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Quant BY NARENDRA SHEKHAWAT

C-25, NEW VIDHAN SABHA ROAD, JAIPUR, Phone number 0141- 4061414 / 4061415, visit us at:

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1., then x is ?

(A) 2(B)5

Sol:

So 2x=10 and x=5

Choice (B)

2.The marked price of a mixie is Rs.1600. The shopkeeper gives successive discounts of 10% and x% to the customer. If the customer pays Rs.1224 for the mixie, find the value of x?

(A) 12%(B)15%

Sol: Shopkeeper is giving discount of Rs.376 so discount percent is Discount percent is given as x+y–xy/100 so putting x=10 in this equation we get 10+y–10y/100=23.5 so y=15% Choice (B)

3.On selling an article for Rs.170, a shopkeeper loses 15%. In order to gain 20% he must sell at what price?

(A) 212.50(B)240

Sol: SP=Choice (B)

4.A shopkeeper marks the price of an article at Rs.80. What will be the selling if he offers two successive discounts of 5% each?

(A) Rs.72(B) Rs.85

(C)Rs.7.2(D) Rs.72.2

Sol: Choice (D)

5.If , the value of is ?

(A) 194(B) 81

Sol: Now squaring both the sides we get ,=14Now squaring again we get so =194Choice(A)

6.Which of the following successive discounts is better to a customer (a) 20%,15%,10% (b) 25%,12%,8%?

(A) a is better(B) b is better

(C)both same(D)None of these

Sol: Let CP be Rs.100 so SP will be

SP1=now

SP2=So customer is getting cheaper in second case so b is better. Also we can directly say that discounts which are widely spaced are better than discounts which are closely spaced for the customers.

Choice (B)

7.A circular road runs around a circular ground. If the difference between the circumferences of the outer and inner circle is 66m, the width of the road is?

(A)7m(B)5.25m

Sol: 2п(r1-r2)=66 so r1-r2=10.5mChoice (D)

8.The degree measure of 1 radian is ?

(A)57˚16’22’’(B) 57˚22’16’’

Sol: Taking п=180˚

1 radian= now this fraction part is in decimal number system and is just greater than ¼ or .25 so angle in minute has to be just greater than ¼ which is greater than 15 minutes(16 here in option A) only option which satisfies the above criteria is A Choice (A)

9.If , the value of x–x2 is?

(A) 1/a(B)–1/a

(C)a(D)–a

Sol: so x2=x–a therefore x2–x =–a

Choice (D)

10.A chord AB of a circle C1of radius (√3+1)cm touches a circle C2which is concentric to C1 . If the radius of C2 is (√3–1)cm, then find the length of AB?

(A)8√3 cm(B)cm

Sol:

A B

Now Applying pythgoras we get AB=2(R2–r2)

2[3+1+2√3–(3+1–2√3)]=8√3cm

Choice (A)

11.Three numbers are in the ratio 1:2:3. By adding 5 to each of them, the new numbers are in the ratio 2:3:4 The numbers are?

(A)15,30,45(B)1,2,3

Sol: Checking the options only C satisfies the above criteria.Choice (C)

12.If secθ+tanθ=2+√5 then the value of sinθ+cosθ is?

(A) √5(B) 7/√5

Sol: secθ+tanθ=√5+2 so

secθ–tanθ=√5–2 (Bigger is always written 1st) now adding these two equations we get 2secθ=2√5 so secθ=√5 now taking hypotenuse as √5 base as 1 and perpendicular comes 2 so sinθ=2/√5 and cosθ=1/√5 adding them we get sinθ+cosθ=3/√5 Choice (D)

13.ABC is an isosceles triangle such that AB=AC and angle B=35˚. AD is the median to the base BC. Find angle BAD?

(A)35˚(B)110˚

Sol: B=C=35 as triangle is isosceles so angle A=110˚ now in isosceles triangle median is also the angle bisector so angle BAD=110˚/2=55˚

Choice (C)

14.Evaluate tan1˚tan2˚tan3˚…….tan89˚?

(A) –1(B) 2

Sol: tan89˚=cot1˚ similarly all pairs willbe cancelled because tanθcotθ=1 leavong only tan45˚ which is 1 only. Choice (D)

15.a2+b2+c2+3=2(a–b–c), then the value of

2a–b–c is?

(A) 4(B)0

Sol: Given equation can be written as

(a–1)2+(b+1)2+(c+1)2=0 so we get a=1,

b= –1 and c= –1 and putting these values we get 2a–b+c=2Choice (C)

16.Out of the 10 teachers of a school, one teacher retires and in his place a new teacher cjoins due to which average age of teachers is reduced by 3 years. Find the age of the retired teacher?

(A) 60(B) 55

Sol: Age=25+3×10=55Choice (B)

17.If p–2q=4, then the value of p3–8q3–24pq-64 is ?

(A)0(B)3

(C)–1(D)2

Sol: Cubing the given eqn. we get p3–8q3–24pq=64 so p3–8q3–24pq-64=0

Choice (A)

18.A man buys 3 cows and 8 goats in Rs.47200. If he had bought 8 cows and 3 goats, he had to pay Rs.53000 more. Cost of one cow is?

(A)Rs.12000(B)Rs.13000

(C)Rs.10000(D)Rs.11000

Sol: 3x+8y=47200 and 8x+3y=100200 adding them we get 11x+11y=147400 and subtracting we get 5x–5y=53000 so x+y=13400 and x–y=10600 adding these two final equations we get 2x=24000 so x=12000

Choice (A)

19.A man can swim 3kmph in still water and the speed of current is 2kmph then how much time it would take to go 10km upstream and back?

(A)10 hours(B)12 hours

Sol: Speed in upstream would be 1kmph and would take 10 hours and speed in downstream would be 5kmph and time would be 10/5=2hours so total time=10+2=12 hours

Choice (B)

20.Equation of the line parallel to x-axis and 3 units below x-axis is?

(A)y=3(B)y= –3

(C)x=3(D)(x= –3

Sol: Equation is y= –3Choice (B)

21.Given A is 50% larger than C and B is 25% larger than C, then A is what percent larger than B?

(A) 50%(B) 75%

Sol: Take C as 100 so A is 150 and B is 125 so required percent is

Choice(C)

22.If ABCD be a rectangle and P, Q, R, S be the mid-points of AB, BC, CD and DA respectively, then the area of PQRS is ?

(A)1/3(AreaABCD)(B)3/4(AreaABCD)

Sol: Area formed by joining midpoints of rectangle is a rhombus and its area is always half of the area of rectangle. Choice (C)

23.A can finish the work in 18 days and B can finish the work in 15 days. B worked for 10 days and then left. Now In how many days A alone can finish the remaining work?

(A)5.5(B)5

Sol: As per question , so x=6

Choice (D)

24.The perimeter of the base of the cone is 8cm. If the height of the cone is 21cm, then its volume is?

(A) 112/п cm2(B) 112п cm2

Sol: 2пr=8 so r=4/п now volume=Choice (A)

25.If 10 men or 20 women or 40 children can do a piece of work in 7 months, then 5 men, 5 women and 5 children can do half of the work in how many months?

(A) 4(B)5

Sol:so whole work can be done in 8 months so half can be done in 4 months Choice (A)

26.If be a perfect square, then the values of t are?

(A) 1,2(B)2,3

(C)±1(D) ±2

Sol: This eqn. becomes a whole square when r=2 and t=±1 moreover eqn. will be (n±1/2)2

Choice(C)

27.The angle of elevation of a tower from a distance 100m from its foot is 30˚. Height of the tower is:

(A) 50√3m(B)200/√3m

Sol: tan30˚=h/100 so h=100tan30˚=100/√3Choice (D)

28.In a triangle ABC, AB=AC, angle BAC=40˚. Then the external angle at B is?

(A)70˚(B)110˚

Sol: Angle BAC=40 so other two angles will be equal and of 70˚ each as triangle is isosceles so external angle will be 180˚–70˚=110˚

Choice (B)

29.The value of sin225+sin265˚ is?

(A)1(B) 0

Sol: Sin265=cos225 so sin2θ+cos2θ=1

Choice (A)

30.The value of cos1˚cos2˚cos3˚……….cos177˚cos178˚cos179˚?

(A) 1/2(B) 1

Sol: This expression will contain cos90˚ which is 0 so anything multiplied by 0 will be 0Choice (D)

31.A chord of length 30cm is at a distance of 8cm from the centre of a circle then the radius of the circle is?

(A) 23(B) 21

Sol:Applying Pythagoras theorem

(30/2)2+82=r2so r=17cm

Choice (D)

32.AB and CD are two parallel chords of a circle such that AB=10cm and CD=24cm. If the chords are on the opposite sides of the centre and distance between them is 17cm, then the radius of the circle is?

(A) 12cm(B) 13cm

Sol: Let the perpendicular distance from centre to 24cm chord be x then perpendicular distance from centre to 10cm will be 17–x. So eqn. will be (24/2)2+x2=r2 and (10/2)2+(17–x)2=r2 equating these two equations we get x=13cm Choice (B)

33.The ratio of inradius and circumradius of a square is?

(A) √2:√3(B) 1:3

Sol: Let the square be of a cm then inradius will be half of the side which is a/2 and circumradius would be half of diagnol which is √2a/2 or a/√2 so ratio becomes 1:√2 Choice (D)

34.If ∆ABC is similar to ∆DEF such that BC=3cm,EF=4cm and area of ∆ABC=54cm2then the area of ∆DEF is?

(A)78cm2(B)96cm2

Sol:

SoArea DEF=96cm2

Choice (B)

35.Number of digits in the square root of 62478078?

(A) 5(B)6

Sol: Digits are find out by making them in pairs of two starting with the last two digits. Number of pairs will be the number of digits in the square root if no number is left unpaired otherwise it will be one more than number of pairs eg. 62 47 80 78 so here 4 pairs are present.

Choice (D)

36.The time in which Rs.80000 amounts to Rs.92610 at 10% p.a. compound interest, interest being compounded annually is?

(A) 2 years(B) 2.5 years

Sol:

So n=3 hence there are 3 cycles of 6 months which means 1.5 yearsChoice (D)

37.P and Q are two points on a circle with centre O. R is a point on the minor arc of the circle, between the points P and Q. The tangents to the circle at the points P and Q meet each other at the points S. If angle PSQ=20˚, Find angle PRQ?

(A) 200˚(B) 160˚

Sol:Angle PSQ=20˚ and OPS=OQS=90˚ so in quadrilateral OPSQ angle OPQ=360–(10+90+90)=160˚ now angle at the point on the circle is half of the angle substended at the centre so angle PRQ=160/2=80˚ Choice (D)

38.A man undertakes to do a certain work in 150 days. He employs 200 men. He finds that only a quarter of the work is done in 50 days then find the number of additional men he need to employ in order to complete the work in time?

(A) 100(B) 125

Sol: so M2=300 so 100 additional men are required.

Choice (A)

39.The value of the machine depreciates 10% every years then find its worth after 2 years if its present value is Rs.50000?

(A) Rs.45000(B) Rs.40005

Sol:Choice (C)

40.The average of 50 numbers is 38. If two numbers namely 45 and 55 are discarded then find the new average?

(A) 32.5(B) 37.5

Sol:Let x be the change in average then

so x= –0.5 Hence new average will be 37.5 This formula is the exclusive copyright of the author. Anyone is free to discuss its numerous applications which makes complex calculations very much easier

Choice (B)

41.A train moving at 36kmph crosses a man in 10 seconds then how long will it take to cross a platform of 55m?

(A)7sec(B)15.5sec

Sol: 36kmph×5/18=10m/s so length of train is 10×10=100m now while crossing platform train has to cover its length and also the length of the platform so time taken will be 155/10=15.5sec

Choice (B)

42.If then the value of x3+1/x3 is?

(A) 110(B) 125

Sol: Dividing the given eqn. by x we get

so x+1/x=5 now cubing both the sides we get x3+1/x3=110

Choice (A)

43.If sinθ+cosecθ=2 then the value of sin9θ+cosec9θ?

(A) 2(B)4

Sol: This eqn. is satisfied only at θ=90˚ so19+19=1+1=2Choice (A)

Directions for 44 to 46 : Pie chart shows the distribution of number of students admitted in different faculties.(in degree)

44.If 1000 students are admitted in science then what is the ratio of students in science and arts?

(A) 7:5(B) 7:6

Sol: Ratio does not need number of students it can even be calculated on the basis of given degrees which is 100:120=5:6 Choice (C)

45.If 1000 students are admitted in science then what is the total ratio of students?

(A) 1800(B) 3600

Sol: Choice (B)

46.If 1000 students are admitted in science then how many students are more in commerce than in law?

(A) 2000(B) 5000

Sol:Choice(D)

Directions for 47 to 50 : Study the following pie charts and answer the questions that follows:

April Month’s salary is Rs.24000. A-Education, B-Savings, C-Grocery, D-Electricity and Phone Bills and E-Miscellaneous

May Month’s salary is Rs.25000 and following is the distribution

47.What is the percent increase in Education in May month than April month?

(A) 12.35%(B) 20%

Sol: April month’s expenses on education are 24000×.47=11280Rs. May month’s expenses on education are 50% of 25000 which is 12500. So increase is Choice(C)

48.The ratio of amount spent for savings in April and miscellaneous of May month?

(A) 217:26(B) 205:13

Sol: Savings in April are 24000×.18=4320 and Miscellaneous in May are 25000×.02=500 so ratio is 4320:500=216:25

Choice(D)

49.From the salary of May, the amount spent on Grocery and Electricity are?

(A) Rs.960,Rs.5040(B)Rs.3500,Rs.2250

(C)Rs.2160,Rs.480(D)Rs.6250,Rs.3360

Sol: Amount on grocery is 25000×.14=3500 so we don’t need to calculate on electricity as only 1 option is there containing grocery Rs.3500 Choice(B)

50.The average amount spent on Education, Grocery and Savings from April month’s salary is?

(A)Rs.6000(B)Rs.6325

(C)Rs.5520(D)Rs.5800

Sol: (47+18+4)/3=23% and 23% of 24000 is Rs.5520Choice(C)

ANSWER KEYS

1. / (B) / 11. / (C) / 21. / (C) / 31. / (D) / 41. / (B)
2. / (B) / 12. / (D) / 22. / (C) / 32. / (B) / 42. / (A)
3. / (B) / 13. / (C) / 23. / (D) / 33. / (D) / 43. / (A)
4. / (D) / 14. / (D) / 24. / (A) / 34. / (B) / 44. / (C)
5. / (A) / 15. / (C) / 25. / (A) / 35. / (D) / 45. / (B)
6. / (B) / 16. / (B) / 26. / (C) / 36. / (D) / 46. / (D)
7. / (D) / 17. / (A) / 27. / (D) / 37. / (D) / 47. / (C)
8. / (A) / 18. / (A) / 28. / (B) / 38. / (A) / 48. / (D)
9. / (D) / 19. / (B) / 29. / (A) / 39. / (C) / 49. / (B)
10. / (A) / 20. / (B) / 30. / (D) / 40. / (B) / 50. / (C)

For SSC Mains(Maths) at Jaipur you can feel free to contact me. Regards Narendra Shekhawat 7790921257

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