Number System
1. Which of the following number is divisible by 24?

A. 35718 B. 63810 C. 537804 D. 3125736

Answer: Option (D)
Explanation: 24 = 3 x 8, where 3 and 8 co-prime. Clearly, 35718 is not divisible by 8, as 718 is not divisible by 8. Similarly, 63810 is not divisible by 8 and 537804 is not divisible by 8. Consider option (D), Sum of digits = (3 + 1 + 2 + 5 + 7 + 3 + 6) = 27, which is divisible by 3. Also, 736 is divisible by 8. 3125736 is divisible by (3 x 8), i.e., 24.
2. (?) + 3699 + 1985 - 2047 = 31111

A. 34748 B. 27474 C. 30154 D. 27574 E. None of these

Answer: Option (B)
Explanation: x + 3699 + 1985 - 2047 = 31111 x + 3699 + 1985 = 31111 + 2047 x + 5684 = 33158 x = 33158 - 5684 = 27474.
3. If the number 481 * 673 is completely divisible by 9, then the smallest whole number in place of * will be:

A. 2 B. 5 C. 6 D. 7 E. None of these

Answer: Option (D)
Explanation: Sum of digits = (4 + 8 + 1 + x + 6 + 7 + 3) = (29 + x), which must be divisible by 9. x = 7.
4. The difference between the local value and the face value of 7 in the numeral 32675149 is

A. 75142 B. 64851 C. 5149 D. 69993 E. None of these

Answer: Option (D)
Explanation: (Local value of 7) - (Face value of 7) = (70000 - 7) = 69993.
5. How many natural numbers are there between 23 and 100 which are exactly divisible by 6?

A. 8 B. 11 C. 12 D. 13 E. None of these

Answer: Option (D)
Explanation: Required numbers are 24, 30, 36, 42, ..., 96 This is an A.P. in which a = 24, d = 6 and l = 96 Let the number of terms in it be n. Then tn = 96 a + (n - 1)d = 96 24 + (n - 1) x 6 = 96 (n - 1) x 6 = 72 (n - 1) = 12 n = 13 Required number of numbers = 13.
Decimals
1. When .36 is written in simplest fractions form, the sum of the numerator and the denominator is:

A. 15 B. 30 C. 45 D. 60

Answer: Option (A)
2. 337.62+8.591+34.4 =?

A. 371.722 B. 391.622 C. 380.611 D. 463.94

Answer: Option (C)
3. 34.95+240.016+23.98=?

A. 298.1057 B. 298.222 C. 298.946 D. 299.09

Answer: Option (C)
4. 3889 + 12.952 –? = 3854.002

A. 47.015 B. 47.641 C. 47.943 D. 47.95

Answer: Option (D)
5. 16.02 X 0.001 =?

A. 0.00 B. 0.01602 C. 0.16020 D. 0.001602

Answer: Option (B)
Fractions
1. What is [latex]\frac{4}{20}[/latex] in its simplest form?

A. [latex]\frac{1}{4}[/latex] B. [latex]\frac{1}{5}[/latex] C. [latex] \frac{2}{10}[/latex] D. [latex] \frac{3}{10}[/latex]

Answer: Option (B)
2. What is [latex]{1}\frac{3}{4}[/latex] as an improper fraction?

A. [latex]\frac{7}{4}[/latex] B. [latex]\frac{8}{4}[/latex] C. [latex]\frac{13}{4}[/latex] D. [latex] \frac{2}{10}[/latex]

Answer: Option (A)
3. What is [latex]\frac{15}{4}[/latex] as a mixed number?

A. [latex]{3}\frac{1}{4}[/latex] B. [latex]{3}\frac{1}{2}[/latex] C. [latex]{3}\frac{3}{4}[/latex] D. [latex] \frac{2}{10}[/latex]

Answer: Option (C)
4. Which is the biggest?

A. [latex]\frac{4}{9}[/latex] B. [latex]\frac{1}{3}[/latex] C. [latex]\frac{4}{7}[/latex] D. [latex] \frac{2}{10}[/latex]

Answer: Option (C)
5. [latex]\frac{3}{4}+\frac{1}{8}[/latex]=?

A. [latex]\frac{7}{8}[/latex] B. [latex] \frac{4}{12}[/latex] C. [latex] \frac{1}{2}[/latex] D. [latex] \frac{2}{10}[/latex]

Answer: Option (A)
LCM
1. The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

A. 9000 B. 9400 C. 9600 D. 9800

Answer: Option (C)
Explanation: The greatest number of 4-digits is 9999. L.C.M. of 15, 25, 40 and 75 is 600. On dividing 9999 by 600, the remainder is 399. Required number (9999 - 399) = 9600.
2. The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

A. 74 B. 94 C. 184 D. 364

Answer: Option (D)
Explanation: L.C.M. of 6, 9, 15 and 18 is 90. Let the required number be 90k + 4, which is multiple of 7. Least value of k for which (90k + 4) is divisible by 7 is k = 4. Required number = (90 x 4) + 4 = 364.
3. Find the lowest common multiple of 24, 36 and 40.

A. 120 B. 240 C. 360 D. 480

Answer: Option (C)
Explanation: 2 | 24 - 36 - 40 -------------------- 2 | 12 - 18 - 20 -------------------- 2 | 6 - 9 - 10 ------------------- 3 | 3 - 9 - 5 ------------------- | 1 - 3 - 5 L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360.
4. The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:

A. 3 B. 13 C. 23 D. 33

Answer: Option (C)
Explanation: L.C.M. of 5, 6, 4 and 3 = 60. On dividing 2497 by 60, the remainder is 37. Number to be added = (60 - 37) = 23.
5. 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.
HCF
1. 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.
2. The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

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

Answer: Option (B)
Explanation: Let the numbers 13a and 13b. Then, 13a x 13b = 2028 ab = 12. Now, the co-primes with product 12 are (1, 12) and (3, 4). [Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ] So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4). Clearly, there are 2 such pairs.
3. The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is:

A. 15 cm B. 25 cm C. 35 cm D. 42 cm

Answer: Option (C)
Explanation: Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm.
4. Find the highest common factor of 36 and 84.

A. 4 B. 6 C. 12 D. 18

Answer: Option (C)
Explanation: 36 = 2[latex]^{2}[/latex] x 3[latex]^{2}[/latex] 84 = 2[latex]^{2}[/latex] x 3 x 7 H.C.F. = 2[latex]^{2}[/latex] x 3 = 12.
5. The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:

A. 504 B. 536 C. 544 D. 548

Answer: Option (D)
Explanation: Required number = (L.C.M. of 12, 15, 20, 54) + 8 = 540 + 8 = 548.

Ratio and Proportions
1. A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1000 more than D, what is B's share?

A. Rs. 500 B. Rs. 1500 C. Rs. 2000 D. None of these

Answer: Option (C)
Explanation: Let the shares of A, B, C, and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively. Then, 4x - 3x = 1000 x = 1000. B's share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000.
2. Salaries of Ravi and Sumit are in the ratio 2 : 3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40: 57. What is Sumit's salary?

A. Rs. 17,000 B. Rs. 20,000 C. Rs. 25,500 D. Rs. 38,000

Answer: Option (D)
Explanation: Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively. Then,[latex]\frac {2x + 4000}{3x + 4000}[/latex] = [latex]\frac {40}{57}[/latex] 57(2x + 4000) = 40(3x + 4000) 6x = 68,000 3x = 34,000 Sumit's present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000.
3. The fourth proportional to 5, 8, 15 is:

A. 18 B. 24 C. 19 D. 20

Answer: Option (B)
Explanation: Let the fourth proportional to 5, 8, 15 be x. Then, 5: 8: 15: x 5x = (8 x 15) x = [latex]\frac {(8 x 15)}{5}[/latex] = 24.
4. Two number are in the ratio 3: 5. If 9 is subtracted from each, the new numbers are in the ratio 12: 23. The smaller number is:

A. 27 B. 33 C. 49 D. 55

Answer: Option (B)
Explanation: Let the numbers be 3x and 5x. Then, [latex]\frac {3x - 9}{5x - 9}[/latex] = [latex]\frac {12}{23}[/latex] 23(3x - 9) = 12(5x - 9) 9x = 99 x = 11. The smaller number = (3 x 11) = 33.
5. The ratio of the number of boys and girls in a college is 7: 8. If the percentage increase in the number of boys and girls be 20% and 10% respectively, what will be the new ratio?

A. 8: 9 B. 17: 18 C. 21: 22 D. Cannot be determined

Answer: Option (C)
Explanation: Originally, let the number of boys and girls in the college be 7x and 8x respectively. Their increased number is (120% of 7x) and (110% of 8x). [latex]\frac {120}{100}[/latex] x 7x and [latex]\frac {110}{100}[/latex] x 8x [latex]\frac {42x}{5}[/latex] and [latex]\frac {44x}{5}[/latex] The required ratio = [latex]\frac {42x}{5}[/latex] : [latex]\frac {44x}{5}[/latex] = 21 : 22.
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: 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. In ABC College, 63% of students are less than 20 years of age. The number of students more than 20 years of age is 2/3 of a number of students of 20 years of age which is 42. What is the total number of students in ABC College?

A. 75 B. 90 C. 130 D. 200

Answer: Option (D)
Explanation: Let the total number of students be x. Then, Number of students more than 20 years of age = (100 - 63)% of x = 37% of x. 37% of x = 42 + 2/3 of 48 37/100x = 74 x = 200
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. The length of a rectangular plot is thrice its breadth. If the area of the rectangular plot is 867 sq m, then what is the breadth of the rectangular plot?

A. 8.5 m B. 17 m C. 34 m D. 51 m E. None of these

Answer: Option (B)
Explanation: Let the breadth of the plot be b m. Length of the plot = 3 b m (3b)(b) = 867 3b[latex]^{2}[/latex] = 867 b[latex]^{2}[/latex] = 289 = 17[latex]^{2}[/latex] (b > 0) b = 17 m.
2. The length of a rectangular floor is more than its breadth by 200%. If Rs. 324 is required to paint the floor at the rate of Rs. 3 per sq m, then what would be the length of the floor?

A. 27 m B. 24 m C. 18 m D. 21 m E. None of these

Answer: Option (C)
Explanation: Let the length and the breadth of the floor be l m and b m respectively. l = b + 200% of b = l + 2b = 3b Area of the floor = 324/3 = 108 sq m l b = 108 i.e., l * l/3 = 108 l[latex]^{2}[/latex] = 324 => l = 18.
3. An order was placed for the supply of a carpet whose breadth was 6 m and length was 1.44 times the breadth. What be 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 sq m?

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
4. 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.
5. 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.
Time and Work
1. A can do a piece of work in 10 days, and B can do the same work in 20 days. With the help of C, they finished the work in 4 days. C can do the work in how many days, working alone?

A. 5 days B. 10 days C. 15 days D. 20 days E. 25 days

Answer: Option (A)
Explanation: Their combined 4 day work = 4(1/10 + 1/15) = 12/20 = 3/5. Remaining work = 1 – 3/5 =2/5. This means C did 2/5 work in 4 days, hence he can finish the complete work in 5/2 × 4 = 10 days.
2. A can do a piece of work in 12 days. B can do this work in 16 days. A started the 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.
3. A and B can do a piece of work in 40 days, B and C can do it in 120 days. If B alone can do it in 180 days, in how many days will A and C do it together?

A. 45 days B. 22.5 days C. 25 days D. 18 days E. 12 days

Answer: Option (A)
Explanation: A + B take 40 days. B alone takes 180 days. ∴ A will take 1/40 – 1/180 = 7/360 ⇒ 360/7 days. B + C take 120 days. ∴ C alone will take 1/120 – 1/180 = 1/360 i.e. 360 days. ∴ A & C together will take 7/360 + 1/360 = 8/360 ⇒ 360/8 = 45 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 the work after 7 days.
Time and Distance
1. 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
2. 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
3. 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.
4. If a train, travelling at a speed of 90 kmph, crosses a pole in 5 sec, then the length of train is?

A. 104 m B. 125 m C. 140 m D. 152 m

Answer: Option (B)
Explanation: D = 90 * 5/18 * 5 = 125 m
5. How many seconds will a train 100 meters long take to cross a bridge 150 meters long if the speed of the train is 36 kmph?

A. 18 B. 22 C. 25 D. 28

Answer: Option (C)
Explanation: D = 100 + 150 = 250 S = 36 * 5/18 = 10 mps T = 250/10 = 25 sec

Simple and Compound Interest
1. A sum of Rs. 25000 becomes Rs. 27250 at the end of 3 years when calculated at simple interest. Find the rate of interest.

A. 5% B. 6% C. 3% D. 4%

Answer: Option (C)
Explanation: Simple interest = 27250 – 25000 = 2250 Time = 3 years. SI = PTR / 100 → R = SI * 100 / PT R = 2250 * 100 / 25000 * 3 → R = 3%.
2. Find the present worth of Rs. 78000 due in 4 years at 5% interest per year.

A. 50000 B. 60000 C. 55000 D. 65000

Answer: Option (D)
Explanation: Amount with interest after 4 years = Rs. 78000 Therefore, simple interest = 78000 – Principal. Let the principal amount be p. 78000 – p = p*4*5/100 → p=13000 Principal = 78000 – 13000 = Rs. 65000.
3. A certain principal amounts to Rs. 15000 in 2.5 years and to Rs. 16500 in 4 years at the same rate of interest. Find the rate of interest.

A. 10% B. 12% C. 8% D. 10%

Answer: Option (C)
Explanation: Amount becomes 15000 in 2.5 years and 16500 in 4 years. Simple interest for (4-2.5) years = 16500 – 15000 Therefore, SI for 1.5 years = Rs. 1500. SI for 2.5 years = 1500/1.5 * 2.5 = 2500 Principal amount = 15000 – 2500 = Rs. 12500. Rate of Interest = 2500 * 100 / 12500 * 2.5 → R = 8%.
4. How long will it take a certain amount to increase by 30% at the rate of 15% simple interest?

A. 2% B. 3% C. 5% D. 6%

Answer: Option (A)
Explanation: Let the principal be Rs. x Simple interest = x*30/100 = 3x/10 T = 100*SI/PR = 100*3x/10 / x*15 = 2% Alternatively, this can be solved by considering principal amount to be Rs. 100. Then simple interest becomes Rs. 30. Then, T = 100*30/100*15 = 2%
5. A money lender lent Rs. 1000 at 3% per year and Rs. 1400 at 5% per year. The amount should be returned to him when the total interest comes to Rs. 350. Find the number of years.

A. 3.5 B. 3.75 C. 4 D. 4.5

Answer: Option (A)
Explanation: (1000*t*3/100) + (1400*t*5/100) = 350 → t =3.5
Profit and Loss
1. A trader mixes 26 kg of rice at Rs. 20 per kg with 30 kg of rice of other variety at Rs. 36 per kg and sells the mixture at Rs. 30 per kg. His profit percent is:

A. No profit, no loss B. 5% C. 8% D. 10%

Answer: Option (B)
Explanation: C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600. S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680. Gain =(80/1600*100) % = 5%.
2. If selling price is doubled, the profit triples. Find the profit percent?

A. 100% B. 200% C. 300% D. 400%

Answer: Option (A)
Explanation: Let the C.P be Rs.100 and S.P be Rs.x, Then The profit is (x-100) Now the S.P is doubled, then the new S.P is 2x New profit is (2x-100) Now as per the given condition; => 3(x-100) = 2x-100 By solving, we get x = 200 Then the Profit percent = (200-100)/100 = 100 Hence the profit percentage is 100%
3. If books bought at prices ranging from Rs. 200 to Rs. 350 are sold at prices ranging from Rs. 300 to Rs. 425, what is the greatest possible profit that might be made in selling eight books?

A. 600 B. 1200 C. 1800 D. none of these

Answer: Option (C)
Explanation: Least Cost Price = Rs. (200 * 8) = Rs. 1600. Greatest Selling Price = Rs. (425 * 8) = Rs. 3400. Required profit = Rs. (3400 - 1600) = Rs. 1800.
4. If the cost price is 25% of selling price. Then what is the profit percent.

A. 150% B. 200% C. 300% D. 350%

Answer: Option (C)
Explanation: Let the S.P = 100 then C.P. = 25 Profit = 75 Profit% = (75/25) * 100 = 300%
5. A man buys oranges at Rs 5 a dozen and an equal number at Rs 4 a dozen. He sells them at Rs 5.50 a dozen and makes a profit of Rs 50. How many oranges does he buy?

A. 30 dozens B. 40 dozens C. 50 dozens D. 60 dozens

Answer: Option (C)
Explanation: Cost Price of 2 dozen oranges Rs. (5 + 4) = Rs. 9. Sell price of 2 dozen oranges = Rs. 11. If profit is Rs 2, oranges bought = 2 dozen. If profit is Rs. 50, oranges bought = (2/2) * 50 dozens = 50 dozens.
Elementary Algebra
1. What is the solution of the equations x - y = 0.9 and 11(x + y)[latex]^{-1}[/latex] = 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
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. 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.
4. If a number is divided by 4, then 3 is subtracted, the result is 0. What is the number?

A. 12 B. 4 C. 3 D. 2

Answer: Option (A)
Explanation: The correct answer is 12, because ( 12 / 4 ) - 3 = 0.
5. 16x - 8 =?

A. 8x B. 8(2x-x) C. 8(2x-1) D. 8(2x-8)

Answer: Option (C)
Explanation: The expression 16x - 8 may be factored as 8(2x - 1).
Geometry and Trigonometry
1. Sine rule for a triangle states that

A. a/sin A = b/sin B = c/sin C B. sin A/a = sin B/b = sin C/c C. a/sin A + b/sin B + c/sin C D. 2a/sin A = 2b/sin B = 2c/sin C

Answer: Option (B)
2. Considering Cosine Rule of any triangle ABC, possible measures of angle A includes

A. angle A is acute B. angle A is obtuse C. angle A is right-angle D. all of above

Answer: Option (D)
3. If cos 55[latex]^{0}[/latex] and sin 55[latex]^{0}[/latex] = 0.8 each then answer of cos 125[latex]^{0}[/latex] + 5 sin 55[latex]^{0}[/latex] is

A. 0.8 B. 2.4 C. 2.8 D. 3.2

Answer: Option (B)
4. How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively 144 cm and 100 cm:

A. 160 B. 240 C. 320 D. 450

Answer: Option (B)
Explanation: Total area of the region =100×144 =14400 cm[latex]^{2}[/latex] Area of one tile =12×5 =60 cm[latex]^{2}[/latex] Number of tiles required =14400/60 =240 Therefore, 240 tiles are required.
5. A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.

A. 1:2 B. 2:1 C. 3:1 D. 3:4

Answer: Option (B)
Explanation: Volume of the cone =πr[latex]^{2}[/latex]h/3 =Volume of a hemisphere =2πr[latex]^{3}[/latex]/3 ⇒ Height of a hemisphere = Radius of its base By the above formula, we can see that h:r = 2:1.
Elementary Statistics etc.
1. Marks obtained by an student in 5 subjects are given below 25, 26, 27, 28, 29 in these obtained marks 27 is

A. mode B. median and mode C. mean and median D. both (A) and (C)

Answer: Option (C)
Explanation: 27 is in the middle, so is the median and its the mean too.
2. A dice is thrown 3 times .what is the probability that atleast one head is obtained?

A. 7/8 B. 6/8 C. 5/8 D. 9/8

Answer: Option (A)
Explanation: Sample space = [HHH, HHT, HTH, THH, TTH, THT, HTT, TTT] Total number of ways = 2 × 2 × 2 = 8. Fav. Cases = 7 P (A) = 7/8 OR P (of getting at least one head) = 1 – P (no head)⇒ 1 – (1/8) = 7/8.
3. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

A. 1/2 B. 3/5 C. 9/20 D. 8/15

Answer: Option (C)
Explanation: Here, S = {1, 2, 3, 4, ...., 19, 20}. Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}. P(E) = n(E)/n(S) = 9/20.
4. Two dice are tossed. The probability that the total score is a prime number is:

A. 5/12 B. 1/6 C. 1/2 D. 7/9

Answer: Option (A)
Explanation: Clearly, n(S) = (6 x 6) = 36. Let E = Event that the sum is a prime number. Then E= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4,3),(5, 2), (5, 6), (6, 1), (6, 5) } n(E) = 15. P(E) = n(E)/n(S) = 15/36 = 5/12.
5. A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?

A. 3/7 B. 4/7 C. 1/8 D. 3/4

Answer: Option (B)
Explanation: Let number of balls = (6 + 8) = 14. Number of white balls = 8. P (drawing a white ball) = 8 /14 = 4/7.