Directions(1-5): LCM & HCF
1. Find the least number which when divided by 2, 3, 4 and 5 leaves a remainder 3. But when divided by 9 leaves no remainder?

A. 33 B. 63 C. 81 D. 123

Answer : Option B
Explanation: LCM of 2, 3, 4 and 5 is 30, let number be 30k + 3 put k = 2, we get 63 which is divisible by 9
2. Find the 4-digit smallest number which when divided by 12, 15, 25, 30 leaves no remainder?

A. 1020 B. 1120 C. 1200 D. 1800

Answer : Option C
Explanation: LCM of 12, 15, 25 and 30 is 300 least number of 4-digit divided by 300 is 1200
3. The HCF and LCM of two numbers is 84 and 840 respectively. If the first number is 168, find the second one.

A. 210 B. 420 C. 225 D. 326

Answer: Option B
Explanation: Product of two numbers = HCF * LCM So 2nd number = 84*[latex]\frac{840}{168}[/latex]
4. Sum of 2 numbers is 128 and their HCF is 8. How many numbers of pairs of numbers will satisfy this condition?

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

Answer: Option D
Explanation: Since HCF = 8 is highest common factor among those numbers So, let first no = 8x, Second number = 8y So 8x + 8y = 128 x + y = 16 the co-prime numbers which sum up 16 are (1,15), (3,13), (5,11), and (7,9) *co-prime numbers are those which do no have any factor in common. These pairs are taken here because in HCF highest common factor is taken, so the remaining multiples x and y must have no common factors. Since there are 4 pairs of co-prime numbers here, there will be 4 pairs of numbers satisfying given conditions. These are (8*1, 8*15), (8*3, 8*13), (8*5, 8*11), (8*7, 8*9)
5. Find the least number which when divided by 10, 15, 18 and 30 leaves remainders 6, 11, 14 and 26 respectively.

A. 72 B. 86 C. 85 D. 90

Answer: Option B
Explanation: Since 10 - 6 = 4, 15 - 11 = 4, 18 - 14 = 4, 30 - 26 = 4 So answer will be the LCM of these numbers – 4 So 90 – 4 = 86

Directions(1-5): Simple Interest
1. Cost of a Mobile Rs.8000. Sudha bought Mobile in EMI. She paid a Down payment of Rs. 2000 and paid rest in 6 equal installments of Rs.1020 for next 6 months. Then what is the SI rate charged?

A. 6.5% B. 6.95% C. 10.5% D. 12.5%

Answer : Option B
Explanation: Balance to be paid in installments = 8000-2000 = 6000 (6000 + 6000*r*[latex]\frac{3}{12}[/latex]*100) = 1020*6 + [latex]\frac{1020r}{12}[/latex]*100(1 + 2 + 3 + 4 + 5) r = 6.95%
2. What amount would Rs.2560 fetch if it is lent at 8% SI for 15 years?

A. Rs.3072 B. Rs.4632 C. Rs.5072 D. Rs.5632

Answer : Option D
Explanation: SI = [latex]\frac{2560*8*15}{100}[/latex] = 3072 Amount = 2560+3072 = 5632 100% = 100*[latex]\frac{13000}{65}[/latex] = 20,000
3. Veena has to pay Rs. 2460 to Sita, 5 Months later at 6% SI per annum, and Gita has to pay Sita same amount at 7.5% SI per annum after certain months. If both took the same amount of loan from Sita then Gita paid loan after how many months?

A. 3 Months B. 4 Months C. 6 Months D. 12 Months

Answer : Option B
Explanation: 2460 = p + p*6*[latex]\frac{5}{12}[/latex]*100 p = 2400 Now Gita 2460 = 2400 +2 400*7.5*[latex]\frac{x}{12}[/latex]*100 x = 4
4. A man invests Rs. 124000 for 9 years at 5% SI. Income tax at the rate of 19% is deducted from interest earned at the end of every year. Find the amount at the end of the [latex]{9}^{th}[/latex] year?

A. Rs. 169198 B. RS. 169918 C. Rs. 196918 D. Rs. 199698

Answer : Option A
Explanation: for one year = 124000*5/100 = 6200 income tax = 6200*[latex]\frac{81}{100}[/latex] = 5022 for 9 years = 45198 Amount = 124000+45198 = 169198
5. Nitin invested an amount of Rs. 24000 at the 4% SI per annum, and another amount at 10% SI per annum. The total interest earned at the end of one year will be same as interest earned when the total amount invested at 6% SI per annum. Find the total amount invested?

A. Rs.12000 B. Rs.24000 C. Rs.30000 D. Rs.36000

Answer : Option D
Explanation: 24000*[latex]\frac{4}{100}[/latex] = 960 x*[latex]\frac{10}{100}[/latex] = 0.1x 960 + 0.1 x = [latex]\frac{(24000+x)*6}{100}[/latex] x = 12000 Total = 24,000 + 12,000 = 36,000

Directions(1-5): Mensuration Problems
1. If the ratio of radius two Cylinders A and B are in the ratio of 2:1 and their heights are in the ratio of 2:1 respectively. The ratio of their total surface areas of Cylinder A to B is?

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

Answer : Option D
Explanation: Cylinder A: 2πr1 (r1 + h1) Cylinder B: 2πr2 (r2 + h2) [latex]\frac{r1}{r2}[/latex] = 2:1; [latex]\frac{h1}{h2}[/latex] = 2:1 [latex]\frac{T A}{T B}[/latex] = [latex]\frac{2πr1 (r1 + h1)}{2πr2 (r2 + h2)}[/latex]
2. The length and breadth of a rectangle are increased by 32% and 25% respectively. Its area will be increased by

A. 55% B. 65% C. 56% D. 50%

Answer : Option B
Explanation: 32 + 25 + ([latex]\frac{32*25}{100}[/latex])
3. After measuring 320m of a rope, it was discovered that the measuring metre rod was 6cm longer. The true length of the rope measured is

A. 320m 860cm B. 329m 60cm C. 320m 60cm D. 230m 860cm

Answer : Option B
Explanation: True length= 320 m + 320* 3 cm 320 m + 960 cm 329 m 60 cm(1 m = 100 cm)
4. Poles are to be fixed along the boundary of a rectangular field in such a way that distance between any two adjacent poles is 2 m. The perimeter of the field is 70m and length and the breadth of the field are in the ratio 4:3 resp. How many poles will be required?

A. 42 B. 40 C. 35 D. 38

Answer : Option C
Explanation: Required between the two poles = (Perimeter/Dist.between any two adjacent poles) = [latex]\frac{70}{2}[/latex] = 35
5. The perimeter of a square is equal to twice the perimeter of a rectangle of length 8 cm and breadth 7 cm. What is the circumference of a semicircle whose diameter is equal to the side of the square ?

A. 35.73 cm B. 33.15 cm C. 36.62 cm D. 38.57 cm

Answer : Option D
Explanation: Perimeter of square = 2 x Perimeter of rectangle = 2 * 2 (8+7) = 60 cm. Side of square = [latex]\frac{60}{4}[/latex] = 15 cm = Diameter of semi-circle Circumference of semi-circle = [latex]\frac{πd}{2}[/latex] + d = ([latex]\frac{22}{7}[/latex]) * 2 * 15 + 15 = 38.57 cm