p>\n

b)10<\/span><\/p>\n

c)20<\/span><\/p>\n

d)20\u221a2<\/span><\/p>\n

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]a)10\u221a2<\/span>[\/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]<\/p>\n

Solution:<\/span><\/p>\n

<\/p>\n

Since each side is an equilateral triangle of side 20 cm , <\/span><\/p>\n

Hence the side of base square will also be 20 each.<\/span><\/p>\n

So in right triangle ABC <\/span><\/p>\n

BC^2=AB^2+AC^2<\/span><\/p>\n

AC= \u00bd * diagonal of square<\/span><\/p>\n

Diagonal of square= 20\u221a2<\/span><\/p>\n

So AC=10\u221a2<\/span><\/p>\n

So AB^2=20^2-(10\u221a2)^2=200<\/span><\/p>\n

AB=10\u221a2<\/span><\/p>\n

[\/bg_collapse]<\/span><\/p>\n

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?<\/strong><\/p>\n

 a)427<\/span><\/p>\n

b)549<\/span><\/p>\n

c)366<\/span><\/p>\n

d)305<\/span><\/p>\n

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]b)549<\/span>[\/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]<\/p>\n

Solution:<\/span><\/p>\n

So roots will be 4a,3a<\/span><\/p>\n

So equation will be x^2-7ax+12a^2=0<\/span><\/p>\n

b^2+c=(7a)^2+12a=49a^2+12a^2=61a^2<\/span><\/p>\n

so answer must be a multiple of 61 multiplied by a perfect square which is only among the 549<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

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<\/strong><\/p>\n

a)10:14<\/span><\/p>\n

b)10:27<\/span><\/p>\n

c)10:15<\/span><\/p>\n

d)10:29<\/span><\/p>\n

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]b)10:27<\/span>[\/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]<\/p>\n

Solution: <\/span><\/p>\n

<\/p>\n

[\/bg_collapse]<\/p>\n

Q.18) Let f be a function such that f (mn) = f (m) f (n) for every positive integers m and n. If f(1), f (2) and f (3) are positive integers, f (1) < f (2), and f (24) = 54, then f (18) equals<\/strong><\/p>\n

a)18<\/span><\/p>\n

b)52<\/span><\/p>\n

c)54<\/span><\/p>\n

d)67<\/span><\/p>\n

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]a)18[\/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]<\/p>\n

Solution:<\/span><\/p>\n

f(24)=54<\/span><\/p>\n

f(24)=f(2*2*2*3)=f(2)*f(2)*f(2)*f(3)=54=3*3*3*2<\/span><\/p>\n

so comparing LHS and RHS<\/span><\/p>\n

f(2)=3 and f(3)=2<\/span><\/p>\n

so<\/span><\/p>\n

f(18)=f(2*3*3)=f(2)*f(3)*f(3)=3*3*2=18<\/span><\/p>\n

54\/f(18)=f(2)(f(2)\/f(3)<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

Q.19) John gets Rs 57 per hour of regular work and Rs 114 per hour of overtime work. He works altogether 172 hours and his income from overtime hours is 15% of his income from regular hours. Then, for how many hours did he work overtime?<\/strong><\/p>\n

a)10<\/span><\/p>\n

b)11<\/span><\/p>\n

c)14<\/span><\/p>\n

d)12<\/span><\/p>\n

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]d)12[\/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]<\/p>\n

Solution:<\/span><\/p>\n

X hrs normally<\/span><\/p>\n

172-x hrs overtime<\/span><\/p>\n

Income from normal working =57x<\/span><\/p>\n

Income from overtime=114(172-x)<\/span><\/p>\n

Given income from overtime hours is 15% of his income from regular hours<\/span><\/p>\n

So 114(172-x)=(15*57x)\/100<\/span><\/p>\n

200(172-x)=15x<\/span><\/p>\n

40(172-x)=3x<\/span><\/p>\n

43x=40*172<\/span><\/p>\n

X=160<\/span><\/p>\n

So he worked overtime for 172-x i.e. 172-160=12 hrs<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

Q.20) In an examination, the score of A was 10% less than that of B, the score of B was 25% more than that of C, and the score of C was 20% less than that of D. If A scored 72, then the score of D was<\/strong><\/p>\n

a)50<\/span><\/p>\n

b)60<\/span><\/p>\n

c)70<\/span><\/p>\n

d)80<\/span><\/p>\n

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]d)80[\/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]<\/p>\n

Solution:<\/span><\/p>\n

Given<\/span><\/p>\n

A=0.9B<\/span><\/p>\n

B=1.25C<\/span><\/p>\n

C=0.8D<\/span><\/p>\n

A=72\u2026putting this in first eq above<\/span><\/p>\n

So B=80<\/span><\/p>\n

C=64<\/span><\/p>\n

D=80<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

Q.21) The number of common terms in the two sequences: <\/strong><\/p>\n

15, 19, 23, 27, . . . . , 415 and 14, 19,24, 29, . . . , 464 is<\/strong><\/p>\n

a)20<\/span><\/p>\n

b)21<\/span><\/p>\n

c)18<\/span><\/p>\n

d)23<\/span><\/p>\n

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]a)20[\/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]<\/p>\n

Solution:<\/span><\/p>\n

Common difference of 1<\/span>st<\/span> series is 4<\/span><\/p>\n

Common difference of 2<\/span>nd<\/span> series is 5<\/span><\/p>\n

So series of common terms will have common difference =LCM(4,5)=20<\/span><\/p>\n

Now find the first common terms in both the series<\/span><\/p>\n

And that is 19<\/span><\/p>\n

So series of common terms will be like 19, 39,, 59\u2026. till 415 because first series has terms only till 415<\/span><\/p>\n

So 415=19+(n-1)20<\/span><\/p>\n

20(n-1)=396<\/span><\/p>\n

n=20.xx<\/span><\/p>\n

so 20 terms will be common  <\/span><\/p>\n

[\/bg_collapse]<\/p>\n

Q.22)<\/strong><\/span><\/p>\n

a)A<1\/6<\/span><\/p>\n

b)A>1\/16<\/span><\/p>\n

c)A=3\/4<\/span><\/p>\n

d)A>=1\/4<\/span><\/p>\n

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]b)A>1\/16[\/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]<\/p>\n

Solution:<\/span><\/p>\n

Real and distinct  i.e b^2-4ac>0<\/span><\/p>\n

(4^2 +4log<\/span>2<\/span>A) >0<\/span><\/p>\n

4+log<\/span>2<\/span>A>0<\/span><\/p>\n

Log<\/span>2<\/span>A>-4<\/span><\/p>\n

A>2^(-4)<\/span><\/p>\n

A>1\/16<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

Q.23) <\/strong>Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm. If AP is <\/strong><\/span>perpendicular on BC, then the maximum possible length of AP, in cm, is<\/strong><\/p>\n

a)15<\/span><\/p>\n

b)20<\/span><\/p>\n

c)25<\/span><\/p>\n

d)10<\/span><\/p>\n

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]d)10[\/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]<\/p>\n

Solution:<\/span><\/p>\n

<\/p>\n

Max value of AP will be when ABC becomes a right isosceles triangle and  AP=BP=PC<\/span><\/p>\n

That means we can put this triangle inside a circle\u2026hence AP,BP,PC become the radius of this circle <\/span><\/p>\n

So AP=20\/2=10 cm<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

Q.24) Mukesh purchased 10 bicycles in 2017, all at the same price. He sold six of these at a profit of 25% and the remaining four at a loss of 25%. If he made a total profit of Rs. 2000, then his purchase price of a bicycle, in Rupees, was<\/strong><\/p>\n

a)1000<\/span><\/p>\n

b)4000<\/span><\/p>\n

c)3000<\/span><\/p>\n

d)2000<\/span><\/p>\n

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]b)4000[\/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]<\/p>\n

Solution:<\/span><\/p>\n

Let CP per cycle=x<\/span><\/p>\n

Total CP=10x<\/span><\/p>\n

Total SP=6*1.25x+4*0.75x=10.5x<\/span><\/p>\n

Total profit=total Sp- total Cp=0.5X<\/span><\/p>\n

0.5x=2000<\/span><\/p>\n

x=4000<\/span><\/p>\n

so CP=4000<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

Q.25)<\/strong><\/span><\/p>\n

a)10<\/span><\/p>\n

b)13<\/span><\/p>\n

c)12<\/span><\/p>\n

d)14<\/span><\/p>\n

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]c)12[\/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]<\/p>\n

Solution:<\/span><\/p>\n

(n+3)(n+4)\/(n-4)(n+3)=(n+4)\/(n-4)=1+8\/(n-4)<\/span><\/p>\n

So for this to be a positive integer, largest value that n can take will be when 8=n-4..<\/span><\/p>\n

so n=12  <\/span><\/p>\n

[\/bg_collapse]<\/p>\n

Q.26) The average of 30 integers is 5. Among these 30 integers, there are exactly 20 which do not exceed 5. What is the highest possible value of the average of these 20 integers?<\/strong><\/p>\n

a)4<\/span><\/p>\n

b)4.5<\/span><\/p>\n

c)5<\/span><\/p>\n

d)5.5<\/span><\/p>\n

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]b)4.5[\/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]<\/p>\n

Solution:<\/span><\/p>\n

Average of 30 integers=5<\/span><\/p>\n

So sum of all 30 = 150<\/span><\/p>\n

20 are less than equal to 5<\/span><\/p>\n

So 10 integers must be greater than 5 i.e 6<\/span><\/p>\n

So 60 will be their sum<\/span><\/p>\n

So sum of rest 20 integers=90<\/span><\/p>\n

So max possible average =90\/20=4.5<\/span> <\/span><\/p>\n

[\/bg_collapse]<\/p>\n

Q.27) A man makes complete use of 405 cc of iron, 783 cc of aluminum, and 351 cc of copper to make a number of solid right circular cylinders of each type of metal. These cylinders have the same volume and each of these has radius 3 cm. If the total number of cylinders is to be kept at a minimum, then the total surface area of all these cylinders, in sq cm, is<\/strong><\/p>\n

a)1026(1+\u220f)<\/span><\/p>\n

b)1026\u220f<\/span><\/p>\n

c)1026<\/span><\/p>\n

d)1000(\u220f+2)<\/span><\/p>\n

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]a)1026(1+\u220f)<\/span>[\/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]<\/p>\n

Solution:<\/span><\/p>\n

Volume of each cylinder= \u220f(3^2)*h = 9\u220fh<\/span><\/p>\n

Hcf(405,783,351)=27=volume of each cylinder<\/span><\/p>\n

So num of iron cylinders=405\/27=15<\/span><\/p>\n

Num of aluminium cylinders=783\/27=29<\/span><\/p>\n

Num of copper cylinders=351\/27=13<\/span><\/p>\n

Total cylinders=15+29+13=57<\/span><\/p>\n

Vol of cylinder=9\u220fh=27<\/span><\/p>\n

So h=3\/\u220f<\/span><\/p>\n

So surface area of all cylinders=57*(2\u220f*9+2\u220f*3*3\/\u220f)=1026(1+\u220f)<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

Q.28) In an examination, Rama’s score was one-twelfth of the sum of the scores of Mohan and Anjali. After a review, the score of each of them increased by 6. The revised scores of Anjali, Mohan, and Rama were in the ratio 11:10:3. Then Anjali’s score exceeded Rama’s score by<\/strong><\/p>\n

a)12<\/span><\/p>\n

b)22<\/span><\/p>\n

c)32<\/span><\/p>\n

d)43<\/span><\/p>\n

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]c)32[\/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]<\/p>\n

Solution:<\/p>\n

Given <\/span><\/p>\n

R=(M+A)\/12<\/span><\/p>\n

M+A=12R\u2026eq1<\/span><\/p>\n

After new scores, ratio was.. (A+6):(M+6):(R+6)=11x:10x:3x<\/span><\/p>\n

A+6=11x\/24<\/span><\/p>\n

M+6=10x\/24<\/span><\/p>\n

R+6=3x\/24<\/span><\/p>\n

Putting values of A,M,R in eq 1<\/span><\/p>\n

We get x=96<\/span><\/p>\n

So A=44<\/span><\/p>\n

M=40<\/span><\/p>\n

R=12<\/span><\/p>\n

So difference in Scores of Rama and  Anjali=44-12=32<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

Q.29) In a six-digit number, the sixth, that is, the rightmost, digit is the sum of the first three digits, the fifth digit is the sum of first two digits, the third digit is equal to the first digit, the second digit is twice the first digit and the fourth digit is the sum of fifth and sixth digits. Then, the largest possible value of the fourth digit is<\/strong><\/p>\n

a)4<\/span><\/p>\n

b)6<\/span><\/p>\n

c)7<\/span><\/p>\n

d)12<\/span><\/p>\n

[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” ]<\/p>\n

Solution:<\/span><\/p>\n

Let the number be abcdef<\/span><\/p>\n

f=a+b+c<\/span><\/p>\n

e=a+b<\/span><\/p>\n

c=a<\/span><\/p>\n

b=2a<\/span><\/p>\n

d=e+f<\/span><\/p>\n

f=a+2a+a=4a<\/span><\/p>\n

e=a+2a=3a<\/span><\/p>\n

d=e+f=7a<\/span><\/p>\n

now d can be a single digit number so d=7a, hence a=1 or d=7 <\/span><\/p>\n

so max value of d=7<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

Q.30) The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. Each of three vessels A, B, C contains 500 ml of salt solution of strengths 10%, 22%, and 32%, respectively. Now, 100 ml of the solution in vessel A is transferred to vessel B. Then, 100 ml of the solution in vessel B is transferred to vessel C. Finally, 100 ml of the solution in vessel C is transferred to vessel A. The strength, in percentage, of the resulting solution in vessel A is <\/strong><\/p>\n

a)18<\/span><\/p>\n

b)14<\/span><\/p>\n

c)15<\/span><\/p>\n

d)13<\/span><\/p>\n

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]b)14[\/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]<\/p>\n

Solution:<\/span><\/p>\n

A \u202610%<\/span><\/p>\n

B\u2026.22%<\/span><\/p>\n

C\u2026.32%<\/span><\/p>\n

Now 100ml of A is transferred to vessel B <\/span><\/p>\n

So using alligations<\/span><\/p>\n

 <\/span><\/p>\n

[\/bg_collapse] <\/span><\/p>\n

Q.31) A cyclist leaves A at 10 am and reaches B at 11 am. Starting from 10:01 am, every minute a motorcycle leaves A and moves towards B. Forty-five such motorcycles reach B by 11 am. All motorcycles have the same speed. If the cyclist had doubled his speed, how many motorcycles would have reached B by the time the cyclist reached B?<\/strong><\/p>\n

a)13<\/span><\/p>\n

b)14<\/span><\/p>\n

c)15<\/span><\/p>\n

d)16<\/span><\/p>\n

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]c)15[\/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]<\/p>\n

solution:<\/span><\/p>\n

As cyclist starts from 10 and reach 11. Let distance be 60km. so the speed of cyclists is 1km\/h. now in this time total 45 bike reaches from a to b. that means a bike takes 15 minute to reach a to b. that means a bike starting from A at 10:01 will reach B at 10:16 so from 10:01 to 10:45 total 45 bikes will leave A and reach B on or before 11. <\/span>Now, if cyclist will double its speed =2 km\/h, so It will take 30 minute to reach B, and so from 10:16 to 10:30. Total 15 bike person will reach B. So answer =15<\/span><\/p>\n

[\/bg_collapse]<\/p>\n

Q.32) How many factors of 2^4 x 3^5 x 10^4 are perfect squares which are greater than 1?<\/strong><\/p>\n

a)40<\/span><\/p>\n

b)42<\/span><\/p>\n

c)44<\/span><\/p>\n

d)46<\/span><\/p>\n

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]b)44[\/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]<\/p>\n

Solution:<\/span><\/p>\n

It can be rewritten as<\/span><\/p>\n

(2^8*3^5*5*4)<\/span><\/p>\n

So perfect squares will be when powers are 0,2,4,6,8<\/span><\/p>\n

So 2s power can be 0,2,4,6,8\u2026..i.e 5 values<\/span><\/p>\n

3s power can be 0,2,4\u2026i.e 3 values<\/span><\/p>\n

5s power can be 0,2,4\u2026.i.e 3 values<\/span><\/p>\n

So total perfect square factors =5*3*3=45<\/span><\/p>\n

Now among these 1 will be a perfect square and we need perfect squares greater than 1 so remove 1<\/span><\/p>\n

Hence we get 44 perfect square factors greater than 1 <\/span><\/p>\n

[\/bg_collapse]<\/p>\n

Q.33)<\/strong><\/span><\/p>\n

a)log<\/span>2<\/span>7\/3<\/span><\/p>\n

b)log<\/span>7<\/span>2\/3<\/span><\/p>\n

c)log<\/span>2<\/span>7<\/span><\/p>\n

d)log<\/span>7<\/span>2<\/span><\/p>\n

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]a)log27\/3[\/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]<\/p>\n

Solution:<\/span><\/p>\n

Let 2^3x=t<\/span><\/p>\n

So we can rewrite the equation as t^2+4t-21=0<\/span><\/p>\n

(t+3)(t-7)=0<\/span><\/p>\n

t=-3 or 7<\/span><\/p>\n

2^3x can never be negative so<\/span><\/p>\n

2^3x=7<\/span><\/p>\n

Taking log both sides<\/span><\/p>\n

3xlog 2=log 7<\/span><\/p>\n

x=log7\/3log2<\/span><\/p>\n

=log<\/span>2<\/span>7\/3 <\/span><\/p>\n

[\/bg_collapse]<\/p>\n

Q.34) A shopkeeper sells two tables, each procured at cost price p, to Amal and Asim at a profit of 20% and at a loss of 20%, respectively. Amal sells his table to Bimal at a profit of 30%, while Asim sells his table to Barun at a loss of 30%. If the amounts paid by Bimal and Barun are x and y, respectively, then (x \u2212y) \/ p equals <\/strong><\/p>\n

a)1.2<\/span><\/p>\n

b)0.7<\/span><\/p>\n

c)0.5<\/span><\/p>\n

d)1<\/span><\/p>\n

[bg_collapse view=”button-blue” color=”#ffffff” expand_text=”Correct Answer” collapse_text=”Show Less” ]d)1[\/bg_collapse] [bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”View Solution” collapse_text=”Show Less” ]<\/p>\n

Solution:<\/span><\/p>\n

Total CP of shopkeeper=2p<\/span><\/p>\n

Total SP for shopkeeper =1.2p+0.8p=2p <\/span><\/p>\n

CP for Amal=1.2p<\/span><\/p>\n

CP for Asim=0.8p<\/span><\/p>\n

SP for Amal= 1.3*1.2p=1.56p=CP for Bimal=x<\/span><\/p>\n

SP for Asim=0.7*0.8p=0.56p=CP for Barun=y<\/span><\/p>\n

x-y=1.56p-0.56p=p<\/span><\/p>\n

so (x-y)\/p=p\/p=1<\/span><\/p>\n

[\/bg_collapse]