Directions(1-5): Number Series
1. In a division sum, the divisor is 10 times the quotient and 5 times the remainder. If the remainder is 46, the dividend is:

A. 4236 B. 4306 C. 4336 D. 5336

Answer : Option D
Explanation: Divisor = (5 x 46) = 230. Also, 10 x Q = 230 Q = 23, R = 46. Dividend = (230 x 23 + 46) = 5336
2. Three consecutive numbers such that twice the first, 3 times the second and 4 times the third together make 182. The numbers in question are

A. 18, 22 and 23 B. 18, 19 and 20 C. 19, 20 and 21 D. 20, 21 and 22

Answer : Option C
Explanation: let number be x, x + 1, x + 2 2x + 3(x + 1) + 4(x + 2) = 180 x = 19 so numbers are 19, 20, 21
3. A two digit number is such that the sum of the digits is 11.When the number with the same digits is reversed is subtracted from this number, the difference is 9.What is the number?

A. 23 B. 24 C. 65 D. 14

Answer: Option C
Explanation: Let the number be written as xy x-ten’s place y units place x + y = 11 _____________ (1) (10x + y) - (10y + x) = 9 10x + y - 10y- x = 9 9x - 9y= 9 x - y = 1 ______ (2) Add (1) & (2) 2x = 12 x = 6 y = 5 the number is 65
4. In a division sum, the divisor is 3 times the quotient and 6 times the remainder.If the remainder is 2 then the dividend is

A. 36 B. 40 C. 45 D. 50

Answer: Option D
Explanation: Divident = Quotient × Divisor +Remainder Divisor = 6×2 = 12 Q = [latex]\frac{12}{3}[/latex] = 4 Divident = 4 × 1 2 + 3 = 48 + 2 = 50
5. If the number obtained on interchanging the digits of a 2 digit number is 18 more than the original number and the sum of the digit is 8.Then what is the original number?

A. 35 B. 53 C. 32 D. 34

Answer: Option A
Explanation: Unit digit= x ; Ten’s digit = (8-x) [10(8 - x) + x] – [10x + (8 - x)] = 18 80 -10x + x- 10x -8 + x = 18 -18x = 18 - 80 +8 18x = 54 X = [latex]\frac{54}{18}[/latex] = 3 8 – x = 8 – 3 =5 Hence the number is 35

Directions(1-5): Percentages
1. 45% of ? = 25% of 355

A. 195 B. 176 C. 127 D. 197.22

Answer : Option D
Explanation: ([latex]\frac{45}{100}[/latex]) × X = ([latex]\frac{25}{100}[/latex]) × 355 ⇒ 264x = [latex]\frac{100}{45}[/latex] x [latex]\frac{25}{100}[/latex] = 197.22
2. From the salary, Akilesh spent 15% for house rent, 5% for children’s education and 15% for Entertainment. Now he left with Rs.13,000. His salary is

A. 19,000 B. 20,000 C. 18,000 D. 15,000

Answer : Option B
Explanation: 10 + 15 + 10 = 35% 100-35 = 65% = 13,000 100% = 100*[latex]\frac{13000}{65}[/latex] = 20,000
3. Out of 500 students of a school 35% students plays football, 25% plays cricket and 20% neither play football nor cricket.How many students play football and cricket ?

A. 200 B. 100 C. 50 D. 94

Answer : Option A
Explanation: Football = n(A) = ([latex]\frac{35}{100}[/latex]) × 500 = 175 Cricket = n(B) = ([latex]\frac{25}{100}[/latex]) × 500 = 125 neither play football nor cricket = ( [latex]\frac{20}{100}[/latex]) × 500 = 100 both football and cricket = n(A U B) = 175 + 125 - 100 = 200
4. The price of sugar is reduced by 3%.How many kg of sugar can now be bought for the money which was sufficient to buy 50kg of rice earlier ?

A. 50kg B. 55kg C. 51.5kg D. 56kg

Answer : Option C
Explanation: Let one kg of sugar earlier = Rs. 100 50 kg of sugar earlier = Rs. 5000 Now 1 kg of sugar = Rs.97 Quantity to buy now = [latex]\frac{5000}{97}[/latex] = 51.5kg
5. Fresh fruits contains 70% of water and dry fruits contain 20% of water.How much dry fruit can be obtained from 100kg of fresh fruits ?

A. 35 B. 37 C. 37.5 D. 40

Answer : Option C
Explanation: Quantity of pulb in 100kg of fresh fruit = (100 – 70) ×100= 30kg Quantity of dry fruit be x kg (100 – 20 ) % of x = 30 ([latex]\frac{80}{100}[/latex]) x = 30 x = ([latex]\frac{30 × 100}{80}[/latex]) = 37.5

Directions(1-5): Mensuration Problems
1. After measuring 100m of a rope, it was discovered that the metre rod was 2cm longer. The true length of the rod is

A. 95m B. 98m C. 96m D. 93m

Answer : Option B
Explanation: 100*2 = 200cm measured long Correct length = 100 – ([latex]\frac{200}{100}[/latex]) = 100 - 2 = 98m
2. Find the cost of carpenting a room 12 m long and 7 m broad with a carpet 10m long and 6m wide and at the rate of Rs.150 per meter.

A. Rs.320 B. Rs.210 C. Rs.210 D. Rs.235

Answer : Option B
Explanation: Area of the room = 12 × 7 => 84 [latex]{m}^{2}[/latex] Area of carpet = 10*6 = 60 [latex]{m}^{2}[/latex] Cost = ([latex]\frac{84}{60}[/latex])*150 = 210
3. 2 cubes have their volume in the ratio 1:8, find the ratio of their surface area ?

A. 2:5 B. 4:1 C. 1:4 D. 3:7

Answer : Option C
Explanation: [latex]{a}^{3}[/latex] : [latex]{b}^{3}[/latex] = 1:8 a:b = 1:2 Surface area = 6[latex]{a}^{2}[/latex] : 6[latex]{b}^{2}[/latex] = 1:4
4. A square field has an area of 7396 sq m,find the cost of fencing around it at Rs.22 per meter ?

A. 6754 B. 7568 C. 8765 D. 7943

Answer : Option B
Explanation: [latex]{a}^{2}[/latex] = 7396 a = 86 length of wire = 4*86 = 344 cost = 344*22 = Rs.7568
5. How many marbles of 10cm length and 7cm width are required to pave the floor of room 7m length and 4m breadth ?

A. 4000 B. 5100 C. 2800 D. 3200

Answer : Option A
Explanation: Area of floor = 700*400 = 280000 Area of marble = 10*7 = 70 N = [latex]\frac{280000}{70}[/latex] = 4000