CAT,MAT,GMAT exam questions and entrepreneurship opportunities discussed

1. My words.... Here I present a few basic questions on CAT,MAT,GMAT, . I wish
that more people should become entrepreneurs. An ordinary Indian entrepreneur
wishes to remain an honest entrepreneur and contribute to the development of
nation but we have to strengthen those institutions which truly promote
entrepreneurship, not just degree granting institutions. Let us work together to
promote knowledge, wisdom, social development and education. We believe in
free education for all, free support for all, entrepreneurship opportunities and
training for all. Let us work together for these goals. ... I alone cant do much, I
need support of perosns like you .......... ...
2. In how many ways can the letters of the word "PROBLEM" be
rearranged to make seven letter words such that none of the letters repeat? We
have 7 letters, so answer = 7! =7*6*5*4*3*2*1 =5040 answer
3. Ten coins are tossed simultaneously. In how many of the outcomes will the third
coin turn up a head? The total number of outcomes are : 2^10 in half the cases,
the third coin will turn up a head. =512 out of total 1024 answer
4. In how many ways can 5 letters be posted in 3 post boxes, if any number of letters
can be posted in all of the three post boxes? First letter can be posted in 3 boxes,
so we have 3 choices, for the 2 nd letter also we have 3 choices and so on ....
=3*3*3*3*3 =243 answer
5. How many different four letter words can be formed (the words need not be
meaningful) using the letters of the word MEDITERRANEAN such that the first
letter is E and the last letter is R? In MEDITERRANEAN we have 13 digits, but
we have 3 E 2 R, 2A, 2N, thus we have only 8 digits left. Thus we have : 1*8*7*1
= 56 choices if the 2 nd digit is E or A or N then we have one more choice so in
that case we will have 3 more choices. Total = 59 choices
6. A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the
probability that he will hit his target? Find the probability of NOT hitting any shot
: ¾ * ¾ * ¾ * ¾ = 81/256 probability of hitting = 1 – 81/256 = 175/256 answer
7. Vertices of a quadrilateral ABCD are A(0, 0), B(4, 5), C(9, 9) and D(5, 4). What
is the shape of the quadrilateral? Rhombus.
8. How many five digit positive integers that are divisible by 3 can be formed using
the digits 0, 1, 2, 3, 4 and 5, without any of the digits getting repeated in a
number? The total of five digits leaving 0 is 15, so any number would be divisible
by 3. we have : 5! =120 if we leave 3 and include 0, (total is 12) we have :
4*4*3*2*1 = 96 options thus we have 120+96 = 216 options
9. In how many ways can 5 different toys be packed in 3 identical boxes such that no
box is empty, if any of the boxes may hold all of the toys? We have following
choices : 5c2+3c2+1c1 = 10+3+1 =14 or 5c3+2c1+1c1 = 10+2+1 = 13 total = 27
answer
10. In how many ways can the letters of the word ABACUS be rearranged such that
the vowels always appear together? We have 4 vowels so let us treat them as 1.
we have 4 digits in all. We can have 4! = 24 options. We can have 3!/2! for 3
vowels (with 2 common) we have : 3 options for this. Thus answer is 24+3 = 27
answer

11. How many integers, greater than 999 but not greater than 4000, can be formed
with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed? For the first digit
we have 3 options (1,2,3) thus we have : 3*4*3*2*1 = 72 options
12. There are 6 boxes numbered 1, 2,...6. Each box is to be filled up either with a red
or a green ball in such a way that at least 1 box contains a green ball and the
boxes containing green balls are consecutively numbered. The total number of
ways in which this can be done is We can put green in only one or upto 6 box. We
can have it in 1 box = 6 options we can have it in 2 boxes = 5 options
(12,23,34,45,56) similarly we can have 4,3,2,1 option total = 21 options answer
13. 1 liter of water is add to 5 liters of a 20 % solution of alcohol in water . Find the
strength of alcohol in new solution? At present we have 1 liter alcohol and 4 liters
water. But now water becomes 5 liters. The strength of alchoal is : 1/(1+5) *100 =
16.66% answer
14. If A earns 33 1/3 more than B .Then B earns less than A by what percent? If B
earns 100, A earns 133.33 B earns 33.33 less than A or in % : 33.33 / 133.33 *100
= 25% answer
15. A school has only three classes which contain 40,50,60 students respectively.The
pass percent of these classes are 10, 20 and 10 respectively . Then find the pass
percent in the school. Number of students passing out : 4,10 and 6 from the 3
classes. Total pass = 20 total students = 150 pass percentage=20/150 *100 =
13.33%
16. There are 600 boys in a hostel . Each plays either hockey or football or both .If
75% play hockey and 45 % play football ,Find how many play both? Add 75+45
= 120, (120 -100 = 20), thus 20% play both the games. 120 students play both
football and hockey. Answer
17. A bag contains 600 coins of 25p denomination and 1200 coins of 50p
denomination. If 12% of 25p coins and 24 % of 50p coins are removed, Find the
percentage of money removed from the bag ? Toal amount = (600 .25) +
(1200* .5) = 750 amount of coins removed = (72*.25) + (288*.5) = 162 162/750 *
100 = 21.6% answer
18. P is six times as large as Q.Find the percent that Q is less than P? Let us assume Q
= X P = 6X Q is less by (6X – X) = 5X 5X/6X * 100 = 83% approximately.
Answer
19. For a sphere of radius 10 cm ,the numerical value of surface area is what percent
of the numerical value of its volume? Volume of shpere = 4/3 * pi * r^3 surphace
area of sphere = 4pi*r^2 ratio of surface area to volume = 4pir^2 / 4/3pir^3 * 100
= 3/10 * 100 = 30% answer
20. A reduction of 21 % in the price of wheat enables a person to buy 10 .5 kg more
for Rs 100.What is the reduced price per kg. Let us assume reduced price = X 100
/1.27X - 100 / X = 10.5 (100 – 127) / 1.27X = 10.5 -27 / (10.5*1.27) =x X = 2.02
answer
21. The average temperature of the town in the first four days of a month was 58
degrees. The average for the second ,third,fourth and fifth days was 60 degree .If
the temperature of the first and fifth days were in the ratio 7:8 then what is the
temperature on the fifth day? 1 st day = 7X, 5 th day = 8x Total for first 4 days :
58*4 = 232 total for 2,3,4,5 day = 60*4 = 240 difference between 5 th and 1 st

day = 8X-7X or 240 -232 X = 8 temperature on fifth day = 8*8 = 64 degrees
answer
22. The average of 5 numbers is 7.When 3 new numbers are added the average of the
eight numbers is 8.5. The average of the three new number is? Total of first 5 =
(5*7) = 35 total of 8 (including 3 new digits) = (8*8.5) = 68 total of 3 new digits
= (68 – 35 = 33 average of 3 new digits=33/3 = 11 answer
23. Of the three numbers the first is twice the second and the second is twice the third
.The average of the reciprocal of the numbers is 7/72,the numbers are Assume 3
rd number = x, 2 nd number = 2x and 1 st number = 4x actually the total is : 7/72
* 3 = 7/24 if we take average of reciprocals :, then 1/x + 1/2x + 1/4x = 7/4x 7/4x
= 7/24 x = 7/24 * 4/7 = 1/6 x = 6, other two numbers are 12 and 24 answer
24. 16 children are to be divided into two groups A and B of 10 and 6 children. The
average percent marks obtained by the children of group A is 75 and the average
percent marks of all the 16 children is 76. What is the average percent marks of
children of groups B? Total of 16 children = (16 * 76) = 1216 total of group A = (
10 * 75)= 750 total of group B = 466 average of B = 466 / 6 = 77.6 answer
25. The mean temperature of Monday to Wednesday was 37 degrees and of Tuesday
to Thursday was 34 degrees .If the temperature on Thursday was 4/5th that of
Monday. The temperature on Thursday was Total from Monday to Wednesday =
(3 * 37) = 111 total from tuesday to thursday = (3* 34) = 102 difference between
Monday and Thursday : 5X – 4X = (111 -102) = 9 X = 9, temperature on tursday
= (4*9) = 36 answer
26. A man whose bowling average is 12.4,takes 5 wickets for 26 runs and there by
decrease his average by 0.4. The number of wickets,taken by him before his last
match is: Let us assume number of wickets taken before start of this match = x
and number of runs = Y Y/X = 12.4 or Y -12.4X = 0 (Y+26) / (X +5) = 12 or
Y-12X=34 X = 85 (number of wickets) Y = 1054 answer
27. A team of 8 persons joins in a shooting competition. The best marksman scored
85 points.If he had scored 92 points ,the average score for the team would have
been 84.The team scored was. Total of team = (8*84) – (92-85) =672 – 7 =665
answer
28. The average weight of 3 men A,B and C is 84 Kgs. Another man D joins the
group and the average now becomes 80 Kgs.If another man E whose weight is 3
Kgs more than that of D ,replaces A then the average weight B,C,D and E
becomes 79 Kgs. The weight of A is. Weight of A+B+C = (3*84) = 252 Weight
of A+B+C+D= ((4*80) = 320 weight of D = 320-252 = 68 weight of E= 71 total
of B+C+D+E = (4*79) = 316 difference of A and E is just 4 Kg. (320-316) weight
of A = 71+4 = 75 kg. Answer
29. Five years ago, the average age of P and Q was 15 years ,average age of P,Q, and
R today is 20 years,how old will R be after 10 years? Total of P & Q = (15+5) * 2
= 40 total of P,Q,R = (20*3) = 60 R = 20 age of R after 10 years = 20+10 = 30
answer
30. The average age of the boys in the class is twice the number of girls in the class
.If the ratio of boys and girls in the class of 36 be 5:1, what is the total of the age
(in years) of the boys in the class? Number of girls = 1/(1+5) * 36 = 6 average age

of boys = 2*6 = 12 Number of boys= 36-6 = 30 total age of boys=30*12 = 360
answer
31. .In a cricket eleven ,the average of eleven players is 28 years .Out of these ,the
average ages of three groups of players each are 25 years,28 years, and 30 years
respectively. If in these groups ,the captain and the youngest player are not
included and the captain is eleven years older than the youngest players , what is
the age of the captain? Total of 11 = (11*28) = 308 total of 3 groups (3*25 + 3*28
+ 3*30) = (75+84+90) = 249 remaining = 59 Captain + Y = 59 C-Y= 11 2C = 70
or C = 35 answer
32. Dhoni scored 87 in his 17 th innings and this increased his average by 3. What is
the new average of Dhoni ? Let us assume that his average is X total runs =
(number of innings * the average run rate) = 17 *x 17x – 87 = 16(x-3) 17x – 87 =
16x – 48 x = 39 answer
33. The average weight of 24 students is 35 kg. If the weight of teacher is also
included, the average rises by 400 grams. What is the weight of the teacher ? Let
us assume the weight of teacher = x total weight earlier : (number of students *
average weight) =24*35 = 840 new total weight = (24+1) * (35+.400) = 885
(difference is due to inclusion of teacher) weight of teacher = 885 – 840 = 45 Kgs.
34. Average earning of 11 days is 50000. the average for first 6 days is 49000 and the
average for last 6 days is 52000. what is the earning on 6 th day? Total for 11 =
(11 * 50000) = 550000 total for first 6 = (6*49000) = 294000 total for last 6 =
(6*52000) = 312000 let us add the last two : 294000+312000 = 606000 find the
difference of this and total of 11 : 606000 – 550000=56000 (which is income of
the 6 th day)
35. Find the average of all these numbers.142,147,153,165,157. Formula for average
= total for all / number of digits First add them all : total = 764 there are 5 digits
so divide it by 5 = 764/5 = 152.8 answer
36. Find the average of all these numbers: 4,10,16,22,28 Total = 80 there are 5 digits
so divide by 5 average = 80 / 5 = 16 answer shortcut = difference of all the
numbers is equal, so mid value (16) is the average – so there is no need for all this
calculation)
37. Find the average of all these numbers: =4,10,16,22,28,34. Here also we have same
difference between continuous digits, so take the middle digits : middle digits
are : 16 and 22. find the mid point of these two digits : (16+22)/2 = 19 or 16 + 3
(half of 6, which is common difference). Answer
38. The average marks of a student in 4 Examinations is 40.If he got 80 marks in 5th
Exam then what is his new average. His total for 4 examinations = 160 total after
5 examinations = 160+80 = 240 new average = 240/5 = 48 answer
39. The average marks of Surabhi Daga in 10 Examinations is 80%.If she got 90
marks in 5th Exam then what is her new average. her total for 10 examinations =
800 total after 11 examinations =800+90 new average = 890/11 = 80.9 answer
40. In a group the average income of 6 men is 500 and that of 5 women is 280, then
what is average income of the group. Total for men = (6*500) = 3000 total for
women = (5*280) = 1400 total for all= 4400 there are 11 persons (6 men + 5
women), so average = 4400/11 = 400 per person.

41. The average weight of a class of 30 students is 40 kgs if the teacher weight is
included then average increases by 2 kgs then find the weight of the teacher?
Total for students = 30 * 40 = 1200 total with teacher (30+1) * (40+2) = 1302 the
difference is due to teacher : 1302– 1200 = 102 kgs. Answer
42. The average age of Mr and Mrs Sharma (Ilahabad Wale) 4 years ago is 28 years .
If the present average age of Mr and Mrs Sharma and their son is 22 years. What
is the age of their son? Age of Mrs and Mr. Sharma : (28+4) * 2 = 64 age with son
: (3*22) = 66 age of son = 66 – 64 = 2 years. Answer
43. The average price of 10 books is increased by 17 Rupees when one of them
whose value is Rs.400 is replaced by a new book. What is the price of new book?
Old price = 400 addition = 10 * 17 = 170 new price = 570 answer
44. The average marks of girls in a class is 62.5. The average marks of 4 girls among
them is 60.The average marks of remaining girls is 63,then what is the number of
girls in the class? Let us assume the number of students = X total = 62.5x
(4*60)+(63 * ( X -4) = 62.5X 240+ 63X- 252 = 62.5x .5X = 12 X = 24 answer
45. Find the average of first 50 natural numbers? Just add 1 and divide by 2, you will
get the mid point. Simply find the mid point (50+1) / 2 = 51/2 = 25.5 answer
46. The average of the first nine prime number is? Add them :
(2+3+5+7+11+13+17+19+23) =100 now divide this total by 9, we get : 11.11
answer
47. The average of 2,7,6 and x is 5 and the average of and the average of 18,1,6,x and
y is 10 .what is the value of y? Total of first 4 digits = (4*5 ) = 20 X = 20 –
(2+7+6) = 5 total next 5 digits = (5*10) = 50 value of Y = 50 - (18+1+6+5) = 20
Y = 20 answer
48. The average of a non-zero number and its square is 5 times the number.The
number is? Let us assume the number to be X X + X^2 = 2(5X) X^2 = 10x- X
X^2 = 9X X = 9 answer
49. P and Q started a business investing Rs 85,000 and Rs 15,000 respectively. In
what ratio the profit earned after 2 years be divided between P and Q
respectively? Profit can be divided either in capital ratio or equally. Here we
assume that profit is divided in capital ratio : 85:15 or 17:3 answer
50. P and Q started a business investing Rs 85,000 each. However, P withdrew 20%
of his capital after 1.5 year and Q doubled his investment after 20 months. . In
what ratio the profit earned after 2 years be divided between P and Q
respectively? Profit can be divided either in capital ratio or equally. Here we
assume that profit for 2 nd year is divided in capital ratio : p = (85*6 + 68*6)
=918 Q = (85*6 + 170*6) =1530 their ratio : 918:1530 or 3:5 answer
51. A,B and C started a business by investing Rs 1,20,000, Rs 1,35,000 and Rs
1,50,000.Find the share of each ,out of an annual profit of Rs 56,700? Their profit
sharing ratio as per their capital : 120: 135:150 or 8:9:10 share of A =8/27 *56700
= 16800 share of B = 9/27 *56700 = 18900 share of C = 10/27 *56700 = 21000
answer
52. A,B and C started a business by investing Rs 2,20,000, Rs 350000 and Rs
4,50,000.Find the share of each ,out of an annual profit of Rs 10200 ? Their profit
sharing ratio as per their capital : 22:35:45 share of A =22/102 *10200 = 2200
share of B = 35/102 *10200 = 3500 share of C =45/102 *10200 = 4500 answer

53. 3 milkman A,B,C rented a pasture A grazed his 45 cows for 12 days B grazed his
36 cows for 15 days and c 60 cows for 10 days. If b's share of rent was Rs
540 What is the total rent? Share of A = 45*12 = 540 share of B = 36*15 = 540
share of C = 60 * 10 = 600 share of B's rent is 540, that means total rent is :
(540+540+600) = 1680 answer
54. 3 milkman A,B,C rented a pasture A grazed his 90 cows for 6 days B grazed his
15 cows for 20 days and c 30 cows for 5 days. If C's share of rent was Rs
300 What is A's rent ? Share of A = 90*6 = 540 share of B = 15*20= 300
share of C = 30*5= 150 share of c's rent is 300, that meansthe rent is twice
the total cow-days. So A's rent should be 1080 answer
55. Ram and Krishn entered into a partnership with Rs 50,000 and Rs 60,000, after 4
months Ram invested Rs 25,000 more while Krishn withdraw Rs 20,000 . Find
the share of Ram in the annual profit of Rs 289,000. Ram's share = (50*4 +
75*8) =800 Krishn's share = (60*4 + 40*8) =560 share of ram = (80)/
(80+56) * 289000 =170000 answer
56. A,B,C enter into partnership .A invests 3 times as much as B invests and B invests
two third of what C invests. At the end of the year ,the profit earned is Rs 6600.
what is the share of B? Let us start from C. Suppose C invests Rs. 3. B will invest
2/3 of 3 = 2. Thus A would invest 6 total investment = 11 share of B = 2/11 *
6600 = 1200 answer
57. A,B,C enter into partnership . A invests two third of what B invests and B invests
two third of what C invests. At the end of the year ,the share of A is 2000, what is
the share of B and C? (start with the last - here C) Suppose C invests (3*3) = 9 B
will invest = 9 * 2/3 = 6 A will invest = 6 * 2/3 = 4 thus if share of A is known,
we can find share of B and C also. B = 6/4 * 2000 = 3000 C = 9/4 * 2000 = 4500
shortcut = simply multiply 2000 *3/2 for B and again by 3/2 for C. You will get
the answer
58. A,B and C enter into a partnership by investing in the ratio of 3:2:4. After one
year ,B invests another Rs 2,70,000 and C,at the end of 2 years, also invests Rs
2,70,000.At the end of 3 years,profit are shared in the ratio of 3:4:5.Find the
initial investment of each? Share of A = 3x*3 = 9x share of B = 2x*3 + 270*2 =
6x+540 Share of C =4x*3+270*1 = 12x+270 based on new ratio : 9x/3 =
(12x+270)/5 3x=2.4x+54 or .6x=54 or x= 54/.6 = 90 so initial investments were :
A : 270, B 180 and C 360 (all in thousands). Answer
59. A,B and C enter into a partnership by investing in the ratio of 3:2:4. After one
year ,B invests another Rs 300,000 and C,at the end of 2 years, also invests Rs
300,000.At the end of 3 years,profit are shared in the ratio of 2:2:3 .Find the
initial investment of each? Share of A = 3x*3 = 9x share of B = 2x*3 + 300*2 =
6x+600 Share of C =4x*3+300*1 = 12x+300 based on new ratio : 9x / 2 =
(12x+300)/3 4.5x = 4x+100 or .5x = 100 or x = 200 initial investment by A = 600,
B = 400 and C =800 (all in '000) answer
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63. 2^(2x-1) = 1/(8^(x-3)), what is x? Options : 3,2,0,-2 put x = 2 and you get the
answer
64. 3^(4x-2) = 729, what is x? Options : 1,1.5,2,2.5, try to put the value x=2, you get
the answer
65. 32^1/3 = 2^x, what is x? Options : 5,3,3/5,5/3 answer :
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AffairsAdvanced Mathematics for GMAT CAT MAT AFTERSCHO ☺ OL –
DEVELOPING CHANGE MAKERS CENTRE FOR SOCIAL
ENTREPRENEURSHIP PGPSE PROGRAMME – World’ Most Comprehensive
programme in social entrepreneurship & spiritual entrepreneurship OPEN FOR
ALL FREE FOR ALL www.afterschoool.tk AFTERSCHO☺OL's
MATERIAL FOR PGPSE PARTICIPANTS
67. Advanced Mathematics for GMAT CAT MAT Dr. T.K. Jain. AFTERSCHO ☺
OL Centre for social entrepreneurship Bikaner M: 9414430763 [email_address]
www.afterschool.tk , www.afterschoool.tk www.afterschoool.tk
AFTERSCHO☺OL's MATERIAL FOR PGPSE PARTICIPANTS
68. There are 2 tangents of 21 inches from point P to the circle with centre C inclined
at an angle of 60 degree. Find the circumference of the circle?
1. W will always be 90 degree. WP = 21

2. WP=sqrt(3)/2 * CP
3. 21=sqrt(3)/2 *CP
4. CP =42/ (sqrt(3) =24.24

P c W 60 degree

69. Solution continued…
1. When the angle is 60 degree, CW should be half of CP. Thus radius
should be ½*24
2. Radius should be 12.12 and thus cirmuference should be : 2* 22/7*12.12
3. = 76.18 ans.
70. In the next graph, you find RS =12, radius of the circle is 70. What is the distance
of WX, when W is intersection of RS and diameter AX.
1. Let us assume the centre to be C. C to R distance is = radius = 70. let us
assume mid point of RS to beW. Now WR= 6.

rsXAcW

71. solution
1. WC^2= 70^2 – 6^2
2. WC^2 = 4900-36 = 4864
3. WC = sqrt (4864)= 69.7
4. We already know that CX is 70 (radius). Thus length of WX = 139.7
72. What is the probability of throwing a number greater than 3 with an ordinary die
whose faces are numbered from 1 to 6 ?
1. There are only 3 digits greater than 3 : 4,5,6 out of 6 digits. Thus
probaility is 3/6 or ½ (probability = the number of favourable outcomes /
total number of outcomes).
73. In a class of 12 students, 5 are boys and the remaining are girls. Find the
probability that a student selected is a girl ?
1. There are 7 girls – so the probability of selection of a girl is 7/12 answer.
74. A bag contains 7 white and 2 black balls. Find the probability of drawing a white
ball.
1. 7/9 answer.
75. In a given race, the odds in favour of four horses, A, B, C, D are 1: 3, 1:4, 1: 5 and
1: 6 respectively. Assuming that a dead heat is impossible, find the chance that
one of them will win the race.
1. Probability that none of them wins :
2. ¾ * 4/5 * 5/6 * 6/7= .428
3. The probability that one of them will win the race = 1- .428 = .572 answer.
76. X^3 + 5x^2 +10k leaves remainder - 2x, when divided by (X^2 +2), what is k?
1. Let us divide this number by (x^2 +2)
2. X^3 + 5x^2 +10k
3. - X^3 - 2x
4. _________________ 5x^2 – 2x +10k -5X^2 – 10
5. -----------------------------

6. - 2x -10 +10k
7. Thus the remainder is :
8. - 2x – 10 + 10k= -2x
9. 10k = 10
10. K = 1 answer.
77. If (x-2) is a factor of (X^2 + 3qx – 2q), what is the value of q?
1. As X – 2 is a factor, so take X = 2
2. (X^2 + 3qx – 2q) = 0, and taking X=2,
3. Thus we will have the following equation:
4. 4 + 6q – 2q = 0
5. 4q = -4
6. q = -1 answer.
78. X^100 + 2x^99 + k is divisible by (x+1), what is the value of k?
1. Let us take X = -1.
2. 1 + 2 + k = 0
3. K = -3 answer.
79. (x-1) is a factor of (X^3 – k), what is the value of k?
1. Taking x = 1, we have following solution:
2. 1 – K = 0
3. K = 1 answer.
80. (X+2) and (X-1) are factors of (x^3 +10x^2 +mx +n), the values of m and n are ?
1. Let us take X = -2 and 1. we have two solutions with these two values:
2. -8 +40 -2m + n =0 or -2m+n = -32
3. 1 + 10 + m +n = 0 or 2m +2n = -22
4. 3n = -54 or n = -18, and m = 7 answer.
81. What will be the remainder when (X^4 – 3x^3 +2x^2 -5x +7) is divided by (x-2)?
1. Let us take X = 2.
2. 16 – 24 + 8 – 10 + 7
3. = -3 answer.
82. What will be the remainder if (5x^3 + 5x^2 – 6x +9) is divided by (x+3)?
1. Let us take X = -3
2. -135 +45 +18 + 9
3. = -63 answer.
83. What will be the remainder if (X^11 + 1) is divided by (x +1)?
2. -1 + 1 = 0
3. Thus the remainder is zero.
4. Answer.
84. In order to make (X^4 - 11X^2Y^2 + Y^4) a perfect square, what should we add
in it?
1. (X^2 – Y^2)^2 = X^4 - 2X^2Y^2 + Y^4,
2. Thus in order to make it perfect square, we have to add 9x^2Y^2 to this
number, so that it becomes a perfect square. Answer
85. What is the sum of (X^2 +1) and reciprocal of (x^2 -1)?
1. We have to obtain sum of :
2. (X^2 +1) + ( 1/ (X^2 -1)

3. We get ( X^4 – 1 + 1) / ( X^2 – 1)
4. Solving further, we get
5. (X^4 / (X^2 – 1))
6. Answer.
86. WHAT IS THE NUMBER OF NON-CONGRUENT RECTANGLES THAT WE
CAN FIND ON A CHESS BOARD?
1. There are 8 rectangles on 8 rows – which are non- congruent. We can
change the sizes of these rectangles. By taking minimum any 2 rectangles,
we can make ½(64-8) = 28 more rectangles. Thus we have total 8 + 28 =
36 non-congruent rectangles.
87. How many words can you make taking all the letters of ZENITH. What will be
the rank of ZENITH in the list of these words?
1. 6! = 720 words can be made. The letters from Z will start from 600
onwards. 4! Or 24 letters start with ZE. Thus there will be 6+6+4 letters
more before ZENITH. Thus its rank will be 616 from the beginning.
88. Example…
1. ZENITH
2. There are 600 words starting with either e,N,I, T,H.
3. Thus there are 120 words starting with Z
4. Out of them first 24 are having ZE. In ZE, we can have words like Zen..
Or Zet.. Or Zei.. Thus counting the words like Zet (2 words) and the words
like ZENTHI, ZENTIH. Thus removing them, we can 616 as the number.
89. What is the number of words fored from : AAAAA, BBBB,CCC,DD,F?
1. There are 15 letters, so we have 15! Ways of making these letters.
2. There are also many similar letters.
3. Thus we have 15!/(5! *4! * 3! *2! * 1!)
90. What is the number of words fored from : AAAAA, BBBB,CCC,DD,F with all
Cs taking different places?
1. There are 12 letters exclusing Cs. Thus Cs can take 12 +1 = 13 places
between or before them. Thus we have 13c3 options. Thus we have 13c3 *
(12! / (5!*4!*2!)) options.
91. The probability of a student A solving a problem is 0.6 and that of another student
B solving the problem is 0.3. What is the probability that at least one will solve it
right?
1. A will not solve = (1-.6)=.4, B will not solve = .7
2. Both of them will not solve = .4 *.7 = .28
3. One of them / both may solve it = 1-.28=.72
92. Veeru and Jai are two friends. If Veeru tells the truth 40% of the time and Jai 30%
of the time, the probability that they will contradict each other when they are
narrating the same incident is?
1. V true, J false = .4*.7 = .28
2. V false, J true = .6 * .3 = .18
3. They will contradict = .28 +.18 = .46 answer
93. If one diagonal of a rhombus is equal to a side, what is the ratio between two
adjacent angles of the rhombus?

1. This will be the situation when we combine two equilateral triangle to
form a rhombus. Thus two angles of the rhomus will be 60 each and
remaining two angles will be 120 each. Thus ratio will be 1:2 ANSWER.
94. PUZZLE
1. A,B,C,D, are 4 villages on a straight road in that order, given that B is
equidistant from A and C while C is equidistant from B and D Two girls
Reena and Sheena decide to watk frorn A to D, covers the distance
walking at 3 km/hr and Sheena covers the first lap (upto B) at either l.5 or
4.5 km/ hr and then walked at the other speed till she meets Reena.
Thereafter they walk together to D at Reena ‘s speed.What is Sheena’s
speed from A to B?
95. Solution…
1. The challenge is to ensure that the speed should be such that average of
which is 3 KM per hour. Let us assume that the distance from A to B and
B to C and C to D is 9 KM each. if the speed is 4.5, it will take her 2 hours
to reach B and then X / 1.5 and Y / 3 (total = 27/3 = 9 hours).
2. Other possibility is : 9/1.5 + 18 (at 4.5 and 3 per hour)
96. About AFTERSCHO☺OL
1. PGPSE - World’s most comprehensive programme on social
entrepreneurship – after class 12 th
2. Flexible – fast changing to meet the requirements
3. Admission open throughout the year
4. Complete support from beginning to the end – from idea generation to
making the project viable.
97. Branches of AFTERSCHO☺OL
1. PGPSE programme is open all over the world as free online programme.
2. Those who complete PSPSE have the freedom to start branches of
AFTERSCHO☺OL
3. A few branches have already started - one such branch is at KOTA
(Rajasthan).
98. Workshop on social entrepreneurship
1. We conduct workshop on social entrepreneurship – all over India and out
of India also - in school, college, club, association or any such place - just
send us a call and we will come to conduct the workshop on social
entrepreeurship.
2. These workshops are great moments of learning, sharing, and
commitments.
99. FREE ONLINE PROGRAMME
1. AFTERSCHO☺OL is absolutely free programme available online – any
person can join it. The programme has four components :
2. 1. case studies – writing and analysing – using latest tools of management
3. 2. articles / reports writing & presentation of them in conferences /
seminars
4. 3. Study material / books / ebooks / audio / audio visual material to
support the study

5. 4. business plan preparation and presentations of those plans in
conferences / seminars
100.Advanced Mathematics for GMAT CAT MAT Dr. T.K. Jain. AFTERSCHO ☺
101.There are 2 tangents of 21 inches from point P to the circle with centre C inclined
3. 21=sqrt(3)/2 *CP
4. CP =42/ (sqrt(3) =24.24

P c W 60 degree

102.Solution continued…
should be ½*24
3. = 76.18 ans.
103.In the next graph, you find RS =12, radius of the circle is 70. What is the distance

rsXAcW

104.solution
1. WC^2= 70^2 – 6^2
2. WC^2 = 4900-36 = 4864
3. WC = sqrt (4864)= 69.7
105.What is the probability of throwing a number greater than 3 with an ordinary die
106.In a class of 12 students, 5 are boys and the remaining are girls. Find the
107.A bag contains 7 white and 2 black balls. Find the probability of drawing a white
ball.
1. 7/9 answer.
108.In a given race, the odds in favour of four horses, A, B, C, D are 1: 3, 1:4, 1: 5
and 1: 6 respectively. Assuming that a dead heat is impossible, find the chance
that one of them will win the race.

2. ¾ * 4/5 * 5/6 * 6/7= .428
109.X^3 + 5x^2 +10k leaves remainder - 2x, when divided by (X^2 +2), what is k?
2. X^3 + 5x^2 +10k
3. - X^3 - 2x
4. _________________ 5x^2 – 2x +10k -5X^2 – 10
5. -----------------------------
6. - 2x -10 +10k
8. - 2x – 10 + 10k= -2x
9. 10k = 10
10. K = 1 answer.
110.If (x-2) is a factor of (X^2 + 3qx – 2q), what is the value of q?
2. (X^2 + 3qx – 2q) = 0, and taking X=2,
4. 4 + 6q – 2q = 0
5. 4q = -4
6. q = -1 answer.
111.X^100 + 2x^99 + k is divisible by (x+1), what is the value of k?
2. 1 + 2 + k = 0
3. K = -3 answer.
112.(x-1) is a factor of (X^3 – k), what is the value of k?
2. 1 – K = 0
3. K = 1 answer.
113.(X+2) and (X-1) are factors of (x^3 +10x^2 +mx +n), the values of m and n are ?
2. -8 +40 -2m + n =0 or -2m+n = -32
3. 1 + 10 + m +n = 0 or 2m +2n = -22
4. 3n = -54 or n = -18, and m = 7 answer.
114.What will be the remainder when (X^4 – 3x^3 +2x^2 -5x +7) is divided by (x-2)?
2. 16 – 24 + 8 – 10 + 7
3. = -3 answer.
115.What will be the remainder if (5x^3 + 5x^2 – 6x +9) is divided by (x+3)?
2. -135 +45 +18 + 9
3. = -63 answer.
116.What will be the remainder if (X^11 + 1) is divided by (x +1)?
2. -1 + 1 = 0

4. Answer.
117.In order to make (X^4 - 11X^2Y^2 + Y^4) a perfect square, what should we add
in it?
1. (X^2 – Y^2)^2 = X^4 - 2X^2Y^2 + Y^4,
118.What is the sum of (X^2 +1) and reciprocal of (x^2 -1)?
2. (X^2 +1) + ( 1/ (X^2 -1)
3. We get ( X^4 – 1 + 1) / ( X^2 – 1)
5. (X^4 / (X^2 – 1))
6. Answer.
119.WHAT IS THE NUMBER OF NON-CONGRUENT RECTANGLES THAT
WE CAN FIND ON A CHESS BOARD?
120.How many words can you make taking all the letters of ZENITH. What will be
121.Example…
1. ZENITH
122.What is the number of words fored from : AAAAA, BBBB,CCC,DD,F?
3. Thus we have 15!/(5! *4! * 3! *2! * 1!)
123.What is the number of words fored from : AAAAA, BBBB,CCC,DD,F with all
(12! / (5!*4!*2!)) options.
124.The probability of a student A solving a problem is 0.6 and that of another
student B solving the problem is 0.3. What is the probability that at least one will
solve it right?

125.Veeru and Jai are two friends. If Veeru tells the truth 40% of the time and Jai
30% of the time, the probability that they will contradict each other when they are
1. V true, J false = .4*.7 = .28
2. V false, J true = .6 * .3 = .18
126.If one diagonal of a rhombus is equal to a side, what is the ratio between two
127.PUZZLE
speed from A to B?
128.Solution…
129.About AFTERSCHO☺OL
130.Branches of AFTERSCHO☺OL
AFTERSCHO☺OL
(Rajasthan).
131.Workshop on social entrepreneurship
entrepreeurship.
commitments.
132.FREE ONLINE PROGRAMME

seminars
support the study
133.100% placement / entrepreneurship
1. AFTERSCHO☺OL has the record of 100% placement / entrepreneurship
till date
2. Be assured of a bright career – if you join AFTERSCHO☺OL
134.Advanced Mathematics for GMAT CAT MAT Dr. T.K. Jain. AFTERSCHO ☺
135.There are 2 tangents of 21 inches from point P to the circle with centre C inclined
3. 21=sqrt(3)/2 *CP
4. CP =42/ (sqrt(3) =24.24

P c W 60 degree

should be ½*24
3. = 76.18 ans.
137.In the next graph, you find RS =12, radius of the circle is 70. What is the distance

rsXAcW

138.solution
1. WC^2= 70^2 – 6^2
2. WC^2 = 4900-36 = 4864
3. WC = sqrt (4864)= 69.7
139.What is the probability of throwing a number greater than 3 with an ordinary die

140.In a class of 12 students, 5 are boys and the remaining are girls. Find the
141.A bag contains 7 white and 2 black balls. Find the probability of drawing a white
ball.
1. 7/9 answer.
142.In a given race, the odds in favour of four horses, A, B, C, D are 1: 3, 1:4, 1: 5
and 1: 6 respectively. Assuming that a dead heat is impossible, find the chance
that one of them will win the race.
2. ¾ * 4/5 * 5/6 * 6/7= .428
2. X^3 + 5x^2 +10k
3. - X^3 - 2x
4. _________________ 5x^2 – 2x +10k -5X^2 – 10
5. -----------------------------
6. - 2x -10 +10k
8. - 2x – 10 + 10k= -2x
9. 10k = 10
10. K = 1 answer.
2. (X^2 + 3qx – 2q) = 0, and taking X=2,
4. 4 + 6q – 2q = 0
5. 4q = -4
6. q = -1 answer.
2. 1 + 2 + k = 0
3. K = -3 answer.
2. 1 – K = 0
3. K = 1 answer.
2. -8 +40 -2m + n =0 or -2m+n = -32
3. 1 + 10 + m +n = 0 or 2m +2n = -22
4. 3n = -54 or n = -18, and m = 7 answer.

2. 16 – 24 + 8 – 10 + 7
3. = -3 answer.
2. -135 +45 +18 + 9
3. = -63 answer.
2. -1 + 1 = 0
4. Answer.
151.In order to make (X^4 - 11X^2Y^2 + Y^4) a perfect square, what should we add
in it?
1. (X^2 – Y^2)^2 = X^4 - 2X^2Y^2 + Y^4,
2. (X^2 +1) + ( 1/ (X^2 -1)
3. We get ( X^4 – 1 + 1) / ( X^2 – 1)
5. (X^4 / (X^2 – 1))
6. Answer.
153.WHAT IS THE NUMBER OF NON-CONGRUENT RECTANGLES THAT
WE CAN FIND ON A CHESS BOARD?
154.How many words can you make taking all the letters of ZENITH. What will be
155.Example…
1. ZENITH

3. Thus we have 15!/(5! *4! * 3! *2! * 1!)
157.What is the number of words fored from : AAAAA, BBBB,CCC,DD,F with all
(12! / (5!*4!*2!)) options.
158.The probability of a student A solving a problem is 0.6 and that of another
student B solving the problem is 0.3. What is the probability that at least one will
solve it right?
159.Veeru and Jai are two friends. If Veeru tells the truth 40% of the time and Jai
30% of the time, the probability that they will contradict each other when they are
1. V true, J false = .4*.7 = .28
2. V false, J true = .6 * .3 = .18
160.If one diagonal of a rhombus is equal to a side, what is the ratio between two
161.PUZZLE
speed from A to B?
162.Solution…

AFTERSCHO☺OL
(Rajasthan).
entrepreeurship.
commitments.
seminars
support the study
1. AFTERSCHO☺OL has the record of 100% placement / entrepreneurship
till date
2. Be assured of a bright career – if you join AFTERSCHO☺OL
168.Advanced Mathematics for GMAT CAT MAT Dr. T.K. Jain. AFTERSCHO ☺ OL Centre for social
entrepreneurship Bikaner M: 9414430763 [email_address] www.afterschool.tk ,
www.afterschoool.tk www.afterschoool.tk AFTERSCHO☺OL's MATERIAL FOR PGPSE
PARTICIPANTS
169.There are 2 tangents of 21 inches from point P to the circle with centre C inclined at an angle of 60
degree. Find the circumference of the circle?
3. 21=sqrt(3)/2 *CP
4. CP =42/ (sqrt(3) =24.24

P c W 60 degree

1. When the angle is 60 degree, CW should be half of CP. Thus radius should be ½*24
3. = 76.18 ans.
171.In the next graph, you find RS =12, radius of the circle is 70. What is the distance of WX, when W
is intersection of RS and diameter AX.
1. Let us assume the centre to be C. C to R distance is = radius = 70. let us assume mid
point of RS to beW. Now WR= 6.

rsXAcW

172.solution
1. WC^2= 70^2 – 6^2
2. WC^2 = 4900-36 = 4864
3. WC = sqrt (4864)= 69.7

173.What is the probability of throwing a number greater than 3 with an ordinary die whose faces are
numbered from 1 to 6 ?
1. There are only 3 digits greater than 3 : 4,5,6 out of 6 digits. Thus probaility is 3/6 or ½
(probability = the number of favourable outcomes / total number of outcomes).
174.In a class of 12 students, 5 are boys and the remaining are girls. Find the probability that a student
selected is a girl ?
175.A bag contains 7 white and 2 black balls. Find the probability of drawing a white ball.
1. 7/9 answer.
176.In a given race, the odds in favour of four horses, A, B, C, D are 1: 3, 1:4, 1: 5 and 1: 6
respectively. Assuming that a dead heat is impossible, find the chance that one of them will win the
race.
2. ¾ * 4/5 * 5/6 * 6/7= .428
2. X^3 + 5x^2 +10k
3. - X^3 - 2x
4. _________________ 5x^2 – 2x +10k -5X^2 – 10
5. -----------------------------
6. - 2x -10 +10k
8. - 2x – 10 + 10k= -2x
9. 10k = 10
10. K = 1 answer.
2. (X^2 + 3qx – 2q) = 0, and taking X=2,
4. 4 + 6q – 2q = 0
5. 4q = -4
6. q = -1 answer.
2. 1 + 2 + k = 0
3. K = -3 answer.
2. 1 – K = 0
3. K = 1 answer.
2. -8 +40 -2m + n =0 or -2m+n = -32
3. 1 + 10 + m +n = 0 or 2m +2n = -22
4. 3n = -54 or n = -18, and m = 7 answer.
2. 16 – 24 + 8 – 10 + 7
3. = -3 answer.
2. -135 +45 +18 + 9
3. = -63 answer.
2. -1 + 1 = 0
4. Answer.
185.In order to make (X^4 - 11X^2Y^2 + Y^4) a perfect square, what should we add in it?
1. (X^2 – Y^2)^2 = X^4 - 2X^2Y^2 + Y^4,
2. Thus in order to make it perfect square, we have to add 9x^2Y^2 to this number, so
that it becomes a perfect square. Answer
2. (X^2 +1) + ( 1/ (X^2 -1)
3. We get ( X^4 – 1 + 1) / ( X^2 – 1)
5. (X^4 / (X^2 – 1))
6. Answer.
187.WHAT IS THE NUMBER OF NON-CONGRUENT RECTANGLES THAT WE CAN FIND ON A
CHESS BOARD?

1. There are 8 rectangles on 8 rows – which are non- congruent. We can change the
sizes of these rectangles. By taking minimum any 2 rectangles, we can make ½(64-8) =
28 more rectangles. Thus we have total 8 + 28 = 36 non-congruent rectangles.
188.How many words can you make taking all the letters of ZENITH. What will be the rank of ZENITH in
the list of these words?
1. 6! = 720 words can be made. The letters from Z will start from 600 onwards. 4! Or 24
letters start with ZE. Thus there will be 6+6+4 letters more before ZENITH. Thus its
rank will be 616 from the beginning.
189.Example…
1. ZENITH
4. Out of them first 24 are having ZE. In ZE, we can have words like Zen.. Or Zet.. Or
Zei.. Thus counting the words like Zet (2 words) and the words like ZENTHI, ZENTIH.
Thus removing them, we can 616 as the number.
3. Thus we have 15!/(5! *4! * 3! *2! * 1!)
191.What is the number of words fored from : AAAAA, BBBB,CCC,DD,F with all Cs taking different
places?
1. There are 12 letters exclusing Cs. Thus Cs can take 12 +1 = 13 places between or
before them. Thus we have 13c3 options. Thus we have 13c3 * (12! / (5!*4!*2!))
options.
192.The probability of a student A solving a problem is 0.6 and that of another student B solving the
problem is 0.3. What is the probability that at least one will solve it right?
193.Veeru and Jai are two friends. If Veeru tells the truth 40% of the time and Jai 30% of the time, the
probability that they will contradict each other when they are narrating the same incident is?
1. V true, J false = .4*.7 = .28
2. V false, J true = .6 * .3 = .18
194.If one diagonal of a rhombus is equal to a side, what is the ratio between two adjacent angles of the
rhombus?
1. This will be the situation when we combine two equilateral triangle to form a rhombus.
Thus two angles of the rhomus will be 60 each and remaining two angles will be 120
each. Thus ratio will be 1:2 ANSWER.
195.PUZZLE
1. A,B,C,D, are 4 villages on a straight road in that order, given that B is equidistant from
A and C while C is equidistant from B and D Two girls Reena and Sheena decide to
watk frorn A to D, covers the distance walking at 3 km/hr and Sheena covers the first
lap (upto B) at either l.5 or 4.5 km/ hr and then walked at the other speed till she meets
Reena. Thereafter they walk together to D at Reena ‘s speed.What is Sheena’s speed
from A to B?
196.Solution…
1. The challenge is to ensure that the speed should be such that average of which is 3
KM per hour. Let us assume that the distance from A to B and B to C and C to D is 9
KM each. if the speed is 4.5, it will take her 2 hours to reach B and then X / 1.5 and Y /
3 (total = 27/3 = 9 hours).
1. PGPSE - World’s most comprehensive programme on social entrepreneurship – after
class 12 th
4. Complete support from beginning to the end – from idea generation to making the
project viable.
2. Those who complete PSPSE have the freedom to start branches of AFTERSCHO☺OL
3. A few branches have already started - one such branch is at KOTA (Rajasthan).
1. We conduct workshop on social entrepreneurship – all over India and out of India also -
in school, college, club, association or any such place - just send us a call and we will
come to conduct the workshop on social entrepreeurship.
2. These workshops are great moments of learning, sharing, and commitments.

1. AFTERSCHO☺OL is absolutely free programme available online – any person can join
it. The programme has four components :
3. 2. articles / reports writing & presentation of them in conferences / seminars
4. 3. Study material / books / ebooks / audio / audio visual material to support the study
5. 4. business plan preparation and presentations of those plans in conferences /
seminars

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