Direction: Read the question and answer accordingly.
1. If we write down all the natural numbers from 259 to 492 side by side get a very large natural number 259260261....491492. How many 8(s) will be used to write this large natural number?

A. 32 B. 43 C. 52 D. 53

Answer: Option D
Explanation: Total 8(s)From 259 to 458 = 40 Total 8(s) From 459 to 492 = 13 Totl 8(s) = 40 + 13 = 53
2. The LCM of two numbers is 280 and their ratio is 7:8. The two numbers are

A. 70, 80 B. 35, 40 C. 42, 48 D. 28,32

Answer: Option B
Explanation: Let the number be 7x and 8x HCF = x Hence, HCF × LCM = 7x × 8x ⇒ 280x = 56x × x ⇒ x = 5 Thus numbers are 35 and 40.
3. If each of the three nonzero numbers a, b and c is divisible by 3, then abc must be divisible by which one of the following the numbers?

A. 8 B. 27 C. 81 D. 121

Answer: Option B
Explanation: Since each one of the three numbers a, b, and c is divisible by 3, the numbers can be represented as 3p,3q and 3r, respectively, where p,q and r are integers. The product of the three numbers is 3p × 3q ×3r = 27(pqr). Since p, q and r are integers, pqr is an integer and therefore abc is divisible by 27.
4. A girl wrote all the numbers from 100 to 200. Then she started counting the number of one's that has been used while writing all these numbers. What is the number that she got?

A. 111 B. 119 C. 120 D. 121

Answer: Option C
Explanation: From 100 to 200 there are 101 numbers. There are 100, 1's in the hundreds place. 10, 1's in tens place and10, 1's in-unit place. Thus, the answer is 100 + 10 + 10 = 120.
5. Which of the following has most number of divisors?

A. 90 B. 101 C. 176 D. 182

Answer: Option C
Explanation: Divisors of 99 = 1, 3, 9, 11, 33, 99 Divisors of 101 = 1, 101 Divisors of 176 = 1, 2, 4, 8, 11, 22, 44, 88, 176 Divisors of 182 = 1, 2, 7, 13, 14, 26, 91, 182

1. The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight of B is:

A. 17 kg B. 20 kg C. 26 kg D. 31 kg

Answer: Option D
Explanation: Let A, B, C represent their respective weights. Then, we have: A + B + C = (45 x 3) = 135 .... (i) A + B = (40 x 2) = 80 .... (ii) B + C = (43 x 2) = 86 ....(iii) Adding (ii) and (iii), we get: A + 2B + C = 166 .... (iv) Subtracting (i) from (iv), we get : B = 31.
2. The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class.

A. 47.55 kg B. 48 kg C. 48.55 kg D. 49.25 kg

Answer: Option C
Explanation: Required average = [latex]\frac{50.25 × 16 + 45.15 × 8}{16 + 8}[/latex] = [latex]\frac{804 + 361.20}{24}[/latex] = [latex]\frac{1165.20}{24}[/latex] = 48.55
3. Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get?

A. 57% B. 60% C. 65% D. 90%

Answer: Option A
Explanation: Total number of votes polled = (1136 + 7636 + 11628) = 20400. Required percentage = [[latex]\frac{11628 }{20400}[/latex] x 100]% = 57%
4. A driver of a auto rickshaw sees a lorry 60 meters ahead of him. After 30 seconds the lorry is 90 meters behind. If the speed of the auto rickshaw is 38 kmph, then what is the speed of the lorry?

A. 23 kmph B. 25 kmph C. 20 kmph D. 18 kmph

Answer: Option C
Explanation: Relative speed = [latex]\frac{Total distance }{Total time}[/latex] = [latex]\frac{60 + 90 }{30}[/latex] = [latex]\frac{60 + 90 }{30}[/latex] = 5 m/s 5 m/s = 5 [latex]\frac{18 }{5}[/latex] = 18 kmph Now relative speed = speed of auto rickshaw – speed of lorry or, 18 = 38 – the speed of the lorry ∴ Speed of lorry = 38 – 18 = 20 km/h
5. The hourly wages of a mason have increased by 25%. Since the increase, the number of hours he works daily has reduced by 16%. If he was earning Rs. 120 per day before the increase, how much (in Rs.) is he earning now?

A. 124.5 B. 115.5 C. 126 D. 120

Answer: Option C
Explanation: Daily wages = hourly wages × work hours. Let the original hourly wages and work hours be Rs. x and y hours respectively. Since he used to earn Rs. 120 earlier, xy = 120 New hourly wages = Rs. (1.25x) and new working hours = 0.84y ∴ New daily wages = (1.25x)(0.84y) = 1.05xy = 1.05 × 120 = Rs. 126

1. A lawn is in the form of a rectangle having its sides in the ratio 2:3 The area of the lawn is 1/6 hectares. Find the length and breadth of the lawn.

A. 10 m B. 20 m C. 30 m D. 50 m

Answer: Option D
Explanation: Let length = 2x meters and the breadth = 3x meters. Now, area = ([latex]\frac{1 }{6}[/latex] x 100 )[latex]{m}^{2}[/latex] = [latex]\frac{5000 }{3}[/latex] [latex]{m}^{2}[/latex] So, 2x × 3x = [latex]\frac{5000 }{3}[/latex] ⇔ [latex]{x}^{2}[/latex] = [latex]\frac{2500 }{9}[/latex] ⇔ x = [latex]\frac{50 }{3}[/latex] So, Length = 2x = [latex]\frac{100 }{3}[/latex] m = 33 [latex]\frac{1 }{3}[/latex] and Breadth = 3x = ( 3 x [latex]\frac{50 }{3}[/latex] ) = 50m
2. The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq. cm. find the length of the rectangle.

A. 5 cm B. 10 cm C. 20 cm D. 40 cm

Answer: Option D
Explanation: Let breadth = x. then, length = 2x. then, (2x - 5) (x +5) – 2x × x = 75 ⇒ 5x – 25 = 75 ⇔ x = 20 So, the length of the rectangle = 2x Therefore, 2 × 20 cm ⇒ 40 cm.
3. Anmol had 10 paise, 25 paise and 50 paise coins in the ratio of 10 : 8 : 9 respectively. After giving Rs. 20 his mother he has Rs. 40. How many 50 paise coins did he have?

A. 72 B. 60 C. 54 D. 35

Answer: Option A
Explanation: Total money = 20 + 40 = 60 In this question, the total money is given so we need to multiply the value of the coins in the ratio. 10 × 0.10 x + 8 × 0.25x + 9 × 0.50x = 60 x + 2x + 4.5x = 60 7.5 x = 60 x = 8 Number of 50 paise coins = 9 × 8 = 72
4. Out of three positive numbers, the ratio of the first and the second numbers is 3 : 4 that of the second and the third numbers is 5 : 6 if the product of the second and the third numbers is 4320. What is the sum of three numbers?

A. 177 B. 165 C. 185 D. 160

Answer: Option A
Explanation: Ratio of 1st and 2nd numbers = 3 : 4 The ratio of [latex]{2}^{nd}[/latex] and [latex]{3}^{rd}[/latex] numbers = 5: 6 Let the 2nd number = 5x, third number = 6x Product of 2nd and 3rd numbers = 4320 5x × 6x = 4320 [latex]{x}^{2}[/latex] = 144 x = 12 [latex]{2}^{nd}[/latex] number = 60, 3rd number = 72, [latex]{1}^{st}[/latex] number = [latex]\frac{60 }{4}[/latex] x 3 = 45 Sum of three numbers = 60 + 72 + 45 = 177
5. The cost price of 2 shirts and 3 jeans is Rs. 2200 and the cost price of 2 jeans and 4 shirts is Rs. 2400. Find the ratio between the cost price of the jeans and the shirt.

A. 8 : 9 B. 10 : 7 C. 6 : 5 D. 11 : 10

Answer: Option B
Explanation: Let the cost price of one shirt = x Rs., the cost price of one jeans = y Rs. According to the question, 2x + 3y = 2200 ....1 2y + 4x = 2400 2x + y = 1200 ....2 After solving these 2 equation, x = 350 Rs. , y = 500 Rs. Ratio = 500 : 350 = 10 : 7