Number System
1. The difference between a two-digit number and the number obtained by interchanging the digits is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2?

A. 4 B. 8 C. 16 D. None of these

Answer: Option (B)
Explanation: Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit. Let ten's and unit's digits are 2x and x respectively. Then, (10 x 2x + x) - (10x + 2x) = 36 9x = 36 x = 4. Required difference = (2x + x) - (2x - x) = 2x = 8.
2. The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number?

A. 3 B. 4 C. 9 D. Cannot be determined E. None of these

Answer: Option (B)
Explanation: Let the ten's digit be x and unit's digit be y. Then, (10x + y) - (10y + x) = 36 9(x - y) = 36 x - y = 4.
3. Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is:

A. 9 B. 11 C. 13 D. 15

Answer: Option (D)
Explanation: Let the three integers be x, x + 2 and x + 4. Then, 3x = 2(x + 4) + 3 x = 11. Third integer = x + 4 = 15.
4. A number consists of two digits. If the digits interchange places and the new number is added to the original number, then the resulting number will be divisible by:

A. 3 B. 5 C. 9 D. 11

Answer: Option (D)
Explanation: Let the ten's digit be x and unit's digit be y. Then, number = 10x + y. The number obtained by interchanging the digits = 10y + x. (10x + y) + (10y + x) = 11(x + y), which is divisible by 11.
5. A number consists of 3 digits whose sum is 10. The middle digit is equal to the sum of the other two and the number will be increased by 99 if its digits are reversed. The number is:

A. 145 B. 253 C. 370 D. 352

Answer: Option (B)
Explanation: Let the middle digit be x. Then, 2x = 10 or x = 5. So, the number is either 253 or 352. Since the number increases on reversing the digits, so the hundred's digits are smaller than the unit's digit. Hence, the required number = 253.
Decimals
1. 3 x 0.3 x 0.03 x 0.003 x 30 =?

A. 0.0000243 B. 0.000243 C. 0.00243 D. 0.0243

Answer: Option (C)
Explanation: 3 x 3 x 3 x 3 x 30 = 2430. Sum of decimal places = 6 Therefore, 3 x 0.3 x 0.03 x 0.003 x 30 = 0.002430 = 0.00243
2. A tailor has 37.5 meters of cloth and he has to make 8 pieces out of a meter of cloth. How many pieces can he make out this cloth?

A. 300 B. 360 C. 400 D. 450

Answer: Option (A)
Explanation: Length of each piece = (1/8) m = 0.125 m Required no of pieces = (37.5/0.125) = (375 * 100) / 125 = 300.
3. The price of commodity X increases by 40 paise every year, while the price of commodity Y increases by 15 paise every year. If in 2001, the price of commodity X was Rs. 4.20 and that of Y was Rs. 6.30, in which year commodity X will cost 40 paise more than the commodity?

A. 2010 B. 2011 C. 2012 D. 2013

Answer: Option (B)
Explanation: Suppose commodity X wil cost 40 paise more than Y after Z years. Then, (4.20 + 0.40Z) - (6.30 + 0.15Z) = 0.40 => 0.25Z = 0.40 + 2.10 => Z= 2.50/0.25 = 10 Therefore, X will cost 40 paise more than Y 10 years after 2001. i.e in 2011.
4. What is the difference between the biggest and the smallest fraction among 2/3, 3/4, 4/5 and 5/6?

A. 1/6 B. 1/12 C. 1/20 D. 1/30

Answer: Option (A)
Explanation: Converting each of the given fractions into decimal form, we get 2/3 = 0.66, 3/4 = 0.75, 4/5 = 0.8, 5/6 = 0.833 Since 0.833>0.8>0.75>0.66 So, 5/6 > 4/5 > 3/4 > 2/3 Therefore, Required Difference = 5/6 - 2/3 = 1/6.
5. 54.327 x 357.2 x 0.0057 is same as

A. 5.4327 x 3.572 x5.7 B. 5.4327 x 3.572 x0.57 C. 54327 x 3572 x 0.0000057 D. None of these

Answer: Option (A)
Explanation: Number of decimal places in the given expression = 8 Number of decimal places in (A) = 8 Number of decimal places in (B) = 9 Number of decimal places in (C)= 7 Clearly, the expression in (A) is the same as the given Expression.
Fractions
1. 0.004 × 0.5 = ?

A. None of these B. 0.02 C. 0.002 D. 0.0002

Answer: Option (C)
Explanation: 0.004 × 0.5 = 0.002
2. How many digits will be there to the right of the decimal point in the product of 89.635 and .02218?

A. 5 B. 6 C. 7 D. 8

Answer: Option (C)
Explanation: Sum of decimal places = 3 + 5 = 8 The last digit in the product(digit at the extreme right) is zero (Since 5 x 8 = 40) Hence, there will be 7 significant digits to the right of the decimal point.
3. 24.39 + 562.093 + 35.96 = ?

A. 622.441 B. 622.243 C. 622.233 D. 622.443

Answer: Option (D)
Explanation: 24.39 + 562.093 + 35.96 = 622.443.
4. What decimal of an hour is a second?

A. 0.00027 B. 0.00025 C. 0.00026 D. 0.00024

Answer: Option (A)
Explanation: 1 hour = 60 minutes = 3600 seconds Hence, required decimal = 1/3600 = 0.00027
5. Which of the following is equal to 0.024 × 10[latex]^{6 }[/latex]?

A. 2400000 B. 240000 C. 24000 D. 2400

Answer: Option (C)
Explanation: 0.024 × 10[latex]^{6 }[/latex] = 0.024 × 1000000 = 24000
LCM
1. Product of two co-prime numbers is 117. Their L.C.M should be

A. 1 B. 117 C. Equal to their H.C.F D. cannot be calculated

Answer: Option (B)
Explanation: H.C.F of co-prime numbers is 1. So, L.C.M = 117/1 =117
2. The ratio of two numbers is 3: 4 and their H.C.F is 4. Their L.C.M is

A. 12 B. 16 C. 24 D. 48

Answer: Option (D)
Explanation: Let the numbers be 3x and 4x. Then their H.C.F = x. So, x=4 Therefore, The numbers are 12 and 16 L.C.M of 12 and 16 = 48
3. L.C.M of two prime numbers x and y (x>y) is 161. The value of 3y-x is :

A. -2 B. -1 C. 1 D. 2

Answer: Option (A)
Explanation: H. C. F of two prime numbers is 1. Product of numbers = 1 x 161 = 161. Let the numbers be a and b. Then, ab= 161. Now, co-primes with product 161 are (1, 161) and (7, 23). Since x and y are prime numbers and x > y, we have x=23 and y=7. Therefore, 3y-x = (3 x 7)-23 = -2
4. The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:

A. 1677 B. 1683 C. 2523 D. 3363

Answer: Option (B)
Explanation: L.C.M. of 5, 6, 7, 8 = 840. The required number is of form 840k + 3 Least value of k for which (840k + 3) is divisible by 9 is k = 2. Required number = (840 x 2 + 3) = 1683.
5. If HCF of two numbers is 8, which of the following can never be their LCM?

A. 32 B. 48 C. 60 D. 152

Answer: Option (C)
Explanation: 60 is not a multiple of 8
HCF
1. The G.C.D of 1.08, 0.36 and 0.9 is

A. 0.03 B. 0.9 C. 0.18 D. 0.108

Answer: Option (C)
Explanation: Given numbers are 1.08, 0.36 and 0.90 H.C.F of 108, 36 and 90 is 18 [ ∵ G.C.D is nothing but H.C.F] Therefore, H.C.F of given numbers = 0.18.
2. Three numbers are in the ratio 1: 2: 3 and their H.C.F is 12. The numbers are

A. 4, 8, 12 B. 5, 10, 15 C. 10, 20, 30 D. 12, 24, 36

Answer: Option (D)
Explanation: Let the required numbers be x, 2x, 3x. Then, their H.C.F =x. so, x= 12 Therefore, The numbers are 12, 24, 36.
3. The maximum number of students among them 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and the same number of pencils is:

A. 91 B. 910 C. 1001 D. 1911

Answer: Option (A)
Explanation: Required number of students = H.C.F of 1001 and 910 = 91.
4. Three number are in the ratio of 3: 4: 5 and their L.C.M. is 2400. Their H.C.F. is:

A. 40 B. 80 C. 120 D. 200

Answer: Option (A)
Explanation: Let the numbers be 3x, 4x and 5x. Then, their L.C.M. = 60x. So, 60x = 2400 or x = 40. The numbers are (3 x 40), (4 x 40) and (5 x 40). Hence, the required H.C.F. = 40.
5. The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:

A. 123 B. 127 C. 235 D. 305

Answer: Option (B)
Explanation: Required number = H.C.F. of (1657 - 6) and (2037 - 5) = H.C.F. of 1651 and 2032 = 127.

Ratio and Proportions
1. If a: b = 3:4, b:c = 7:9, c:d = 5:7, find a:d?

A. 5:12 B. 7:12 C. 3:11 D. 5:11

Answer: Option (A)
Explanation: a/d = (3/4)*(7/9)*(5/7) => 5/12.
2. The ratio of the number of ladies to gents at a party was 1:2 but when 2 ladies and 2 gents left, the ratio became 1:3. How many people were at the party originally?

A. 36 B. 24 C. 12 D. 6

Answer: Option (C)
Explanation: x, 2x (x-2):(2x-2) = 1:3 3x-6 = 2x-2 x = 4 x+2x = 3x => 3*4 = 12
3. A 70 cm long wire is to be cut into two pieces so that one piece will be 2/5th of the other, how many centimeters will the shorter piece be?

A. 10cm B. 20cm C. 25cm D. 30cm

Answer: Option (B)
Explanation: 1: 2/5 = 5: 2 2/7 * 70 = 20
4. 1000 men have provisions for 15 days. If 200 more men join them, for how many days will the provisions last now?

A. 10.5 B. 11.5 C. 12.5 D. 10.4

Answer: Option (C)
Explanation: 1000*15 = 1200*x x = 12.5
5. A garrison of 400 men had a provision for 31 days. After 28 days 280 person re-enforcement leave the garrison. Find the number of days for which the remaining ration will be sufficient?

A. 3 days B. 8 days C. 10 days D. 6 days

Answer: Option (C)
Explanation: 400 --- 31 400 --- 3 120---? 400*3 = 120*x => x =10 days
Percentage
1. Two friends, Akash & Beenu had some candies each. One of them had 15 candies more than the other. The candies with Akash was 60% of the total candies with them. How many candies did each have?

A. 40, 25 B. 47, 32 C. 45, 30 D. 49, 34

Answer: Option (C)
Explanation: Let the candies with be (x + 15) and x. Therefore, x + 15 = 60/100(x + 15 + x) (x + 15) = 3/5(2x + 15) 5x + 75 = 6x + 45 x = 30 So, the marks of two students are 45 and 30.
2. A fruit seller had some oranges. He sells 30% of oranges and still has 140 mangoes. Originally, he had:

A. 288 oranges B. 300 oranges C. 672 oranges D. 200 oranges

Answer: Option (D)
Explanation: Suppose originally he had x oranges. Then, (100 - 30)% of x = 140. 70/100 x = 140 x = (140 x 100)/70 = 200.
3. What percentage of numbers from 1 to 30 has 1 or 9 in the unit's digit?

A. 12 B. 15 C. 20 D. 22

Answer: Option (C)
Explanation: Such numbers from 1 to 30 are 1, 9, 11, 19, 21, 29 Number of such numbers =6 Required percentage is (6/20 * 100) % = 20%
4. If M = y% of z and B = z% of y, then which of the following must be true?

A. M is lesser than N. B. M is more than N C. The relation between M and N cannot be determined. D. None of these

Answer: Option (D)
Explanation: y% of z = (y/100 x z) = (z/100 x y) = z% of y A = B.
5. A student erroneously multiplied a number by 2/5 instead of 5/2. What is the percentage error in the calculation?

A. 24% B. 54% C. 74% D. 84%

Answer: Option (D)
Explanation: Let the number be 100. 2/5 of 100 is 40 while 5/2 of 100 is 250. Now the difference is 210 on a base of 250. Therefore, percentage difference is 210/250 into 100 = 84%.
Mensuration
1. An order was placed for the supply of a carpet whose breadth was 6 m and length was 1.44 times the breadth. What is the cost of a carpet whose length and breadth are 40% more and 25% more respectively than the first carpet? Given that the ratio of carpet is Rs. 45 per sqm?

A. Rs. 3642.40 B. Rs. 3868.80 C. Rs. 4216.20 D. Rs. 4082.40 E. None of these

Answer: Option (D)
Explanation: Length of the first carpet = (1.44)(6) = 8.64 cm Area of the second carpet = 8.64(1 + 40/100) 6 (1 + 25/100) = 51.84(1.4)(5/4) sq m = (12.96)(7) sq m Cost of the second carpet = (45)(12.96 * 7) = 315 (13 - 0.04) = 4095 - 12.6 = Rs. 4082.40.
2. What will be the cost of building a fence around a square plot with an area equal to 289 sq ft, if the price per foot of building the fence is Rs. 58?

A. Rs. 3944 B. Rs. 3828 C. Rs. 4176 D. Cannot be determined E. None of these

Answer: Option (A)
Explanation: Let the side of the square plot be an ft. a[latex]^{2}[/latex] = 289 => a = 17 Length of the fence = Perimeter of the plot = 4a = 68 ft. Cost of building the fence = 68 * 58 = Rs. 3944.
3. The area of a square is equal to five times the area of a rectangle of dimensions 125 cm * 64 cm. What is the perimeter of the square?

A. 600 cm B. 800 cm C. 400 cm D. 1000 cm E. None of these

Answer: Option (B)
Explanation: Area of the square = s * s = 5(125 * 64) => s = 25 * 8 = 200 cm Perimeter of the square = 4 * 200 = 800 cm.
4. The parameter of a square is double the perimeter of a rectangle. The area of the rectangle is 480 sq cm. Find the area of the square.

A. 200 sq cm B. 72 sq cm C. 162 sq cm D. Cannot be determined E. None of these

Answer: Option (D)
Explanation: Let the side of the square be a cm. Let the length and the breadth of the rectangle be l cm and b cm respectively. 4a = 2(l + b) 2a = l + b l . b = 480 We cannot find ( l + b) only with the help of l. b. Therefore a cannot be found. Area of the square cannot be found.
5. A cube of side one-meter length is cut into small cubes of side 10 cm each. How many such small cubes can be obtained?

A. 10 B. 100 C. 1000 D. 10000 E. None of these

Answer: Option (C)
Explanation: Along one edge, the number of small cubes that can be cut = 100/10 = 10 Along each edge, 10 cubes can be cut. (Along length, breadth, and height). Total number of small cubes that can be cut = 10 * 10 * 10 = 1000.
Time and Work
1. A can do a piece of work in 12 days. B can do this work in 16 days. A started work alone. After how many days should B join him, so that the work is finished in 9 days?

A. 2 days B. 3 days C. 4 days D. 5 days E. 1 days

Answer: Option (D)
Explanation: A's work in 9 days = 9/12 = 3/4. Remaining work = 1/4. This work was done by B in 1/4 × 16 = 4 days. ∴ B would have joined A after 9 – 4 = 5 days.
2. A and B can do a piece of work in 4 days, while C and D can do the same work in 12 days. In how many days will A, B, C and D do it together?

A. 12 days B. 4 days C. 3 days D. 2 days E. None of these

Answer: Option (C)
Explanation: A, B, C, and D will together take 1/4 + 1/12 = 4/12 = 1/3 ⇒ 3 days to complete the work.
3. A, B, C, and D can do a piece of work in 20 days. If A and B can do it together in 50 days, and C alone in 60 days, find the time in which D alone can do it.

A. 120 days B. 200 days C. 150 days D. 90 days E. 75 days

Answer: Option (E)
Explanation: D alone will take 1/20 – 1/50 – 1/60 = 4/300 = 1/75 ⇒ 75 days to complete the work.
4. A, B, and C can do a piece of work in 8 days. B and C together do it in 24 days. B alone can do it in 40 days. In what time will it be done by C working alone?

A. 25 days B. 24 days C. 60 days D. 20 days E. 30 days

Answer: Option (C)
Explanation: B & C do this work in 24 days. B alone does this work in 40 days. C alone will take 1/24 – 1/40 = 2/120=1/60 ⇒ 60 days to finish the work.
5. Daku and Tamatar can do a piece of work in 70 and 60 days respectively. They began the work together, but Daku leaves after some days and Tamatar finished the remaining work in 47 days. After how many days did Daku leave?

A. 14 days B. 16 days C. 18 days D. 10 days E. 7 days

Answer: Option (E)
Explanation: Tamatar would have done 47/60 work in 47 days. The remaining work i.e. 13/60 must have been done by Daku and Tamatar together. They can do the whole work in 60 × 70 / (60 + 70) = 60 × 70/130 = 420/13 days. So, they would have done 13/60 work in 420/13 × 13/60 = 7 days. Therefore, Daku left work after 7 days.
Time and Distance
1. A train 240 m in length crosses a telegraph post in 16 seconds. The speed of the train is?

A. 50 kmph B. 52 kmph C. 54 kmph D. 56 kmph

Answer: Option (C)
Explanation: S = 240/16 * 18/5 = 54 kmph
2. The speed of a car is 90 km in the first hour and 60 km in the second hour. What is the average speed of the car?

A. 72 kmph B. 75 kmph C. 30 kmph D. 80 kmph

Answer: Option (B)
Explanation: S = (90 + 60)/2 = 75 kmph
3. Two cars cover the same distance at the speed of 60 and 64 kmps respectively. Find the distance traveled by them if the slower car takes 1 hour more than the faster car.

A. 906 km B. 960 m C. 960 km D. 966 km

Answer: Option (C)
Explanation: 60(x + 1) = 64x X = 15 60 * 16 = 960 km
4. If I walk at 3 kmph, I miss the train by 2 min, if however, I walk at 4 kmph. I reach the station 2 min before the arrival of the train. How far do I walk to reach the station?

A. 4/5 km B. 5/4 km C. 6/5 km D. 3/4 km

Answer: Option (A)
Explanation: x/3 – x/4 = 4/60 x = 4/5 km
5. Two trains each 250 m in length are running on the same parallel lines in opposite directions with the speed of 80 kmph and 70 kmph, respectively. In what time will they cross each other completely?

A. 10 sec B. 11 sec C. 12 sec D. 14 sec

Answer: Option (C)
Explanation: D = 250 m + 250 m = 500 m RS = 80 + 70 = 150 * 5/18 = 125/3 T = 500 * 3/125 = 12 sec

Simple and Compound Interest
1. A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:

A. 650 B. 690 C. 698 D. 700

Answer: Option (C)
Explanation: S.I. for 1 year = Rs. (854 - 815) = Rs. 39. S.I. for 3 years = Rs.(39 x 3) = Rs. 117. Principal = Rs. (815 - 117) = Rs. 698
2. Simple interest on a certain sum is 16/25 of the sum. Find the rate percent and time, If both are numerically equal.

A. Rate = 7% and Time = 7 years. B. Rate = 8% and Time = 8 years. C. Rate = 6% and Time = 6 years. D. Rate = 5% and Time = 5 years.

Answer: Option (B)
Explanation: Let sum = X. Then S.I = 16x/25 Let rate = R% and Time = R years. Therefore, (x * R * R)/100 = 16x/25 => R = 40/5 = 8 Therefore, Rate = 8% and Time = 8 years.
3. A sum was put at simple interest at a certain rate for 10 years. Had it been put at 5% higher rate, it would have fetched Rs.600 more. What was the Sum?

A. Rs.1200 B. Rs.1300 C. Rs.1400 D. Rs.1500

Answer: Option (A)
Explanation: At 5% more rate, the increase in S.I for 10 years = Rs.600 (given) So, at 5% more rate, the increase in SI for 1 year = 600/10 = Rs.60/- i.e. Rs.60 is 5% of the invested sum So, 1% of the invested sum = 60/5 Therefore, the invested sum = 60 × 100/5 = Rs.1200.
4. Simple interest on a sum at 4% per annum for 2 years is Rs.80. The C.I. on the same sum for the same period is?

A. Rs.81.60 B. Rs.160 C. Rs.1081.60 D. Rs.99

Answer: Option (A)
Explanation: SI = 40 + 40 CI = 40 + 40 + 1.6 = 81.6
5. The C.I. on a certain sum for 2 years Rs.41 and the simple interest is Rs.40. What is the rate percent?

A. 4% B. 5% C. 6% D. 8%

Answer: Option (B)
Explanation: SI = 20 + 20 CI = 20 + 21 20 ---- 1 100 ---- ? => 5%
Profit and Loss
1. Two shopkeepers announce the same price of Rs 700 for a sewing machine. The first offers successive discounts of 30% and 6% while the second offers successive discounts of 20% and 16%. The shopkeeper that offers a better discount, charges ……… less than the other shopkeeper.

A. Rs 9.80 B. Rs. 16.80 C. Rs. 22.40 D. Rs. 36.40

Answer: Option (A)
2. The marked price of a watch was Rs 720. A man bought the same for Rs. 550.80 after getting two successive discounts, the first being 10%. What was the second discount rate?

A. 12% B. 14% C. 15% D. 18%

Answer: Option (B)
3. A shopkeeper purchased 150 identical pieces of calculators at the rate of Rs 250 each. He spent an amount of Rs. 2500 on transport and packaging. He fixed the labeled price of each calculator at Rs 320. However, he decided to give a discount of 5% on the labeled price. What is the percentage profit earned by him?

A. 14% B. 15% C. 16% D. 25%

Answer: Option (A)
4. A trader marked the price of his commodity so as to include a profit of 25%. He allowed a discount of 16% on the marked price. His actual profit was:

A. 5% B. 9% C. 16% D. 25%

Answer: Option (A)
5. A tradesman marks his goods 30% above the C.P. If he allows a discount of 6(1/4) %, then his gain percent is:

A. 21(7/8)% B. 22% C. 23(3/4)% D. None of these.

Answer: Option (A)
Elementary Algebra
1. The sum of two numbers is 24 and their product is 143. The sum of their squares is

A. 296 B. 295 C. 290 D. 228

Answer: Option (C)
Explanation: Let the two numbers be P and Q. According to given question, Sum of two numbers = 24 ∴ P + Q = 24 ...................... (1) Product of two numbers = 143 and, PQ = 143 .......................... (2) As we know the formula, ∴ P[latex]^{2}[/latex] + Q[latex]^{2}[/latex] = (P + Q)[latex]^{2}[/latex] – 2PQ Put the value from the equation (1) and (2), We will get ∴ P [latex]^{2}[/latex] + Q[latex]^{2}[/latex] = (24)[latex]^{2}[/latex] – 2 × 143 ∴ P [latex]^{2}[/latex] + Q[latex]^{2}[/latex] = 576 – 286 = 290.
2. The product of two alternate odd integers exceeds three times the smaller by 12. What is the larger integer?

A. 9 B. 7 C. 3 D. 5

Answer: Option (B)
Explanation: Let two alternate odd integers odd integers be (2x+1) and (2x+5). Then according to the question, (2x + 1) (2x + 5) = 3(2x + 1) + 12 ⇒ (2x + 1) (2x + 5 - 3) = 12 ⇒ 2x[latex]^{2}[/latex] + 3x - 5 = 0 On solving this quadratic equation,we get x = 1 and x = -5/2 x = -5/2 is not a integer ∴ x = 1 Then, larger integer = 2x + 5 = 2 x 1 + 5 = 7.
3. What is the solution of the equations x - y = 0.9 and 11(x + y)-1 = 2 ?

A. x = 3.2, y = 2.3 B. x = 1, y = 0.1 C. x = 2, y = 1.1 D. x = 1.2, y = 0.3

Answer: Option (A)
Explanation: x - y = 0.9 ...(i) and 11(x + y)[latex]^{-1}[/latex]=2 ⇒ 11/ (x + y) = 2 ⇒ 2(x + y) =11 ⇒ x + y = 11/2 ...(ii) On solving Eqs.(i) and (ii),we get x = 3.2 and y = 2.3
4. The degree f polynomial p(x) = x[latex]^{3}[/latex] + 1 + 2x = 6x + 1/x is?

A. 2 B. 4 C. 3 D. 5

Answer: Option (B)
Explanation: x[latex]^{3}[/latex] + 1 + 2x =6x + 1/x x[latex]^{3}[/latex] + 1 + 2x = (6x[latex]^{2}[/latex] + 1)/x (x[latex]^{3}[/latex] + 1 + 2x)x = 6x[latex]^{2}[/latex] + 1 x[latex]^{4}[/latex] - 4x[latex]^{2}[/latex] - 6x[latex]^{2}[/latex] -1 =0 x[latex]^{4}[/latex] - 4x[latex]^{2}[/latex] + x - 1 =0 Degree of polynomial is highest exponent degree term i.e.,4.
5. If x2 - x - 6 = 0, then x is

A. -2 or 3 B. -1 or 6 C. 1 or -6 D. 2 or -3

Answer: Option (A)
Explanation: If x 2 - x - 6 = 0, then (x + 2)(x - 3) = 0, so x = -2 or x = 3.
Geometry and Trigonometry
1. ABC is a right angled triangle with a right angle at A. Points D, E are the middle points of AB and AC respectively. Which of the following relations is correct ?

A. 3 (BE2 + CD2) = 4 BC2 B. 4(BE2 + CD2) = 5 BC2 C. 4(BE2 + CD2) = 3 BC2 D. None of these

Answer: Option (B)
2. Two circles whose radii are 10 cm and 8 cm, intersect each other and their common chord is 12 cm long. What is the distance between their centers?

A. 11.27 cm B. 12.29 cm C. 12.27 cm D. 13.29 cm

Answer: Option (D)
3. In D ABC, AB = 6 cms, BC = 10 cms, AC = 8cm and AD ^ BC. Find the value of the ratio of BD: DC.

A. 3: 4 B. 9: 16 C. 4: 5 D. 16: 25

Answer: Option (B)
4. In D ABC, a line parallel to BC intersects AB and AC at D and E. If AE = 3 AD, find the ratio BD: EC.

A. 1: 3 B. 1: 2 C. 2: 3 D. 3: 2

Answer: Option (A)
5. PQRS is a cyclic quadrilateral. The bisectors of the angles ÐP and ÐR meet the circle ABCD at A and B respectively. If the radius of the circle be r units, then AB =?

A. r B. 2r C. 3r D. 4r

Answer: Option (B)
Elementary Statistics etc.
1. If a dice is thrown twice, what is the probability of not getting a one on either throw?

A. 1/12 B. 1/18 C. 25/36 D. 12/36

Answer: Option (C)
Explanation: Out of total 36 possible outcomes, getting 1 on either draw can happen in = 11 ways Reqd. Probability = 1 – (Probability of Getting one on either draw) = 1 - 11 / 36 = 25 / 36.
2. A card is drawn from a pack of 52 cards. What is the probability that the card is a Queen?

A. 1/52 B. 1/4 C. 1/16 D. None of these

Answer: Option (D)
Explanation: Total possible outcomes = 52. Favorable outcomes = 4. Probability = 4 / 52 = 1 / 13.
3. Two cards are drawn in succession from a pack of 52 cards, without replacement. What is the probability, that the first is a Queen and the second is a Jack of a different suit?

A. 1/52 B. 1/13 C. 4/13 D. 1/221

Answer: Option (D)
Explanation: The probability of first Queen = 4 / 52 The probability of Second Jack of different suit = 3 / 51 Reqd. Probability = (4/52) x (3/51) = (1/13) x (1/17) = (1/221)
4. Two dice are thrown, what is the probability that both the dices are not having the same number.

A. 1/4 B. 5/6 C. 1/9 D. 1/12

Answer: Option (B)
Explanation: Total possible outcomes = 36. Favorable outcomes of having same number = 6 [(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)]. Probability of both the dices having same number = 6 / 36 = 1 / 6 Required probability = 1 - 1 / 6 = 5 / 6
5. Determine the probability that a number is chosen at random from the digits 1, 2, 3, ……., 10 will be a multiple of.

A. 1/4 B. 1/3 C. 1/5 D. 1/2

Answer: Option (C)
Explanation: Total possible outcomes = 10. Favorable outcomes = 2. (i.e. 4,8). Probability = 2 / 10 = 1 / 5