Q1. How many numbers from 90 to 261 are divisible by 13?

A. 13 B. 14 C. 15 D. 16

Answer: Option B
Explanation: = [latex]\frac {260 - 91}{13} + 1[/latex] = 14
Q2. How many of the following numbers are divisible by 132 ? 264, 396, 462, 792, 968, 2178, 5184, 6336

A. 4 B. 5 C. 6 D. 7

Answer: Option A
Explanation: 132 = 4 x 3 x 11 So, if the number divisible by all the three number 4, 3 and 11, then the number is divisible by 132 also. 264 → 11,3,4 (/) 396 → 11,3,4 (/) 462 → 11,3 (X) 792 → 11,3,4 (/) 968 → 11,4 (X) 2178 → 11,3 (X) 5184 → 3,4 (X) 6336 → 11,3,4 (/) Therefore the following numbers are divisible by 132 : 264, 396, 792 and 6336. Required number of number = 4.
Q3. If a number is as much as greater than 44 as it is less than 94, then what is that number?

A. 25 B. 50 C. 69 D. 72

Answer: Option C
Explanation: [latex]44 + x = 94 - x [/latex]
[latex]2x = 50[/latex]
[latex]x = 25[/latex]
No. is 44 + 25 = 69
Q4. What is the value of [latex] [({279}^{3}) + ({372}^{3})] ÷ [({279}^{2}) - 279 \times 372 + ({372}^{2})] [/latex]

A. 755 B. 651 C. 572 D. 482

Answer: Option B
Explanation: [latex]\Rightarrow \frac {[(279) + (372)][({279}^{2}) - 279 \times 372 + ({372}^{2})]}{[({279}^{2}) - 279 \times 372 + ({372}^{2})]}[/latex]
[latex]\Rightarrow 279 + 372[/latex]
[latex]\Rightarrow 651 [/latex]
Q5. N is a number divisible by 17. if (N + 2) (N + 7) is divided by 17, then what will be the remainder?

A. 2 B. 16 C. 7 D. 14

Answer: Option D
Explanation: [latex]\Rightarrow \frac {(N + 2)(N + 7)}{17} [/latex]
[latex]\Rightarrow \frac {{N}^{2}}{17} + \frac {9N}{17} + \frac {14}{17}[/latex]
[latex]\Rightarrow [/latex] 14 is remainder

1. Alfred buys an old scooter for Rs. 4700 and spends Rs. 800 on its repairs. If he sells the scooter for Rs. 5800, his gain percent is:

A. 4 [latex]\frac {4}{7} [/latex]% B. 5[latex]\frac {5}{11} [/latex]% C. 10% D. 12%

Answer: Option B
Explanation: Cost Price (C.P.) = Rs. (4700 + 800) = Rs. 5500. Selling Price (S.P.) = Rs. 5800. Gain = (S.P.) - (C.P.) = Rs.(5800 - 5500) = Rs. 300. Gain % = [[latex]\frac {300}{5500} [/latex] x 100]% = 5[latex]\frac {5}{11} [/latex]%
Q2. In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit?

A. 30% B. 70% C. 100% D. 250%

Answer: Option B
Explanation: Let C.P.= Rs. 100. Then, Profit = Rs. 320, S.P. = Rs. 420. New C.P. = 125% of Rs. 100 = Rs. 125 New S.P. = Rs. 420. Profit = Rs. (420 - 125) = Rs. 295. [[latex]\frac {295}{420}[/latex] x 100] x 100]% [latex]\frac {1475}{21}[/latex] % = 70%
Q3. A vendor bought toffees at 6 for a rupee. How many for a rupee must he sell to gain 20%?

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

Answer: Option C
Explanation: C.P. of 6 toffees = Re. 1 S.P. of 6 toffees = 120% of Re. 1 = Rs. [latex]\frac {6}{5}[/latex] For Rs. [latex]\frac {6}{5}[/latex] , toffees sold = 6. For Re. 1, toffees sold = [6 x [latex]\frac {5}{6}[/latex]] = 5
Q4. A shopkeeper offers 25% discount on balls which have been marked 20% above the cost price. Ram bought the ball for Rs. 2700 what was the coast price of the ball?

A. 2900 B. 3150 C. 3050 D. 3000

Answer: Option D
Explanation: Let CP = 100
MRP = 100 [latex] \times \frac {120}{100} = 120 [/latex]
Sp = 120 [latex] \times \frac {75}{100} = 90 [/latex]
Value of 90 unit = 2700
1 Unit = 30
100 Unit = 3000 (CP)
Q5. On selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls. The cost price of a ball is:

A. Rs. 45 B. Rs. 50 C. Rs. 55 D. Rs. 60

Answer: Option D
Explanation: (C.P. of 17 balls) - (S.P. of 17 balls) = (C.P. of 5 balls) C.P. of 12 balls = S.P. of 17 balls = Rs.720. C.P. of 1 ball = Rs. [latex]\frac {720}{120}[/latex] = Rs. 60.

Q1. The product of two numbers is 120 and the sum of their squares is 289. The sum of the number is

A. 20 B. 19 C. 23 D. None of these

Answer: Option C
Explanation: Let the number be x and y
Then, xy = 120 and [latex]{x}^{2} [/latex]+ [latex]{y}^{2} [/latex] = 289
therefore, [latex] {(x + y)}^{2}[/latex] = [latex]{x}^{2} [/latex]+ [latex]{y}^{2} [/latex] + 2xy = 289 + (2 x 120) = 529
x + y = [latex]\sqrt {529}[/latex] = 23
Q2. The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

A. 276 B. 322 C. 276 D. 385

Answer: Option B
Explanation: Clearly, the numbers are [latex] (23 \times 13) and (23 \times 14) [/latex]
∴ Larger number = [latex] (23 \times 14) [/latex] = 322 Gain % =
Q3. In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay's speed is:

A. 5 kmph B. 6 kmph C. 6.25 kmph D. 7.5 kmph

Answer: Option A
Explanation: Let Abhay's speed be x km/hr. Then, [latex] \frac {30}{x} [/latex] - [latex] \frac {20}{x} [/latex] = 3 6x = 30 x = 5 km/hr.
Q4. 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. B's weight = 31 kg.
Q5. A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?

A. 34 B. 40 C. 68 D. 88

Answer: Option D
Explanation: Amount, he would pay = We have: l = 20 ft and lb = 680 sq. ft. So, b = 34 ft. Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.